Considering Figure 2.6, find out the maximum shear stress and angle of twist at B where a torque of 4 kN.m is applied as shown. Position AB is solid and made up of aluminium with G = 28 GPa, whereas position BC is hollow and made up of steel with G = 84 GPa. Both ends of the stepped shaft are fixed

Question

Considering Figure 2.6, find out the maximum shear stress and angle of twist at B where a torque of 4 kN.m is applied as shown. Position AB is solid and made up of aluminium with G = 28 GPa, whereas position BC is hollow and made up of steel with G = 84 GPa. Both ends of the stepped shaft are fixed to walls.

Accepted Answer

Let the resistive torques developed at the two rigid supports be [latex] T_1 ext { and } T_2 [/latex] as shown in Figure 2.7.

Now, as equilibrium prevails, we can write

[latex] T_1+T_2=4000 Nm [/latex] (1)

This is the only equilibrium equation that can be formulated. But, it is clear that there are two unknowns. Hence, it is a statically indeterminate system. So, what we need is another equation. This compatibility equation can be written from the consideration of angular deformation.
If we approach from the left end, we get a torque [latex] T_1 [/latex] at A and an equal and opposite resistive torque at B.

Similarly, if we approach from the right, we get [latex] T_2 [/latex] to be applied at C and resistive [latex] T_2 [/latex] at B, as shown in Figure 2.8(b). For the left as well as the right portion, angle of twist of B with respect to the respective fixed ends (A and C) must be same. So, [latex] heta_1 [/latex] (angle of twist of B with respect to A) [latex] = heta_2 [/latex] (angle of twist of B with respect to C) or

[latex] \begin{aligned} \frac{T_1 L_1}{G_1 J_1} & =\frac{T_2 L_2}{G_2 J_2} \ \frac{T_1(1.2) imes 1000}{28 imes 10^3 imes \frac{\pi}{32} imes 45^4} & =\frac{T_2(1.8) imes 1000}{84 imes 10^3 imes \frac{\pi}{32}\left(60^4-30^4 ight)} \end{aligned} [/latex]

or [latex] T_1=0.187 T_2 [/latex] (2)

Solving Eqs. (1) and (2), we get [latex] T_1=630 Nm ext { and } T_2=3370 Nm [/latex]. Now, maximum shear stress in left portion, which is a solid shaft, is

[latex] au_{\max }=\frac{630 imes 22.5 imes 1000}{\frac{\pi}{32} imes 45^4}=35.2 MPa [/latex]

Similarly, maximum shear stress in right portion, which is a hollow shaft, is

[latex] au_{\max }=\frac{3370 imes 1000 imes \frac{60}{2}}{\frac{\pi}{32}\left(60^4-30^4 ight)}=85 MPa [/latex]

So, the maximum shear stress is 85 MPa and angle of twist at B,

[latex] heta =\frac{T_1 L_1}{G_1 J_1}=\frac{T_2 L_2}{G_2 J_2} [/latex]

Considering any one of the two portions (say left portion), we get

[latex] heta=\frac{630 imes 1000 imes 1.2}{280 imes 10^3 imes \frac{\pi}{32} imes 45^4}=0.675 imes 10^{-3} rad =0.039^{\circ} [/latex]

Therefore, angle of twist at B is 0.039° and maximum shear stress is 85 MPa.

Question 2.5: Considering Figure 2.6, find out the maximum shear stress an...

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