Question 7.2: Derive the moment expression for a uniformly loaded circular...
Derive the moment expression for a uniformly loaded circular plate of radius a that is fixed at the edge. For 𝜇 =0.3, plot the moment diagram and determine the maximum deflection and stress values.
From Example 7.1,
w=\frac{P r^{4}}{64 D}+\frac{C_{1} r^{2}}{4}+C_{2} \ln r+C_{3} (1)
and
\theta=\frac{P r^{3}}{16 D}+\frac{C_{1} r}{2}+\frac{C_{2}}{r} (2)
At the center of the plate, r =0 and the slope is zero due to symmetry. Hence, from Eq. (2), C_{2} =0.
At r =a, the slope is zero and from Eq. (2),
C_{1}=\frac{-P a^{2}}{8 D}Also at r =a, the deflection is zero and from Eq. (1),
C_{3}=\frac{P a^{4}}{64 D}Hence, Eq. (1) becomes
w=\frac{P r^{4}}{64 D}-\frac{P a^{2} r^{2}}{32 D}+\frac{P a^{4}}{64 D}= \frac{P}{64 D}\left(a^{2}-r^{2}\right)^{2} (3)
The maximum value of deflection occurs at r =0:
w=\frac{P a^{4}}{64 D}=\frac{P a^{4}}{64} \frac{12\left(1-\mu^{2}\right)}{E t^{3}}and for 𝜇 =0.3,
w_{\max }=0.171\left(\frac{P a^{4}}{E t^{3}}\right)The moment expression is obtained by substituting Eq. (3) into Eqs. (7.3a), (7.3b), and (7.4). Hence,
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)
M_{ t }=-D\left(\frac{\phi}{r}+\mu \frac{ d \phi}{ d r}\right) (7.4)
M_{ r }=\frac{P}{16}\left[a^{2}(1+\mu)-r^{2}(3+\mu)\right] (4)
and
M_{ t }=\frac{P}{16}\left[a^{2}(1+\mu)-r^{2}(1+3 \mu)\right] (5)
A plot of Eqs. (4) and (5) for 𝜇 =0.3 is shown in Figure 7.5. The plot indicates that the maximum moment occurs at the edge and is given by
M_{\max }=\frac{-P a^{2}}{8}and
\sigma_{\max }=\frac{6 M}{t^{2}}=\frac{-0.75 P a^{2}}{t^{2}}