Question 15.CA.2: The simply supported prismatic beam AB carries a uniformly d...
The simply supported prismatic beam AB carries a uniformly distributed load w per unit length (Fig. 15.10a). Determine the equation of the elastic curve and the maximum deflection of the beam.
Draw the free-body diagram of the portion AD of the beam (Fig. 15.10b) and take moments about D for
M=\frac{1}{2} w L x-\frac{1}{2} W x^2 (1)
Substituting for M into Eq. (15.4) and multiplying both members of this equation by the constant EI gives
\frac{d^2 y}{d x^2}=\frac{M(x)}{E I} (15.4)
E I \frac{d^2 y}{d x^2}=-\frac{1}{2} w x^2+\frac{1}{2} w L x (2)
Integrating twice in x,
E I \frac{d y}{d x}=-\frac{1}{6} w x^3+\frac{1}{4} w L x^2+C_1 (3)
EI y=-\frac{1}{24} w x^4+\frac{1}{12} w L x^3+C_1 x+C_2 (4)
Observing that y = 0 at both ends of the beam (Fig. 15.10c), let x = 0 and y = 0 in Eq. (4) and obtain C_2 = 0. Then make x = L and y = 0 in the same equation, so
\begin{gathered}0=-\frac{1}{24} w L^4+\frac{1}{12} w L^4+C_1 L \\C_1=-\frac{1}{24} w L^3\end{gathered}Carrying the values of C_1 and C_2 back into Eq. (15.4), the elastic curve is
EI y=-\frac{1}{24} w x^4+\frac{1}{12} w L x^3-\frac{1}{24} w L^3 xor
y=\frac{w}{24 E I}\left(-x^4+2 L x^3-L^3 x\right) (5)
Substituting the value for C_1 into Eq. (3), we check that the slope of the beam is zero for x = L/2 and, thus, that the elastic curve has a minimum at the midpoint C (Fig. 15.10d). Letting x = L/2 in Eq. (5),
y_C=\frac{w}{24 E I}\left(-\frac{L^4}{16}+2 L \frac{L^3}{8}-L^3 \frac{L}{2}\right)=-\frac{5 w L^4}{384 E I}The maximum deflection (the maximum absolute value) is
|y|_{\max }=\frac{5 w L^4}{384 E I}