Question 5.1: For the flat steel spring shown in Figure 5-6, compute the m...

Objective           Compute the maximum, minimum, inean, and altemafing tensile stresses in the flat spring. Compute the stress ratio, R.
Given                Layout shown in Figure 5-6. The spring is steel: L = 65 mm.
Spring cross section dimensions: t = 0.80 mm and b = 6.0 mm.
Maximum deflection ofthe spring at the follower = 8.0 mm.
Minimum deflection ofthe spring at the follower = 3.0 mm.
Analysis Point A at the base of the spring experiences the maximum tensile stress. Determine the force exerted on the spring by the follower for each level of deflection using the formulas from Table A14-2, Case (a). Compute the bending moment at the base ofthe spring for each deflection. Then compute the stresses at point A using the bending stress formula, σ = Mc/I. Use Equations (5-1), (5-2), and (5-3) for the mean and alternating stresses and R.

\sigma _m=(\sigma _{max}+\sigma _{min})/2                    (5-1)
\sigma _a=(\sigma _{max}-\sigma _{min})/2                    (5-2)
Stress ratio R =\frac{minimum \ stress}{maximum \ stress} =\frac{\sigma _{min}}{\sigma _{max}}
Stress ratio A=\frac{alternating \ stress}{mean \ stress} =\frac{\sigma _{a}}{\sigma _{m}}                     (5-3)

Results       Case (a) of Table Table A14-2 gives the following formula for the amount of deflection of a cantilever for a given applied force:

y = PL^3/3EI

Solve for the force as a function of deflection:

P = 3EIy/L^3

Appendix 3 gives the modulus of elasticity for steel to be E = 207 GPa. The moment of inertia,I, for the spring cross section is found from

I= bt^3/12 = (6.00mm)(0.80mm)^3/12 = 0.256 \ mm^4

Then the force on the spring when the deflection y is 3.0 mm is

P=\frac{3(207 \times 10^9 \ N/m^2)(0.256 \ mm^4)(3.0 \ mm)}{(65 \ mm)^3}\frac{(1.0 \ m^2)}{(10^6 \ mm^2)}=1 .74 \ N

The bending moment at the support is

M = P.L = (1.74 N)(65 mm) = 113 N.mm

The bending .stress at point A caused by this moment is

\sigma =\frac{Mc}{I} =\frac{(113 \ N.mm)(0.40 \ mm)}{0.265 \ mm^4} =176 \ /mm^2=176 / MPa

This is the lowest stress that the spring sees in service, and therefore \sigma_{min} = 176 MPa.
Because the force on the spring is proportional to the deflection, the force exerted when the deflection is 8.0 mm is

P = (1.74  \ N)(8.0 \ mm)/(3.0 \ mm) = 4.63 N

The bending moment is

M = PL = (4.63 \ N)(65 \ mm) = 301 N.mm

The bending stress at point A is

\sigma =\frac{Mc}{I} =\frac{(301 \ N.mm)(0.40 \ mm)}{0.256 \ mm^4} =470 \ N/mm^2 = 470 \ MPa

This is the maximum stress that the spring sees, and therefore \sigma_{max} =470 MPa.

Now the mean .stress can be computed:

\sigma_{m} = ( \sigma_{max} + \sigma_{min})/2 = (470 + 176)/2 = 323 MPa

Finally, the alternating stress is

\sigma_{a} = ( \sigma_{max} – \sigma_{min} )/2= (470 – 176)/2 = 147 MPa

The stress ratio is found using Equation (5-3):

Stress ratio R =\frac{minimum \ stress}{maximum \ stress} =\frac{\sigma _{min}}{\sigma _{max}}=\frac{176 \ MPa}{471 \ MPa}=0.37

Comments The sketch of stress versus time shown in Figure 5-4(a) illustrates the form of the fluctuating stress on the spring. In Section 5-9, you will see how to design parts subjected to this kind of stress.