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1a. TorsionCHAPTER OBJECTIVES

Shafts of circular section are generally used for power transmission from prime mover (such as engine, electric motor) to operating machines such as lathe, shaper or any manufacturing machine. In this chapter students are made to learn about the development of shear stress and angular twist in a shaft or a combination of shafts due to twisting moment applied on them. Then, the shafts are connected to gears and pulleys with the help of keys and how stresses are developed in the keys. Moreover, students will learn about the variation of stress along the radius and angular twist along the length of the shafts.

Introduction

Shafts are generally made of ductile materials like mild steel. Due to the application of a twisting moment, the development of stresses, strains and angular twist will be discussed with in this chapter. Single shaft and a combination of shafts as ‘shafts in series’ and compound shafts will also be discussed here. How much horse-power is transmitted by a shaft transmitting torque at a particular speed is the important criterion in the design of shafts.

Stress concentration due to key ways in shaft for providing key so that a shaft is connected to a hub of machine component as gears/pulleys will also be discussed.

Development of Shear Stress and Angular Twist in a Shaft Due to Twisting Moment

A relationship between the twisting moment, shear stress, angular twist, length of the shaft and polar moment of inertia of shaft will be derived. For that the following assumptions are taken into consideration:

1. Material is homogeneous and isotropic.2. Shaft is not initially distorted.3. Displacement at a point in the shaft is proportional to its

distance from centre of the shaft, that is, shear strain at any point is proportional to its distance from the centre of the shaft.

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Figure 12.1 (a) Shaft subjected to twisting moment (b) variation of shear strain and (c) variation of angular twist

4. Cross-sections perpendicular to the axis of the shaft which are plane before the shaft is subjected to twisting moment remain plane after the application of twisting moment.

5. Angular twist is uniformly distributed along the length of the shaft.

6. Shear stress developed in the shaft is within the proportional limit, that is, shear stress α is shear strain.

Figure 12.1(a) shows a shaft of radius R, length L fixed at one end and a twisting moment T applied at the other end. Say, a clockwise twisting moment T is applied at A and a resisting moment T (ccw) develops at the other end. Axis of the shaft is O′O. If a line CA, parallel to the axes O′O is drawn on the shaft before the application of the twisting moment, it is twisted to the position CA′, making ∠ACA′ = ø (shear angle). Since the angle ø is extremely small, shear angle ø is also known as shear strain ø.

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At radius R (outer radius of shaft),

AA′ = Rθ, where θ is angle of twist.

At any radius r, shift BB′ = rθ

Shear strain ϕr at radius

or, shear strain ø α a radius r,

Similarly, AA′ = Rθ

Shear strain,

or maximum strain, øαR = τ /G, where τ is maximum shear stress at radius R.

Taking Eq. (12.1) again,

Shear strain,

But,

or,




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From Eq. (12.2),

or,

From this relation, we learn that the shear stress at any point is directly proportional to its distance from centre of the shaft.

Consider an elementary area at radius r and thickness dr (Fig. 12.2)

Area = 2πrdr

Shear stress = τr

Now, τ is the maximum shear stress at radius R,

Shear force in small area,




Figure 12.2

Twisting moment of δQ about the centre of the shaft,

Total twisting moment,

Resulting twisting moment,

where πR4/2 = J, polar moment of inertia of the section of shaft

Therefore,

or,

Combining the Eq. (12.5) and Eq. (12.8), we get,

The students must learn about the significance of each term as follows:

T = twisting moment applied on the shaft

= resisting twisting moment in shaft

J = polar moment of inertia of section of shaft

= (for a solid shaft)

= for a hollow shaft of external radius R2 and internal radius R1

τ = maximum shear stress at radius R in the case of solid shaft and at external radius R2 in the case of hollow shaft

G = shear modulus of the maternal

θ = angular twist in shaft over length L of the shaft

τr = shear stress at any radius r of the shaft

Figure 12.3(a) shows the shear stress distribution in a solid circular shaft and Figure 12.3(b) show the shear stress distribution in a hollow circular shaft, the minimum stress occurs at inner radius R1 and maximum shear stress occurs at outer radius R2.

Example 12.1 A hollow circular steel shaft of inner radius 30 mm and outer radius 50 mm is subjected to a torque of 10 kN m. Determine (a) the maximum and minimum shear stresses developed in shaft and (b) the angular twist over 1 m length of shaft. Given shear modulus, G = 82 kN/mm2 for steel.





Figure 12.3 Shear stress distribution in (a) solid circular shaft and (b) hollow circular shaft

Solution

Outer radius, R2 = 50 mm

Inner radius, R1 = 30 mm

Polar moment of inertia,

Torque, T = 10 kN m = 10 × 106 Nmm

Maximum shear stress,

Minimum shear stress,

Note that shear stress is proportional to radius.

Length of shaft, L = 1 m = 1,000 mm

Angular twist,

Exercise 12.1 A circular steel shaft of 30 mm diameter is subjected to a torque of 0.56 kN m. Determine (a) the maximum shear stress developed in shaft, (b) the angular twist over 1m long shaft and (c) the shear stress at a point 10 mm away from the centre. Given G for steel = 82 kN/mm2.

Modulus of Rupture

Torsion formula, T/J = Gθ/L = τ/R, has been derived on the assumption that shear stress is proportional to shear strain but when a specimen is tested in torsion, it fails by fracture and stress developed in this material goes beyond the proportional limit, that is, in the plastic stage. If shear stress and shear strain is drawn along radius of the shaft, shear strain at any point remains proportional to radius but shear stress at any point is no longer proportional to its radius,rather the curve becomes non-linear as shown in Fig. 12.4. A torsion test piece is continuously twisted, that is,θ is increased gradually till the shaft breaks. Say, Tmax is the maximum torque at which shaft fails by fracture.


Figure 12.4 Shear stress-shear strain curves


This maximum shear stress calculated by using torsion formula is known as modulus of rupture. It can be noted that it is not the actual shear stress on the surface of the shaft but it is a hypothetical stress taking straight line relation between shear stress and shaft radius. However, it is a useful information to know the limiting torque at which the specimen would fail in torsion.

Example 12.2 A steel sample of a gauge length of 250 mm and a diameter of 25 mm is tested under torsion and it was observed that sample fractured at the torque of 920 N m. Determine the modulus of rupture of the material.

Solution

Maximum torque, Tmax = 920 N m = 9,20,000 Nmm

Shaft radius, R = 12.5 mm

Modulus of rupture,

= 300 N/mm2

Exercise 12.2 A specimen of copper of a gauge length of 200 mm and a diameter of 25 mm is tested under torsion. It was observed that the specimen was fractured at the torque of 640 Nm. Determine the modulus of rupture of copper.

Horse-Power Transmitted by a Shaft

A shaft is transmitting torque T at N rpm, then

Power transmitted in 1 min = 2πNT

Horse-power transmitted by shaft

if T is in N m as 1 W = 1 N m/s

So,

If torque is in kgf m

Metric , where Torque T is in kgf m

There is a minor difference between metric HP and HP.

If HP transmitted is known, then torque,

Maximum stress developed on the surface of the shaft is τ =

2T/πR3, in case of solid shaft of radius,R, , in case of hollow shaft of outer radius, R2 and inner radius, R1.

Example 12.3 A hollow circular shaft is transmitting power at 300 rpm. Outer diameter of shaft is 80 mm and inner diameter is 50 mm. The maximum shear stress developed in shaft section is 65 N/mm2. Determine the horse-power transmitted by the shaft.

Solution



rpm = 300

Maximum shear stress, τ = 65 N/mm2

Torque,

HP transmitted

Exercise 12.3 The maximum shearing stress developed in 60 mm steel shaft is 60 N/mm2. If the shaft rotates at 360 rpm, find the horse-power transmitted by the shaft.

Shafts of Varying Diameters

(a) Uniformly tapered circular shaft:

Let us consider a circular tapered shaft of diameter d2, gradually decreasing to diameter d1 over the axial length L of the shaft. Consider a small disc of length dx at a distance x from the end A of the shaft as shown in Fig. 12.5.


Figure 12.5 Tapered circular shaft

Diameter of the shaft at the section

Polar moment of inertia of the disc,

Angular twist over this length dx, i.e.,

Total angular twist,

Figure 12.6 Stepped shaft

(b) Stepped shaft:

Consider a shaft with different diameters along different axial lengths as shown in Fig. 12.6. Say, there are three steps of diameter, d3, d2 and d1 with axial length L3, L2 and L1, respectively, as shown inFig. 12.6. Polar moment inertia of these sections are:

Torque on each part of the stepped shaft is subjected to same twisting moment, this is known asshaft in series. Maximum shear



stresses on surface in three steps are τ1 = 16T/πd13, τ2 =

16T/πd23 andτ3 = 16T/πd3

3 this shows that portion AB with minimum diameter d1 is subjected to maximum shear stress. Angular twist between ends A and D is:

Example 12.4 A tapered circular shaft of 1 m long having a diameter of 60 mm at one end gradually reduces to a diameter of 36 mm at the other end is subjected to a torque such that an angular twist of 1° develops between the ends. Determine the torque applied on the shaft.

G = 82,000 N/mm2

Solution

Length, L = 1,000 mm

d1 = 36 mm, d2 = 60 mm

or torque,

Exercise 12.4 A circular stepped shaft of a length of 750 mm, with diameters 80 and 40 mm over AB and AC, respectively, is shown in Fig. 12.7. This shaft is subjected to a torque T = 1.5 kN m. Determine (a) maximum shear stress developed in shaft and (b) angular twist between C and A, if G = 80 kN/mm2.

Compound Shaft

In a compound shaft, there are two distinct portions of shaft subjected to different values of twisting moment but angular twist in the two shafts is the same. Fig. 12.8 shows two shafts A and B, of diameter dA and dB and length LA and LB, respectively, subjected to a torque T at the junction of the two as shown in Fig. 12.8.

Torque T is shared by two shaft A and B,




Figure 12.7

T = TA + TB,

but angular twist

θA = θB

Say GA = shear modulus of shaft A

GB = shear modulus of shaft B

JA = polar moment of inertia of the shaft A

JB = polar moment of inertia of the shaft B

Figure 12.8

or,

or,

For the case shown in Fig. 12.8,

Another example of a compound shaft is shown in Fig. 12.9. A solid circular shaft of material A is encased in a hollow circular shaft of material B. Both the shafts are subjected to a twisting momentT, which is shared by A and B, that is, total twisting moment T = TA + TB, in this case LA = LB = L.

θA = θB

so

or

Figure 12.9

where GA, GB = shear modulus of materials A and B, respectively.

Example 12.5 A composite shaft is made by joining an 800 mm long solid steel shaft with 800 mm long hollow copper shaft as



shown in Fig. 12.10. The diameter of solid shaft is 40 mm, while internal and external diameters of hollow shaft are 25 and 50 mm, respectively. Determine the maximum shear stresses developed in steel and copper shaft, if torque T applied at junction is 4 kN m. What is the angular twist at the junction? Given G steel = 2.G copper = 82,000 N/mm2.

Solution

θCA = θCB

Torque, T = TA + TB

LA = LB = 800 mm

So,

Figure 12.10


So

But TA + TB = 4 kN m

0.8738 TB = 4

TB = 2.1.5 kN m

TA = 1.865 kN m

Shear stresses

Maximum shear stress in steel,

Maximum shear stress in copper,

Angular twist at junction

Exercise 12.5 A solid circular steel shaft is encased in a copper hollow shaft so as to make a compound shaft. The diameter of steel shaft is 40 mm and the outside diameter of copper shaft is60 mm. The length of compound shaft is 2 m and is subjected to an axial torque of 3.2 kN m. Determine (a) the maximum shear stress in steel and copper shafts and (b) the angular twist per unit length, if Gsteel = 2Gcopper = 80 kN/mm2.

Stresses in a Shaft Subjected to Twisting Moment

A shaft in transmitting power, that is, the torque, is under pure torsion. Due to the torque T, the maximum shear stress developed on the surface of the shaft is τ = 16T/πd3, where d is the diameter of a solid shaft. Figure 12.11(a) shows shear stress τ at the surface of the shaft and a complementary shear stress τ at right angles (i.e., in longitudinal direction). State of stress on a small element on the surface of the shaft is shown in Figure 12.11(a). Figure 12.11(b) shows complementary shear stress τ at right angles, variation of shear stress along the radius of the shaft. Figure 12.11(c) shows the enlarged view of the stresses on a small element. At angle θ = ±45°, principal stresses are developed, p1 = ±τand p2 = –τ as shown (derivation of principal stresses) in a shaft has been dealt in Chapter 3 on principal stresses. Figure 12.11(d) shows the development of principal stresses on a small element with direction of principal stress ± 45° to the axes of the shaft.

So, principal stresses at a point are +τ, –τ, 0.

Principal strains,







Volumetric strain, ɛv = ɛ1 + ɛ2 + ɛ3 = 0

This shows that if a shaft is subjected to a pure twisting moment there will not be any change is its volume. However, the shafts are subjected to bending moments also due to the forces on elements mounted on shaft as gears, pulleys and due to reactions from the supports. Therefore, a shaft is subjected simultaneously to a bending moment M and a twisting moment T at any section shown inFig. 12.12.

Figure 12.11

Shafts Subjected to T and M

When a shaft transmits torque T, it is simultaneously subjected to bending moments also due to belt tensions on a pulley, tooth load on gears and support reactions in bearings. Figure 12.12 shows a shaft subjected to twisting moment T and a bending moment M. If



we take small elements on the surface of the shaft, stresses acting on the element will be as shown in Figure 12.13.

Figure 12.12 Twisting and bending moments on a shaft

Figure 12.13 Stresses on a small element of shaft

σ = Normal stress due to bending moment M = 32 M/πd3

τ = Shear stress due to twisting moment M = 16 T/πd3

Maximum principal stress at element

or,

or,


The bending moment due to maximum principal stress in shafts is known as equivalent bending moment, Me.

or,

Similarly, maximum shear stress developed on the surface of the shaft,

or, equivalent twisting moment.

The twisting moment corresponding to maximum most shear stress on the surface of the shaft is known as equivalent twisting moment, Te.

Example 12.6 A solid shaft of diameter d is subjected to a bending moment, M = 15 kN m and a twisting moment, T = 25 kN m. What is the maximum diameter of the shaft if the maximum most shear stress in shaft is not to exceed 160 N/mm2 and the maximum direct stress is not to exceed 200 N/mm2.

Solution

Bending moment, M =15 kN m

Twisting moment, T = 25 kN m

Equilateral bending moment,

Equilateral twisting moment,

Now

Maximum diameter of the shaft is 104 mm.

Exercise 12.6 A solid shaft of a diameter of 180 mm is transmitting 700 kW at 200 rpm. It is also subjected to a bending

moment of 10 kN m. Determine (a) the equivalent bending moment, (b) the equivalent twisting moment, (c) the maximum principal stress on the surface of shaft and (d) the maximum most shear stress.

[Hint: 1 W = 1 Nm/s]

Torsional Resilience of a Shaft

A solid shaft of radius R and length L is subjected to a twisting moment, T.

Strain energy per unit volume τr2/2G

where

τr = shear stress at any radius

G = modulus of rigidity.

Consider an elementary area at radius r and of thickness dr,

Area, dA = 2πrdr

Volume, dv = 2πrdrL

Shear stress, τr = (τ/R)r, where τ is the maximum shear stress at outer radius R as shown in Figure 12.14.

Figure 12.14

Strain energy per unit volume =


Strain energy for an elementary volume dv,

Strain energy for the solid shaft,

Note that τ is the maximum shear stress on the surface of the shaft.

Hollow shaft (inner radius R1 and outer radius R2)

τr, at any radius r = (τ/R2)r, R2 is maximum radius.

Strain energy for an elementary volume,

Example 12.7 A 1 m long hollow circular shaft of an outer diameter of 60 mm and an inner diameter of 20 mm is subjected to a twisting moment of 4 kN m. Determine the strain energy absorbed by the shaft, if G = 80 kN/mm2.

Solution

R2 = 30 mm, T = 4 × 106 Nmm

R1 = 10 mm

L = 1,000 mm


Volume of the shaft,

V = π (R22 − R1

2) × L

= π (302 − 102) × 1,000

= 8 π × 105 mm3

Ratio,

Strain energy,

Exercise 12.7 A solid circular steel shaft of diameter of 50 mm and length of 1 m is subjected to a twisting moment of 2.5 kN m. Determine the strain energy absorbed by the shaft. G = 78.4 kN/mm2.

Stresses Developed in a Key

Shafts transmit power through element like pulley, gear and flywheel. These elements are keyed to the shaft with the different types of keys, but most common key is rectangular section key as shown inFigure 12.15. The breadth and thickness of the key are b and t, respectively, as shown. Half of the key is embedded in keyway on shaft and another half fits into the keyway provided in hub of pulley, flywheel, gear etc.

Say, torque transmitted = T

Shaft radius = R


The tangential force Q or shear force acting on the periphery of the shaft, Q = T/R

Say, length of key = L

Section of the key in shear = bL

Shear stress developed in key = Q/bL = T/RbL

Same force Q acts as a bearing force for half of the key in shaft

Section of the key under compression = L(t/2)

Bearing pressure or stress developed in key,

σp = 2Q/Lt

Figure 12.15 Key fitting the shaft and hub

Example 12.8 A shaft of a diameter of 60 mm is transmitting 20 HP at 300 rpm. The shaft is connected to a pulley of an axial width of 120 mm with the help of a key 14 mm × 12 mm (deep). Determine the shear and compressive stresses developed in the key. Take 1 HP = 746 W/s.

Solution

HP = 20rpm = 300

Angular speed,

Torque,

Radius of the shaft, R = 30 mm = 0.03 m

Tangential force on periphery,

Length of the key = axial width of pulley = 120 mm

Breadth, b = 14 mm

Thickness, t = 12 mm

Shear stress developed in key,

Bearing pressure developed in key,

Exercise 12.8 A shaft of 50 mm in diameter and 1.5 m in length is transmitting 15 HP at 200 rpm. Shaft is connected to the pulley of an axial length of 100 mm by a key of breadth 12 mm and depth 10 mm. Determine the shear and compressive stresses developed in the section of the key.

Stress Concentration in Torsional Loading

Using the torsion formula, we determine the shear stress developed in a shaft of uniform diameter. However, if the shaft has abrupt change in section such as fillet radius, groove or notch, stress concentration will occur near the abrupt change in diameter of shaft.


Where,

Ri = radius of smaller diameter at abrupt change.

T = torque.

J = polar moment of inertia for section for smaller diameter.

K = stress concentration factor due to abrupt change.

Magnitude of stress concentration factor K depends upon the ratio of D/d and the ratio of r/d at the fillet as shown in Figure 12.16(a).

Similarly, for a grooved shaft the value of K depends on the ratio of D/d and r/d where r is the groove radius as shown in Figure 12.16(b).





For shaft with keyways, which are generally used to connect gears, pulleys and flywheels on shaft, ASME allows 25 per cent reduction in allowable stress using the equation, τ = TR2/T, where R2 is outer radius of the shaft.

Figure 12.16 Stress concentration factors

Example 12.9 A stepped shaft transmits 100 HP as it rotates at 400 rpm. The allowable shear stress in section of shaft is 50 MPa. The larger shaft has a diameter of 100 mm and smaller shaft has a diameter of 75 mm. The fillet radius at the junction of the shafts is 6 mm. Use the stress concentration factor value, to determine the maximum shear stress in the smaller shaft.

Solution

Figure 12.17 shows the stepped shaft with a fillet, the dimensions are:

Figure 12.17

Fillet radius, r = 6 mm


Ratio,

From Figure 12.16 (a),

Now,

HP = 100rpm = 400

Torque,

Smaller diameter, d = 75 mm

Shear stress,

Maximum shear stress, τmax = K × t = 1.45 × 21.5 = 31.2

Exercise 12.9 A transmission shaft has a semi circular groove with a radius of 5 mm machined into a diameter of 100 mm. The maximum shear stress in the shaft must not exceed 70 MPa, determine the maximum torque that can be transmitted by the shaft (Refer to Figure 12.18)

Problem 12.1 A torsion bar, 1 m long, is to be designed. Shear modulus of the bar = 84,000 N/mm2. Determine the required diameter of the shaft, so that resisting torsional spring constant (or torsional stiffness) of the bar is 36 Nm/degree of angular trust.



Solution

Figure 12.18

Twisting moment, T = 36 Nm

or,

or,

Required shaft diameter, d = 22.36 mm.

Problem 12.2 A hollow circular steel shaft is required to transmit 300 HP at 200 rpm. The maximum torque developed is 1.5 times of the mean torque. Determine the external diameter of the shaft if it is double the internal diameter. The maximum shear stress is not to exceed 80 MPa. GivenG = 82 kN/mm2.

Solution

HP = 300

= 300 × 746 Nm/s

N = 200 rpm

Angular speed,

Mean torque,

Maximum torque,

Tmax = 1.5 Tmean = 1.5 × 10,685.66

= 16,028.5 Nm

= 16.0285 − 106 Nmm

Say, external diameter = D

internal diameter = 0.5D

J = polar moment of inertia

Maximum shear stress, τ= 80 MPa, at radius D/2

or,

or,

Problem 12.3 A vessel having a single propeller shaft 200 mm in diameter running at 300 rpm is re-engined to two propeller shafts of equal cross-section and producing 40 per cent more HP at 500 rpm. If the working shear stress in re-engined shafts is 15 per cent more than in the single shaft, determine diameter of these shafts.

Solution

Say T1, torque transmitted by a single shaft

T2, torque transmitted by each of two shafts

H2 = 1.4 H1 (as given)

Now,

Diameter of each re-engined shaft = 143 mm.

Problem 12.4 A solid steel shaft of 60 mm diameter is made of low-carbon steel which is assumed to be elasto-plastic with τy = 160 MPa and G = 84,000 N/mm2. Determine the maximum shear stress due to an applied torque of (a) 2.5 kN m and (b) 7.5 kN m.

Solution

1. Shaft diameter, d = 60 mm

Torque, T = 2.5 kN m

= 2.5 × 106 Nmm

2. Maximum shear stress,

3.4. Torque,

T = 7.5 kN m = 7.5 × 106 NmmMaximum shear stress,

But the behaviour of the material is elasto-plastic, that is, O–A, elastic and A–B, fully plastic as shown in Figure 12.19.

So, maximum shear stress in shaft is only 160 N/mm2 but not 176.8 N/mm2 (as calculated).

Figure 12.19

Problem 12.5 A solid marine propeller shaft is transmitting power at 1,200 rpm. The vessel is being propelled at a speed of 18 km/h, for the expenditure of 4,000 HP. If the efficiency of propeller is 60 per cent, and the greatest shear stress is not to exceed 50 MPa, calculate the shaft diameter and maximum shearing stress developed in the shaft.


Solution

Say, thrust = P Newton

Useful work done per second

Efficiency = 60 per cent

Input work = 5P/0.6 = 8.33P

Work input per second = 4,000 × 746 = 298.4 × 104 Nm

= 8.33P

Thrust,

Allowable direct stress, σ = 50 N/mm2

Speed,

Torque,


Problem 12.6 A solid circular copper shaft is required to transmit 30 HP at 200 rpm. Determine the diameter of the shaft if the maximum shear stress is not to exceed 60 N/mm2 in shaft.

The solid shaft is now replaced by a hollow copper shaft with the internal diameter 0.7 times the external diameter. Determine the external diameter of the shaft, if it is required to transmit same horse-power, at same rpm and maximum stress produced in shaft also remains the same. Find the percentage saving of material of shaft by using hollow shaft in place of solid shaft.

Solution

Torque,

Shear stress, τ = 16T/πd3 where T is shaft diameter

Hollow shaft

Inner diameter, d = 0.7D


So,

D3 =

D3 = 119.365 × 103

D = 49.23 mm

d = 0.7 D = 34.46 mm

Area of cross-section,

Area of cross-section of solid shaft,

Percentage saving of material

Problem 12.7 A steel shaft of a diameter of 100 mm runs at 300 rpm. This steel shaft has a 20-mm-thick bronze bushing shrunk over its entire length of 2 m. If the maximum shearing stress in steel shaft is not to exceed 40 N/mm2, find (a) power of the engine and (b) torsional rigidity of the shaft.

Given Gsteel = 84 kN/mm2, Gbronze = 42 kN/mm2

Solution

Maximum shear stress in steel shaft, τs = 40 N/mm2

Diameter of steel shaft, d = 100 mm

Torque transmitted by steel shaft,

Steel shaft has bronze sleeve of thickness 20 mm, say torque transmitted by bronze shaft is TB, then

where polar moment of inertia of bronze shaft,

So,

TB = 1.421 × TS = 1.421 × 7.854 × 106 = 11.16 × 106 Nmm

Total torque, T = TS + TB = (7.854 + 11.16) × 106 Nmm

= 19.014 × 106 Nmm = 19.014 × 103 Nm

Now, θs = θB, in a compound shaft

Length of shaft = 2 m = 2,000 mm

Torsional rigidity of shaft =

= 1,000.74 × 106 Nmm/rad

= 1,000.74 kN m/rad

Power of the engine = 2πNT Nm/min

= 2π × 300 × 19.014 × 103 Nm/min

= 35,840.5 × 103 Nm/min

= 597.74 × 103 Nm/s = 597.74 kW

Problem 12.8 A compound drive shaft consists of a 4-m-long solid circular steel shaft of 50 mm in diameter surrounded by a 2-m-long aluminium tube as shown in Fig. 12.20. Tube and shaft are both rigidly fixed to a machine. Pin D of diameter 15 mm fills a hole drilled completely through tube and the shaft. Shearing deformation in pin and deformation in between pin and shaft can be neglected. Calculate the maximum torque T which can be applied to the steel shaft without exceeding an average shearing stress of 8 MPa in the pin.

Gsteel = 80 GPa, Galuminium = 28 GPa

Solution

Pin diameter, dp = 15 mm

Shearing stress in pin = 8 MPa

Shearing force in pin, (Shearing at two ends)

Figure 12.20 Problem 12.8

https://www.safaribooksonline.com/library/view/strength-of-materials/9789332503519/xhtml/chapter012.xhtml#pro-12.8


Radius of shaft B, R = 25 mm

Torque on shaft B in portion AB (Compound shaft)

= Q × R = 2,827.4 × 25

= 70, 686 Nmm = 70, 686 Nm

Compound shaft

TA = torque shared by aluminium

TB = torque shared by steel

Now TB = 70.686 Nm

TA = 1.4218 × 70.686 = 100.506 Nm

T = TA + TB = 70.686 + 100.506

= 171.192 Nm

Problem 12.9 A circular stepped shaft shown in Fig. 12.21 is subjected to a torque TA = 500 Nm andTB = 2,000 Nm. Determine


the maximum shear stress in both steel and brass section of the shaft. The shaft is fixed at left end.

Solution

For equilibrium, clockwise resisting couple is 2,000 + 500 = 2,500 Nm (cw). Torque in two portions AB and BC will be as shown in Fig. 12.22.



Note that net torque at B is 2,500 – 500 = 2,000 Nm (ccw)

Stress

Steel shaft

Brass shaft

Problem 12.10 A 5-HP motor shown in Fig. 12.23 delivers 3 and 2 HP, respectively, to accessory gears B and C. Determine (a) maximum shear stress in steel shaft drive and (b) angular twist





between C and A, if the angular speed of the shaft is 1,000 rpm and G = 80 kN/mm2.

Figure 12.23

Solution

HP delivered by motor = 5 HP

HP consumed by gear B = 3 HP

HP received by gear C = 2 HP (as shown in Fig. 12.23)

Portion AB transmits 5 HP power.

Speed = 1,000 rpm

Angular speed,

Torque,

Diameter, d = 20 mm


Problem 12.11 A horizontal shaft AB, 1 m long, rigidly fixed at both the ends, A and B, is subjected to axial twisting moments of T1 = 2.5 kN m and T2 = 4 kN m at distances of 0.25 and 0.75 m from end A and are in anticlockwise direction looking from end A as shown in Fig. 12.24. Determine end fixing couples.

Solution

Say, fixing couples at ends A and B are TA and TB, respectively.

Torque on portion CD = TA –2.5 kNm

Torque on portion DB = TA – 2.5 – 4 = TA – 6.5

Total angular twist between A and B

θAC + θCD + θDB = 0 as both the ends are fixed

G and J of shaft is the same throughout, so


or, TA × 0.25 + (TA − 2.5)0.5 + (TA − 6.5)0.25 = 0

or, 0.25TA + 0.5TA − 1.25 + 0.25TA − 1.625 = 0

TA − 2.875 = 0

TA = 2.875 kN m = TAC

TA − 2.5 = + 0.375 kN m = TCD

TA − 6.5 = 2.875 − 6.5 = −3.625 kN m

Figure 12.24 shows the torque distribution diagram.

Figure 12.24 Torque distribution diagram

Problem 12.12 A pipe cantilever is subjected to a load at the end of a lever as shown in the Fig. 12.25. The outer diameter of pipe is 100 mm and the inner diameter of the pipe is 80 mm. Determine principal stresses at point D at top edge of cantilever as shown in the figure.



Solution

Bending moment at point D,

M = 2 × 1.2 = 2.4 kN m

Twisting moment at point D,

T = 2 × 0.6 = 1.6 kN m

Outer diameter of pipe, D = 100 mm

Inner diameter, d = 80 mm

Moment of inertia,

Polar moment of inertia, Izz = 2Izz = 579.6 × 104 mm4

Shear stress at D,

Figure 12.25

Bending stress at D

State of stress at the point D is shown in Fig. 12.26.

Principal stress


Figure 12.26

Problem 12.13 A solid shaft of a diameter of 80 mm is transmitting 250 kW at 200 rpm. It is also subjected to a bending moment of 5 kN m and an end thrust. If the maximum principal stress developed in shaft is 200 N/mm2, determine the magnitude of end thrust.

Solution

Power transmission = 250 kW = 250 kN m/s

rpm = 200

Speed,

Torque,

Shaft diameter, d = 80 mm

Shear stress developed in shaft,

Bending moment, M = 5 kN m = 5 × 106 Nmm

σb = direct stress due to bending moment

Say, direct stress due to end thrust is σ

Resultant direct stress, σ = σb − σt = 99.47 − σt

= σb + σt = 99.47 + σt

Maximum principal stress,

Say σb and σt are both compressive

129.5 = 99.47 + σt (both are compressive)

σt = 129.5 – 99.47 = 30.03 N/mm2

Area of cross-section of shaft

End thrust,

P = σt × A = 5,026.55 σ 30.03

= 1,50,947 N

= 150.947 kN

Considering σ = σb – σt (tensile and compressive)

129.5 = 99.47 − σt which is not possible as σt will also be in tension

so, P = 150.947 kN

Problem 12.14 A flanged coupling has n bolts of a diameter of 19 mm arranged symmetrically along a bolt circle of diameter of 300 mm. If the diameter of the shaft is 80 mm and it is stressed up to 80 N/mm2, determine the value of n if the shear stress in bolts is not to exceed 35 N/mm2.

Figure 12.27 Bolts joining flanges

Solution

A flange coupling is used to connect two shafts, so that the power can be transmitted from one shaft to the other. Flanges connecting the shafts are joined together with the help of bolts (Fig. 12.27)

Maximum stress in shaft, τ = 80 N/mm2

Shaft diameter, d = 80 mm

Torque transmitted,

Bolt circle radius, R = 150 mm

Shear force is both,

Bolt diameter, db = 19 mm

Area of cross-section of each bolt

Allowable stress in each bolt = 35 N/mm2

Shear force transmitted by each bolt = 35 × 283.53

F = 9923.6 N

or number of bolts


n = 6 bolts to connect the two flanges of coupling.

Problem 12.15 A splined shaft connection is shown in Fig. 12.28 is 60 mm long and is used to permit axial movement of the shaft relative to the hub during torque transmission in order to facilitate axial movement in the connection, side pressure on spline is not to exceed 10 N/ mm2. Calculate the power that can be transmitted by shaft at 300 rpm.

Figure 12.28 Splined shaft

Solution

Length of splines, L = 60 mm

Outer diameter of splines = 60 mm

Inner diameter of splines = 48 mm

Width of each spline, w = 12 mm (as shown in Fig. 12.28)

Area under shear = w.L = 12 × 60 = 720 mm2

Side pressure = 10 N/mm2

Compressive force per spline




Angular speed,

Number of splines = 6

Mean radius of spline,

Torque transmitted = nPRm

= 6 × 3,600 × 27

= 583.2 Nm

Power, P = ω.T = 31.416 × 589.2

= 18,322 Nm/s

= 18.322 kW

Shear stress in splines,

Key Points to Remember

o Torsion formula, where

T = torque on shaft

J = polar moment of inertia of shaft section

G = shear modulus

θ = angular twist over length L

τ = maximum shear stress at radius R

τr = shear stress at any radius r.

o Modulus of rupture, for a solid shafto HP transmitted by a shaft

where

N = speed in rpm

T = Torque in Nm

o If several shaft are in series, then same twisting moment acts on all the shafts, but angular twist will be different and total angular twist will be the sum of all the angular twists.

o It two shafts are in parallel, then angular twist in both shafts will be the same. Total torque T will be shared by the two, that is, T1 + T2 = T.

o Strain energy per unit volume = τ2/4G, where τ is the maximum shear stress (for a solid shaft). For hollow shaft shear strain energy per unit volume τ2/4G (D2

2 + D12/D2

2), where D2 and D1 are outer and inner diameters of the shaft.

o If a key of breadth b, thickness t and length L connects a shaft and hub for power transmission, shear stress in key = 2T/DbL

o Bearing stress in key = 4T/DtL where D is shaft diameter and T is torque transmitted by shaft.

Review Questions

1. A shaft is subjected to a twisting moment, show how shear strain varies along the radius of the shaft.

2. A hollow shaft of outer radius R2 and inner radius R1 is subjected to a twisting moment T. Derive expression for maximum shear stress in shaft.

3. A solid shaft is subjected to a twisting moment T, such that maximum shear stress developed on the surface of the shaft is τ, if G is shear modulus, prove that strain energy per unit is τ2/4G.

4. Take a hollow shaft subjected to twisting moment and with the help of a sketch show the variation of shear stress along radial thickness of shaft.

5. What are equivalent twisting moment and equivalent bending moment in a shaft? How these are obtained?

6. Make a simple sketch of a shaft subjected to twisting moment. Take a small element on the surface of the shaft and mark directions of principal stresses.

7. Show that volumetric strain for a shaft subjected to pure torsion is 0.

Multiple Choice Questions

1. A shaft of 20 mm diameter and length 1,000 mm is subjected to twisting moment such that θ = 0.1 rad. What is the shear strain in the shaft at outer surface?

1. 0.001 rad2. 0.0001 rad3. 0.0005 rad4. None of these

2. A hollow shaft of inner radius 30 mm and outer radius 50 mm is subjected to a twisting moment. If the shear stress developed at inner radius of shaft is 60 N/mm2. What is the maximum shear stress in shaft?

1. 60 N/mm2

2. 75 N/mm2

3. 100 N/mm2

4. None of these3. Torsional rigidity of a shaft is given by

1. T/G2. T/J3. JG4. None of these

where T is torque, G is shear modulus and J is polar moment of inertia.

4. A solid steel shaft is surrounded by a copper shaft, such that, Gsteel, what is the ratio of TS/TC if compound shaft is subjected to a twisting moment?

1. 22. 13. 0.54. None of these

5. A shaft of diameter d and length L is subjected to twisting moment T, shear angle developed in shaft is 0.001 rad. Now, the length of the shaft is doubled, but the diameter and torque remain the same. What will be the shear angle?

1. 0.0005 rad2. 0.001 rad3. 0.002 rad4. None of these

6. A shaft is transmitting 2 HP at 100 rpm, maximum shear stress in shaft is 50 MPa. Now the shear stress remains the same, how much HP will be transmitted at 200 rpm?

1. 1 HP2. 2 HP3. 3 HP4. None of these

7. What is the shear modulus of steel?1. 40–42 GPa2. 60–63 GPa3. 70–72 GPa4. None of these

8. A solid shaft of diameter d is subjected to twisting moment T, such that maximum shear stress developed is τ. Now a hole of 0.5d diameter is drilled throughout the length of the shaft along the axis. How much torque T must be reduced so that shear stress remain the same?

1. T/22. T/43. T/84. T/16

9. A solid circular steel shaft is subjected to twisting moment T, such that maximum shear stress developed on shaft is 200 MPa. If E = 200 GPa and n = 0.3, what is the maximum principal strain developed on the shaft?

1. Zero2. 1,000 micro strain3. 1,300 micro strain4. None of these

10. A shaft of 100 mm diameter is keyed to a pulley transmitting power. The dimension of key are b= 20 mm, t = 20 mm and L = 100 mm. Twisting moment of shaft is 1,000 Nm. What is the shear stress developed in key?

1. 10 N/mm2

2. 20 N/mm2

3. 40 N/mm2

4. None of these

Practice Problems

1. A torsion bar 1.2 m long is to be designed. Shear modulus of bar is 82 kN/mm2. Determine the required diameter of the shaft so that the resulting torsional spring constant (or torsional stiffness) of the bar is 40 Nm for 1° of angular twist.

2. A hollow circular steel shaft is required to transmit 200 HP at 360 rpm. The maximum torque developed is 1.3 times the mean torque. Determine the external diameter of the shaft if it is 1.6 times the internal diameter and the maximum shear stress in shaft is not to exceed 75 MPa. GivenG = 82 kN/mm2.

3. A vessel having a single propeller shaft 250 mm in diameter running at 200 rpm is re-engined to two propeller shafts of equal cross-section and producing 50 per cent more horse-power at 500 rpm. If the working shear stress in these shafts is 20 per cent more than that in the single shaft, determine diameter of these shafts.

4. A solid steel shaft of 50 mm diameter is made of a low carbon steel which is assumed to be elasto-plastic with Ty = 150 MPa, and G = 84 × 103 N/mm2. Determine the maximum shear stress due to an applied torque of (a) 2 kN m and (b) and 4 kN m.

5. A solid marine propeller shaft is transmitting power at 1,000 rpm. The vessel is being propelled at a speed of 20 km/h for the expenditure of 5,000 HP. If the efficiency of propeller is 70 per cent, and the greatest thrust is not to exceed 60 MPa, calculate the shaft diameter and maximum shearing stress developed in the shaft.

6. A solid circular steel shaft is required to transmit 60 HP at 200 rpm. Determine the diameter of the shaft, if the maximum shear stress is not to exceed 60 N/mm2 in shaft.

The solid shaft is now replaced by a hollow steel shaft with the internal diameter equal to 75 per cent of the external diameter. Determine external diameter of the shaft if it is required to transmit the same horse-power at same rpm and the maximum shear stress produced is also the same. Find the percentage saving of the material by using hollow shaft in place of solid shaft.

7. A solid alloy shaft of diameter 60 mm is coupled to a hollow steel shaft of same external diameter. If the angular twist per unit length of hollow steel shaft is 80 per cent of the angular twist of alloy shaft, determine the internal diameter of the steel shaft. At what speed, the shafts will transmit power of 200 kW. The maximum shearing stress in steel shaft is not to exceed 100 MPa

and in alloy shaft it is not to exceed 60 MPa, G for steel = 2G for alloy.

[Hint: It is a question of two shafts in series]8. A compound drive shaft of solid circular steel of 40 mm diameter

and 5 m long surrounded by a copper tube 3 m long, outer diameter 80 mm as shown in Fig. 12.29. Both the tube and shaft are rigidly fixed to a machine. Pin C of diameter 12 mm fills a hole drilled completely through tube and shaft. Shearing deformation in pin and shearing deformation in between pin and shaft can be neglected. Calculate the maximum torque T which can be applied to the steel shaft without exceeding an average shearing stress of 10 MPa in the pin. Gsteel = 80 GPa, Gcopper = 40 GPa

Figure 12.29

9. Figure 12.30 shows a hollow shaft subjected to torque TB = 3.5 kN m and Tc = 1.5 kN m. Determine the location and magnitudes of maximum shear stress if the outer diameter of the hollow shaft is 60 mm and inner diameter is 40 mm.

[Hint: TCB = 1.5 kN m (ccw), TBC = 2 kN m (ccw)]

Figure 12.30



10. A 7.5-HP motor shown in the Fig. 12.31 delivers 5 and 2.5 HP responsively to accessory gears B and C, respectively. Determine (a) maximum shear stress in steel shaft drive, and (b) angular twist between C and A, if the angular speed of the shaft is 500 rpm.

G = 84 kN/mm2

Figure 12.31

11. A horizontal shaft AD, 1.4 m long, rigidly fixed at both the ends A and D is subjected to axial twisting moments of T1 = 3 kN m and T2 = 6 kN m at distances of 0.4 and 1.0 m from end A and are in anticlockwise direction when looking from end A. Determine end couples Fig. 12.32.

Figure 12.32

12. A cantilever made from a solid circular rod of diameter 50 mm is subjected to a load P kN at point C as shown in the Fig. 12.33. Determine the magnitude of P such that maximum principal stress at point is not to exceed 100 MPa.

13. A solid shaft of diameter 11 cm is transmitting 700 kW at 200 rpm. It is also subjected to a bending moment of 15 kN m and an end thrust P. If the maximum principal stress developed in the shaft is 200 N/mm2, determine the magnitude of end thrust, P.

14. A flanged coupling has n bolts of 25 mm diameter arranged symmetrically along a bolt circle of diameter 300 mm. If the




diameter of the shaft is 100 mm and it is stressed up to 100 N/mm2, determine the value of n, if the shear stress in the bolt is not to exceed 50 N/mm2.

Figure 12.33

Special Problems

1. A hollow shaft of external diameter 120 mm is transmitting 400 HP at 200 rpm. Determine the internal diameter if maximum shear stress in shaft is not to exceed 60 N/mm2.

2. How much torque is required to twist a 2-m-long steel shaft of an inside diameter of 80 mm and an outside diameter of 120 mm by an angle of 1.5°? Calculate the shear stress at outer and inner radii of hollow shaft.

3. A composite is fixed at one end and a torque of 11 kN m acts at the other end. Determine the torque resisted by outer shaft. The outer radius is 40 mm and inner radius is 25 mm. Shear modulus of outer shaft is 1.6 times the shear modulus of inner shaft. What are the shear stresses developed in outer and inner shaft?

Figure 12.34 Special Problem 5

Figure 12.35

4. A circular steel shaft is transmitting 20 HP at 300 rpm. Determine the diameter of the shaft if the maximum shear stress is not to exceed 80 N/mm2 and angular twist per meter length of the shaft is not to exceed 1°. Given G = 80 kN/mm2.

5. A solid shaft is transmitting 2,000 kW at 200 rpm. The maximum torque developed in shaft is 1.8 times of the mean torque. The distance between bearings is 1.8 m as shown in Fig. 12.34. Weight of flywheel is 5,000 kg midway between bearings. Determine shaft diameter if: (a) maximum permissible tensile stress is 60 N/mm2, and (b) maximum permissible shearing stress is 40 N/mm2.

6. A torque tube consists of two sections which are riveted together as shown in Fig. 12.35 by 36 rivets of 4 mm diameter each pitched uniformly and the radius of the mating surface of the tubes is 100 mm. If the limiting shear stress for rivets is 50 MN/m2, determine the maximum torque that can be transmitted through the joint.

7. A motor shaft consists of a steel tube of 40 mm internal diameter and 5 mm thick. The engine develops 15 HP at 2,000 rpm. What will be the maximum shear stress in the tube when the power is transmitted through 4:1 gearing?

[Hint: N for tube is 500 rpm]



8. Figure 12.36 shows a vertical shaft with pulleys keyed on it. The shaft is rotating with uniform velocity of 2,000 rpm. The belt tensions are indicated and three pulleys are rigidly keyed to the shaft. If the maximum shear stress in the shaft is not to exceed 50 N/mm2, determine the necessary diameter of the solid circular shaft. Draw the torque distribution diagram along OABCD.

Figure 12.36

Answers to Exercises

Exercise 12.1: J = 7.9521 × 104 mm4, 105.63 N/mm2, 4.92°; 70.42 N/mm2

Exercise 12.2: 208.6 N/mm2

Exercise 12.3: 128.6

Exercise 12.4: 119.4 N/mm2; 0.021 rad; 1.2°

Exercise 12.5: 84.03, 63.0 N/mm2, 6.02°

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Exercise 12.6: Me = 22.44 kN m, Te = 34.88 kN m,p1 = 39.2 N/mm2,tmax = 30.46 N/mm2

Exercise 12.7: 13.233 kN m

Exercise 12.8: 17.8 N/mm2, 42.74 N/mm2

Exercise 12.9: 6.42 kN m

Answers to Multiple Choice Questions

1. (a) 2. (c) 3. (c) 4. (b) 5. (b)

6. (d) 7. (d) 8. (d) 9. (c) 10. (a)

Answers to Practice Problems

1. 23.1 mm2. D = 74.45 mm3. 157.5 mm4. (a) 81.5 N/mm2, and (b) τ′ = 162.97 N/mm2 but shear

stress is 150 MPa5. 99.8 mm, 182.5 N/mm2

6. 56.6 mm, 64.25 mm, 43.6 per cent7. Internal diameter = 46.95 mm, N=750 rpm8. 384.53 Nm9. τmax = 58.76 in portion AB10. 34.83 MPa, 0.0276 rad, 1.58°11. TAB = 3.857 kN m, TBC = 0.857 kN m, TCD = –5.143 kN

m12. 1.105 kN13. 33.8 kN14. n = 6

Answers to Special Problems

1. Internal diameter = 88.8 mm





2. 17.1 kN m, 41.87 MPa, 62.8 MPa3. T0 = 9.887 kN m, 116 N/mm2, 45.35 N/mm2

4. 43.14 mm5. 280.5 mm6. 2,262 Nm7. 14.75 N/mm2

8. d = 27.76 mm, TAB = 210 Nm, TBC = 210 – 100 = 110 Nm, TCD = 110 Nm

1b. SpringsCHAPTER OBJECTIVES

A spring is the most important element in a system or a mechanism. A mechanism cannot work without the application of a spring which restores the mechanism to its original configuration. In this chapter, students will learn about:

o Stiffness of a spring, most useful characteristic.o Stresses developed in a spring due to applied axial load and or an axial moment.o An open-coiled helical spring subjected to combined effect of bending moment and twisting moment.o Angular rotation or angular twist in a spring.o Importance of graduated leaves in a leaf spring or a carriage spring.o Storage of energy in a plane spiral spring for the functioning of a system as a time-watch.

Introduction

To start with, a close-coiled helical spring subjected to an axial load is considered and relations for stress and stiffness are derived. The effect of curvature and direct shear stress is taken into account while calculating shear stress in spring wire section. The effect of axial moment on a close-coiled helical spring is also studied.

Open-coiled helical spring with a large angle of helix subjected to an axial load and an axial moment is discussed. In this case, the axial moment or a moment due to an axial load each has two components, that is, a component of twisting moment and another component of bending moment. The effect of these moments on the stresses in spring wire, on axial deflection and on axial rotation, is discussed.

Plane spiral springs used in watches or toys for storing energy for slow and steady operation is also analysed.

Finally, carriage springs or leaf springs which are used in automobiles to absorb shocks during motion of a vehicle over uneven roads and derivation of expressions for deflection and stresses.

Helical Springs

When the axis of the wire of the spring forms a helix on the surface of a right circular cylinder or a right circular cane, a helical spring is obtained.


Figure 13.1 Coil of a helical spring

Figure 13.1 shows a coil of a helical spring subjected to an axial load W or an axial moment M. Then ABCDE is the helix made by the spring wire. The axis of the spring wire forms a right circular cylinder of radius R and axes of the cylinder is OO′, ∠BPL = α, helix angle.

R = mean radius of the coil of spring (radius of right circular cylinder)




α = helix angle

n = number of coils in the spring

Length of spring wire in one coil = 2πR/cos α

Length of spring wire in n coils = 2πnR/cos α

In the close-coiled helical spring, α is very small, cos α &cong; 1,

Therefore, the length of spring wire = 2πnR

Close-Coiled Helical Spring

When the coils of the spring are very close to each other then the coils are regarded as lying in planes at right angle to the axis of the helix, α very small (Fig. 13.2).

Say, wire diameter = d

Polar moment of inertia of section, J= πd4/32

Direct shear force at any section = W

Direct shear stress, τ0=4W/αd2

Torsional shear stress due to twisting moment WR,τ = 16WR /αd3

As shown in Fig. 13.3, direct shear stress τ0 is added to τ at inner radius and is subtracted from τ at outer radius. So, the resultant shear stress at the inner radius of the coil is more than the resultant shear stress at the outer radius of the coil.

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Figure 13.2

The coil of a spring is not contained in a plane as shown in Fig. 13.2, but the coil is curved throughout its length. Shearing strain at the inner surface of the coil is more than the shearing strain at the outer surface of the coil. Wahl studied the effect of continuous curvature on spring and presented a research paper in a conference and observed that the maximum shear stress in a spring is given by the equation,

where 16WR/αd3 = τ, torsional shear stress

Spring index,


Figure 13.3 Resultant shear stress in spring wire section





Example 13.1 A close-coiled helical spring of mean coil radius R = 32 mm is made of a steel wire of diameter d = 8 mm. If the axial load on spring is 80 N, then determine the maximum shearing stress developed in spring wire.

Solution

Mean coil radius, R = 32 mm

Mean coil diameter, D = 64 mm

Wire diameter, d = 8 mm

Spring index, C = 64/8 = 8

Shear stress, τ = 16WR/πd3

= 25.46 (1.107 + 0.077)

Maximum shear stress = 25.46 (1.184) = 30.14 N/mm2

Exercise 13.1 A close-coiled helical spring of mean coil diameter 50 mm is made of steel wire of 6 mm diameter. If the maximum shear stress developed is 90 MPa, what is the axial load applied on the spring?

Close-Coiled Helical Spring Subjected to an Axial Load

Consider a small element mnn′m′ of the spring wire subtending an angle δø at the centre of the spring axis. Say, under the action of twisting moment T = WR (acting on every wire section along this coil), angular twist is δθ as shown in Fig. 13.4.

Figure 13.4

Vertical deflection, dδ = Rδθ

Angular twist, δθ = TRd/GJ

Because Rdφ = dL = length of wire considered cc′

T, twisting moment at any section is same, that is, WR

Total axial deflection,

n = number of turns of the spring

Total angle subtended by coils at the axis of the spring, φ = 2πn

Polar moment of inertia of wire, J = πd4/32


So, axial deflection,

Stiffness of the spring, k=W/δ = Gd4/64nR3, i.e. an axial force required to extend or compress the spring by a unit axial deflection.

Example 13.2 A close-coiled helical spring made of round steel wire is required to carry an axial load of 800 N for a maximum stress not to exceed 150 N/mm2 in shear. Determine the wire diameter if the stiffness of the spring is 10 N/mm and diameter of helix is 80 mm and also calculate the number of turns required in the spring. Neglect the correction due to spring index. Given G for steel = 84 kN/mm2.

Solution


Maximum shear stress, τ = 150 N/mm2

Axial load, W = 800 N

τ = 16WR/πd3, substituting the values

d = 10.28 mm.

Stiffness, k = 10 N/mm

k = Gd4/64nR3, substituting the values

, number of coils.

Exercise 13.2 A close-coiled helical spring made of round steel wire with mean coil radius 40 mm and the number of turns equal to 10 is subjected to an axial load of 200 N. Determine (a) wire diameter, if the maximum shear stress is 160 N/mm2 and (b) axial deflection. Given G = 82 kN/mm2.

Closed-Coiled Helical Spring Subjected to an Axial Moment

Figure 13.5 shows a close-coiled helical spring subjected to an axial couple M. The effect of the couple is to rotate end B of the spring with respect to end A.

Figure 13.5

Helix angle α is very small.

φ is the total angle through which one end of the spring is turned relative to the other, when coupleM is applied.

Work done on spring = (1/2)Mφ

Figure 13.6 shows a bar AB of length or length of spring wire subjected to bending moment M.



ρφ = L

φ = L/ρ, where ρ is radius of curvature

From Flexure formula,

So, φ = ML/EI

Therefore, work done

L = 2πnR, where n = number of coils, R = radius of spring

I = πd4/64 for spring wire

Maximum stress in spring wire due to axial couple M is

where d = spring wire diameter.

Figure 13.6

Example 13.3 A close-coiled helical spring made of round steel wire 5 mm diameter having 10 complete turns is subjected to an axial moment M. Determine the magnitude of axial couple M if the maximum bending stress in spring wire is not to exceed 240

N/mm2 and also calculate the angle through which one end of the spring is turned relative to the other end, if the mean coil radius is 35 mm. Given E for steel = 200 kN/mm2

Solution


Number of turn, n = 10


Maximum bending stress,

Angular rotation,

= 1.0555 rad = 60.47°

Exercise 13.3 A close-coiled helical spring made of round steel wire of diameter d, having 12 complete turns is subjected to an axial moment of 3.6 Nm. Determine wire diameter if the maximum bending stress in wire section is not to exceed 210 N/mm2 and also calculate the strain energy absorbed by spring when couple M is applied. Take the mean coil radius as 40 mm.Given, E = 210 kN/mm2.

Open-Coiled Helical Spring

In the case of open-coiled helical spring, helix angle α is sufficient and effect of α on twisting moment or bending moment cannot be neglected. We have to clearly identify the following four parameters in open-coiled helical spring.

1. Spring axis and coil axis, both inclined at helix angle, α.2. Plane of spring and plane of coil, both are inclined to each

other at helix angle, α as shown inFig. 13.7.

Say, an open-coiled helical spring of mean coil radius R is subjected to an axial load W (along axis YY of spring)

WR = total moment on spring wire.

WRcos α = T ′ = Twisting moment on a small element of length, δL at the centres of coil (about y′y′ axis).

WRsin α = M ′, bending moment on small element of length, δL in the plane of coil.

Due to twisting moment T ′, there will be an angular twist δø ′ and due to bending moment M ′, there will be an angular rotation δø′.

δθ ′ is shown by ln and it has two components δθ ′cos α and δθ ′sin α.


Figure 13.7

δφ′= shown by pq has two components,

δφ′ cos α and δφ′ sin α.

δ θ = angular twist about XX axis

= δθ′ cos α + δφ′ sin α

δ ø = angular rotation about YY axis

= mm − qr = δθ′ sin α − δφ′ cos α

Moreover, δθ′ = T′δL/GJ and δφ′ = M′δL/EI

Total angular twist θ, about XX axis,

When L = length of spring wire

= 2πnR sec α

So angular twist,

Axial deflection, δ = Rθ

Total angular rotation about YY axis,

Substituting the values of T ′ and M ′

or

Length of spring wire,

Strain Energy Method

Total axial deflection of a spring can also be obtained by strain energy principle.

As

Example 13.4 An open-coded helical spring made from steel wire of circular cross-section is to carry an axial load of 120 N. The wire diameter is 8 mm and mean coil radius is 40 mm. Calculate the (a) axial deflection and (b) angular rotation of the free-end with respect to the fixed-end if helix angle of spring is 25°, and number of turns is 12. Given G = 84 kN/mm2, E = 210 kN/mm2

Solution

Helix angle, α = 25°

sin α = 0.4226, cos α = 0.9063

sin2α = 0.1786, cos2α = 0.8214

sin α cos α = 0.3830

d = 8 mm, number of coils, n = 12, W = 120 N

Angular rotation,

Exercise 13.4 An open-coiled helical spring made from steel wire of circular cross-section is to carry a load of 100 N. The wire diameter is 8 mm and the mean coil radius is 40 mm. Calculate (a) the axial deflection and (b) the angular rotation of the free end with respect to the fixed end if the helix angle of the spring is 30° and number of turns is 12. Given E = 80 kN/mm2, E = 200 kN/mm2.

Open-Coiled Helical Spring Subjected to Axial Moment

An open-coiled helical spring of mean coil radius R, helix angle α and number of turns n is subjected to an axial moment M as shown in Fig. 13.8. Components of M are M′ = M cos α = bending moment and T′ = M sin α = twisting moment. Take a small length, δL along coil of the spring.


Figure 13.8

δθ′

= angular twist due to T ′

δθ′

= T′δL/GJ, about x′x′

δø = angular rotation due to M′ about y′y′ axis = M′δL/EI

Taking components of δθ′ and δφ′ along xy direction

δθ

= angular twist about XX axis

= δφ′ sin α − δθ′ cos α, substituting the value of δθ′, δφ′, T′, M

δφ′ = angular rotation about YY axis

= δφ′ cos α+ δθ′ sin α

Total angular twist,

where L = spring wire length = 2πnR sec α

Total angular rotation,

as L = 2πnR sec α

Axial deflection,

Substituting the value of L = 2πnRsec α

Minus sign indicates winding couple, reduces the axial length of spring.

Example 13.5 An open-coiled helical spring made of 5 mm diameter steel wire, 25 mm mean coil radius, and 23° angle of helix is subjected to an axial moment of 2 Nm. Determine (a) angular rotation of one end with respect to the other end and (b) axial deflection, if number of coils in the spring is 15. Given E = 210 GPa, G = 82.7 GPa.

Solution


Wire diameter, d = 5 mm, n = 15


Moment of inertia, I = J/2=30.68mm4

Moment, M = 2 Nm = 2 × 103 N mm

EI = 2,10,000 × 30.68 = 6.44 × 106 N mm2

GJ = 82,700 × 61.36 = 5.0744 × 106 N mm2

α = 23°

sin α = 0.390, sin2α = 0.1526

cos α = 0.920, cos2α = 0.8473, sec α = 1.087

1. Angular rotation,

2. Axial deflection,

Exercise 13.5 An open-coiled helical spring made of steel wire 6 mm diameter and 36 mm mean coil radius with 65° inclination of coils with the spring axis is subjected to an axial moment M. Determine the magnitude of M if the number of turns in the spring increase by 1/8 and also calculate the change in the axial length of spring if the original number of turns are 10. GivenGsteel = 84 kN/mm2, Esteel = 210 kN/mm2

Open-Coiled Helical Spring—Stresses Developed in Spring Wire

(a) Axial load W on spring

Total torque, T = WR (Fig. 13.9)

Twisting moment, T ′ = WR cos α

Bending moment, M ′ = WR sin α

Wire diameter = d

Maximum torsional shear stress, τs = 16T ′/πd3

Direct shear stress, τd = 4W/πd2

Maximum shear stress at inner coil radius, Ri

Minimum shear stress at outer coil radius, Ro

Maximum stress due to bending,

Principal stresses at inner coil radius

(b) Axial couple M acting on spring

Bending moment,


M′ = M cos α

Twisting moment, T ′ = M sinα (Fig. 13.10)

Torsional shear stress on spring wire, τs = 16 M sin α/πd3

Maximum stress due to bending moment,

Principal stresses at extreme radius,

Figure 13.9

Figure 13.10


Example13.6 An open-coiled helical spring made of steel wire of 15 mm diameter, mean coil radius 90 mm with helix angle 30° is subjected to an axial moment of 40 Nm. Determine the shear and direct stresses developed in the section of the wire of the spring.

Solution

Axial moment, M = 40 Nm = 40 × 103 N mm

Helix angle, α = 30°


Bending moment, M′ = Mcos α = 40 × 103 × 0.866

= 34.64 × 103 N mm

Twisting moment, T′ = M sin α = 40 × 103 × 0.5

= 20 × 103 N mm

Shear stress, τs =

= 30.18 N/mm2

Direct stress, σ =

= 104.54 N/mm2

Exercise 13.6 An open-coiled helical spring, wire diameter 10 mm, mean coil radius 70 mm and helix angle 20°, carries an axial

load of 400 N. Determine the shear stress and direct stress at inner radius of coil.

Plane Spiral Spring

A plane spiral spring consists of a thin uniform strip wound in the form of spirals as shown in Fig. 13.11. One end of the strip is connected to winding spindle A and other end is hinged at point B. Winding couple is applied at spindle and the number of turns of spirals are increased.

Figure 13.11 Plane spiral spring

Reaction at hinged end B has two components RH, along BA (joining hinged end and centre of spindle) RV, perpendicular to line BA.

Consider a small element PQ of length δs and the co-ordinate of centroids of PQ are x and y, taking origin at B and x along the line BA (Fig. 13.12).



Figure 13.12

Angle between the tangents P′P′ at P and Q′Q′ at Q is φ.

After applying the winding couple, the number of turns are increased or φ increases to φ + dφ, radiusr1 decreases to r2.

Bending moment at element, Me = xRV − yRH

= EI × change of curvature

By integrating both the sides

Assuming that centre of the spring lies at the centre of the spindle.

when ∫ xds ≃ L × R, where L is the length of the strip making the spiral spring and R is the outer radius of spring.

But RvR = M, winding couple

φ = ML/EI

Energy stored in spring, U = (1/2) Mφ = M2l/2EI

Bending moment at any element, Mc = xRv − yRH

Me will be maximum, where y = 0 along line BA and extended up to C. At the point C, x = 2R

Mmax = maximum bending moment = 2RRV

= 2M

Maximum bending stress, σmax = 2 M/Z

where Z = bt2/6, section modulus of strip about plane of bending.

b = breadth of strip t = thickness of strip

σmax = 12 M/Bt2

Energy stored,

Example 13.7 A plane spiral spring is made of 5 mm wide and 0.25 mm thick steel strip. Torque applied at winding spindle is 5 N mm. Determine (a) the number of turns if the length of the strip is10 m and (b) the maximum stress developed at the point of greatest bending. Given E = 210 GPa.

Solution

Winding moment, M = 5 N mm

Width, b = 5 mm

Thickness, t = 0.25 mm

L = 10,000 mm = 10 m

E = 2,10,000 N/mm2

Angle of rotation,

= 36.57 rad

Number of turns,

Maximum stress at the point of greatest bending

moment,

Exercise 13.7 A flat spiral spring is made of a strip 5 mm wide, 0.25 mm thick and 12 m long. The torque is applied at the winding spindle and seven complete turns are given. Calculate the torque and the energy stored and the maximum stress developed at the point of greatest bending moment. Given E = 210 kN/mm2.

Conical Spring

A conical spring is generally used for large deflections (i.e., very small stiffness), when one coil sets into another coil and load is applied, as in the case of sofa sets where deflection of top of a sofa set is of the order of several centimeters. Figure 13.13 shows a conical spring with minimum radius R1 is increasing to R2 over n number of turns.

From A to B, total angle turned through by spring wire = 2πn

Consider a small element ab of length ds at radius R.

where


Figure 13.13

where φ is the angle turned through by spring wire from A to element ab of length ds and dθ is the angular twist produced by an axial load W, over length ds.

dθ = WRds/GJ

Strain energy of the length ds of spring

when

So,

Total strain energy,

Axial deflection,

Length of this spring wire, L = πn(R2 + R1)

Axial deflection, δ = WL(R22 + R1

2)/2GJ

Polar moment of inertia, J = πd4/32, where d is the spring wire diameter

Maximum shear stress, (as per assumption)

= maximum torsional shear stress + × direct shear stress

Example 13.8 A conical spring of minimum diameter 50 mm and maximum diameter 100 mm is made up of wire of 6 mm diameter. If the resultant shear stress in wire is not to exceed 200 MPa, determine the load which the spring can carry if it has six coils. What is axial deflection in spring? Given G = 84 kN/mm2.

Solution

R1 = 25 mmR2 = 50 mm

Length of wire = πn(R1 + R2)

Axial deflection,

Exercise 13.8 A conical spring of minimum diameter 60 mm and maximum diameter 120 mm of steel wire 10 mm diameter. If the

maximum stress in the wire is not to exceed 180 MPa, determine the load which the spring can carry if it has six coils. What is axial deflection in spring. Given G = 82,000 N/mm2.

Leaf Spring

These springs are used in vehicles such as cars, trucks, railway wagons, trolley, etc. and are connected between chassis and wheels. They are also termed as Carriage springs. Generally they are of two types: (1) semi elliptic and (2) quarter elliptic shapes. A number of plates of rectangular section of same width but different lengths are clamped together with the help of bolts and nuts. At the ends of largest leaf, there are eyes which connect the chassis to the spring with the help of shackles in between.

Figure 13.14 Equilvalent carriage spring semi elliptical spring

A simple theory for deflection, stress and radius of curvature of a leaf spring is developed taking the following assumptions:

1. Centre line of all the leaves is initially a circular arc of same radius R, so that contact between the plates is only at the ends.

2. Each plates is of uniform thickness and each plate overlaps the plate below it by an amountp=L/2n, where L is the length of longest leaf, n is the number of plates.

3. The overlaps are tapered in width to a triangular shape. Figure 13.14 shows the Equivalent Carriage spring of width B = nb.

Since each plate is initially of same radius of curvature, each plate will touch the one above it only at its ends when unloaded. After the application of a central load, the change in radius of curvature is uniform and radius of curvature becomes the same for all leaves (plates). They will continue to contact at the ends only.

Consider only two plates, top one of length L and bottom one of length (L – 2p), loaded at ends byW/2 load. Bending moment diagram varies linearly from A to B and C to D, while bending moment is uniform from B to C and equal to (W/2)p. This shows that each triangular overhanging end (refer toFig. 13.15) is loaded as a cantilever while portion of uniform width carries uniform bending momentWp/2. So on this tapered portion, M and I (moment of inertia) vary linearly and proportional to this distance from end, so M/I remains constant. Moreover in the central portion M and I are constant. This proves that M/I remains constant throughout, the length of the plate (with triangular ends as shown).



Figure 13.15

Now, M/EI = 1/R′–1/R, change in curvature. If M/EI is constant, then 1/R is also constant. This 1/R′ is also constant with changed radius of curvature R′, contact between the leaves continues to be at ends only. Taking the friction between the plates as negligible, each plate is of same radius of curvature. So, these plates can be considered to be arranged side by side forming a beam of same thickness t with width B = nb at centre as shown in Fig. 13.14.

In a carriage spring as shown in Fig. 13.14

Maximum bending moment, Mmax = WL/4

Maximum width, B = nb

Moment of inertia,



Bending stress,

Initial central deflection, y0 = L2/8R (using properties of a circle),

Final radius of curvature,

(a constant)

R′ is constant, if all the plates are bent into the arcs of a circle of radius R′.

Say R′ = ∝ (infinity) plates will become straight under Proof load, W0

Example 13.9 A carriage spring centrally loaded has eight steel plates, 5 mm thick and 50 mm wide. If the largest plate is 800 mm long, find the initial radius of curvature if the maximum stress is 150 N/mm2, when the plates become straight under the central load. Given, E = 210 GPa.

Solution

Initial radius of curvature,

Exercise 13.9 A carriage spring centrally loaded has six steel plates, 6 mm thick and 60 mm wide. If the largest plate is 960 mm long and the load required to straighten the spring is 3 kN. Determine (a) the initial radius of curvature, (b) the initial central deflection and (c) the bending stress under proof load.

Cantilever Leaf Spring (Quarter Elliptic Spring)

Proceeding in the same manner as for semi elliptic spring (Fig. 13.16), we get:

Bending moment = WL

Moment of inertia, I = nbt3/12

Bending stress, σ = 6WL/nbt2



Figure 13.16 Equivalent cantilever spring

Say, R′ = final radius of curative

R = initial radius of curative

y0 = L2/2 (using property of a circle)

y = final deflection under load

or,

Say, 1/R′=α (infinity)

From Eq. (1),


Example 13.10 A cantilever leaf spring of length 0.6 m has six leaves of thickness 8 mm each. The width of each plate is 48 mm. If an end load of 1 kN is applied at its end, determine (a) the end deflection under this load, (b) the initial radius of curvature if deflection provided is 100 mm and (c) the bending stress developed under the load. Given E = 210 kN/mm2.

Solution

W = 1,000 N

L = 600 mm

b = 48 mm

t = 8 mm

n = 6

1. Deflection under load,

2. Initial radius of curvature,

3. Bending stress,

Exercise 13.10 A cantilever leaf spring of length 500 mm has five leaves of thickness 10 mm. If an end load of 2 kN produces a deflection of 30 mm, find the width of the leaves. Given E = 200 GPa.

Problem 13.1 Design a close-coiled helical spring to have a mean coil diameter 120 mm and an axial deflection of 110 mm under axial load of 2,750 N, so that the maximum shear stress developed in the spring is not to exceed 330 N/mm2. The steel wires are available in the following diameter: 10, 12, 14 and 16 mm. Determine the most suitable diameter of the wire and the number of coils required. Also calculate the maximum shear stress in the designed spring (neglect direct stress). Given G = 84,000 N/mm2.

Solution

Axial load, W = 2,750 N

Mean coil radius = 60 mm

τmax, allowable = 330 N/mm2

Wire diameter, d = 13.65

= 14 mm available size of the wire

Number of coils,

Actual shear stress,

Problem 13.2 A close-coiled helical spring is made of round steel wire. It carries an axial load of 200 N and is to just get over a rod of 35 mm. Deflection in the spring is not to exceed 20 mm. The maximum allowable shearing stress developed in spring wire is 165 N/mm2. G for steel = 80 kN/mm2. Find the mean coil diameter, wire diameter and number of turns in spring (neglect direct stress in spring).

Solution

Say wire diameter = d

Spring is just get over a rod of 35 mm, i.e.,

Mean coil diameter, D = 35 + d

W = 200 N

δ = 20 mm

Now, δ = 8WD3n/Gd4, substituting the value

or

From Eqs. (13.2) and (13.4),

From Eq. (13.3),

D = d3(0.3239)

35 + d = d3(0.3239)

0.3239d3 − d − 35

= 0

d3 − 3.0873d = 108.058

d3 = 3.0873d + 108.058

If, d = 5,

LHS = 125, RHS = 123.4945

d = 4.95,




LHS = 121.87375, RHS = 123.5

d = 4.98 mm,

LHS = 123.5, RHS = 123.47

LHS = RHS

Wire diameter,

d = 4.98 mm

D = 0.3239, d3 = 0.3239 × 4.983 = 40 mm

n =

So, wire diameter = 4.98 mm, D = 40 mm, n = 9.8

Problem 13.3 A close-coiled helical spring is made of round wire with n turns and spring index is eight. Show that the stiffness of such a spring is (D/n)× constant. Determine the constant, if G = 84,000 N/mm2; such a spring is required to support a load of 500 N with an extension of 40 mm, and maximum shear stress is not to exceed 230 N/mm2. Determine the mean coil diameter and the number of turns.

Solution

Stiffens, k = GD4/8nD3

where D is mean coil diameter.

Now, W = 500 N,

or

Problem 13.4 A safety valve 60 mm diameter is to blow off at a pressure of 3 N/mm2 gauge. The safety valve is held by a close-coiled helical spring of steel with a mean coil radius equal to 60 mm. Determine the diameter of the steel wire and number of coils necessary if the maximum shear stress in wire is not to exceed 220 N/mm2, and spring is initially compressed by 30 mm. Given G = 84 kN/mm2. Take into account the effect of direct shear stress also.

Solution

Diameter of safety valve = 60 mm

Pressure (at which valve open) = 3 N/mm2

Total force on spring wire,

Say wire diameter = d mm


Compression in spring, δ = 30 mm

Value is to open, when extension in spring is 30 mm, due to shear pressure.

Maximum shear,

d3 = 49.09d + 11,782

For d = 25,

RHS = 13,009, LHS = 15,625

d = 24,

LHS = 13,824, RHS = 129

d = 23.5,

LHS = 12,978, RHS = 12,935

By trial and error, d = 23.5 mm

(Wire diameter)

Number of turns

Problem 13.5 A weight of 200 N is dropped onto a close-coiled helical spring from a height of 600 mm which produces a maximum instantaneous stress of 120 N/mm2 in the spring. If the mean radius of coil is six times the wire diameter, determine (a) instantaneous compression in spring and (b) number of coils in the spring. Given wire diameter is 15 mm and G for steel = 84 kN/mm2.

Solution

W = 200 N

Height, h = 600 mm

Maximum instantaneous stress, τ = 120 N/mm2


R = 6d = 6 × 15 = 90 mm

We = Equivalent static load on the spring

Say, instantaneous compression in spring is δ mm.

Potential energy lost by weight = W (h + δ)

Energy stored in spring = (1/2)Weδ

Instantaneous compression in spring = 496.3 mm.

Number of turns

Problem 13.6 A semi elliptical spring, 960 mm long, carries a central load of 4,000 N. Determine the number, breadth and thickness of the leaves, if central deflection is 45 mm. Assume breadth is eight times the thickness. The leaves are available in breadths in multiples of 5 mm and thickness in multiples of 1 mm. The maximum bending stress is also limited to 320 N/mm2. Given E = 210 kN/mm2.

Solution

Substituting the value,

Bending stress,

From Eqs. (13.9) and (13.10),

Available thickness, t = 8 mm (13.11)

Breadth, b = 8 × 8 = 64 mm

= 65 mm available width

so, b = 65 mm, t = 8 mm

Now, nbt2 = 18,000

Substituting the value of b and t,

Let us check for maximum stress,



Problem 13.7 A rigid bar AB, weighing 150 N carries a load W = 40 N as shown in Fig. 13.17. The bar rests on three springs of stiffnesses k1 = 15 N/mm, k2 = 7 N/mm and k3 = 10 N/mm as shown in the figure. If the unloaded springs are of the same length, determine the value of distance x such that the bar remains horizontal.

Solution

Say, the bar moves downwards by deflection δ, reactions from spring will be k1δ, k2δ and k3δ, respectively.

Figure 13.17

Taking moments of forces about A,

400 × (250 − x) + 150 × 250 = k2δ × 250 + k2δ × 500

Substituting the values of k2 and k3

1,00,000 − 400x + 37,500 = 7δ × 250 + 10δ × 500

= 1,750δ + 5,000δ = 6,750δ

1,37,500 − 400x = 6,750δ


or 1,375 − 4x = 67.5δ (13.12)

Taking the moment of forces about point B

k2 × δ × 250 + k1 × δ × 250 = 150 × 250 + (250 + x) × 400

Substituting the value of k2 and k1, we get

7 × 250δ + 15δ × 500 = 37,500 + 1,00,000 + 400x

9,250δ = 1,37,500 + 400x

or, 1,375 + 4x = 92.5δ (13.13)

By adding the Eqs (13.12) and (13.13),

2,750 = 160δ

Deflection, δ = 2,750/160 = 17.1875

Substituting the value of d in Eq. (13.13),

Problem 13.8 A rigid chute plate, 2 m in length, is supported horizontally at a height of 250 mm above a hopper by a spring at one end of stiffness 23 N/mm and a second spring at mid length of stiffness 15 N/mm as shown in Fig. 13.18. Determine the position on the chute to which a component of 2 kN reaches when


https://www.safaribooksonline.com/library/view/strength-of-materials/9789332503519/xhtml/chapter013.xhtml#para-c13e013



the unsupported end of the chute just touches the edge of the hopper.

Solution

When the chute touches the hopper, extension in spring 1 is δ1 and compression in spring 2 is δ2 as shown in Fig. 13.18(b).

As it is obvious from the figure

(250 + δ1) = 2 (250 − δ2)

2δ2 + δ1 = 500 − 250 (13.14)




Say that the load shared by spring 1 is W1 and load shared by spring 2 is W2

−W1 + W2 = 1.5 kN= 1,500 N

(W1 is in tension and W2 is in compression).

k1 = 23 N/mmk2 = 15 N/mm

Numerically,

From Eq. (13.14),

23 × 2W2 + 15W1 = 15 × 23 × 250

46W2 + 15W1 = 86,250 (13.16)

But −W1 + W2 = 1,500

W2 = (1,500 + W1) (13.17)

Substituting the values in Eq. (13.16),

46(1,500 + W1) + 15W1 = 86,250

69,000 + 46W1 + 15W1 = 86,250

61W1 = 17,250



W1 = 2,828 N

W2 = 1,500 + 282.8 = 1782.8 N

Taking moments about A,

At this position, chute just touches the edge of the hopper.

Problem 13.9 In designing a valve spring, it is estimated that the mass of the valve is 0.8 kg and it requires an acceleration of 200 m/s2, when lifting through a height of 6 mm. The free length of the spring is 210 mm and the axial length of the spring is 170 mm, when the valve is shut. If the total lift is 10 mm, determine the maximum force on the spring.

The diameter of the spring wire is 2.8 mm and the maximum shear stress is not to exceed 320 MPa. Determine the mean coil diameter and the number of coils. Given G = 84 kN/mm2.

Solution

Mass of the valve = 0.8 kg

Acceleration = 200 m/s2

Force on valve, F = 0.8 × 200 = 180 N

Total valve left = 10 mm

Valve lift through opening and closing of valve = 6 mm

Free length of the spring = 210 mm

Axial length when valve is shut = 170 mm

Initial compression in length of the spring = 210 − 170 = 40 mm

Further compression when valve is shut = 6 mm

Total compression in length = 40 + 6 = 46 mm

For a force of 100 N,

Stillness of the spring = 160/46 = 3.48 N/mm

Maximum valve lift = 10 mm

Maximum change in the length of the spring = 40 + 10 = 50 mm

Maximum force on the spring Wmax = 3.48 × 50 = 174 N


Mean coil radius,

Mean coil diameter, D = 2R = 15.86 mm

Number of coils,

Problem 13.10 A close-coiled helical spring of 17 mm mean coil diameter and 10 turns is arranged with in and concentric with an outer spring. The free length of the inner spring is 5 mm more than that of the outer spring. The outer spring has 12 coils of mean diameter 25 mm and a wire diameter of 3 mm. The spring load against which a valve is opened is provided by the inner spring. The initial compression in outer spring is 5 mm when the valve is closed, find the stiffness of inner spring if the greatest force required to open the valve by 8 mm is 138 N. Also, find the wire diameter of the inner spring. Given G = 80,000 N/mm2.

Solution

Initial compression in outer spring = 6 mm

Initial compression in inner spring = 6 + 5 = 11mm

(As the free length of the inner spring is 5 mm more than the free length of the outer spring)

Say, k1 = stiffness of inner spring in N/mm k2 = stiffness of outer spring in N/mm

Fi, initial load on valve = 11k1 + 6k2

Outer spring

The valve is opened by 8 mm, additional force required to open the valve, Fo = 8k1 + 8k2 Total load to lift the valve,

Ft = Fo + Fi = 11k1 + 6k2 + 8k1 + 8k2

= 19k1 + 14k2

So, 19k1 + 14k2 = 138 N

19k1 + 14 × 4.32 = 138 N

d1 = 2.116 mm (wire diameter of the inner spring).


o For a close-coiled helical spring, W = axial load, R = mean radius, C = spring index and D/d = ratio of mean coil diameter and wire diameter.

o Stiffness of a close-coiled helical spring,G = shear modulus, n = number of coils,

δ = axial change in length.o In a close-coiled helical spring, angular rotation due

to axial moment M,

where E = Young’s modulus.o For a open-coiled helical spring, W = axial load, R =

mean coil radius, Twisting moment, T ′ =WRcos α and Bending moments, M′ = WRsin α, where α = helix angle.

o Axial deflection, where J = 2I = Polar moment of inertia = πd4/32

Angular rotation, o Open-coiled helical string, subjected to an axial moment M,

Twisting moment = M sin α, Bending moment = M cos α

Angular rotation, Axial deflection,

o Plane spiral spring, b = breadth, t = thickness, M = momentMaximum stress, σmax = 12M/bt2

Energy stored × volume of tripo Carriage spring, n = number of leaves, b = breadth, t =

thickness, L = Length of longest leaf; R = radius of curvature. Initial central deflection, y0 = L2/8R

Proof load, W0 = Enbt3/3LR

σmax = 3WL/2nbt2.

Review Questions

1. What is the effect of spring index on the stresses developed in spring wire?

2. What is Wahl’s factor and how it accounts for the curvature effect on stress in spring wire?

3. Show by a sketch, how resultant shear stress at inner coil radius is maximum.

4. Derive relation ϕ = ML/MI, for a close-coiled helical spring subjected to axial moment M.

5. What are the twisting moment and bending moment components on an open-coiled helical spring subjected to axial load W?

6. What are the applications of a plane spiral spring? Make a simple sketch and mark the point where bending stress is maximum.

7. Where the conical springs are used? Derive the expression for maximum shear stress in a wire of conical spring.

8. In a leaf spring, explain the assumptions that a leaf touches the adjoining leaf only at ends.

9. What is proof load in a cantilever spring?10. What is meant by resilience of a spring?


1. A close-coiled spring absorbs 50 N mm energy is extending by 5 mm, what is the stiffness of spring?

1. 10 N/mm2. 5 N/mm3. 2 N/mm4. None of these

2. A carriage spring is with longest leaf 800 mm long and radius of curvature is 2,000 mm. What is the central deflection?

1. 80 mm2. 40 mm3. 20 mm4. None of these

3. A carriage spring subjected to a central load such that leaves become straight. What is this load called?

1. Safe load2. Proof load3. Ultimate load4. None of these

4. A close-coiled helical spring of stiffness 4 N/mm is in series with another spring of stiffness 6 N/mm. What is the stiffness of composite spring?

1. 5 N/mm2. 4 N/mm3. 2.4 N/mm4. None of these

5. An open-coiled helical spring with α = 45° is subjected to an axial load W such that shear stress due to twisting moment is 100 MPa. What is the bending stress due to bending moment?

1. 50 MPa

2. 100 MPa3. 200 MPa4. None of these

6. A flat spiral spring is subjected to winding couple producing σon at the point of greatest bending. What is the strain energy per unit volume?

1.2.3.4. None of these

7. A closed-coil helical spring of wire, diameter 6 mm, is made by taking mean coil radius as 16 mm, what is its spring index?

1. 2.662. 5.333. 84. None of these

8. Stiffness of a close-coiled helical string of wire diameter d, modulus of rigidity G, number of coils n, mean coil radius R, is


9. A close-coiled helical spring is subjected to an axial moment M, producing an angle of rotation 90° at free end with respect to fixed end, the strain energy absorbed is 100p N mm, what is M ?

1. 100 Nm

2. 200 N mm3. 300 N mm4. None of these

10. Wahl’s factor takes into account1. Curvature of the helical wire2. Direct shear stress3. Both curvature and direct shear effect4. Neither (a) nor (b)

Practice Problems

1. Design a close-coiled helical spring to have a mean coil diameter of 120 mm and an axial deflection of 150 mm under an axial load of 4,050 N, so that the maximum shear stress developed in the spring is not to exceed 320 N/mm2. Steel wires are available in the following diameters: 10, 12 and 16 mm. Determine the most suitable diameter of the wire and number of coils required. Also calculate the maximum shear stress developed in designed spring. Given, G = 84,000 N/mm2.

2. A close-coiled helical spring is made of round steel wire. It carries an axial load of 150 N and is just to get over a rod of 36 mm. The deflection in the spring is not to exceed 25 mm. The maximum allowable shear stress developed in spring wire is 200 N/mm2 (neglecting the effect of direct shear stress). G for steel = 80,000 N/mm2. Find the mean coil diameter, wire diameter and number of turns.

3. A close-coiled helical spring is made of round wire having n turns and mean coil radius is five times the wire diameter. Show that stiffness of wire spring is (R/n) × constant. Determine constant, if G = 82 kN/mm2. Such a spring is required to support a load of 1 kN with 100 mm compression and maximum shear stress is 245 N/mm2, determine (a) mean coil radius and (b) number of turns.

4. A safety valve 80 mm diameter is to blow off at a pressure of 2 N/mm2 gauge. The safety valve is held by a close-coiled helical spring of steel with a mean coil radius equal to 75 mm. Determine the diameter of the steel wire and the number of turns necessary, if the maximum shear stress in the wire is not to exceed 200 MPa

and the spring is initially compressed by 25 mm. Given, G = 84 kN/mm2. Take into account the effect of direct shear stress also.

5. A weight of 250 N is dropped onto a close-coiled helical spring through a height of 800 mm, which produces a maximum instantaneous stress of 200 N/mm2 in the spring. If the mean radius of the coil is five times the wire diameter, determine (a) instantaneous compression in the spring and (b) number of coils in the spring. Given, wire diameter, d = 20 mm and G = 8,410 N/mm2.

6. A laminated carriage spring made of 12 steel plates is 1 m long. The maximum central load is 6 kN. If the maximum allowable stress in steel is 200 MN/m2 and the maximum deflection is 40 mm, determine the thickness and width of plates. Given, E = 200 GPa.

7. A rigid bar AB, weighing 100 N carries a load W = 300 N as shown in Fig. 13.19. The bar rests on three springs of stiffnesses: k1 = 20 N/mm, k2 = 8 N/mm and k3 = 10 N/mm as shown in Fig. 13.19.If the unloaded springs are of the same length, determine the value of distance x such that the bar remains horizontal.

[Hint: Reaction k1δ, k2d, k3δ]

Figure 13.19

8. A rigid chute plate, 2.4 m in length, is supported horizontally at a height of 300 mm above a hopper by a spring at one end of stiffness 30 N/mm and a second spring of stiffness 20 N/mm. Determine the position on the chute to which a component of 2



kN reaches the unsupported end of chute just touches the edge of the hopper (Fig. 13.20).

9. In designing a valve spring, it is estimated that the mass of valve is 1 kg and it requires an acceleration of 150 m/s2; when lifting through a height of 5 mm. The free length of the spring is 200 mm and the axial length of the spring is 160 mm when the valve is shut. If the total valve lift is 10 mm, determine the maximum force on the spring. The diameter of the spring wire is 3 mm and the maximum shear stress is not to exceed 300 MPa, determine the mean coil diameter and number of coils. Given, G = 84 kN/mm2.

Figure 13.20

Special Problems

1. A close-coiled cylindrical spring is of 80 mm mean coil diameter. The spring extends by 40 mm when axially loaded by a weight of 530 N. When the same spring is subjected to an axial couple, M= 28 Nm, there is an angular rotation of free end by 60°. Determine Poisson’s ratio of the spring.


2. Determine the stiffness of a close-coiled helical spring consisting of 10 turns of 4 mm diameter steel wire coiled on a mandrel of 60 mm in diameter. Given, G = 84 GPa.

[Hint: D = 60 + d = 64 mm]3. An open-coiled helical spring made of 10 mm diameter steel rod,

45 mm mean coil radius and 20° angle of helix is subjected to an axial load W. Determine the magnitude of W if the maximum shear stress in the wire is limited to 135 MPa. Calculate the number of turns in the spring if axial extension in the spring under load is 40 mm. Given, G = 80 kN/mm2 and E = 200 kN/mm2 for steel.

4. Two close-coiled helical springs wound from the same wire but with different core radii are compressed between two rigid plates at their ends. Calculate the maximum shear stress induced in each spring, if the wire diameter is 10 mm and the load applied between the rigid plates is 500 N. The core radii of the springs are 75 and 100 mm, respectively. The number of turns in both the springs is the same.

[Hint: k1 + k2 = k, W = W1 + W2]5. A semi-elliptical carriage spring made of steel leaves, 1 m long is

to support a central load of 8 kN with a maximum deflection of 60 mm and maximum bending stress of 320 N/mm2. E = 200 GPa. Calculate the thickness of the leaves and decide their number and breadth.

6. An open-coiled helical spring made of round steel bar 10 mm diameter has 10 coils of 80 mm mean diameter and the pitch is 60 mm. If the axial moment is 40 Nm, find the deflection in spring. Given, E = 210 kN/mm2, v = 0.28.

7. A wagon weighing 3,000 kg is moving at a speed of 1 m/s. How many springs each of 20 coils will be required in a buffer stop to absorb energy of impact during a compression of 150 mm? The mean coil radius is 100 mm and diameter of steel rod comprising the coil is 25 mm. Given, G = 84 kN/mm2.


Exercise 13.1: 129.8 N

Exercise 13.2: d = 6.34 mm, 61.83 mm, axial deflection

Exercise 13.3: d = 5.6 mm, 1.928 Nm (strain energy)

Exercise 13.4: 16.45 mm, 2.1 °

Exercise 13.5: 4.029 Nm, d = 2.585 mm

Exercise 13.6: 5.093 + 134 = 139.093 N/mm2,97.54 N/mm2

Exercise 13.7: 5.01 N mm, 110.2 N mm, 192.4 N/mm2

Exercise 13.8: 501.62 N, 23.78 mm

Exercise 13.9: (a) 1.89 m, (b) 60.95 m, (c) 333.33 N/mm2

Exercise 13.10: 50 mm


1. (d) 2. (b) 3. (b) 4. (c) 5. (c)

6. (c) 7. (b) 8. (a) 9. (d) 10. (c)


1. 16 mm, 14.75 turns, 302.15 N/mm2

2. 40.14 mm, 4.248 mm, 8.4 turns3. constant 2.05, R = 51 mm, n = 10.464. d = 27.6 mm, n = 4.55. 151.4 mm; n = 10.126. t = 6.25 mm, b = 96 mm7. x = 122.8 mm8. x = 1.65 m, W1 = −750 N, W2 = 2,750 N











9. 19.08 mm, 36.7 turns


1. 0.26122. 1.025 N/mm3. 592 N, 8.92 turns4. W1 = 351.63 N, W2 = 148.37 N 134.3 N/mm2, 75.56 N/mm2

5. t = 6.66 mm, n = 10, b = 85.54 mm6. 6.56 mm7. 6 springs

2a. Columns and Struts CHAPTER OBJECTIVES

Any strut or a column member in a machine, or a structure is designed on the basis of failure against buckling. Professor Euler initially provided a theory of buckling of such members. In this chapter, students will learn about:

o Theory of buckling developed by Euler for long columns.o Euler considered the columns in the buckled state and then

derived expressions for buckling loads. Professor. Rankine modified the Euler’s buckling load formula by combining the crushing load for short columns and Euler’s buckling load for long columns.

o Johnson’s parabolic formula for buckling load—an empirical formula.

o Secant formula providing information on maximum stress developed in a column section due to eccentric load on column.

o Professor Perry’s approximate formula for eccentrically loaded columns.

o Long columns with eccentricity in its geomentry.o Professor Perry Robertson’s formula for combining

eccentricities in geometry of columns and in loading of columns.

o Columns subjected to lateral loads in addition to axial buckling loads.

o Energy approach for determination of buckling loads in columns with different end conditions.

Introduction

Collapse of long structural members under compressive axial loads is a common occurrence. Through an efficient design of strut or column members in a structure, the collapse must be prevented; otherwise, in addition to damage to structure, there will be loss of human lives.

Euler has developed a buckling theory for long column members; however, for short columns, the theory gives absurd results. However, he provided the basic formula for buckling of columns. Rankine modified the Euler’s buckling load formula by combining the crushing load for a short column and the buckling load (provided by Euler) for a long column. In Rankine’s formula, there are two experimentally determined constants. This formula is applicable for long as well as medium-long columns.

Many a times, load is eccentrically applied on a column and buckling takes place at a smaller load than the load given by Euler or Rankine. For eccentrically loaded columns, expressions for maximum stress have been derived in the form of a secant formula. Many researchers such as Professor Perry and Perry–Robertson have modified the secant formula for eccentrically loaded columns.

Higher-order differential equations have been used for determining the buckling loads of columns with different end conditions. Moreover, energy approach is briefly discussed to determine the buckling loads of columns, taking various shape functions for the deflection curve.

Euler’s Theory of Buckling

Euler has developed a theory of buckling for columns or struts, but considered the column/strut already buckled under a

particular load while developing the theory of buckling. He has made the following assumptions:

1. material of the column is homogeneous and isotropic,2. compressive load acting on the column is fully axial,3. column has failed only due to buckling,4. weight of the column is negligible,5. column is initially straight and buckles suddenly at a

particular load,6. fixed ends are rigid and7. hinged ends are frictionless.

Example 14.1 Consider a column AB with one end fixed and the other end free, buckled under the load P as shown in the Fig. 14.1. Length of the column is L. Deflection at free end B is a as shown, and the buckling load is P. Consider a section of the column at a distance x from end A.

Solution

Bending moment at section XX = P(a − y),

Solution of this differential equation is

y = A sin mx + B cos mx + a

where and A and B are constants.

(One can verify the correctness of the solution by differentiating y two times with respect to x.)

End conditions



y = 0, x = 0, dy/dx = 0 at x = 0, at end A.

y = a, x = L, at end B

so, 0 = A sin 0 + B cos 0 + a

0 = 0 + B + a

Figure 14.1 Column fixed at one end and free at the other end

or constant, B = −a

dy/dx = Am cos mx − Bm sin mx

0 at x = 0

0 = A mx cos 0 − B × sin 0

0 = Am

m # 0, because then the load P will become zero, if m = 0

so, constant A = 0

Finally, y = −a cos mx + a

At end B, x = L, y = a

a = −a cos mL + a

or,

a cos mL = 0

a # 0, because if a = 0, then no buckling takes place

so,

mL=π/2 (minimum significant value)

Euler’s buckling load, Pe = π2EI/4L2.

Example 14.2 Let us take another case of a column hinged at one end and fixed at the other end as shown in Fig. 14.2. The column is under buckled condition. P is the buckling load, H is the horizontal reaction at B. Considering a section at a distance x from fixed end A as shown.


Figure 14.2 Column with one end fixed, other end hinged

Solution

Bending moment at sections XX


End conditions x = 0, y = 0

Constant, B = −HL/P (14.4)

Then,

So, Am = H/P

Constant, A = H/mP (14.5)

Putting the values of A and B in Eq. (14.3)

At the end B, x = L, y = 0,

so

or tan mL = mL (14.6)

Exercise 14.1 A column of length L and flexural rigidity EI is hinged at both the ends. Derive the expression of Euler’s buckling load.

[Hint: M = −Py, y = A sin mx + B cos mx, End conditions x = 0, y = 0, x = L, y = 0]

Exercise 14.2 Derive the expressions for Euler’s buckling load of a column fixed at both the ends. Length of column L, buckling


load P (Fig. 14.3). (Fixing moments at fixed ends are MA and MB, respectively)

M at section shown = –Py + MA

Figure 14.3

y = A sin mx+ B cos mx + MA/ P

End conditions x = 0, y = 0, dy/dx = 0

x = L, y = 0, dy/dx = 0

Equivalent Length

Buckling load of a column is given by Euler’s formula and it depends on the end conditions of a column. However, to express the buckling load formula in a simplified manner, equivalent length of column is determined that is equivalent to the length of column with both ends hinged. This length can be obtained by completing the bending curve of the column with different end conditions similar to the bending curve of a column with both the ends hinged as shown in Fig. 14.4(a–d).

Pe = Euler’s buckling load = π2EImin/Le2



Figure 14.4 (a) A column with both ends hinged, Le = L; (b) a column with both ends fixed, Le = L/2; (c) a column with one end fixed and the other end hinged, and (d) a column with one end fixed and the other end free, Le = 2L.

Example 14.3 An allowable axial load for a 3-m-long column with hinged ends is 30 kN. Another column of the same material, same cross-section and same length, but with one end fixed and the other end hinged, suffers buckling; what is buckling load for the column?

Solution

Euler’s buckling load,

where Le = L, both ends hinged.

other column with one end fixed and the other end hinged

Exercise 14.3 A column of a circular section, diameter d and length L, buckles at a load of 25 kN, when the column is fixed at one end and free at the other end. If both the ends of the column are now fixed, what will be its buckling load?

Limitations of Euler’s Theory of Buckling

Euler’s theory of buckling is applicable for long columns. If there is a limit on the slenderness ratio of the column and it is less than this limiting value, then Euler’s formula gives the value of buckling load that is greater than the crushing load, which is not possible.


where

E = Young’s modulus

Imin = Ak2min, where A is the area of cross-section

Le = equivalent length, kmin = the minimum radius of gyration

so

Pe/A < σc (the crushing stress of a short column)

Le/Kmin = the slenderness ratio of a column

Mild stress

E = 208 GPa = 2,08,000 N/mm2

σc = crushing strength

= 320 N/mm2

For mild steel column, the slenderness ratio should be greater than 80, so that Euler’s formula can be used to predict the buckling load of a column.

Example 14.4 For what length of a mild steel bar of 60 mm in diameter used as strut, the Euler’s theory is applicable, if the ultimate compressive strength is 0.33 kN/mm2 and E = 210 kN/mm2, when one end of the strut is hinged and the other end is fixed?

Solution

End conditions, one end hinged and the other end fixed

Equivalent length,

kmin = the minimum radius of gyration

Length of the column,

Length should be greater than 1.68 m.

Exercise 14.4 For what length of a cast iron column of 80 mm in diameter, the Euler’s theory is applicable, if σc = 550 N/mm2 for CI and E = 102 kN/mm2, the column is hinged at both the ends?

Higher-Order Differential Equation

For any end conditions, the buckling load of a column/strut can be determined by a higher order differential equation as follows:


y = C1sin kx + C2cos kx + C3x + C4 (14.7)

Constants C1, C2, C3 and C4 are determined using end conditions.

Example 14.5 Let us consider a column with both the ends fixed as shown in Fig. 14.5.

Figure 14.5

Solution

Column AB of length L is fixed at ends A and B. At the ends, both slope and deflection are zero, that is, the boundary conditions of the column are:

At end A, x = 0, y = 0, dy/dx = 0At end B, x = L, y = 0, dy/dx = 0

Differentiating Eq. (14.7) with respect to x, we get

dy/dx = C1k cos kx − C2k sin kx + C3 (14.8)

Taking x = 0, y = 0,x = L, y = 0, we get,

0 = C1sin 0 + C2cos 0 + C3x 0 + C4



0 = 0 + C2 + 0 + C4 (14.9)

0 = C1sin kL + C2cos kL + C2L + C4 (14.10)

Taking dy/dx = 0 at x = 0, and x = L we get,

0 = C1k cos 0 − C2k sin 0 + C3

0 = C1k − 0 + C3 (14.11)

0 = C1k cos kL − C2sin kL + C3 (14.12)

Using Eqs (14.9), (14.10), (14.11) and (14.12), the following matrix is made:

Let us take 1 − cos kL = 0,

cos kL = 0





kL = 0, 2π, 4π - - -

So, kL = 2π (minimum significant value)

or,

Squaring both the sides,

Buckling load,

Moreover, sin kL = sin 2π = 0(also), which will also give P = 4π2EI/L2.

Example 14.6 Let us take another case of a column fixed at one end and hinged at the other end as shown in Fig. 14.6. End A is fixed as shown. End B is hinged.

Figure 14.6

Solution

Boundary conditions

at x = 0, end A, y = 0, dy/dx = 0

at x = L, end B, y = 0, d2y/dx2 = 0 (moment at hinged end is zero).


Equation of deflection,

y = C1sin kx + C2cos kx + C3x + C4 (14.3)

At x = 0, y = 0

0 = C1 × sin 0 + C2cos 0 + C3 × 0 + C4

= 0 + C2 + 0 + C4 (14.16)

At

x = 0, dy/dx = 0

0 = C1kcos 0 − C2ksin 0 + C3

0 = C1k + C3 (14.17)

At

x = L, y = 0

0 = C1sin kL + C2cos kL + C3L + C4 (14.18)

At

x = L,d2y/dx2 = 0, that is, bending moment is zero at the hinged end

0 = −C2k2sin kL − C2k2cos kL (14.19)

Using Eqs. (14.16)–(14.19), the following matrix is made:



or 0 = kL coskL − sin kL

or tan kL = kL

or tan 0 = θ

Buckling load,

Exercise 14.5 Using higher-order differential equation, determine the buckling load for a column hinged at both the ends.

Exercise 14.6 Using higher-order differential equation, determine the buckling load for a column fixed at one end and free at the other end.

[Hint: say, y = C1sin kx + C2cos kx + C3x + a, where a = deflection at free end]

Rankine Gordon Formula

Euler has developed a theory for the buckling load of long columns, but short columns get crushed under the compressive load and crushing stress (σc) is much greater than the stress (σe)

given by Euler’s buckling load. Some columns are of medium length and cannot be classified as short columns. Rankinehas combined the two loads, that is, the crushing load for short column and the Euler’s buckling load for long columns to determine the buckling load for any long or short column.

Rankine’s load, PR is

or

Rankine’s load,

where σc = crushing stress,

Rankine has taken σc/π2E a constant that is determined experimentally

Rankine’s load,

both σc and a are determined experimentally.

Le = equivalent length of column

kmin = minimum radius of gyration of the section of the column

Le/Kmin is known as slenderness ratio of the column, an important parameter, and the buckling load depends on the slenderness ratio; more the slenderness ratio, less is the buckling load. Table 14.1 gives the values of σc and a for various materials.

https://www.safaribooksonline.com/library/view/strength-of-materials/9789332503519/xhtml/chapter014.xhtml#para-8306


Table 14.1 Rankine’s constants for different materials

Example 14.7 A cast iron column of a hollow circular section with an external diameter of 250 mm and a wall thickness of 45 mm is subjected to an axial compressive load. The column is 7 m long with both ends hinged. Taking factor of safety (FOS) as 8, determine safe value of P.

Rankine’s constants are, σc = 560 N/mm2, a = 1/1,600

Solution

External diameter, D = 250mm

Wall thickness, t = 45mm

Internal diameter, d = 250 − 2 × 45 = 160 mm


Length, L = 7 m = 7,000 mm

End condition: both the ends are hinged

Rankine’s load,

Safe load, P′R = 286.6/8 = 35.83 kN

Exercise 14.7 A hollow cast iron column has 200 mm outside diameter, 150 mm inside diameter and is 6 m long with both the ends fixed. It is subjected to an axial compressive load P. Taking a FOS as 6, determine safe Rankine’s buckling load.

Constants are σc = 550 N/mm2, a = 1/1,600 for both ends hinged.

Johnson’s Parabolic Formula

We have learnt so far that buckling loads depend upon the slenderness ratio (Le /kmin) of the column. As the slenderness ratio increases, buckling required for the column decreases. Based on this principle, Johnson has suggested a formula:

Working stress, σw = σc′[1 − φ(Le/k)] as function of Le/k, slenderness ratio

Taking φ(Le/k) = b(Le2/k2), where b is a constant

σw = σc′[1 − b(L2e/k2)], an equation of a parabola.

σ’c = allowable stress in compression taking into account the

FOS (factor of safety)

= 110 N/mm2 for mild steel.

Constant, b = 0.00003 for hinged ends

= 0.00002 for fixed ends.

Straight line formula

σw′ = σc′[1 − C(Le/k)], applicable for column for which slenderness ratio is greater than 90.

For structural steel,

σc′ = 140 N/mm2 in compression

C = 0.0054 for hinged ends

= 0.0038 for riveted ends.

Example 14.8 A strut is built-up of two 100 mm × 45 mm channels placed back-to-back at a distance of 100 mm apart and riveted to two flange plates each 200 mm × 10 mm symmetrically, properties of one channel are:

Area, A = 7.41 cm2,Ixx = 123.8 cm4,Iyy = 14.9 cm4, and = 1.4 cm (distance of CG from outer edge of web). If the effective length is 5 m, calculate the working load for the strut using Johnson’s parabolic formula:

where σw = σc′[1 − b(Le/k2)], where constant b = 0.00003 for hinged ends and

σc′ = 110 N/mm2

Solution

Area of cross-section of built-up section, A

= 2 × 200 ×10 + 2 × 741

= 4,000 + 1,482 = 5,482 mm2

Ixx of built-up section

= 2 ×123.8 ×104 + 2 × 200 ×10 × (55)2

= 247.6 ×104 +1210 ×104

= 1457.6 ×104 mm4

Iyy of built-up section (Fig. 14.7)

Iyy = 2 ×14.9 ×104 + 2 × 741× (50 +14)2 + 2 ×10 × (2002/12)

= 29.6 ×104 + 607 ×104 +1,333.33×104

= 1,970 ×104 mm4

Figure 14.7 Example 14.8

https://www.safaribooksonline.com/library/view/strength-of-materials/9789332503519/xhtml/chapter014.xhtml#exe-14.8


Now,

Exercise 14.8 A stanchion is built up of 3 − 200 × 100 mm Rolled Steel Joist (RSJ) as shown inFig. 14.8. If the length of the stanchion is 6 m, calculate the working load. The working stress is given by:

Properties of one RSJ, area = 2,527 mm2, Ixx = 1,696.6 ×104 mm4, Iyy = 115.4 ×104 mm4 and web thickness = 5.4 mm.


Figure 14.8 Exercise 14.8

Eccentric Loading of Columns

A column AB of length L loaded eccentrically e (eccentricity with respect to axis of column) buckled under load P as shown in the Fig. 14.9. Say, deflection at end B is a, when the column has buckled. Consider a section at a distance x from end A, where deflection is y.

Figure 14.9

Bending moment at the section,

M = P(a + e − y)

or, EI(d2y/dx2) = P(a + e − y)

or,

Solution of this differential equation is,

y = A sin mx + Bcos mx + (a + e)


https://www.safaribooksonline.com/library/view/strength-of-materials/9789332503519/xhtml/chapter014.xhtml#exc14.8

where m = A and B are constants

At end A, x = 0, y = 0 dy/dx = 0

Pulling x = 0, y = 0

0 = Asin 0 + Bcos 0 + (a + e)

0 = 0 + B + (a + e)

or constant, B = −(a + e)

Moreover, dy/dx = Am cos mx − Bm sin mx

dy/dx = 0, at x = 0

0 = Am cos 0 − Bm sin 0

0 = Am

m ≠ 0, because m =

therefore, constant A = 0.

Finally, y = −(a + e) cos mx + (a + e)

At the end B, x + L, y = a, putting this boundary condition

a = −(a + e)cos mL + (a + e)

or, −e = −(a + e) cos mL

(a + e) = e sec mL

Maximum bending moment occurs at fixed end A.

Mmax = P(a + e) = Pe sec mL

σb = maximum stress due to bending

= P e sec mL/ , secant formula,

where = section modulus = I /yc,

I = moment of inertia

yc = distance of extreme layer in compression from neutral layer

σc = direct compressive stress.

Resultant stress at fixed end, = P/A

In this case, we have considered that one end of the column is fixed and the other end is free. In this case, Le = 2L, that is, equivalent length, Le = 2L. Secant formula can be generalized for any column with any type of end conditions as follows:

σr = resultant maximum stress

where Le is the equivalent length.

Example 14.9 A steel tube of 80 mm outer diameter, 50 mm inner diameter and 3 m long is used as a strut with both ends hinged. The load is parallel to the axis of the strut but is eccentric. Find the maximum value of eccentricity so that the crippling load on strut is equal to 50 per cent of the Euler’s crippling load.

Yield strength = 320 N/mm2, E = 210 kN/mm2.

Solution

End conditions: both ends hinged

Length, Le = L = 3 m

E = 210 kN/mm2

Steel tube


Crippling load due to eccentricity, P = Pe /2 = 1,96,184.7 N

Eccentricity, e = (320 - 64.05)/10.37 = 24.68 mm

Exercise 14.9 A steel tube of 80 mm outer diameter, 60 mm inner diameter and 2.8 m long is used as a strut with the ends hinged. The load is parallel to the axis of the strut but is eccentric. Find the maximum value of the eccentricity so that the crippling load on the strut is 60 per cent of the Euler’s buckling load. Given yield strength = 320 N/mm2.

E = 210 kN/mm2

Professor Perry’s Approximate Formula

Professor Perry has given an approximate formula for the buckling load of eccentrically loaded columns.

In the secant formula for maximum stress due to buckling load,

If Euler’s buckling load, Pe = π2EI/L2e

putting the value in secant formula,

Professor Perry found through experiments that the expression,

where

Pe = Euler’s buckling load (without eccentricity)

P = buckling load of column with eccentricity.

Moreover, σe = Pe /A, σo = P/A

where A is the cross-sectional area of the column.

Now,

Section modulus, = Ak2/yc, where yc is distance of extreme layer in compression from neutral layer.

However,

This is Professor Perry’s approximate formula, knowing the value e, yc and k, one can determine the stress σo due to load P on the eccentrically loaded column.

Example 14.10 A column is of I-section, 125 mm × 250 mm. Find the safe load for this column of length of 5 m, hinged at both the ends using Professor Perry’s formula if maximum compressive stress is limited to 80 N/mm2 (Fig. 14.10).

For RSJ,

Ixx = 37.18 ×106mm4, A = 3,553 mm2, Iyy = 1.93×106mm4

E = 200 GPa, eccentricity from y-axis is 30 mm.

Solution


Eccentricity is about x-x axis, e = 30 mm, we will take Iyy as moment of inertia

Length, L = 5,000 mm

Iyy = 1.93 × 106 mm4

Figure 14.10

Safe load = 38.10 × 3,553

= 1,35,369 N

= 135.37 kN

Exercise 14.10 A 400 mm × 140 mm RSJ is used as a sturt with hinged ends, having a length of 6 m. Using Prof. Perry’s formula, determine the safe load for: (a) eccentricity along x-x axis, e = 24 mm, (b) minimum allowable permissible compressive stress is 75 MPa and (c) for the joist A = 7,846 mm2, Ixx = 20,458.4 × 104 mm4,

Iyy = 622.1×104 mm4, E = 210 kN/mm2

Long Columns with Eccentricity in Geometry

A column AB of length L with both the ends hinged has initial eccentricity e′ in the centre. Assuming deflection curve to be sinusoidal (Fig. 14.11).

Eccentricity at any section,

e′ = maximum initial eccentricity at centre.

Taking a section at a distance of x from end A, considering column buckled under load P.

Final deflection = y

Change in deflection = y-y′ (as shown)

BM at the section at a distance x from A as shown

Figure 14.11


as given in Eq. (14.21)

Putting this value in Eq. (14.22)

Let us assume the solution as

Putting this value in Eq. (14.24),

Putting the assumed value of y




However, π2EI/L2 = Pe, Euler’s buckling load

Finally, the deflection curve equation will be,

Maximum deflection occurs at the centre, where

Maximum bending moment,

σmax = maximum compressive stress at central section of column

where

k = radius of gyration

yc = distance of extreme layer

However, Pe = σeA, Euler’s load.

P = σeA, putting these values; load applied.

Example 14.11 A 5-m-long hollow circular steel strut having an outside diameter of 120 mm and an inside diameter of 80 mm with both the ends hinged is initially bent. Assume that the centre line of the strut as sinusoidal with maximum deviation of 6 mm. Determine the maximum stress developed due to an axial load of 100 kN.

E = 210 kN/mm2

Solution

Length, L = 5 m

Eccentricity, e′ = 6 mm

Area of cross-section, A = (π /4)(1202 - 802) = 6283.2mm2

Moment of inertia, I = (π /64)(1204 — 804)

= 816.816 × 104 mm4

Distance, yc = 120/2 = 60 mm

e′ = 6 mm

Exercise 14.11 A 4-m-long hollow circular aluminium strut having an outside diameter of 150 mm and an inside diameter of 100 mm with both ends hinged is initially bent. Assuming that the centre line of the strut as a sinusoidal curve with maximum deviation of 8 mm. Determine the maximum stress developed due to an axial load of 60 kN.

E = 70 kN/mm2

Professor Perry Robertson Formula

Professor Perry gave the relationship between σe, σo and σmax (discussed in the last section ‘Long Columns with Eccentricity in Geometry’) for a long column with eccentric loading as follows:

where = eccentricity in loading (load is not axial)

Moreover,

where e′ = eccentricity in geometry of strut

say, e1 = 1.2e + e′

For an initially bent and eccentrically loaded column, formula has been modified as

where σmax = allowable stress, σ

say e1 yc/k2 = λ, then

Rearranging the terms, we get,

σo2 − σo [σ + σe(1 + λ)] + σ σe = 0, (14.29)

a quadratic in σo.

where λ = e1 yc / k2 and σe = πE/(L2 k2)for both ends hinged.

Professor Robertson after having experimental observations came to the conclusion that λ = 0.003(L/k) is valid for a large number of observations.

However, λ depends on allowable stress in tons/in2 and σe = π2E/(L/k)2 in tons/in2.

However, in the Eq. (14.28), there are ratios of σ/σ0, and σo/σe, dimensionless λ = e1 yc/k2 also dimensionless, the expression for λ remains the same for stresses in N/mm2.

Example 14.12 Two 200 × 70 mm 6-m-long MS channels are welded together at their toes to form a box section of 200 × 140 mm.The box section is used as a strut with both the ends hinged. Estimate the safe load for this strut using Prof. Perry Robertson’s formula taking allowable stress, σ = 250 MPa, λ = 0.003(L/K)(Fig. 14.12).

Figure 14.12 Built-up section

For each channel




Area of cross-section, A = 1,777 mm2

Ixx = 1,161.9 × 104 mm4, Iyy = 84.2 × 104 mm4

= 19.7 mm, E = 210 kN/mm2

Solution

Total, A = 2 × 1,777 = 3,554 mm2

Ix′x′ = 2 × 1,161.9 × 104 mm4

= 2,323.8 × 104 mm4

Iy′y′ = 2 × 84.2 × 104 + 2 × 1,777(70 − 19.7)2

= 168.4 × 104 + 899.2 × 104

= 1,067.6 × 104 mm4

Note that

Length of the strut,

Allowable stress,

Exercise 14.12 A stanchion is built up of a 500 mm × 180 mm RSJ with 200 mm × 20 mm plates riveted to each flange as shown in Fig. 14.7 of example 14.8. Estimate the safe load for this stanchion of a length of 5 m, with both the ends hinged using Prof. Perry Robertson’s formula, taking allowable stress, σ = 280 MPa, λ = 0.003(L/k) for the joist.

A = 9,550 mm2, Ixx = 3.8529 × 106 mm4

Iyy = 10.639 × 106 mm4

E = 208 GPa

Hint: [Taking Iyy, calculate, k and σe]

Lateral Loading of Strut with Point Load

A strut or column has buckled under the combined action of axial thrust P and transverse load (at centre) as shown in Fig. 14.13. Lateral loading produces deflection in the strut and axial thrust produces a bending moment on account of deflection. Figure





14.13 shows a column AB of length L, hinged at both the ends and carrying a central transverse load W as shown. Due to W at C, there are horizontal support reactions, W / 2each.

Consider a section at a distance x from end As

Bending moment at the section, M = −Py − (Wx/2)

or

Figure 14.13 Lateral point load on a strut

or


where

Boundary conditions are x = 0, y = 0,

0 = A cos 0 + B sin 0 − 0


0 = A + 0 − 0

or constant, A = 0

Moreover,

Equation of deflection curve becomes

Maximum deflection occurs at centre at x = L/2

Maximum bending moment occurs at the centre

putting the value of m

yc = distance of extreme layer in compression from neutral layer

or,

Example 14.13 A 4-m-long horizontal pin-ended strut is formed from a standard T-section of 150 mm × 100 mm × 12.5 mm. The axial compressive load is 60 kN. A lateral concentrated load of 6 kN acts at the centre of the strut. Find the maximum stress developed if the xx axis is horizontal and the table of the T-section forms the compressive face. The centre of gravity is 24 mm away from the edge of the table (Fig. 14.14).

Given Ixx = 250 × 104mm4, A = 3,100 mm2, E = 200 GPa

Figure 14.14

Solution

Axial load, P = 60 kN


Lateral load, W = 6 kN

Ixx = 250 × 104 mm2

A = 3,100 mm2

Length, L = 4 m

T-section, yc = 24 mm

Exercise 14.13 A circular steel strut of a diameter of 25 mm and length of 1 m is subjected to an axial thrust of 12 kN. In addition a lateral load W acts at the centre of the strut. If the strut is to fail at σmax = 320 N/mm2, determine the magnitude of W. Given E = 210 kN/mm2.

Strut with an Uniformly Distributed Lateral Load

A strut subjected to axial thrust P and lateral load w per unit length is shown in Fig. 14.15. The length of the strut is L and it is hinged at both the ends A and B. Consider a section at a distance x from end A. The bending moment at the section,

Figure 14.15

Solution of this differential equation, y = complementary function + particular integral.

Complementary function = Acos mx + Bsin mx, where .

Particular integral,


or constant, A = wEI / P2 (14.31)

Then,

dy/dx = 0 at x = L/2, at the centre of the beam due to symmetrical loading about centre C,

However, A = wEI / P2

So constant,

Finally, the equation of y

At the centre, deflection is maximum, therefore, taking

Bending moment at the centre,

Maximum stress,

Example 14.14 A circular rod of diameter 50 mm is supported horizontally through pin joints at its ends and carries an uniformly distributed load of 1 kN/m run throughout its length and an axial thrust of 25 kN. If its length is 2.4 m, estimate the maximum stress induced in the rod, E = 200 GPa.

Solution

Axial thrust, P = 25,000 N

Diameter, d = 50 mm

Area, A = (π/4)d2 = 1963.5 mm2

w = 1 kN/m = 1 N/mm

E = 2,00,000 N/mm2

yc = 25 mm

Length, L = 2.4 m

Exercise 14.14 A rod of rectangular section 80 mm × 40 mm is supported horizontally through pin joints at its end and carries a vertical load of 3,000 N/m and an axial thrust of 100 kN. If its length is 2.0 m, estimate the maximum stress induced.

E = 208 × 103 N/mm2

Energy Approach

To determine the critical load at which a column ceases to be in a stable equilibrium, energy criterion can be used. Moreover, in a situation where the exact solution of a differential equation is not possible or difficult to obtain, energy approach can be used to get a good approximate solution. However, this approach converges to the exact solution if a large number of terms are taken to represent the deflection curve.

Now,

Energy stored in a system = work done by external loads

δU + δVe = 0

δVe = external work done due to virtual displacement.

δ(U+Ve) = 0 δ(Kp) = 0

U + Ve = Kp = a constant referred to as total potential of the system.

Energy stored,

External work done,

For small values of P, Kp is positive for any non-trivial admissible function y(x).

The critical load can be obtained by assuming suitable admissible function.

Critical condition is reached when constant Kp = 0

Example 14.15 Let us consider a case of column that is fixed at one end and free at the other end, with boundary conditions (Fig. 14.16).

Figure 14.16

Solution



Deflection, y = 0 and x = 0

slope, dy/dx = 0, x = 0

Moment at end B is zero, that is,

Let us take a shape function,

Putting these values in Eq. (14.32)

Pcr = 3EI / L2, but the exact value given by Euler’s theory is π2EI / 4L2 = 2.467(EI /L2)

Let us take more terms in the deflection shape function


Putting these values in the equation of total potential Kp,

Exercise 14.15 Consider a column with both ends hinged as shown in Fig. 14.17. Take the shape function y = A(x4 –


2Lx3 + L3x), using energy approved, that is, total potential, derive expression for its buckling load.

Problem 14.1 A column made of an aluminium alloy of rectangular section b × d and a length of 600 mm is fixed at end B as shown in Fig. 14.18. Two smooth and rounded fixed plates at the end A restraint the end A to move in one of the vertical planes of symmetry of the column. Determine the ratio of b/d of two sides of cross-section for the efficient design against buckling.

E = 67 GPa, P = 15 kN, FOS = 2


Solution

For the efficient design, the buckling load for two possible modes of failure should be the same. There are two types of end conditions in the present case:

1. one end is fixed and the other end is free, I1 = bd 3/12.2. one end is fixed and the other end is hinged, I2 = db 3/12.



Figure 14.18

Taking Euler’s formula for buckling

Design load = 2 × 15 = 30 kN

Length = 600 mm

Dimensions of the section are 13.63 mm × 8.6 mm.

Problem 14.2 A long strut AB of length L is of uniform section throughout. A thrust P is applied at the ends eccentrically on the same side of the centre line with eccentricity at the end B twice than that at end A (Fig. 14.19). Show that the maximum bending moment occurs at a distance x from end A, where

Figure 14.19

Solution

The strut is buckled under the thrust load P.

P at two ends produce a couple P(2e − e) (cw) that is to be balanced by anticlockwise couple of horizontal reactions F each at A and B, as shown F × L (ccw).

Force F is unknown.

so 2Pe − Pe − FL = 0

or,


Consider a section at a distance x from end A,

Bending moment at the section, M = –P(2e + y) + F(L – x)

or

Solution of this differential equation is,

where A and B are constants,

End conditions

At end A, x = 0, y = 0, putting this in Eq. (14.36)

0 = A cos 0 + B cos 0 – e – 0

Constant, A = e

Moreover, at end B, k = L, y = 0

0 = Acos kL+ Bsin kL – 2e

However, A = e

so, 0 = e cos kL + Bsin kL – 2e

Constant,


Finally, the equation of deflection becomes,

Bending moment at the section,

For maximum bending moment, dMx /dx = 0

Problem 14.3 A strut of length L is fixed at its lower end, its upper end is eccentrically supported against a lateral deflection (through a spring), so that the resisting force is k times the end deflection, kis spring constant. Show that the crippling load P is given by 1 – (P/kL) = tan mL/mL.

where,

Figure 14.20

Solution

Figure 14.20 shows a strut AB of length L, fixed at end A and free at end B. At the end B, a horizontal reaction H = ka, given by a spring and P is the crippling load. Consider a section, at a distance x from end A.

Bending moment at the section, M = P(a − y) − ka(L − x)

Solution of the differential equation

At end A, fixed end, x = 0, y = 0


Then,

or constant, B = –(ka/mP)

Finally, the equation of deflection curve is

At the end B,

or

Problem 14.4 A thin steel bar of a rectangular section 6 mm × 4 mm is axially compressed by 200-N load between two plates that are fixed at a constant distance of 160 mm apart as shown in Fig. 14.21. The assembly is made at 24°C, how high can the temperature of the bar rise so as to have an FOS of 2 with respect to buckling. E = 200 GPa, α = 15 × 10−6/°C?

Figure 14.21

Solution

Section = 4 × 6 mm



End conditions: both ends hinged.

L = 160 mn


Initial compressive load = 200 N

Balance load to be applied through expansion of column (prevented by fixed ends)

= 1,233.7 − 200 = 1,033.7 N

Say temperature rise is ∆T

Thus

αΔT EA = 1033.7 N

where A = area of cross-section of column

15 × 10−6 × ΔT × 2,00,000 × 24 = 1,033.7 N

Final temperature, = 24 + 14.36 = 38.36°C.

Problem 14.5 A 6-m-long RSJ of a rectangular section of 325 mm × 165 mm is used as a strut, with one end is fixed and the other end is hinged. Calculate the crippling load by Rankine’s formula. Compare this with the load obtained by Euler’s formula. For what length of this strut will the two formulae give the same crippling load for the joist?

Area of cross-section, A = 5,490 mm2

Ixx = 9,874.6 × 104 mm4

Iyy = 510.8 × 104 mm2

E = 210 kN/mm2

σc = 320 N/mm2

a = for both hinged ends.

Solution

Iyy < Ixx

L = 6 m

(one end is fixed and the other end is hinged)

Euler’s buckling load.

Rankine’s load,

Equating two loads,

Putting the values,

L = 8.48 × 103m, for this length both Pe and PR are equal

L = 8.48 m

Problem 14.6

A round bar in a vertical position is clamped at the lower end and is free at the other end. The effective length is 2 m. If a horizontal force of 400 N at the top produces a horizontal deflection of 15 mm, what is the axial buckling load for the bar under the given conditions?


Solution


The bar is fixed as a cantilever beam (as shown Fig. 14.22).

Deflection at the top

Buckling load, Pe = π2 × EI /4L2, as one end is fixed and the other end is free

Problem 14.7

A 3-m-long straight steel bar of a rectangular section of 30 mm × 20 mm is used as a strut with both the ends hinged. Assuming that Euler’s formula is applicable and the material attains its yield strength at the time of buckling, determine the central deflection. E = 210 kN/mm2. Yield strength of steel is 320 N/mm2.

Solution

Section 30 × 20 mm2, A = 600 mm2

E = 2,10,000 N/mm2


L = 3,000 mm (both the ends are hinged)


Say, central deflection, ymax.

Problem 14.8

A steel column of a hollow circular section of 200 mm external diameter, 160 mm internal diameter and 7-m-long has to take a 20-kN load at an eccentricity of 30 mm from the geometrical axis. If the ends are fixed, calculate the maximum and minimum stress intensities induced in the section, taking E= 210 kN/mm2. Moreover, calculate the maximum permissible eccentricity so that no tension is induced any where in the section.

Solution

Section

D = 200 mm

d = 160 mm

A = (2002 − 1602) = 1.131 × 104mm2

I = (2004 − 1604) = 0.4637 × 108mm4

P = 200 kN

E = 2,10,000 N/mm2

L = 7 m

Ends are fixed, Le = L/2 = 3.5 m = 3,500 mm

yc = 100 mm

Eccentricity, e = 30 mm

For no tension

Eccentricity,

= 34.3 mm


o Euler’s buckling load, Pe = π2EImin /Le2

where Imin = minimum moment of inertiaLe = equivalent length of a column depending upon end conditions.

o Euler’s formula is applicable for the slenderness ratio , where σc is ultimate compressive strength of the material.

o Higher-order differential equation

o Solution, y = C1sin kx + C2 cos kx + C3x + C4

using different end conditions, the buckling load is determined.

o Rankine’s load = where

σc = crushing strength of column Le = equivalent length, kmin = minimum radius of gyration

o Johnson’s parabolic formula for working stress

where σc′ = allowable stress in compressive taking into account FOS

b = constant

o Eccentrically loaded column

where P = axial applied load Z = section modulus e = eccentricity I = moment of inertia

o Professor Perry’s formula

where σ = maximum stress allowed σ0 = P/A, applied load/area σe = Pe /A, Euler’s load/area e = eccentricity yc = distance of extreme layer in compression from

neutral layer

o If the column has initial eccentricity e′ along central section

o Energy approach, total potential, Kp = 0

to determine Pcr, critical load.

Review Questions

1. What are the drawbacks of Euler’s theory of buckling?2. What is limiting value of the slenderness ratio beyond

which Euler’s formula is applicable?3. What are the merits of Rankine’s load over Euler’s load in

buckling?4. What do you mean by equivalent length of a column?5. Discuss the effect of the slenderness ratio of a column over

buckling load?6. Why the value of ‘a’ Rankine’s constant varies for different

materials?7. What is straight line formula for working stress under

buckling, where it is used?8. What approximation is taken by Professor Perry to modify

the secant formula for eccentrically loaded column?9. What do you understand by total potential constant in

energy approach?


1. A column fixed at one end and free at the other end buckles at a load P. Now, both the ends of the column are fixed. What is the buckling load for these end conditions?

1. 16 P2. 8 P3. 4 P4. 2 P

2. In Rankine’s formula ‘a’ is used, what is its value for cast iron?


3. What is the approximate value of σc for cast iron/σc for mild steel?

1. 0.62. 1.03. 1.74. None of these

4. A hollow circular column, with D = 100 mm, d = 80 mm, what is radius of gyration?

1. 322. 243. 19.44. None of these

5. In Johnson’s parabolic formula, allowable stress in compression for mild steel is

1. 270 N/mm2

2. 110 N/mm2

3. 80 N/mm2

4. None of these6. A 3-m-long column is hinged at one end and fixed at the

other end, what is its equivalent length?1.2. 3 m3.4. None of these

7. A column of a length of 2.4 m, an area of cross-section of 2,000 mm2 and moment of inertia of Ixx= 720 ×104mm4 and Iyy = 80×104mm4 is subjected to buckling load. Both the ends of the column are fixed. What is the slenderness ratio of column?

1. 1202. 803. 604. 40

8. The ratio of equivalent length of a column with one end fixed and the other end free to its own length is

1. 22. 1.03. 0.54. None of these

9. Euler’s buckling theory is applicable for1. Short columns2. Long columns3. Medium long columns4. All of them

10. In Johnson’s parabolic formula, what is constant b for hinged ends, for mild steel

1. 2 × 10−5

2. 3 × 10−5

3. 0.0054. None of these

Practice Problems

1. What size of a steel pipe should be used for the horizontal member of a jib crane shown in Fig. 14.23 for supporting the maximum force of 20 kN. Use an FOS of 2.5. The internal diameter of the pipe is 0.8 times the external diameter. E = 200 GPa.

Figure 14.23 Practice problems 14.12. Find the shortest length for a steel column with hinged ends

having a cross-sectional area of 30 mm × 60 mm for which the elastic Euler’s formula is applicable. E = 200 GPa and assume proportional limit to be 250 MPa.

3. A short-length of a tube of 30 mm internal diameter and 40 mm external diameter failed under a compressive load of 180 kN. When a 2-m length of the same tube is tested as a strut with both the ends hinged, the buckling load was found to be 40.8 kN. Assuming that σc, for the Rankine’s constant is given by first test, determine Rankines’ constant ‘a’. Moreover, estimate the crippling load for a piece of 3-m-long tube when used as strut with fixed ends.

4. A strut of a 2-m-long aluminium alloy has a rectangular section of 20 mm × 50 mm. A bolt through each end secures the strut so that it acts as a hinged column about an axis perpendicular to the 50-mm dimension and as fixed-ended column about an axis

https://www.safaribooksonline.com/library/view/strength-of-materials/9789332503519/xhtml/chapter014.xhtml#pro.14.1


perpendicular to the 20-mm dimension. Determine the safe central load using a FOS of 2. E = 70 GPa.

5. A simple beam of flexural rigidity EIB is propped up at the middle by a slender rod of flexural rigidity EIC. Estimate the deflection of the beam at the centre if a force P double the Euler’s buckling load for the column is applied to the system as shown in Fig. 14.24.

Figure 14.24

6. A thin vertical strut of uniform section and length L is rigidly fixed at its bottom end and its top end is free. At the top end, there is a horizontal force H and a vertical load P acting through the centroid of the section as an axial load. Prove that horizontal deflection at the top is

7. A horizontal bar CD is supported by a pin-ended column AB, 2.5 m long and 40 mm diameter as shown in Fig. 14.25. Calculate the allowable load P, if FOS with respect to buckling of column is 3.0.

E = 200 GPa




Figure 14.25

8. A hollow cast iron column of external diameter of 200 mm and length of 4 m with both ends fixed supports an axial load of 800 kN. Find the thickness of the metal required, use Rankine’s constants,a = 1/1,600 for both ends hinged, working stress, σw = 80 N/mm2.

9. A 1.2-m-long hollow circular steel pipe of 60 mm outside diameter and 50 mm inside diameter is fixed at both the ends so as to prevent any expansion in its length. The pipe is unstressed at the normal temperature. Calculate the temperature stress in the pipe and the FOS against failure as a strut if the temperature rises by 40°C. Use Rankine’s formula.

10. A 5-m-long steel strut, with I = 50 × 104 mm4, carries thrust load P = 20 kN, with eccentricities e= 10 mm on one side and 2e = 20 mm on the other side. E = 200 GPa. Determine the distance from one end, where the bending moment on strut is maximum.


Exercise 14.1:

Exercise 14.2:



Exercise 14.3: 400 kN

Exercise 14.4: L > 0.856 m

Exercise 14.5:

Exercise 14.6:

Exercise 14.7: 516.3kN


Exercise 14.9: 12.07 mm


Exercise 14.11: 8.05 N/mm2

Exercise 14.12: 2,180 kN

Exercise 14.13: W = 1.340 kN

Exercise 14.14: 161.70 N/mm2

Exercise 14.15: 9.88 EI/L2


1. (a) 2. (c) 3. (c) 4. (a) 5. (b)

6. (c) 7. (c) 8. (a) 9. (b) 10. (b)


1. D = 55.47 mm; d = 44.37 mm2. L = 769 mm3.














4. safe load = 11.50 kN

5.6. 5.29 kN7. 32.3 mm8. 92.35 MPa, 3.179. x = 2.859 m

2b. Bending of Curved BarsCHAPTER OBJECTIVES

There are many engineering components which are not straight but curved and possessing large initial curvatures. Flexure formula of theory of simple bending cannot be used to determine stresses in such components. A theory has been developed for the determination of stresses and deflections in such components. In this chapter, students will learn about stresses and deflections in components as:

o Curved bars of different sections as rectangular, circular, trapezoidal and triangular.o Chain links.o Rings.o Crane hooks.o Frames of power presses, C-clamps.o Components with hollow section.

Since the derivations are lengthy and cumber some, students are advised to derive the relationships by themselves after studying the chapter.

Introduction

Components with a large curvature are of engineering use in lifting machine and conveyor equipments. Components such as crane hook, chains and links used for lifting machines are

designed with safety considerations. Therefore, determination of stresses in such components and location of critical sections are of utmost importance.

The analysis of stresses in such components is quite complex. The theory of simple bending cannot be used for such components because in simple bending, the component is considered as initially straight.

In this chapter, we will analyse stresses and deflection in components such as rings, links, crane hooks, frame of punch presses and eccentrically loaded members with large initial curvature, etc.

In the chapter on ‘Theory of simple bending’, we assumed the beam to be initially straight before application of a bending moment and derived the relationship M/I = E/R = σ/y, and studied about the stresses and deflections developed in beams. But, in this chapter, we will study about the effect of bending moment on bars of large initial curvature.

Stresses in a Curved Bar

Figure 17.1 shows a portion of a curved bar of initial radius of curvature R, subtending an angle q at the centre of curvature O. This curved bar is subjected to a bending moment M tending to increase the curvature of the bar. To find out the stresses developed in the bar, a relationship between bending moment M, radius of curvature R and dimensions of the section of the bar are derived by taking the following assumptions:





Figure 17.1

1. The transverse sections of the bar, which are plane before application of a bending moment, remain plane after the application of bending moment.

2. The material obeys Hooke’s law and stress is directly proportional to strain.

Consider a small portion IJHG of the curved bar in its initial unstrained position, where AB is a layer at a radial distance of y from the centroidal layer CD, that is, a layer passing through the centroidal axis of the sections. At layer AB, stresses due to the bending moment M are to be determined.

After application of the bending moment, say, I′J′H′G′ is the final shape of the bar. The centroidal layer is now C′D′ and the layer AB takes the new position A′B′. Say, the final centre of curvature is O1and final radius of curvature is R1 and θ1 is the angle subtended by the length C′D′ at the centre.

Say, σ is the stress in the strained layer A′B′ under the bending moment M tending to increase the curvature (or tending to

reduce the radius of curvature), and e is the strain in the same layer.

Strain,

where y1 is the distance between centroidal layer C′D′ and layer A′B′, in the final position.

or,

Moreover, ε0 is the strain at the centroidal layer, that is, when y = 0

or,


and

Dividing Eq. (17.1) by Eq. (17.2),

or,





or,

Now, y1 ≅ y considering that change in thickness is negligible.

Therefore, strain,

Here, we have added and subtracted a term

or,

The stress in the layer AB, which is tensile as is obvious from the diagram, that is, layers below the centroidal layer are in tension and layers above the centroidal layer are in compression for the bending moment shown.

Stress,

where E is the Young’s modulus of the material.

Total force on the section, F = ∫σ dA

Considering a small strip of elementary area dA, at a distance of y from the centroidal layer CD.

where A is the area of cross-section of the bar.

Now, the total resisting moment will be given by

Because ∫ ydA = 0, that is, first moment of any area about its centroidal layer is zero.

Therefore,

Let us assume that a quantity which depends upon the disposition of section and the radius of curvature.

Therefore,

From Eq. (17.4),

because the bar is in equilibrium and the net force on the section is zero as no force is applied, only moment is applied.

Now,

Considering Eq. (17.4) again,

Substituting this value of in Eq. (17.5)




Substituting the value of ε0 in the equation for stress,

Substituting the value of ε0 again from Eq. (17.7),

On the other side of the centroidal layer y will be negative as for the layer EF shown in the figure.

σ′ = stress where y is negative

The expressions given in Eqs (17.8) and (17.9) are for the stresses due to the bending moment which tends to increase the curvature. If the bending moment tends to straighten the bar or tends to decrease the curvature, then θ1 < θ and R1 >R and the stresses will be reversed.

Bending moment tending to decrease the curvature

For y to be positive (away from centre of curvature)




On the other side of the centroidal layer, where y is negative (towards centre of curvature)

Ah2 for a Rectangular Section

Figure 17.2 shows the rectangular cross-section of breadth B and depth D of a curved bar with radius of curvature R, i.e., the radius from the centre of curvature C to the centroid G of the section. Consider a strip of thickness dy at a distance y from the centroidal layer. Area of the strip, dA = Bdy.

Now, area, A = BD


Figure 17.2

where R is the radius of curvature up to the centroidal layer, R1 is the radius up to the inner surface of the curved bar and R2 is the radius up to the outer surface of the curved bar.

Example 17.1 A circular ring of rectangular section with a slit is loaded as shown in Fig. 17.3. Determine the magnitude of the force P if the maximum resultant stress along the section ab does not exceed 100 N/mm2. Draw the stress distribution diagram along ab.

Figure 17.3

Solution

Mean radius of curvature, R = 11 cm


Radius of curvature of inner surface, R1 = 8 cm

Radius of curvature of outer surface, R2 = 14 cm

Breadth, B = 4 cm

Depth, D = 6 cm

Maximum resultant stress will occur at the inner radius along section ab, that is, at the point b.

Bending moment, M = P × R = 11P N-cm

Direct stress,

Resultant stress at the point

Therefore,

Stress distribution (along Gb)

where y varies from 0 to 3, compressive stressees

Stresses distribution Along Ga where y varies from 0 to 3

Resultant stress, tensile stresses

Since, the direct stress is compressive

Figure 17.4 shows the stress distribution along the radial thickness ab of the section which has maximum bending moment PR. In this case the resultant stress at centroid G is zero.


Figure 17.4

Exercise 17.1 A curved bar of rectangular cross-section 40 mm × 60 mm is subjected to a bending moment, tending to increase the curvature of the bar. Radius of curvature of the bar is 200 mm. If the maximum stress developed in the bar is 100 N/mm2, what is the magnitude of bending moment? Due to this bending moment what is the stress at the CG of the section?

Value of h2 for Sections Made Up of Rectangular Strips

Sections such as T, I and channel section are made up of rectangular strips, the value of h2 for each section can be determined by considering each strip separately. For a single strip,

where B is the breadth, R2 is the radius at outer fibres and R1 is the radius of the inner fibres of the section from the centre of curvature. Using this expression, let us determine h2/R2 for various sections.

1. T-section: Figure 17.5 shows a T section with following dimensions:Breadth of the flange = BBreadth of the web = bRadius of curvature up to centroid G of the section = RRadius up to extreme outer edge of web = R1


Radius up to inner edge of flange = R2Radius up to outer edge of flange = R3Area of cross-section of T section = A

Figure 17.5

2. I-section: Figure 17.6 shows an I section with flange and web of breadths B and b, respectively.R = Radius of curvature up to centroid G of the sectionR1 = Radius up to outer edge of inner flangeR2 = Radius up to inner edge of inner flangeR3 = Radius up to inner edge of outer flangeR4 = Radius up to outer edge of outer flangeA = area of cross-section = B(R4 − R3) + b(R3 − R2) + B(R2 − R1)


Figure 17.6

3. Channel Section: Figure 17.7 shows a channel section withB = breadth of webb = breadth of flangesR = Radius of curvature up to centroid G of the sectionR1 = Radius up to inner surfaceR2 = Radius up to outer edge of webR3 = Radius up to the outer edge of flangeA = Area of cross-section = B(R2 − R1) + 2b(R3 − R2)


Figure 17.7

Example 17.2 A curved beam whose centroidal line is a circular arc of 12 cm radius. The cross-section of the beam is of T shape with dimensions as shown in Fig. 17.8. Determine the maximum tensile and compressive stresses set up by a bending moment of 70,000 N cm; tending to decrease the curvature.

Solution

Figure 17.8 shows the curved bar with T section subjected to a bending moment M tending to decrease the curvature. Therefore, there will be tensile stresses between A to G, and compressive stresses between G to B.

Let us first calculate the distance of centroid from the outer edge of web.

Figure 17.8

Radius of curvature, R = 12 cm (given).

Radius up to inner surface, R1 = 12 −1.864 = 10.136 cm



Radius up to outer edge of flange, R2 = 11.136 cm.

Radius up to outer edge of web, R3 = R1 + 6 = 10.136 + 6 = 16.136 cm

Maximum compressive stress at point B

Maximum tensile stress at point A

Example 17.3 Figure 17.9 shows a press applying a 150-kN force on a job. Determine the stresses at the points a and b. The section is hollow as shown.


Figure 17.9

Solution

Let us first determine the position of the centroid

Radius of curvature, R = 24 + 13 = 37 cm

Area of cross-section, A = 24 × 6 + 2 × 4 × 20 + 4 × 16 = 398 cm2

Bending moment, M = Force × (60 + R) = 150 × 97 kN cm, where R = 37 cm

Direct tensile stress,

Bending stress due to M at

Bending stress due to M at

Resultant stress at the point a = 6.869 + 0.408 = 7.277 kN/cm2

= 72.77 N/mm2 (tensile)

Resultant stress at the point b = 5.682 − 0.408 = 5.274 kN/cm2

= 52.74 N/mm2 (compressive)

Exercise 17.2 An open ring of channel section is subjected to a compressive force of 25 kN as shown in Fig. 17.10. Determine the maximum tensile and maximum compressive stress along the section ab.

Exercise 17.3 A load P = 5 kN is applied on a C-clamp as shown is Fig. 17.11. Determine the stresses at the points a and b.


https://www.safaribooksonline.com/library/view/strength-of-materials/9789332503519/xhtml/chapter017.xhtml#exa17.2




Ah2 for a Trapezoidal Section

Figure 17.12 shows a trapezoidal section of a curved bar with breadths B1 and B2, depth D and radius of curvature R. Say, C is the centre of curvature and G is the centroid of the section. Then,

Figure 17.12

Consider a strip of depth dy at a distance of y from the centroidal layer.

If b is the breadth of the strip

Area of the strip,

Now



Example 17.4 Determine the maximum compressive and tensile stresses in the critical section of a crane hook lifting a load of 40 kN. The dimensions of the hook are shown in Fig. 17.13. The line of application of the load is at a distance of 8 cm from the inner fibre (rounding off the corners of the cross-section are not taken into account).


Solution



Figure 17.13 shows a crane hook and the trapezoidal section. The load line KK′ is away from the centre of the curvature C.

Position of CG of the section

So,

Radius of curvature,


Now,

Substituting the values,

Distance,


Bending moment,

The bending moment tends to reduce the curvature, so the portion GA will be in compression and portion GB will be in tension.

Direct stress,

Maximum compressive stress at A,

Maximum tensile stress at B

Exercise 17.4 The section of a crane hook is a trapezium. At the critical section, the inner and outer sides are 40 mm and 25 mm, respectively and depth is 75 mm. Centre of curvature of the section is at a distance of 60 mm from the inner fibres and the load line is 50 mm from the inner fibres. Determine the maximum load the hook can carry if the maximum stress does not exceed 120 N/mm2.

Ah2 for a Circular Section

Figure 17.14 shows the circular section of diameter d of a curved bar of radius of curvature R, from the centre of curvature C up to the centroid G of the section.

Area of cross section,

Consider a strip of depth dy at a distance of y from the centroidal layer as shown.

Breadth of the layer,

Figure 17.14

Area of the strip,

Now,

or,


Example 17.5 A curved bar is formed of a tube of an outside diameter of 8 cm and a thickness of 0.5 cm. The centre line of this beam is a circular arc of radius 15 cm. Determine the greatest tensile and compressive stresses set up by a bending moment of 1.2 kN m tending to increase its curvature.

Solution

Figure 17.15 shows the cross-section of a curved bar of radius of curvature R = 15 cm.


Area of inner circle,

Area of outer circle,

Bending moment, M = 1.2kNm = 1.2 × 105 N cm



For a circular section

For inner circle,

For outer circle,


Maximum tensile stress at b,

Maximum compressive stress at a,


Exercise 17.5 A bar of circular cross-section is bent in the shape of a horse shoe. The diameter of the section is 8 cm and mean radius R is 8 cm as shown in Fig. 17.16. Two equal and opposite forces P = 15 kN each are applied so as to straighten the bar. Determine the maximum tensile and compressive stresses along the central section.

Ring Subjected to a Diametral Load

Figure 17.17 shows a circular ring of mean radius R subjected to a diametral pull P. Consider a section CD at an angle θ from the line of application of the load, that is, Y1Y2, and determine the bending moment and stresses in this section. Due to symmetry, the ring can be divided into four equal quadrants. Say, M1 is the bending moment on the section AB along the line of symmetry X1X2.

Taking moments about CD,

Bending moment at the section CD,

From Eq. (17.5) of the section ‘Stresses In A Curved Bar’,





Figure 17.17

where E is the Young’s modulus, ε0 is the strain in centroidal layer, R is the initial radius of curvature and R1 is the radius of curvature after bending

Therefore,

Multiplying this equation throughout by Rdθ and integrating for one quadrant

That is,

However, [Eq. (17.2) of the section ‘Stresses in a Curved Bar’]

Now for one quadrant, initial angle θ = 90° = π/2 and final angle θ1 = 90° = π/2 due to symmetry, that is, ∠Y1OX1 remains 90° even after the application of diametral load.

Therefore

Substituting this in Eq. (17.12), we get,

Again by the Eq. (17.4) of the section ‘Stresses in a Curved Bar’

Normal force on the section,

where

Therefore, normal force,




Normal force on the section CD,

(refer to Eq. (17.5) of the section ‘Stresses in a Curved Bar’ again)

Therefore,

Substituting the value of ε0 in Eq. (17.13)

Now,



M will be maximum when θ = 0°.

M will be zero when

So, there will be four sections, one in each quadrant where the bending moment M will be zero and consequently the stress due to bending moment will be zero.

Now, substituting the value of M1 in the following equation.

and

Stress, [Refer to Eq. (17.3) of the section ‘Stresses in a Curved Bar’]


Direct stress at any section,

Resultant stress at any point on the section

Stress along Y1Y2axis, where θ = 0°

at the point K, where d = diameter of the rod of the ring

stress,

at the point J,

stress,

It can be observed that the maximum stress occurs at the point J, where the diametral load is applied. Stresses along X1X2axis, where θ = 90°

At the point B,

At the point A,

Example 17.6 A ring is made of round steel bar, 2 cm diameter and the mean diameter of the ring is 12 cm. Determine the greatest intensities of tensile and compressive stresses along a diameter XXif the ring is subjected to a pull of 5 kN along diameter YY.

Solution

Figure 17.18 shows a ring of a mean diameter of 12 cm, a bar diameter of 2 cm, subjected to a diametral pull P.

Radius of curvature, R = 6 cm

Bar diameter, d = 2 cm

Pull, P = 10 kN



Figure 17.18

Stresses

Exercise 17.6 A ring is made of round steel rod of a diameter of 24 mm. The mean diameter of the ring is 240 mm. The ring is pulled by a force of 1 kN. Determine the greatest intensities of tensile and compressive stresses along the diameter of the loading.

Chain Link Subjected to a Tensile Load

Figure 17.19 shows a chain link of mean radius R, length of the straight portion l, subjected to a pullP. Consider a section CD at an angle θ from the line of application Y1Y2 of the pull P. Determine the bending moment and stresses in this section. Due to symmetry, the ring can be divided into four equal parts as shown. Say, M1 is the bending moment on the section AB along the line OX1.

Taking moments at the section CD,

From Eq. (17.5) of the section ‘Stresses In A Curved Bar’.



where E is the Young’s modulus, ε0 is the strain in the centroidal layer, R is the initial radius of curvature and R1 is the final radius of curvature

Figure 17.19

Therefore,

Multiplying throughout by Rdθ and integrating from 0 to π/2

Now, [Eq. (17.2) of the section ‘Stresses In A Curved Bar’]

In this case, initial angle θ = ∠ X1OY1 = 90°

The final angle θ1 will be a slight change from 90°.

Slope at X1

where I is the moment of inertia of the section.

Therefore,

Substituting in Eq. (17.17)

Again by Eq. (17.4) of the section ‘Stresses In A Curved Bar’.

Normal force on the section CD,

Now,




Therefore, normal force,

Therefore,

(by Eq. (17.5) of the section Stresses In A Curvedd Bar’)

Therfore,

Substituting the values of ε0 in Eq. (17.18)

where I = Ak2; k = radius of gyration of the section



Dividing throughout by π/2

However,

Substituting the value of M1 in Eq. (17.20)

Stress at any layer at a distance of y from the neutral layer is

Moreover,


Therefore,

Substituting the values of M and ε0, stress due to bending moment

Direct stress due to F,

Resultant stress, σR = σb + σd

This is an equation for resultant stress in any section along the curved portions X2Y1X1 and X2′Y2X1′ of the chain link.

The bending moment M1 on the straight portion X1X1′ and X2X2′ will remain constant and for the straight portion bending stress will be found with the help of general flexural formula. To obtain the resultant stress in this straight portion, direct tensile stress P/2A will be added to the bending stress.

On the inner surface of the ring which is also called intrados stress can be obtained by substituting y= −d/2 in Eq. (17.21) where d is the diameter of the bar of the chain link. Similarly, for the outer surface which is also known as extrados the resultant stress is obtained by replacing y by +d/2 in Eq. (17.22).

Maximum stress along Y1OY2axis,

At the intrados,

At the extrados,

Maximum stresses along X1OX2axis,



At the intrados,

At the extrados,

Maximum stress in straight portion,

X1X1′ or X2X2′

Bending moment,

Stress due to bending,

Direct stress,

Resultant stress at intrados and extrados, respectively,

Example 17.7 A chain link is made of round steel rod of 1 cm diameter. If R = 3 cm and l = 5 cm, determine the maximum stress along the section where tensile load is applied. If P = 0.5 kN.

Solution

R = 3 cm, d = 1 cm, l = 5 cm, and P = 1 kN

Area,

Radius of gyration,

Now,

Then,

Maximum stress at intrados,

Maximum stress at extrados,

Maximum stress occurs at the intrados, i.e., where the load is applied.

Exercise 17.7 A chain link is made of round steel rod, 10 mm diameter. If R = 30 mm and L = 50 mm, determine the maximum stresses along the section at the end of the straight portion. A load of 1 kN (tensile) is applied on the chain link.

Deflection of Curved Bar

In order to estimate the stiffness of a curved beam, subjected to a bending moment. It is necessary to determine the deflection of the curved beam and in such cases the influence of the initial curvature of the beam on its deflection is considerable. Figure 17.20 shows the centre line ABCD of a curved bar subjected to variable bending moment. Consider a small portion BC of length ds along the centre line. Say, the bending moment at B is M and at C is M + δM. Due to the bending moment say the centre line of the curved beam takes new position ABF and the element BC rotates by an angle dɸ = DBF at the point B. The angular rotation is small and the displacement of the point D is also small.

Displacement DF ≅ BD dϕ∠ BDF = 90° for very small displacement DF

Components of the displacement are DE perpendicular to the chord AD and EF parallel to the chordAD, that is, the line joining the ends of the centre line of the curved beam considered. FE shows negative displacement towards the point A.

Deflection of the point D with respect to A is δDA and considering the small length ds only, say the deflection is

ΔδDA = −EF = −DF cas α where ∠DEF = α

= −(BD dϕ) cos α



∠ADF = α

Figure 17.20 Deflection of curved bar

Therefore, ∠BDG = 90° − α or ∠DBG = α

and BD cos α = BG

Therefore, ΔδDA = −(BD cas α)dϕ = − BG dϕ = −h dϕwhere h is the perpendicular distance of the point B from the

chord AD.

Moreover,

Therefore,

Total defelection of D with respect to A

(1) Deflection of a closed ring

Figure 17.21 shows the quadrant of a ring of mean radius of curvature R subjected to diametral pull Palong OY1. We have to


determine the deflection along the load line or along the chord Y2Y1. (Point Y2not shown in Fig. 17.19).

Figure 17.21 Closed ring-quadrant

OY1 is half of the chord Y2Y1. Consider a small length ds at C at an angular displacement θ.

CC1 = perpendicular distance on chord from the point C

= h = Rsin θ

Bending moment at the section C,

[see Eq. (17.15) of the section ‘Ring Subjected To A Diametral Load’]



Note that we have considered only one quadrant and when we consider the complete ring, the deflection along the load will be δY1Y2

(2) Deflection perpendicular to load line

Refer to Fig. 17.19 again, now the chord is OX1 and perpendicular distance CC2 = h on the chord OX1from C; h = Rcos θ

δx10 = deflection perpendicular to load line

Total deflection,


Example 17.8 A ring with a mean diameter of 120 mm and a circular cross-section of 40 mm diameter is subjected to a diametral compressive load of 20 kN. Calculate the deflection of the ring along the load line. E = 200 GN/m2.

Solution

Since the diametral load is compressive, there will be reduction in diameter along the load line and increase in diameter perpendicular to the load line.

R = 60 mm = 6 cm; d = 40 mm = 4 cm; P = 20 × 103 NE = 2,000 GN/m2 = 200 × 105 N/cm2

Deflection along the load line

Exercise 17.8 A Ring with a mean diameter of 150 mm and a circular cross-section of 30 mm diameter is subjected to a diametral tensile load of 8 kN. Calculate the deflection of the ring along the direction perpendicular to load line E = 200 kN/mm2.

Deflection of a Chain Link

Figure 17.22 shows the quarter of a chain link subjected to axial tensile load P. The radius of curvature of the link is R and length of the straight portion is l. Consider a section at an angle θ from the axis OY1. We have to determine the deflection along the load P or along the chord Y2Y1.

CC1 = h = perpendicular from C to the chord

= Rsinθ

Bending moment at the section


Figure 17.22

[Refer to Eq. (17.6) of article 19.7.]

Over the length l/2, the bending moment is constant and is equal to M1 = PK′ where the value of K′ is as above.

Deflection along the line of loading

(with the help of theory of simple bending)


Deflection due to direct load P/2 is

Total deἀection,

where

(2) Deflection perpendicular to load line

In this case chord is XX and CC2 = h = Rcos θ + l/2

From theory of simple bending deflection in straight portion

where M1 is bending moment on straight portion and l is the length of the straight portion.

where

Example 17.9 A chain link is made of a steel rod of 12 mm diameter. The straight portion is 60 mm in length and the ends are 60 mm in radius. Determine the deflection of the link along the load line when subjected to a load of 1 kN. Given E = 200 × 103 N/mm2.

Solution

Rod diameter, d = 1.2 cm

Area of cross-section, A = π2/4 = 1.31 cm2

Length of straight portion, l = 6 cm

Radius of curvature, R = 6 cm

Load, P = 1.0 kN, E = 200 × 105 N/cm2

Radius of gyration, k = d/4 = 0.3 cm

Deflection along the load line

Exercise 17.9 A chain link is made of steel rod, 12 mm diameter. The straight portion is 60 mm in length and the ends are 60 mm in radius. Determine the deflection in the link along the direction perpendicular to the load line if the chain link is subjected to a load of 1 kN. Given E = 200 kN/mm2.

Problem 17.1 For the frame of a punching machine shown in Fig. 17.23. Determine the circumferential stresses at A and B on a section inclined at an angle θ = 45° to the vertical.



Force, P = 200 kN.

Solution

Force, P = 200 kN.

Perpendicular force on the section AB = Psin 45° =

Tangential force on the section AB = Pcos 45° =

Figure 17.23

Area of cross-section, A = 30 × 10 + 5 × 20 + 15 × 10 = 550 cm2

Location of G

Radius of curvature, R = R1 + y2 = 20 + 15.9 = 35.9 cm

Bending moment on the section,

Direct force on the section,

Tensile stress at point A

Compressive stress at point B

Problem 17.2 The radius of the inner fibres of a curved bar of trapezoidal section is equal to the depth of the cross-section. The base of the trapezium on the concave side is four times the base on the convex side. Determine the ratio of the stresses in the extreme fibres of the curved bar to the stresses in the same fibres of a straight bar subjected to the same bending moment.

Solution

See Fig. 17.24

Now, R2 = D

Therefore, R1 = 2R2 = 2D

Therefore, y2 = 0.4D

Area,


Figure 17.24

Let us consider that this curved bar is subjected to a bending moment, M tending to reduce the curvature.

The maximum tensile stress,

(when y = −y2)

The maximum compressive stress,

(when y = +y1)

Stresses in the straight bar

Let us divide the section in two triangles as shown, so as to calculate the moment of inertia IYY.

Area of triangle I,

Area of triangle II,

IYY of triangle I, about its CG =

Distance,

IYY of triangle II, about its CG

Distance,

IYY of the whole section

The maximum tensile stress,

The maximum compressive stress,

(y = + y1)

Therefore, the ratios

and, the ratios

Problem 17.3 A chain link is made of round steel rod, diameter 12 mm. If R = 40 mm and l = 60 mm, determine the extreme stresses along the intrados if the link is subjected to a tensile load of 1 kN.

Solution

Bar diameter, d = 12 mm

Radius of curvature, R = 40 mm

Length of straight portion, l = 60 mm

Load, P = 1,500 N


Radius of gyration,

Moreover,

Along the intrados y = −d/2, the equation for the resultant stress is

Substituting the values

Problem 17.4 A chain coupling is made of a 20-mm-diameter steel rod bent into an ‘S’ shape as shown in Fig. 17.25. If the coupling is attached to a tensile load of 1.0 kN as shown; calculate the maximum stress developed in chain coupling.

Solution

Maximum tensile stress will occur either at point A of smaller loop or at point B of larger loop.

Rod diameter, d = 20 mm

Radius of curvature of smaller loop, R1 = 40 + 10 = 50 mm

Radius of curvature of longer loop, R2 = 60 + 10 = 70 mm

Load, W = 1.0 kN = 1,000 N

Area of cross-section of rod,


Figure 17.25

and

Stress at A,

Stress at B,


1. A bar of square section 6 cm × 6 cm is curved to a mean radius of 12 cm. A bending moment Mis applied on the bar. The moment M tries to straighten the bar. If the stress at the innermost fibres is 60 N/mm2 tensile, then the stress at the outermost fibres is

1. 60 N/mm2 (compressive)2. 60 N/mm2 (tensile)3. More than 60 N/mm2 (compressive)4. Less than 60 N/mm2 (compressive).

2. The most suitable section of a crane hook is1. Square

2. Round3. Hollow round4. Trapezoidal

3. A bar of square section 4 cm × 4 cm is curved to a mean radius of 80 m. A bending moment M, tending to increase the curvature is applied on the bar. If the stress at the

outermost fibres is 80 MPa tensile, then the stress at the innermost fibres will be approximately equal to

1. 120 MPa (compressive)2. 100 MPa (compressive)3. 90 MPa (compressive)4. 80 MPa (compressive)

4. A ring is subjected to a diametral tensile load. The variation of the stress at the intrados surface from the point of loading up to the section of symmetry is

1. Maximum tensile stress to maximum compressive stress.2. Throughout tensile stress.3. Maximum compressive stress to maximum tensile

stress.4. Throughout compressive stress.

5. The distribution of stress along a section of a curved bar subjected to a bending moment tend to increase its curvature is

1. Linear2. Uniform3. Parabolic4. Hyperbolic

Practice Problems1. A sharply curved beam of rectangular section is 10 mm thick and

50 mm deep. If the radius of curvature, R = 60 mm, compute the stress in terms of the bending moment M at a point 20 mm from the outer surface.

2. Determine the diameter d of a round steel rod that is used as a hook to lift a 9-kN load acting through the centre of curvature of the centroidal axis of the hook. Assume that , and the maximum stress permitted is 125 N/mm2.

3. The cross-section of a triangular hook has a base of 5 cm and altitude of 7.5 cm and a radius of curvature of 5 cm at the inner face of the shank. If the allowable stress in tension is 100

N/mm2and in compression is 80 N/mm2, what load can be applied along a line 7.5 cm from the inner face of the shank?

4. Three plates are welded to form the curved beam of or, I-section shown in Fig. 17.26. If the moment M = 1 kN m, determine the stresses at points A and B and at the centre of the section (Fig. 17.26).

5. A steel link of rectangular section 24 mm × 8 mm is shown in Fig. 17.27. If the angle β = 90° and the allowable stress in link is 100 MPa, determine the largest value of P which can be applied on the link.

6. A curved bar of rectangular section with breadth B and depth D = 2B, is bent to a radius of curvature equal to 1.2D. It is subjected to a bending moment of 1 kN m, tending to increase its curvature. Determine the size of the section if the maximum stress does not exceed 80 N/mm2.

Figure 17.26

Figure 17.27





7. Determine the maximum stress along, the section A-A in the crane hook. Section of the hook is circular of diameter 25 mm. Load 2 kN (Fig. 17.28).

8. Section of a punch press is in a T section of dimensions shown in Fig. 17.29. Centre of curvature of section is at a distance of 400 mm from inner side and load line passes through O, at a distance of 600 mm from inner side. Punch press frame is made of CI. Determine the safe load for a factor of safety of 3 based on ultimate strength.

σut of CI = 150 MPa

σuc of CI = 600 MPa

9. A proving ring is 250 mm diameter, 40 mm wide and 6 mm thick. The maximum allowable stress in ring is 200 N/mm2. Find the load to cause this stress and the load should give 1 mm deflection of the ring in the direction of loading. (E = 200 GPa).

10. The curved beam with a circular centre line has a trapezoidal cross-section as shown in Fig. 17.30 and is subjected to pure bending in its plane of symmetry. The face b1 is on concave side of the beam. If h = 100 mm and a = 100 mm, find the ratio of b1/b2 of base widths so that the extreme fibre stresses in tension and compression are numerically equal.

11. A chain link is made of round steel rod of diameter 12 mm. If R = 40 mm and l = 60 mm draw the stress distribution diagram along the extrados for an angle of 90° starting from the outermost edge (along the direction of loading) of the link, if the link is subjected to a tensile load of 10 kN.




Figure 17.28

Figure 17.29

Figure 17.30


Exercise 17.1: 2.16 kN m, +4.50 N/mm2


Exercise 17.2: −93.8; +84.05 MPa

Exercise 17.3: 86.70, −37.873 MPa

Exercise 17.4: 30.60 kN

Exercise 17.5: −39.22, +99.83 MPa

Exercise 17.6: +26.1 N/mm2, −30.33 N/mm2

Exercise 17.7: 42.99, −30.26




1. (d) 2. (d) 3. (d) 4. (c) 5. (d)


1. 72.92 M2. 43.52 mm3. 15.7 kN4. +22.755, −16.71,−3.55 MPa5. 4.37 kN6. 29.7, 59.4 mm7. 93.6 MPa8. 148.433 kN9. 375 N, 496 N10. 1.8711. 79.56 to −17.37 MPa









4a. Unsymmetrical Bending and Shear CentreCHAPTER OBJECTIVES

When a section of a beam is not symmetrical about the plane of bending, an unsymmetrical bending takes place, i.e., in addition to bending, due to applied loads twisting is observed in the beam. Then there are principal axes of the section where the product of inertia is zero. In this chapter students will learn about:

o Principal axes and their directions with respect to centroidal axes of the section.

o Moment of inertia of the section about the principal axes.o Stresses developed at various points of the section due to

unsymmetrical bending.o Product of inertia of the section about any co-ordinate

axes.o Position of neutral axis with respect to centroidal axes of

the unsymmetrical sections.o Deflection in a beam due to unsymmetrical bending.o Location of shear center of sections which are symmetrical

about one centroidal axes and sections which are not symmetrical about any centroidal axes.

Introduction

Every section is not symmetrical about both the centroidal axes. Some sections are symmetrical only about one axis, whereas many sections as angle sections are not symmetrical about both the centroidal axes. In theory of simple bending, the section of the beam is symmetrical about the plane of bending. The simple flexural formula derived in theory of simple bending is not applicable when the section is not symmetrical about the plane of bending. In such sections, the principal axes and principal moments of inertia and the product of inertia are determined. Stresses developed in such sections of a beam are dependent on these parameters.

If the load line on a beam does not coincide with one of the principal axes of the section, the bending takes place in a plane different from the plane of principal axes. This type of bending is known as unsymmetrical bending. The two reasons of unsymmetrical bending are as follows:

1. The section is symmetrical about two axes like I-section, rectangular section, circular section but the load-line is inclined to both the principal axes.

2. The section itself is unsymmetrical like angle section or a channel section (with vertical-web) and load-line along vertical any centroidal axes.

Figure 18.1(a) shows a beam with I-section with load-line coinciding with YY principal axis. I-section has two axes of symmetry and both these axes are the principal axes. Section is symmetrical aboutYY plane, i.e., the plane of bending. This type of bending is known as symmetrical bending.

Figure 18.1(b) shows a cantilever with rectangular section, which has two axes of symmetry which are principal axes but the load-line is inclined at an angle α with the YY axis. This is the first type of unsymmetrical bending. Then, Fig. 18.1(c) shows a cantilever with angle-section which does not have any axis of symmetry but the load-line is coinciding with the YY axis. This is the second type of unsymmetrical bending.




Figure 18.1(d) shows a channel section subjected to a vertical load passing through its centroid G. This member has been subjected to bending and twisting under the applied vertical load W. Now, the question arises; is it possible to apply the vertical load W in such a way that the channel member will bend without twisting and, if so, where the load W should be applied?





Figure 18.1 (a) Symmetrical bending, (b) unsymmetrical bending symmetrical section but oblique load, (c) unsymmetrical bending (unsymmetrical section) and (d) unsymmetrical bending(section not symmetrical about bending plane) (e) Channel section(not symmetrical about yy axis, (f) Bending without twisting

Shear force in the flanges and web of the channel section is F1, F2 and F1, respectively, as shown in Fig. 18.1(e). Forces F1 constitute a couple F1 × h about centroid G. This couple is responsible for twisting of the member. Now, if the vertical load W or the shear force in the section is shifted from G, such thatW × e = F1 × h, then the twisting couple is eliminated. So, it can be concluded that if the vertical loadW, or vertical shear F is moved to the left in the channel section through a distance e, such that, F1 ×h = We = Fe, the member will bend without twisting as shown in Fig.18.1(f).

Before proceeding further, let us study about the principal axes of a section.

Principal Axes

Figure 18.2 shows a beam section which is symmetrical about the plane of bending Y–Y, a requirement of the theory of simple bending or symmetrical bending. G is the centroid of the section.XX and YY are the two perpendicular axes passing




through the centroid. Say, the bending moment on the section (in the plane YY of the beam) is M, about the axis XX. Consider a small element of area dA with (x, y) co-ordinates.


Figure 18.2 Plane of bending yy

Stress on the element,

Force on the element,

Bending moment about YY axis,

Total moment,

If no bending take place about YY axis, then

M1 = 0



or

or

or

The expression is called the product of inertia of the area about XX and YY axes, represented byIxy. If the product of inertia is zero about the two co-ordinate axes passing through the centroid, then the bending is symmetrical or pure bending. Such axes (about which product of inertia is zero) are called principal axes of the section and moment of inertia about the principal axes are called principal moments of inertia.

The product of inertia may be positive, negative or zero depending upon the section and co-ordinate axes. The product of inertia of a section with respect to two perpendicular axes is zero if either one of the axis is an axis of symmetry.

Example 18.1 Show that product of inertia of a T-section about a centroidal axis is zero.

Solution

Figure 18.3 shows a T-section with flange B × h1 and web b × h2. The section is symmetrical about YYaxis. Say G is the centroid of the section on the axis YY, and XX and YY are the centroidal axes.

Ixy = I′xy for flange + I″xy for web

For flange, x varies from –(B/2)to +(B/2)

For web, x varies from –(b/2) to +(b/2)

Now,



Example 18.2 Determine the product of inertia about axes X and Y for a triangular section shown in Fig.18.4.


Solution

Consider a small element of area dA at co-ordinates x, y.

Product of inertia about XY axis,

Note that limiting value of x = 40 mm = 2 × limiting value of y





Exercise 18.1 Consider an I-section with flanges B × t1, and web H × t2 and show that the product of inertia about the centroidal axes is zero.

Exercise 18.2 Figure 18.5 shows a rectangular section with breadth, b and altitude, h. Determine the product of inertia of the section about X–Y axes.

Parallel Axes Theorem for Product of Inertia

Figure 18.6 shows a section with its centroid at G, and GX′ and GY′ are the two rectangular co-ordinates passing through G. Say, the product of inertia about X′Y′ is . Let us determine the product of inertia about the axis OX and OY, i.e., Ixy.

Say, distance of G from OX axis = , and distance of G from OY axis = .

Consider a small element of area




dA = dxdy.

Figure 18.6

Say, co-ordinates of the element about the centroidal axis GX′, GY′ are x′, y′.

Then, co-ordinates of the element about X–Y axis are,

x = + x′ and y = + y′

Therefore, the product of inertia,

i.e., the product of inertia of any section with respect to any set of co-ordinate axes in its plane is equal to the product of inertia of

the section with respect to the centroidal axes parallel to the co-ordinate axes plus the product of the area and the co-ordinates of the centroid of the section with respect to the given set of co-ordinate axes.

Example 18.3 Figure 18.7 shows an unequal channel section, determine its product of inertia Ixy and .

Solution

Let us break up the section into three rectangular strips I, II and III as shown in the figure and write the coordinates of their centroids with respect to the given set of axes YOX.

Remember that the product of inertia of these rectangular strips about their principal axes passing through the respective centroids is zero, because these rectangular strips have two axes of symmetry.

(Ixy) I = 0 + 81 cm4 (using the parallel axis theorem for product of inertia)



(Ixy) II = 0 + 24 cm4

(Ixy) III = 0 + 386 cm4

Ixy = 501 cm4

To determine , let us first determine the position of the centroid of the section.


A = 18 + 8 + 12 = 38 cm2

= Ixy − A = 501 − 38 × 3.184 × 5.21

= 501 − 630.37 = −129.37 cm4



Exercise 18.3 Figure 18.8 shows one unequal angle section, determine its product of inertia Ixyand (through the centroidal axes).

Determination of Principal Axes

In the section ‘Introduction’, we have learnt that principal axes pass through the centroid of a section and product of inertia of the section about principal axes is zero. Figure 18.9 shows a section with centroid G and XX and XX are two co-ordinate axes passing through G. Say, UU and VV is another set of axes passing through the centroid G and inclined at an angle θ to the X–Y co-ordinate. Consider an element of area dA at point P having co-ordinates (x, y). Say, u, v are the co-ordinates of the point P in U–V co-ordinate axes.




Figure 18.9 Co-ordinates along principal axes

So, u = GA′ = GD + DA′ = GD + AE

where GD = GA cosθ = x cosθ

AE = DA′ = y sinθ

or, u = x cosθ + y sinθ

v = GB′ = PA′ = PE − A′E

= PE − AD since A′E = AD

= PA cosθ − x sinθ = y cosθ − x sinθ

Similarly, x, y co-ordinates can be written in terms of u, v co-ordinates.

x = GC − AC = GC − A′F = u cosθ − v sinθ

(as PA′ = v and GA′ = u)

y = GB = PA = AF + FP = A′C + FP = u sinθ + v cosθ

Second moment of area about U–U axis,

Second moment of area about V–V,

From Eqs.(18.4) and (18.5),

Iuu + Ivv = Ixx (sin2θ + cos2θ) + Iyy (sin2θ + cos2θ) = Ixx + Iyy (18.6)

Product of inertia about UV axes,

However, as per the condition of pure bending or symmetrical bending Iuv = 0, then U and V will be the principal axes

or, 2Ixy cos 2θ + (Ixx − Iyy)sin 2θ = 0

or,

Say, θ1 and θ2 are two values of θ given by Eq. (18.7)




Substituting these values of sin 2θ1 and cos 2θ1 in Eq. (18.4)

Similarly,

Now, for θ2 = θ1 + (π/2)

sin 2θ2 = sin(2θ1 + π) = −sin 2θ1

cos 2θ2 = cos(2θ1 + π) = −cos 2θ1.

Substituting these values in Eqs.(18.4) and (18.5)

From Eqs.(18.8) to (18.11), we learn that

Maximum and minimum values of Iuu and Ivv

For maximum value of Iuu,






This shows that the values of (Iuu)θ1 and (Iuu)θ2 are the maximum and minimum values of Iuu and Ivv.These values are called the principal values of moment of inertia as Iuv = 0. The directions θ1 and θ2are called the principal directions.

Moment of Inertia About Any Axis

If the principal moments of inertia Iuu and Ivv are known, then moment of inertia about any axis inclined at an angle θ to the principal axes can be determined. Say u, v are the co-ordinates of an element of area dA in the U–V principal axes system. X and Y are the co-ordinate axes inclined at an angle θ to the U–V axes.

x co-ordinate of element = u cos θ − v sin θ

y co-ordinate of element = u sin θ + v cos θ

Moment of inertia,

Similarly,

Ixx = Iuu cos2θ + Ivv sin2θ (18.13)

From Eqs.(18.12) and (18.13),

Ixx + Iyy = Iuu + Ivv = J,

where J is polar moment of inertia about an axis passing through G and normal to the section.

Example 18.4 Determine the principal moments of inertia for the equal angle shown in Fig.18.10.


Solution

Let us consider the angle section in two portions, I and II, as shown and determine the position of the centroid

(due to symmetry )

Moment of inertia, Ixx = Iyy

Co-ordinates of centroid of portion I





= [(5 − 2.87) − (2.87 − 0.5)]

= (2.13, − 2.37)

Co-ordinates of centroid of portion II

= [−(2.87 − 0.5), (5.5 − 2.87)]

= (−2.37, 2.63)

Product of inertia,

Ixy = 10(2.13)(−2.37) + 9(2.63)(−2.37)

(as the product of inertia about their own centroidal axis is zero, since portions I and II are rectangles).

Therefore, Ixy = – 50.481– 56.098 = –106.579 cm4

If θ = angle of principal axis UU with respect to XX-axis

Principal angles are θ1 = 45°, θ2 = 90° + 45° = 135°

Principal moments of inertia

Example 18.5 Figure 18.11 shows an I-section 15 cm × 20 cm. Axis X′X′ and Y′Y′ are inclined at an angle of 30° to the axis of symmetry. Determine the moment of inertia about these axes. Calculate also the product of inertia Ix′y′.


Solution

The I-section shown has two axes of symmetry, i.e., UU and VV passing through the centroid G. Therefore, these are the principal axes and Iuu and Ivv are the principal moments of inertia. The angle of inclinations of UU and VV axes with respect to X ′X ′ and Y ′Y ′ axes is θ = 30°.

sin2θ = 0.25, cos2θ = 0.75

Iy′y′ = Ivv cos2 θ + Iuu sin2θ



since Iuv = 0

Ix′x′ = Iuu cos2θ + Ivv sin2θ

Now,

Therefore, Iy′y′ = 1,126.333 × 0.75 + 5,221.333 × 0.25

= 844.749 + 1,305.333 = 2,150.082 cm4

Ix′x′ = 5,221.333 × 0.75 + 1,126.33 × 0.25

= 3,915.999 + 281.583 = 4,197.582 cm4

Now,


Exercise 18.4 Consider a rectangular section of 6 cm width and 12 cm depth. Determine Ixx, Iyyand Ixy about XX and YY axes inclined at an angle of 45° to the principal axes.

Exercise 18.5 Determine the principal angles and principal moments of inertia of a z-section shown in Fig.18.12.

Stresses due to Unsymmetrical Bending

When the load-line on a beam does not coincide with one of the principal axes of the section, unsymmetrical bending takes place. Figure 18.13(a) shows a rectangular section, symmetrical aboutXX and YY axes or with UU and VV principal axes. Load-line is inclined at an angle ø to the principal axis VV, and passing through G (centroid) or C (shear centre) of the section.

Figure 18.13




Figure 18.13(b) shows an angle section which does not have any axis of symmetry. Principal axes UUand VV are inclined to axes XX and YY at an angle θ. Load-line is inclined at an angle ø to the vertical or at an angle (90 − ø − θ) to the axis UU. Load-line is passing through G (centroid of the section).

Figure 18.13(c) shows a channel section which has one axis of symmetry, i.e., XX. Therefore, UU andVV are the principal axes. G is the centroid of the section while C is the shear centre. Load-line is inclined at an angle ø to the vertical (or the axis VV) and passing through the shear centre of the section.

Shear centre for any transverse section of a beam is the point of intersection of the bending axis and the plane of transverse section. If a load-line passes through the shear centre there will be only bending of the beam and no twisting will occur. If a section has two axes of symmetry, then shear centre coincides with the centre of gravity or centroid of the section as in the case of a rectangular, circular or I-section. For sections having one axis of symmetry only, shear centre does not coincide with centroid but lies on the axis of symmetry, as shown in the case of a channel section.

For a beam subjected to symmetrical bending only, following assumptions are made:

1. The beam is initially straight and of uniform section throughout.

2. Load or loads are assumed to act through the axis of bending.

3. Load or loads act in a direction perpendicular to the bending axes and load-line passes through the shear centre of transverse section.

Figure 18.14, shows the cross-section of a beam subjected to bending moment M, in the plane YY. Gis the centroid of the section and XX and YY are the two co-ordinate axes passing through G. Moreover, UU and VV are the principal axes inclined at an angle θ to the XX and YY axes, respectively. Let us determine the stresses due to bending at the point P having the co-




ordinates u, vcorresponding to principal axes. Moment applied in the plane YY can be resolved into two components M1 and M2.

Figure 18.14

M1, moment in the plane UU = M sin θ

M2, moment in the plane VV = M cos θ

The components M1 and M2 have their axis along VV and UU, respectively.

Resultant bending stress at the point P,

The exact nature of stress (whether tensile or compressive) depends upon the quadrant in which the point P lies. In other words sign of co-ordinates u and v is to be taken into account while determining the resultant bending stress.

The equation of the neutral axis can be determined by considering the resultant bending stress. At the neutral axis bending stress is zero, i.e.,

or,

where

This is the equation of a straight line passing through the centroid G of the section. All the points of the section on one side of the neutral axis have stresses of the same nature and all the points of the section on the other side of the neutral axis have stresses of opposite nature.

Example 18.6 A 40 mm × 40 mm × 5 mm angle section shown in Fig.18.15 is used as a simply supported beam over a span of 2.4 m. It carries a 0.100-kN load along the line YG, where G is the centroid of the section. Determine the resultant bending stresses on points A, B, C, i.e., outer corners of the section, along the middle section of the beam.




Solution


Moment of inertia,

Co-ordinates of G1 (centroid of portion I)

= +(20 − 11.83), −(11.83 − 2.5) = (8.17, −9.33) mm

Co-ordinates of centroid G2

= −(11.83 − 2.5), +(22.5 − 11.83) = −9.33, +10.67 mm

Product of inertia, Ixy = 40 × 5(8.17) × (−9.33) + 35 ×5(−9.33) × (10.67)

(Product of inertia about their own centroidal axes is zero because portions I and II are rectangular strips.)

Ixy = −15,245.22 − 17,421.44 = −32,666.6 mm4

= −3.266 × 104 mm4.

Principal angle, θ


Bending moment,

Components of bending moment,

M1 = M sin 45° = 0.060 × 0.707 × 106 = 42.42 × 103 N mm

M2 = M cos 45° = 0.060 × 0.707 × 106 = 42.42 × 103 N mm

u–v co-ordinates of the points

Point A, x = −11.83, y = 40 − 11.83 = 28.17 mm

u = x cos θ + y sin θ = −11.83 × 0.707 + 28.17 × 0.707 = 11.55 mm

v = y cos θ − x sin θ = 28.17 × 0.707 + 11.83 × 0.707 = 28.28 mm

Point B, x = −11.83, y = −11.83

u = −11.0083 × 0.707 − 11.83 × 0.707 = −16.727 mm, v = 0

Point C, x = 40 − 11.83 = 28.17, y = −11.83

u = 28.17 × cos 45° − 11.83 sin 45°

= 28.17 × 0.707 − 11.83 × 0.707 = 11.55 mm

v = −11.83 × 0.707 − 28.17 × 0.707 = −28.28 mm.

Resultant bending stresses at points A, B and C


Exercise 18.6 Figure 18.16 shows I-section of a cantilever1.2 m long subjected to a load W = 40 N at free end along the direction Y′G inclined at 15° to the vertical. Determine the resultant bending stress at corners A and B, on the fixed section of the cantilever.



Deflection of Beams Due to Unsymmetrical Bending

Figure 18.17 shows the transverse section of a beam with centroid G. XX and YY are two rectangular co-ordinate axes and UU and VV are the principal axes inclined at an angle θ to the XY set of co-ordinate axes. Say the beam is subjected to a load W along the line YG. Load can be resolved into two components, i.e.,

Wu = W sin θ (anlong UG direction)Wv = W cos θ (anlong VG direction)

Say, deflection due to Wu is GA in the direction GU

i.e.,

where K is a constant depending upon the end conditions of the beam and position of the load along the beam.

Deflection due to Wv is GB in direction GV


Figure 18.17

i.e.,

Total deflection,

Total deflection δ is along the direction GC, at angle γ to VV axis.

Comparing this with Eq. (18.15) of the section ‘Stresses Due to Unsymmetrical Bending’

where α is the angle of inclination of the neutral axis with respect to UU axis

and

where γ is the angle of inclination of direction of δ with respect to VV axis.

γ = α, showing thereby that resultant deflection δ takes place in a direction perpendicular to the neutral axis.

Example 18.7 A simply supported beam of a length of 2 m carries a central load of 4 kN inclined at 30° to the vertical and passing through the centroid of the section. Determine (a) maximum tensile stress, (b) maximum compressive stress, (c) deflection due to the load and (d) direction of neutral axis. Give, E = 200 ×105 N/cm2.

Solution

Let us first determine the position of the centroid of the T-section shown in Fig.18.18



The section is symmetrical about vertical axis; therefore, the principal axes pass through the centroidG and are along U–U and V–V axes shown.

Therefore,


Load W = 4,000 N

Components of W,

Wv = 4,000 × cos 30° = 4,000 × 0.866 = 3,464 N


Wu = 4,000 × sin 30° = 4,000 × 0.500 = 2,000 N

Bending moment,

Bending moment,

Due to Mv, there will be maximum compressive stresses at A and B and maximum tensile stress at Cand D.

Due to Mu, there will be maximum compressive stresses at B and D and maximum tensile stresses atA and D.

Therefore, maximum compressive stress at B,

Maximum tensile stress at C,

Deflection,

where K = 1/48, as the beam is simply supported and carries a concentrated load at its centre.

Therefore,

Now, sin θ = 0.5, sin2 θ = 0.25, cos θ = 0.866, cos2 θ = 0.75

Position of the neutral axis

Exercise 18.7 A cantilever 2.8 m long having T-section with flange 12 cm × 2 cm and web 13 cm × 2 cm carries a concentrated load W at its free end but inclined at an angle of 45° to the vertical. Determine the maximum value of W if the deflection at the free end in not to exceed 2 mm. Given that E = 200 kN/mm2, what is the direction of neutral axis with respect to the vertical axis.

Shear Centre

We have studied about the distribution of shear stresses in the transverse section of a beam subjected to a bending moment M and a shear force F. Summation of the shear stresses over the section of the beam gives a set of forces which must be in equilibrium with the applied shear force F. In case of symmetrical sections such as rectangular and I-sections, the applied shear force is balanced by the set of shear forces summed over the rectangular section or over the flanges and the

web of I-section and the shear centre coincides with the centroid of the section. If the applied load is not placed at the shear centre, the section twists about this point and this point is also known as centre of twist. Therefore, the shear centre of a section can be defined as a point about which the applied shear force is balanced by the set of shear forces obtained by summing the shear stresses over the section.

For unsymmetrical sections such as angle section and channel section, summation of shear stresses in each leg gives a set of forces which should be in equilibrium with the applied shear force.

Figure 18.19(a) shows an equal angle section with principal axis UU. We have learnt in previous examples that a principal axis of equal angle section passes through the centroid of the section and the corner of the equal angle as shown in the figure. Say this angle section is subjected to bending about a principal axis UU with a shearing force F at right angles to this axis. The sum of the shear stresses along the legs, gives a shear force in the direction of each leg as shown. It is obvious that the resultant of these shear forces in legs passes through the corner of the angle and unless the applied force F is applied through this point, there will be twisting of the angle section in addition to bending. This point of the equal angle section is called its shear centre or centre of twist.

For a beam of channel section subjected to loads parallel to the web, as shown in Fig. 18.19(b), the total shearing force carried by the web must be equal to applied shear force F, then in flanges there are two equal and opposite forces say F1 each. Then, for equilibrium, F × e is equal to F1 × h and we can determine the position of the shear centre along the axis of symmetry, that is, e = (F1 × h/F).



Figure 18.19

Similarly, Fig. 18.19(c) shows a T-section and its shear centre. Vertical force in the web F is equal to the applied shear force F and horizontal forces F1 in two portions of the flange balance each other at shear centre.

Example 18.8 Figure 18.20(a) shows a channel section, determine its shear centre.

Solution

Figure 18.20(a) shows a channel section with flanges b × t1 and web h × t2. XX is the horizontal symmetric axis of the section. Say, F is the applied shear force, vertically downwards. Then, shear force in the web will be F upwards. Say, the shear force in the top flange = F1.

Shear stress in the flange at a distance of x from right hand edge




Figure 18.20 Shear stresses in channel sections (a) and (b)

where F = applied shear force

first moment of area about axis X–X

t = t1 (thickness of the flange)

Shear stress

Figure 18.20(b) shows, the variation of shear stress in flanges and web. Shear force in elementary area (t1 dx = dA)

= τdA = τt1 dx

Total shear force in top flange , where b = breadth of the flange

(say)


There will be equal and opposite shear force in the bottom flange.

Say shear centre is at a distance of e from web along the symmetric axis XX.

Then, for equilibrium

Moment of inertia,

in which, the expression (2bt13/12) is negligible when comparing

to other terms

Substituting this in the expression for e

if we take,

bt1 = area of flange = Af

ht2 = area of web = Aw

Then,

Exercise 18.8 A channel section has flanges 6 mm × 1 mm and web 8 cm × 0.5 cm. Determine the position of its shear centre.

Problem 18.1 Find the product of inertia of a quadrant of a circle about axes X and Y as shown inFig. 18.21.


Solution

Figure 18.21 shows the quadrant of a circle of radius R. Consider a small element at radius r, radial thickness dr and subtending an angle dθ at the centre.

Area of the element,

dA = rdθdr

Co-ordinates of the element

x = r cos θ and y = r sin θ

Product of inertia,


https://www.safaribooksonline.com/library/view/strength-of-materials/9789332503519/xhtml/chapter018.xhtml#pra18.1


Problem 18.2 A beam of angle section shown in Fig. 18.22 is simply supported over a span of 1.6 m with 15 cm leg vertical. A uniformly distributed vertical load of 10 k N/m is applied throughout the span. Determine (a) maximum bending stress, (b) direction of neutral axis and (c) deflection at the centre. Given E = 210 k N/mm2.

Solution



Moment of inertia



Co-ordinates of G2 and G1: [−1.875(8 − 4.875) and (5 − 2.375) − 4.375]

Ixy = 14(8 − 4.875)(− 1.875) + 10(5 − 2.375) (− 4.375)

= −82.031 − 114.843 = −196.874 cm4

(Note that parallel axes theorem for product of inertia is used here and product of inertia of each rectangular strip about its own centroidal axis is zero.)

Directions of principal axes


(a) Maximum bending stress

w = rate of loading = 10 k N/m = 10,000 N/m = 100 N/cm

Components,

wu = w sin θ = 100 × 0.4065 = 40.65 N/cm

wv = w cos θ = 100 × 0.9138 = 91.38 N/cm.

The beam is simply supported and carries uniformly distributed load, the maximum bending moment occurs at the centre of the beam.

Bending moment Mu = (wul2/8) = (40.65 × 160 × 160/8) = 130,080 N cm (as span length I = 160 cm)

Bending moment Mu = (wul2/8) = (91.38 × 160 × 160/8) = 292,416 N cm

As it is obvious, maximum bending stress occurs at the point A with co-ordinates,

x = −2.375, y = 15 − 4.875 = 10.125.

Co-ordinates u = x cos θ + y sin θ = −2.375 × 0.9138 + 10.125 × 0.4065

= −2.170 + 4.116 = +1.946 cm

v = y cos θ − x sin θ = 10.125 × 0.9138 + 2.375 × 0.4065

= 9.252 + 0.965 = 10.217 cm.

Maximum bending stress,

Direction of neutral axis,

Deflection at the centre

where w = rate of loading

Constant,

Span length, l = 160 cm. E = 210 × 105 N/cm2

Deflection,

(in the direction perpendicular to the neutral axis)

Problem 18.3 A 3-m-long cantilever of I-section carries a load of 2 kN at the free end and 3 kN at its middle. Line of load 2 kN is passing through the centroid of the section and inclined at an angle of 30° to the vertical and the line of application of a load of 3 kN is also passing through the centroid but inclined at 45° to the vertical on the other side of a load of 2 kN as shown in Fig. 18.23. I-section has two flanges 12 cm × 2 cm and web 16 cm ×



1 cm. Determine the resultant bending stress at the corners A, B, C and D.


Solution

Moment of inertia,

I-section shown is symmetrical about XX and YY axes, so principal axes UU and VV passing through the centroid of the section are along XX and YY axes.

Loads sved into components along U and V directions.

Components of 2 kN load,

Wu1 = 2,000 × sin 30° = 1,000 NWu1 = 2,000 × cos 30° = 1,732 N


Components of 3 kN load,

Wu2 = 3,000 × sin 45° = 3,000 × 0.707 = 2,121 NWu2 = 3,000 × cos 45° = 3,000 × 0.707 = 2,121 N

Bending moments at the fixed end,

Mu = Wu1 × 3 + Wu2 × 1.5 = −1,000 ×3 + 2,121 ×1.5 Nm = 181.5 Nm = 0.18 × 105 N cm

Mv = Wv1 × 300 + Wv2 × 150 = 1,732 × 300 + 2,121 ×150 = 8.38 × 105 N cm

Resultant bending stresses

Due to Mu, there will be tensile stresses at points B and C and compressive stresses at points D and A.

Due to Mv, there will tensile stress at points A and B, and compressive stress at points C and D.

Stress,

Stress,

Problem 18.4 For an extruded beam, the cross-section is shown in Fig. 18.24. Determine (a) location of the shear centre O and (b) the distribution of the shearing stresses caused by vertical shearing force, F = 35 kN applied at O. Given Iz = 0.933 × 106 mm4.


Solution

Shear force, F = 35,000 N

Moment of inertia, Iz = 0.933 × 106 mm4

Shear stress at A,



Shear stress at D,

Shear stress at C,

(at the centre of web)

Shear force in the segment AB,

Shear force in the segment DE,

Taking moments about the centre of the web

Problem 18.5 Determine the location of the shear centre O of a beam of a uniform thickness of 3 mm having the cross-section shown in Fig. 18.25.



Solution

Shear force in segment AB and EF will be equal and let us say it is equal to F1. Shear force in segmentBC will pass through the point C. Therefore, we need not calculate F2.

Shear force

Let us take dimensions in centimetre. Say, F is the vertical shear force on the section applied at O.




Shear force,

We need not calculate the shear force F2 in segment BD as it is passing through the point D.

Moment of inertia, Iz

Shear force,

Taking moments of the forces about the point D

or e = 1.454 cm = 14.54 mm

Problem 18.6 Determine the location e of the shear centre, point O for the thin-walled member having the cross-section

shown in Fig. 18.27. The member segments have the same thickness t.


Solution

Moment of inertia of the section

Let us calculate the shear force in members AB or DE due to the vertical shear force F at shear centreO, then



substituting the value of Ioz

Taking moments of the forces about the point C

Problem 18.7 A force P is applied to the web of the beam as shown in Fig. 18.28. If e = 250 mm, determine the height h of the right side flange so that the beam will deflect without any warping. The member segments have the same thickness t.



Solution

Moment of inertia about xx axis

neglecting web.

Shear force in the vertical segment CD (as in Fig.18.29),

Now, e = 250 mm = 25 cm

Taking moment of the forces about the point E



Figure 18.29

Problem 18.8 Determine the location of the shear centre O of a thin-walled beam of uniform thickness having the cross-section as shown in Fig. 18.30.

Solution

Take the dimensions in centimetre. Let us calculate the shear forces in vertical segments DE and FG.


t = thickness of each segment

Say, the shear force on beam = F at O.

SF in vertical number DE,



where Iz = moment of inertia of the section about OZ axis

Similarly, shear force in vertical member FG,

Moment of inertia,

The moments of the shear forces about the point H is taken as shown in Fig. 18.31.

Figure 18.31


Problem 18.9 A beam with thin-walled semi-circular section is shown in Fig. 18.32. It is loaded in a principal plane xy, so as to produce simple bending in this plane. Find the distance e for the shear centre O as shown in figure.

Solution

The shear stress τ at each section s = r·ø along the centre line of radius r will be in the direction of the tangent to the line. τ is tangential to the centre line at s′ or perpendicular to the line GS′ as shown. The magnitude of shear stress at s′,

We note that shear stress is maximum at (ø = π/2) and is zero at ø = 0 and π.

dF, elementary shear force = τb ds = τ br dø, dT, Moment of dF about centre G = δ br2 dδ.Total moment



The horizontal component of the elemental shear force, τbr dø above the neutral axis OZ, cancel the horizontal components of those below the neutral axis, hence the shear stress resultant is a vertical force equal to the shear force F at the section. To produce twisting moment T, calculated above, this force must act through a point O such that,

or,

where

Substituting the value of Iz,


Problem 18.10 Determine the location e of the shear centre point O for the thin-walled open pipe with the cross-section shown in Fig. 18.33.

Solution

The shear stress τ at any section, s = rφ along the centre line of radius r will be in a direction tangential to centre line as shown.

Magnitude of shear stress


where,



Torque about the centre,

For the moment of inertia, consider an element at an angular displacement of θ,

Therefore, twisting moment T,

Problem 18.11 Determine the position of the shear centre of the section of a beam shown in Fig. 18.34

Solution

Figure 18.34 shows the section for which the shear centre is to be determined. In the diagram, direction of shear flow is given. Due to symmetry shear forces, F1 = F5, shear forces, F2 = F4. The section is symmetrical about the axis XX; therefore, shear centre will lie on this axis.

Let us determine shear force F1 or F5.

Shear stress in the vertical portion AB,




where F is the applied shear force on the section.

Now, dA = t1dy

Shear force


Shear force in the horizontal portion BC,

Taking moments of the shear forces about the centre O of the vertical web,

Moment of inertia

Problem 18.12 For a section shown in Fig. 18.35, determine the position of the shear centre. The thickness of the section is t throughout.


Solution

Due to symmetry



Shear force in portion AB, F1 = F4, shear force in portion CD

Shear force in portion BO, F2 = F3, shear force in portion OC

Shear force F1

Shear stress, where F = applied shear force

Shear force,

where a = z t (as shown) z is along AB

Therefore, dA = t dz

Distance,

Shear force,

shear forces F2 and F3 are passing through the point O and will not produce any moment about O.

Moment of inertia Ixx

Moment of inertia of AB, about their principal axes

(as shown in Fig. 18.35)

substituting cos θ = sin θ = 0.707


Similarly, the moment of inertia of BO

Now, θ = 45°



as

Total moment of inertia of section,

Taking moments of the shear forces about the point O

or,


o Unsymmetrical bending occurs in a beam: (i) if the section is symmetrical but load-line is inclined to the principal axes or (ii) if section itself is unsymmetrical.

o Product of inertia, Ixy = ∫ xydAProduct of inertia of a section about its principal axes is zero.

o For a symmetrical section, principal axes are along the axes of symmetry.

o Parallel axes theorem for product of inertia,

whereIxy = Product of inertia about any co-ordinates axes X–Y

= Product of inertia about centroidal axes

= Coordinate of the centroid of the section about XY co-ordinates.

o If Ixy, Iyy, Ixy are moments of inertia about any co-ordinates

axes X–Y passing through the centroid of the section. Inclination of principal axis with respect to X–X axes.


o If principal moments of inertia of a section are Iuu, Ivv, then moment of inertia about an axis X–Xinclined at angle θ to U–V axis is

Ixx = Iuu cos2θ + Ivv sin2θ.

o Stresses due to unsymmetrical bending, if u, v are the co-ordinates of a point and M is the bending moment applied on the section and θ is the angle of inclination of axis of M, with respect to the principal axes UU. Resultant bending stress at the point

o Angle of inclination of neutral axis with respect to principal axis UU

o Deflection of a beam under load W causing unsymmetrical bending

whereK = Constant depending upon end conditions of the

beam and position of the load

θ = Angle of inclination of load W with respect to VV principal axis.

o If the direction of the applied load on a beam passes

through the shear centre of the section, no twisting takes place in the beam.

o For a section symmetrical about two axes, shear centre lies at the centroid of the section.

o For a section symmetrical about one axis only, shear centre lies along the axis of symmetry.

o About the shear centre, the moment due to the applied shear force is balanced by the moment of the shear forces obtained by summing the shear stresses over the various portions of the section.


1. The product of inertia of a rectangular section of a breadth of 4 cm and a depth of 6 cm about its centroidal axis is

1. 144 cm4

2. 72 cm4

3. 36 cm4

4. None of the above2. The product of inertia of a rectangular section of a breadth

of 3 cm and a depth of 6 cm about the co-ordinate axes passing at one corner of the section and parallel to the sides is

1. 81 cm4

2. 72 cm4

3. 54 cm4

4. None of these3. For an equal angle section, co-ordinate

axes XX and YY passing through centroid are parallel to its length. The principal axes are inclined to XY axes at an angle

1. 22.5°

2. 45.0°3. 67.5°4. None of the above

4. For an equal angle section, moment of inertia Ixx and Iyy are both equal to 120 cm4. If one principal moment of inertia is 180 cm4, the magnitude of other principal moment of inertia is

1. 180 cm4

2. 120 cm4

3. 60 cm4

4. 30 cm4

5. For a section, principal moments of inertia are Iuu = 360 cm4 and Ivv = 160 cm4. Moment of inertia of the section about an axis inclined at 30° to the U–U axis, is

1. 310 cm4

2. 260 cm4

3. 210 cm4

4. 120 cm4

6. For an equal angle section, Ixx = Iyy = 32 cm4 and Ixy = −20 cm4. The magnitude of one principal moment of inertia is

1. 52 cm4

2. 42 cm4

3. 32 cm4

4. 16 cm4

7. For a T-section, shear centre is located at1. Centre of the vertical web2. Centre of the horizontal flange3. At the centroid of the section4. None of the above.

8. For an I-section (symmetrical about X–X and Y–Y axis) shear centre lies at

1. Centroid of top flange

2. Centroid of bottom flange3. Centroid of the web4. None of the above.

9. For a channel section symmetrical about X–X axis, shear centre lies at

1. The centroid of the section2. The centre of the vertical web3. The centre of the top flange4. None of the above.

10. If the applied load passes through the shear centre of the section of the beam, then there will be

1. No bending in the beam2. No twisting in the beam3. No deflection in the beam4. None of these

Practice Problems

1. Figure 18.37 shows Z-section of a beam simply appalled over a span of 2 m. A vertical load of 2 kN acts at the centre of the beam and passes through the centroid of the section. Determine the resultant bending stress at points A and B.

2. Figure 18.38 shows a section of a beam subjected to shear force F. Locate the position of the shear centre as defined by e.



Figure 18.37 Practice Problem 1

Figure 18.38 Practice Problem 23. Determine the location e of the shear centre point C for the thin-

walled member having the cross-section shows in Fig. 18.39, where b2 > b1, the member segments have the same thickness t.

4. For an extruded beam having the cross-section shown in Fig. 18.40, determine (a) location of shear centre and (b) distribution of shear stresses caused by vertical shear force F = 12 kN.


https://www.safaribooksonline.com/library/view/strength-of-materials/9789332503519/xhtml/chapter018.xhtml#p-591






Figure 18.40 Practice Problem 45. Determine the location e of the shear centre for the thin-walled

member having the cross-section shown in Fig. 18.41. The member segments have the same thickness.

6. Determine the location of shear centre O of a thin-walled beam of uniform thickness having the equilateral triangular section as shown in Fig. 18.42.






Figure 18.42 Practice Problem 67. Determine the location of shear centre of a thin-walled beam of

uniform thickness having the cross-section shown in Fig. 18.43.



Exercise 18.2:

Exercise 18.3: 147.25 × 104 mm4, +32.30 × 104 mm4.

Exercise 18.4: 540 cm4, 540 cm4, −324 cm4

Exercise 18.5: 28° 31′, 118° 31′, 360 cm4, 38.33 cm4

Exercise 18.6: σA = –491.67 N/cm2, σB = +2,812.5 N/cm2









Exercise 18.7: 221.2 N, 15° 13′

Exercise 18.8: 2.7 cm


1. (d) 2. (a) 3. (b) 4. (c) 5. (a)

6. (a) 7. (b) 8. (c) 9. (d) 10. (b)


1. 147.44 MPa, 17.42 MPa

2.

3.4. 23.2 mm, 12.36 MPa at B, 25.2 MPa at neutral axis

5.6. e = 0.1443a7. e = 2a



4b. Combined Bending and Direct StressesCHAPTER OBJECTIVES

The columns and struts are subjected to axial compressive load and if the load becomes eccentric; then, in addition to direct compressive stress, there will be a bending stress developed in the column/strut and these bending stresses are compressive and tensile as well. However many columns made of brick/stone masonry are incapable of withstanding tensile stress.

o Students will learn, through the chapter, to calculate the resultant stresses due to eccentric load and to mark the area of cross section, on which if the load is applied at any point there will not be any tensile stress anywhere in the section.

o Similarly, frame of manufacturing machines are subjected to eccentric load and section of frame has to be designed so as to withstand the compressive and/or tensile stress.

Introduction

In this chapter, we will take various sections of columns/struts such as circular, rectangular, I-section and hollow section subjected to eccentric loading causing a bending moment in addition to a compressive force on struts/columns. For each section, a core will be established, and if load is applied within the core, there will not be any tensile stress developed anywhere in column section.

Wind loads on walls and chimneys cause bending moment on wall/chimney due to which tensile and compressive stresses are developed in section of wall/chimney. Depending upon the stress due to self-weight of the wall/chimney, resultant stresses at the base of wall/chimney will be determined.

Eccentric Axial Thrust on a Column

A column of rectangular section B × D is shown in Fig. 10.1. G is the centroid of the section abcd. A vertical load P is applied at G along xx axis such that GG′ = e, eccentricity.


If this load acts at the CG of the section, then a direct compressive stress is developed in section. Effect of the eccentric load is to produce bending moment, M = Pe, on the section producing tensile and compressive stresses in the section. Along the edge ad, there will be maximum tensile stress due to bending and along edge bc, there will be maximum compressive stress due to bending.

Figure 10.1 Column under eccentric load

Say, Iyy = moment of inertia of section about yy axis

Direct compressive load = P

Direct stress,

Section modulus,

Maximum bending stresses developed in section,

Resultant stress at edge bc

Resultant stress at edge ad

If σb > σd, then resultant stress at edge ad will be tensile. Resultant stress distribution along xx axis is shown in the Fig. 10.2.

Figure 10.2

Along edge bc

σb + σd = compressive stress

Along edge ad



Note that if

σb − σd = 0, no tensile stress is developed along edge ad

Therefore,

eccentricity must be less than for no tensile stress to develop in section, when load is applied along xx axis

Example 10.1 A cast iron column of a diameter of 250 mm carries a vertical load of 600 kN at a distance of 50 mm from the centre of circular section. Determine the maximum compressive and tensile stresses along the diameter passing through the point of loading.

Solution

Compressive load, P = 600 kN = 6,00,000 N

Diameter of column, d = 250 mm

Area of cross section,

Section modulus,

= 1,533,980 mm3


Moment, M = Pe = 6,00,000 × 50

= 30 × 106 N mm

Direct stress, = 12.223 N/mm2 (Compressive)

Bending stress,

= ±19.557 N/mm2

Maximum compressive stress = 19.557 + 12.223 = 31.78 N/mm2

Maximum tensile stress = 19.557 − 12.223 = 7.334 N/mm2

Figure 10.3 shows the resultant stress distribution along xx axis

Figure 10.3

Exercise 10.1 A cast iron column of rectangular section 150 mm × 250 mm is subjected to an eccentric vertical load of 750 kN. The vertical load is applied at a point 40 mm away from centroid of the section along xx axis, which is parallel to longer side. Determine maximum compressive and maximum tensile stresses in section along the centroidal axis xx.


Load Eccentric to Both Axes (Rectangular Section)

Consider a column of rectangular section B × D as shown in Fig. 10.4. Centroid of the section abcd is atG but a vertical load P is applied at G′ such that GG′ = e, eccentricity. There are two components of this eccentricity along xx and yy axes, that is, ex and ey as shown is the Fig. 10.4.

Figure 10.4 Load eccentric to both axes

Direct vertical load on column = P

Bending moment in xx plane, Mx = Pex




Bending moment in yy plane, My = Pey

Section modulus,

Section modulus,

Bending stresses along xx axis, edges ad and bc =

(Compressive along edge bc and tensile along edge ad)

Bending stresses along yy axis,

(Compressive along edge cd and tensile along edge ab)

Direct compressive stress,

Resultant stresses at corners

Note that we have taken compressive stress as positive stress in a column as the tensile stress is undesirable.

Example 10.2 A cast iron column of a section of 200 mm × 250 mm is subjected to a vertical load of 300 kN acting at a point 40 mm away (along the diagonal) from the centre. Determine the resulting stress at the corners a, b, c and d of the section (Fig. 10.5).

Solution

Sides B = 250 mm

D = 250 mm

Diagonal =


= 320 mm


Eccentricity,

= 31.25 mm




= 25 mm

Bending moment,

Mx = Pex = 300,000 × 31.25 = 9.375 × 106 NmmMy = Pey = 300,000 × 25 = 7.5 × 106 Nmm

Section modulus,

Direct stress,

Bending stress

Resultant stresses

σa = σd − σbx − σby = 6 − 4.5 − 4.5 = −3 N/mm2 (tensile)

σb = σd + σbx − σby = 6 + 4.5 − 4.5 = +6 N/mm2 (compressive)

σc = σd + σbx + σby = 6 + 4.5 + 4.5 = +15 N/mm2 (compressive)

σd = σd − σbx + σby = 6 − 4.5 + 4.5 = +6 N/mm2 (compressive)

Note that only at corner a, there is −3 N/mm2 (tensile stress) at all other corners resultant stress is compressive.

Exercise 10.2 A cast iron column of square section 400 × 400 mm is subjected to a compressive load of 500 kN acting at a point that is 60 mm away from xx axis and 80 mm away from yycentroidal axis. Determine the resultant stresses at the extreme corners of the section.

[Hint: ex = 80 mm, ey = 60 mm]

Core of Rectangular Section

The core of a rectangular section is the part of the column section in which the load can be applied without causing tensile stress anywhere in the section. Since the columns are made of cast iron, brick work, concrete etc., which are not supposed to withstand tensile stresses because tensile stress may develop cracks in column leading to its failure.

Consider a rectangular section B × D of a column. Say a load P is applied at point G′, such that GG′ = ex(Fig. 10.6).

Bending moment, Mx = Pex

Section modules,

Stress due to bending,

Direct stress,


If the resultant stresses σd ± σbx has to be only compressive and no tensile stress is permitted, then

Figure 10.6 Core of rectangular section

or ,

Similarly, load can be considered on the other side of G, and

In the figure, dimension

If the load is applied within the line ln along x axis, then there will not be any tensile stress anywhere in the section.

Eccentricity along yy axis = ey

Bending moment, My = Pey

Section modulus,

Bending stress,

Resultant stress,

If no tensile stress is permitted in the section, then

line mo shows that

If load is applied anywhere within line mo, no tensile stress is produced in the section. Joining the endsl, m, n and o makes a rhombus of diagonals B/3 and D/3. This rhombus is termed as core of a rectangular section.

Core of Circular Section

Let us consider that the section of a column is circular of diameter d as shown in Fig. 10.7, xx and yyare centroidal axes, with G as centroid of the section. Say load P is applied along xx axis at G′, such thatGG′ = e, eccentricity

Bending moment, Mx = Pe

Section modulus,

Direct stress,

Bending stress,


If the resultant stress is not to be tensile anywhere in the section, then:

or eccentricity,

Similarly, load can be considered along yy axis and on the other

side of yy axis, the

Figure 10.7 Core of a circular section

Core of the circular section is a circle of radius d/8 or diameter d/4 as shown in figure. Area covered by a circle of diameter d/4 is called CORE or KERNEL of circular section. If a load on column is applied within the core, then no tensile stress will be developed anywhere in the section.

Core of Any Section

For any section, core can be marked if we know the area of section, Zx and Zy of the section. Figure 10.8shows an I-section, D × B, with moment of inertia Ixx and Iyy about xx and yy axes. Section modulus,


Maximum eccentricity along x axis,

where A = area of only section

Similarly, eccentricity along yy axis,

Core is a rhombus with diagonals 2ex and 2ey. A simple example will further clarify the dimensions of core of an I-section.

Figure 10.8

Example 10.3 An ISLB 150 joist has following properties, area A = 1,808 mm2, depth D = 150 mm, width, B = 80 mm, Ixx = 688.2 × 104 mm4 and Iyy = 55.2 × 104 mm4. Mark the core of the I-section.

Figure 10.9

Solution

Allowable eccentricity,

Diagonals of rhombus of core are 108.14 and 15.26 mm as in fig 10.9.



Exercise 10.3 Mark the core of I-section with the following properties: Area = 4,320 mm2, depth = 203 mm, width = 106 mm, Ixx = 27 × 106 mm4, Zx = 266 × 103 mm3, Iyy = 1.81 × 106 mm4, Zy = 34.2 × 103 mm3.

Example 10.4 A short hollow pier 1.6 m × 1.6 m outer sides and 1 m × 1 m inner sides supports a vertical load of 3,000 kN at a point located on a diagonal 0.5 m from vertical axis of the pier (Fig 10.10). Neglecting self-weight of the pier, calculate the normal stress at four outside corners on a horizontal section of the pier.


Solution

Outer side = 1.6 m

Inner side = 1.0 m

Area, A = 1.62 − 12

= 1.56 m2

Eccentricity about x axis,



Eccentricity about y axis,

Bending moment,

Mx = 3,000 × 0.353 = 1,059 kN mMy = 3,000 × 0.353 = 1,059 kN m

Direct stress,

Bending stress,

Resultant stresses

σa = 1,923 − 1,830.6 − 1,830.6 = −1,738.2 kN/m2 (tensile)

σb = 1,923 − 1,830.6 + 1,830.6 = +1,923 kN/m2 (compressive)

σc = 1,923 + 1,830.6 + 1,830.6 = +5,584.2 kN/m2 (compressive)

σd = 1,923 + 1,830.6 − 1,830.6 = +1,923 kN/m2 (compressive)

Exercise 10.4 A short cast iron column is of hollow circular section with an outer diameter of 1.5 m and an inner diameter of 1.0 m as shown in Fig. 10.11. It is subjected to a vertical load of 2,500 kN at G′, such that GG′ = 0.25 m. Determine resultant stress at abcd edges of the section.


Figure 10.11

Wind Pressure on Walls

A masonry wall of breadth B, length L and height H is shown in Fig. 10.12.

Base of wall is abcd of area B × L. Horizontal wind pressure p acts on front face of the wall.

Figure 10.12 Wind pressure on wall

Total wind pressure = p × H × L = Pw.


C.G of wind load Pw lies at H/2 from base.

Bending moment due to wind load,

Section modulus,

Bending stress due to M,

Say weight density of masonry = w

Direct compressive stress due to self-weight, σd = wH

Resultant stress along edge (compressive)

Resultant stress along edge ad =

Note that wH should be to avoid tension in base.

Example 10.5 A 10-m-high brick masonry wall of rectangular section 4 m × 0.5 m is subjected to a horizontal wind pressure of 150 kN/m2 on the 4 m side. Find the maximum and minimum stress intensities induced in the base, weight density of masonry is 22 kN/m3.

Solution

Wall Height = 10 m

Length, L = hm

Width, B = 0.5 m

w = 22 kN/m3, p = 150 kN/m2

Direct stress, σd = 22 × 10 = 220 kN/m2

Bending stress,

= 180 kN/m2

σmax = 220 + 180

= 400 kN/m2 (compressive) = 0.4 N/mm2

σmin = 220 − 180 kN/m2

= 40 kN/m2 (compressive)

= 0.04 N/mm2

Exercise 10.5 A masonry brick wall of height 3 m, length 6 m and width 0.4 m is subjected to a wind pressure p on the face of length 6 m and height 3 m. Weight density of masonry structure is 21.8 kN/m3. What is the maximum value of p if tensile stress is not to be developed in base due to wind ressure?

Wind Pressure on Chimney Shafts

High chimneys are subjected to wind pressures which may cause overturning of the chimney at the base; therefore, these chimneys must be properly grouted in the ground. Consider a chimney of heightH, internal diameter d and external diameter D as shown in Fig. 10.13.


Figure 10.13 Wind pressure on chimney

Say weight density of chimney material = w

Direct compressive stress at the base due to self-weight, σd = wH.

Consider a small strip of width Rdθ, subtending arc angle dθ at the centre 0, and making an angle θwith the axis xx as shown.

δP = wind force reaching the small strip.

δP = p Rdθ H cos θ = pHR cos θ dθ

Component of the force δP, normal to the strip, δPn = δP cos θ = pHR cos2θ dθ

Horizontal component of δP, δPHx = δPncos θ

= pHR cos3θ dθ (10.4)

Another horizontal component of δPn, that is, δpHy = δPsin θ = pHR cos2θ sin θ dθ is cancelled out when we consider a strip in other quadrant (as shown)

Total force component in xx direction = 2pHR cos3θ dθ

Integrating over the whole exposed surface from θ = 0 to

Total wind force,

where k = coefficient of wind resistance, D = 2R

DH = projected area of curved surface of chimney

CG of the force lies at a distance of

From the base, in the case of uniform cylindrical chimney, M bending moment due to wind force =

Section modulus,

Bending stress,

Generally the coefficient of wind resistance is taken as 0.6 for cylindrical chimneys.

Example 10.6 A 20-m-high cylindrical chimney shaft is of hollow circular section, 2.4 m external diameter and 1.2 m internal diameter. The intensity of horizontal wind pressure varies as y2/3, where yis the height above the ground. Determine the maximum

and minimum intensity of stress at the base (Fig 10.14). Given that:

1. Density of masonry structure is 22.2 kN/m3

2. Coefficient of wind resistance, k = 0.63. Wind pressure at a height of 27 m is 1.8 kN/m2

Figure 10.14

Solution

Say intensity of wind pressure, p = Cy2/3

At y = 27 m and p = 1.8 kN/m2

Therefore, 1.8 = C × 272/3

9C = 1.8

C = 0.2

Pressure at any height, y = 0.2y2/3

Area, dA = 2.4dy

p = 0.2y2/3

dP = kP dA = k × 0.2y2/3 × 2.4 dy


= 0.48 ky2/3 dy

However, k = 0.6

dP = 0.6 × 0.48 × y2/3dy = 0.288y2/3dy

Moment of the force from the base

dM = dPy = 0.288 y5/3dy

Total moment,

Section modulus,

Bending stress,

Height, H = 20 m

Direction stress, σd = wH = 22.2 × 20 = 444 kN/m2

Maximum stress at base = 444 + 248.50 = 692.5 kN/m2 (compressive)

Minimum stress at base = 444 − 248.5 = 195.5 kN/m2 (compressive)

Exercise 10.6 A 20-m-high masonry chimney of uniform circular section, 5 m external diameter and 3 m internal diameter has to withstand a horizontal wind pressure of 2 kN/m2 of the projected area. Find the maximum and minimum stress intensity at the base. Density of masonry structure w= 21 kN/m3.

Problem 10.1 A short column of hollow circular section of internal diameter d and external diameterD is loaded with a compressive load W. If D = 1.5d, determine the diameter of the core (Fig 10.15).

Figure 10.15

Solution

Outer diameter = 1.5d

Inner diameter = d

Area of a cross section

Moment of inertia,

Section modulus,


Radius of core,

Eccentricity

2e = 0.271 × 2d = 0.542d

Diameter of the core = 0.542d

Problem 10.2 The cross section of a short column is of the shape of an arrow. Find the position of the vertical downward force on the line of symmetry of the section so that at edge A stress is just zero (Fig. 10.16).

Solution

AB is line of symmetry or the axis yy.

Let us locate the position of G with respect to lower edge B,

Moment of inertia,

Direct stress,


Bending stress at A,

GG′ = e, as shown in Fig. 10.16.

Figure 10.16

Problem 10.3 To avoid interference, the cross-sectional area of a link in a machine is reduced in section as shown is Fig. 10.17. The thickness of the link is 20 mm. Determine the maximum force that can be applied if the maximum normal stress in section AB is limited to 100 MPa.

Solution

Say P is the force on N.



Area, A = 20 × 50

= 1,000 mm2

Figure 10.17

Eccentricity of load for section AB

Moment = Pe

= 25 P N mm

Section modulus,

Direct stress,

Bending stress,

Maximum normal stress = σd + σb

or P + 3P = 1,00,000

4P = 1,00,000

P = 25,000 N = 25 kN

Problem 10.4 The cross section of a short column is shown in Fig. 10.18. Load of 160 kN is applied atP, 75 mm from edge AD. Section is symmetrical about xx axis, determine the stresses at corners A, B, Cand D of the section.


Solution

Position of G along xx axis

Eccentricity, e = 75 − 71.82 = 3.18 mm



Area, A = 11,000 mm2

P = 1,60,000 N


Moment of inertia,

Edge BC will be in compression, edge AD will be in tension

Resultant stresses

σAD = 14.54 − 1.322 = 13.218 N/mm2 (compressive)

σBC = 14.54 − 1.44 = 15.98 N/mm2 (compressive)

Problem 10.5 A steel rod with a diameter of 20 mm passes through a copper tube, 30 mm internal diameter and 40 mm external diameter. Rigid cover plates are provided at each end of the tube and steel rod also passes through these cover plates as shown in Fig.10.19. Nuts are screwed on the projecting ends of the rod, so that the cover plates put pressure on the ends of the


tube. Determine maximum and minimum stresses in copper tube if the nut is tightened to produce a linear strain of 0.001 in the rod. The centre of the rod is 3 mm out of the centre of the tube. Given ES = 2EC = 210 GPa.

Solution

Strain in steel rod = 0.001 tensile

Es of steel = 2,10,000 N/mm2

Stress in rod, σs = 2,10,000 × 0.001

= 210 N/mm2 (tensile)

Area of cross section of steel rod

= 314.16 mm2

Area of cross section of the copper tube =

= 175p mm2 = 549.78 mm2


Pull in steel rod = push in copper tube

P = 210 × 314.16 = sc × 549.78 = 65973.6 N

Stress, σc = 120 N/mm2 (compressive)

Now rod centre is 3 mm out of the centre of the tube (Fig. 10.19)

Moment of inertia, Iyy = = 8.59 × 104 mm2

e = eccentricity = 3 mm

Bending stress in tube,



Maximum stress in copper tube, σmax = 120 + 460 8


σmin = 120 − 46.08


Problem 10.6 A large C clamp is shown in Fig. 10.20. As the screw is tightened down, strain observed in vertical direction at point B is 400 μ strain. What is the load on the screw if E = 200 GPa.

Figure 10.20 C clamp

Solution

Section is symmetrical about xx axis, G can be located along xx axis


Moment of Inertia,

Eccentricity, e = 300 + x1 = 300 + 25 = 325 mm

Moment, M = P e = P × 325 N mm.

Area, A = 800 + 800 = 1,600 mm2

Bending stress at B,

At point B, direct stress and bending stress both are compressive so

Problem 10.7 A machine part is transmitting a pull of 3 kN. It is offset as shown in the Fig. 10.21. Determine the largest normal stress in the offset position.

Solution

Eccentricity of the load = 20 mm

Moment = Pe

= 3,000 × 20

= 60,000 Nmm


Figure 10.21 Machine part

Edge A comes in compression, edge BC comes is tension.

Direct stress,

Bending stress,

Resultant stresses

σA = 53.33 − 6.667 = 46.666 N/mm2 (compressive)

σBC = −26.667 − 6.667 = −33.334 N/mm2 (tensile)

Maximum normal stress is 46.666 N/mm2 (compressive)

Problem 10.8 A tapering chimney of hollow circular section is 45 m high. Its external diameter at the base is 3.6 m and at the top it is 2.4 m. It is subjected to a wind pressure of 22 kN/m2 of the projected area. Calculate the overturning moment at the base. If the weight of the chimney is 6,000 kN, and the internal diameter at the base is 1.2 m, determine the maximum and minimum stress at the base.

Solution

Weight of chimney = 6,000 kN

Area at base

σd = direct camp.

Stress at base

Projected area

= 135 m2

Total wind load = 135 × 22 = 2,970 kN

CG of the load from the base

Overturning moment at base, M = 21 × 2,970 = 62,370 kN m

Figure 10.22 Tapering chimney

Section modulus, base section,

Bending stress,

σmax = 13,786.78 + 663.15 = +14,450 kN/m2 (compressive)

σmax = −13,786.7 + 663.15 = −13,123.6 kN/m2 (tensile)

or σmax = 14.45 N/mm2 (compressive)

σmin = −13.1236 N/mm2 (tensile)

Problem 10.9 A masonry pillar of diameter D in m is subjected to a horizontal wind pressure of intensity p kN/m2. If the coefficient of wind resistance is k, prove that the maximum permissible heightH of the pillar so that no tension is induced at the base is given by:

where w = weight density of masonry

Solution

Weight density of masonry = w kN/m2

Say permissible height = Hm

Direct stress due to self-weight, σd = wH (compressive)

Intensity of wind pressure = p kN/m2

Bending moment,

Section modulus,

Bending stress,

For no tension,

or,


o A short column of rectangular section, , carries eccentric load along x axis, with ex as eccentricity

Resultant stress, o A short column of circular cross section of diameter D,

supports an eccentric load P at an eccentricity e, Resultant stress,

o The core or the kernel of a section is a small area located around the centroid of the section of a column and if any vertical load is applied on the column within this area of core, there will not be any tensile stress developed anywhere in the section.

o Core of a rectangular section B × D, is a rhombus of diagonals B/3 and D/3.

o Core or kernel of a circular section of diameter D is a circular area of diameter D/4.

o For a wall of rectangular section B × D, wind pressure p acting on the face of breadth B and heightH, stress due to bending moment created by the wind

pressure is at the base of the wall.o For a chimney of outside diameter D, height H, coefficient

of wind resistance k, wind pressure p, and inner diameter d. Total wind force, Pw = kp DH

Moment,

Section modulus,

Bending stress,


Review Questions

1. What is the core or kernel of a section? What is its importance?

2. Mark the core of following sections.1. A rectangular section

2. A circular section3. A hollow circular section4. I-section.

3. How the wind pressure on a wall produces an overturning moment on the wall?

4. What do you mean by coefficient of wind resistance for a chimney?

5. How the overturning moment due to wind load on a chimney is calculated?

6. Consider a C-clamp and show how the direct and bending stresses are developed in the critical section of the frame.

7. A cast iron column of hollow circular section is made by casting. What is the effect on stresses developed in column due to an axial compressive load if the core is offset during casting?


1. A cast iron column is of circular section of a diameter of 200 mm. What is the diameter of core of the column?


2. A steel column has an outside diameter of 80 mm and an inside diameter of 60 mm. What is the diameter of core of the column?

1. 62 mm2. 31.25 mm3. 30 mm4. None of these

3. A column with I-section has following properties: Zx = 90 × 103 mm3, Zy = 54 × 103 mm3, A = 1,800 mm2. What is the allowable eccentricity ey for no tension in column?


4. A short column is of hollow square section with outer side 2a and inner side a. A load acts at a distance of 0.25a from CG of the section, along one diagonal. The maximum and minimum stresses developed at corners of the section are 4.8 and −1.2 N/mm2, respectively. The bending stress introduced at the extreme corners of the section by the eccentric load are

1. ±3.6 N/mm2

2. ±2.4 N/mm2

3. ±1.2 N/mm2

4. None of these5. For a cylindrical chimney of hollow circular section,

subjected to wind pressure, the coefficient of wind resistance is generally taken as

1. 0.3–0.52. 0.45–0.55

3. 0.6–0.74. None of these

6. A column is of hollow circular section. Due to an eccentric load, maximum and minimum stresses are 7 and 1 N/mm2 (both compressive), respectively. Now, the eccentricity is doubled, what will be the maximum stress developed in column section?

1. 14 N/mm2

2. 10 N/mm2

3. 8 N/mm2

4. None of these7. A short masonry square section of 1 m side is 10 m high.

Wind pressure of intensity 2kN/m2 acts on a vertical face of column. Weight density of masonry is 20 kN/m3 The maximum stress at the base of the column is

1. 800 kN/m3

2. 400 kN/m2

3. 200 kN/m2

4. None of these8. A chimney is of brick masonry of a weight density of 22

kN/m3. Height of chimney is 10 m. Outside diameter of chimney is 1 m while inside diameter is 0.5 m. What is the compressive stress developed at base, due to self-weight?

1. 0.275 N/mm2

2. 0.22 N/mm2

3. 0.20 N/mm2

4. None of these9. A cast iron column of circular section carries an eccentric

load due to which maximum and minimum stresses developed in column section are +420 kN/m2 and +120 kN/m2 (compressive), respectively. If Z = 1.2 m3, what is the bending moment due to eccentric load?

1. 270 kN m2. 225 kN m

3. 180 kN m4. None of these

10. A cylindrical chimney of hollow circular cross section is subjected to wind pressure p. Density of masonry work is 20 kN/m3. The maximum and minimum stresses developed at the base of chimney are 650 and 150 kN/m2, respectively. If intensity of wind pressure is increased by 50 per cent, what is the maximum stress developed in section of column?

1. 850 kN/m2

2. 775 kN/m2

3. 700 kN/m2

4. None of these

Practice Problems

1. A short column of hollow circular section of internal diameter d and external diameter D is subjected to a vertical load (eccentric). If d = 0.8 D, determine the diameter of the core.

2. A short block has cross-sectional area of a triangle as shown in Fig. 10.23.

Determine the range along the axis yy over which the downward vertical force could be applied at the top of the block without causing any tension anywhere in the base. Neglect the weight of the block.

3. A flat plate of section of a thickness of 20 mm and a width of 60 mm, which is placed in a testing machine, is subjected to 60 kN of load acting along line AB as shown in Fig. 10.24, an extensometer adjusted along the line of the load recorded an extension of 0.078 mm on a gauge length of 150 mm. Determine (i) minimum and maximum stresses set up in plate, and (ii) Young’s modulus of the plate.




Figure 10.24 Practice Problem 34. The cross section of a short column is shown in Fig. 10.25. Load

of 625 kN is applied at P, 150 mm from edge AB, on yy axis. Determine the stresses developed at corners A, B, C and D of the section.

5. A short cast iron column has an external diameter of 200 mm and an internal diameter of 160 mm. The distance between the centre of outer and inner circles due to the displacement of core during casting is 6 mm. A load of 400 kN acts through a vertical centre line passing through the centre of the outer circle. Calculate the values of greatest and least compressive stresses in a horizontal cross section of the column.

[Hint: Locate G along x-axis, determine x1 and x2, calculate Iyy.]




Figure 10.25 Practice Problem 46. A C clamp is shown in the Fig. 10.26. As the screw is tightened

on wooden blocks, it was observed that a compressive force of 2 kN is applied on blocks. Section of C clamp is given in Fig. 10.26. Determine the stresses at points A and B of the section.

7. A bar of circular section with a diameter of 20 mm is subjected to a load P, 10 mm off the centre as shown in Fig. 10.27. If the maximum stress in bar is not to exceed 80 MPa, calculate the value of P.







Figure 10.27 Practice Problem 78. A tapering chimney of hollow circular section is 30 m high. Its

external diameter at the base is 3 m and at the top it is 2 m. It is subjected to a wind pressure of 2 kN/m2 of the projected area. Calculate the overturning moment at the base, if the weight of the chimney is 5,000 kN and its internal diameter at the base is 1.2 m. Determine the maximum and minimum stresses at the base.

9. A masonry pillar of diameter 0.8 m is subjected to a horizontal wind pressure of 1.2 kN/m2. If the coefficient of wind resistance is 0.6, determine the maximum permissible height of the pillar so that no tension is induced at the base. Given w = 23.2 kN/m3

Special Problems

1. Mark the core of an I-section with the following properties: area = 12,700 mm2, depth = 229 mm, width = 210 mm, Ixx = 113 × 106 mm4, Iyy = 36.6 × 106 mm4.

2. The cross section of a short column is as shown in Fig. 10.28. A vertical load of W is applied at G′,GG′ = 10 mm. Determine (a) magnitude of W if maximum stress set up in the cross section is not to exceed 75 MPa and (b) stress distribution along the edge AD.



Figure 10.28 Special Problem 23. A short hollow cylindrical column carries a vertical load of 400

kN. Its external diameter is 200 mm and internal diameter is 120 mm. Find the maximum permissible eccentricity of the load if the allowable stresses are 60 MPa in compression and 25 MPa in tension.

[Hint: Calculate σd, σd − σb = −25, σd + σb = 60]4. An 80-mm (t = 6 mm) steel pipe is used as a support for a basket

ball backboard as shown in Fig. 10.29. It is securely fixed into the ground. Calculate the maximum normal stress that would be developed if a 750-N player hangs on the base of the rim of the basket.

[Hint: (e = 1.2 m) (80-mm pipe inside diameter)]



Figure 10.29 Special Problem 4


Exercise 10.1: 39.2 N/mm2 (compressive), no tensile stress

Exercise 10.2: +4.063, 9.687, 2.187, −3.437 MPa (tensile)

Exercise 10.3: ey < 60.87 mm, ex < 7.83 mm, Rhombus with diagonals 15.66 mm and 121.74 mm

Exercise 10.4: σa = 196.1 kN/m2 (compressive), σb = σd = 2546.6 kN/m2, σc = 4897.1 kN/m2(compressive)

Exercise 10.5: (0.3875 kN/m2)

Exercise 10.6: 607.26 kN/m2 (compressive), 232.74 kN/m2 (compressive)


1. (c) 2. (b) 3. (c) 4. (a) 5. (c)

6. (b) 7. (a) 8. (b) 9. (c) 10. (b)


1. (0.41 D)2. yG = 100 mm, Gy = 200 mm, e1 = 25 mm, e2 = 50 m,

range = 75 mm below AB to 150 mm below edge AB3. −40 MPa, 140 MPa, 200 GPa4. σA = σB = +8.775 N/mm2 (compressive), σD = σC = 11.225

N/mm2 (compressive)5. 46.01, 26.78 MPa6. 61.62 N/mm2 at B, −31.36 N/mm2 at A (tensile)7. 6.2832 kN8. 2,100 kN m; 1653.1 kN/m2 (compressive), 29.1

kN/m2 (compressive)

https://www.safaribooksonline.com/library/view/strength-of-materials/9789332503519/xhtml/chapter010.xhtml#exer-10.6







9. 4.05 m


1. Rhombus of diagonals 54.88 and 155.4 mm2. 237.2 kN; 75 MPa (compressive), 17.4 MPa (compressive)3. 48.50 mm4. 27.952 N/mm2

5a. Thin Cylindrical and Spherical ShellsCHAPTER OBJECTIVES

Thin cylindrical and spherical shells are used mainly for storage of gas, petrol, liquid, chemicals, grains and so on. Some are subjected to internal/external pressures and the order of pressure is low (10–30 atmospheres). Their D/t ratio, that is, the ratio of diameter to wall thickness, is large, that is, D/t is greater than 20. Because in comparison to diameter, thickness is very small, so the variation of stresses along the thickness is taken negligible. In this chapter, we will learn about:

o How to determine hoop stress and axial stress in a thin shell subjected to internal pressure.o How to determine axial, circumferential and volumetric strains in shell subjected to internal pressures.o The hoop stress developed is double the axial stress in thin cylindrical shell. To reduce the effect of hoop stress; cylinder is wound with a wire under tension, with the purpose of reducing the effect of hoop stress.

o A thin spherical shell is equally strong in two hoop directions; therefore, generally the cylindrical shells are provided with hemispherical ends.o Some pressure vessels have double curved wall, that is, at a particular element radii of curvature in two directions are ρ1 and ρ2 (radius of curvature). So pressure vessels with double curved wall will also be analysed.o Conical tanks are available for storage of water. The hoop stress developed is proportional to the square of the depth of element from free water surface. Stress in conical water tanks will also be analyzed.

Introduction

Thin pressure vessels are subjected to internal uniform pressure or hydrostatic pressure of low magnitude. Development of axial and hoop stresses in the wall thickness of the pressure vessel provides useful data for the designer. Stresses in thin shells are dependent on the D/t ratio, that is, the ratio of shell diameter to wall thickness.

Derivation of axial and hoop stresses, strains in cylindrical pressure vessel and derivation of hoop stresses, hoop strain and volumetric strain in both shells forms the text of this chapter.

The effect of wire winding on the reduction of hoop stresses (which is greater than axial stress) will be analysed.

Pressure vessels with the double curved surface of wall and the conical surface of wall will also be analysed. In such shells, equations for hoop stresses will be derived.

Cylindrical pressure vessel can be conveniently used to determine experimentally E and v, that is, elastic constants of material of pressure vessel.

Thin Cylinder Subjected to Internal Pressure

Consider a thin cylindrical shell of inner diameter D, length L, with wall thickness equal to t. The shell is full of liquid. The volume of

the liquid inside the shell is [π/4 (D2L)]and the pressure of the liquid is atmospheric. Now, the additional volume of the liquid is pumped under pressure inside the shell; as a result, there is overall expansion in the shell, that is, expansion in both diameter and length. However, the expansion of liquid is partly prevented by the wall of the metallic cylinder, which offers equal and opposite reaction and compresses the liquid inside the cylinder (Fig 5.1).

Figure. 5.1 Small element of cylinder subjected to internal pressure

Mathematically

δV = additional volume of the liquid pumped inside.

= δv1 + δv2

= expansion in the volume of cylinder + contraction in the volume of the liquid.

Since the liquid exerts pressure on cylinder wall, the length and diameter of the cylinder get increased introducing axial and circumferential stresses, in the cylinder.

Figure 5.2 shows a small element of the cylinder subjected to internal pressure p; σc is the circumferential a hoop stress developed (along circumference) and σa is the axial stress developed (along the axes); at outer surface pa is atmospheric pressure, such that p >>pa.

Axial stress



Diameter of cylinder = D

Wall thickness = t

Axial stress = σa

Figure. 5.2 Small element under internal pressure

Figure. 5.3 Axial bursting force

Axial bursting force on cylinder in axial direction, [Pa = π/4(D2p)] acting on the end plates as shown inFig. 5.3

Resisting area = πDt

Axial stress developed,

Circumferential stress (σc)


Consider a small portion along length δL and inner radius D/2, subjected to internal pressure p, Fig 5.4Take a small element at an angle θ from x-axis, subtending an angle dθ at inner surface.

Area of the small element

Force on the small element,

Vertical component of

Horizontal component of

The horizontal component of the force is cancelled out when the force is integrated over the semicircular portion of the cylinder.

Figure. 5.4 Diametral bursting force

Therefore, total diametral bursting force,


(= p × projected area of the curved surface)

Area of cross-section resisting the diametral bursting force = 2t dl

Circumferential stress developed,

Strains

Consider Fig. 5.2, σc, σa and p stresses are perpendicular to each other. Note that p acts as a compressive radial stress on inner surface of cylinder. Stresses σc and σa are much larger than p; Therefore in calculation of strains, the effect of internal pressure p is neglected. Say E is Young’s modulus and v is Poisson’s ratio of the material.

Circumferential strain,

Axial strain,


Note that εc = circumferential strain

εa = axial strain

Change in the diameter of cylinder, δD = εcD

Change in the length of cylinder, δL = εaL

Volumetric strain

Volume of cylinder [v = π/4 (D2L)], taking partial derivative

or

= 2 × circumferential strain + axial strain

Change in volume

Change in volume of cylinder, δV1 = ev V

Contraction in volume of liquid,

where K is bulk modulus of liquid.

Additional volume of liquid pumped inside the cylinder, δv = δv1 + δv2.

Note that if, in any problem, K for the liquid is not given that δv2, that is, reduction in volume of liquid can be taken as negligible.

Example 5.1 A thin cylindrical shell made of 5-mm-thick steel plate is filled with water under pressure of 3 N/mm2. The internal diameter of the cylinder is 200 mm and its length is 1.0 m. Determine the additional volume of the water pumped inside the cylinder to develop the required pressure. Given for steel E = 208 kN/mm2 and v = 0.3, and for water K = 2,200 N/mm2.

Solution

p = 3 N/mm2, E = 208 kN/mm2

D = 200 mm, v = 0.3

t = 5 mm

L = 1.0 m

Volume,

Volumetric strain, , putting the values

= 0.548 × 10−3

Change in volume of the cylinder, δV1 = εvV

= 0.548 × 10−3 × π × 107

= 1.72 × 104 mm3

Change in volume of water, , where P/K is volumetric strain on water

Additional volume of water pumped in shell, δV = δV1 + δV2

= 1.72 × 104 + 4.284 × 104 mm3 = 6.004 × 104 mm3

= 60 cm3 = 60 cc of water.

Exercise 5.1 A thin cylindrical shell made of 4-mm-thick copper plate is filled with oil under a pressure of 2.4 N/mm2. The internal diameter of the cylinder is 200 mm and its length is 800 mm. Determine the additional volume of oil pumped inside the cylinder so to develop the required pressure. Given E for copper = 104 kN/mm2, v for copper = 0.32, and K for oil = 2,800 N/mm2.

Thin Spherical Shell

Consider a thin spherical shell of diameter D and wall thickness t subjected to internal fluid pressure pas shown in Fig. 5.5(a). The internal pressure p acts throughout the inner curved surface and thediametral bursting force tends to break the thin



shell in two halves, that is, two hemispherical halves, as shown in Fig. 5.5(b).

Figure. 5.5 Thin spherical shell

Diametral bursting force, PD = p × projected area of curved surface

Say σc = hoop stress developed in the wall of the shell.

Area resisting the diametral bursting force = πDt

So

Strains

Consider a small element of thin spherical shell subjected to internal radial stress p, due to which circumferential stress σc is developed in shell, as shown in Fig. 5.6.



Figure. 5.6 Small element of thin spherical shell

pa = atmosphere pressure on outer surface.

If E and v are Young’s modulus and Poisson’s ratio of the material, respectively, then

Circumferential strain

= , putting value of σc

Internal pressure p << σc; therefore, the effect of p is not considered in determining the strain εc.

Volume of the shell,

as diametral strain = circumferential strain

Volumetric strain, , putting the value of εc

Note that circumferential stress in thin spherical shell is only PD/4t; but in thin cylindrical shell, circumferential stress is PD/2t, that is, two times, the stress which is developed in a spherical shell, so it is advantageous to use a thin spherical shell, but economically it is costlier to make a thin spherical shell than a thin cylindrical shell. To make a thin spherical shell, two hemispherical portions are made and then joined. To make a hemispherical portion, several segments are made and joined to develop the hemispherical part.

Example 5.2 A thin spherical shell of a wall thickness of 2.5 mm and a diameter of 500 mm is subjected to an internal pressure p. What is the magnitude of p if diametral strain in the shell is limited to 0.0005 only. E = 200 GPa, v = 0.3

Solution

Diameter strain, εD = εc circumferential strain

or

Exercise 5.2 A thin spherical shell of a diameter of 300 mm and wall thickness t is subjected to internal pressure of 2.4 N/mm2. Determine the wall thickness t if the hoop stress in the shell is limited to 60 MPa. What is the change in diameter of the shell, if E = 105 kN/mm2 and v = 0.32.

Cylindrical Shell with Hemispherical Ends

For thin cylindrical shell subjected to internal pressure, we have derived expressions for axial strain, εa= [(PD/4tE)(1 − 2v)] and we have assumed that along the length of the shell, axial strain remains constant, meaning that there is no distortion of end plates. But in actual practice there is distortion of end plates. Moreover, we have learnt that hoop stress developed in spherical shell is only half the hoop stress developed in cylindrical shell for same values of p, D and t. Therefore, it is advantageous to use hemispherical ends for thin cylindrical shell as shown in Fig. 5.7. Figure shows a cylindrical shell ofdiameter D, length L and wall thickness t, with hemispherical ends of diameter D and wall thickness t2. Obviously, the thickness of hemispherical portion will be less than the wall thickness of cylindrical part.

Figure. 5.7

Say, σc′ = hoop stress in cylindrical part

σc″ = hoop stress in hemispherical part

Similarly hoop strain

εc′ = hoop strain in cylindrical part

εc″ = hoop strain hemispherical part


There can be two criteria for design in this case:

1. Hoop stress developed in both portions is the same.

or t2 = 0.5t1 (5.4)

2. Hoop strain in both the portions is the same, so that there is no distortion at the junction of two portions.

Example 5.3 A thin cylindrical steel shell of a diameter of 150 mm and a wall thickness of 5 mm has hemispherical ends. Determine the thickness of hemispherical ends, if there is no distortion of the junction under pressure. E = 208 GPa, v = 0.3

Solution

Thickness of the cylindrical portion, t1 = 5 mm

Poisson’s ratio, v = 0.3

For no distortion of junction under pressure,

or,

= thickness of the hemispherical end.

Exercise 5.3 A thin cylindrical copper shell of a diameter of 125 mm and a wall thickness of 4 mm has hemispherical ends. Determine the thickness of hemispherical ends, if hoop stress developed in both portions is the same. What is the ratio of their hoop strains given v = 0.32?

Wire Winding of Thin Cylindrical Shells

We have learnt that due to internal pressure, hoop stress developed in the cylindrical shell is double the axial stress. Under pressure, there are chances of development of longitudinal crack in the cylinder, if circumferential stress exceeds the safe limit. Moreover, due to high hoop stress, the pressure-bearing capacity of cylindrical shell is reduced. Therefore, to reduce the intensity of hoop stress and to strengthen the cylinder in longitudinal direction, a wire under tension is wound around the circumference of the cylinder, as shown in Fig. 5.8 tensile stress in winding wire puts pressure on cylinder from outside and a negative hoop stress is developed in the cylinder wall.

Figure. 5.8

Consider a thin cylinder of diameter D, length L, and wall thickness t. It is wound with a single layer of wire of diameter dw under an initial tension of σw.


Numbers of turns over the cylinder,

Tensile force exerted by the wire over cylinder

(Note that there are two sections)

Compressive force developed in cylinder = σc 2Lt

For equilibrium

Putting the value of

This is the initial compressive hoop stress developed in cylinder wall due to wire winding.

Figure 5.8 shows, the development of σc, in cylinder

Internal pressure

Now the wire-wound cylinder is subjected to internal pressure p. Due to this say the stresses developed in cylinder are σa′, σc′ and σw′.

Axial bursting force in cylinder,

Axial stress developed,


Dimensional bursting force, PD = pDL

Say, the stresses due to internal pressure developed in cylinder and wire are, respectively, σc′, σw′.

Then,

Putting the value of

or

There are two unknowns σc′ and σw′ but only one equation, so let us consider compatibility condition, taking strain into account.

Circumferential strain in wire = circumferential strain in cylinder.

Let us take Ec and vc as Young’s modulus and Poisson’s ratio for cylinder, respectively, and Ew and vw as Young’s modulus and Poisson’s ratio of wire, respectively.

From Eqs (5.9) and Eqs (5.10), σc′ in cylinder and σw′ in wire can be determined.

Resultant stresses

https://www.safaribooksonline.com/library/view/strength-of-materials/9789332503519/xhtml/chapter005.xhtml#tab-c5e010


In cylinder, σcr = σc + σc′ (tensile)

In wire, σwr = σw + σw′ (tensile)

Example 5.4 A thin bronze cylinder of 250 mm internal diameter and 6 mm thick is wound with a single layer of 1.5-mm-thick steel tape under a tensile stress of 100 N/mm2. Find the maximum internal pressure if the hoop stress in the cylinder is not to exceed 50 N/mm2. Determine also the tensile stress in steel tape. For bronze E = 117 N/mm2, v = 0.33 and for steel E = 208 kN/mm2.

Solution

Let us consider only a unit length of cylinder

σct × 2 × 1 = σt t′ 2

where t = thickness of cylinder = 6 mm

t′ = thickness of tape = 1.5 mm

Say internal pressure is p, then

p · D · 1 = σ′c × 2t × 1 + σ′t × 2t′ × 1

p × 250 = σ′c × 2 × 6 + σ′t × 2 × 1.5

250 p = 12 σ′c + 3σ′t (5.8)

where σc′ and σt′ are the additional stresses in cylinder and tape, respectively due to internal pressure in cylinder.

Axial stress in cylinder,

using strain compatibility

Putting the values

Putting the value σt′ of from Eq. (5.9) into Eq. (5.8)

250p = 12σ′c + 3(1.77 σ′c − 6.11p)

= 12σ′c + 5.333 σ′c − 18.33p

17.333 σ′c = 268.33p

σ′c = 15.480p (5.10)

Resultant stress in cylinder = 50 N/mm2

−σc + σ′c = 50 N/mm2

−25 + σ′c = 50

σ′c = 75 = 15.480p

or internal pressure,

σ′t = 1.77σ′c − 6.11 × p

= 1.777 × 15.480p − 6.11 × p

= (27.5 − 6.11)p = 21.39p



= 21.39 × 4.845 = 103.63 N/mm2

(putting the value of p)

Resultant stress in tape, σtr = σt + σ′t

= 100 + 103.63

= 203.63 N/mm2

Exercise 5.4 A thin cylindrical shell of an internal diameter of 400 mm and a wall thickness of 10 mm is closely wound around its circumference by a 3-mm-diameter steel wire under an initial tension of 10 N/mm2. The cylinder is further subjected to an internal pressure of 2.4 N/mm2. Determine the resultant hoop stress developed in the cylinder and wire. Cylinder is also made of steel. E = 208 Gpa, v = 0.3.

Pressure Vessel with a Double Curved Wall

Sometimes, the cylinder heads of a pressure vessel have double curved wall, that is, a particular element of wall has two radii of curvatures in two planes.

Let us consider a pressure vessel, the form of whose wall is a surface of revolution as shown in Fig. 5.9(a). Thickness of the wall is very small in comparison to principal radii of curvatures r1, r2 as shown in Fig. 5.9(b), so the wall can be treated as a membrane. The stresses are assumed to act on the middle surface of the membrane. An elementary part shown in Fig. 5.9(a) is enlarged in Fig. 5.9(b), element bounded by two circumferential arcs and two meridional arcs. Say





σ1 = stress in circumferential direction

σ2 = stress along the meridional direction

r1 = radius of curvature of circumferential arc

r2 = radius of curvature of meridional arc

δθ1 = angle subtended by circumferential arc

δθ2 = angle subtended by meridional arc

Figure. 5.9 (a) Element of double curved wall (b) Stresses on a small element

Forces acting along edges of element are

σ1ds2t and σ2ds1t (refer to Fig. 5.9(a,b))

Let us resolve these forces along the normal to the element abcd as shown in Fig. 5.10.



Figure. 5.10

But , putting these values

Simplifying the expression

For a thin spherical shell,

For a thin cylindrical shell,

Circumferential stress

Conical Water Tank

Figure 5.11 shows an open conical tank uniformly suspended in brackets around its curved conical surface. Say, H is the depth of water in tank, from bottom. Take a section X–X at a distance of y from the bottom apex point.

Radius of curvature,

Pressure, p = w(H − y)

Using the equation , for double curved surface

or

where w = weight density of water

or


Figure. 5.11

For the hoop stress to be maximum

Meridional stress σ2 can be calculated by considering the total weight of the water (shown by dotted lines) which is equal to (w (H – y)π y2 tan2α + wπy tan2α(1/3)) The component of force due to σ2 in the vertical direction is equal to (σ2 (2πy tan α) t cosα).

So for equilibrium

For meridional stress, σ2 to be maximum

Example 5.5 A conical water storage tank has an included angle of 60° and water is filled up to a vertical depth of 3 m. If the wall thickness is 5 mm, determine the circumferential and meridional stresses at a depth of 2 m from top.

Solution

Water density, w = 10 kN/m3.

Then, α = 30°, semicone angle

t = 5 mm, wall thickness = 0.005 m

y = 3 − 2 = 1 m

H = 3 m

w = 10 kN/m3

Circumferential stress,

tan α = 0.577

cos α = 0.866

= 2.665 N/mm2

Meridional stress,

Putting the values

= 1,554,657.4 N/m2

= 1.55 N/mm2

Exercise 5.5 A conical water storage tank has an included angle of 60° and a vertical depth of 4 m. If the wall thickness is 6 mm, determine the location and magnitude of maximum hoop and maximum meridional stresses.

Water density, w = 10 kN/m3.

Problem 5.1 A stream boiler of an internal diameter of 1.6 m is subjected to an internal pressure of 1.5 N/mm2. What is the tension per linear metre of the longitudinal joint in the boiler shell. Calculate the thickness of the plate if the maximum tensile stress in the plate section is not to exceed 100 N/mm2, taking an efficiency of the longitudinal riveted joint equal to 75 per cent.

Solution

Diameter D = 1,600 mm

Internal pressure, p = 1.5 N/mm2


Length = 1,000 mm (to calculate tension per linear metre)

Thickness = t mm

Area = 1,000t mm2

Tension per linear metre = σc × 1,000t

Plate thickness

Efficiency of longitudinal joint, ηl = 0.75

σc = 100 N/mm2 as given in problem.

Problem 5.2 A closed pressure vessel of a length of 400 mm, a thickness of 5 mm and an internal diameter of 120 mm is subjected to an internal pressure of 8 N/mm2. Determine the normal and shear stresses on an element of the cylindrical wall on a plane at 30° to the longitudinal axis of cylinder (Fig 5.12).


Solution

O–O—longitudinal axis of cylinder = AC

B–A—inclined plane

A–C—reference plane (Fig. 5.12)

Figure. 5.12

Problem 5.3 The ends of a thin cylindrical shell are closed by flat plates. It is subjected to an internal fluid pressure, but the ends of the cylinder are rigidly stayed and no axial movement is permitted. Determine the increase in the volume of the shell. Take Poisson’s ratio as 0.30.

Solution


Due to internal pressure, the cylinder tries to expand longitudinally. However, in this case, the ends are rigidly stayed, which means that the ends exert axial compressive force on cylinder, so that the axial strain in cylinder becomes zero.

Due to internal pressure

Hoop stress

Axial stress

Due to resistance from end plate

Say σa′ is the axial stress on cylinder

Resultant axial stress, σar = σa − σa′

Axial strain,

or

or

or

Volumetric strain, εv = 2εc + εa but εa = 0

Where

But v = 0.3

Volume

δV = change in volume

Problem 5.4 A thin-walled aluminium alloy pressure vessel of a diameter of 200 mm and a wall thickness of 3 mm is subjected to an internal pressure of 4 MPa. Strain gauges that are bonded on the outer surface in the hoop and axial directions give readings of 1,620 and 380 μ strain respectively at full pressure respectively. Determine E and v for the material.

Solution

p = 4 N/mm2

D = 200 mm

t = 3 mm

Circumferential strain, (5.13)

Axial strain, (5.14)

Dividing Eq. (5.13) by Eq. (5.14)

From Eq. (5.13)

Problem 5.5 A 15-cm-long bronze tube with closed ends has a diameter of 75 mm and a wall thickness of 3 mm. With no internal pressure, the tube just fits between the two rigid end walls. Calculate the longitudinal and tangential stresses in the cylinder for an internal pressure of 2.5 N/mm2 (Fig 5.13).

E = 100 kN/mm2, v = 0.32

Solution

Internal pressure





The rigid wall exerts axial pressure σa′ to keep axial strain zero as shown in Fig. 5.13. Net axial stress σa– σa′ = 15.625 – σa′

Axial strain

or σ′a = 15.625 − 10 = 5.625 N/mm2

Tangential stress in cylinder, σc = 31.25 N/mm2

Axial stress, σar = σa − σ′a = 15.625 − 5 625

= 10 N/mm2

Figure. 5.13

Problem 5.6 A gun metal tube of 50-mm bore and a wall thickness of 1.5 mm is closely wound externally by a steel wire 1.0 mm diameter. Determine the tension under which the wire must be wound on the tube if an internal pressure of 1.8 N/mm2 is required before the tube is subjected to tensile stress in the circumferential direction. For gun metal E = 105 GPa and v = 0.34 and for E = 210 GPa.


Solution

Let us first determine the stresses σa′, σc′ and σw′ developed in cylinder and the wire due to the internal pressure.

Take the unit length of cylinder, ℓ = 1 mm

Number of wires per millimetre length of tube,

Axial stress,

Compatibility equation

1.68 (57.325 − 1.91σ′c) = 2σ′c

96.308 = (2 + 3.2088)σ′c = 5.2088σ′c

σ′c = 18.5 N/mm2

σ′w = 57.325 − 1.91σ′c

= 57.325 − 1.91(18.5)

= 57.325 − 35.335

= 21.99 N/mm2

Now, σc – σc′ = 0 (as given in the problem)

σc = σc′ = 18.5N/mm2 (the initial compressive stress in the cylinder in hoop direction)

But

Problem 5.7 A 200-mm-long copper tube closed at its ends has an outer diameter of 90 mm and a thickness of 4 mm. It fits without clearance in a 90-mm hole in a rigid block. The tube is then subjected to internal pressure of 5 N/mm2. Taking v = 0.34 and E = 105 kN/mm2. Determine the final hoop stress in the tube (Fig 5.14).

Solution

The block is rigid so diametral strain due to internal pressure p will be zero.

Do = outer diameter = 90 mm.

Di = inner diameter = 90 − 2 × 4 = 82 mm.

The rigid block will offer radial pressure p′ to prevent diametral strain.


Internal pressure

Stresses

Figure. 5.14

Due to external pressure, p′

Diametral strain

Final hoop stress, σcr = 51.25 − 3.78 × 11.25 = 8.71 N/mm2

Problem 5.8 A thin spherical shell of steel has an internal diameter of 3 m and a wall thickness of 6 mm and is just filled with water at 20°C and at atmospheric pressure. Find the rise in

water pressure if the temperature of the water and the shell rises to 60°C, then determine the volume of water that would escape if small leak is developed at the top of the vessel. For steel E = 200 GPa, αs = 11 × 10−6/°C,v = 0.3 and for water K = 2.2 × 103 N/mm2, α = 0.207 × 10−3/°C (coefficient of volumetric expansion).

Solution

For thin spherical shell

Say, p = pressure developed

Because α for water >>α for steel

Hoop stress in shell,

Volumetric strain in sphere, εvs = volumetric strain in water, εvw

Equating

εvs = εvw

1312.5 p × 10−6 + 1320 × 10−6 = −0.4545 p × 10−3 + 8.20 × 10×3

1.3125 p × 10−3 + 1.32 × 10−3 = −0.4545 p × 10−3 + 8.28 × 10×3

Pressure,

The volume of water which escapes through the leak is simply the difference of free expansion of the water and the vessel, as after leakage there is no pressure remaining in the shell.

Problem 5.9 A brass hoop of an inside diameter of 40 cm and a thickness of 1 cm fits snugly at 180°C over a steel hoop which is 1.5 cm thick. Both the hoops are 5 cm wide. If the temperature drops to 20°C, determine the circumferential stress in each hoop and the radial pressure at common radius. For steelE = 200 GPa, α = 12 × 10−6/°C and for brass E = 100 GPa, α = 20 × 10−6/°C (Fig 5.15).


Figure. 5.15

Solution

Brass hoop initial temperature = 180°C

Final temperature = 20°C

Reduction in temperature = 160°C

As the brass hoop cools down, brass will try to contract, exerting pressure on steel hoop, say the radial pressure developed at common radius of 200 mm is p′ as shown.

Due to p′, compressive hoop stress in steel hoop will be developed.

At the same time steel hoop exerts outer pressure p′ on brass hoop and tensile hoop stress will be developed in brass hoop.

In brass

In steel

Total strain

Problem 5.10 A bronze sleeve of an internal diameter of 200 mm and a thickness of 5 mm is pressed over a steel liner of an external diameter of 200 mm and a thickness of 10 mm with a force fit allowance of 0.06 mm on diameter. Considering both the bronze sleeve and the steel liner as thin cylinder, determine (a) radial pressure at the common radius, (b) hoop stresses in sleeve and liner, and (c) percentage of fit allowance met by the sleeve.

Given for bronze E = 120 kN/mm2, vb = 0.33 and for steel E = 208 kN/mm2, vs = 0.3.

Solution

Figure 5.16 shows the bronze sleeve under an initial pressure p and the sleeve liner under an external pressure p. The internal pressure p is developed due to force fit between the bronze sleeve and the steel liner.

Figure. 5.16


(a) Stresses

Bronze,

Steel,

Radial strain in bronze sleeve

where σcb is tensile and p is compressive

Radial strain in steel liner

Total radial clearance, δR = 0.06mm

Common radius, R = 100 mm

So,

Radial pressure, = 2.78 N/mm2 at common surface

(b) Hoop stresses

σcb = 20 p = 2.78 × 20 = 55.6 N/mm2

σcs = −10 p = −2.78 × 10 = −27.8 N/mm2

(c) Radial clearance

On bronze sleeve

On steel liner

Percentage of fit allowance of sleeve

Problem 5.11 A steel tyre of a thickness of 10 mm, a width of 80 mm and an inside diameter of 1,500 mm is heated and shrunk on to a steel wheel of a diameter of 1,501 mm. If the coefficient of static friction between the tyre and the wheel is 0.3, what twisting moment is required to twist the tyre relative to the wheel? Neglect the deformation of the wheel. E = 200 GPa.

Solution

Width of tyre, b = 80 mm

Thickness of tyre, t = 10 mm

Tyre diameter, D = 1,500 mm

Wheel diameter, D′ = 1,501 mm

Change in the inner diameter of tyre = 1,501 − 1,500 = 1 mm

Circumferential or diametral strain,

E = 2,00,000 N/mm2

Hoop stress in tyre,

where p is junction pressure between the tyre and the wheel

or,

Twisting moment

Total radial force on tyre, R = b × πDp

= 80 × π × 1,500 × 1.77

= 6,67,274 N

Coefficient of friction, μ = 0.3

Force of friction, F = μR = 0.3 × 667,274

= 2,00,182 N

Twisting moment, T = F × radius of wheel

= 2,00,182 × 750

= 1,50,136,500 N mm

= 150.136 kN m


o For a shell, D/t ratio is greater than 20, it is classified as a thin shell.

o For a cylindrical thin shell is subjected to internal pressure, p.

Hoop stress,

Axial stress,

Hoop strain,

Axial strain,

Volumetric strain,

o For a thin spherical shell subjected to internal pressure, p.

Hoop stress,

Diametral strain,

Volumetric strain,

o δV, addition of volume of liquid pumped inside the cylinder δV1 (increase in the volume of cylinder) + δV2 (decrease in the volume of liquid)

o Due to wire winding, initial compressive hoop stress is developed in cylinder. Pressure bearing capacity of cylinder is increased.

o In double curved wall of a shell under pressure

o

o In a cylindrical shell, σ1 = σc, σ2 = σa, r2 = ∝ (infinity)o In a conical water tank, stresses

Where α is semi-cone angle of tank, t is wall thickness, w is specific weight and H is depth of liquid.

Review Questions

1. What is a thin cylindrical/spherical shell? On what D/t ratio it is classified as a thin shell?

2. Derive the expressions for hoop and axial stresses developed in a thin cylindrical shell subjected to internal pressure p.

3. Take a small element of a thin spherical shell and show the stresses acting on this element.

4. Derive the expression for the volumetric strain of a thin spherical shell subjected to internal pressure p.

5. A cylinder of diameter D and wall thickness t is wound with a single layer of wire under tensionσw. Derive the expression for hoop stress developed in cylindrical shell.

6. Derive the following expression for a double curved surface under pressure p.

7. Derive the expression for maximum hoop stress developed in a conical water tank.

8. How the strain gauges mounted on an external surface of a thin shell subjected to an internal pressure can be used to measure E and v of the material of the shell.


1. A thin cylindrical shell with D/t = 30 is subjected to an internal pressure of 3 N/mm2. What is the hoop stress developed in shell?

1. 90 MPa2. 45 MPa3. 22.5 MPa4. None of these

2. A thin spherical shell of an inner diameter of 400 mm is subjected to an internal pressure of 2.5 N/mm2. If the hoop stress is not to exceed 100 MPa, what is the thickness of shell?

1. 2.5 mm2. 5 mm3. 10 mm4. None of these

3. A thin cylindrical shell is made of steel with v = 0.3. It is subjected to an internal pressure p. What is the ratio of hoop strain to axial strain?

1. 1.02. 2.03. 3.0

4. 4.254. A thin spherical shell is subjected to an internal pressure p.

What is the ratio of volumetric strain to circumferential strain in shell?

1. 1.02. 2.03. 3.04. None of these

5. A cylindrical tank of an inside diameter of 1 m and a height of 20 m is filled with water of a specific weight of 10 kN/m3. If the thickness of tank is 25 mm, then the maximum stress developed in wall of the tank is

1. 4 N/mm22. 2 N/mm23. 1 N/mm24. None of these

6. A thin cylindrical shell is made of steel with v = 0.30. It is subjected to an internal pressure. What is the ratio of volumetric strain to circumferential strain?

1. 2.02. 2.2353. 9.54. None of these

7. A closed pressure vessel of a length of 400 mm, a wall thickness of 5 mm and an internal diameter of 100 mm is subjected to an internal pressure of 8 N/mm2. The normal stress on an element of the cylinder on a plane 30° to the longitudinal axis will be

1. 140 MPa2. 70 MPa3. 77.32 MPa4. None of these

8. A steam boiler of an internal diameter of 1.5 m is subjected to an internal pressure of 2 N/mm2. If the efficiency of the longitudinal joint is 80 per cent and the maximum tensile stress in the plate section is not to exceed 125 MPa, what is the thickness of the plate?


Practice Problems

1. A steam boiler of an internal diameter of 1.5 m is subjected to an internal pressure of 1.2 N/mm2. What is the tension per linear metre of the longitudinal joint in the boiler shell? Calculate the thickness of the plate if the maximum tensile stress in the plate section is not to exceed 90 N/mm2, taking efficiency of longitudinal joint as 70 per cent.

2. A single strain gauge making an angle of 15° with the horizontal plane is used to determine the gauge pressure in a cylindrical tank with its axis vertical as shown in Fig. 5.17. The tank is 6 mm in thickness and 500 mm in diameter. It is made of steel with E = 200 GPa and v = 0.29. The strain gauge reading is 350 m strain. Determine the pressure inside the tank.

3. The ends of a thin cylindrical shell are closed by flat plates. It is subjected to an internal fluid pressure of 3 N/mm2, but the ends of the cylinder are rigidly stayed and no axial movement is permitted. The diameter of cylinder is 200 mm and its length is 800 mm, while the wall thickness is 5 mm. Determine the change in the volume of cylinder, if E = 200 GPa and v = 0.3.


Figure. 5.17

4. A thin-walled copper alloy pressure vessel of a diameter of 250 mm and a wall thickness of 5 mm is subjected to an internal pressure of 5 MPa. The strain gauges that are bonded on the surface in the hoop and axial directions give readings of 988 and 191 μs at full pressure, respectively. Determine Eand n for the material.

5. A 20-cm-long copper tube with closed ends is 100 mm in diameter and has a wall thickness of 4 mm with no internal pressure. The tube just fits between the two rigid end walls. Calculate the axial and hoop stresses in the cylinder for an internal pressure of 3 N/mm2. E = 105 kN/mm2 and n = 0.33.

6. A 50-mm bore gun metal tube of a wall thickness of 1.25 mm is closely wound externally by a steel wire of a diameter of 0.5 mm. Determine the tension under which the wire must be wound on the tube if an internal pressure of 1.5 N/mm2 is required before the tube is subjected to a tensile stress in the circumferential direction. For gun metal E = 102 GPa and v = 0.35 and for steel E = 210 GPa

7. A 150-mm-long bronze tube closed at its ends is 80 mm in outer diameter and has a thickness of 3 mm. It fits without clearance in an 80-mm hole in a rigid block. The tube is then subjected to an internal pressure of 4 N/mm2. Take v = 1/3 and E = 83 kN/mm2, determine the tangential stress in the tube.

8. A thin spherical shell of copper has an internal diameter of 2.5 m and a wall thickness of 5 mm and is just filled with water at 25°C and at atmospheric pressure. If the temperature of water and shell rises to 60°C, determine the volume of water that would escape if a small leak is developed at the top of the vessel. For copper E = 105 GPa, αcu = 18 × 10−6/°C and v = 0.34 and for water K = 2,200 N/mm2, α = 0.207 × 10−3/°C (for volumetric expansion).

9. A steel tube of an internal diameter of 150 mm and a wall thickness of 8 mm in a chemical plant is lined internally with a well-fitting copper sleeve of a wall thickness of 2 mm. If the composite tube is initially unstressed, calculate the hoop stresses set up assumed to be uniform throughout the wall thickness, in a unit length of each part of the tube due to increase in temperature of 100°C. Neglect the temperature effect in axial direction.

[Hint: compressive stress in copper, αc > tensile stress in steel, αs]

For steel, E = 208 GPa, α = 11 × 10−6/′C

For copper, E = 104 GPa, α = 18 × 10−6/′C

10. A copper sleeve of an internal diameter of 150 mm and a thickness of 2.5 mm is pressed over a steel liner of an external diameter of 150 mm and a thickness of 2.5 mm with a force fit allowance of 0.04 mm on diameter. Considering both the copper sleeve and the steel liner as thin cylinders, determine (a) radial

pressure at common radius, (b) hoop stresses in sleeve and liner (c) percentage of fit allowance met by the sleeve.

For copper, E = 105 kN/mm2, vc = 0.34For steel, E = 210 kN/mm2, vs = 0.3

11. A thin tyre is to be shrunk onto a rigid wheel of a diameter of 1 m. Determine the necessary internal diameter of the tyre if the maximum hoop stress in the tyre is 10 N/mm2. Also determine the least temperature to which the tyre must be heated above that of wheel before it could be slipped on.

α for tyre = 11 × 10−6/°C

E = 204 GPa

Special Problems

1. A closed cylindrical vessel made of steel plate of a thickness of 5 mm with plane ends carries fluid under a pressure of 4 N/mm2. The diameter of the cylinder is 250 mm and its length is 750 mm. Calculate the longitudinal and hoop stresses developed in the cylinder. What are the changes in the diameter, length and volume of the cylinder?

E = 210 GPa, v = 0.3

2. One method of determining Poisson’s ratio for a material is to subject a cylinder to internal pressure and to measure axial strain εa and hoop strain εc on the outer surface of cylinder show that

3. A cylindrical tank of an inside diameter of 2 m and a height of 20 m is filled with water of a specific weight of 10,000 N/mm3. The material of the tank is structural steel with yield strength of 250 N/mm2. What is the minimum thickness required at the bottom of the steel tank if the efficiency of the longitudinal seams is 80 per cent. Take factor of safety (FOS) as 4.

4. A thin spherical shell made of copper alloy is 300 mm in diameter and 1.5 mm in wall thickness. It is full of water at an atmospheric pressure. Find by how much the internal pressure will increase if 20 cc of water is pumped inside the shell. E = 100 GPa, v = 0.29

For water, K = 2,200 N/mm2

5. A pressurized steel cylinder tank has an inner radius of 600 mm and a thickness of 16 mm. The tank is subjected to a pressure p = 1,750 kPa and an axial force P = 125 kN. The butt weld seam forms an angle of 54° with the longitudinal axis of the tank. Determine (a) the normal stress perpendicular to weld and (b) the in-plane shear stress parallel to weld.

, net stress in axial directionσa = 54° with the plane of σc

6. A spherical steel vessel is made of two hemispherical portions fitted together at flanges. The inner diameter of the sphere is 600 mm and the wall thickness is 6 mm. Assuming that the vessel is a homogeneous sphere, what is the maximum working pressure for an allowable tensile stress in a shell of 150 MPa. If 20 bolts of a diameter of 16 mm are used to hold flanges together, what is the tensile stress in bolts when the sphere is under full pressure?

7. For a hydraulic test, a steel tube of an internal diameter of 80 mm, a wall thickness of 2 mm and a length of 1.2 m is fitted with end plugs and filled with oil at a pressure of 2 MPa. Determine the volume of oil leakage which would cause the pressure to fall to 1.5 MPa.

For oil, K for oil = 2.8 GN/m2,For steel, E = 208 GPa, v for steel = 0.29


Exercise 5.1: 48.51 cc

Exercise 5.2: 3mm, ec = 0.3888 × 10−3, dD = 0.1166 mm

Exercise 5.3: t2 = 2 mm, 1.235

Exercise 5.4: σcr = +37.864 N/mm2

σwr = 43.02 N/mm2

Exercise 5.5:


1. (b) 2. (a) 3. (d) 4. (c) 5. (a)

6. (b) 7. (b) 8. (c)

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1. 400 kN/linear metre, t = 14.3 mm2. p = 2.07 N/mm2

3. 14.17 cc4. E = 105 kN/mm2, n = 0.345. 37.5 N/mm2, 12.335 N/mm2

6. 64.52 N/mm2

7. 8.72 N/mm2

8. 43,810 cm3

9. σcs = 16.18 N/mm2; (tensile)= 57.325 – 1.91(18.5)

10. (a) 1.24 N/mm2; (b) 37.2 N/mm2 (tensile), 37.2 N/mm2 (compressive); (c) 67.14%

11. 44.56°C, 999.51 mm


1. 50 MPa, 100 MPa, 0.101 mm, 0.0714 mm, 33.3 × 103 mm3

2. 4 mm3. 0.93 MPa4. 42.46 MPa, 14.51 MPa5. p = 12 N/mm2; 211 Mpa6. δv = 1.63 ccd

5b. Thick ShellsCHAPTER OBJECTIVES

For low pressures, thin pressure vessels are used where D/t ratio is greater than 20. For high internal pressure as in the case of tubes used for high pressure gauge and bulk compression gauge, tubes with considerable thickness are used where D/t << 20. The objectives of this chapter are to make the students learn about:

o Derivation of Lame’s equations for radial and circumferential stresses.o Variation of circumferential and radial stresses along the thickness of the shell.o There is large variation of hoop stress along the thickness of the shell. The maximum hoop stress cannot be more than the allowable stress; therefore, two cylinders are compounded.o How compressive hoop stress in inner cylinder and tensile hoop stress in outer cylinder are developed in a compound cylinder.o Calculation of proper shrinkage/force fit allowance between two compounded cylinders.o Calculation of hoop and radial stresses in a hub and shaft assembly.o Derivation of radial and hoop stresses in a thick spherical shell.

Introduction

In a thin pressure vessel, the hoop and axial stresses are much greater than the intensity of fluid pressure and in many applications such as tube (for high-pressure gauge) it is subjected to high internal pressure. The tube section can be designed so that the ratio of hoop stress/internal pressure is not large.

In this chapter we will deal with Lame’s equations for hoop and axial stresses in terms of internal pressure. To increase the pressure-bearing capacity of a thick cylinder, it is compounded with another cylinder which introduces external junction pressure on inner cylinder.

We will study about stresses developed due to shrinkage and shrinkage allowance in compound cylinders.

Stresses developed in thick spherical shell will also be briefly discussed.

Lame’s Equations

Stresses in a thick cylinder are determined on the assumption that plane sections (perpendicular to the axes) of the cylinder remain plane after the application of internal pressure. In other words, axial strain in cylinder remains constant along its lengths and there is no distortion of end plates.

Figure 6.1 shows a section of a thick cylinder of which thickness t = R2 − R1 is considerable. It is subjected to internal pressure p such that axial and circumstantial stresses develop in the cylinder. Radial stress σr varies from p (compressive) at radius R1 to zero stress at radius R2, a nonlinear variation ofσr. Consider a thin circular cylinder of inner radius r and outer radius r + dr, that is, radial thickness drsubjected to radial stress σr at inner radius r and σr + dσr at outer radius r + dr. Consider the equilibrium of half of the cylinder, that is, XAX as shown in Fig. 6.1, σc is the circumstantial stress (taking force equal to stress × projected area with cylinder length equal to unity).



2(σr + dσr) (r + dr) + 2σc dr = 2rσr

2σcdr + 2rdσr + 2σrdr + 2dσrdr = 0

But, 2dσrdr is a negligible quantity,

So, σcdr + σrdr + rdσr = 0

or,

Say E and v are elastic constants

Axial stress developed in cylinder,

remains constant.

Figure 6.1 Radial and hoop stresses on thin cylinder

Figure 6.2 Stresses on a small element

Stresses on a small element of thick cylinder are shown in Fig. 6.2, that is, σr, σc and σa in radial, circumstantial and axial directions. Note that σr is compressive stress, while σa and σc are tensile stresses.

Axial strain,

But stress σa is constant axial stress and E and v are elastic constants; therefore,

σc − σr = constant = 2A (say)

From Eqs (6.3) and (6.4)

σc − σr = 2A (6.4)

From Eqs (6.1)


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Subtracting Eq. (6.4) from Eq. (6.5)

or

or

or ln (A + σr) = −2 ln r + ln B, where ln B is another constant

or

or

Radial stress,

From Eq. (6.4)


Boundary conditions [Refer to Fig. 6.1(a)]

Radial stress at





Radial stress at

or

putting the value of A in Eq. (6.9)

or constant

constant

putting the values of A and B in equation for σc we get,

at


Figure 6.3 Variation of radial and circumferential stresses

at

Figure 6.3 shows the variation of radial and circumstantial stresses in a thick cylinder subjected to internal pressure p.

Example 6.1 A thick cylinder of inner radius 150 mm and outer radius 210 mm is subjected to internal premiere p such that the maximum hoop stress developed in cylinder is 154.16 N/mm2. Draw the hoop stress and radial stress distribution along the thickness of cylinder. If E = 200 GPa, what is circumstantial strain in cylinder at the outer surface? Given v = 0.3.

Solution

Internal radius, R1 = 150 mm



Internal pressure = p

Lame’s constants

Radial stress σr

At r = R1, σr = p = +50 N/mm2 (compressive)

r = 180 mm,

= 70.891 − 52.083

= 18.80 N/mm2

r = 210 mm, σr = 0

Hoop stress

at r = R1, σcmax = 154.16 N/mm2 (given)

r = 180 mm,

= 70.891 + 52.083

= 122.974 N/mm2

At r = 210 mm,

= 52.083 + 52.083

= 104.166 N/mm2

Figure 6.4 shows the distribution of σc and σr stresses along the thickness (R2 − R1) of cylinder.

Axial stress

Figure 6.4


At the outer surface of the cylinder, there are only two principal stresses, that is, σcmin and σa, and the radial stress is zero.

Circumstantial strain at outer surface of cylinder

Exercise 6.1 A cylindrical shell of inner radius 60 mm and outer radius 100 mm is subjected to internal fluid pressure of 64 N/mm2. Draw the distribution of σc and σr along the thickness of cylinder. If E = 105 kN/mm2 and v = 0.34, what are the axial and circumstantial strains at the outer surface of cylinder?

Thick Cylindrical Shell Subjected to External Pressure

In the previous chapter, we have studied about the thick cylinder subjected to an internal pressure which causes tensile hoop and axial stresses in the cylinder. Yet, in many applications, a cylinder may be subjected to an external pressure as in the case of the submarine tank subjected to external pressure and the inner cylinder of a compounded cylinder subjected to external pressure which causes compressive hoop and axial stresses in the cylinder.

Figure 6.5 shows a thick cylinder of inner radius R1, outer radius R2 and length L subjected to an external pressure p.


σr = p at r = R2

σr = 0 = at r R1

Using Lame’s equations,

or, A = B/R12, putting the value in Eq.(6.12)

Figure 6.5 Thick cylinder subjected to external pressure

or constant,


Radial stress,

at r = R1, σr = 0

at

Hoop stress

at

Note that 2R22 > (R2

2 + R12)

Therefore, the maximum hoop stress occurs at inner radius.

Axial stress,

as the pressure acts on the outer surface.

Stresses on an element of a cylinder subjected to pressure p are shown in Fig. 6.6. Note that all the stresses σc, σr and σa are compressive. The maximum hoop stress (though compressive in this case) occurs at inner radius.


Figure 6.6

Example 6.2 A thick cylinder of inner radius 85 mm and outer radius 120 mm is subjected to an external pressure of 50 N/mm2. Determine σcmax and σcmin and show variation of σc and σr along the thickness of the cylinder.

Solution

Figure 6.7

σr = p, at outer radiusσr = 0, at inner radius

Figure 6.7 shows the variation of σc and σr in the cylinder due to external pressure.

Exercise 6.2 A thick cylinder with an external diameter of 240 mm and an internal diameter of Dis subjected to an external pressure of 50 MPa. Determine the diameter D of the maximum hoop stress in the cylinder that is not to exceed 200 MPa.

Compound Cylinder

To increase the pressure-bearing capacity of a cylinder and to decrease the variation in hoop stress developed in cylinder, another cylinder is force fitted or shrink fitted over the cylinder. This process is called compounding of cylinders. Therefore, for compounding, the inner diameter of the outer cylinder is slightly less than the outer diameter of the inner cylinder so as to provide the force fit allowance or shrinkage allowance.

In shrinkage fitting, the outer cylinder is heated such that its inner radius R3″ expands to R3′ and then the outer cylinder is slipped over the inner cylinder. When the outer cylinder cools, it exerts a radial pressure p′ on the outer surface of the inner cylinder. In reaction, the inner cylinder exerts a pressure p′ on the inner surface of the outer cylinder as shown in Fig. 6.8(c). The outer pressure p′ on the inner cylinder develops compressive hoop stresses in the inner cylinder. Similarly the inner pressure p′ on the outer cylinder develops tensile hoop stresses in the outer cylinder.



The final junction radius between two cylinders is R3 such that R3 < R3′ but R3 > R3″

The difference R3″ ‒ R3′ is termed as shrinkage allowance.

Stresses due to junction pressure

Inner cylinder

(expression derived earlier in case of cylinder with external pressure)

Outer cylinder

Figure 6.8

Figure 6.9 shows the variation of hoop stresses developed in the inner and outer cylinders due to junction pressure.

Figure 6.9

Compound cylinder

Say the compound cylinder is now subjected to an internal pressure p.

Say A and B are Lame’s constants, thus

or

constant,


Stresses at R1, R2 and R3 due to internal pressure

Resultant stresses

Inner cylinder

Outer cylinder

Figure 6.10 shows the variation of resultant hoop stresses developed due to junction pressure p′ and internal pressure p on the inner and outer cylinders.

Figure 6.10 Resultant hoop stress in the inner and outer cylinders of a compound cylinder

Example 6.3 A compound cylinder is made by shrinking one cylinder over another such that the outer diameter is 200 mm, the inner diameter is 100 mm and the junction diameter is 150 mm. If the junction pressure developed between two cylinders is 10 N/mm2 and the internal pressure is 50 N/mm2, what are the hoop stresses at inner and outer radii of both the cylinders?

Solution

Radii R1 = 50 mm

R2 = 100 mm


R3 = 75 mm

Pressure p′ = 10 N/mm2

p = 50 N/mm2

Resultant hoop stresses

Inner cylinder

Putting the values

Outer cylinder

Exercise 6.3 A steel cylinder of an outer diameter of 180 mm is shrunk on another cylinder of an inner diameter of 120 mm. The common diameter is 150 mm. If after shrinkage the radial pressure at the common surface is 6 N/mm2, determine the magnitude of internal pressure to which the compound cylinder can be subjected so that the maximum hoop tension in the inner and outer cylinders is equal.

Then,

Shrinkage Allowance

In a compound cylinder at the junction radius due to junction pressure, tensile hoop stress in the outer cylinder and compressive hoop stress in the inner cylinder are developed. At common radius R3

Inner cylinder

Junction pressure, p′ (compressive)

Radial or circumferential strain,

When vi and Ei are the Poisson’s ratio and the elastic modulus of the inner cylinder, respectively

Outer cylinder

Junctions pressure, p′ (compressive)

where v0 and E0 are the Poisson’s ratio and the elastic modulus of the outer cylinder.

Shrinkage allowance (R′3 − R′′3 from Eqs (16) and (18)

If both the cylinders are of different materials, then the shrinkage allowance is calculated from Eq. (6.19).

If Ei = E0 = E,vi = v0 = v, that is, both the cylinders are of same material.

The shrinkage allowance on common radius R3 is

where R1, R3 and R2 are inner radius, junction radius and outer radius of compound cylinder.




Remember that (numerical sum of hoop stresses developed at the common radius of two cylinders).

Example 6.4 A compound cylinder is formed by shrinking one outer steel cylinder over bronze cylinder. The final dimensions are internal diameter 100 mm, external diameter 200 mm and junction diameter 160 mm, and the shrinkage pressure at the common surface is 12 N/mm2. Calculate the necessary difference in radii of two cylinders at the common surface.

For steel, E0 = 200 GPa, ni = 0.3

For bronze, Ei = 100 GPa, n0 = 0.32

What is the maximum temperature through which the outer cylinder should be heated before it can be slipped on?

For steel α = 11 × 10−6/°C

Solution

Junction pressure, p′ = 12 N/mm2

R1 = 50 mm, R2 = 100 mm, R3 = 80 mm

Putting the values

The outer cylinder should be heated by 47.84°C so that it can be slipped over inner cylinder.

Exercise 6.4 A steel tube of an outside diameter of 220 mm is shrunk on another steel tube of an inside diameter of 140 mm. The diameter at junction is 180 mm after shrinking on. The shrinkage allowance provided on the radius of the inner tube is 0.08 mm. Determine (a) the junction pressure, (b) the hoop stresses at the inner and outer radii of the inner tube and (c) the hoop stress at the inner and outer radii of the outer tube if E = 210 kN/mm2.

Hub and Shaft Assembly

In a power transmission system, the shaft is connected to the hub with the help of a key inserted in keyways cut on the shaft and in the hub. These keys severely reduce the strength of shaft and hub due to the keyways having sharp corners. To avoid keyways, the hub can be shrunk fitted on the shaft as in the case of compound cylinders. Due to shrink fitting the junction pressure p′ is developed between the shaft and the hub. Let as consider a shaft of radius R1 fitted in a hub of outer radius R2 and inner radiusR1 as shown in Fig. 6.11.


Figure 6.11

Shaft

Let as take Lame’s constants A and B for shaft.

Radial stress,


But material exists at centre, that is, at r = 0, and the stress cannot be infinite; therefore, constant B = 0.

In shaft σr = − A = p′ (junction pressure) or A = −p′

σc = + A = − p′

While deriving Lame’s equations, we have considered σr as negative and σc as positive. So now σc and σrboth are compressive and equal to junction pressure p′.

Radial strain in shaft at radius R1

negative strian

Hub

Due to junction pressure p′, the hoop stress developed

Radial strain in hub, positive strain

Because p′ is compressive

Shrinkage allowance at radius R1

δR1 = ɛcsR1 + ɛchR1

If both shaft and hub are of same material such that Es = Eh = E; Vs = Vh = V then

Shrinkage allowance on radius

Example 6.5 A steel shaft of a diameter of 100 mm is driven into a bronze hub. The driving allowance provided is on diameter 1/1,000 of the shaft diameter. Determine the thickness of the hub, if the maximum bursting stress in the hub is limited to 80 N/mm2.

Given

Es = 200 GPa, Vs = 0.3

EB = 100GPa, VB = 0.34

Solution

Say junction pressure is p′

Hub

Shaft

But total strain = εch + εcs

As given in the problem

putting the values of ES and EB

Multiplying throughout by 1,00,000, we get

80 + 0.34 p‴ + 0.5 p ′ − 0.15 p′ = 100

0.69 p′ = 20

Junction pressure,

Outer radius of hub

Thickness of hub = R2 − R1 = 73.08 − 50 = 23.08 mm

Exercise 6.5 A bronze liner of an outside diameter of 60 mm and an inside diameter of 39.94 mm is forced over a steel shaft of 40 mm in diameter. Determine (a) the radial pressure between the shaft and the liner and (b) the maximum hoop stress in the liner.

For steel, E = 208 GPa, v = 0.29

For steel, E = 125 GPa, v = 0.33

Thick Spherical Shell

To determine radial and hoop stresses in a thick spherical shell subjected to an internal pressure, let us first learn about the radial and circumferential strains in such axi-symmetric cases.

Consider a thin disc of inner radius R1 and outer radius R2 subjected to an internal pressure p. A small element abcd is subtending an angle δθ at the centre is considered as shown in Fig. 6.12.

Figure 6.12

Radius oa = r

Radius od = r + δr before the application of internal pressure.

After the application of internal pressure, say

r → changes to r + u

δr → changes to δr + δu


Circumferential strain in element,

Radial strain in element,

These strains are tensile if positive.

On the element of a spherical shell, say stresses are σr + δσr at radius r + δr

Circumferential stress on element = σc

Bursting force P on the elementary shell (taking the projected area of hemispherical portion)

πr2σr − π (r + δr)2 (σr + δσr)

Resisting force = σc × 2 πrδr , where δr is the radial thickness.

For equilibrium

πr2σr − π (r + δr)2 (σr + δσr) = σc × 2πrδr

2σrδr − rδσr = 2σcδr (neglecting the small quantities of higher order terms)

Three principal stresses at any point in the elementary shell are:

1. radial pressure, σr (compressive),2. hoop stress, σc (tensile), and3. hoop stress, σc on another plane (at right angle to the first

plane).

Radial strain at any point

or

Circumferential strain at the point

From initial Eqs. (22) and (23)

Substituting for εr and εc from Eqs. (25) and (26)

After simplification we get

From Eq. (6.24)






Differentiating Eq. (6.29) we get

Substituting for σc and dσc/dr in Eq. (6.24) and after multiplication we get

Let us put dσr/dr = k, then we have

So ln k = −4 ln r + ln C1

where C1 is a constant of integration

or

Integrating Eq. (6.30),

σr = , where C2 is another constant of integration.

Now σc = (from Eq. 6.29)

or σc = (using Eq. 6.30)






σc =

Let us put constants C1 = 6B and C2 = ‒A

Then,

Radial stress, (compressive)

Hoop stress, (tensile)

For a thick spherical shell subjected to an internal pressure p, the boundary conditions are

Constants,

Example 6.6 Calculate the thickness of a spherical shell of an inside diameter of 120 mm to withstand an internal pressure of 50 MPa, if the maximum permissible tensile stress in the shell is 120 MPa.

Solution

Putting the values

or 120 (R23 − 1203) = 25R2

3 + 50 × 603

95 R23 = 206.905 × 104

= 2 06905 × 106

R2 = 127.42mm

Thickness of shell = 127.42 − 60 = 67.42 mm

Exercise 6.6 Determine the maximum shear stress at the inner surface of a spherical shell, having a diameter ratio of 1.5 for an internal pressure of 7 N/mm2.

Problem 6.1 Two thick cylinders are of the same dimensions. The external diameter is 1.5 times the internal diameter. Cylinder A is subjected to an internal pressure p1, while cylinder B is subjected to an external pressure p2. Find the ratio of p1/p2 if the greatest hoop strain in both cylinders is the same.

Poisson’s ratio = 0.3

Solution

Cylinder A

Inner radius = R1

Outer radius, R2 = 1.5 R1

Internal pressure = P1

At inner radius,

Axial stress,

σr = p1 (compressive)

Hoop strain,

Cylinder B

External pressure = p2

At inner radius,

Axial stress,

Hoop strain,

However, ɛc′ = ɛc″

Problem 6.2 A steel cylinder of an inside diameter of 800 mm and a length of 6 m is subjected to an internal pressure of 40 MPa. Determine the thickness of the cylinder if the maximum shear stress of cylinder is not to exceed 65 MPa. What is the increase in the volume of the cylinder? E = 200 GPa, v = 0.3.

Solution


p = 40 MPa

τmax = 65MPa

where

So

or

Thickness of cylinder, t = R2 − R1 = 645 − 400 = 245 mm

At the inner radius

At the inner surface

σc = 90MPa

σa = 25MPa

p = −40MPa

E = 200,000 N/mm2, v = 0 3

Strains

Volumetric strain,

Volume of cylinder,

Change in volume,

Problem 6.3 Strain gauge are mounted on the outer surface of a thick cylinder with a diameter ratio of 2.5. The cylinder is subjected to an internal pressure of 150 MPa. The recorded strains are

1. longitudinal strain = 60 × 10‒6 and2. circumferential strains = 241 × 10‒6

Determine E and v of the material.

Solution

Internal radius = R1

Outer radius = R2

Outer surface

Strains

or 57.14 − v × 28057 = 57 = 241 × 10−6 × E (6.31)

28.57 − v × 57.14 = 60 × 10−6E (6.32)

Dividing Eq. (6.31) by Eq. (6.32)

Poisson’s ratio,

Putting the value of v is Eq. (6.31)




Young’s modulus,

= 203,000 N/mm2

= 203 kN/mm2

Problem 6.4 A thick cylinder is subjected to both internal and external pressures. The internal diameter of the cylinder is 120 mm and the external diameter is 200 mm. If the maximum permissible stress in the cylinder is 25 N/mm2 and the external radial pressure is 5N/mm2, determine the intensity of internal fluid pressure (Fig. 6.13).

Figure 6.13

Solution

R1 = 60mmR2 = 100mm

Radial stress


σr = 5 N/mm2

at r = R2

σr = p (say) at r = R1 as shown in Fig. 6.13

Using Lame’s equation,

From Eqs. (6.33) and (6.34)

from Eq. (6.33)

= (p − 5) × 0.5625 −5 (6.36)

Circumferential stress

Putting the value of σcmax, 25 = (p − 5) (2.125) − 5





Internal pressure = 19.117 N/mm2

Problem 6.5 A compound cylinder is formed by shrinking one cylinder over another. The outer diameter of the compound cylinder is 240 mm, the inner diameter is 100 mm and the diameter at the common surface is 170 mm. The junction pressure due to shrinkage is 5 N/mm2. If the compound cylinder is now subjected to an internal fluid pressure of 60 N/mm2, determine the maximum hoop tension in the cylinder. How much heavier a single cylinder of an internal diameter of 100 mm would be if it is subjected to the same internal pressure in order to withstand the same maximum hoop tension?

Solution

R1 = 50 mm, R3 = 85mm, R2 = 120 mmp = 60 N/mm2, p′ = 5 N/mm2,

Inner cylinder

Resultant at R1,

Outer cylinder

Resultant at R3,

At inner radius of inner cylinder, σc is maximum.

Single cylinder

p = 60 N/mm2

R1 = 50 mm

R′2 = ?

σcmax = 69.91

69.91

1.165

1.165 R′22 − 1.165 × 502 = R22 + 502

0.165 R22 = 5412.5

R′2 = 181.11 mm

Single cylinder, A′ π (R′22 − 502) = π (181.112 − 502)

= π × 30303.5

Compound cylinder, A = π (1202 − 502) = π (11, 900)

Problem 6.6 A compound cylinder has a bore of 120 mm, an outer diameter of 240 mm and the diameter at common surface is 180 mm. Determine the radial pressure at the common surface which must be provided by shrinkage fitting if the resultant maximum hoop stress in the inner cylinder under the superimposed internal pressure of 60 N/mm2 is to be 60 per cent of the value if the maximum hoop tension at the inner radius of overall cylinder subjected to an internal pressure of 50 N/mm2 (Fig. 6.14).

Figure 6.14

Solution



Junction radius, R3 = 90 mm


Internal pressure, p = 60 N/mm2

Considering overall cylinder

Compound cylinder

Inner cylinder subjected to junction pressure p′ and internal pressure p, with σcmax equal to 100 × 0.6 = 60 N/mm2 as given in the problem. This stress occurs at inner radius R1 (Fig. 6.14).

σr = 50at r R1 = 60 mmσr = p ′ at r = R3 = 90 mm

Moreover

at r = R1, σc = 60 N/mm2 as given

From Eqs. (6.37) and (39)




2A = 60 − 50 = 10A = 5 N/mm2

From Eq. (6.37)

Problem 6.7 A thick cylinder of an internal diameter of 100 mm is subjected to an internal pressure of 20 N/mm2. If the allowable stress for cylinder is 110 N/mm2, determine the wall thickness of the cylinder. The cylinder is now strengthened by wire winding so that it can be safely subjected to an internal pressure of 25 N/mm2. Find the radial pressure exerted by wire winding.

Solution

Radius R1 = 50mm

R2 = to be calculated

Internal pressure, P = 20 N/mm2

Allowable maximum stress,

or 5.5 (R22 − 502) = R2

2 + 502

4.5 R22 = 6.5 × 502


R2 = 1.2018 × 50 = 60.1 mm

Internal pressure

σ″c, reduction in σ′cmax, due to wire winding.

= 137.5 − 110= 27.5 N/ mm2

Pressure exerted by wire winding, p′ = 4.233 N/mm2.

Problem 6.8 A compound cylinder consists of a steel cylinder of an internal diameter of 120 mm and an external diameter of 180 mm and a bronze liner of a thickness of 10 mm with an external diameter of 120 mm as shown in Fig. 6.15. Assuming liner to be a thin cylinder and that there is no stress in the compound cylinder due to shrink fitting, determine the maximum hoop stress in the liner due to an internal pressure of 40 N/mm2. Ignore the longitudinal stress and strain.

Solution

For compound cylinder

R1 = 50 mm


R2 = 90 mmp = 40N/mm2

Figure 6.15

Liner

Problem 6.9 A steel cylinder has an internal diameter of 100 mm and an external diameter of 200 mm. Another cylinder of same steel is to be shrunk over the first cylinder so that shrinkage stresses just produce a condition of yield at the inner surface of each cylinder. Determine the necessary difference in diameters of the cylinder at the mating surface before shrinking and the required external diameter of the outer cylinder, assuming that yielding occurs according to the maximum shear stress criteria’s and that no axial stresses are set up due to shrinking. Yield stress in simple tension or compression = 270 N/mm2 and E = 200 GPa.

Solution

In a simple tensile test on a bar, at yield point stress σyp, the maximum shear stress occurs at ±45° to the axis of the bar and it is equal to 0.5 σyp. This can be obtained by drawing a Mohr’s stress circle.

Maximum shear stress in simple tension or compression

Say, junction pressure is p′

R1 = inner radius of liner cylinder = 50mm

R3 = junction radius = 100mm

R2 = outer radius of outer cylinder = ?

Inner cylinder

σcmax occurs at inner radius due to p′ at outer radius

σr = 0, at inner radius of inner cylinder.

So, 1.333 p′ = 135

p ′ 101.25 N/mm2

Outer cylinder

Junction pressure or σr = p′ (compression)

or R22 + 1002 = 1.667 R2

2 − 1.667 × 1002

0.667 R22 = 2.667 × 1002

R22 = 2.667 × 1002

Shrinkage allowance

Shrinkage allowance on diameter, δD3 = 2δR3 = 0.337 mm

Problem 6.10 A bronze sleeve is pressed onto a steel shaft of 60 mm in diameter. The radial pressure between steel shaft and bronze sleeve is 14 N/mm2 and the hoop stress at the inner surface of the sleeve is 50 N/mm2. If an axial compressive force of 60 kN is now applied to the shaft, determine the change in radial pressure.

For steel, E = 200 Gpa, v = 0.3

For bronze, E = 100 Gpa, v = 0.33

Solution

Bronze sleeve

Junction pressure, p′ = 14 N/mm2

Hoop stress at inner surface = +50 N/mm2

Diameter of shaft = 60mm

Radius, R1 = 30mm

When compressive force is applied onto the shaft, axial compressive stress will be developed in the shaft, which will cause positive radial strains (lateral strain in shaft) putting

internal pressure p″ on sleeve. In reaction, sleeve will exert p″ radial pressure on the shaft.

Additional junction pressure p″

Additional hoop stress in sleeve,

Additional circumferential strain in sleeve

Shaft

Axial compressive stress,

Additional stresses in shaft are

p″, radial compressive

p″, circumferential compressive

σα axial compressive

Circumferential strain in shaft

Strain compatibility, ε″cb = ε″cs

Problem 6.11 A steel rod of a diameter of 50 mm is forced into a bronze sleeve of an outside diameter of 80 mm, thereby producing a tension of 40 N/mm2 at the outer surface of sleeve. Determine (a) the radial pressure between the bronze sleeve and the steel rod and (b) the rise in temperature which would eliminate the force fit.

For steel, E = 210 Gpa, v = 0.25, a = 11.2 × 10−6/°C

For bronze, E114 Gpa, v = 0.33, α = 18 × 10−6/°C

Solution

Bronze sleeve

R1 = 25 mmR2 = 40 mm

Tension on outer surface, , where p′ is junction pressure

Radial pressure between bronze sleeve and rod

At junction

Circumferential strain,

Shaft

σ″c = − p′ (compressive) both are compressive

σ″c = − p′


Total numerical sum of circumferential strains

αb > αs(on heating bronze sleeve will expand more than steel shaft.)

(αb − as) ΔT = ɛc

(18 − 11.2)× 10−6 × ΔT = 82.631 × 10−6

Rise in temperature

Problem 6.12 A steel plug of a diameter of 50 mm is forced into a copper ring of an external diameter of 80 mm and a width of 30 mm. From a strain gauge fixed on the outer surface of the ring in the circumferential direction, the strain is found to be 64 × 10‒6. Considering that the coefficient of friction between the mating surfaces is 0.25, determine the axial force required to push the plug out of the ring.

For bronze, E = 100 × 103 N/mm2

For steel, E = 200 × 103 N/mm2

Solution

Bronze ring

Bronze ring

R1 = 25 mm

R2 = 40 mm

At outer surface, ɛc = 64 × 10−6

At outer surface, σc = 64 × 10−6 × EB

= 64 × 10−6 × 100,000

= 6.4 N/mm2

Say p′ = junction pressure

Normal force on shaft, N = 2πR1 × B × p′,where B is the width of ring

N = 2π × 25 × 30 × 4.99= 23,515 N

Tangential force, F = μN = 0.25 × 23515 = 5878.7 N (along the length of shaft)

Axial force required to push the plug out of the ring = 5878.7 N


o In a thick shell, the wall thickness is significant and stresses vary along the thickness of the cylinder and cannot be assumed to be uniform.

o In a thick cylindrical shell subjected to internal pressure, Lame’s equations can be used to determine radial and hoop stresses.

Constants A and B are determined by using boundary conditions.

o In a thick cylindrical shell with inner radius (R1) and outer radius (R2) subjected to internal pressurep At inner radius, Axial stress,

o In a thick cylindrical shell with inner radius (R1) and outer radius (R2) subjected to external pressurep

At inner radius, Constant axial stress,

o In a compound cylinder, the outer cylinder is shrink fitted over the inner cylinder. The junction pressure is developed which causes tensile hoop stress in the outer cylinder but compressive hoop stress in the inner cylinder.R1 = inner radius, R2 = outer radius and R3 = junction radius.If both cylinders are made of same material, that is, same EδR3 = shrinkage allowance

o In a hub and shaft assembly, junction pressure introduces tensile hoop stress in the hub but compressive hoop stress in the shaft.In shaft, σc = −p′, σr = p′ (junction pressure)

o In a thick spherical shell subjected to internal pressure,

With the help of boundary conditions, the values of Lame’s constants can be determined.

Review Questions

1. In a thick cylinder subjected to internal pressure, explain the assumption that plane sections remain plane after the application of internal pressure.

2. Show the variation of σc and σr in a thick cylinder along its thickness due to external pressure p.

3. In a compound cylinder, in which one cylinder is shrink fitted over another cylinder, show the variation of hoop stresses in outer and inner cylinders along the wall thickness.

4. Derive expressions of shrinkage allowance in a hub and shaft assembly.

5. Explain the purpose of compounding two cylinders.6. In an assembly of bronze sleeve and steel shaft, if the

temperature of the assembly is raised, at a particular temperature, the junction pressure becomes zero, why?


1. In a thick cylindrical shell with R2 = 2R1 subjected to external pressure of 45 N/mm2, what is the maximum hoop stress developed in the cylinder?

1. 120 N/mm2

2. 75 N/mm2

3. 60 N/mm2

4. None of these2. In a compound cylinder at junction, the sum of

circumferential strains in outer and inner cylinders is 120 × 10‒6. If the junction’s diameter is 200 mm, what is the shrinkage allowance?

1. 0.048 mm2. 0.024mm3. 0.0024 mm4. None of these

3. In a thick cylindrical shell, σcmax = 1.25p, what is the ratio of R2/R1?

1. 1.52. 2.03. 3.04. None of these

4. In a shaft and hub assembly with shaft diameter 50 mm, hub diameter 100 mm and junction pressure 30 N/mm2, what is the hoop stress in the shaft?

1. +50 N/mm2

2. +30 N/mm2

3. −30 N/mm2

4. None of these5. In a thick cylinder, R2/R1 = 2, the internal pressure is 60

N/mm2. What is the maximum shear stress at inner radius?1. 20 N/mm2

2. 30 N/mm2

3. 80 N/mm2

4. None of these6. In a thick cylindrical shell, σcmax = 120 p = 50 N/mm2, what

is the value of Lame’s constant A?1. 30 MPa2. 35 MPa3. 70 MPa4. None of these

7. In a compound cylinder, the hoop stresses developed at the junction in outer and inner cylinders are +84 MPa and −66 MPa. If E = 200 GPa, the junction radius is 100 mm, what is the shrinkage allowance as diameter?

1. 0.3 mm2. 0.15 mm3. 168 mm4. None of these

8. The variation of hoop stress across the thickness of a thick cylinder is

1. Linear2. Uniform3. Parabolic4. None of these

9. Purpose of compounding cylinder is1. To increase pressure-bearing capacity of a single

cylinder2. To reduce the variation in hoop stress distribution3. To increase the strength of the cylinder4. All the above

10. A bronze sleeve of an outer diameter of 100 mm is forced over a solid steel shaft of a diameter of 60 mm. If the junction pressure is 32N/mm2, the hoop strain at outer radius of sleeve is given by (if E = 100 GPa)

1. 320 mstrain2. 360 mstrain3. 720 mstrain4. None of these

Practice Problems

1. Two thick cylinders A and B are of the same dimensions. The external diameter is double the internal diameter. A is subjected to an internal pressure p1, while B is subjected to an external pressure p2. Find the ratio of p1 / p2 if the greatest hoop strain developed in both cylinders is the same. The Poisson’s ratio of the material is 0.3.

2. A steel cylinder of an inside diameter of 1 m and a length of 4 m is subjected to an internal pressure of 60 MPa. Determine the thickness of the cylinder if the maximum shear stress in the cylinder is not to exceed 80 MPa. What is the change in the volume of the cylinder if E = 210 GPa, and v = 0.3?

3. A pressure vessel of an internal diameter of 200 mm, an external diameter of 250 mm and a length of 1 m is tested under a hydraulic pressure of 20 N/mm2. Determine the change in its internal and external diameters if E = 208 kN/mm2 and v = 0.3.

If a strain gauge is mounted on the surface of the pressure vessel in an axial direction, what will be the strain reading?

4. A thick cylinder is subjected to both internal and external pressures. The internal diameter of the cylinder is 150 mm and the external diameter is 200 mm. If the maximum permissible stress in the cylinder is 20 N/mm2 and the external radial pressure is 4 N/mm2, determine the intensity of internal radial pressure.

5. A compound cylinder is formed by shrinking one cylinder over another. The outer diameter of the compound cylinder is 200 mm, the inner diameter is 120 mm and the diameter at the common surface is 160 mm. The junction pressure developed due to shrinkage is 4 N/mm2.

The compound cylinder is now subjected to an internal pressure of 50 N/mm2, determine themaximum hoop tension in the cylinder. How much heavier a single cylinder of an internal diameter of 120 mm would be if it is subjected to the same internal pressure in order to withstand the same maximum hoop stress.

6. A compound cylinder has a bore of 120 mm and an outer diameter of 200 mm and the diameter at the common surface is 160 mm. Determine the radial pressure at the common surface which must be provided by the shrinkage fitting, if the resultant maximum hoop stress in the inner cylinder under a superimposed internal pressure of 50 N/mm2 is to be half the value of the maximum hoop tension in the inner cylinder if this cylinder alone is subjected to an internal pressure of 50 N/mm2.

7. A thick cylinder of an internal diameter of 160 mm is subjected to an internal pressure of 5 N/mm2. If the allowable stress for cylinder is 25 N/mm2, determine the wall thickness of the cylinder. The cylinder is now strengthened by wire winding so that it can be safely subjected to an internal pressure of 8 N/mm2. Find the radial pressure exerted by wire winding.

8. A compound cylinder consists of a steel cylinder of an internal diameter of 180 mm and an external diameter of 250 mm and bronze liner of an external diameter of 180 mm and an internal diameter of 170 mm as shown in Figure 6.16. Assuming the liner to be thin cylinder and that there is no shrinkage stress in compound cylinder, determine the hoop stress developed in the liner due to an internal pressure of 50 N/mm2. Ignore the longitudinal stress and strain.

For steel, E = 208 kN/mm2, v = 0.28

For bronze, E = 112 kN/mm2, v = 0.309. A steel tube has an internal diameter of 25 mm and an external

diameter of 50 mm. Another tube of the same steel is to be shrunk over the first tube so that shrinkage stresses just produce a condition of yield at the inner surface of each tube. Determine the necessary difference in the diameter of the tubes at the mating surface before shrinking and the required external diameter of the outer tube. Assuming that yielding occurs according to the maximum shear stress condition and that no axial stresses are set up due to shrinkage. The yield stress in simple tension or compression is 414 N/mm2, E = 207 GPa.


Figure 6.16 Compound cylinder10. A steel sleeve is pressed onto a steel shaft of a diameter of 50

mm. The radial pressure between the steel shaft and the steel sleeve is 20 N/mm2 and the hoop stress at the inner surface of sleeve is 56 N/mm2. If a compressive force of 50 kN is now applied to the shaft, determine the change in radial pressure. E = 210 Gpa, v = 0.3.

11. A steel rod of a diameter of 60 mm is forced into an aluminium sleeve of an outside diameter of 80 mm, thereby producing a tension of 54 N/mm2 at the outer surface of sleeve. Determine (a) the radial pressure between the bronze sleeve and the steel rod and (b) the rise in temperature which would eliminate the force fit.

For steel, E = 200 Gpa, v = 0.3, α = 11.2 × 10−6/°C

For aluminum, E = 68 Gpa, v = 0.33, α = 22 × 10−6/°C12. A steel plug of a diameter of 80 mm is forced into a steel ring of

an external diameter of 120 mm and a width of 50 mm. From a strain gauge fixed on the outer surface of the ring in the circumferential direction, the strain is found to be 41 × 10‒6. Considering that the coefficient of friction between the mating surface is 0.2, determine the axial force required to push the plug out of the ring. E = 210 GPa.

[Hint: Normal force, N = p′ × 20 × 40 × 50, F = μN]

Special Problems

1. A compound cylinder is made by shrinking a cylinder of an outer diameter of 200 mm over another cylinder of an inner diameter of 100 mm. If the numerical value of maximum hoop stress developed due to shrinkage fitting in both the cylinders is the same, find the junction diameter.

2. A thick cylinder of internal diameter D and wall thickness t is subjected to an internal pressure ofp. If the maximum hoop stress developed in the cylinder is 2.5p, determine the ratio of t/D.

3. A steel shaft of a diameter of 120 mm is force fitted in a steel hub of an external diameter of 200 mm, so that the radial pressure developed at the common surface is 12 MPa. If E = 200 GPa, determine the force fit allowance on the diameter. What is the maximum hoop stress developed in the hub?

4. A steel ring of internal radius r and external radius R is shrunk onto a solid steel shaft of radius r + dr. Prove that the intensity of

pressure at the mating surface is equal to , where E is the modulus of elasticity of steel.

[Hint: use shrinkage formula for shaft + hub]

5. Two thick cylinders A and B are of same dimensions. The external diameter is double the internal diameter. Cylinder A is subjected to an internal pressure p1, while cylinder B is subjected to an external pressure p2. Find the ratio of pressure p1 to p2 if the greatest circumferential stress developed in both the cylinders is the same.

6. A steel cylinder of an internal diameter of 100 mm and an external diameter of 150 mm is strengthened by shrinking

another cylinder onto it, the internal diameter of which before heating is 149.92 mm. Determine the outer diameter of the outer cylinder at the junction if the pressure at the junction after shrinking is 20 N/mm2.

E = 210 GPa.

[Hint: use expressions of shrinkage allowance are taking R1 = 50 mm and R3 = 75 mm]

7. A compound cylinder is made by shrinking a tube of an outer diameter of 150 mm over another tube of an inner diameter of 100 mm. Find the common diameter if the greatest hoop stress in the inner tube is numerically 0.7 times that of outer tube.

8. A thick cylinder of an internal diameter of 120 mm and an external diameter of 180 mm is used for a working pressure of 15 N/mm2. Because of external corrosion, the outer diameter of the cylinder is machined to 178 mm. Determine by how much the internal pressure is to be reduced so that the maximum hoop stress in the cylinder remains the same as before machining.


Ecercise 6.1: 109.7, 56.9 ms

Exercise 6.2: D = 169.7 mm

Exercise 6.3: p = 102.78 N/mm2

Exercise 6.4: 20.485, −103.70, −83.22, +103.45, + 82.96 MPa.

Exercise 6.5: (a) 55.576 N/mm2, (b) 144.5 N/mm2






Exercise 6.6: 7.46 MPa


1. (a) 2. (b) 3. (c) 4. (c) 5. (c)

6. (b) 7. (b) 8. (c) 9. (d) 10. (b)


1.2. t = 0.5m, δv = 5,206cc3. 0.1707mm, 0.0727 mm, 68.375 μ strain

= 68.375 × 10−6μ strain4. 10.72 N/mm2

5. σcmax = 100.71 N/mm2 at inner radius, single cylinder is 41.5 per cent heavier than the compound cylinder

6. p′ = 3.71 N/mm2 due to shrinkage7. p′ = 2 5 . N/mm2

8. 130.74 N/mm2

9. R2 = 50 mm, D2 = 100 mm, δD3 = 0.125 mm10. 2.01 N/mm2

11. (a) 21 N/mm2, (b) Rise in temperature 118.4°C12. 13.52 kN


1. R3 = 75.12 mm D3 = 150.24 mm2. 0.26353. 0.0225 mm, 25.5 MPa4. p1 = 1.6p25. 212 mm6. 129.92 mm7. 0.373 N/mm2


(som notes vtu link - simply civil -...

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