Let us draw the free-body diagram of the shaft as shown in Figure 9.53:

In the figure, we isolate a differential element on the shaft outer surface aligned along the axis of the shaft [refer to Figure 9.53(a)]. Clearly due torque T, the element will be in a state of pure shear and if the element is rotated along 20°, it will be subjected to the stress state as shown in Figure 9.53(b).
From the Mohr’s circle of stresses, we can say that

\sigma=\tau \sin 40^{\circ} \quad \text { where } \quad \tau=\frac{16 T}{\pi d^3}

Obviously, ∈_{x^{\prime} x^{\prime}}=∈_1 is the gauge reading and it is given by

∈_{x^{\prime} x^{\prime}}=∈_1=\frac{1}{E}\left[\sigma_{x^{\prime} x^{\prime}}-ν\left(\sigma_{z^{\prime} z^{\prime}}+\sigma_{y^{\prime} y^{\prime}}\right)\right]

For the plane-stress condition, \sigma_{z^{\prime} z^{\prime}} = out of plane stress = 0. Therefore,

∈_{x^{\prime} x^{\prime}}=\frac{1}{E}(\sigma+ν \sigma) \quad\left(\text { as } \sigma_{y^{\prime} y^{\prime}}=-\sigma\right)

or          ∈_{x^{\prime} x^{\prime}}=\frac{1+ν}{E} \sigma=\frac{\sigma}{2 G} \quad\left[\text { as } G=\frac{E}{2(1+ν)}\right]

hence,

\sigma=2 G \in_{x^{\prime} x^{\prime}}=2 G ∈_1

or        \left\lgroup \frac{16 T}{\pi d^3}\right\rgroup \sin 40^{\circ}=2 G \in_1 \quad\left[\text { as } \sigma=\tau \sin 45^{\circ} \text { and } \tau=16 T / \pi d^3\right]

Therefore,

T=\left\lgroup \frac{\pi d^3}{8} \right\rgroup \frac{G\in_1}{\sin 40^{\circ}}

=\frac{\pi(0.096)^3}{8} \frac{(27)\left(10^9\right)(400)\left(10^{-6}\right)}{\sin 40^{\circ}}  N m

or      T = 5.8375 kNm

Therefore, the required torque applied on the shaft is 5.84 kN m.