Question 16.12: A laminated spring is to carry a central load of 5 kN. It is...
A laminated spring is to carry a central load of 5 kN. It is made up of 16 plates 75 mm wide, 6 mm thick. If the central dip is 40 mm and the permissible bending stress is 190 MPa, determine the length and curvature for the leaf spring.
Using Eq. (16.12)
W=\frac{8nEbt^3\delta _0}{3L^3} (16.12)
\sigma _0=\frac{3W_0}{2nbt^2}L=\frac{190\times2\times16\times75\times6^2}{3\times5000} =1094.9 mm say 1100 mm
Radius of curvature is given by Eq. (16.9)
\frac{1}{R}-\frac{1}{r}=\frac{M}{EI} =\frac{6Wa}{Ebt^3} (16.9)
R=\frac{L^2}{8\delta _0} =\frac{1100}{8\times40} =3.78 mRelated Answered Questions
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