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Chapter Objectives

To determine the torsional deformation of a perfectly

elastic circular shaft.

To determine the support reactions when these

reactions cannot be determined solely from the

moment equilibrium equation.

To determine the maximum power that can be

transmitted by a shaft.

Copyright © 2011 Pearson Education South Asia Pte Ltd

1. Reading Quiz

2. Applications

3. Torsion Formula

4. Angle of Twist

5. Statically indeterminate torque-loaded members

6. Solid non-circular shafts

7. Stress concentration

8. Concept Quiz

In-class Activities


READING QUIZ

1) Given the angle of rotation is small and the

materials remain linear elastic, which

statement below is incorrect for the torsional

behavior of a long straight circular shaft.

a) Section-shape remains unchanged

b) Straight member remains straight

c) Plane section remains plane

d) End of member may wrap


READING QUIZ (cont)

2) The unit for a shaft’s polar moment of inertia

J is:

a) kPa

b) m4

c) m2

d) m3


READING QUIZ (cont)

3) Which one of the statements below is

incorrect?

Inelastic torsion of a circular shaft leads to

a) non-linear distribution of shear stress

b) plane section remaining plane

c) change of sectional shape

d) wrapping at the end of member


APPLICATIONS


APPLICATIONS (cont)


TORSION FORMULA


• Assumptions:

– Linear and elastic deformation

– Plane section remains plane and undistorted

TORSION FORMULA (cont)


• Linear distribution of stress:

• Torsion – shear relationship:

J

T

J

Tc

dAc

T

dAc

dAT

A

AA

,Similarily

max

2max

max

max

c

TORSION FORMULA (cont)


• Polar moment of inertia – For solid shaft:

– For tubular shaft:

4

0

4

0

3

0

22

2

4

1222

cJ

dddAJ

CCC

A

44

2io ccJ

EXAMPLE 1


The pipe shown in Fig. 5–12a has an inner diameter of 80

mm and an outer diameter of 100 mm. If its end is tightened

against the support at A using a torque wrench at B,

determine the shear stress developed in the material at the

inner and outer walls along the central portion of the pipe

when the 80-N forces are applied to the wrench.

EXAMPLE 1 (cont)


• The only unknown at the section is the internal torque T

• The polar moment of inertia for the pipe’s cross-sectional area is

• For any point lying on the outside

surface of the pipe,

Solutions

4644m 10796.504.005.0

2

J

mN 40

02.0803.080

;0

T

T

M y

m 05.00 c

(Ans) MPa 345.010796.5

05.0406

00

J

Tc

EXAMPLE 1 (cont)


• And for any point located on the inside surface,

• The resultant internal torque is equal but opposite.

Solutions

m 04.0 ic

(Ans) MPa 276.010796.5

04.0406

J

Tcii

ANGLE OF TWIST


• For constant torque and cross-sectional area:

dxGxJ

xTd

dxd

JG

TL

ANGLE OF TWIST (cont)


• Sign convention for both torque and angle of twist – positive if (right hand) thumb directs outward from the shaft

EXAMPLE 2


The 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.

EXAMPLE 2 (cont)


• From free body diagram,

• Angle of twist at C is

• Since the gears at the end

of the shaft are in mesh,

Solutions

Nm 5.22075.0300

N 30015.0/45

xDT

F

rad 0269.01080001.02

5.15.2294

JG

TLDCC

rad 0134.0075.00269.015.0 B

EXAMPLE 2 (cont)


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

caused by the torque 45 Nm,

• The rotation of end A is therefore

Solutions

rad 0716.01080010.02

24594/

JG

LT ABABBA

(Ans) rad 0850.00716.00134.0/ BABA

STATICALLY INDETERMINATE TORQUE-

LOADED MEMBERS


• Procedure for analysis: – use both equilibrium and compatibility equations

Equilibrium

• Draw a free-body diagram of the shaft in order to identify all the torques that act on it. Then write the equations of moment equilibrium about the axis of the shaft.

Compatibility

• To write the compatibility equation, investigate the way the shaft will twist when subjected to the external loads, and give consideration as to how the supports constrain the shaft when it is twisted.

Please refer to the website for the animation: Statically Indeterminate Torque-Loaded Members

STATICALLY INDETERMINATE TORQUE-

LOADED MEMBERS (cont)


• Express the compatibility condition in terms of the

rotational displacements caused by the reactive

torques, and then use a torque-displacement relation,

such as Φ = TL/JG, to relate the unknown torques to

the unknown displacements.

• Solve the equilibrium and compatibility equations for

the unknown reactive torques. If any of the magnitudes

have a negative numerical value, it indicates that this

torque acts in the opposite sense of direction to that

indicated on the free-body diagram.

EXAMPLE 3


The solid steel shaft shown in Fig. 5–23a has a diameter of

20 mm. If it is subjected to the two torques, determine the

reactions at the fixed supports A and B.

EXAMPLE 3 (cont)


• It is seen that the problem is statically indeterminate since there is only

one available equation of equilibrium and there are 2 unknowns

• Since the ends of the shaft are fixed, the angle of twist of one end of the

shaft with respect to the other must be zero.

Solution

(1) 0500800

0

Ab

x

TT

M

0/ BA

EXAMPLE 3 (cont)


• Using the sign convention established,

• Using Eq. 1,

• The negative sign indicates that acts

in the opposite direction of that shown

in Fig. 5–23b.

Solution

(Ans) mN 645

03.03005.18002.0

B

BBB

T

JG

T

JG

T

JG

T

mN 345 AT

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