Shear and Bending Moment Diagrams using Relations between Shear

and Bending Moment

*Example: (continued)

Strategy: dM/dx = V(x) establishes the slope of the moment diagram. Also note that the area under the shear diagram is M(x) =∫ V(x) dx . So the area under the shear diagram represents the change in the bending moment along the beam.

In this example the bending moment at x = 0 is zero. From the shear diagram the slope

Starts out at + 3wL/8, decreases to zero at some value of x < L/2. For 0 ≤ x ≤ L/2
dV/dx = ˗ w so V(x) = ˗ wx + C. At x = 0 , V(0) = 3wL/8. So C = 3wL/8.
Thus for 0 ≤ x ≤ L/2 V(x) = ˗ wx + 3wL/8 and at x = L/2, V(L/2) = ˗ wL/8. Next use
similar triangles, ABE and CDE, to determine the location on the beam where the shear force
is zero.

So (3wL/8) / (L/2 ˗ x) = (wL/8) / x . Solve for x. x = (1/8) L . Thus the shear force

is zero at x = (3/8) L and the slope of the moment, dM/dx = V = 0 at this location.

Click here to continue with this example.

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