PPB 25403 Strength of

Materials

Lecture 5: Torsion

Learning Outcomes

Torsion twisting of an object due to an applied torque

Torsional Deformation of Circular Shaft

Torsion Formula

Angle of Twist

Transmission of Power

Torsional Deformation of a Circular Shaft

Torque is a moment that twists a member about its

longitudinal axis.

If the angle of rotation is small, the length of the shaft

and its radius will remain unchanged.

The Torsion Formula

When material is linear-elastic, Hooke’s law applies.

A linear variation in shear strain leads to a

corresponding linear variation in shear stress

along any radial line on the cross section.

J

Tp

J

Tcor max

= maximum shear stress in the shaft

= shear stress

= resultant internal torque

= polar moment of inertia of cross-sectional area

= outer radius of the shaft

= intermediate distance

max

TJc

p

The Torsion Formula

If the shaft has a solid circular cross section,

If a shaft has a tubular cross section,

4

2cJ

44

2io ccJ

Example 5.2The solid shaft of radius c is subjected to a torque T. Find the fraction of T that is

resisted by the material contained within the outer region of the shaft, which has an

inner radius of c/2 and outer radius c.

Solution:

dcdAdT 2' max

maxc

For the entire lighter-shaded area the torque is

(1) 32

152' 3

max

2/

3max cdc

T

c

c

Stress in the shaft varies linearly, thus

The torque on the ring (area) located within

the lighter-shaded region is

Using the torsion formula to determine the maximum stress in the shaft, we have

Solution:

(Ans) 16

15' TT

3max

4max

2

2

c

T

c

Tc

J

Tc

Substituting this into Eq. 1 yields

Example 5.3The shaft is supported by two bearings and is subjected to three torques. Determine

the shear stress developed at points A and B, located at section a–a of the shaft.

Solution:From the free-body diagram of the left segment,

mm 1097.4752

74J

kNmm 1250030004250 ;0 TTM x

The polar moment of inertia for the shaft is

Since point A is at ρ = c = 75 mm,

(Ans) MPa 89.11097.4

7512507J

TcB

Likewise for point B, at ρ =15 mm, we have

(Ans) MPa 377.01097.4

1512507J

TcB

Power Transmission

Power is defined as the work performed per unit of

time.

For a rotating shaft with a torque, the power is

Since , the power equation is

For shaft design, the design or geometric parameter

is

dtdTP / locity,angular veshaft where

f2rad 2cycle 1

fTP 2

allow

T

c

J

Example 5.5A solid steel shaft AB is to be used to transmit 3750 W from the motor M to which it

is attached. If the shaft rotates at w =175 rpm and the steel has an allowable shear

stress of allow τallow =100 MPa, determine the required diameter of the shaft to the

nearest mm.

Solution:The torque on the shaft is

Nm 6.20460

21753750 TT

TP

Since

mm 92.10100

10006.20422

2

3/13/1

4

allow

allow

Tc

T

c

c

c

J

As 2c = 21.84 mm, select a shaft having a diameter of 22 mm.

Angle of Twist

Integrating over the entire length L of the shaft, we have

Assume material is homogeneous, G is constant, thus

Sign convention is

determined by right hand rule,

L

GxJ

dxxT

0

Φ = angle of twist

T(x) = internal torque

J(x) = shaft’s polar moment of inertia

G = shear modulus of elasticity for the material

JG

TL

Example 5.8The two solid steel shafts are coupled together using the meshed gears. Determine

the angle of twist of end A of shaft AB when the torque 45 Nm is applied. Take G to

be 80 GPa. Shaft AB is free to rotate within bearings E and F, whereas shaft DC is

fixed at D. Each shaft has a diameter of 20 mm.

Solution:From free body diagram,

Nm 5.22075.0300

N 30015.0/45

xDT

F

Angle of twist at C is

rad 0269.01080001.02

5.15.2294

JG

TLDCC

Since the gears at the end of the shaft are in mesh,

rad 0134.0075.00269.015.0B

Solution:

Since the angle of twist of end A with respect to end B of shaft AB caused by the

torque 45 Nm,

rad 0716.01080010.02

24594/

JG

LT ABABBA

The rotation of end A is therefore

(Ans) rad 0850.00716.00134.0/ BABA

Example 5.10The tapered shaft is made of a material having a shear modulus G. Determine the

angle of twist of its end B when subjected to the torque.

Solution:From free body diagram, the internal torque is T.

L

ccxcc

x

cc

L

cc 122

212

Thus, at x,

4

122

2 L

ccxcxJ

For angle of twist,

(Ans) 3

223

2

3

1

2

121

2

2

0

4

122

cc

cccc

G

TL

L

ccxc

dx

G

TL

Example 5.11The solid steel shaft has a diameter of 20 mm. If it is subjected to the two torques,

determine the reactions at the fixed supports A and B.

Solution:By inspection of the free-body diagram,

(1) 0500800 ;0 Abx TTM

Since the ends of the shaft are fixed, 0/ BA

Using the sign convention,

(2) 7502.08.1

03.05.15002.0

BA

AAB

TT

JG

T

JG

T

JG

T

Solving Eqs. 1 and 2 yields TA = -345 Nm and TB = 645 Nm.

Solid Noncircular Shafts

The maximum shear stress and the angle of twist for

solid noncircular shafts are tabulated as below:

Example 5.13The 6061-T6 aluminum shaft has a cross-sectional area in the shape of an

equilateral triangle. Determine the largest torque T that can be applied to the end of

the shaft if the allowable shear stress is τallow = 56 MPa and the angle of twist at its

end is restricted to Φallow = 0.02 rad. How much torque can be applied to a shaft of

circular cross section made from the same amount of material? Gal = 26 GPa.

Solution:By inspection, the resultant internal torque at any cross section

along the shaft’s axis is also T.

(Ans) Nm 12.24102640

102.146.020 ;

46

Nm 2.177940

2056 ;

20

34

3

4

33

TT

Ga

T

TT

a

T

al

allow

allow

By comparison, the torque is limited due to the angle of twist.

Solution:

(Ans) Nm 10.33102685.142/

102.102.0 ;

Nm 06.28885.142/

85.1456 ;

34

3

4

TT

JG

TL

TT

J

Tc

al

allow

allow

For circular cross section, we have

mm 85.1460sin40402

1c ; 2 cAA trianglecircle

The limitations of stress and angle of twist then require

Again, the angle of twist limits the applied torque.

Thin-Walled Tubes Having Closed Cross Sections

Shear flow q is the product of the tube’s thickness and

the average shear stress.

Average shear stress for thin-walled tubes is

For angle of twist,

tq avg

m

avgtA

T

2

τavg = average shear stress

T = resultant internal torque at the cross section

t = thickness of the tube

Am = mean area enclosed boundary

t

ds

GA

TL

m

24

Example 5.14Calculate the average shear stress in a thin-walled tube having a circular cross

section of mean radius rm and thickness t, which is subjected to a torque T. Also,

what is the relative angle of twist if the tube has a length L?

Solution:The mean area for the tube is

2

mm rA

For angle of twist,

(Ans) 22 2

mm

avgtr

T

tA

T

(Ans) 24 32 Gtr

TL

t

ds

GA

TL

mm

Example 5.16A square aluminum tube has the dimensions. Determine the average shear stress in

the tube at point A if it is subjected to a torque of 85 Nm. Also compute the angle of

twist due to this loading. Take Gal = 26 GPa.

Solution:By inspection, the internal resultant torque is T = 85 Nm.

(Ans) N/mm 7.12500102

1085

2

23

m

avgtA

T

For average shear stress,

22 mm 250050mAThe shaded area is

Solution:

dsds

t

ds

GA

TL

m

1-4

32

33

2mm 10196.0

10102625004

105.11085

4

For angle of twist,

Integral represents the length around the centreline boundary of the tube, thus

(Ans) rad 1092.350410196.0 34

Stress Concentration

Torsional stress concentration factor, K, is used to

simplify complex stress analysis.

The maximum shear stress is then determined from the

equation

J

TcKmax

Example 5.18The stepped shaft is supported by bearings at A and B. Determine the maximum

stress in the shaft due to the applied torques. The fillet at the junction of each shaft

has a radius of r = 6 mm.

Solution:By inspection, moment equilibrium about the axis

of the shaft is satisfied

15.0202

6 ;2

202

402

d

r

d

D

(Ans) MPa 10.3020.02

020.0303.1

4maxJ

TcK

The stress-concentration factor can be determined

by the graph using the geometry,

Thus, K = 1.3 and maximum shear stress is

Inelastic Torsion

Considering the shear stress acting on an element of

area dA located a distance p from the center of the

shaft,

Shear–strain distribution over a radial line on a shaft is

always linear.

Perfectly plastic assumes the shaft will continue to twist

with no increase in torque.

It is called plastic torque.

dTA

2

2

Example 5.20A solid circular shaft has a radius of 20 mm and length of 1.5 m. The material has an

elastic–plastic diagram as shown. Determine the torque needed to twist the shaft Φ

= 0.6 rad.

Solution:The maximum shear strain occurs at the surface of

the shaft,

rad 008.002.0

5.16.0 ; max

maxL

mm 4m 004.0008.0

02.0

0016.0Y

Y

Based on the shear–strain distribution, we have

The radius of the elastic core can be obtained by

(Ans) kNm 25.1004.002.046

10754

6

336

33

YY cT