The bending moment diagram of a beam ABCDE is zero over AB and DE, rises linearly from zero at B to a peak at C, and falls linearly back to zero at D (a symmetric triangle). The corresponding shear force diagram is:
Try it first: pick an option, then read the explanation.
Asked in SSC JE 03/03/2017 (Morning)
Detailed Explanation
The SFD is the slope of the BMD (V = dM/dx). Read the slope of each straight segment of the BMD and you have the SFD. (In the exam the options were sketches; they are described in words here.)
Why “zero over AB and DE; a constant positive value over BC and an equal constant negative value over CD” Is Correct
Over AB and DE the BMD is flat, so the shear is zero. From B to C the moment rises at a steady rate — constant positive shear; from C to D it falls at the same rate — an equal constant negative shear. That is a rectangular block above the axis followed by an equal block below it — option (a). It is the SFD of a beam with a point load at C between supports B and D.
Why the Other Options Are Wrong
- B. a straight line falling from a positive value at B to a negative value at D, crossing zero at C: A sloping SFD belongs to a parabolic BMD (UDL), not a triangular one.
- C. a constant positive value over the whole length AE: A constant shear everywhere would give a BMD rising all the way from A to E.
- D. zero over BC and CD, with non-zero values only over AB and DE: The BMD is flat over AB and DE, so the shear there is zero — not the other way round.
Key Point
V = dM/dx: straight BMD ⇒ constant SF; peak of BMD ⇒ SF changes sign.
Key Concepts for Students
- A kink in the BMD marks a point load.
- A parabolic BMD means a UDL (linear SFD).
- The jump in the SFD at C equals the load P.
Practice more: this is question 38 in Bending Moment and Shear Force MCQs – Part 3. Try the whole set and check every answer instantly.