A Belleville spring is made of silicon steel. The spring is compressed completely flat when it is subjected to an axial force of 4500 N. The corresponding maximum stress is \left(1375 \times 10^{6}\right) N / m ^{2}. Assume,
\frac{d_{o}}{d_{i}}=1.75 \quad \text { and } \quad \frac{h}{t}=1.5Calculate
(i) thickness of the washer;
(ii) free height of the washer minus thickness (h);
(iii) outer diameter of the washer; and
(iv) inner diameter of the washer.
\text { Given } P=4500 N \quad \sigma_{\max .}=1375 \times 10^{5} N / m ^{2} .
Step I Thickness of washer
When the spring is compressed completely flat,
\delta=h .
From Eqs (10.44), (10.45) and (10.46),
M=\frac{6}{\pi \log _{e}\left(d_{o} / d_{i}\right)}\left[\frac{\left(d_{o} / d_{i}\right)-1}{\left(d_{o} / d_{i}\right)}\right]^{2} (10.44)
C_{1}=\frac{6}{\pi \log _{e}\left(d_{o} / d_{i}\right)}\left[\frac{\left(d_{o} / d_{i}\right)-1}{\log _{e}\left(d_{o} / d_{i}\right)}-1\right] (10.45)
C_{2}=\frac{6}{\pi \log _{e}\left(d_{o} / d_{i}\right)}\left[\frac{\left(d_{o} / d_{i}\right)-1}{2}\right] (10.46).
M=\frac{6}{\pi \log _{e}\left(d_{o} / d_{i}\right)}\left[\frac{\left(d_{o} / d_{i}\right)-1}{\left(d_{o} / d_{i}\right)}\right]^{2} .
=\frac{6}{\pi \log _{e}(1.75)}\left[\frac{1.75-1}{1.75}\right]^{2}=0.6268 .
C_{1}=\frac{6}{\pi \log _{e}\left(d_{o} / d_{i}\right)}\left[\frac{\left(d_{o} / d_{i}\right)-1}{\log _{e}\left(d_{o} / d_{i}\right)}-1\right] .
=\frac{6}{\pi \log _{e}(1.75)}\left[\frac{1.75-1}{\log _{e}(1.75)}-1\right]=1.161 .
=\frac{6}{\pi \log _{e}(1.75)}\left[\frac{1.75-1}{\log _{e}(1.75)}-1\right]=1.161 .
C_{2}=\frac{6}{\pi \log _{e}\left(d_{o} / d_{i}\right)}\left[\frac{\left(d_{o} / d_{i}\right)-1}{2}\right] .
=\frac{6}{\pi \log _{e}(1.75)}\left[\frac{1.75-1}{2}\right]=1.28 .
Dividing Eq. (10.42) by Eq. (10.43)
P=\frac{E \delta}{\left(1-\mu^{2}\right) M\left(d_{o} / 2\right)^{2}}\left[(h-\delta / 2)(h-\delta) t+t^{3}\right] (10.42).
\sigma=\frac{E \delta}{\left(1-\mu^{2}\right) M\left(d_{o} / 2\right)^{2}}\left[C_{1}(h-\delta / 2)+C_{2} t\right] (10.43).
\frac{P}{\sigma}=\frac{(h-\delta / 2)(h-\delta) t+t^{3}}{C_{1}(h-\delta / 2)+C_{2} t}=\frac{t^{3}}{C_{1}(h-\delta / 2)+C_{2} t} .
\text { because }(h=\delta) .
Substituting values,
\frac{4500}{1375\left(10^{6}\right)}=\frac{t^{3}}{(1.161)(1.5 t-1.5 t / 2)+(1.28) t}=\frac{t^{2}}{2.15} .
t=2.653\left(10^{-3}\right) m =2.653 mm =2.65 mm (i).
Step II Free height of washer minus thickness (h)
h = 1.5 t = 1.5 (2.65) = 3.98 mm = 4 mm (ii)
Step III Outer diameter of washer
From Eq. (10.42),
P=\frac{E \delta}{\left(1-\mu^{2}\right) M\left(d_{o} / 2\right)^{2}}\left[(h-\delta / 2)(h-\delta) t+t^{3}\right] (10.42).
P=\frac{E \delta}{\left(1-\mu^{2}\right) M\left(d_{o} / 2\right)^{2}}\left[(h-\delta / 2)(h-\delta) t+t^{3}\right] .
\text { or } P=\frac{E \delta}{\left(1-\mu^{2}\right) M\left(d_{o} / 2\right)^{2}}\left[t^{3}\right] \quad \text { because }(h=\delta) .
E=207000 N / mm ^{2}=\left(207000 \times 10^{6}\right) N / m ^{2} .
μ = 0.3.
t=\left(2.65 \times 10^{-3}\right) m \quad h=\delta=\left(4 \times 10^{-3}\right)=0.004 m .
Substituting these values,
\text { or } 4500=\frac{\left(207000 \times 10^{6}\right)(0.004)}{\left(1-0.3^{2}\right)(0.6268)\left(d_{o} / 2\right)^{2}}\left[\left(2.65 \times 10^{-3}\right)^{3}\right. .
d_{o}=154.96 \times 10^{-3} m =155 mm (iii).
Step IV Inner diameter of washer
d_{i}=\frac{d_{o}}{1.75}=\frac{155}{1.75}=88.57 mm (iv).