Question

A beam is supported at the two ends and is uniformly loaded. The bending moment M at a distance x from one end is given by
(i) $M=\frac{WL}{2}x-\frac{W}{2}{x}^2$

(ii) $M=\frac{Wx}{3}x-\frac{W}{3}\frac{x^3}{L^2}$

Find the point at which M is maximum in each case.

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right) \\ \mathrm{Given}:\hspace{0.17em}M=\frac{WL}{2}x-\frac{W}{2}{x}^2 \\ \Rightarrow \frac{dM}{dx}=\frac{WL}{2}-2\times \frac{Wx}{2} \\ \Rightarrow \frac{dM}{dx}=\frac{WL}{2}-Wx \\ \mathrm{For}\mathrm{maximum}\mathrm{or}\mathrm{minimum}\mathrm{values}\mathrm{of}M,\mathrm{we}\mathrm{must}\mathrm{have} \\ \frac{dM}{dx}=0 \\ \Rightarrow \frac{WL}{2}-Wx=0 \\ \Rightarrow \frac{WL}{2}=Wx \\ \Rightarrow x=\frac{L}{2} \\ \mathrm{Now}, \\ \frac{d^{2}M}{d{x}^2}=-W<0 \\ \text{So,}\mathit{\text{M}}\text{is maximum at}x=\frac{L}{2}. \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ii}\right) \\ \mathrm{Given}:\hspace{0.17em}M=\frac{Wx}{3}-\frac{W{x}^3}{3L^2} \\ \Rightarrow \frac{dM}{dx}=\frac{W}{3}-3\times \frac{W{x}^2}{3L^2} \\ \Rightarrow \frac{dM}{dx}=\frac{W}{3}-\frac{W{x}^2}{L^2} \\ \mathrm{For}\mathrm{maximum}\mathrm{or}\mathrm{minimum}\mathrm{values}\mathrm{of}M,\mathrm{we}\mathrm{must}\mathrm{have} \\ \frac{dM}{dx}=0 \\ \Rightarrow \frac{W}{3}-\frac{W{x}^2}{L^2}=0 \\ \Rightarrow \frac{W}{3}=\frac{W{x}^2}{L^2} \\ \Rightarrow x=\frac{L}{\sqrt{3}} \\ \mathrm{Now}, \\ \frac{d^{2}M}{d{x}^2}=-\frac{2Wx}{L^2}<0 \\ \text{So,}\mathit{\text{M}}\text{is maximum at}x=\frac{L}{\sqrt{3}}. \end{gathered}$$