Question 15.7: Determine the deflection curve and the mid-span deflection o...

The support reactions are each wL/2 and the bending moment, M, at any section Z, a distance z from the lefthand support is

M=-\frac{w L}{2} z+\frac{w z^{2}}{2}  (i)

Substituting for M in the second of Eqs. (15.31), we obtain

u^{\prime \prime}=-\frac{M_{y}}{E I_{y y}}, \quad v^{\prime \prime}=-\frac{M_{x}}{E I_{x x}}  (15.31)

E I v^{\prime \prime}=\frac{w}{2}\left(L z-z^{2}\right)  (ii)

Integrating, we have

From symmetry, it is clear that, at the mid-span section, the gradient v^{\prime}=0. Hence,

which gives

Therefore,

E I v^{\prime}=\frac{w}{24}\left(6 L z^{2}-4 z^{3}-L^{3}\right)  (iii)

Integrating again gives

Since v=0 when z=0 (or since v=0 when z=L), it follows that C_{2}=0 and the deflected shape of the beam has the equation

v=\frac{w}{24 E I}\left(2 L z^{3}-z^{4}-L^{3} z\right)  (iv)

The maximum deflection occurs at mid-span, where z=L/2 and is

v_{\text {mid-span }}=-\frac{5 w L^{4}}{384 E I}  (v)