Question 15.8: Figure 15.19(a) shows a simply supported beam carrying a con...

The support reactions are eachW/2 and the bending moment Mat a section Z a distance z(\leq L / 2) from the lefthand support is

M=-\frac{W}{2} z  (i)

From the second of Eqs. (15.31), we have

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} z  (ii)

Integrating we obtain

From symmetry, the slope of the beam is zero at mid-span, where z = L/2. Thus, C_{1}=-W L^{2} / 16 and

E I v^{\prime}=\frac{W}{16}\left(4 z^{2}-L^{2}\right)  (iii)

Integrating Eq. (iii), we have

and, when z = 0, v = 0, so that C_{2}=0. The equation of the deflection curve is, therefore,

v=\frac{W}{48 E I}\left(4 z^{3}-3 L^{2} z\right)  (iv)

The maximum deflection occurs at mid-span and is

v_{\text {mid-span }}=-\frac{W L^{3}}{48 E I}  (v)

Note that, in this problem, we could not use the boundary condition that v = 0 at z = L to determine C_{2}, since Eq. (i) applies only for 0 \leq z \leq L / 2; it follows that Eqs. (iii) and (iv) for slope and deflection apply only for 0 \leq z \leq L / 2, although the deflection curve is clearly symmetrical about mid-span. Examples 15.5–15.8 are frequently regarded as ‘standard’ cases of beam deflection.