A simply supported beam of length 1 m is subjected to a uniformly distributed bending moment of 1 N m per m throughout the length as shown in the figure given below. The bending moment at the mid-point of the beam is _____N m (rounded off to the nearest integer).

The correct answer is 0 (or 0 N m).

1. Understanding the System and Reactions
Let us consider a simply supported beam $AB$ of length $L=1\text{m}$ subjected to a uniformly distributed clockwise bending moment of intensity $m=1\text{N m/m}$ along its entire length. Support A is a hinge/pin support, and Support B is a roller support.
The total external moment applied to the beam is:
$M_{\text{total}}=m\times L=1\text{N m/m}\times 1\text{m}=1\text{N m}$

For the beam to be in static equilibrium, the support reactions $R_A$ (at support A) and $R_B$ (at support B) must form a couple that balances this total applied moment.
Taking the sum of moments about support A:
$\sum M_A=0\Rightarrow R_B\times L+mL=0$
$\Rightarrow R_B=-m=-1\text{N}$ (downward reaction of 1 N)
Since the sum of vertical forces must be zero:
$\sum F_y=0��R_A+R_B=0\Rightarrow R_A=-R_B=1\text{N}$ (upward reaction of 1 N)

2. Bending Moment at Any Section $x$
Let us determine the bending moment $M\left(x\right)$ at a distance $x$ from the left support A. We take a section at $x$ and sum the moments of all forces and distributed moments acting on the left portion of the beam:
The moment due to the reaction force $R_A$ about the section $x$ is:
$M_{\text{force}}=R_A\times x=1\times x=x\text{(clockwise)}$
The total distributed moment acting on the portion of length $x$ is:
$M_{\text{distributed}}=m\times x=1\times x=x\text{(counter-clockwise)}$

Summing these moments to find the net internal bending moment at section $x$:
$M\left(x\right)=M_{\text{force}}-M_{\text{distributed}}=x-x=0$

Since the bending moment $M\left(x\right)=0$ at every point along the beam, the bending moment at the mid-point of the beam ($x=0.5\text{m}$) is also:
$M\left(0.5\right)=0\text{N m}$