Question

A beam of length \[l\] is supported at one end. If w is the uniform load per unit length the bending moment \[M\] at a distance $x$ from the end is given by $M = \dfrac{1}{2}\ln - \dfrac{1}{2}w{n^2}$, find the point on the beam at which bending moment has maximum value.

Answer 1

Hint: When we are solving problems of maximum and minimum we have to remember that this can be solved by only formulas of derivatives.
for maximum value, we should differentiate it concerning $x$.
When we find the maximum value we have to show the second-order derivative must be less than zero.

Complete step by step solution
Bending moment $M = \dfrac{1}{2}{\text{ or }}\dfrac{{ - 1}}{2}w{n^2}$ for bending moment to be maximum \[M\], $\dfrac{{dM}}{{dx}} = 0$
$ \Rightarrow \dfrac{{dm}}{{dx}} = 0$
\[ \Rightarrow \dfrac{{dm}}{{dn}} = \dfrac{1}{2}l - \dfrac{1}{2}w\left( {2n} \right)\]
So to find the critical point,
$ \Rightarrow \dfrac{{dm}}{{dx}} = 0$
$ \Rightarrow \dfrac{1}{2} - wn = 0$
$x = \dfrac{1}{{2w}}$..
This is the critical point of the solution.
Now we have to find the second-order derivative.
$\dfrac{{({d^2}m)}}{{(d{n^2})}} = - w$
Which shows that the value is negative..that is this is less than zero.
So we can say this is maximum.

Note: In these types of questions, we have to differentiate the moment concerning \[x\]. As here we have to find the maximum or minimum value we have to find the derivatives..then we can solve the problem. I have already said that the maximum will be held only when the second-order derivative is less than zero.so have to find the second-order derivative.