Mechanical Engineering Principles

10

Twisting of shafts

At the end of this chapter you should be

able to:

• appreciate practical applications where
  torsion of shafts occur


• prove that

τ

r

=

T

J

=

Gθ

L

• calculate the shearing stressτ due to a
  torque,T


• calculate the resulting angle of twist,θ,
  due to torque,T


• calculate the power that can be transmitted
  by a shaft

10.1 Introduction

The torsion of shafts appears in a number of differ-

ent branches of engineering including the following:

(a) propeller shafts for ships and aircraft

(b) shafts driving the blades of a helicopter

(c) shafts driving the rear wheels of an automobile

(d) shafts driving food mixers, washing machines,
tumble dryers, dishwashers, etc.

If the shaft is overstressed due to a torque, so that

the maximum shear stress in the shaft exceeds the

yield shear stress of the shaft’s material, the shaft

can fracture. This is an undesirable phenomenon and

normally, it should be designed out; hence the need

for the theory contained in this chapter.

10.2 To prove that

τ

r

=

T

J

=

Gθ

L

In the formula

τ

r

=

T

J

=

Gθ

L

:

τ=the shear stress at radius r

T=the applied torque

J=polar second moment of area of the shaft

(note that for non-circular sections,Jis the

torsional constant and not the polar second

moment of area)

G=rigidity or shear modulus

θ=angle of twist, in radians, over its lengthL

Prior to proving the above formula, the following

assumptions are made for circular section shafts:

(a) the shaft is of circular cross-section

(b) the cross-section of the shaft is uniform along

its entire length

(c) the shaft is straight and not bent

(d) the shaft’s material is homogeneous (i.e. uni-

form) and isotropic (i.e. exhibits properties

with the same values when measured along dif-

ferent axes) and obeys Hooke’s law

(e) the limit of proportionality is not exceeded and

the angles of twist due to the torque are small

(f) plane cross-sections remain plane and normal

during twisting

(g) radial lines across the shaft’s cross-section

remain straight and radial during twisting.

Consider a circular section shaft, built-in at one end,

namelyA, and subjected to a torqueTat the other

end, namelyB, as shown in Figure 10.1.

Letθ be the angle of twist due to this torqueT,

where the direction ofT is according to the right

hand screw rule. N.B. The direction of a couple,

according to the right hand screw rule, is obtained

by pointing the right hand in the direction of the

double-tailed arrow and rotating the right hand in a

clockwise direction.

From Figure 10.1, it can be seen that:γ=shear

strain, and that

γL=Rθ ( 10. 1 )

providedθis small.