Question 7.3: Determine the maximum moment in a simply supported rectangul...
Determine the maximum moment in a simply supported rectangular plate of length a and width b if the applied load is expressed as
q=q_{0} \sin \frac{\pi x}{a} \sin \frac{\pi y}{b}Let a=60 in., b=25 in., q_{0} =3 psi, and 𝜇 =0.3
Assume w to be of the form
w=C \sin \frac{\pi x}{a} \sin \frac{\pi y}{b}This expression satisfies the boundary conditions of w=0 and M =0 at all four edges. Substituting this expression into Eq. (7.17) gives
\frac{\partial^{4} w}{\partial x^{4}}+\frac{2 \partial^{4} w}{\partial x^{2} \partial y^{2}}+\frac{\partial^{4} w}{\partial y^{4}}=\frac{q(x, y)}{D} (7.17)
C=\frac{q_{0}}{D \pi^{4}\left(1 / a^{2}+1 / b^{2}\right)^{2}}and the deflection expression becomes
w=\frac{q_{0}}{D \pi^{4}\left(1 / a^{2}+1 / b^{2}\right)^{2}} \sin \frac{\pi x}{a} \sin \frac{\pi y}{b} .Substituting this expression into Eq. (7.18) gives
M_{x}=-D\left(\frac{\partial^{2} w}{\partial x^{2}}+\mu \frac{\partial^{2} w}{\partial y^{2}}\right) (7.18a)
M_{y}=-D\left(\frac{\partial^{2} w}{\partial y^{2}}+\mu \frac{\partial^{2} w}{\partial x^{2}}\right) (7.18b)
M_{x y}=D \frac{\partial^{2} w}{\partial x \partial y} (7.18c)
M_{x}=\frac{q_{0}}{\pi^{2}\left(1 / a^{2}+1 / b^{2}\right)^{2}}\left(\frac{1}{a^{2}}+\frac{\mu}{b^{2}}\right) \sin \frac{\pi x}{a} \sin \frac{\pi y}{b} ,
M_{y}=\frac{q_{0}}{\pi^{2}\left(1 / a^{2}+1 / b^{2}\right)^{2}}\left(\frac{\mu}{a^{2}}+\frac{1}{b^{2}}\right) \sin \frac{\pi x}{a} \sin \frac{\pi y}{b} ,
M_{x y}=\frac{q_{0}(1-\mu)}{\pi^{2}\left(1 / a^{2}+1 / b^{2}\right)^{2} a b} \cos \frac{\pi x}{a} \cos \frac{\pi y}{b} .
The maximum values of M_{x} and M_{y} occur when x=a/2 and y=b/2. A comparison of the factors in parentheses in the expressions for M_{x} and M_{y} indicates that M_{y} will always give a larger value of M_{x} for the given values of a and b. Accordingly, the maximum value of M is given by
M_{\max }=\frac{q_{0}}{\pi^{2}\left(1 / a^{2}+1 / b^{2}\right)^{2}}\left(\frac{\mu}{a^{2}}+\frac{1}{b^{2}}\right)M_{\max }=\frac{3.0}{\pi^{2}\left(1 / 60^{2}+1 / 25^{2}\right)^{2}}\left(\frac{0.3}{60^{2}}+\frac{1}{25^{2}}\right)
= 145.1 in.-lb∕in.
The maximum value of M_{xy} occurs when x=0 and y=0. Hence, the maximum value of M_{xy} is given by
M_{x y}=\frac{3.0(0.7)}{\pi^{2}\left(1 / 60^{2}+1 / 25^{2}\right)^{2}(60)(25)}= 40.2 in.-lb∕in.