A semi-elliptical spring of length 500 mm, thickness 10 mm, and 40 mm broad is subjected to a load of 5 kN and get straightened. If the stress in spring should not cross 250 MPa, then calculate the number of plates, the initial radius of curvature, and central deflection. Take E as 200 GPa.
Answers
Answer: The primary function of a spring is to deflect or distort under load and to recover its original
shape when the load is released. During deflection or distortion, it absorbs energy and
release the same as and when required. Springs are used in many engineering applications
such as automobiles and railway buffers in order to cushion, absorb or control energy due to
shock and vibrations. Springs will suffer a sizeable change in form without being distorted
permanently when the loads are applied. Springs are generally classified as leaf springs or
helical springs. Leaf springs consist of a number of thin curved plates, each of same
thiclcness and width but of different lengths, all bent to the same curvature. Helical springs
are formed by coiling thick spring wire into a helix. Helical springs are classified into two
groups. When the helix angle is less than about lo", it is named as close-coiled helical
spring. In such springs, the wire experiences too little bending or direct shear stress and their
effect is neglected. Torsional stresses are predominant in such springs. If, however the helix
angle is significant, then the wire experiences both torsional and bending stresses. Such type
of spring is termed as open-coiled helical spring.