A flat-leaf spring is used to retain an oscillating flat-faced follower in contact with a plate cam. The follower range of motion is 2 in and fixed, so the alternating component of force, bending moment, and stress is fixed, too. The spring is preloaded to adjust to various cam speeds. The preload must be increased to prevent follower float or jump. For lower speeds the preload should be decreased to obtain longer life of cam and follower surfaces. The spring is a steel cantilever 32 in long, 2 in wide, and \frac {1}{4} in thick, as seen in Fig. 6–30a. The spring strengths are S_{ut} = 150 kpsi, S_{y} = 127 kpsi, and S_{e} = 28 kpsi fully corrected. The total cam motion is 2 in. The designer wishes to preload the spring by deflecting it 2 in for low speed and 5 in for high speed.
(a) Plot the Gerber-Langer failure lines with the load line.
(b) What are the strength factors of safety corresponding to 2 in and 5 in preload?
Question 6.11: A flat-leaf spring is used to retain an oscillating flat-fac...
We begin with preliminaries. The second area moment of the cantilever cross section is
I =\frac{bh^{3}}{12} =\frac{2(0.25)^{3}}{12} = 0.00260 in^{4}
Since, from Table A–9, beam 1, force F and deflection y in a cantilever are related by F = 3E I y/l^{3}, then stress σ and deflection y are related by
Table A–9 Shear, Moment, and Deflection of Beams (Note: Force and moment reactions are positive in the directions shown; equations for shear force V and bending moment M follow the sign conventions given in Sec. 3–2.)
| 1 Cantilever—end load |
R_{1} = V = F M_{1} = Fl
M = F(x − l) y =\frac {Fx^{2}}{6E I}(x − 3l)
y_{max} = −\frac{Fl^{3}}{3EI} |
σ =\frac {Mc}{I} =\frac {32Fc}{I} =\frac{32(3E I y)}{l^{3}} \frac{c}{I} =\frac{96Ecy}{l^{3}} = K_{y}
where K =\frac{96Ec}{l^{3}} =\frac{96(30 · 10^{6})0.125}{32^{3}}= 10.99(10^{3}) psi/in = 10.99 kpsi/in
Now the minimums and maximums of y and σ can be defined by
y_{min} = δ y_{max} = 2 + δ
σ_{min} = Kδ σ_{max} = K(2 + δ)
The stress components are thus
σ_{a} =\frac{K(2 + δ) − Kδ}{2} = K = 10.99 kpsi
σ_{m} =\frac{K(2 + δ) + Kδ}{2} = K(1 + δ) = 10.99(1 + δ)
For δ = 0, σ_{a} = σ_{m} = 10.99 = 11 kpsi
For δ = 2 in, σ_{a} = 11 kpsi, σ_{m} = 10.99(1 + 2) = 33 kpsi
For δ = 5 in, σ_{a} = 11 kpsi, σ_{m} = 10.99(1 + 5) = 65.9 kpsi
(a) A plot of the Gerber and Langer criteria is shown in Fig. 6–30b. The three preload deflections of 0, 2, and 5 in are shown as points A, A′, and A′′. Note that since σ_{a} is constant at 11 kpsi, the load line is horizontal and does not contain the origin. The intersection between the Gerber line and the load line is found from solving Eq. (6–42) for S_{m} and substituting 11 kpsi for S_{a} :
\frac{S_{a}}{S_{e}} +\left(\frac{S_{m}}{S_{ut}}\right)^{2}= 1 (6–42)
S_{m} = S_{ut} \sqrt{1 −\frac{S_{a}}{S_{e}} }= 150 \sqrt {1 −\frac {11}{28}} = 116.9 kpsi
The intersection of the Langer line and the load line is found from solving Eq. (6–44) for S_{m} and substituting 11 kpsi for S_{a} :
S_{a} + S_{m} = S_{y} (6–44)
S_{m} = S_{y} − S_{a} = 127 − 11 = 116 kpsi
The threats from fatigue and first-cycle yielding are approximately equal.
(b) For δ = 2 in,
n_{f} =\frac {S_{m}}{σ_{m}} =\frac{116.9}{33} = 3.54 n_{y}=\frac{116}{33} = 3.52
and for δ = 5 in,
n_{f} =\frac {116.9}{65.9} = 1.77 n_{y} =\frac {116}{65.9} = 1.76
