GME asked 7 months ago
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.
ਸਾਜਿਦ ਅਹਿਮਦ answered 7 months ago
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.