Question 9.15: A simple beam AB of a length L supports a uniform load of an...

Use a four-step problem-solving approach. Combine steps as needed for an efficient solution.
Part (a): Strain energy from the bending moment.
1, 2. Conceptualize, Categorize: The reaction of the beam at support A is qL/2, so the expression for the bending moment in the beam is

M=\frac{q L x}{2}-\frac{q x^{2}}{2}=\frac{q}{2}\left(L x-x^{2}\right)             (a)

3. Analyze: The strain energy of the beam [from Eq. (9-95a)] is

U=\int_{0}^{L} \frac{M^{2} d x}{2 E I}=\frac{1}{2 E I} \int_{0}^{L}\left[\frac{q}{2}\left(L x-x^{2}\right)\right]^{2} d x=\frac{q^{2}}{8 E I} \int_{0}^{L}\left(L^{2} x^{2}-2 L x^{2}+x^{4}\right) d x            (b)

which leads to

U=\frac{q^{2} L^{5}}{240 E I}              (9-98)

Note that the load q appears to the second power, which is consistent with the fact that strain energy is always positive. Furthermore, Eq. (9-98) shows that strain energy is not a linear function of the loads, even though the beam itself behaves in a linearly elastic manner.

Part (b): Strain energy from the deflection curve.

1, 2. Conceptualize, Categorize: The equation of the deflection curve for a simple beam with a uniform load is given in Case 1 of Table H-2, Appendix H, as

v=-\frac{q x}{24 E I}\left(L^{3}-2 L x^{2}+x^{3}\right)             (c)

3. Analyze: Taking two derivatives of this equation gives

Substitute the latter expression into the equation for strain energy [Eq. (9-95b)] to obtain

= \frac{q^{2}}{8 E I} \int_{0}^{L}\left(L^{2} x^{2}-2 L x^{3}+x^{4}\right) d x               (d)

4. Finalize: The final integral in this equation is the same as the final integral in Eq. (b) which leads to the same result as before [Eq. (9-98)].

Table H-2
Deflections and Slopes of Simple Beams
		Notation:
		v = deflection in the y direction (positive upward)
		v′ = dv/dx = slope of the deflection curve
		\delta_{C}=-v(L / 2)= deflection at midpoint C of the beam (positive downward)
		x_{1} = distance from support A to point of maximum deflection
		\delta_{\max }=-v_{\max }= maximum deflection (positive downward)
		\theta_{A}=-v^{\prime}(0)= angle of rotation at left-hand end of the beam (positive clockwise)
		\theta_{B}=v^{\prime}(L)= angle of rotation at right-hand end of the beam (positive counterclockwise)
		EI = constant
		v=-\frac{q x}{24 E I}\left(L^{3}-2 L x^{2}+x^{3}\right)
		v^{\prime}=-\frac{q}{24 E I}\left(L^{3}-6 L x^{2}+4 x^{3}\right)
		\delta_{C}=\delta_{\max }=\frac{5 q L^{4}}{384 E I} \quad\theta_{A}=\theta_{B}=\frac{q L^{3}}{24 E I}
		v=-\frac{q x}{384 E I}\left(9 L^{3}-24 L x^{2}+16 x^{3}\right) \quad\left(0 \leq x \leq \frac{L}{2}\right)
		v^{\prime}=-\frac{q}{384 E I}\left(9 L^{3}-72 L x^{2}+64 x^{3}\right) \quad\left(0 \leq x \leq \frac{L}{2}\right)
		v=-\frac{q L}{384 E I}\left(8 x^{3}-24 L x^{2}+17 L^{2} x-L^{3}\right) \quad\left(\frac{L}{2} \leq x \leq L\right)
		v^{\prime}=-\frac{q L}{384 E I}\left(24 x^{2}-48 L x+17 L^{2}\right) \quad\left(\frac{L}{2} \leq x \leq L\right)
		\delta_{C}=\frac{5 q L^{4}}{768 E I} \quad \theta_{A}=\frac{3 q L^{3}}{128 E I} \quad \theta_{B}=\frac{7 q L^{3}}{384 E I}
		v=-\frac{q x}{24 L E I}\left(a^{4}-4 a^{3} L+4 a^{2} L^{2}+2 a^{2} x^{2}-4 a L x^{2}+L x^{3}\right) \quad(0 \leq x \leq a)
		v^{\prime}=-\frac{q}{24 L E I}\left(a^{4}-4 a^{3} L+4 a^{2} L^{2}+6 a^{2} x^{2}-12 a L x^{2}-4 L x^{3}\right) \quad(0 \leq x \leq a)
		v=-\frac{q a^{2}}{24 L E I}\left(-a^{2} L+4 L^{2} x+a^{2} x-6 L x^{2}+2 x^{3}\right)             (a \leq x \leq L)
		v^{\prime}=-\frac{q a^{2}}{24 L E I}\left(4 L^{2}+a^{2}-12 L x+6 x^{2}\right)              (a \leq x \leq L)
		\theta_{A}=\frac{q a^{2}}{24 L E I}(2 L-a)^{2}   \quad \theta_{B}=\frac{q a^{2}}{24 L E I}\left(2 L^{2}-a^{2}\right)
		v=-\frac{P x}{48 E I}\left(3 L^{2}-4 x^{2}\right) \quad v^{\prime}=-\frac{P}{16 E I}\left(L^{2}-4 x^{2}\right) \quad\left(0 \leq x \leq \frac{L}{2}\right)
		\delta_{C}=\delta_{\max }=\frac{P L^{3}}{48 E I} \quad \theta_{A}=\theta_{B}=\frac{P L^{2}}{16 E I}
		v=-\frac{P b x}{6 L E I}\left(L^{2}-b^{2}-x^{2}\right) \quad v^{\prime}=-\frac{P b}{6 L E I}\left(L^{2}-b^{2}-3 x^{2}\right) \quad(0 \leq x \leq a)
		\theta_{A}=\frac{P a b(L+b)}{6 L E I} \quad \theta_{B}=\frac{P a b(L+a)}{6 L E I}
		\text { If } a \geq b, \quad \delta_{C}=\frac{P b\left(3 L^{2}-4 b^{2}\right)}{48 E I} \quad \text { If } a \leq b, \quad \delta_{C}=\frac{P a\left(3 L^{2}-4 a^{2}\right)}{48 E I}
		\text { If } a \geq b, \quad x_{1}=\sqrt{\frac{L^{2}-b^{2}}{3}} \quad \text { and } \quad \delta_{\max }=\frac{P b\left(L^{2}-b^{2}\right)^{3 / 2}}{9 \sqrt{3} L E I}
		v=-\frac{P x}{6 E I}\left(3 a L-3 a^{2}-x^{2}\right) \quad v^{\prime}=-\frac{P}{2 E I}\left(a L-a^{2}-x^{2}\right) \quad(0 \leq x \leq a)
		v=-\frac{P a}{6 E I}\left(3 L x-3 x^{2}-a^{2}\right) \quad v^{\prime}=-\frac{P a}{2 E I}(L-2 x) \quad(a \leq x \leq L-a)
		\delta_{C}=\delta_{\max }=\frac{P a}{24 E I}\left(3 L^{2}-4 a^{2}\right) \quad \theta_{A}=\theta_{B}=\frac{P a(L-a)}{2 E I}
		v=-\frac{M_{0} x}{6 L E I}\left(2 L^{2}-3 L x+x^{2}\right) \quad v^{\prime}=-\frac{M_{0}}{6 L E I}\left(2 L^{2}-6 L x+3 x^{2}\right)
		\delta_{C}=\frac{M_{0} L^{2}}{16 E I} \quad \theta_{A}=\frac{M_{0} L}{3 E I} \quad \theta_{B}=\frac{M_{0} L}{6 E I}
		x_{1}=L\left(1-\frac{\sqrt{3}}{3}\right) \text { and } \delta_{\max }=\frac{M_{0} L^{2}}{9 \sqrt{3} E I}
		v=-\frac{M_{0} x}{24 L E I}\left(L^{2}-4 x^{2}\right) \quad v^{\prime}=-\frac{M_{0}}{24 L E I}\left(L^{2}-12 x^{2}\right) \quad\left(0 \leq x \leq \frac{L}{2}\right)
		\delta_{C}=0 \quad \theta_{A}=\frac{M_{0} L}{24 E I} \quad \theta_{B}=-\frac{M_{0} L}{24 E I}
		v=-\frac{M_{0} x}{6 L E I}\left(6 a L-3 a^{2}-2 L^{2}-x^{2}\right) \quad(0 \leq x \leq a)
		v^{\prime}=-\frac{M_{0}}{6 L E I}\left(6 a L-3 a^{2}-2 L^{2}-3 x^{2}\right) \quad(0 \leq x \leq a)
		\text { At } x=a: \quad v=-\frac{M_{0} a b}{3 L E I}(2 a-L) \quad v^{\prime}=-\frac{M_{0}}{3 L E I}\left(3 a L-3 a^{2}-L^{2}\right)
		\theta_{A}=\frac{M_{0}}{6 L E I}\left(6 a L-3 a^{2}-2 L^{2}\right) \quad \theta_{B}=\frac{M_{0}}{6 L E I}\left(3 a^{2}-L^{2}\right)
		v=-\frac{M_{0} x}{2 E I}(L-x) \quad v^{\prime}=-\frac{M_{0}}{2 E I}(L-2 x)
		\delta_{C}=\delta_{\max }=\frac{M_{0} L^{2}}{8 E I} \quad \theta_{A}=\theta_{B}=\frac{M_{0} L}{2 E I}
		v=-\frac{q_{0} x}{360 L E I}\left(7 L^{4}-10 L^{2} x^{2}+3 x^{4}\right)
		v^{\prime}=-\frac{q_{0}}{360 L E I}\left(7 L^{4}-30 L^{2} x^{2}+15 x^{4}\right)
		\delta_{C}=\frac{5 q_{0} L^{4}}{768 E I} \quad \theta_{A}=\frac{7 q_{0} L^{3}}{360 E I} \quad \theta_{B}=\frac{q_{0} L^{3}}{45 E I}
		x_{1}=0.5193 L \quad \delta_{\max }=0.00652 \frac{q_{0} L^{4}}{E I}
		v=-\frac{q_{0} x}{960 L E I}\left(5 L^{2}-4 x^{2}\right)^{2} \quad\left(0 \leq x \leq \frac{L}{2}\right)
		v^{\prime}=-\frac{q_{0}}{192 L E I}\left(5 L^{2}-4 x^{2}\right)\left(L^{2}-4 x^{2}\right) \quad\left(0 \leq x \leq \frac{L}{2}\right)
		\delta_{C}=\delta_{ max }=\frac{q_{0} L^{4}}{120 E I} \quad \theta_{A}=\theta_{B}=\frac{5 q_{0} L^{3}}{192 E I}
		v=-\frac{q_{0} L^{4}}{\pi^{4} E I} \sin \frac{\pi x}{L} \quad v^{\prime}=-\frac{q_{0} L^{3}}{\pi^{3} E I} \cos \frac{\pi x}{L}
		\delta_{C}=\delta_{ max }=\frac{q_{0} L^{4}}{\pi^{4} E I} \quad \theta_{A}=\theta_{B}=\frac{q_{0} L^{3}}{\pi^{3} E I}