Question

A steel beam (G=77 GPa) is applied with torques as shown below, calculate the
maximum angle of twist.
2 kN-m
4 kN-m
50 mm
30 mm
2 m
2 m​

Answer 1

Answer:

Given G = 77 GPa

Outer diameter of shaft = 30 mm

Inner diameter of shaft = 20 mm

Length of shaft = 1.8 m

a)

Solution

d = 30 mm

Therefore radius c = 15 mm

From Torsional equation

T/J = G0/I

For solid steel shaft angle of twist at 0 = TI/GJ

= (250*1.8)/(77*109*3.14/2*154*10-12)

= 0.073491*(180/3.14)

= 4.210

Hence angle of twist = 4.210

b)

Solution

d = 30 mm

c = 15 mm

Hence inner radius ci = 10 mm

Polar moment of inertia of hollow cylinder J = 3.14/2(c4-c4i)

Hence; from torsional equation, angle of twist 0H = (TI/GJ)

= - (TI)/(G*3.14/2(c2-c4i)

= (250*1.8*2)/(77*109*3.14*(154-104)*10-12

= 5.2470

Therefore the angle of twist is 5.2470