Question 15.4: The simply supported prismatic beam AB carries a uniformly d...

Since ω = constant, the first three of Eqs. (15.33) yield

$EI \frac{d^{4}y}{dx^{4}} = -w(x)$

$EI \frac{d^{3}y}{dx^{3}} = V(x) = -\int{w(x) dx} + C_{1}$

$EI \frac{d^{2}y}{dx^{2}} = M(x) = -\int{dx} \int{w(x) dx} + C_{1}x + C_{2}$                                      (15.33)

$EI \frac{d^{4}y}{dx^{4}} = -w$

$EI \frac{d^{3}y}{dx^{3}} = V(x) = -wx + C_{1}$

$EI \frac{d^{2}y}{dx^{2}} = M(x) = -\frac{1}{2}  wx^{2} + C_{1}x + C_{2}$                                (15.34)

Noting that the boundary conditions require that M = 0 at both ends of the beam (Fig. 15.22), we first let x = 0 and M = 0 in Eq. (15.34) and obtain $C_{2}$ = 0. We then make x = L and M = 0 in the same equation and obtain $C_{1}= \frac{1}{2} wL$ .

Carrying the values of $C_{1}   and   C_{2}$ back into Eq. (15.34), and integrating twice, we write

$EI \frac{d^{2}y}{dx^{2}} = – \frac{1}{2} wx^{2} + \frac{1}{2} wL x$

$EI \frac{dy}{dx} = – \frac{1}{6} wx^{3} + \frac{1}{4} wL x^{2} + C_{3}$

$EI  y = – \frac{1}{24} wx^{4} + \frac{1}{12}  wL x^{3} + C_{3} x + C_{4}$                                 (15.35)

But the boundary conditions also require that y = 0 at both ends of the beam. Letting x = 0 and y = 0 in Eq. (15.35), we obtain $C_{4} = 0$; letting x = L and y = 0 in the same equation, we write

$0 = – \frac{1}{24} wL^{4} + \frac{1}{12} wL^{4} + C_{3}L$

$C_{3} = – \frac{1}{24} wL^{3}$

Carrying the values of $C_{3}   and   C_{4}$ back into Eq. (15.35) and dividing both members by EI, we obtain the equation of the elastic curve:

$y =\frac{w}{24EI} (-x^{4} + 2L x^{3} – L^{3}x)$                           (15.36)

The value of the maximum deflection is obtained by making x = L/2 in Eq. (15.36). We have

$|y|_ {max} = \frac{5wL^{4}}{384EI}$