Lecture mechanics of materials chapter three mechanical properties of materials

August 2012 5-1Torsion of Shafts the largest cross-sectional dimension used in transmitting torque from one plane to another.. August 2012 5-3Internal Torque 5.1 equiv-alency between she

August 2012 5-1

Torsion of Shafts

the largest cross-sectional dimension used in transmitting torque from one plane to another

Learning objectives

and analysis of Torsion of circular shafts

and the surface on which it acts

C5.1 Three pairs of bars are symmetrically attached to rigid discs at

August 2012 5-3

Internal Torque

5.1

equiv-alency between shear stress and internal torque on a cross-section

T ρd V

A∫ ρτxθd A

A∫

C5.2 A hollow titanium (GTi = 36 GPa) shaft and a hollow

Deter-mine the equivalent internal torque acting at the cross-section

Fig C5.2

dTi

dAl

x

θ

di Titanium Aluminum

ρ

August 2012 5-5

Theory for Circular Shafts

Theory Objective

internal torque T

torque T.

T2 T

x2

x z

y

r ␪

Kinematics

Assumption 1 Plane sections perpendicular to the axis remain plane during

deformation (No Warping) Assumption 2 On a cross-section, all radials lines rotate by equal angle during

deformation.

Assumption 3 Radials lines remain straight during deformation.

Assumption 4 Strains are small

Deformed Grid Original Grid

␾

A 1

B 1

A o

B o

A o ,B o —Initial position

A 1 ,B 1 —Deformed position

Δx x

y

z

B1

γxθ

C

ρ

Δφ

γxθ

ρ

γmax

γxθ

ρ R

x d

dφ

=

August 2012 5-7

Material Model

Assumption 5 Material is linearly elastic.

Assumption 6 Material is isotropic.

Sign Convention

x d

dφ

=

x

τxθ

τθx

θ

Failure surface in wooden shaft due to τθx Failure surface in aluminum shaft due to τxθ

x

Positive T

Outward normal

Outward normal

Positive τxθ Positive τxθ

Positive T

Torsion Formulas

Assumption 7 Material is homogenous across the cross-section

Assumption 8 Material is homogenous between x1 and x2

Assumption 9 The shaft is not tapered.

Assumption 10 The external (hence internal) torque does not change with x

between x1 and x2.

x d

dφd A

A∫ d dφx Gρ2d A

A∫

x d

GJ

-=

2

32

-D4

ρ

τxθ

τmax

J

-=

φ1

φ2

x1

x2

∫

GJ

-=

August 2012 5-9

Two options for determining internal torque T

outward normal of the imaginary cut on the free body diagram

Direction of τxθ can be determined using subscripts.

Positive is counter-clockwise with respect to x-axis.

is positive counter-clockwise with respect to x-axis

direc-tion to equilibrate the external torques

Direction of τxθ must be determined by inspection.

Direction of must be determined by inspection.

Direction of must be determined by inspection.

Torsional Stresses and Strains

zero Shear strain can be found from Hooke’s law

φ

φ

Torsional Shear Stress

C5.3 Determine the direction of shear stress at points A and B (a) by inspection, and (b) by using the sign convention for internal torque

Class Problem 1

C5.4 Determine the direction of shear stress at points A and B (a)

by inspection, and (b) by using the sign convention for internal torque

x

y

A

A

B

x

x

x

y

B

B

A

A T

x

x

August 2012 5-11

C5.5 Determine the internal torque in the shaft below by making imaginary cuts and drawing free body diagrams

A B

C D

0.5 m

1.0 m

0.4 m

10 kN-m

12 kN-m

18 kN-m

20 kN-m

Torque Diagram

crosses the external torque from left to right

A template is a free body diagram of a small segment of a shaft created

by making an imaginary cut just before and just after the section where the external torque is applied

C5.6 Determine the internal torque in the shaft below by drawing the torque diagram

Template 2 Equation Template 1 Equation

A B

C D

0.5 m

1.0 m

0.4 m

10 kN-m

12 kN-m

18 kN-m

20 kN-m

August 2012 5-13

C5.7 A solid circular steel (Gs = 12,000 ksi) shaft BC is securely attached to two hollow steel shafts AB and CD as shown Determine: (a) the angle of rotation of section at D with respect to section at A (b) the maximum torsional shear stress in the shaft (c) the torsional shear stress

at point E and show it on a stress cube Point E is on the inside bottom surface of CD

Statically Indeterminate Shafts

we have only on moment equilibrium equation

with respect to the left wall is zero

reac-tion torque of the left (or right) wall Add all the relative rotareac-tions and equate to zero to obtain reaction torque

C5.8 Two hollow aluminum (G = 10,000 ksi) shafts are securely fastened to a solid aluminum shaft and loaded as shown Fig C5.8 Point

E is on the inner surface of the shaft If T= 300 in-kips in Fig C5.8, Determine (a) the rotation of section at C with respect to rotation the wall

at A (b) the shear strain at point E

Fig C5.8

T

4 in 2 in

24 in

36 in

24 in

D E A

August 2012 5-15

C5.9 Under the action of the applied couple the section B of the two tubes shown Fig C5.9 rotate by an angle of 0.03 rads Determine (a) the magnitude maximum torsional shear stress in aluminum and copper (b) the magnitude of the couple that produced the given rotation

Fig C5.9

aluminum

copper A

B

F

F