_________ is the maximum bending moment of a simply supported beam (length = L metre) with a point load of "W" kN at its center?
W × \((\frac{L}{4})\) kNm
Let's find the maximum bending moment for a simply supported beam carrying a point load at its center.
A simply supported beam is supported at two points and is free to rotate at the supports. A point load is a concentrated load applied at a single point on the beam.
For a simply supported beam of length \(L\) metres with a point load \(W\) kN acting exactly at the center (at a distance \(L/2\) from each support), we can determine the support reactions and the bending moment.
Let the supports be at points A and B. Due to the symmetry of the beam and the central load, the reactions at both supports will be equal.
The bending moment at any section of the beam is calculated by considering the forces to one side of the section. For a simply supported beam with a central point load, the maximum bending moment occurs at the point where the shear force is zero, which is under the central point load.
Consider a section at a distance \(x\) from support A, where \(0 \le x \le \frac{L}{2}\). The bending moment \(M(x)\) at this section is due to the reaction at A:
\(M(x) = R_A \times x = \frac{W}{2} \times x\)
The bending moment increases linearly from 0 at support A to its maximum value at the center.
The maximum bending moment occurs at the center of the beam, where \(x = \frac{L}{2}\).
Maximum bending moment = \(M\left(\frac{L}{2}\right) = \frac{W}{2} \times \frac{L}{2}\)
Maximum bending moment = \(\frac{W \times L}{4}\)
So, the maximum bending moment is \(\frac{WL}{4}\) kNm.
Comparing this with the given options:
The derived formula matches Option 4.
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