Question 12.S-P.5: Sketch the shear and bending-moment diagrams for the cantile...
Sketch the shear and bending-moment diagrams for the cantilever beam shown.
Shear Diagram. At the free end of the beam, we find V_{A} = 0. Between A and B, the area under the load curve is \frac{1}{2} w_{0} a; we find V_{B} by writing
V_{B} – V_{A} = -\frac{1}{2} w_{0} a V_{B} = -\frac{1}{2} w_{0} a
Between B and C, the beam is not loaded; thus V_{C} = V_{B}. At A, we have w = w_{0} and, according to Eq. (12.5), the slope of the shear curve is dV/dx = -w_{0}, while at B the slope is dV/dx = 0. Between A and B, the loading decreases linearly, and the shear diagram is parabolic. Between B and C, w = 0, and the shear diagram is a horizontal line.
\frac{dV}{dx} = -w (12.5)
Bending-Moment Diagram. The bending moment M_{A} at the free end of the beam is zero. We compute the area under the shear curve and write
M_{B} – M_{A} = -\frac{1}{3} w_{0} a^{2} M_{B} = -\frac{1}{3} w_{0} a^{2}
M_{C} – M_{B} = -\frac{1}{2} w_{0} a(L – a)
M_{C} = -\frac{1}{6} w_{0} a(3L – a)
The sketch of the bending-moment diagram is completed by recalling that dM/dx = V. We find that between A and B the diagram is represented by a cubic curve with zero slope at A, and between B and C by a straight line.