Plates and Shells

Abstract

Plates are flat structures whose planar dimensions are much larger than the thickness. Shells are curved structures whose thickness is much smaller than the other structural dimensions.

Appendix: Solutions

• 14.7.1 Show that Eq. (14.7) is correct.

  $$\nabla^{4} w(x,y) \equiv \frac{{\partial^{4} w}}{{\partial x^{4} }} + 2\frac{{\partial^{2} w}}{{\partial x^{2} }}\frac{{\partial^{2} w}}{{\partial y^{2} }} + \frac{{\partial^{4} w}}{{\partial y^{4} }} = \frac{q(x,y)}{D}$$

Solution

Substituting Eq. (14.3):

$$\begin{aligned} M_{x} & = - D\left( {\frac{{\partial^{2} w}}{{\partial x^{2} }} + \nu \frac{{\partial^{2} w}}{{\partial y^{2} }}} \right) \\ M_{y} & = - D\left( {\frac{{\partial^{2} w}}{{\partial y^{2} }} + \nu \frac{{\partial^{2} w}}{{\partial x^{2} }}} \right) \\ M_{xy} & = - \left( {1 - \nu } \right)D\left( {\frac{{\partial^{2} w}}{\partial x\partial y}} \right) \\ \end{aligned}$$

into (14.6)

$$\frac{{\partial^{2} M_{x} }}{{\partial x^{2} }} + 2\frac{{\partial^{2} M_{xy} }}{\partial x\partial y} + \frac{{\partial^{2} M_{y} }}{{\partial y^{2} }} + q(x,y) = 0$$

gives:

$$\begin{aligned} & \frac{{\partial^{2} \left( {\frac{{\partial^{2} w}}{{\partial x^{2} }} + \nu \frac{{\partial^{2} w}}{{\partial y^{2} }}} \right)}}{{\partial x^{2} }} + 2\frac{{\partial^{2} \left( {1 - \nu } \right)\left( {\frac{{\partial^{2} w}}{\partial x\partial y}} \right)}}{\partial x\partial y} + \frac{{\partial^{2} \left( {\frac{{\partial^{2} w}}{{\partial y^{2} }} + \nu \frac{{\partial^{2} w}}{{\partial x^{2} }}} \right)}}{{\partial y^{2} }} = \frac{q(x,y)}{D} \\ & \frac{{\partial^{4} w}}{{\partial x^{4} }} + \nu \frac{{\partial^{4} w}}{{\partial x^{2} \partial y^{2} }} + 2\left( {1 - \nu } \right)\frac{{\partial^{4} w}}{{\partial x^{2} \partial y^{2} }} + \frac{{\partial^{4} w}}{{\partial y^{4} }} + \nu \frac{{\partial^{4} w}}{{\partial x^{2} \partial y^{2} }} = \frac{q(x,y)}{D} \\ & \frac{{\partial^{4} w}}{{\partial x^{4} }} + 2\frac{{\partial^{4} w}}{{\partial x^{2} \partial y^{2} }} + \frac{{\partial^{4} w}}{{\partial y^{4} }} = \frac{q(x,y)}{D} \equiv \nabla^{4} w \\ \end{aligned}$$

• 14.7.2 Show that Eq. (14.10) gives the unknown constants for a uniformly loaded plate in cylindrical bending.

  $$\begin{aligned} C_{1} & = - \frac{{q_{ \circ } a}}{2D} \\ C_{2} & = 0 \\ C_{3} & = \frac{{q_{ \circ } a^{3} }}{24D} \\ C_{4} & = 0 \\ \end{aligned}$$

Solution

From Eq. (14.9), the general solution is:

$$w(x) = \frac{{q_{ \circ } x^{4} }}{24D} + \frac{{C_{1} x^{3} }}{6} + \frac{{C_{2} x^{2} }}{2} + C_{3} x + C_{4}$$

From Eq.  (14.8), the boundary conditions are:

$$\begin{aligned} & w(0) = w(a) = 0 \\ & M_{x} (0) = M_{x} (a) = 0 \\ \end{aligned}$$

Substitutions give:

$$\begin{aligned} & w(0) = 0\quad \Rightarrow C_{4} = 0 \\ & M_{x} (0) = 0\quad \Rightarrow - D\left( {\frac{{\partial^{2} w}}{{\partial x^{2} }}} \right) = 0\quad \Rightarrow C_{2} = 0 \\ & M_{x} \left( a \right) = 0\quad \Rightarrow \frac{{\left( {q_{ \circ } a^{2} } \right)}}{2D} + C_{1} a = 0\quad \Rightarrow C_{1} = - \frac{{q_{ \circ } a}}{2D} \\ & w(a) = 0\quad \Rightarrow \frac{{q_{ \circ } a^{4} }}{24D} + \left( { - \frac{{q_{ \circ } a}}{2D}} \right)\frac{{a^{3} }}{6} + C_{3} a = 0\quad \Rightarrow C_{3} = \frac{{q_{ \circ } a^{3} }}{24D} \\ \end{aligned}$$

• 14.7.3 Show that Eq. (14.13) is a solution of Eq. (14.12) for a circular plate under uniform loading.

Recall Eq. (14.12):

$$\frac{1}{r}\frac{d}{dr}\{ r\frac{d}{dr}[\frac{1}{r}\frac{d}{dr}(r\frac{dw}{dr})]\} = \frac{q(r)}{D}$$

and (14.13):

$$w(r) = \frac{{qr^{4} }}{64D} + C_{1} \frac{{r^{2} }}{4} + C_{2} \log \frac{r}{a} + C_{3}$$

Solution

Substitution using the known values of the constants $C_{2} = C_{4} = 0;\quad C_{1} = \frac{{ - q_{ \circ a} }}{2D};\quad C_{3} = \frac{{q_{ \circ } a^{2} }}{24D}$

$$\begin{aligned} & \frac{dw}{dr} = \frac{{q_{ \circ } r^{3} }}{16D} + \frac{{C_{1} r}}{2} \\ & \frac{1}{r}\frac{{d\left( {r\left( {\frac{{q_{ \circ } r^{3} }}{16D} + \frac{{C_{1} r}}{2}} \right)} \right)}}{dr} = \frac{1}{r}\frac{{d\left( {\left( {\frac{{q_{ \circ } r^{4} }}{16D} + \frac{{C_{1} r^{2} }}{2}} \right)} \right)}}{dr} = \frac{1}{r}\left( {\left( {\frac{{q_{ \circ } r^{3} }}{4D} + C_{1} r} \right)} \right) \\ & r\frac{{d\left[ {\frac{1}{r}\left( {\frac{{q_{ \circ } r^{3} }}{4D} + C_{1} r} \right)} \right]}}{dr} = r\frac{{d\left( {\left( {\frac{{q_{ \circ } r^{2} }}{4D} + C_{1} } \right)} \right)}}{dr} = \left( {\frac{{q_{ \circ } r^{2} }}{2D}} \right) \\ & \frac{1}{r}\frac{{d\left( {\frac{{q_{ \circ } r^{2} }}{2D}} \right)}}{dr} = \frac{1}{r}\frac{{q_{ \circ } r}}{D} = \frac{{q_{ \circ } }}{D} \\ \end{aligned}$$

• 14.7.4 What is the maximum vertical deflection of the uniformly loaded plate in cylindrical bending (Fig. 14.4)?

Solution

From symmetry, the maximum deflection is at the center $x = \frac{a}{2}$. The plate deflection equation with known constants is:

$$w(x) = \frac{{q_{ \circ } x^{4} }}{24D} - \frac{{q_{ \circ } ax^{3} }}{12D} + \frac{{q_{ \circ } a^{3} }}{24D}x$$

At $x = \frac{a}{2}$:

$$\begin{aligned} w(\frac{a}{2}) & = \frac{{q_{ \circ } \left( {\frac{a}{2}} \right)^{4} }}{24D} - \frac{{q_{ \circ } a\left( {\frac{a}{2}} \right)^{3} }}{12D} + \frac{{q_{ \circ } a^{3} }}{24D}\frac{a}{2} \\ w(\frac{a}{2}) & = \frac{{q_{ \circ } a^{4} }}{12D}\left( {\frac{1}{32} - \frac{4}{32} + \frac{8}{32}} \right) = \frac{{q_{ \circ } a^{4} }}{12D}\left( {\frac{5}{32}} \right) = \frac{{5q_{ \circ } a^{4} }}{384D} \\ \end{aligned}$$

Note that for negative $q_{ \circ }$, the deflection is negative.