A simply supported beam which carries a uniformly distributed load has two equal overhangs. To have maximum B.M. produced in the beam least possible, the ratio of the length of the overhang to the total length of the beam, is A. 0.207 B. 0.307 C. 0.407 D. 0.508

0.207
0.307
0.407
0.508

The correct answer is $\boxed{\text{A. }0.207}$.

The maximum bending moment in a simply supported beam with two equal overhangs occurs at the support, and is equal to $wL^2/8$, where $w$ is the uniform load per unit length and $L$ is the total length of the beam. The ratio of the length of the overhang to the total length of the beam is $a/L$.

To minimize the maximum bending moment, we need to minimize $a/L$. The minimum value of $a/L$ is 0.207, which occurs when $a=0.207L$.

Here is a diagram of a simply supported beam with two equal overhangs:

[Diagram of a simply supported beam with two equal overhangs]

The maximum bending moment occurs at the support, and is equal to $wL^2/8$. The ratio of the length of the overhang to the total length of the beam is $a/L$.

The following are brief explanations of each option:

  • Option A: $a/L=0.207$. This is the minimum value of $a/L$, and it minimizes the maximum bending moment.
  • Option B: $a/L=0.307$. This is a larger value of $a/L$ than Option A, and it will result in a larger maximum bending moment.
  • Option C: $a/L=0.407$. This is an even larger value of $a/L$ than Option B, and it will result in an even larger maximum bending moment.
  • Option D: $a/L=0.508$. This is the largest value of $a/L$, and it will result in the largest maximum bending moment.
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