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: Since the beam is simply supported with a single point load of 50 kN at C, use the direct reaction-sharing rule \( R_A = \dfrac{Pb}{L} \), \( R_B = \dfrac{Pa}{L} \), where a, b are C's distances from A and B.
Step 2: The given geometry works out to \( R_A = 0.75(50) = 37.5\ \text{kN} \) and \( R_B = 0.25(50) = 12.5\ \text{kN} \).
Step 3: No load acts between A and C, so the shear force stays constant and equal to \(R_A\) all along that stretch, right up to just before C.
\[ \boxed{V_C = 37.5\ \text{kN}} \]

Answer 2

A third check is to work from the right-hand portion of the beam (from C to support B) instead of the left portion, and use overall vertical equilibrium to relate the two results.

Isolating the segment from just to the right of C to support B, the only external vertical force acting on it is the upward reaction \(R_B = 12.5\ \text{kN}\); for this piece to be in equilibrium, the internal shear force transmitted across the cut at C from the rest of the beam must balance this reaction, giving a shear magnitude of \(12.5\ \text{kN}\) in that segment.

Now apply overall vertical equilibrium to the whole beam: the total downward load is \(50\ \text{kN}\), and the shear diagram must jump down by this full applied load exactly at C, moving from the value just left of the cut to the value just right of it.

1. 17.5 kN: Combined with a 50 kN drop at C, the value just before C would need to be inconsistent with either support reaction found, so this is ruled out.
2. 27.5 kN: Similarly, this does not correspond to either \(R_A = 37.5\ \text{kN}\) (just before C) or \(R_B = 12.5\ \text{kN}\) (just after C, in magnitude) once the 50 kN drop is accounted for; ruled out.
3. 37.5 kN: This matches the shear on the A-side of the cut exactly, and is consistent with a jump of \(37.5 - (-12.5) = 50\ \text{kN}\) in the shear diagram right at C, matching the applied point load.
4. 47.5 kN: This does not correspond to either support reaction and does not produce a consistent 50 kN jump at C; ruled out.

Balancing the shear jump across C against the two support reactions confirms the value on the left side of the cut.

Therefore, the correct answer is 37.5 kN.