A freely supported beam at its ends is loaded by a central concentrated load then maximum moment is M. If the same weight be equally distributed over the beam, then its maximum moment will be
Beam Bending Moment Analysis
Understanding the maximum bending moment in a simply supported beam under different loading conditions is fundamental in structural mechanics. We will analyze two specific scenarios: a central concentrated load and an equally distributed load, both with the same total weight.
Central Concentrated Load on Beam
When a simply supported beam of span length \(L\) is subjected to a central concentrated load \(W\) at its mid-span, the maximum bending moment occurs at the center of the beam. The formula for this maximum moment (\(M_{max,1}\)) is:
\[ M_{max,1} = \frac{WL}{4} \]
According to the question, this maximum moment is given as \(M\). Therefore, we have:
\[ M = \frac{WL}{4} \quad \text{(Equation 1)} \]
Equally Distributed Load on Beam
Now, consider the same simply supported beam with the same total weight, but this weight is equally distributed over the entire span. If the total weight is \(W\), and it's uniformly distributed, the intensity of the uniformly distributed load (UDL) per unit length, \(w\), can be calculated as:
\[ w = \frac{\text{Total Weight}}{\text{Span Length}} = \frac{W}{L} \]
For a simply supported beam subjected to a uniformly distributed load \(w\) over its entire span \(L\), the maximum bending moment (\(M_{max,2}\)) also occurs at the center and is given by the formula:
\[ M_{max,2} = \frac{wL^2}{8} \]
Substitute the expression for \(w\) (\(w = \frac{W}{L}\)) into the formula for \(M_{max,2}\):
\[ M_{max,2} = \frac{\left(\frac{W}{L}\right)L^2}{8} \]
\[ M_{max,2} = \frac{WL}{8} \quad \text{(Equation 2)} \]
Maximum Moments Comparison
We need to find the relationship between \(M_{max,2}\) and \(M\). From Equation 1, we know that \(M = \frac{WL}{4}\). We can rewrite Equation 2 by observing the common term \(\frac{WL}{4}\):
\[ M_{max,2} = \frac{1}{2} \times \frac{WL}{4} \]
Substitute \(M\) from Equation 1 into this expression:
\[ M_{max,2} = \frac{1}{2} \times M \]
So, the maximum moment when the same weight is equally distributed over the beam is \(M\over 2\).
| Loading Condition | Total Load | Maximum Bending Moment Formula | Maximum Moment in terms of M |
|---|---|---|---|
| Central Concentrated Load | \(W\) | \(M = \frac{WL}{4}\) | \(M\) (given) |
| Equally Distributed Load (UDL) | \(W\) | \(M_{UDL} = \frac{wL^2}{8} = \frac{(W/L)L^2}{8} = \frac{WL}{8}\) | \(M_{UDL} = \frac{1}{2} M\) |
Therefore, if the maximum moment for a simply supported beam with a central concentrated load is \(M\), the maximum moment for the same total weight equally distributed over the beam will be \(M\over 2\).
Important Questions from Bending Moment
A simply supported beam of 4 m span carries a uniformly distributed load all over the span of 20 kN. The maximum bending moment is
When a beam is subjected to a bending moment, the strain in a layer is _____ the distance from the neutral axis.
- A cantilever and a simply supported beam have the same length and are subject to the same uniformly distributed load. The ratio of their maximum bending moments is
For a cantilever beam of length 2 m, under load of 1 kN/m, the maximum bending moment is
The bending moment diagram for simply supported beam loaded in its center is
