Question

Show that the hallow cylinder has a greater torsional rigidity than a solid cylinder of mass length material​

Answer 1

Answer:

Torsional rigidity of solid shaft is more than the torsional rigidity of the hollow shaft

Hence more torque is required in the solid shaft to produce the same deformation.

Answer 2

Answer:

For a circular shaft the shear stress induced due to torque is directly proportional to the radius. Hence, the maximum resistance to torque is offered by the outer layers of the shaft. The inner portion of the solid shaft which is not effective to resist the torque may be removed without considerable reduction in the torque carrying capacity of the shaft. Imagine a shaft made by removing some material surrounding the axis (thus making it hollow) and adding this removed material to the outer surface. This shaft definitely can resist more torque, since the outer layers are more effective in offering resistance. Hence, for the same cross sectional area a hollow shaft is stronger than solid shaft and can resist more torque.

To prove a hollow shaft is stronger than solid shaft, we have to show that hollow shaft can transmit more torque by keeping the maximum shear stress induced on both to same value while transmitting these torque.

The torsion equation for circular shafts considering the strength of the shaft is given by

TJ=τmaxR

Rearranging we get,

T=JRτmax

Here, JR is known as polar modulus. Hence the torque carrying capacity of a circular shaft depends on the polar modulus for the same maximum shear stress ( same material)

Hence if we have to prove a shaft is stronger, which are made up of same material, we have to only show that it has higher polar modulus.

Proof

The hollow shaft and solid shaft are having same length and mass, hence their cross sectional areas must be equal.

Equating the cross sectional areas,

π4d2=π4(d2o−d2i)

d2=(d2o−d2i)

d2=d2o(1−k2)

d=do(1−k2)12−−−−−−−(1)

where k is the diameter ratio.

We have to prove that polar modulus of hollow shaft is higher than solid shaft. That is,

π16d3o(1−k4)>π16d3

d3o(1−k4)>d3

Substituting the result (1) we get,

d3o(1−k4)>d3o(1−k2)32

(1−k2)(1+k2)>(1−k2)(1−k2)12

(1+k2)>(1−k2)12

LHS is always greater than RHS for any value of k (k is the diameter ratio and its value lies between 0 & 1). Hence a hollow shaft is stronger than a solid shaft for the same length, weight and material.