Question 16.7: Pump Diaphragm Stress Analysis Determine deflection and stre...
Pump Diaphragm Stress Analysis
Determine deflection and stress for a clamped circular plate of radius a, representing a pump diaphragm, subjected to uniformly distributed load p_{o} (Figure 16.16).
Assumptions: Presuppositions of the thin-plate bending theory given in Section 4.10 applies.
Design Decision: The boundary conditions are
w=0, \quad \frac{ d w}{ d r}=0, \quad(r=a)
The terms involving logarithms in Equation 16.60 give an infinite displacement at r = 0 for all values of c_{1} and c_{2} except 0, so c_{1} = c_{2} = 0. Satisfying the preceding conditions, we have
c_3=\frac{p_o a^2}{32 D}, \quad c_4=\frac{p_o a^4}{64 D}
The deflection is then
w=\frac{p_o}{64 D}\left(a^2-r^2\right)^2 (16.61)
The maximum displacement, occurring at the center of the plate, is
w_{\max }=\frac{p_o a^4}{64 D} (16.62)
Expressions for the bending moments may be determined by Equation 16.61 in the
form
\begin{array}{l} M_r=\frac{p_o}{16}\left[(1+\nu) a^2-(3+\nu) r^2\right] \\ M_\theta=\frac{p_o}{16}\left[(1+\nu) a^2-(1+3 \nu) r^2\right] \end{array} (16.63)
Algebraically extreme values of the moments are found at the center and at the edge. At the edge (r = a), Equations 16.63 result in
M_r=-\frac{p_o a^2}{8}, \quad M_\theta=-\frac{\nu p_o a^2}{8}
while at r=0, M_r=M_\theta=(1+\nu) p_o a^2 / 16 . We observe that the maximum moment occurs at the edge. Hence, we have the value of the maximum stress as
\sigma_{r, \max }=\frac{6 M_r}{t^2}=-\frac{3 p_o}{4}\left\lgroup \frac{a}{t} \right\rgroup ^2
Comment: The minus sign means a compressive stress in the bottom half of the plate.