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,
$\frac{d_{o}}{d_{i}}=1.75 \quad \text { and } \quad \frac{h}{t}=1.5$
Calculate
(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.

Question

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 [latex] \left(1375 imes 10^{6} ight) N / m ^{2}[/latex]. Assume,

[latex] \frac{d_{o}}{d_{i}}=1.75 \quad ext { and } \quad \frac{h}{t}=1.5 [/latex]

Calculate
(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.

Accepted Answer

[latex] ext { Given } P=4500 N \quad \sigma_{\max .}=1375 imes 10^{5} N / m ^{2} [/latex].

Step I Thickness of washer
When the spring is compressed completely flat,

[latex] \delta=h [/latex].

From Eqs (10.44), (10.45) and (10.46),

[latex] M=\frac{6}{\pi \log _{e}\left(d_{o} / d_{i} ight)}\left[\frac{\left(d_{o} / d_{i} ight)-1}{\left(d_{o} / d_{i} ight)} ight]^{2} [/latex] (10.44)

[latex] C_{1}=\frac{6}{\pi \log _{e}\left(d_{o} / d_{i} ight)}\left[\frac{\left(d_{o} / d_{i} ight)-1}{\log _{e}\left(d_{o} / d_{i} ight)}-1 ight] [/latex] (10.45)

[latex] C_{2}=\frac{6}{\pi \log _{e}\left(d_{o} / d_{i} ight)}\left[\frac{\left(d_{o} / d_{i} ight)-1}{2} ight] [/latex] (10.46).

[latex] M=\frac{6}{\pi \log _{e}\left(d_{o} / d_{i} ight)}\left[\frac{\left(d_{o} / d_{i} ight)-1}{\left(d_{o} / d_{i} ight)} ight]^{2} [/latex].

[latex] =\frac{6}{\pi \log _{e}(1.75)}\left[\frac{1.75-1}{1.75} ight]^{2}=0.6268 [/latex].

[latex] C_{1}=\frac{6}{\pi \log _{e}\left(d_{o} / d_{i} ight)}\left[\frac{\left(d_{o} / d_{i} ight)-1}{\log _{e}\left(d_{o} / d_{i} ight)}-1 ight] [/latex].

[latex] =\frac{6}{\pi \log _{e}(1.75)}\left[\frac{1.75-1}{\log _{e}(1.75)}-1 ight]=1.161 [/latex].

[latex] C_{2}=\frac{6}{\pi \log _{e}\left(d_{o} / d_{i} ight)}\left[\frac{\left(d_{o} / d_{i} ight)-1}{2} ight] [/latex].

[latex] =\frac{6}{\pi \log _{e}(1.75)}\left[\frac{1.75-1}{2} ight]=1.28 [/latex].

Dividing Eq. (10.42) by Eq. (10.43)

[latex] P=\frac{E \delta}{\left(1-\mu^{2} ight) M\left(d_{o} / 2 ight)^{2}}\left[(h-\delta / 2)(h-\delta) t+t^{3} ight] [/latex] (10.42).

[latex] \sigma=\frac{E \delta}{\left(1-\mu^{2} ight) M\left(d_{o} / 2 ight)^{2}}\left[C_{1}(h-\delta / 2)+C_{2} t ight] [/latex] (10.43).

[latex] \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} [/latex].

[latex] ext { because }(h=\delta) [/latex].

Substituting values,

[latex] \frac{4500}{1375\left(10^{6} ight)}=\frac{t^{3}}{(1.161)(1.5 t-1.5 t / 2)+(1.28) t}=\frac{t^{2}}{2.15} [/latex].

[latex] t=2.653\left(10^{-3} ight) m =2.653 mm =2.65 mm [/latex] (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),

[latex] P=\frac{E \delta}{\left(1-\mu^{2} ight) M\left(d_{o} / 2 ight)^{2}}\left[(h-\delta / 2)(h-\delta) t+t^{3} ight] [/latex] (10.42).

[latex] P=\frac{E \delta}{\left(1-\mu^{2} ight) M\left(d_{o} / 2 ight)^{2}}\left[(h-\delta / 2)(h-\delta) t+t^{3} ight] [/latex].

[latex] ext { or } P=\frac{E \delta}{\left(1-\mu^{2} ight) M\left(d_{o} / 2 ight)^{2}}\left[t^{3} ight] \quad ext { because }(h=\delta) [/latex].

[latex] E=207000 N / mm ^{2}=\left(207000 imes 10^{6} ight) N / m ^{2} [/latex].

μ = 0.3.

[latex] t=\left(2.65 imes 10^{-3} ight) m \quad h=\delta=\left(4 imes 10^{-3} ight)=0.004 m [/latex].

Substituting these values,

[latex] ext { or } 4500=\frac{\left(207000 imes 10^{6} ight)(0.004)}{\left(1-0.3^{2} ight)(0.6268)\left(d_{o} / 2 ight)^{2}}\left[\left(2.65 imes 10^{-3} ight)^{3} ight. [/latex].

[latex] d_{o}=154.96 imes 10^{-3} m =155 mm [/latex] (iii).

Step IV Inner diameter of washer

[latex] d_{i}=\frac{d_{o}}{1.75}=\frac{155}{1.75}=88.57 mm [/latex] (iv).

Question 10.26: A Belleville spring is made of silicon steel. The spring is ...

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