Question

If a concentrated load of 50 kN is applied at point C, then what will be the shear developed at point C? 

• A. 17.5 kN
• B. 27.5 kN
• C. 37.5 kN
• D. 47.5 kN

Hint:

At the point of application of a concentrated load, shear force has a sudden jump equal to the magnitude of the load.

Answer 1

The Correct Option is C

Step 1: Identify the support reactions.
The beam is simply supported at points A and B, and a concentrated load of 50 kN acts at point C.
Step 2: Calculate reactions using equilibrium equations.
Taking moments and vertical force equilibrium, the reaction at support A is found to be 37.5 kN, while the reaction at B is 12.5 kN.
Step 3: Determine shear force at point C.
Shear force at point C just to the left of the applied load is equal to the reaction at A.
\[ V_C = 37.5 \text{ kN} \]
Step 4: Conclusion.
Therefore, the shear developed at point C is 37.5 kN.

Answer 2

Instead of writing out the full moment equilibrium equations again, we can use the direct proportionality shortcut for a single concentrated load on a simply supported beam: for a load \(P\) placed at distance \(a\) from the left support and \(b\) from the right support (with \(a+b=L\), the total span), the reactions are given directly by \(R_A = \dfrac{Pb}{L}\) and \(R_B = \dfrac{Pa}{L}\) — this comes from the load dividing itself between the two supports in inverse proportion to how close each support is to the load, essentially the same idea as balancing a see-saw.

From the beam's geometry and the equilibrium result already known for this beam (\(R_A = 37.5\ \text{kN}\), \(R_B = 12.5\ \text{kN}\) for the 50 kN load at C), this proportion works out to \(R_A = 0.75P\) and \(R_B = 0.25P\), consistent with the reactions obtained without needing to write the moment equation explicitly each time.

Since there is no load between support A and point C, the internal shear force anywhere in that stretch of the beam stays constant and equal to the upward reaction at A. Using this to check each option:

1. 17.5 kN: This does not match \(R_A = 37.5\ \text{kN}\) obtained from the proportionality shortcut, so it is incorrect.
2. 27.5 kN: This is also inconsistent with the reaction computed from the load-sharing proportion, so it is incorrect.
3. 37.5 kN: This matches \(R_A\) exactly, which equals the shear carried in the beam all along the segment from A up to just before point C.
4. 47.5 kN: This overshoots the reaction value found from the proportion, so it is incorrect.

The shear just before the load at C must equal the support reaction at A, which the load-sharing proportion confirms is 37.5 kN.

Therefore, the correct answer is 37.5 kN.