The maximum bending moment of a simply supported beam of span $ l $ and carrying a point load $ W $ at the centre of the beam is

Question

A rectangular section has width $b$ and depth $d$ and a symmetrical triangular section has base $b$ and depth $d$. The ratio of moment of inertia of rectangular section to that of triangular section with respect to their respective centroidal axis will be:

Answer 1

Step 1: The shear force diagram for this beam is constant at $ +\dfrac{W}{2} $ from the left support up to mid-span, then drops to $ -\dfrac{W}{2} $ for the remaining half.
Step 2: Bending moment at any section equals the area under the shear force diagram up to that section, so at mid-span this area is $ \dfrac{W}{2}\times\dfrac{l}{2} $.
Step 3: Evaluating this area:
\[ M_{\max} = \dfrac{W}{2}\times\dfrac{l}{2} = \dfrac{Wl}{4} \]

Answer 2

A quick way to confirm the peak bending moment is to compare with the well-known standard result for this exact loading case (a central point load on a simply supported beam), which structural engineering references consistently tabulate as $ \dfrac{Wl}{4} $. Testing the given options against this standard case:

$ \dfrac{Wl}{2} $: This value is twice the tabulated standard result for a central point load on a simply supported span, so it does not match the known case.
$ \dfrac{Wl}{4} $: This exactly matches the standard tabulated maximum bending moment for a simply supported beam of span $ l $ carrying a central point load $ W $, confirming it as correct.
$ Wl $: This is the standard tabulated result for the fixed-end moment of a cantilever of length $ l $ carrying a load $ W $ at its free end, a different support and loading case entirely, not a simply supported beam with a central load.
$ 2Wl $: This value does not correspond to any standard tabulated case for this type of loading and support condition, and is far larger than reasoning would allow for a moderate central load.

Comparing against the standard reference case confirms the peak moment for a centrally loaded simply supported beam is $ \dfrac{Wl}{4} $.

Therefore, the correct answer is $ \dfrac{Wl}{4} $.

Answer 3

A quick way to confirm the peak bending moment is to compare with the well-known standard result for this exact loading case (a central point load on a simply supported beam), which structural engineering references consistently tabulate as $ \dfrac{Wl}{4} $. Testing the given options against this standard case:

$ \dfrac{Wl}{2} $: This value is twice the tabulated standard result for a central point load on a simply supported span, so it does not match the known case.
$ \dfrac{Wl}{4} $: This exactly matches the standard tabulated maximum bending moment for a simply supported beam of span $ l $ carrying a central point load $ W $, confirming it as correct.
$ Wl $: This is the standard tabulated result for the fixed-end moment of a cantilever of length $ l $ carrying a load $ W $ at its free end, a different support and loading case entirely, not a simply supported beam with a central load.
$ 2Wl $: This value does not correspond to any standard tabulated case for this type of loading and support condition, and is far larger than reasoning would allow for a moderate central load.

Comparing against the standard reference case confirms the peak moment for a centrally loaded simply supported beam is $ \dfrac{Wl}{4} $.

Therefore, the correct answer is $ \dfrac{Wl}{4} $.

The maximum bending moment of a simply supported beam of span \( l \) and carrying a point load \( W \) at the centre of the beam is

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The Correct Option is B

Approach Solution - 1

Approach Solution -2

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