For a cantilever beam of length l carrying a load that varies linearly from zero at the free end to w per unit length at the fixed end, the shear force diagram (SFD) has what shape?
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AHorizontal straight line
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BVertical straight line
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CInclined straight line
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DParabolic curve
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ECubic curve
Answer
Correct Answer: Parabolic curve
Explanation
Introduction / Context:Being able to sketch the qualitative shapes of shear-force and bending-moment diagrams for different loadings is a core skill in structural analysis and machine design.
Given Data / Assumptions:
- Cantilever beam, length l.
- Linearly varying distributed load: w(x) increases from 0 at free end to w at fixed end.
- Small-deflection beam theory; sign conventions standard.
Concept / Approach:The differential relationships are:dV/dx = −w(x)dM/dx = V(x)With w(x) linear in x, integrating once gives V(x) quadratic in x (a parabola), and integrating again gives M(x) cubic.
Step-by-Step Solution:
Let w(x) = (w/l) * x (taking x from free end).Integrate: dV/dx = −(w/l) * x → V(x) = −(w/(2l)) * x^2 + C.At free end (x = 0), V(0) = 0 → C = 0.Hence V(x) = −(w/(2l)) * x^2 → a parabola opening downward.Verification / Alternative check:Total shear at the fixed end equals the resultant of the triangular load: W_total = (1/2) * w * l. Evaluating V(l) = −(w/(2l)) * l^2 = −(w l)/2 matches the resultant, confirming the parabolic SFD.
Why Other Options Are Wrong:Horizontal/vertical/inclined lines correspond to zero/impulse/constant load distributions, not a linearly varying load.Cubic curve describes the bending moment diagram for this case, not the SFD.
Common Pitfalls:Using the wrong origin for x; forgetting that V is the integral of −w(x), hence one degree higher in polynomial order than w(x).
Final Answer:
Parabolic curve