Question 12.5: Sketch the shear and bending-moment diagrams for the cantile...
Sketch the shear and bending-moment diagrams for the cantilever beam shown in Fig. 1.
STRATEGY: Because there are no support reactions until the right end of the beam, you can rely solely on the equations from this section without needing to use free-body diagrams and equilibrium equations. Due to the non-uniform distributed load, you should expect the results to involve equations of higher degree, with a parabolic curve in the shear diagram and a cubic curve in the bending-moment diagram.
MODELING and ANALYSIS:
Shear Diagram. At the free end of the beam, V_A = 0. Between A and B, the area under the load curve is \frac{1}{2} w_0 a. Thus,
V_B-V_A=-\frac{1}{2}w_0a \quad \quad V_B=-\frac{1}{2}w_0a
Between B and C, the beam is not loaded, so V_C = V_B. At A, w = w_0.
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.
Bending-Moment Diagram. The bending moment M_A at the free end of the beam is zero. Compute the area under the shear curve to obtain.
\begin{matrix} M_B-M_A&=&-\frac{1}{3}w_0a^2 & M_B=-\frac{1}{3}w_0a^2 \\ M_C-M_B&=&-\frac{1}{2}w_0a(L-a) \\ M_C&=&-\frac{1}{6}w_0a(3L-a) \end{matrix}
The sketch of the bending-moment diagram is completed by recalling that dM/dx = V. 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.
REFLECT and THINK: Although not strictly required for the solution of this problem, determination of the support reactions would serve as an excellent check of the final values of the shear and bending-moment diagrams.