sa1 torsion - ch 4a - vtctycnw01.vtc.edu.hk/cbe2027/sa1_torsion.pdf · 2009-09-21 · when a steel...

CBE2027 Structural Analysis I Chapter 4 – Torsion

HD in Civil Engineering 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.



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.



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


HD in Civil Engineering -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



A B C D

8 kNm

3 kNm

-4 kNm



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 φ relative to the fixed end which does not

rotate.

L

A

A'

O

Mxφ



Geometrically,

R

rx-axis

A

C

B

D

B'

D'

a bb'

dd'

c

dφ

dx

dx

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

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

deformation must be accompanied by shear stresses around the element.

A B

C DD'

B'

dx

γ

BB’ = shear displacement of B

BB’/dx = shear angle γ at the surface.



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

has the similar deformation.

a b

c dd'

b'

dx

γr

γ r(dx)

bb’ = γr dx = shear displacement of b

γr = shear angle

On the other hand, bb’ = rdφ

dx

dr

r

φγ =

Physically, G

τγ =

Therefore, γτ G=

dx

dGr

r

φτ =

In which dx

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

Equation of equilibrium,

r

dA τr



∫∫ ⋅=⋅= dArdx

dGdArT

r

2φτ

G & dx

dφ are the constants in this integration and ∫ ⋅= dArJ

2

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

center & 2

4R

Jπ

=

dx

dGJT

φ=

GJ

T

dx

d=

φ

From dx

dGr

r

φτ =

Therefore, J

TR

J

Trr

== max and ττ

τmax

τmax

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

=τ is also valid

for hollow circular sections.

τmax

τmax



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

adjacent cross-sections are under pure shear.

τ

τ

τ

τ

τ

σ

σσ 12

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

τ

τ

τ

τ

σ

σ σ

σ21

1 2

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

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

chalk.

mx

mx σσ1

1



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



Torsional Rotation

L

A

A'

O

Mxφ

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

cross-section.

From GJ

T

dx

d=

φ

∫=

=L

xGJ

TL

GJ

Tdx

0

= thereforeφφ

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

rotate.



Summary of Formulae for Torsion of Circular Section

Maximum Shear Stress:

J

TR=maxτ

Shear Stress at any radial position r:

J

Trr

=τ

Polar Moment of Inertia for Solid Circular Section:

232

44RD

Jππ

==

D

Polar Moment of Inertia for Hollow Circular Section:

( ) ( )

232

4444ioio

RRDDJ

−=

−=

ππ

Do

iD

Angle of Twist for Circular Section:

GJ

TL=φ



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=

⋅==

πτ

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

=⋅

⋅

⋅==

πφ

x

x

GJ

TL



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 =

⋅==

ππ

Nmm 1013.3375

)1097.4(50 6

7max

max xx

R

JT

J

TR=

⋅=

⋅=⇒=

ττ

∴T = 33.13 kNm

The angle of twist is given by,

( )

rad 1033.81097.41080

10001013.33 3

73

6−=

⋅

⋅== x

xx

x

GJ

TLφ



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

πφ

( ) 4644

mm 10664.132

7080xJ =

−=

π

( ) ( )Nmm 10647.4

1500

05236.010664.11080 6

63

xxx

L

GJT

GJ

TL=

⋅⋅==⇒=

φφ

∴ T = 4.647 kNm



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 =

−=

π

( )

10811.6

2/175100

7maxx

T

J

TR=⇒=τ

kNm 77.84 Nmm 1084.77 6 ==⇒ xT



(b) Solid shaft

Area of hollow shaft = Area of solid shaft

( ) mm 5.122 4

1251754

12

122 =⇒⋅=− dd

ππ

( ) 4744

1 mm 10211.232

5.122

32x

dJ =

⋅==

ππ

( )

10211.2

2/5.122100

7maxx

T

J

TR=⇒=τ

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

ππ

( ) 4844

mm 10622.132

9.217250xJ =

−=

π

( )

10622.1

2/250100

8maxx

T

J

TR=⇒=τ

kNm 129.76 Nmm 1076.129 6 ==⇒ xT



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

d s

s

T

T

steel

alloy

Solution:

Talloy = Tsteel = T

and a

a

s

s

LL

φφ⋅= 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

dDxx

D −=

⋅ ππ

1004 = 1.4*(1004 – ds4)

⇒ ds = 73.1 mm



The torque that can be carried by the alloy is

kNm 14.73 Nmm 1073.1450

32

10075

6

4

max ==

⋅⋅

=⋅

= xR

JT

πτ

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

πτ

Hence the maximum allowable torque is 14.03 kNm



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

ΦΦ

1 BC

Φ2

Since A does not rotate, section B will rotate an angle φ1 and it is a rotation

in a absolute sense

( ) ( )

radkNm

mxkNm03.0

100

5.1221 ==φ



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

( ) ( )

radx

BC096.0

100

2.18==φ

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

angle of

φ2 = φ1 - φBC = 0.03 – 0.096 = -0.066 rad.



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 CΦ

∑Mx = 0, RA + RB – T = 0

Geometrical compatibility

φCA = φCB = φC

Physically, GJ

LT

GJ

LTBCCA

2211 and == φφ

Equation of compatibility, GJ

LT

GJ

LT 2211 =

Therefore, 1

221

L

LTT =



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

TTL

LT =+∴ 2

1

22

TL

LT =

+⇒

1

22 1

TL

LT =

⇒

12

TL

LTR

B⋅== 1

2

TL

LTR

A⋅== 2

1

A C

B

T

T

1

2



Tutorial 4 (Torsion)

Q1. (a) Determine the torque T which causes an angle of twist of 3o in

the hollow cylindrical steel shaft as shown below (G = 77 GPa).

(b) Determine the angle of twist caused by the same torque T in a

solid cylindrical shaft of the same cross-sectional area.

45mm30 mm

2.4

m

T

Q2. A solid steel shaft is loaded as shown below. Using G = 83 GPa,

determine the required diameter of the shaft if the shearing stress is

limited to 60 MPa and the angle of rotation at the free end is not to

exceed 4o.

2.5 m 2.5 m

750 Nm 1200 Nm



Q3. A compound shaft consisting of a steel segment and an aluminum

segment is acted upon by two torques as shown. Determine the

maximum permissible value of T subject to the following conditions:

τst ≤ 100 MPa, τal ≤ 60 MPa, and the angle of rotation of the free end

is limited to 5o. For steel, G = 83 GPa and for aluminum, G = 28

GPa.

50 mm dia.

steel40 mm dia.

aluminum

1000 mm 600 mm

T 2T

Q4. A circular bar ABC, 3 m long, is rigidly fixed at its ends A and C.

The portion AB is 1.8 m long and of 50 mm diameter and BC is 1.2

m long and of 25 mm diameter. If a twisting moment of 1 kNm is

applied at B, determine the values of the resisting moments at A and

C and the maximum stress in each section of the shaft. What will be

the angle of twist of each portion? For the material of the shaft G =

80 GPa.

ABC

1.2m 1.8m

25501 kNm



Q5. The solid cylinders AB and BC are bonded together at B and

attached to fixed supports at A and C. Knowing that AB is made of

aluminum (G = 26 GPa) and BC of brass (G = 39 GPa), determine for

the loading (acted at B) shown:

(a) the reaction at each support;

(b) the maximum shearing stress in AB;

(c) the maximum shearing stress in BC.

125 mm

75 mm

300 mm 200 mm

12.5 kNm

A

B

C

sa1 torsion - ch 4a - vtctycnw01.vtc.edu.hk/cbe2027/sa1_torsion.pdf · 2009-09-21 · when a steel...

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