Question 15.CA.1: The cantilever beam AB is of uniform cross section and carri...

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)

Integrating 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

which 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),

Carrying the value of $C_2$ back into Eq. (4), the equation of the elastic curve is

or

$y=\frac{P}{6 E I}\left(-x^3+3 L^2 x-2 L^3\right)$      (5)

The deflection and slope at A are obtained by letting x = 0 in Eqs. (3) and (5).