Consider a beam of length 'AB' with 'C' as the midpoint. What is the magnitude of the shear force (sign convenction to be neglected) in the span 'AC' of this simply supported beam with a central point load 'P' kN?

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RRB JE CBT 2 (Civil Engineering) ReExam Official Paper (Held On: 04 Jun, 2025)

Answer 1

Concept:

The beam is simply supported at ends A and B.
A point load P is applied at the midpoint C of the beam.
Because the load is placed symmetrically, the reactions at supports A and B are equal.
Shear force is the internal force that resists the sliding of one part of a beam past the other.
In span AC (left of point C), the shear force remains constant, since there are no loads between A and C.

Calculation:
Let total length of the beam be L

Reaction at support A = \(R_A=\frac{P}{2}\)

Now, consider a section in span AC, just to the left of point C:

Shear force = Upward reaction at A = \(\frac{P}{2}\)

Therefore, the magnitude of the shear force in span AC is: \(\frac{P}{2}\)