Question

A simply supported beam with an overhang has experienced the bending moment as shown below. The corresponding concentrated load is 

 

• A. 5 kN at mid span of PR
• B. 10 kN at Q
• C. 10 kN at mid span of RS
• D. 5 kN at S

Hint:

When analyzing beams with overhangs, start with the overhang section. It is statically determinate, so you can often find forces and moments there directly without needing to calculate support reactions first. A linear BMD implies constant shear, and in an overhang, constant shear is caused by a point load at the end.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
This problem requires interpreting a Bending Moment Diagram (BMD) to determine the loading on a beam. Key relationships are that the shear force \(V\) is the slope of the bending moment diagram (\(V = dM/dx\)), and concentrated loads cause jumps in the shear force diagram and kinks (changes in slope) in the bending moment diagram.
Step 2: Key Formula or Approach:
1. Analyze the features of the given BMD.
2. Pay attention to the values of the moment at key points (supports, ends) and the shape of the diagram (linear, curved).
3. Use the relationship \(M = F \times d\) for moments caused by concentrated loads.
4. Evaluate the options based on the analysis.
Step 3: Detailed Explanation or Calculation:
The beam has simple supports at P and R, and an overhang from R to S.
Let's analyze the BMD:
- At the free end S, the bending moment is 0, which is expected.
- At the support R, the bending moment has a value of 10 kN-m (the diagram shows the value, and it's on the tension side, conventionally negative, so \(M_R = -10\) kN-m).
- The BMD between R and S is a straight line. This indicates that the shear force in this section is constant. A constant shear force in an overhang section is caused by a concentrated load at the free end.
Let's assume there is a downward concentrated load, F, at the end S. The bending moment at any point in the overhang RS, at a distance \(x\) from S, is given by \(M(x) = -F . x\).
At support R, the distance from S is 2 m. The moment at R would be: \[ M_R = -F \times 2 \text{ m} \] From the diagram, we know the magnitude of the moment at R is 10 kN-m. \[ 10 \text{ kN-m} = F \times 2 \text{ m} \] Solving for F: \[ F = \frac{10 \text{ kN-m}}{2 \text{ m}} = 5 \text{ kN} \] So, a concentrated load of 5 kN at point S would produce the moment shown in the overhang section RS.
Let's check the other options:
- (A) or (B): A concentrated load between P and R would cause a triangular BMD between P and R, with the peak at the point of application. The given diagram shows a single straight line from P to R, which is inconsistent with this loading (and also inconsistent with the reaction from the overhang load, but it's the most plausible cause). - (C) A load at the mid-span of RS would create a kink in the BMD at that point. The diagram is a straight line from R to S, so this is incorrect.
The most consistent explanation for the key features of the BMD (especially in the overhang section which is statically determinate) is a 5 kN load at S.
Step 4: Final Answer:
The corresponding concentrated load is 5 kN at S.
Step 5: Why This is Correct:
A 5 kN downward force at the free end S creates a linearly varying bending moment in the overhang, from 0 at S to \(-5 \text{ kN} \times 2 \text{ m} = -10 \text{ kN-m}\) at support R. This exactly matches the critical features of the provided bending moment diagram in the overhang section, making option (D) the correct answer. The linear moment diagram between P and R is likely an inaccurate representation of the moment due to the reactions, but the load at S is unambiguously determined by the moment at R.