Question 15.CA.4: 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. 15.13a). Determine the equation of the elastic curve and the maximum deflection of the beam. (This is the same beam and load as in Concept Application 15.2.)
Because w = constant, the first three of Eqs. (15.9) yield
\begin{aligned}&E I \frac{d^4 y}{d x^4}=-w(x) \\&E I \frac{d^3 y}{d x^3}=V(x)=-\int w(x) d x+C_1 \\&E I \frac{d^2 y}{d x^2}=M(x)=-\int d x \int w(x) d x+C_1 x+C_2 \\&E I \frac{d y}{d x}=E I \theta(x)=-\int d x \int d x \int w(x) d x+\frac{1}{2} C_1 x^2+C_2 x+C_3 \\&E I y(x)=-\int d x \int d x \int d x \int w(x) d x+\frac{1}{6} C_1 x^3+\frac{1}{2} C_2 x^2+C_3 x+C_4 \end{aligned} (15.9)
\begin{aligned}&E I \frac{d^4 y}{d x^4}=-w \\&E I \frac{d^3 y}{d x^3}=V(x)=-w x+C_1 \\&E I \frac{d^2 y}{d x^2}=M(x)=-\frac{1}{2} w x^2+C_1 x+C_2\end{aligned} (1)
Noting that the boundary conditions require that M = 0 at both ends of the beam (Fig. 15.13b), let x = 0 and M = 0 in Eq. (1) and obtain C_2 = 0.
Then make x = L and M = 0 in the same equation and obtain
C_1=\frac{1}{2} wLCarry the values of C_1 and C_2 back into Eq. (1) and integrate twice to obtain
\begin{aligned}&E I \frac{d^2 y}{d x^2}=-\frac{1}{2} w x^2+\frac{1}{2} w L x \\&E I \frac{d y}{d x}=-\frac{1}{6} w x^3+\frac{1}{4} w L x^2+C_3 \\&EI y=-\frac{1}{24} w x^4+\frac{1}{12} w L x^3+C_3 x+C_4\end{aligned} (2)
But the boundary conditions also require that y = 0 at both ends of the beam. Letting x = 0 and y = 0 in Eq. (2), C_4 = 0. Letting x = L and y = 0 in the same equation gives
\begin{gathered}0=-\frac{1}{24} w L^4+\frac{1}{12} w L^4+C_3 L \\C_3=-\frac{1}{24} w L^3\end{gathered}Carrying the values of C_3 and C_4 back into Eq. (2) and dividing both members by EI, the equation of the elastic curve is
y=\frac{w}{24 E I}\left(-x^4+2 L x^3-L^3 x\right) (3)
The maximum deflection is obtained by making x = L/2 in Eq. (3).
|y|_{\max }=\frac{5 w L^4}{384 E I}