Question 15.CA.1: The cantilever beam AB is of uniform cross section and carri...
The cantilever beam AB is of uniform cross section and carries a load P at its free end A (Fig. 15.9a). Determine the equation of the elastic curve and the deflection and slope at A.
Using the free-body diagram of the portion AC of the beam (Fig. 15.9b), where C is located at a distance x from end A,
M = −Px (1)
Substituting for M into Eq. (15.4) and multiplying both members 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}=-P xIntegrating in x,
E I \frac{d y}{d x}=-\frac{1}{2} P x^2+C_1 (2)
Now observe the fixed end B where x = L and θ = dy/dx = 0 (Fig. 15.9c). Substituting these values into Eq. (2) and solving for C_1 gives
C_1=\frac{1}{2} P L^2which we carry back into Eq. (2):
E I \frac{d y}{d x}=-\frac{1}{2} P x^2+\frac{1}{2} P L^2 (3)
Integrating both members of Eq. (3),
EI y=-\frac{1}{6} P x^3+\frac{1}{2} P L^2 x+C_2 (4)
But at B, x = L, y = 0. Substituting into Eq. (4),
\begin{gathered}0=-\frac{1}{6} P L^3+\frac{1}{2} P L^3+C_2 \\C_2=-\frac{1}{3} P L^3\end{gathered}Carrying the value of C_2 back into Eq. (4), the equation of the elastic curve is
EI y=-\frac{1}{6} P x^3+\frac{1}{2} P L^2 x-\frac{1}{3} P L^3or
y=\frac{P}{6 E I}\left(-x^3+3 L^2 x-2 L^3\right) (5)
The deflection and slope at A are obtained by letting x = 0 in Eqs. (3) and (5).
y_A=-\frac{P L^3}{3 E I} \quad \text { and } \quad \theta_A=\left(\frac{d y}{d x}\right)_A=\frac{P L^2}{2 E I}