Design of a Belleville Spring for Static Loading Problem Design a Belleville spring to give a constant load over a given deflection. Given An end seal on a shaft requires a nearly constant load over small motions associated with temperature change. The spring must apply a nominal force of 10 lb ±5%

Question

Design of a Belleville Spring for Static Loading

Problem Design a Belleville spring to give a constant load over a given deflection.

Given An end seal on a shaft requires a nearly constant load over small motions associated with temperature change. The spring must apply a nominal force of 10 lb ±5% over a range of ±0.006 in at any convenient nominal deflection. The spring must fit in a 1.25-in-diameter hole.

Assumptions Assume a diameter ratio [latex]R_{d}[/latex] = 2. Use unset carbon spring steel 50HRC.

Accepted Answer

1 Assume an outside diameter [latex]D_{o}[/latex] of 1.2 in to allow some clearance in the hole.

2 Since a constant-force spring is needed, the [latex]h/t[/latex] ratio is 1.414 (see Figure 14-30).

3 The required force variation of not more than ±5% can be met by choosing an appropriate deflection range to operate in from Figure 14-31. If the deflection is kept between about 65% and 135% of the flat deflection, this tolerance will be achieved. The nominal force will then occur at the flat position and the spring must operate on both sides, so it must be mounted in similar fashion to that shown in Figure 14-33.

4 Use the above assumptions and the specified nominal force in equation 14.38us to find an appropriate spring thickness [latex]t[/latex]:

[latex] t=\sqrt[4]{\frac{F_{ ext {flat }}}{19.2 E 7} \frac{D_o^2}{h / t}}=\sqrt[4]{\frac{10}{19.2 E 7} \frac{(1.2)^2}{1.414}}=0.015 in [/latex] (a)

5 The height [latex]h[/latex] can now be found:

[latex] h=1.414 t=1.414(0.015)=0.021 ext { in } [/latex] (b)

6 Based on the choices in step 3 above, find the minimum and maximum deflections:

[latex] \begin{array}{l} y_{\min }=0.65 h=0.65(0.021)=0.014 ext { in } \ y_{\max }=1.35 h=1.35(0.021)=0.029 ext { in } \end{array} [/latex] (c)

The difference between these distances is greater than the required deflection range, so the force tolerance can be met over that range.

7 Figure 14-34 shows that the worst stress state will occur at the largest deflection [latex]y_{max}[/latex], so solve equations 14.36b and 14.37 for stresses at that deflection:

[latex] K_1=\frac{6}{\pi \ln R_d}\left[\frac{\left(R_d-1 ight)^2}{R_d^2} ight] \quad ext { and } \quad R_d=\frac{D_o}{D_i} [/latex] (14.36b)

[latex] \sigma_c=-\frac{4 E y}{K_1 D_o^2\left(1- u^2 ight)}\left[K_2\left\lgroup h-\frac{y}{2} ight group +K_3 t ight] [/latex] (14.37a)

[latex] \sigma_{t_i}=\frac{4 E y}{K_1 D_o^2\left(1- u ^2 ight)}\left[-K_2\left\lgroup h-\frac{y}{2} ight group +K_3 t ight] [/latex] (14.37b)

[latex] \sigma_{t_o}=\frac{4 E y}{K_1 D_o^2\left(1- u^2 ight)}\left[K_4\left\lgroup h-\frac{y}{2} ight group +K_5 t ight] [/latex] (14.37c)

[latex] K_2=\frac{6}{\pi \ln R_d}\left\lgroup\frac{R_d-1}{\ln R_d}-1 ight group \quad ext { and } \quad R_d=\frac{D_o}{D_i} [/latex] (14.37d)

[latex] K_3=\frac{6}{\pi \ln R_d}\left\lgroup\frac{R_d-1}{2} ight group [/latex] (14.37e)

[latex] K_4=\left[\frac{R_d \ln R_d-\left(R_d-1 ight)}{\ln R_d} ight]\left[\frac{R_d}{\left(R_d-1 ight)^2} ight] [/latex] (14.37f)

[latex] K_5=\frac{R_d}{2\left(R_d-1 ight)} [/latex] (14.37g)

[latex] K_1=\frac{6}{\pi \ln R_d}\left[\frac{\left(R_d-1 ight)^2}{R_d^2} ight]=\frac{6}{\pi \ln 2}\left[\frac{(2-1)^2}{2^2} ight]=0.689 [/latex] (d)

[latex] K_2=\frac{6}{\pi \ln R_d}\left\lgroup \frac{R_d-1}{\ln R_d}-1 ight group =\frac{6}{\pi \ln 2}\left\lgroup \frac{2-1}{\ln 2}-1 ight group =1.220 [/latex] (e)

[latex] K_3=\frac{6}{\pi \ln R_d}\left(\frac{R_d-1}{2} ight)=\frac{2}{\pi \ln 2}\left(\frac{2-1}{2} ight)=1.378 [/latex] (f)

[latex] K_4=\left[\frac{R_d \ln R_d-\left(R_d-1 ight)}{\ln R_d} ight]\left[\frac{R_d}{\left(R_d-1 ight)^2} ight]=\left[\frac{2 \ln 2-(2-1)}{\ln 2} ight]\left[\frac{2}{(2-1)^2} ight]=1.115 [/latex] (g)

[latex] K_5=\frac{R_d}{2\left(R_d-1 ight)}=\frac{2}{2(2-1)}=1 [/latex] (h)

[latex] \begin{aligned} \sigma_c &=-\frac{4 E y}{K_1 D_o^2\left(1- u^2 ight)}\left[K_2\left\lgroup h-\frac{y}{2} ight group+K_3 t ight] \ &=-\frac{4(30 E 6)(0.029)}{0.689(1.2)^2\left(1-0.3^2 ight)}\left[1.220\left\lgroup 0.021-\frac{0.029}{2} ight group+1.378(0.015) ight] \ \sigma_c &=-112227 psi \end{aligned} [/latex] (i)

[latex] \begin{aligned} \sigma_{t_i} &=\frac{4 E y}{K_1 D_o^2\left(1- u^2 ight)}\left[-K_2\left\lgroup h-\frac{y}{2} ight group +K_3 t ight] \ &=\frac{4(30 E 6)(0.029)}{0.689(1.2)^2\left(1-0.3^2 ight)}\left[-1.220\left\lgroup 0.021-\frac{0.029}{2} ight group +1.378(0.015) ight] \ \sigma_{t_i} &=46600 psi \end{aligned} [/latex] (j)

[latex] \begin{aligned} \sigma_{t_o} &=\frac{4 E y}{K_1 D_o^2\left(1- u^2 ight)}\left[K_4\left\lgroup h-\frac{y}{2} ight group +K_5 t ight] \ &=\frac{4(30 E 6)(0.029)}{0.689(1.2)^2\left(1-0.3^2 ight)}\left[1.115\left\lgroup 0.021-\frac{0.029}{2} ight group +(1)(0.015) ight] \ \sigma_{t_o} &=87628 psi \end{aligned} [/latex] (k)

8 Table 14-5 (p. 795) gives [latex]S_{ut}[/latex] = 246 kpsi for this material. Table 14-17 indicates that 120% of this value can be used for an unset spring. The safety factor for static loading is then

[latex] N_s=\frac{1.2 S_{u t}}{\sigma_c}=\frac{1.2(246000)}{112227}=2.6 [/latex] (l)

which is acceptable.

9 A summary of this spring design is

[latex] D_o=1.2 \quad D_i=0.6 \quad t=0.015 \quad h=0.021 [/latex] (m)

10 The files EX14-07 can be found on the CD-ROM.

Table 14-5 Typical Properties of Spring Temper Alloy Strip Source: Reference 1
Materia	Sut MPa (ksi)	Rockwell Hardness	Elongation %	Bend Factor	E GPa (Mpsi)	Poisson’s Ratio
Spring steel	1 700 (246)	C50	2	5	207 (30)	0.30
Stainless 301	1 300 (189)	C40	8	3	193 (28)	0.31
Stainless 302	1 300 (189)	C40	5	4	193 (28)	0.31
Monel 400	690 (100)	B95	2	5	179 (26)	0.32
Monel K500	1 200 (174)	C34	40	5	17.9 (26)	0.29
Inconel 600	1 040 (151)	C30	2	2	214 (31)	0.29
Inconel X-750	1 050 (152)	C35	20	3	214 (31)	0.29
Beryllium copper	1 300 (189)	C40	2	5	128 (18.5)	0.33
Ni-Span-C	1 400 (203)	C42	6	2	186 (27)	–
Brass CA 260	620 (90)	B90	3	3	11 (16)	0.33
Phosphor bronze	690 (100)	B90	3	2.5	103 (15)	0.20
17-7PH RH950	1 450 (210)	C44	6	flat	203 (29.5)	0.34
17-7PH Cond. C	1 650 (239)	C46	1	2.5	203 (29.5)	0.34

Table 14-17 Maximum Recommended Compressive Stress Levels [latex]\sigma _{c}[/latex] for Belleville Washers in Static Applications, Assuming [latex]S_{uc} = S_{ut}[/latex] Source: Reference 1
Material	Maximum Percent of Ultimate Tensile Strength
Material	Before Set Removed	After Set Removed
Carbon or alloy steel	120%	275%
Nonferrous and austenitic stainless steel	95	160

Question 14.7: Design of a Belleville Spring for Static Loading Problem Des...

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