Question 9.27: A transmission shaft with a uniformly distributed load of 10...

\text { Given } \delta_{\max .}=0.003 L \quad E=207000 N / mm ^{2} .

Step I Permissible deflection

\delta_{\max .}=0.003 L=0.003(800)=2.4 mm                   (i).

Step II Deflection at centre of span
The deflection is maximum at the centre of the span length. An imaginary force Q is assumed at the centre as shown in Fig. 9.48 (b). By symmetry, the reactions are given by

R_{A}=R_{E}=\left(3000+\frac{Q}{2}\right) .

The bending moments in the regions AB and BC are given by,

\left(M_{b}\right)_{A B}=\left(3000+\frac{Q}{2}\right) x=3000 x+\frac{Q x}{2} .

\left(M_{b}\right)_{B C}=\left(3000+\frac{Q}{2}\right) x-\frac{10}{2}(x-100)(x-100) .

=\frac{Q x}{2}-5 x^{2}+4000 x-50000 .

Differentiating partially with respect to Q,

\frac{\partial}{\partial Q}\left[\left(M_{b}\right)_{A B}\right]=\frac{x}{2}           (a).

\frac{\partial}{\partial Q}\left[\left(M_{b}\right)_{B C}\right]=\frac{x}{2}          (b).

The total strain energy stored in the shaft is given by,

U=2 U_{A B}+2 U_{B C} .

From Eq. (9.56),

U=\int \frac{\left(M_{b}\right)^{2} d x}{2 E I}                     (9.56).

U=2 \int_{0}^{100} \frac{\left[\left(M_{b}\right)_{A B}\right]^{2} d x}{2 E I_{1}}+2 \int_{100}^{400} \frac{\left[\left(M_{b}\right)_{B C}\right]^{2} d x}{2 E I_{2}} .

From Eq. (9.53),

\delta_{i}=\frac{\partial U}{\partial P_{i}}               (9.53).

\delta_{\max .}=\frac{\partial U}{\partial Q} .

+2 \int_{100}^{400}\left(\frac{2\left(M_{b}\right)_{B C}}{2 E I_{2}}\right)\left(\frac{\partial}{\partial Q}\left[\left(M_{b}\right)_{B C}\right]\right) d x .

\text { Substituting values of }\left[\left(M_{b}\right)_{A B}\right],\left[\left(M_{b}\right)_{b c}\right] \text {, (a) and } (b) in the above expression,

+\frac{2}{E I_{2}} \int_{100}^{400}\left(\frac{Q x}{2}-5 x^{2}+4000 x-50000\right)\left(\frac{x}{2}\right) d x .

Substituting (Q = 0) and integrating,

\delta_{\max .}=\frac{10^{9}}{E I_{1}}+\frac{48.375\left(10^{9}\right)}{E I_{2}}                 (ii).

Step III Shaft diameter
Equating (i) and (ii) and substituting values,

+\frac{48.375\left(10^{9}\right)}{(207000)\left(\pi d^{4} / 64\right)} .

d = 37.99 mm.