Question 9.6: A cantilever beam AB supports a uniform load of intensity q ...
A cantilever beam AB supports a uniform load of intensity q acting over part of the span and a concentrated load P acting at the free end (Fig. 9-18a).
Determine the deflection δ_{B} and angle of rotation θ_{B} at end B of the beam (Fig. 9-18b). Note: The beam has length L and constant flexural rigidity EI.
Use a four-step problem-solving approach.
1. Conceptualize: Obtain the deflection and angle of rotation at end B of the beam by combining the effects of the loads acting separately.
2. Categorize: If the uniform load acts alone, the deflection and angle of rotation (obtained from Case 2 of Table H-1, Appendix H) are
If the load P acts alone, the corresponding quantities (from Case 4, Table H-1) are
\left(\delta_{B}\right)_{2}=\frac{P L^{3}}{3 E I} \quad\left(\theta_{B}\right)_{2}=\frac{P L^{2}}{2 E I}3. Analyze: Therefore, the deflection and angle of rotation due to the combined loading (Fig. 9-18a) are
\delta_{B}=\left(\delta_{B}\right)_{1}+\left(\delta_{B}\right)_{2}=\frac{q a^{3}}{24 E I}(4 L-a)+\frac{P L^{3}}{3 E I} (9-61)
\theta_{B}=\left(\theta_{B}\right)_{1}+\left(\theta_{B}\right)_{2}=\frac{q a^{3}}{6 E I}+\frac{P L^{2}}{2 E I} (9-62)
4. Finalize: The required quantities were found by using tabulated formulas and the method of superposition. Note that you would have to integrate over two separate regions (0 ≤ x ≤ a, a ≤ x ≤ L) to find the deflection curve shown in Fig. 9-18b.
| Table H-1 | |
| Deflections and Slopes of Cantilever Beams | |
 |
Notation: |
| v = deflection in the y direction (positive upward) | |
| v′ = dv/dx = slope of the deflection curve | |
| \delta_{B}=-v(L)= deflection at end B of the beam (positive downward) | |
| \theta_{B}=-v^{\prime}(L)= angle of rotation at end B of the beam (positive clockwise) | |
| EI = constant | |
|
v=-\frac{q x^{2}}{24 E I}\left(6 L^{2}-4 L x+x^{2}\right) \quad v^{\prime}=\frac{q x}{6 E I}\left(3 L^{2}-3 L x+x^{2}\right) |
| \delta_{B}=\frac{q L^{4}}{8 E I} \quad \theta_{B}=\frac{q L^{3}}{6 E I} | |
 |
v=-\frac{q x^{2}}{24 E I}\left(6 a^{2}-4 a x+x^{2}\right) \quad(0 \leq x \leq a) |
| v^{\prime}=-\frac{q x}{6 E I}\left(3 a^{2}-3 a x+x^{2}\right) \quad(0 \leq x \leq a) | |
| v=-\frac{q a^{3}}{24 E I}(4 x-a) \quad v^{\prime}=-\frac{q a^{3}}{6 E I} \quad(a \leq x \leq L) | |
| At x=a: v=-\frac{q a^{4}}{8 E I} \quad v^{\prime}=-\frac{q a^{3}}{6 E I} | |
| \delta_{B}=\frac{q a^{3}}{24 E I}(4 L-a) \quad \theta_{B}=\frac{q a^{3}}{6 E I} | |
 |
v=-\frac{q b x^{2}}{12 E I}(3 L+3 a-2 x) (0 \leq x \leq a) |
| v^{\prime}=-\frac{q b x}{2 E I}(L+a-x) (0 \leq x \leq a) | |
| v=-\frac{q}{24 E I}\left(x^{4}-4 L x^{3}+6 L^{2} x^{2}-4 a^{3} x+a^{4}\right) \quad(a \leq x \leq L) | |
| v^{\prime}=-\frac{q}{6 E I}\left(x^{3}-3 L x^{2}+3 L^{2} x-a^{3}\right) \quad(a \leq x \leq L) | |
| At x=a: v=-\frac{q a^{2} b}{12 E I}(3 L+a) \quad v^{\prime}=-\frac{q a b L}{2 E I} | |
| \delta_{B}=\frac{q}{24 E I}\left(3 L^{4}-4 a^{3} L+a^{4}\right) \quad \theta_{B}=\frac{q}{6 E I}\left(L^{3}-a^{3}\right) | |
 |
v=-\frac{P x^{2}}{6 E I}(3 L-x) \quad v^{\prime}=-\frac{P x}{2 E I}(2 L-x) |
| \delta_{B}=\frac{P L^{3}}{3 E I} \quad \theta_{B}=\frac{P L^{2}}{2 E I} | |
 |
v=-\frac{P x^{2}}{6 E I}(3 a-x) \quad v^{\prime}=-\frac{P x}{2 E I}(2 a-x) \quad(0 \leq x \leq a) |
| v=-\frac{P a^{2}}{6 E I}(3 x-a) \quad v^{\prime}=-\frac{P a^{2}}{2 E I} \quad(a \leq x \leq L) | |
| At x=a: \quad v=-\frac{P a^{3}}{3 E I} \quad v^{\prime}=-\frac{P a^{2}}{2 E I} | |
| \delta_{B}=\frac{P a^{2}}{6 E I}(3 L-a) \quad \theta_{B}=\frac{P a^{2}}{2 E I} | |
 |
v=-\frac{M_{0} x^{2}}{2 E I} \quad v^{\prime}=-\frac{M_{0} x}{E I} |
| \delta_{B}=\frac{M_{0} L^{2}}{2 E I} \quad \theta_{B}=\frac{M_{0} L}{E I} | |
 |
v=-\frac{M_{0} x^{2}}{2 E I} \quad v^{\prime}=-\frac{M_{0} x}{E I} \quad(0 \leq x \leq a) |
| v=-\frac{M_{0} a}{2 E I}(2 x-a) \quad v^{\prime}=-\frac{M_{0} a}{E I} \quad(a \leq x \leq L) | |
| At x=a: \quad v=-\frac{M_{0} a^{2}}{2 E I} \quad v^{\prime}=-\frac{M_{0} a}{E I} | |
| \delta_{B}=\frac{M_{0} a}{2 E I}(2 L-a) \quad \theta_{B}=\frac{M_{0} a}{E I} | |
 |
v=-\frac{q_{0} x^{2}}{120 L E I}\left(10 L^{3}-10 L^{2} x+5 L x^{2}-x^{3}\right) |
| v^{\prime}=-\frac{q_{0} x}{24 L E I}\left(4 L^{3}-6 L^{2} x+4 L x^{2}-x^{3}\right) | |
| \delta_{B}=\frac{q_{0} L^{4}}{30 E I} \quad \theta_{B}=\frac{q_{0} L^{3}}{24 E I} | |
 |
v=-\frac{q_{0} x^{2}}{120 L E I}\left(20 L^{3}-10 L^{2} x+x^{3}\right) |
| v^{\prime}=-\frac{q_{0} x}{24 L E I}\left(8 L^{3}-6 L^{2} x+x^{3}\right) | |
| \delta_{B}=\frac{11 q_{0} L^{4}}{120 E I} \quad \theta_{B}=\frac{q_{0} L^{3}}{8 E I} | |
 |
v=-\frac{q_{0} L}{3 \pi^{4} E I}\left(48 L^{3} \cos \frac{\pi x}{2 L}-48 L^{3}+3 \pi^{3} L x^{2}-\pi^{3} x^{3}\right) |
| v^{\prime}=-\frac{q_{0} L}{\pi^{3} E I}\left(2 \pi^{2} L x-\pi^{2} x^{2}-8L^{2} \sin \frac{\pi x}{2 L}\right) | |
| \delta_{B}=\frac{2 q_{0} L^{4}}{3 \pi^{4} E I}\left(\pi^{3}-24\right) \quad\theta_{B}=\frac{q_{0} L^{3}}{\pi^{3} E I}\left(\pi^{2}-8\right) | |