Question 9.29: Consider the cantilever beam illustrated in Example 9.28. Us...

\text { Given } \delta_{\max }=0.05 mm \quad E=207000 N / mm ^{2} .

P = 500 N
Step I Deflection by Castigliano’s theorem
Let us denote by P the force at the free end of the cantilever beam. The bending moment at a distance x from the fixed end is given by,

M_{b}=P(200-x)                  (i).

The above relationship can be used for any point from A to C. Differentiating with respect to P,

\frac{\partial}{\partial P}\left(M_{b}\right)=(200-x)                (ii).

The total strain energy stored in the beam is given by

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

From Eq. (9.56),

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

U=\int_{0}^{100} \frac{\left(M_{b}\right)^{2} d x}{2 E I^{\prime}}+\int_{100}^{200} \frac{\left(M_{b}\right)^{2} d x}{2 E I}           (iii).

By Castigliano’s theorem,

\delta_{\max }=\frac{\partial U}{\partial P}                 (iv).

From (iii) and (iv),

+\int_{100}^{200}\left(\frac{2\left(M_{b}\right)}{2 E I}\right)\left(\frac{\partial\left(M_{b}\right)}{\partial P}\right) d x .

Substituting (i) and (ii) in the above expression,

+\frac{P}{E I} \int_{100}^{200}(200-x)^{2} d x .

On integration, we get,

\delta_{\max .}=\frac{\left(7 \times 10^{6}\right) P}{3 E I^{\prime}}+\frac{\left(10^{6}\right) P}{3 E I} .

Substituting the values of P and I′ in the above equation,

\delta_{\max }=\frac{\left(729.19 \times 10^{6}\right)}{E I} mm .

Step II Diameter of beam
The permissible deflection is 0.05 mm. Therefore,

0.05=\frac{\left(729.19 \times 10^{6}\right)}{(207000)\left(\pi d^{4} / 64\right)} .

∴          d = 34.61 mm.