Question 10.25: A semi-elliptic multi-leaf spring is used for the suspension...

$\text { Given } 2 P=30 kN \quad 2 L=1.2 m \quad n_{f}=2 \quad n_{g}=10$.

$E =207000 N / mm ^{2} \quad S_{y t}=1500 N / mm ^{2} \quad(f s)=2.5$.

Step I Cross-section of the leaves
2 P = 30 kN or           P = 15 000 N
2 L = 1.2 m or             L = 600 mm.

$\sigma_{b}=\frac{S_{y t}}{(f s)}=\frac{1500}{2.5}=600 N / mm ^{2}$.

From Eq. (10.41),

$\sigma_{b}=\frac{6 P L}{n b t^{2}}$            (10.41).

$\sigma_{b}=\frac{6 P L}{n b t^{2}} \quad \text { or } \quad(600)=\frac{6(15000)(600)}{(2+10) b t^{2}}$.

$b t^{2}=7500 mm ^{3}$.

Assuming a standard width of 60 mm,

$t^{2}=\frac{7500}{60} \text { or } t=11.18 \text { or } 12 mm$.

Cross-section of the leaves = 60 × 12 mm                (i)
Step II Deflection at the end of the spring

From Eq. (10.38),
$\delta=\frac{12 P L^{3}}{E b t^{3}\left(3 n_{f}+2 n_{g}\right)}$              (10.38).

$\delta=\frac{12 P L^{3}}{E b t^{3}\left(3 n_{f}+2 n_{g}\right)}$.

$=\frac{12(15000)(600)^{3}}{(207000)(60)(12)^{3}(3 \times 2+2 \times 10)}$.

= 69.68 mm             (ii).