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

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

2    Since a constant-force spring is needed, the h/t 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 t:

t=\sqrt[4]{\frac{F_{\text {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         (a)

5    The height h can now be found:

h=1.414 t=1.414(0.015)=0.021  \text { in }       (b)

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

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 y_{max}, so solve equations 14.36b and 14.37 for stresses at that deflection:

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

\sigma_c=-\frac{4 E y}{K_1 D_o^2\left(1-\nu^2\right)}\left[K_2\left\lgroup h-\frac{y}{2}\right\rgroup +K_3 t\right]        (14.37a)

\sigma_{t_i}=\frac{4 E y}{K_1 D_o^2\left(1- \nu ^2\right)}\left[-K_2\left\lgroup h-\frac{y}{2} \right\rgroup +K_3 t\right]      (14.37b)

\sigma_{t_o}=\frac{4 E y}{K_1 D_o^2\left(1-\nu^2\right)}\left[K_4\left\lgroup h-\frac{y}{2}\right\rgroup +K_5 t\right]       (14.37c)

K_2=\frac{6}{\pi \ln R_d}\left\lgroup\frac{R_d-1}{\ln R_d}-1 \right\rgroup \quad \text { and } \quad R_d=\frac{D_o}{D_i}      (14.37d)

K_3=\frac{6}{\pi \ln R_d}\left\lgroup\frac{R_d-1}{2} \right\rgroup         (14.37e)

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

K_5=\frac{R_d}{2\left(R_d-1\right)}     (14.37g)

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

K_2=\frac{6}{\pi \ln R_d}\left\lgroup \frac{R_d-1}{\ln R_d}-1 \right\rgroup =\frac{6}{\pi \ln 2}\left\lgroup \frac{2-1}{\ln 2}-1 \right\rgroup =1.220        (e)

K_3=\frac{6}{\pi \ln R_d}\left(\frac{R_d-1}{2}\right)=\frac{2}{\pi \ln 2}\left(\frac{2-1}{2}\right)=1.378      (f)

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

K_5=\frac{R_d}{2\left(R_d-1\right)}=\frac{2}{2(2-1)}=1      (h)

8    Table 14-5 (p. 795) gives S_{ut} = 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

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

which is acceptable.

9    A summary of this spring design is

D_o=1.2 \quad D_i=0.6 \quad t=0.015 \quad h=0.021      (m)

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