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

Question

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 preloadmust 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 [latex]\frac{1}{4}[/latex] in thick, as seen in Fig. 6–30a. The spring strengths are [latex]S_{u t}=150 kpsi , S_{y}=127 kpsi , ext { and } S_{e}=[/latex] 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?

Accepted Answer

We begin with preliminaries. The second area moment of the cantilever cross section is

[latex]I=\frac{b h^{3}}{12}=\frac{2(0.25)^{3}}{12}=0.00260 in ^{4}[/latex]

Since, from Table A–9, beam 1, force F and deflection y in a cantilever are related by [latex]F=3 E I y / l^{3}[/latex], then stress σ and deflection y are related by

[latex]\sigma=\frac{M c}{I}=\frac{32 F c}{I}=\frac{32(3 E I y)}{13} \frac{c}{I}=\frac{96 E c y}{{ }^{3}}=K y[/latex]

[latex] ext { where } K=\frac{96 E c}{l^{3}}=\frac{96\left(30 \cdot 10^{6} ight) 0.125}{32^{3}}=10.99\left(10^{3} ight) psi / ext { in }=10.99 kpsi / in[/latex]

Now the minimums and maximums of y and [latex]\sigma[/latex] can be defined by

[latex]y_{\min }=\delta \quad y_{\max }=2+\delta[/latex]

[latex]\sigma_{\min }=K \delta \quad \sigma_{\max }=K(2+\delta) [/latex]

The stress components are thus

[latex]\sigma_{a}=\frac{K(2+\delta)-K \delta}{2}=K=10.99 kpsi[/latex]

[latex]\sigma_{m}=\frac{K(2+\delta)+K \delta}{2}=K(1+\delta)=10.99(1+\delta)[/latex]

[latex] ext { For } \delta=0, \quad \sigma_{a}=\sigma_{m}=10.99=11 kpsi[/latex]

[latex] ext { For } \delta=2 ext { in, } \quad \sigma_{a}=11 kpsi , \sigma_{m}=10.99(1+2)=33 kpsi[/latex]

[latex] ext { For } \delta=5 in , \quad \sigma_{a}=11 kpsi , \sigma_{m}=10.99(1+5)=65.9 kpsi[/latex]

(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 [latex]A, A^{\prime}, ext { and } A^{\prime \prime}[/latex]. Note that since [latex]\sigma_{a}[/latex] 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 [latex]S_{m}[/latex] and substituting 11 kpsi for [latex]S_{a}[/latex]:

[latex]\frac{S_{a}}{S_{e}}+\left(\frac{S_{m}}{S_{u t}} ight)^{2}=1[/latex] (6–42)

[latex]S_{m}=S_{u t} \sqrt{1-\frac{S_{a}}{S_{e}}}=150 \sqrt{1-\frac{11}{28}}=116.9 kpsi[/latex]

The intersection of the Langer line and the load line is found from solving Eq. (6–44) for [latex]S_{m}[/latex] and substituting 11 kpsi for [latex]S_{a}[/latex]:

[latex]S_{a}+S_{m}=S_{y}[/latex] (6–44)

[latex]S_{m}=S_{y}-S_{a}=127-11=116 kpsi[/latex]

The threats from fatigue and first-cycle yielding are approximately equal. (b) For [latex]\delta=2[/latex] in,

[latex]n_{f}=\frac{S_{m}}{\sigma_{m}}=\frac{116.9}{33}=3.54 \quad n_{y}=\frac{116}{33}=3.52[/latex]

and for [latex]\delta=5[/latex] in,

[latex]n_{f}=\frac{116.9}{65.9}=1.77 \quad n_{y}=\frac{116}{65.9}=1.76[/latex]

Question 6.11: A flat-leaf spring is used to retain an oscillating flat-fac...

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