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) $M=\frac{W L}{2} x-\frac{W}{2} x^{2}$
(ii) $M=\frac{W_{x}}{3} x-\frac{W}{3} \frac{x^{3}}{L^{2}}$
Find the point at which M is maximum in each case.
(i)
Given : $M=\frac{W L}{2} x-\frac{W}{2} x^{2}$
$\Rightarrow \frac{d M}{d x}=\frac{W L}{2}-2 \times \frac{W x}{2}$
$\Rightarrow \frac{d M}{d x}=\frac{W L}{2}-W x$
For maximum or minimum values of $M$, we must have
$\frac{d M}{d x}=0$
$\Rightarrow \frac{W L}{2}-W x=0$
$\Rightarrow \frac{W L}{2}=W x$
$\Rightarrow x=\frac{L}{2}$
Now,
$\frac{d^{2} M}{d x^{2}}=-W<0$
So, $M$ is maximum at $x=\frac{L}{2}$.
(ii)
Given : $M=\frac{W x}{3}-\frac{W x^{3}}{3 L^{2}}$
$\Rightarrow \frac{d M}{d x}=\frac{W}{3}-3 \times \frac{W x^{2}}{3 L^{2}}$
$\Rightarrow \frac{d M}{d x}=\frac{W}{3}-\frac{W x^{2}}{L^{2}}$
For maximum or minimum values of $M$, we must have
$\frac{d M}{d x}=0$
$\Rightarrow \frac{W}{3}-\frac{W x^{2}}{L^{2}}=0$
$\Rightarrow \frac{W}{3}=\frac{W x^{2}}{L^{2}}$
$\Rightarrow x=\frac{L}{\sqrt{3}}$
Now,
$\frac{d^{2} M}{d x^{2}}=-\frac{2 W x}{L^{2}}<0$
So, $M$ is maximum at $x=\frac{L}{\sqrt{3}}$.