Question 15.9: Determine the position and magnitude of the maximum upward a...

A consideration of the overall equilibrium of the beam gives the support reactions; thus,

Using the method of singularity functions and taking the origin of axes at the left-hand support, we write an expression for the bending moment, M, at any section Z between D and F, the region of the beam furthest from the origin:

M=-R_{ A } z+W[z-a]+W[z-2 a]-2 W[z-3 a]  (i)

Substituting for M in 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{3}{4} W z-W[z-a]-W[z-2 a]+2 W[z-3 a]  (ii)

Integrating Eq. (ii) and retaining the square brackets, we obtain

E I v^{\prime}=\frac{3}{8} W z^{2}-\frac{W}{2}[z-a]^{2}-\frac{W}{2}[z-2 a]^{2}+W[z-3 a]^{2}+C_{1}  (iii)

and

E l v=\frac{1}{8} W z^{3}-\frac{W}{6}[z-a]^{3}-\frac{W}{6}[z-2 a]^{3}+\frac{W}{3}[z-3 a]^{3}+C_{1} z+C_{2}  (iv)

in which C_{1} and C_{2} are arbitrary constants. When z = 0 (at A), v = 0 and hence C_{2}=0 . Note that the second, third, and fourth terms on the right-hand side of Eq. (iv) disappear for z<a . also v = 0 at z = 4a (F), so that, from Eq. (iv), we have

which gives

Equations (iii) and (iv) now become

E I v^{\prime}=\frac{3}{8} W z^{2}-\frac{W}{2}[z-a]^{2}-\frac{W}{2}[z-2 a]^{2}+W[z-3 a]^{2}-\frac{5}{8} W a^{2}  (v)

and

E I v=\frac{1}{8} W z^{3}-\frac{W}{6}[z-a]^{3}-\frac{W}{6}[z-2 a]^{3}+\frac{W}{3}[z-3 a]^{3}-\frac{5}{8} W a^{2} z  (vi)

respectively. To determine the maximum upward and downward deflections, we need to know in which bays v^{\prime}=0 and thereby which terms in Eq. (v) disappear when the exact positions are being located. One method is to select a bay and determine the sign of the slope of the beam at the extremities of the bay. A change of sign indicates that the slope is zero within the bay. By inspection of Fig. 15.20, it seems likely that the maximum downward deflection occurs in BC. At B, using Eq. (v),

which is clearly negative. At C,

which is positive. Therefore, the maximum downward deflection does occur in BC and its exact position is located by equating v^{\prime} to zero for any section in BC. Thus, from Eq. (v),

or, simplifying,

0=z^{2}-8 a z+9 a^{2}  (vii)

Solution of Eq. (vii) gives

z = 1.35 a

so that the maximum downward deflection is, from Eq. (vi),

that is,

In a similar manner it can be shown that the maximum upward deflection lies between D and F at z = 3.42a and that its magnitude is

An alternative method of determining the position of maximum deflection is to select a possible bay, set v^{\prime}=0 for that bay and solve the resulting equation in z. If the solution gives a value of z that lies within the bay, then the selection is correct; otherwise, the procedure must be repeated for a second and possibly a third and a fourth bay. This method is quicker than the former if the correct bay is selected initially; if not, the equation corresponding to each selected bay must be completely solved, a procedure clearly longer than determining the sign of the slope at the extremities of the bay.