Question 15.10: Determine the position and magnitude of the maximum deflecti...

Following the method of Example 15.9, we determine the support reactions and find the bending moment,M, at any section Z in the bay furthest from the origin of the axes. Then,

M=-R_{ A } z+w \frac{L}{4}\left[z-\frac{5 L}{8}\right]  (i)

Examining Eq. (i), we see that the singularity function [z-5 L / 8] does not become zero until z \leq 5 L / 8, although Eq. (i) is valid only for z \geq 3 L / 4. To obviate this difficulty, we extend the distributed load to the support D while simultaneously restoring the status quo by applying an upward distributed load of the same intensity and length as the additional load (Fig. 15.22).

At the section Z, a distance z from A, the bending moment is now given by

M=-R_{ A } z+\frac{w}{2}\left[z-\frac{L}{2}\right]^{2}-\frac{w}{2}\left[z-\frac{3 L}{4}\right]^{2}  (ii)

Equation (ii) is valid for all sections of the beam if the singularity functions are discarded as they become zero. Substituting Eq. (ii) into 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{3}{32} w L z-\frac{w}{2}\left[z-\frac{L}{2}\right]^{2}+\frac{w}{2}\left[z-\frac{3 L}{4}\right]^{2}  (iii)

Integrating Eq. (iii) gives

E I v^{\prime}=\frac{3}{64} w L z^{2}-\frac{w}{6}\left[z-\frac{L}{2}\right]^{3}+\frac{w}{6}\left[z-\frac{3 L}{4}\right]^{3}+C_{1}  (iv)

E I v=\frac{w L z^{3}}{64}-\frac{w}{24}\left[z-\frac{L}{2}\right]^{4}+\frac{w}{24}\left[z-\frac{3 L}{4}\right]^{4}+C_{1} z+C_{2}  (v)

where C_{1} and C_{2} are arbitrary constants. The required boundary conditions are v = 0 when z = 0 and z = L. From the first of these, we obtain C_{2}=0, while the second gives

from which

Equations (iv) and (v) then become

E I v^{\prime}=\frac{3}{64} w L z^{2}-\frac{w}{6}\left[z-\frac{L}{2}\right]^{3}+\frac{w}{6}\left[z-\frac{3 L}{4}\right]^{3}-\frac{27 w L^{3}}{2,048}  (vi)

and

E I v=\frac{w L z^{3}}{64}-\frac{w}{24}\left[z-\frac{L}{2}\right]^{4}+\frac{w}{24}\left[z-\frac{3 L}{4}\right]^{4}-\frac{27 w L^{3}}{2,048} z  (vii)

In this problem, the maximum deflection clearly occurs in the region BC of the beam. Therefore, equating the slope to zero for BC, we have

which simplifies to

z^{3}-1.78 L z^{2}+0.75 z L^{2}-0.046 L^{3}=0  (viii)

Solving Eq. (viii) by trial and error, we see that the slope is zero at z \simeq 0.6 L. Hence, from Eq. (vii), the maximum deflection is