chapter3 torsion final

Torsion of Circular Bars

Torsion refers to the twisting of a structural member when it is loaded by moments/torques that produce rotation about the longitudinal axis of the member Recall that the sense or direction of a torque is determined using

the right hand rule

The problem of transmitting a torque or rotary motion from one plane to another is frequently encountered in machine design. Normally circular bars are used for such transmissions chiefly because, in these bars, a plane section before twisting remains plane after twisting, i.e. there is no warping of the section after loading.

Circular Bars

We will be interested in determining the maximum load a circular bar is capable of sustaining, the stresses in the bar as well as the deformation or twist of the bar. To do this, we need to establish the relationship between the following quantities:

the applied torque, T, the shear stress developed in the material by

the applied torque, T (since this will cause material failure if the limiting stress is exceeded [Strength Condition], and

the deformation/angle of twist produced by the applied torque [Serviceability Condition]

Consider a bar of circular cross-section twisted by couples T at the ends. Because the bar is subjected to torsion only, it is said to be in pure torsion.

Assuming that the end A is fixed, then the torque will cause end B to rotate through a small angle Ф, known as the angle of twist. Thus the longitudinal line mn on the surface of the bar will rotate through a small angle to position mn'

Torsional Deformation of Circular Bars

Consider a small element abcd of length dx within the elemental segment of the bar, the magnitude of shear strain , is given by:

where r is the radius of the circular bar and dФ is the angle of rotation of one cross-section with respect to the other. dФ/dx is the rate of change of the angle of twist and representing this by θ, we obtain

where is the rate of change of with respect to length, i.e. angle of twist per unit length.

Torsional Deformation of Circular Bars Contd...

For the case of pure torsion, the angle of twist per unit length is constant along the length of the bar, hence

Note that the above expression is based only on geometric concepts and is therefore valid for any circular bar in pure torsion, irrespective of its material behaviour (elastic or inelastic, linear or non-linear).

Also since is constant for pure shear, the expression above implies that is linearly proportional to the radial distance r. Thus at any interior section at radial distance , we have that = and maximum occurs at the outside of a circular bar.


d

Lrr and

Ldxd

According to Hooke’s law, for linear elastic materials, shear stresses are proportional to shear strains and the constant of proportionality is the modulus of rigidity, G. Hence

ie.

Torsional Deformation of Circular Bars Contd...Torsion Formula for Linearly Elastic Materials

To determine the relationship between the applied torque T and the stresses it produces, we consider equilibrium of the internal forces and the externally applied torque, T. Considering an elemental area dA within an elemental ring of thickness dρ situated at radius ρ from the centre:

If the shear stress in the elemental area is , then the shear force dF on this area is dA. Moment of this force about the axis of the shaft/bar is dF ρ = ( dA)ρ. That is: dF = dA

dM = dF = ( dA)ρ = (G dA) = (G dA) = Gθρ2dA

Torsion Formula: Relationship between T and


Total resisting torque about the axis of the shaft is the summation taken over the entire x-sectional area, of the moments of all individual elements, i.e.

pA

GdAGdA I GdMTM 2

A A

2

i.e.

rG

I

TIGT

pp

where dAIAp

2 Polar moment of inertiayx II

Relationship between T and ...

For a circular x-section of radius r and considering the

elemental shaded ring shown:

dAI p2

32162

4

22

4

222

2

44

4

0

4

0

32

d

x

d

dr

ddI

ddA

r r

p

32

4dI p

Summary

Units: T is in Nm Ip is in m4 is in Pa (i.e. N/m2) r is in m G is in Pa θ is in radian/m

G

rI

T

p

We have obtained that

Observe that is maximum at the outer boundary and zero at the centre of the bar.

Computation of Angle of Twist

The product GIp is known as torsional rigidity while GIp/L is

called the torsional stiffness, defined as the torque required to produce a unit angle of rotation of one end of the bar with respect to the other end. The torsional flexibility is the reciprocal of the torsional stiffness and is defined as the angle of rotation required to produce a unit torque.

pp GI

TL

GI

T

L

Note that:

Hollow Circular BarsObserve that the shear stress distribution in a solid bar varies from zero at the centre of the bar to a maximum value at the outer boundary of the bar. Hence material near the centre of the bar will be grossly under-stressed and is not being used to full capacity. Where weight reductions and savings of material are important, it is advisable to use hollow shafts to transmit torque.

Hollow Circular Bars Contd...

The analysis of the torsion of a hollow circular bar is essentially the same as that for a solid bar except that the limits of integration for determining Ip changes from 0 to r and becomes r1 to r2 where r1 is the internal radius and r2 is the outer radius. Hence, for a hollow circular bar of internal radius r1 and outer radius r2, the polar moment of inertia is given by: 4

142

41

42 322

ddrrI p

Example Problem 3.2

A solid steel shaft of diameter d = 60 mm and length L = 4 m is to be designed using allowable = 40 MPa and allowable angle of twist per unit length = 1 per m. (a) Determine the maximum permissible

torque that may be applied to the shaft assuming that G = 80 GPa.

(b) Determine the angle of twist .

Solution

d = 60 mm

L = 4 m

Non Uniform Torsion

Non-uniform torsion arises when a bar made up of two or more segments of different diameters is subjected to torsion at several cross-sections. For this type of case, each region of the bar between applied torques or between changes in x-sections is considered to be in pure torsion and the effect of the applied torque can be determined for each part or segment of the bar. The same equations obtained earlier are used.

Non Uniform Torsion Contd...

The total angle of twist Ф of one end of the bar with respect to the other end is obtained by summation, using the general formula:

If the torque and/or the x-sectional area of the bar changes continuously along the axis of the bar, then the total angle of twist Ф is given by:

Example 3.3 A tapered bar AB of solid circular x-section is twisted by

torques T applied at the ends. The diameter of the bar varies linearly from da at the left-hand end to db at the right-hand end. Derive an expression for the angle of twist of

the bar.

Solution

3.4 Pure Shear A object that is subjected to only shear stresses is said to

be in pure shear. Consider the small volume element that is subjected to pure shear as shown

Force on positive x-face = 1bc

Force on negative x-face = 2bc

Force on positive y-face = 3ac

Force on negative y-face = 4ac

Pure ShearConsidering equilibrium of forces

ΣFx = 0 => 3ac = 4ac = 3 = 4

ΣFy = 0 => 1bc = 2bc = 1 = 2

ΣMz = 0 => 1abc = 3abc = 1 = 3

Therefore 1 = 2 = 3 = =

Pure ShearSince 1 = 2 = 3 = 4 = we conclude that:

(1) Shear stresses on opposite faces of an element are equal in magnitude but opposite in direction.

(2) Shear stresses on perpendicular faces of an element are equal in magnitude and have directions such that both stresses point toward, or both point away from, the point of intersection of the faces. Such shear stresses are known as complementary shear stresses.

Stresses on an Inclined Plane of Shafts Subjected to Pure Shear For a body is pure shear, there is no

normal stress on the cross-section of the body. However, on oblique planes inclined at an angle of to the cross-section, both normal and shear stresses will be present, as shown.

Normal and Shear Stresses on Inclined Plane

Determining the expressions for normal and shear stresses on an oblique plane whose normal is inclined at to the longitudinal-axis of the member which is subjected to pure shear:

The stresses on the inclined plane can be determined by considering equilibrium of all forces on the three sides of the triangular element. If A0 is the cross-sectional area and A is the oblique/inclined plane area, we have that:

Ao/A = cos A = A0sec

Also the area of the bottom face of the triangle Ab is obtained from

Ab/Ao = tan Ab = A0tan

Considering equilibrium of forces in a direction normal to the inclined plane,

A0sec = A0sin + A0tancos = 2A0sin

= 2sincos = sin2

Normal and Shear Stresses on Inclined Plane Contd...

Considering equilibrium of forces in a direction parallel to the inclined plane.

A0sec = A0cos - A0tansin = A0cos - sin2 / cos = cos2sin2 = cos2

Hence for an object that is in a state of pure shear, normal and shear stresses and will act on a plane inclined at to the x-section where

= sin2

= cos2

Note that for = 0

= 0 and =

For = 90, i.e. the top face of the element, = 0 and = -.

The normal stress reaches a maximum value at = 45 where = (tensile) and = 0

Similarly, the normal stress is minimum (maximum compressive value) at = -45 where = - (compressive) and = 0. Hence an element oriented at an angle of 45; is acted upon by equal tensile and compressive stresses in perpendicular directions but no shear stresses are present.


Note that the foregoing implies that the state of pure shear at the surface or x-section of an object is equivalent to equal tensile and compressive stresses acting on an element of the object which is oriented at an angle of 45 to the cross-section. If a material is weaker in shear on longitudinal plane than on cross-sectional planes, e.g. a circular bar made of wood with grains in the longitudinal direction, the first cracks due to twisting will be as a result of shear failure and will occur on thesurface in the longitudinal direction. (Qn: Why on the Surface and in the longitudinal Direction). However, if the material is very brittle (ie. much weaker in tension than in shear) failure will occur along the plane of maximum tensile stress, ie.e. along a helix inclined at 45 to the longitudinal axis. (Qn: Why along a helix inclined at 45 to the longitudinal axis; Sketch a typical failure surface). This type of failure is easily demonstrated by twisting a piece of classroom chalk.


Mohr’s Circle for Pure Shear


2cos

2sin

Recall that for a bar subjected to pure shear, the stresses acting on a plane whose normal is inclined at θ to the positive axis is given by the expressions:

These equations can be shown to be the parametric equations of a circle, with radius , which is centered at the origin and has 2 as the parameter. To eliminate the parameter 2, square both sides of equations A and B and sum up the resulting expressions.

222

2222222 2cos2sin

Equation is the standard form of the equations of a circle on the

axes whose centre is at the origin (0,0) and has radius r =

represent the co-ordinates of pints lying on the Mohr’s circle and fire the normal and shear stresses acting on inclinded planes.

222

and

direction.

clockwisecounter in the positive measured 2 is circle sMohr' on the

perameter hat theremember t Always circle. sMohr' on the with

correspondmust 0 hence and 0at occurs that Note

max

max

0 Planeon StressShear Postive A.

0

ve

minmax

90 planeon occurs

0 planeon occurs

45 planeon occurs

45 planeon occurs

min

max

min

max

0 planeon StressShear Negative B.

ve

0

0 planeon occurs

90 planeon occurs

45 planeon occurs

45 planeon occurs

min

max

min

max

maxmin

θ

τ

τ

τ

τ

0ve

Coordinates of point is)2 cos ,2sin(

Mohr’s Circle in Pure Shear con’t

Normal Strains in Pure Shear

Next, consider the strains associated with pure shear. Recall that the effect of shear is to produce a shear distortion (change in shape) but no change in volume. Considering a stress element that is oriented at 45. Only tensile and compressive stresses and - respectively, act on this stress element. The effect of the tensile stress is to elongate the element in the 45 direction and to shorten it in the perpendicular direction (135 or -45) directions. Similarly, the compressive stress -, will shorten the element in the 135 or -45 direction and elongate it in the 45 direction.

Normal Strains in Pure Shear Contd...

If the material is linearly elastic, the shear strain for the element at

= 0 is given by = /G.

For the element at = 45,

Normal strain due to max = (acting along 45 and 225 direction):

x = /E = /E (at 45 direction) and

y = -x = - /E (at 135 or -45 direction).

Normal strains due to min = -(acting along -45 and 135 direction):

y = /E = -/E (at 135 or -45 direction) and

x= = -y = /E (at 45 direction)

Observe that the shortening in the lateral direction resulting from the tensile force is exactly equal to the elongation in the 45 direction resulting from the compressive force.

Maximum and minimum normal strains can be obtained by summing up the strains in the x- and y-directions, respectively.

Hence the resultant normal strain at 45 direction.

max = /E + /E = /E(1+)

Similarly resultant normal strain at 135or -45 direction.

min = -/E - /E = -/E(1+)

Hence pure shear produces elongation in the 45 direction and shortening in the 135 or -45 direction. This is consistent

with the deformed shape of the element in pure shear.

Normal Strains in Pure Shear Contd...

Example 3.4 A hollow circular steel bar (G = 80 GPa) is

twisted by a torque T that produces a maximum shear strain max = 750 x 10-6 rad. The bar has outside and inside radii of 75 and 60 mm, respectively. What is the maximum tensile stress max in the bar. What is the magnitude of the applied torque, T?

Solution

Example 3.5 A hollow circular shaft of aluminum (G =

28 GPa) has an outside diameter of 100 mm and an inside diameter of 50 mm. When twisted by a torque T, the bar has an angle of twist per unit length = 0.03 rad/m. What is the maximum tensile stress max in the shaft? What is the magnitude of the applied torque T?

Solution

3.5 Relationship between the Elastic Constants E,

G, and Consider the square stress element of side

h. The effect of pure shear is to distort the square element into a rhombus as shown.

Relationship between the Elastic Constants E, G, and Contd...

Recall that the diagonal bd will elongate while ac will shorten. If Emax is the normal strain is the 45 direction, then the new length of the diagonal bd, Lbd is given by

Lbd = L0 + = L0 + L0 = L0 (1+)

where h 2 is the original length (h 2 )max is the elongation. Also the shear distortion reduces the angle adc from to where is the shear strain. Now adb = abd = _adc.

sin121

sin2

cossincesin12

cos11

2cos1212

2cos2

2maxmax

2max

22max

2

2222

Hence

hh

hhhLbd

Using the Cosine Rule:

Because max and are very small quantities, e2max is

neglectedand hence: 1+ 2max=1+

max = /2


This shows that the normal strain max in the 45 direction is pure shear is equal to one-half the shear strain . Recall that = /G. Hence max = /2 = /2G.

But the normal strain in the 45° direction max is given by max = /E + /E = /E (1 + )

Hence max = /2G = /E (1 + ) G = E/2(1 + ) Thus, the the elastic constants E, G, and V are not idependant

properties as one of them can be determined, using the equation above, if the other two are known.


Summary

Note that this expression implies that the three elastic constants, the Young’s modulus E, the modulus of rigidity G and the Poisson’s ratio are not independent properties as one of them can be determined, using the equation above, if the other two are known.

Example 3.6 A hollow shaft of outside diameter 80 mm

and inside diameter 50 mm is made of aluminum having G = 27 GPa. When the shaft is subjected to a torque T = 52 kN.m, what are the values of the maximum shear strain max and the maximum normal strain max.

Solution

Example 3.7 A circular bar is subjected to a torque that

produces a tensile stress = 8000 psi at 45° to the longitudinal axis. Determine the maximum normal strain and maximum shear strain in the bar assuming that E = 11.5 x 106 psi and = 0.30.

Solution

3.6 Transmission of Power by Circular Shafts The work done by a torque T is equal to the product of T

and the angle through which it rotates , i.e. W = T where = angular displacement (measured in radians). Hence:

where (omega) is the angular speed, i.e. the rate of change of angular displacement with time

Tdt

dT

dt

dWP

Note that the angular speed is often expressed in terms of the frequency, f of revolution, i.e. the number of revolutions per unit time.

= 2 f where f is in Hetz (Hz) = s-1

Hence P = 2 fT

SI Units for Power

Power is measured in watts (W) Torque is measured in N.m Angular speed is measured in radian/second (rad/s)

1 Watt = 1 N.m/s = 1 x 103 N.mm/s 1 kW = 106 N.mm/s

Popular English Unit for Power is the Horse Power 1 Hp = 550 ft.lb/sec ~ 746 W

The angular speed is often expressed as the frequency, f, of rotation, i.e. the number of revolutions per second. This means that

= 2 f where f is measured in Hetz (Hz) = s-1;

Hence, P = 2 f T

Also, may be expressed in revolutions per minute (rpm), denoted by n. If the angular speed is expressed in rpm, then n = 60 f and this implies that f = n/60 and

P = 2 fT = 2 n T/60 (where n is in rpm)

Observe that these expressions relate Power transmitted to the Torque, T (hence corresponding can be computed) and that Power is a function of the frequency of rotation, f, or the rpm, n.

Example 3.8 A hollow propeller shaft of outside diameter 6.0

inches is required to transmit 4000 hp while rotating at 1500 rpm. The material is steel with G = 11.5 x 106 psi. The allowable shear stress is 7500 psi and the allowable angle of twist per unit length is 0.01 degrees per inch. (a) Calculate the required inside diameter d of the propeller shaft. (b) specify to the nearest inch the actual inside diameter to be used.

Solution

Example 3.9 How much power P(kW) may be

transmitted by a solid circular shaft of diameter 80 mm turning at 0.75 Hz if the shear stress is not to exceed 40 MPa?

Solution

Example 3.10 A circular shaft ABC is driven by a motor at A, which

delivers 300 kW at a rotational speed of 3.2 Hz. The gears at B and C take out 120 and 180 kW, respectively. The lengths of the two parts of the shaft are LAB = 1.5 m and LBC = 0.9 m respectively. Calculate the required diameter of the shaft if the allowable shear stress is 50 MPa, the allowable angle of twist in the shaft between points A and C

is 0.02 rad, and G = 75 GPa.

Solution

3.7 Statically Indeterminate Torsional

Members Consider a shaft consisting of two prismatic

sections, loaded by a torque To at C with the ends fixed at A and B. The applied Torque To will give rise to reactive torques Ta and Tb at ends A and B respectively. Because only one fixed end is needed to ensure equilibrium, the extra fixed end makes the problem to be statically indeterminate to the first degree (since there is one extra/redundant reactive torque that is present but is not needed for equilibrium).

Statically Indeterminate Torsional Members Contd...

Using the flexibility method, the statically indeterminate structure is "released", i.e. split-up into a statically determinate primary structure with the externally applied load acting on it and a secondary structure with only the redundant force acting on it.

pb

a

pa

a

pb

oA GI

bT

GI

aT

GI

bT0

pb

b

pa

aBCAC GI

bT

GI

aT

Different Possible Compatibility Conditions:

1) If end A is released, then

2). If end B is released, as shown subsequently, then

3). Alternatively, we can use C for compatibility

Statically Indeterminate Torsional Members Contd...

pb

b

pa

b

pa

oB GI

bT

GI

aT

GI

aT0

Statically Indeterminate Torsional Members

Equilibrium Equation: 0TTT ba

Compatibility Equation

for end B released:

bTbTb 0

2). Let end B be released, then


Considering the reactive torque at end B to be the redundant force, the structure is released as shown below:

Given Structure

Released Structure


Statically Determinate Primary Structure

Statically Determinate Secondary Structure


The angle of twist b produced by To at end B is given by:

Considering the secondary structure, the angle of twist produced by Tb at end B is given by:

pb GI

aT00)(

pb

b

pa

brb GI

bT

GI

aT)(


We know that the angle on twist at end B is equal to zero, since this end is fixed. Hence, for compatibility at end B, (b)o = (b)r; i.e

papb

pbb

pa

o

pbpa

papbb

pb

b

pa

b

pa

bIaI

aITT

GI

aT

II

bIaI

G

T

GI

bT

GI

aT

GI

aT

0

0


From equilibrium Ta + Tb = T0 Ta = T0 + Tb

Substituting Tb from above, we obtain:

Having obtained Ta, we can obtian Tb from Ta

+ Tb = T0

papb

paa bIaI

bITT 0


Note that the shear stress on each part of the bar is different and is given by:

pb

bbcb

pa

aaac I

rTand

I

rT

pa

pbab

pa

pbab

pb

b

pa

a

bI

aITT

bI

aITT

GI

bT

GI

aT

An alternative approach to solving the same problem is to consider that the angle of twist oneach part of the bar must be the same, i.e. fa = fb = f0


Using Equilibrium:

papb

paa

pc

pba

pa

pbaa

ba

bIaI

bITT

TbI

aIT

TaI

aITT

TTT

0

0

0

0

1

which is the same as was obtained previously. Note that we have utilized both equilibrium and compatibility equations except that we enforced compatibility at C instead of at B.


If the composite bar is acted upon by a total torque To, reactive torques Ta and Tb will be developed in the inner core and outer tube, respectively, from static equilibrium.

T0 = Ta + Tb -------------(1)

From compatibility of rotation,

------------ (2)

pbb

b

paa

a

IG

LT

IG

LT


Using these two equations, Ta and Tb are determined and the maximum shear stresses in each bar is given by:

pb

bbb

pa

aaa I

rT

I

rT ;

Example 3.11 A solid shaft is formed of two materials, an outer sleeve of

steel (Gs = 80 GPa) and an inner rod of brass (Gb = 36 GPa). The outer diameters of the two parts are 75 mm and 60 mm. Assuming that the allowable shear stresses are s = 80 MPa and b = 48 MPa in the steel and brass respectively, determine the maximum permissible torque T that may be

applied to the shaft.

Solution

3.8 Strain Energy In Pure Shear

When a load is applied to a structure, the structure deforms. The applied load does work as the structure deforms and strain energy is absorbed by the structure. The strain energy for a bar in torsion is equal to the area under the torque - rotation diagram. If the bar is linearly elastic, then the torque T will be proportional to the angle of twist .

U = Area under the T - diagram

3.8 Strain Energy In Pure Shear

For linearly elastic bars, the strain energy stored in the bar is equal to the area under the torque – rotation (i.e T-) diagram, i.e.

U = ½ T

L

GI

GI

LTTU

GI

TL

andIL

GT

p

p

p

p

22

222

21

noting that:

3.8 Strain Energy Density, U

Similarily, the strain energy density U is equal to the area under the shear stress – shear strain diagram, i.e. U= ½ Noting that = G = /G we have that:

222

22 G

GU

Shear Force The shear force V corresponding to the shear

stress is given by V = ht. The top face of the element is displaced horizontally by , where = h

The work done by the shear force on the top face of the element is equal to the area below the load-displacement diagram and this is equal to the strain energy absorbed. For linearly elastic materials,

222

2thhhtVWU

Observe that h2t is the volume of the element. Hence the strain energy density:

222

22 G

GU

To obtain the total amount of strain energy stored in a circular bar subjected to pure torsion, consider an elemental circular section of radius and thickness d, extending along the entire length of the bar. The strain energy in the elemental

section dU is given by L

GI

GI

LTU

IdASinceGI

LTU

dAGI

LTdA

GI

LTU

LdAGI

T

LdAG

UldAdU

p

p

A

pp

A App

p

22

2

22

2

2

22

22

22

22

2

2

2

22

2

Shear Force continued...

Strain Energy In Pure Shear

Corresponding expression for non-uniform torsion and continuously varying x-sectional area is:

Lp

L

p

n

i

p

pi

ii

dxdx

dyxGI

xGI

dxxTU

L

GI

GI

LTU

0

2

0

2

1

22

22

22

Example 3.12 How much strain energy is stored in the stepped steel shaft

shown in the figure when the angle of twist of equals 0.015 radians. Use G = 11.8 x 106 psi.

Please Note: to use the second expression, you'd need to know 1 and 2 and the angle of twist will be different for the two

prismatic sections, where 1 + 2 = = 0.015 radian.

Solution

Thin-walled Tubes of Arbitray x-sectional shape (Closed sections)

Circular x-sectional shape is the most efficient shape for resisting torsion and transmitting power and is frequently utilized in machine parts. For complex structures however, the need may arise to use structural members with non-circular x-sections, eg. the wings of an airplane.

Thin-walled Tubes of Arbitray x-sectional shape (Closed sections)

Consider a cylindrical tube of arbitrary x-sectional area (the tube has a longitudinal axis of which is a straight line and the x-section at every point along this axis is identical). The tube is subjected to pure torsion and its thickness t is small compared to its width and t may vary around the x-section.

Thin-walled Tubes of Arbitray x-sectional

shape (Closed sections)Consider the rectangular element abcd where ab and cd are located || to the longitudinal axis and bc and ad are perpendicular to the axis. Let us assume that the shear stress varies from b at b to c at c since t is free to vary around the x-section. From equilibrium, identical shear stresses must act on the opposite face ad in the opposite direction. At corners b and c, the shear stresses or orthogonal faces must be b and c, respectively. Because ab and cd are || to the longitudinal axis and since the x-section is identical at every point along this axis, the thickness t is constant along ab and also constant (different value) along cd. The shear stresses on the face ab and cd are thus constant and have values b and c, respectively.

Thin-walled Tubes of Arbitray x-sectional

shape (Closed sections) The forces associated with the shear stresses b and c are: fb = btbdx and fc = ctcdx

where tb and tc are the thickness points b and c, respectively. For equilibrium in the longitudinal direction, fb = fc ie:

btbdx = ctcdx == btb = ctc

hence at every point on the x-section, the product of the shear stress and the tube thickness is a constant called the shear flow, f, ie:

shear flow, f= t

Relationship between Torque and Stress

Consider an element of length ds from one arbitrary x-sectional shape:

This element is located at s from an arbitrarily chosen co-ordinate established through point O. On the elemental area, f = t == = f/t

if dV is the shear force on this elemental area, then dV = A = f/t . tds = fds

Moment dM of the shear force dV about point O is given by dM = rdV = rfds

Relationship between Torque and Stress

The total moment of the shear forces in all elemental areas must be equal to the applied Torque T, where Lm is the length of the midline.

Observe that the area enclosed by ds at O is 1/2 rds. Hence the integeral above can be interpreted geometrically as twice the area enclosed by the midline.

Note that Am is not the x-section if t varies.

Strain Energy of Thin-walled Tubes

Consider again an elemental volume of dimensions t, ds and dx:

Strain energy density for element Hence, strain energy for element

Example 3.13 A uniformly tapered tube AB of hollow circular cross-section

of uniform wall thickness t and length L is loaded by a distributed torque having intensity t(x) per unit length that varies linearly from a maximum value tA at end A to 0 at end B. The bar is fixed at end A and free at end B. (a) What is the shear stress at the fixed end A in the bar? )b) What is the angle of twist at end B? The average diameters are dA and dB = 1/2 dA. The polar moment of inertia may be

represented by approximate formula

Solution

Example 3.14 A uniformly tapered tube AB of hollow circular cross section

is shown below. The tube has constant wall thickness t and length L. The average diameters at the ends are dA and dB = 2dA. The polar moment of inertia may be represented by the approximate formula (Eq. 3-18). Derive a formula for the angle of twist of the tube when it is subjected to torques T acting at the ends.

Solution

chapter3 torsion final

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