A flat steel strip is used as a spring to maintain a force against part of a cabinet latch in a

commercial printer, as shown in Figure P5–9. When the cabinet door is open, the spring is deflected

by the latch pin by an amount y1 = 0.25 mm. The pin causes the deflection to increase to 0.40 mm

when the door is closed. Young Modulus for steel E=207 GPa.

**Solution:**

This is a problem related to Beam-deflection for statically indeterminate beams.

Here we have a force ‘P’ acting variable distance from both sides of a spring which is a supported cantilever

Assuming that

Deflection is proportional and the bending moment are proportional to the force P

We have

L = 40 mm

b = 5 mm

d = 0.6 mm

y1= 0.25 mm

y2 = 0.40 mm

Factor of safety, n = 3

Modulus of Elasticity of spring, E = 209 GPa = 209 x 103 N/mm2

General representation of these conditions is shown in Figure.1

Deflection at point B due to force P is given by

We have

■ Moment of Inertia, I is

■ Section modulus, Z is

Solving for P

Substituting values

For yb = y1 = 0.25 mm, the force P acting is

For yb = y2 = 0.40 mm, the force P acting is

**■ Calculating moments at A and B at this condition**

Moment at A is

Moment at B is

Therefore at P1 = 11.80 N, the moment at A is given by

■ Stress at this point σA is

At P2 = 18.89 N, the moment at A is given by

■ Stress at this point σA is

■ The mean average stress σm is given by

■ The alternating stress σa is given by

The Stress ratio R is given by

From the σm and σa values the suitable material for the spring is **SAE 4140 OQT 400**

Tensile strength = 2000 MPa

Yield strength = 1730 MPa

It has the highest ultimate strength with > 10% elongation for good ductility.

** **