51

Chapter

5

TORSION: 5.1 Introduction 5.2 Deformation of a Circular Shaft 5.3 The Torsion Formula 5.4 Axial and Transverse Shear Stresses 5.5 Stresses on Inclined Planes 5.6 Angle of Twist 5.7 Statically Indeterminate Shafts 5.8 Design of Circular Shafts 5.9 Stress Concentrations

5.1 Introduction: �� = �� × �� ,�� = �� × �� units of T is N-m, lb-ft

Figure 5.1

Figure 5.2( a) &(b)

In many engineering applications, members are required to transmitting a

torque from one plane to a parallel plane and to carry these torsional loads.

The simplest members for accomplished this function is called a shafts.

Shafts are commonly used to connect an engine or motor to a pump,

compressor, axle, or similar members. Shafts connecting gears and pulleys

are a common application involving torsion members.

Torque is moment that tends to twist a member about the longitudinal axis.

Figure 5.1 illustrated produced couples T1 and T2 that are called torques,

twisting couples or twisting moments.

· Consider a long, circular shaft (circular cross section is an efficient shape

for resisting torsional loads), of length L and radius c that is fixed at one

end A, as shown in Figure 5.2a.

· When an external torque T is applied to the free end of the shaft at B, the

shaft is deforms as shown in Figure 5.2b. All cross sections of the shaft

are subjected to the same torque T, therefore, the shaft is said to be in

pure torsion.

· Longitudinal lines in Figure 5.2b are twisted into helixes as the free end

of the shaft rotates through an angle ∅. The angle of rotation is called as

the angle of twist or angular displacement. The angle of twist changes

along the length L of the shaft. For prismatic shaft, ∅ will vary linearly

between the ends of the shafts.

· The twisting deformation does not distort cross sections of the shaft in any way and the overall length remains constants. The following assumptions can be applied to torsion of shafts that have circular-either solid or hollow- cross-sections:

52

1) The plane sections before twisting remain plane after twisting (points in a given plane remain in that plane after twisting). In other words, circular cross sections do not warp as they twist. 2) Cross sections rotate about and remain perpendicular to the

longitudinal axis of the shaft.

3) Furthermore, each cross section remains undistorted as it rotates

relative to neighboring cross sections. In other words, the cross sections

remains circular and the expansion or contraction of a cross section does

not occur. Thus all normal strains are zero and redial lines remain straight

and redial as the cross section rotates.

4) The distances between cross sections remain constant during the

twisting deformation. In other words, no axial strain occurs in a round

shaft as it twists.

5) The material is homogeneous and isotropic.

· In this chapter, we will develop formulas for the stresses and

deformations produced in circular bars subjected to torsion, such as

drive shafts, thin-walled members. This chapter covers several

additional topics related to torsion: statically indeterminate members,

torsional strain energy, stress concentration, and non-circular bars.

5.2 Torsional Shear Strain (γ):

Figure 5.3a

· To investigate the deformations that occur during twisting, a short segment ∆� of the shaft shown in Figure 5.2 is isolated in Figure 5.3a. The shaft radius C ; however, for more generally, an interior cylindrical portion for an arbitrary radial

distance ρ (where � < � ≤ �) from the center of the shaft will be examined as

shown in Figure 5.3b .

· As the shaft twist, the two cross sections of the segment rotate about the x-axis, and the line element CD on the undeformed shaft is twisted into helix C'D'

· The angular difference between the rotation of the two cross sections is equal

to ∆∅ , which creates a shear strain γ in the shaft.

· The shear strain γ is equal to the angle between line elements C'D' and C'D'',

as shown in Figure 5.3b, which is given by:

53

Figure 5.3b

tan � = � = �′�′′∆�

· The distance �′�′′ can also be expressed by the arc length ����� = �Δ� ,

which gives:

tan � = � = �′�′′∆� = �Δ�∆�

� = � Δ�∆�

· As the length ∆� of the shaft segment decrease to zero, the shear strain

becomes:

� = � ���� … … (5.1)

· The quantitydϕ ��� , represents the rate of change of the angle of twist, is

the angle of twist per unit length. Note that Eq. (5.1) show that the shear

strain in a circular shaft varies linearly with respect to the redial

coordinate ρ; therefore, the shear strain γ at the shaft centerline (i.e., ρ = 0

) is zero while the largest shear strain occurs for the largest value of ρ (i.e.,

ρ = c ), which occurs on the outermost surface of the shaft. Then the

maximum shear strain ���� at the outer radius c is given by:

���� = � ���� … … (5.2)

· Eqs. (5.1) and (5.2) can be combined to express � at any radial coordinate ρ in terms of ���� as:

� = �� ���� … … (5.3)

· The above equations are valid for a circular bar of any material, elastic

or inelastic, linear or nonlinear.

54

5.2 Deformation of a Circular Shaft (another procedure):

Deformation of a Circular Shaft: Consider a bar or shaft of circular cross section twisted by a

couple T, assume the left-hand end is fixed and the right-hand end will rotate a small angle,

called angle of twist.

• By observation, if angle of rotation is small, length of shaft and its radius remain unchanged.

· If every cross section has the same radius and subjected to the same torque, the angle ∅ (x)

will vary linearly between ends under twisting deformation.

· Important assumptions: 1. The plane cross sections perpendicular to the axis of the bar remain plane after the

application of a torque: points in a given plane remain in that plane after twisting.

2. Furthermore, expansion or contraction of a cross section does not occur, nor does a

shortening or lengthening of the bar. Thus all normal strains are zero.

3. The material is homogeneous and isotropic.

4. If ∅ is small neither the length L nor its radius will change.

· The torque T twist the bar (Figure 5.3a), the longitudinal line AB deform into the helix

AB', and the angle BOB' represent the total angle of twist ∅.

· The turning of the right end through a small angle of twist d∅ relative to the left end

(that is m = m') of an element of length dx in such a shaft (Figure 5.3b).

55

· Note that the straight line mn, initially parallel to the axis of the bar, distorts onto line

mn', which is approximately straight.

· The magnitude of ���� is given by angle nmn'.

· Consider an element of the bar dx, on its outer surface we choose an small element

mmnn, during twisting the element rotate a small angle d∅, the element is in a state of

pure shear, and deformed into mmnn', observe that the length of nn' is: ��� = ��∅… … . (�)

· And the shear strain may be taken: tan ���� = ���� … … . (�) ���� = ����� = ��∅�� … … … (�)

· �∅�� , represents the rate of change of the angle of twist, denote as the angle of twist per

unit length or the rate of twist. Then the maximum shear strain at the outer radius c is

given by: ���� = �∅� … … … (5.1)

· Additionally, the shear strain inside the bar or for an arbitrary radial distance ρ from

the center of the bar is: � = �∅� … … … (5.2)

· Eq. (5.2), show that the shear strain in a circular shaft varies linearly with the distance

from the axis of the shaft.

· Eliminating∅, from eqs. (5.1) and (5.2), we have: � = �� ���� … … … (5.3)

5.3 The Torsion Formula:

· Material Behaviour: Shear stress t in the bar of a linear elastic material

is: ���� = ����� = � �∅� … … . . (5.4) § For solid shaft, shear stress varies from zero at shaft’s longitudinal

axis to maximum value at its outer surface.

§ t and γ in circular bar vary linearly with the radial distance from

the center, the maximum values tmax and γmax occur at the outer

surface..

§ Due to proportionality of triangles, or using Hooke’s law and Eq.

56

5.4, yields: � = �� ���� … … … (5.5)

§ Consider a bar subjected to pure torsion, the shear force acting on

an element area dA is t dA, the moment of this force about the axis

of bar is (t dA)ρ. The applied torque is: � = � (���)�� = � ��� �������� �

∵ ����� �� ��������, � = ����� � �� ��� … … … … . . (�)

· Hear the integral part is called the polar moment of inertia J of the

entire cross section of a bar. For solid circular and hollow cross sections, J

is given by the formulas: ������ = ���� = 0.5��� … . . (5.6), where c is the radius of the bar

In terms of the bar diameter, ������ = ����� = 0.03125��� … . (�.�) ������� = �� (�� − ��) … . . (5.8) , where c and b are the inner and outer

radii of the bar.

In terms of the diameters,

������� = ��� (d�� − d��) = 0.03125π(do4 – di

4)… . . (5.9)

For thin-walled circular members (i.e., r/t ≥ 10), an approximate

formula for J is given by: � = 2����… … … . (5.10) where r is the mean radius and t is the thickness of the tube,

respectively. The maximum shear stress on the outer surface is:

421 cJ p=

( )41

422

1 ccJ -= p

57

⇒ ���� = ��� … … . (5.11) · Shear stress at an arbitrary radial distance ρ from the center of the

bar is: � = ��� … … . (5.12)

· Because Hook´s law was used in the derivation of Eqs.(5.4),

(5.11), and (5.12), these formulas are valid if the shear stresses do not

exceed the proportional limit of the material shear. The above

relationships are referred to as the torsion formula and based only upon

geometric concepts; they are valid if shaft is circular, its material

homogenous, and it behaves in a linear-elastic manner.

· Furthermore, The shear stress acting on the plane of the cross section are

accompanied by shear stresses of the same magnitude acting on longitudinal

plane of the bar as depicted in Figure 5.6.

· If the material is weaker in shear on longitudinal plane than on cross-

sectional planes, as in the case of a circular bar made of wood, the first crack

due to twisting will appear on the surface in longitudinal direction will be

subjected to tensile and compressive stresses.

• Sign Convention: T is positive, by right-hand rule, is directed

outward from the shaft.

5.4 Stresses on inclined planes:

The elastic strain formula eq. (5.12) can be used to calculate the absolute maximum torsional stress on

a transverse section in a circular shaft subjected to pure torsion. Thus we need to find location where

ratio ��/� is maximum, and to draw a torque diagram (internal torque T vs. X along shaft). Once

internal torque throughout shaft is determined, maximum ratio of ��/� can be identified.

· For this investigation, the stresses at a point in the shaft of Figure 5.7a will be analyzed.

· If the equations of equilibrium are applied to the free body diagram of Figure 5.7b, the following results

are obtained: ���� = ������� − ���(�� cos �) cos � + ���(�� sin �) sin � = 0

From which: ����� = ���(cos� � − sin� �) = ��� cos 2� … … … (5.13)

Figure 5.5

Figure 5.6

58

and : ���� = ����� − ���(�� cos �) sin � − ���(�� sin �) cos � = 0

From which: ��� = 2��� cos � sin � = ��� sin 2� … … . . (5.14)

These results are shown in the graph of Figure 5.8. from which, we observe ���� = ��� = ��� , and ����� = 0, when � = 45�, and ���� = −��� = − ��� ,

and ����� = 0, when � = 135� . thus the

The state of pure shear stress is

equivalent to equal tensile and compressive stresses on

an element rotation through an angle of 45o.

If a twisted bar is made of material that is

weaker in tension than in shear, failure will occur in

tension along a helix inclined at 45o, such as chalk.

Example 5.1(P (5.1) ): Given: A hollow steel shaft of outer radius c is fixed at one end and subjected to a torque T at the

other end. If c = 30 mm , T = 2 kN × m, allt = 80 MPa.

Find: If the shearing stress is not to exceed allt , what is the required inner radius b?

SOLUTION:

4 4( )2

allTc

c bt

p=

-

or 3

64 4

2(2 10 )(0.03)80(10 )[(0.03) ]bp

´=

-

or 6 6 480(10 )[0.81(10 ) ] 38.197b- - =

Solving, 0.024 24b m mm= =

c b.

59

Example 5.2 Given: A hollow steel shaft with an outside diameter of 100 mm and a wall thickness of 10 mm is

subjected to a pure torque of T = 5,500 N-m. Find: (a) The maximum shear stress in the hollow shaft. (b) The minimum diameter of a solid steel shaft for which the maximum shear stress is the same as in part (a) for the same torque T. SOLUTION:

Example 5.3 Given: A solid 0.75-in.-diameter shaft is

subjected to the torques shown in Fig. P6.8. The

bearings shown allow the shaft to turn freely.

Find: (a) Plot a torque diagram showing the internal

torque in segments (1), (2), and (3) of the shaft.

60

Use the sign convention presented.

(b) The maximum shear stress magnitude in the shaft.

SOLUTION: by equilibrium:

Example 5.4:

Given: A compound shaft (Fig. P6.7) consists of brass segment (1) and aluminum

segment (2). Segment (1) is a solid brass shaft with an allowable shear stress of 60

MPa. Segment (2) is a solid aluminum shaft with an allowable shear stress of 90 MPa.

The torque of T = 23,000 N-m is applied at C.

Find: the minimum required diameter of (a) the brass shaft and (b) the aluminum

shaft.

61

SOLUTION:

Example 5.5(P-5.4):

Given: The circular shaft is subjected to the torques shown in Fig. P5.4.

Find: What is the largest shearing stress in the member and where does it occur?

SOLUTION: Apply the method of sections between the changes of load points:

50EFT N m= × ↠ 30DET N m= × ↞

120CD BCT T N m= = × ↠ 80ABT N m= × ↞

We have

4 4 6 42 [(0.025) (0.015) ] 0.534 10hJ mp -= - = ´

Thus,

max :Tc Jt =

3 32 2(80) 3.26

(0.025)ABT MPac

tp p

= = =

3 62(120) 120(0.025)4.89 , 5.62(0.025) 0.534 10BC CDMPa MPat t

p -= = = =´

Example 5.6(P-5.12):

62

Given: A hollow shaft is made by rolling a plate of thickness t into a cylindrical shape and

welding the edges along the helical seams oriented at an angle f to the axis of the member (Fig.

P5.12). The permissible tensile and shear stresses in the weld are alls = 100 MPa, allt = 55

MPa, respectively, and D = 120 mm, t = 5 mm, f = 60o.

Find: The maximum torque that can be applied to the shaft?

SOLUTION:

60 90 150oq = + =

4 4 6 4[(0.06) (0.055) ] 5.984 102

J mp -= - = ´

From Eq. (5.14a) at 150oq = :

' sin 300 0.866oxs t t= = -

66

(0.06)100 10 0.866 0.8665.984 10

Tc TJ -´ = =

´

or 11.52T kN m= ×

Similarly, Eq. (5.14b):

' ' cos300 0.5ox yt t t= =

gives

66

(0.06)55 10 0.5 , 10.975.986 10

T T kN m-´ = = ×´

Thus,

10.97allT kN m= ×

5.5 Torsional Deformation (Angle of twistf ):

i. Prismatic Shafts: let us consider the deformation of a bar of

length L, of uniform circular cross section, fixed at one end and

subjected to torque T at the free end as shown in Figure 5.13. The rotation of the cross section at the free end of the shaft,

called the angle of twist θ , is obtained as:

Geometry of deformation: ���� = �∅� … … … (5.1) Equilibrium conditions:

63

���� = ��� … … . (5.11)

Material Behaviour ( Hook's law): ���� = ����� ⇒ ���� = ����� = ���� … … . . (�) Equations (5.1) and (a) can then be combined to obtain the following expression of the angle of twist f of the

shaft:

f = ���� … … … (5.15) Where f is measured in radians. The direction of the angle of twist is the same as that of the torque T.

Torsional spring stiffness: � = �f

= ��� … … … (5.16) ii. Multiple Prismatic Shafts (Stepped): in this case of shaft made of several segments with various

cross sections subjected to torques at a number of locations

as shown in Figure 5.14, Eq. (5.15) may applied to each part

separately. ¨ For this purpose, internal reaction torque Ti

and the angle of twist fi for each part must be calculated as, �� = �������� … . (�)

¨ In general, the total angle of twist f of the

shaft is then obtained by the algabric sum as:

f = ������� = ����������

��� … … … (5.17) Here, the subscript i is an index identifing the various parts of the shaft and n is the total number of segments,

each having constant values for Gi, Ji,Li, and Ti.

v In the U.S. Customary system, the consistent units are G

[ psi ], T [ lb．in ] , and L [ in.], and J [ in4]

v in the SI system, the consistent units are G [ Pa ], T [

N．m ], L [ m ], and J [ m4].

v The unit of θ in Eq. (3.15) is radians, regardless of

which system of unit is used in the computation.

v Sign convention: Represent torques as vectors using the right-hand rule, the same sign convention applies

to both T & f which are taken as + if a right hand screw advances with the direction of the vector when

64

twisted in the direction indicated by the torque as illustrated in Figure 5.15.

iii. Non-prismatic Shafts: ç In this case, either the torque or the section or both changes continuously along the axis of the shaft,

such as depicated in Figure 5.16a.

ç In this situation, the torque Tx of a cross-section located

at a distance x from the end of the shaft may be obtained from

the conditions of equilibrium (Figure 5.16b).

ç Hence assuming the taper is small and gradual, the

maximum shearing stress at a cross section can be detrmine by

the application of the tortional formula.

ç Equation (5.15) may be applied to disk element of thickness ��. The differential angle of rotation ��

for the element equals: �� = �� �����

ç Integration of this expression results in the total angle of twist for a nonprismatic shaft as, � = � �� ����� … … . (5.18)��

5.6 Power transmission: Shafts are used to transmit power as shown in Figure 5.17. The power P is transmited by a torque T

rotating at the angular speed ω is given by � = ��, where ω is measured in radians per unit time.

If the shaft is rotating with a frequency of ¦ revolutions per unit time, then � = 2� � , which gives � = � (2� � ). Therefore, the torque can be expressed as: � = � (2� � )� … … . . (5.19) In SI units, P usually measured in watts (1 � = 1 �. ‧�/�) and � in hertz (1 �� = 1 ���/�); Eq.

(5.19) then determines the torque T in N‧m

In U.S. Customary units with P in lb‧in./s and f in hertz, Eq. (5.19) calculates the torque T in lb‧in.

Because power in U.S. Customary units is often expressed in horsepower (1 ℎ� = 550 �� • ��/� = 396 ×10� �� • ��./���), a convenient form of Eq.(5.19) is: �(��. ��) = � (ℎ�)2������ ���� � × 396 × 10�(��. ��/���)1(ℎ�)

which simplifies to : �(��. ��) = 63.0 × 10� � (ℎ�)����� ���� �

65

(1 ℎ� = 745.7 � = 745.7�.� �� .

5.7 Statically indeterminate problems: Note: Analogous to the case of axially loaded members. Needed torques cannot be determined using static

equilibrium equations alone. Additional “compatibility” conditions are needed.

1. Draw the required free-body diagrams and write the equations of equilibrium.

2. Derive the compatibility equations from the restrictions imposed on the angles of twist. Satisfies structural

integrity (or geometric continuity) written in terms of displacements at a point in a given direction.

3. Use the torque- twist relationships in Eq.(5.15) to express the angles of twist in the compatibility equations

in terms of the torques. Angle of twist at a given X-section is the same (i.e., unique) regardless of which

segment joined at X-section is used to determine the angle of twist.

4. Solve the equations of equilibrium and compatibility for the torques.

Example 5.7

Given: A compound shaft supports several pulleys as

shown in Figure E7. Segments (1) and (4) are solid

25-mm-diameter steel [G = 80 GPa] shafts. Segments

(2) and (3) are solid 50-mm-diameter steel shafts. The

bearings shown allow the shaft to turn freely.

Find: (a) The maximum shear stress in the compound shaft.

(b) The rotation angle of pulley D with respect to pulley B.

(c) The rotation angle of pulley E with respect to pulley A.

SOLUTION:

66

67

Example 5.8

68

Given: Figure (a) shows a steel shaft (G =80GPa) of length L = 1.5 m and diameter d = 25 mm that

carries a distributed toque of intensity (torque per unit length)� = �� (�/�), where tB = 200N· m/m.

Find: (1) the maximum shear stress in the shaft; and

(2) The angle of twist.

SOLUTION:

Example 5.9

69

Given: The shaft in Figure (a) consists of a 3-in diameter

aluminum segment (G = 4×106 psi) that is rigidly joined to a

2-in diameter steel segment (G = 12×106 psi). The ends of

the shaft are attached to rigid supports, the torque � = 10 ���. �� is applied.

Find: The maximum shear stress developed in each

segment?

SOLUTION: by equilibrium:

∑�� = 0, (10 × 10�) − ��� − ��� = 0,⟹ This

problem is statically indeterminate.

Example 5.10 Given: The four rigid gears, loaded as shown in Figure (a), are

attached to a 2-in.-diameter steel shaft. Compute: The angle θ of rotation of gear A relative to gear D. Use

G = 12×106 psi for the shaft.

SOLUTION: It is convenient to represent the torques as vectors

(using the right-hand rule) on the FBDs in Fig. (b).

70

Example 5.11(P5.48): Given: Two transmission shafts-one a hollow tube with an outer diameter of 100 mm and an

inner diameter of 60 mm, the other solid with a diameter of 100 mm are each to transmit 200

kW. If both operate at f = 6 Hz, compute the highest shearing stresses in each.

SOLUTION:

159 159(200) 5.36

PT kN mf

= = = ×

4 4 6 4[(0.1) (0.06) ] 8.54 1032hJ mp -= - = ´

Thus,

3

65.3 10 (0.05) 31.03

8.54 10hh

Tc MPaJ

t -

´= = =

´

3

3 316 16(5.3 10 ) 27

(0.1)sT MPa

dt

p p´

= = =

1.15 1.15h s h st t t t= =

5.8 stress concentrations:

The derivation of the torsion formula, J

Tcmax =t , assumed a circular shaft with uniform cross-

section loaded through rigid end plates.

The use of flange couplings, gears and pulleys attached to shafts by keys in keyways, and

cross-section discontinuities can cause stress concentrations.

Experimental or numerically determined concentration factors are applied as,

71

JTcKK nommax == tt .

Here the nominal or average shear stress is obtained for the smaller diameter shaft.

5.9 Torsion of Noncircular Members:

• Previous torsion formulas are valid for

axisymmetric or circular shafts.

• Planar cross-sections of noncircular shafts

do not remain planar and stress and strain

distribution do not vary linearly

• For uniform rectangular cross-

section(table 5.1),

GabTLand,

abT

32max bf

at ==

Example 5.12: Given: A stepped shaft with a major diameter of D = 20 mm and a minor diameter of d = 16 mm

is subjected to a torque of 25 N-m. A full quarter-circular fillet having a radius of r = 2 mm is

used to transition from the major diameter to the minor diameter.

Find: The maximum shear stress in the shaft?

SOLUTION:

72

Example 5.13: Given: A 40-mm-diameter shaft contains a 10-mm-deep U-shaped groove that has a 6-mm radius

at the bottom of the groove. The maximum shear stress in the shaft must be limited to 60 MPa.

Find: If the shaft rotates at a constant angular speed of 22 Hz, determine the maximum power

that may be delivered by the shaft.

SOLUTION:

73

Example 5.14:

Given: A torque of magnitude T = 1.5 kip-in. is

applied to each of the bars shown in Fig below.

The allowable shear stress is specified as ������ = 8 ���. Find: The minimum required dimension b for

each bar.

SOLUTION: