torsion - vtctycnw01.vtc.edu.hk/con4331/8- tosion.pdfstructural analysis i chapter 4 - torsion page...

Structural Analysis I

Chapter 4 - Torsion

Page 4-1

TORSION

Torsional stress results from the action of torsional or twisting moments

acting about the longitudinal axis of a shaft. The effect of the application of

a torsional moment, combined with appropriate fixity at the supports, is to

cause torsional stresses within the shaft. These stresses, which are

effectively shear stresses, are greatest on the outer surface of the shaft.


Chapter 4 - Torsion

Page 4-2

Internal Resisting Torsional Moment (T)

When a steel shaft (a circular section straight bar) with one end fixed and the

other end free is subject to a concentrated moment Mx at the free end. This

Mx is rotating about the longitudinal axis, x-axis of the shaft.

x-axis

Mx

x

The action of this moment Mx which is sometime called torque, will then

cause a twist (torsional rotation) of the shaft. The internal torsional

moment in the member under torsional load can be found by using the

equation of equilibrium.

Mx = 0, T = Mx

x

T

Mx

x-axis

It is seen in this example that the internal torsional moment T in all

cross-sections of this member remains constant.


Chapter 4 - Torsion

Page 4-3

x

T

Mx

x-axis

Mx

T-Diagram

L

T = Mx

In the case when more than one external torsional moment (load) is acting

along the member as shown, for example

x-axis 1.5m

0.5m

1.2m

A

B

C

D

m = 5 kNm

m = 7 kNm

m = 4 kNm

1

2

3


Chapter 4 - Torsion

-4

Between C & D

T

x-axis

m 3

CD

TCD = m3 = 4 kNm

Between B & C

T

x-axis

m 3

BC

m 2

TBC = 7 – 4 = 3 kNm

Between A & B

T

x-axis

m 3

AB

m 2

m 1

TAB = 5 + 7 – 4 = 8 kNm


Chapter 4 - Torsion

Page 4-5

A B C D

8 kNm

3 kNm

-4 kNm


Chapter 4 - Torsion

Page 4-6

Torsional Shear Stress in a Circular Section

Assumptions:

i) The material is homogeneous.

ii) The material is elastic and obeys Hooke’s Law.

iii) The stress does not exceed the elastic limit or limit of

proportionality.

iv) Circular sections remain circular.

v) Plane sections remain plane.

vi) All diameters of the cross-section which are original straight

remain straight and their magnitudes do not change.

The above assumptions are based on the observation of the behaviour of

circular section shafts under torsion.

The observation also confirms that during the twist, a longitudinal straight

line OA on the surface takes up new position as OA’ while the section at the

free end rotates an angle of twist relative to the fixed end which does not

rotate.

L

A

A'

O

Mx


Chapter 4 - Torsion

Page 4-7

Geometrically,

R

rx-axis

A

C

B

D

B'

D'

a bb'

dd'

c

d

dx

dx

It is observed that the element ABCD on the surface has been distorted to

AB’CD’ due to the torsional rotation d over a distance dx. This kind of

deformation must be accompanied by shear stresses around the element.

A B

C DD'

B'

dx

BB’ = shear displacement of B

BB’/dx = shear angle at the surface.


Chapter 4 - Torsion

Page 4-8

Somewhere between the centre (x-axis) & the surface, an element abcd also

has the similar deformation.

a b

c dd'

b'

dx

r

r(dx)

bb’ = r dx = shear displacement of b

r = shear angle

On the other hand, bb’ = rd

dx

drr

Physically, G

Therefore, G

dx

dGrr

In which dx

d is the torsional rotation of the shaft per unit length.

Equation of equilibrium,

r

dA r


Chapter 4 - Torsion

Page 4-9

dArdx

dGdArT r

2

G & dx

d are the constants in this integration and dArJ 2

J = Polar moment of inertia of the circular section about its

center & 2

4RJ

dx

dGJT

GJ

T

dx

d

From dx

dGrr

Therefore, J

TR

J

Trr max and

max

max

It can be seen that torsional shear stress varies linearly from zero at the

centre to maximum at the outside fibre. The formula J

Trr is also valid

for hollow circular sections.

max

max


Chapter 4 - Torsion

Page 4-10

When a circular shaft is subjected to torsion, the elements between two

adjacent cross-sections are under pure shear.

12

From Mohr’s circle, it can be seen that the principal stresses, 21 .

21

1 2

The occurrence of max1 at 45o to longitudinal axis causes cracking of

brittle shaft because brittle material is very weak in tension, like cast iron &

chalk.

mx

mx 1

1


Chapter 4 - Torsion

Page 4-11

If this shaft is made of mild steel which is ductile material strong in tension

& compression but is comparatively weak in shear. This shaft will

eventually fail in shear when the external torsional load keeps on increasing.

The failure section is the cross-section of the shaft perpendicular to its

longitudinal axis.

Ductile Material


Chapter 4 - Torsion

Page 4-12

Torsional Rotation

L

A

A'

O

Mx

The deformation of a shaft under pure torsion is the torsional rotation of a

cross-section.

From GJ

T

dx

d

L

xGJ

TL

GJ

Tdx

0

= therefore

The is then the torsional rotation of the free end as the fixed end does not

rotate.


Chapter 4 - Torsion

Page 4-13

Summary of Formulae for Torsion of Circular Section

Maximum Shear Stress:

J

TRmax

Shear Stress at any radial position r:

J

Trr

Polar Moment of Inertia for Solid Circular Section:

232

44 RDJ

D

Polar Moment of Inertia for Hollow Circular Section:

232

4444ioio RRDD

J

Do

iD

Angle of Twist for Circular Section:

GJ

TL


Chapter 4 - Torsion

Page 4-14

Example 1

A solid steel shaft 50 mm in diameter is subjected to a torsional moment of 2

kNm. Determine the maximum shear stress in the shaft.

Solution:

The maximum shear stress on the periphery of the shaft is

2

4

6

max N/mm 5.81

32

50

25102

xx

J

TR

Example 2

Determine the angle of twist between two sections 500 mm apart in a steel

rod having a diameter of 25 mm with a torque of 300 Nm is applied. Shear

modulus G of steel is 80 GPa.

Solution:

The angle of twist of the steel rod is

rad 0489.0

32

251080

500103004

3

3

x

x

GJ

TL


Chapter 4 - Torsion

Page 4-15

Example 3

Find the torque that a solid circular shaft with a diameter of 150 mm can

transmit if the maximum shearing stress is 50 MPa. What is the angle of

twist per meter of length if the shear modulus G is 80 GPa?

Solution:

The polar moment of inertia J is given by,

4744

mm 1097.432

150

32x

DJ

Nmm 1013.3375

)1097.4(50 6

7max

max xx

R

JT

J

TR

T = 33.13 kNm

The angle of twist is given by,

rad 1033.81097.41080

10001013.33 3

73

6

x

xx

x

GJ

TL


Chapter 4 - Torsion

Page 4-16

Example 4

What torque should be applied to the end of the steel shaft to produce a twist

of 3o? Use the value G = 80 GPa for the shear modulus of steel shaft. The

outside diameter Do = 80 mm and the inside diameter Di = 70 mm.

1.5m

Solution:

G = 80x103 N/mm2 L = 1.5 m = 1500 mm

rad 05236.0360

32

o

o

4644

mm 10664.132

7080xJ

Nmm 10647.4

1500

05236.010664.11080 6

63

xxx

L

GJT

GJ

TL

T = 4.647 kNm


Chapter 4 - Torsion

Page 4-17

Example 5

The preliminary design of a large shaft connecting a motor to a generator

calls for the use of a hollow shaft with inner and outer diameters of 125 mm

and 175 mm respectively. Knowing that the allowable shearing stress is 100

MPa, determine the maximum torque which may be transmitted (a) by the

shaft as designed, (b) by a solid shaft of the same weight, (c) by a hollow

shaft of the same weight and 250 mm outside diameter.

Solution:

(a)

(b)

(c)

125

175

d1

d2

250

(a) Hollow shaft as Designed

47

44

mm 10811.632

125175xJ

10811.6

2/175100

7maxx

T

J

TR

kNm 77.84 Nmm 1084.77 6 xT


Chapter 4 - Torsion

Page 4-18

(b) Solid shaft

Area of hollow shaft = Area of solid shaft

mm 5.122 4

1251754

12

122 dd

4744

1 mm 10211.232

5.122

32x

dJ

10211.2

2/5.122100

7maxx

T

J

TR

kNm 36.1 Nmm 1010.36 6 xT

(c) Enlarged hollow shaft

Area of hollow shaft = Area of enlarged hollow shaft

mm 9.217 2504

1251754

22

2222 dd

4844

mm 10622.132

9.217250xJ

10622.1

2/250100

8maxx

T

J

TR

kNm 129.76 Nmm 1076.129 6 xT


Chapter 4 - Torsion

Page 4-19

Example 6

A solid alloy shaft of 100 mm diameter is to be friction-welded

concentrically to the end of a hollow steel shaft of the same external

diameter. Find the internal diameter of the steel shaft if the angle of twist

per unit length is to be 70% of that of the alloy shaft.

What is the maximum torque that can be transmitted if the limiting shear

stresses in the alloy and the steel are 75 MPa and 100 MPa respectively?

Given that Gsteel = 2Galloy.

L La s

D = Da

ds

s

T

T

steel

alloy

Solution:

Talloy = Tsteel = T

and a

a

s

s

LL

7.0

Hence, aa

a

ss

s

JG

T

JG

T 7.0

Since Ts = Ta and Gs = 2Ga

Ja = 0.7*2*Js

32

27.032

444ssa dD

xxD

1004 = 1.4*(1004 – ds4)

ds = 73.1 mm


Chapter 4 - Torsion

Page 4-20

The torque that can be carried by the alloy is

kNm 14.73 Nmm 1073.1450

32

10075

6

4

max

xR

JT

The torque that can be carried by the steel is

kNm 14.03 Nmm 1003.14

50

32

1.73100100

6

44

max

xR

JT

Hence the maximum allowable torque is 14.03 kNm


Chapter 4 - Torsion

Page 4-21

Example 7

M1 = 10 kNm, M2 = 8 kNm and GJ = 100 kNm2

A is a fixed end which does not rotate.

Find the torsional rotations of sections B & C.

Solution:

AB C

M =10 kNm M =8 kNm1 2

1.5m 1.2m

T =8 kNm

T =2 kNmAB

BC

A

B C

Plan View of the Shaft

1 BC

2

Since A does not rotate, section B will rotate an angle 1 and it is a rotation

in a absolute sense

radkNm

mxkNm03.0

100

5.1221


Chapter 4 - Torsion

Page 4-22

If section B did not rotate, then section C rotates relative to B an angle of

radx

BC 096.0100

2.18

Since section B does not rotate, therefore section C rotates, relative to A, an

angle of

2 = 1 - BC = 0.03 – 0.096 = -0.066 rad.


Chapter 4 - Torsion

Page 4-23

Example 8

A B

RR

L

AB

1 2L L

C

T

A circular shaft with two ends fixed is subjected to a torsional moment T in

its span as shown. Find the torsional reactions RA & RB and draw a torsional

moment diagram for the shaft.

Solution:

AC

B C

Mx = 0, RA + RB – T = 0

Geometrical compatibility

CA = CB = C

Physically, GJ

LT

GJ

LTBCCA

2211 and

Equation of compatibility, GJ

LT

GJ

LT 2211

Therefore, 1

221

L

LTT


Chapter 4 - Torsion

Page 4-24

Note that L1 + L2 = L, RA = T1 and RB = T2

TTL

LT 2

1

22

TL

LT

1

22 1

TL

LT

12

TL

LTRB 1

2

TL

LTRA 2

1

A C

B

T

T

1

2


Chapter 4 - Torsion

Page 4-25

Strain Energy in Torsion

L

A

A'

O

Mx

dx

d

xext MW2

1

intWWE extstrain

Within elastic limit, dTdW 2

1int

GJ

Tdxd

GJ

LT

GJ

dxTW

L

x22

2

0

2

int

therefore,

GJ

LTEstrain

2

2

torsion - vtctycnw01.vtc.edu.hk/con4331/8- tosion.pdfstructural analysis i chapter 4 - torsion page...

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