A bar is shown in (a) in pure shear due
to torsion. Pure shear means that the
only forces acting on the element are
shear stresses. The directions of these
stresses depend on the direction of the
applied torques. Assuming the right
hand rule end cb will rotate in the
sense shown in (b). The shear
stresses shown in (b) are by sign
convention positive.
A shear stress on a positive face is
positive if it acts in the direction of
one of the positive axes.
To determine the state of stress as a function of angle we will cut an oblique section
at an angle  this results in a triangle with AB unchanged but all other lengths
altered. The state of pure shear is shown in (a). In (b) we take an oblique section
and resolve the resultant force into two components  and . The resolved
components can be rewritten as shown in (c) where we have equated the forces and
used the relationship that F= A= .L.t since t is constant for the element A=A0.L
(NB Secant=h/a)
  A0 sec   A0 sin   A0 tan  cos 
   2 sin  cos    sin 2
  A0 sec   A0 cos  A0 tan  sin 
2
2
    cos   sin     cos 2
    cos 2
    sin 2
The stress as a function of angle is shown below
in the figure. Maximum shear stress occurs at
0°±90° and the maximum normal stresses at
±45°.
At 45° there are equal and opposite normal stresses acting on the element.
This often leads to torsion failures at 45° to the torsion axis.
To determine the deformation of the bar we need to consider shear strains and
normal strains. The shear distortion tends to deform a rectangle to a parallelepiped.
The normal stresses at 45° will elongate the section but lateral contraction will also
occur due to poisson contraction. Similarly the compressive stress at 135° will lead
to a compression of the element but will also extend the element at 45°.
The total normal strain in the 45° direction is given by
 max 

E


E


E
(1  )
After deformation
Lbd  2h(1  max )
Using cosine rule c2=a2+b2-2abcos 


L bd  h  h  2h cos   
2



2
1   max   1  cos   
2

2
2
2
2
2
1  2 max   max
 1  sin  
For small strains
 max 

2


2G
Earlier we derived
 max 

E


E
Therefore
E
G
21   


E
(1  )
Work Done = Force x distance = torque x angle moved
W  T
Power =Rate of Work Done
dW
d
P
T
 T
dt
dt
 Is in rads.s-1
Very often rate is given in r.p.m and Power is given in hp
1 hp = 550 ft-lb/s.
A solid steel shaft AB is shown in the Figure. It is to
be used to transmit 5 hp from the motor M to which
it is attached. If the shaft rotates at =175 rpm and
the steel has an allowable shear stress of
allow=14.5 ksi, determine the required diameter of
the shaft to the nearest 1/8 in.
A solid steel shaft AB is shown in the Figure. It is to
be used to transmit 5 hp from the motor M to which
it is attached. If the shaft rotates at =175 rpm and
the steel has an allowable shear stress of
allow=14.5 ksi, determine the required diameter of
the shaft to the nearest 1/8 in.
2 

P  T  5 * 550  2750 ft.lb / s   175
  18.33rad / s
 60 
2750
J   c4 
T
T
 150.1 ft.lb
   
18.33
c 2  c   allow
1/ 3
 2T 

c  
 allow 
1/ 3
 2(150.1)(12) 

 
  (14500) 
 0.429in
Therefore d of the shaft should be greater than 0.858in i.e.
7/8=0.875in
A solid shaft ABC of 50 mm diameter is driven at A by a motor that transmits 50 kW to
the shaft at 10 Hz. The gears at B and C drive machinery requiring power equal to 35
kW and 15 kW respectively. Compute the maximum shear stress in the shaft and the
angle of twist ac between the motor at A and the gear at C. (Use G=80 GPa).