Question 14.3: Design of a Helical Compression Spring for Static Loading Pr...

See Table 14-10

1    Assume a trial wire diameter of 0.162 in from the available sizes in Table 14-2 (p. 791).

2    Assume a spring index of 8, which is in the middle of the recommended range, and calculate the mean coil diameter D from equation 14.5 (p. 797).

C=\frac{D}{d}      (14.5)

D=C d=8(0.162)=1.30 \text { in }      (a)

3    Find the direct shear factor K_{s} and use it to calculate the shear stress in the coil at the larger force.

K_s=1+\frac{0.5}{C}=1+\frac{0.5}{8}=1.06     (b)

\tau=K_s \frac{8 F D}{\pi d^3}=1.06 \frac{8(150)(1.30)}{\pi(0.162)^3}=123714  psi       (c)

4    Find the ultimate tensile strength of this wire material from equation 14.3 and Table 14-4 (p. 792) and use it to find the torsional yield strength from Table 14- 8 (p. 803), assuming that the set has been removed and using the low end of the recommended range.

S_{u t} \cong A d^b     (14.3)

S_{u t}=A d^b=141040(0.162)^{-0.1822}=196503  psi     (d)

S_{y s}=0.60 S_{u t}=0.60(196503)=117902  psi       (e)

5    Find the safety factor against yielding at this working deflection from equation 14.15 (p. 809).

N_s=\frac{S_{y s}}{\tau}     (14.15)

N_s=\frac{S_{y s}}{\tau}=\frac{117902  psi }{123714  psi }=0.95      (f)

This is obviously not acceptable, so the design must be iterated with some parameters changed.

6    Try increasing the wire diameter slightly, perhaps to 0.192 in, keeping the same spring index. Recalculate the coil diameter, stress, strength, and safety factor.

D=C d=8(0.192)=1.54 \text { in }     (g)

\tau=K_s \frac{8 F D}{\pi d^3}=1.06 \frac{8(150)(1.54)}{\pi(0.192)^3}=88074  psi        (h)

S_{u t}=A d^b=141040(0.192)^{-0.1822}=190513  psi         (i)

S_{y s}=0.60 S_{u t}=0.60(190513)=114308  psi      (j)

N_s=\frac{S_{y s}}{\tau}=\frac{114308  psi }{88074  psi }=1.30        (k)

This appears to be acceptable, so we can go on to design the other spring parameters.

7    The spring rate is defined in this problem because of the two specified forces at a particular relative deflection.

k=\frac{\Delta F}{y}=\frac{150-100}{0.75}=66.7  lb / in     (l)

8    To achieve this spring rate, the number of active coils must satisfy equation 14.7 (p. 797):

k=\frac{F}{y}=\frac{d^4 G}{8 D^3 N_a}       (14.7)

k=\frac{d^4 G}{8 D^3 N_a} \quad \text { or } \quad N_a=\frac{d^4 G}{8 D^3 k}=\frac{(0.192)^4 11.5 E 6}{8(1.54)^3 66.67}=8.09 \equiv 8        (m)

Note that we round it to the nearest 1/4 coil, as the manufacturing tolerance cannot achieve better than that accuracy. This makes the spring rate k = 67.4 lb/in.

9    Assume squared and ground ends, making the total number of coils from Figure 14-9 (p. 796):

N_t=N_a+2=8+2=10     (n)

10    The shut height can now be determined.

L_s=d N_t=0.192(10)=1.92  \text { in }          (o)

11    The initial deflection to reach the smaller of the two specified loads is

y_{\text {initial }}=\frac{F_{\text {initial }}}{k}=\frac{100}{67.4}=1.48  \text { in }        (p)

12    Assume a clash allowance of 15% of the working deflection:

y_{\text {clash }}=0.15 y=0.15(0.75)=0.113  \text { in }     (q)

13    The free length (see Figure 14-8, p. 796) can now be found from

L_f=L_s+y_{\text {clash }}+y_{\text {working }}+y_{\text {initial }}=1.92+0.113+0.75+1.48=4.26  \text { in }       (r)

14    The deflection to the shut height is

y_{\text {shut }}=L_f-L_s=4.26-1.92=2.34  \text { in }     (s)

15    The force at this shut-height deflection is

F_{\text {shut }}=k y_{\text {shut }}=67.4(2.34)=158  lb       (t)

16    The shut-height stress and safety factor are

\tau_{\text {shut }}=K_s \frac{8 F D}{\pi d^3}=1.06 \frac{8(158)(1.54)}{\pi(0.192)^3}=92794  psi     (u)

N_{s_{\text {shut }}}=\frac{S_{s y}}{\tau_{\text {shut }}}=\frac{114308  psi }{92794  psi }=1.2      (v)

which is acceptable.

17    To check for buckling, two ratios need to be calculated, L_{f} / D and y_{max} / L_{f}

Take these two values to Figure 14-14 (p. 801) and find that their coordinates are safely within the zones that are stable against buckling for either end-condition case.

18    The inside and outside coil diameters are

19    The smallest hole and largest pin that should be used with this spring are

20    The total weight of the spring is

W_t=\frac{\pi^2 d^2 D N_t \rho}{4}=\frac{\pi^2(0.192)^2(1.54)(10)(0.28)}{4}=0.40  lb      (z)

21    We now have a complete design specification for this A227 wire spring:

d = 0.192 in    OD = 1.73 in    N_{t} 10, sq & g      L_{f} = 4.26 in       (aa)

Other calculated spring parameters for this example are shown in Table 14-10.

22    Models for this example are on the CD-ROM as EX14-03. There are also alternate approaches to the solution of this example shown in separate Mathcad and TKSolver files on the CD-ROM with a letter added to their name.