Question 7.1: Derive the moment expression for a uniformly loaded, simply ...
Derive the moment expression for a uniformly loaded, simply supported circular plate of radius a. For 𝜇 =0.3, plot the moment diagram and determine the maximum deflection, rotation, and stress values.
From Figure 7.3, the shear Q at any radius r is given by 2 \pi r Q=\pi r^{2} P , or
Q=\frac{P r}{2}Therefore, from Eqs. (7.7a) and (7.7b),
\frac{ d ^{3} w}{ d r^{3}}+\frac{1}{r} \frac{ d ^{2} w}{ d r^{2}}-\frac{1}{r^{2}} \frac{ d w}{ d r}=\frac{Q}{D} (7.7a)
\frac{ d }{ d r}\left[\frac{1}{r} \frac{ d }{ d r}\left(r \frac{ d w}{ d r}\right)\right]=\frac{Q}{D} (7.7b)
\frac{ d }{ d r}\left[\frac{1}{r} \frac{ d }{ d r}\left(r \frac{ d w}{ d r}\right)\right]=\frac{P r}{2 D}Integrating both sides gives
\frac{ d }{ d r}\left(r \frac{ d w}{ d r}\right)=\frac{P r^{3}}{4 D}+C_{1} rA second integration gives the expression for the slope:
\theta=\frac{ d w}{ d r}=\frac{P r^{3}}{16 D}+\frac{C_{1} r}{2}+\frac{C_{2}}{r} , (1)
and the third integration gives the deflection w as
w=\frac{P r^{4}}{64 D}+\frac{C_{1} r^{2}}{4}+C_{2} \ln r+C_{3} . (2)
At the center of the plate, r =0 and the slope is zero due to symmetry. Hence, from Eq. (1), C_{2} =0 .
At r =a, the moment M_{r} =0 and Eqs. (7.3a) and (7.3b) give
M_{ r }=-D\left(\frac{ d ^{2} w}{ d r^{2}}+\frac{\mu}{r} \frac{ d w}{ d r}\right) (7.3a)
=D\left(\frac{1}{r} \frac{ d w}{ d r}+\mu \frac{\phi}{r}\right) (7.3b)
C_{1}=-\frac{P a^{2}}{8 D}\left(\frac{3+\mu}{1+\mu}\right)At r =a, the deflection is zero, and Eq. (2) gives
C_{3}=\frac{P a^{4}}{64 D}\left(\frac{6+2 \mu}{1+\mu}-1\right)Hence, the deflection as expressed by Eq. (2) becomes
w=\frac{P r^{4}}{64 D}-\frac{r^{2}}{4}\left(\frac{3+\mu}{1+\mu}\right) \frac{P a^{2}}{8 D} +\frac{P a^{4}}{64 D}\left(\frac{6+2 \mu}{1+\mu}-1\right)or
w=\frac{P}{64 D}\left(a^{2}-r^{2}\right)\left(\frac{5+\mu}{1+\mu} \cdot a^{2}-r^{2}\right) (3)
and
\theta=\frac{ d w}{ d r}=\frac{P r}{16 D}\left(r^{2}-\frac{3+\mu}{1+\mu} \cdot a^{2}\right) (4)
The maximum deflection occurs in the middle where r =0. Hence,
w_{\max }=\frac{P a^{4}}{64 D}\left(\frac{5+\mu}{1+\mu}\right)= \frac{P a^{4}}{64}\left(\frac{5+\mu}{1+\mu}\right) \frac{12\left(1-\mu^{2}\right)}{E t^{3}}
and with 𝜇 =0.30
w_{\max }=\frac{0.696 P a^{4}}{E t^{3}}The maximum rotation occurs at the edge where r =a. Hence,
\theta_{\max }=\frac{-P a^{3}}{8 D(1+\mu)}=\frac{-3}{2} P a^{3} \frac{1-\mu}{E t^{3}}and with 𝜇 =0.30
\theta_{\max }=-1.05 \frac{P a^{3}}{E t^{3}}The moment expression is obtained by substituting Eq. (3) into Eqs. (7.3a), (7.3b), and (7.4). Hence,
M_{ t }=-D\left(\frac{\phi}{r}+\mu \frac{ d \phi}{ d r}\right) (7.4)
M_{r}=\frac{P}{16}(3+\mu)\left(a^{2}-r^{2}\right) (5)
M_{t}=\frac{P}{16}\left[a^{2}(3+\mu)-r^{2}(1+3 \mu)\right] (6)
A plot of Eqs. (5) and (6) for 𝜇 =0.3 is shown in Figure 7.4. The plot indicates that the maximum moment occurs in the center and is given by
M_{\max }=\frac{3.3 P a^{2}}{16}or
\sigma_{\max }=\frac{6 M}{t^{2}}=\frac{1.24 P a^{2}}{t^{2}}
