Question 15.2: The simply supported prismatic beam AB carries a uni- 611 fo...

Drawing the free-body diagram of the portion AD of the beam (Fig. 15.13) and taking moments about D, we find that

M = \frac{1}{2} wL x – \frac{1}{2} wx^{2}                       (15.12)

Substituting for M into Eq. (15.4) and multiplying both members of this equation by the constant EI, we write

EI \frac{d^{2}y}{dx^{2}} = \frac{M(x)}{EI}                    (15.4)

EI \frac{d^{2}y}{dx^{2}} = – \frac{1}{2} wx^{2} + \frac{1}{2} wL x                           (15.13)

Integrating twice in x, we have

EI \frac{dy}{dx}= – \frac{1}{6} wx^{3} + \frac{1}{4} wL  x^{2} + C_{1}                                    (15.14)

EI  y = – \frac{1}{24}  wx^{4}+ \frac{1}{2}  wL  x^{3} + C_{1}x + C_{2}                      (15.15)

Observing that y = 0 at both ends of the beam (Fig. 15.14), we first let x = 0 and y = 0 in Eq. (15.15) and obtain C_{2} = 0. We then make x = L and y = 0 in the same equation and write

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

C_{1} = – \frac{1}{24} wL^{3}

Carrying the values of C_{1}   and   C_{2} back into Eq. (15.15), we obtain the equation of the elastic curve:

EI  y = – \frac{1}{24} wx^{4} + \frac{1}{12}  wL  x^{3} – \frac{1}{24} wL^{3}x

or

y = \frac{w}{24EI} (-x^{4} + 2Lx^{3} – L^{3}x)                                (15.16)

Substituting into Eq. (15.14) the value obtained for C_{1}, we check that the slope of the beam is zero for x = L/2 and that the elastic curve has a minimum at the midpoint C of the beam (Fig. 15.15). Letting x = L/2 in Eq. (15.16), we have

y_{C} =\frac{w}{24EI} \left(- \frac{L^{4}}{16} + 2L \frac{L^{3}}{8} – L^{3} \frac{L}{2} \right) = – \frac{5wL^{4}}{384EI}

The maximum deflection or, more precisely, the maximum absolute value of the deflection, is thus

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