Question 13.5: Figure 13.6(a) shows a simply supported beam carrying a conc...

The support reactions are each W/2 and the bending moment M at a section X a distance x (≤ L/2) from the left-hand support is

M = \frac{W}{2}x                                              (i)

From Eq. (13.3) we have

\frac{d^2v}{dx^2} =\frac{M}{EI}                                       (13.3)

EI\frac{d^2v}{dx^2} =\frac{W}{2}x                                                            (ii)

Integrating we obtain

EI\frac{dv}{dx} =\frac{W}{2} \frac{x^2}{2} +C_1

From symmetry the slope of the beam is zero at mid span where x = L/2. Thus C_1 = −WL^2/16 and

EI\frac{dv}{dx} =\frac{W}{16} (4x^2-L^2)                                                   (iii)

Integrating Eq. (iii) we have

EIv = \frac{W}{16} \left(\frac{4x^3}{3} – L^2x \right) +C_2

and when x = 0, v = 0 so that C_2 = 0. The equation of the deflection curve is, therefore

v=\frac{W}{48EI} (4x^3-3L^2x)                                                (iv)

The maximum deflection occurs at mid-span and is

v_{\text{mid-span}}= – \frac{WL^3}{48EI}                                              (v)

Note that in this problem we could not use the boundary condition that v = 0 at x = L to determine C_2 since Eq. (i) applies only for 0 ≤ x ≤ L/2; it follows that Eqs (iii) and (iv) for slope and deflection apply only for 0 ≤ x ≤ L/2 although the deflection curve is clearly symmetrical about mid span.