Chapter 10, Problem 31P

Solve Prob. 10-30 using the Goodman-Zimmerli fatigue-failure criterion.

To determine

The design parameters for the spring.

Answer to Problem 31P

The specifications of the spring are A313 stainless steel wire.

The wire diameter for the spring is $\text{0}\text{.0915}\text{in}$.

The outer diameter for the spring is $\text{0}\text{.903}\text{in}$.

The free length for the spring is $4.64\text{in}$.

The total number of the coils for the spring is $19.3$.

Explanation of Solution

Write the expression for the amplitude of alternating component of force.

$F_a=\frac{F_{\max }-F_{\min }}{2}$ (I)

Here, the maximum load on the spring is $F_{\max }$, the minimum load on the spring is $F_{\min }$ and the amplitude of alternating component of force is $F_a$.

Write the expression for the midrange steady component of the force.

$F_m=\frac{F_{\max }+F_{\min }}{2}$ (II)

Here, the midrange steady component of the force is $F_m$.

Write the expression for ultimate tensile strength.

$S_{ut}=\frac{A}{d^m}$ (III)

Here, the intercept constant is $A$ and the slope constant is $m$ and the wire diameter is $d$.

Write the expression for the maximum allowable stresses for helical springs.

$S_{sy}=0.35S_{ut}$ . (IV)

Here, the allowable yield stress for helical springs $S_{sy}$.

Write the expression for the shearing ultimate strength.

$S_{su}=0.67S_{ut}$ . (V)

Here, the shearing ultimate strength is $S_{su}$.

Write the expression for the slope of the load line using the goodman fatigue failure criterion.

$r=\frac{{\tau }_a}{{\tau }_m}$ . (VI)

Here, the load line slope is $r$, the alternating shear stress component is ${\tau }_a$ and the midrange component is ${\tau }_m$.

Write the expression for the goodman ordinate intercept.

$S_{se}=\frac{S_{sa}}{1-\left(\frac{S_{sm}}{S_{su}}\right)}$ (VII)

Here, the ordinate intercept for shear is $S_{se}$, the amplitude component of the strength is $S_{sa}$.

Write the expression for the amplitude component of the strength.

$s_{sa}=\frac{r^2{S}^2{}_{su}}{2s_{se}}\left[-1+{\sqrt{1+\left(\frac{2S_{se}}{r{S}_{su}}\right)}}^2\right]$ (VIII)

Write the expression for the back angle.

$\alpha =\frac{S_{sa}}{n_f}$ (IX)

Here, the back angle is $\alpha$ and the factor of safety is $n_f$.

Write the expression for the free end location angle.

$\beta =\frac{8F_a}{\pi d^2}$ (X)

Here, the free end location angle $\beta$.

Write the expression for the spring index.

$C=\frac{2\alpha -\beta }{4\beta }+\sqrt{{\left[\frac{2\alpha -\beta }{4\beta }\right]}^2-\frac{3\alpha }{4\beta }}$ (XI)

Here, the spring index is $C$.

Write the expression for the mean coil diameter.

$D=Cd$ (XII)

Here, the mean coil diameter is $D$.

Write the expression for the Bergstrasser factor to compensate the curvature effect.

$K_B=\frac{4C+2}{4C-3}$ (XIII)

Here, the Bergstrasser factor is $K_B$.

Write the expression for the alternating shear stress component.

${\tau }_a=K_B\left(\frac{8F_{a}D}{\pi d^3}\right)$ (XIV)

Write the expression for the fatigue factor of the safety.

$n_f=\frac{S_{sa}}{{\tau }_a}$ (XV)

Here, the fatigue factor of the safety is $n_f$.

Write the expression for the number of the active coils.

$N_a=\frac{G{d}^4}{8k{D}^3}$ (XVI)

Here, the number of the active coils is $N_a$ and the spring rate is $k$.

Write the expression for the total number of the coils.

$N_t=N_a+2$ (XVII)

Here, the total number of the coils is $N_t$.

Write the expression for the maximum deflection of the spring.

$y_{\max }=\frac{F_{\max }}{k}$ (XVIII)

Here, the maximum deflection of the spring is $y_{\max }$.

Write the expression for the deflection under the steady load.

$y_s=\left(1+\xi \right)y_{\max }$ (XIX)

Here, the fractional overrun to closure is $\xi$ and the deflection under the steady load is $y_s$.

Write the expression for the solid length of the spring.

$L_s=d{N}_t$ (XX)

Here, the solid length of the spring is $L_s$.

Write the expression for the free length of the spring.

$L_o=y_{\max }+y_s$ (XXI)

Here, the free length of the spring is $L_o$.

Write the expression for the critical free length of the spring.

${\left(L_o\right)}_{cr}=2.63\frac{D}{\alpha }$ (XXII)

Here, the critical free length of the spring is ${\left(L_o\right)}_{cr}$.

Write the expression for the shear force of the spring.

${\tau }_s=1.15\left(\frac{F_{\max }}{F_a}\right){\tau }_a$ (XXIII)

Here, the shear force of the spring is ${\tau }_s$.

Write the expression for the factor of the safety.

$n_s=\frac{S_{sy}}{{\tau }_s}$ (XXIV)

Here, the factor of the safety is $n_s$.

Write the expression for the frequency of the fundamental wave.

$f=\sqrt{\frac{kg}{{\pi }^2{d}^{2}D{N}_a\gamma }}$ (XXV)

Here, the acceleration due to gravity is $g$, the frequency of the fundamental wave is $f$ and the specific weight is $\gamma$.

Write the expression for the outer diameter of the spring.

$D_o=D+d$ . (VVVI)

Here, the outer diameter of the spring is $D_o$.

Conclusion:

Substitute $\text{18}\text{lbf}$ for $F_{\max }$ and $4\text{lbf}$ for $F_{\min }$ in Equation (I).

$\begin{array}{c}{F}_a=\frac{\text{18}\text{lbf}-4\text{lbf}}{2}\\ =\frac{\text{14}\text{lbf}}{2}\\ =7\text{lbf}\end{array}$

Substitute $\text{18}\text{lbf}$ for $F_{\max }$ and $4\text{lbf}$ for $F_{\min }$ in Equation (II).

$\begin{array}{c}{F}_m=\frac{\text{18}\text{lbf}+4\text{lbf}}{2}\\ =\frac{22\text{lbf}}{2}\\ =11\text{lbf}\end{array}$

Refer to table 10-4 “for estimating minimum tensile strength of the spring wires” to obtain the intercept and slope constants $169\text{kpsi}\cdot \text{in}$ for $A$ and $0.146$ for $m$ at the wire diameter $\text{0}\text{.080}\text{in}$.

Substitute $0.080\text{in}$ for $d$, $169\text{kpsi}\cdot \text{in}$ for $A$ and $0.146$ for $m$ in Equation (III).

$\begin{array}{c}{S}_{ut}=\frac{169\text{kpsi}\cdot \text{in}}{{\left(0.080\text{in}\right)}^{0.146}}\\ =\frac{169\text{kpsi}}{0.6915}\\ =244.36\text{kpsi}\end{array}$

Substitute $244.36\text{kpsi}$ for $S_{ut}$ in Equation (IV).

$\begin{array}{c}{S}_{sy}=0.35\times 244.36\text{kpsi}\\ =\text{85}\text{.52}\text{kpsi}\end{array}$

Substitute $244.36\text{kpsi}$ for $S_{ut}$ in Equation (V).

$\begin{array}{c}{S}_{su}=0.67\times 244.36\text{kpsi}\\ =163.7\text{kpsi}\end{array}$

Refer to Zimmerli’s endurance data to obtain the amplitude component of the strength and mid range component of the strength as $\text{35}\text{kpsi}$ for $S_{sa}$ and $55\text{kpsi}$ for $S_{sm}$.

Substitute $163.7\text{kpsi}$ for $S_{su}$, $\text{35}\text{kpsi}$ for $S_{sa}$ and $55\text{kpsi}$ for $S_{sm}$ in Equation (VII).

$\begin{array}{c}{S}_{se}=\frac{35\text{kpsi}}{1-\left(\frac{55\text{kpsi}}{163.7\text{kpsi}}\right)}\\ =\frac{35\text{kpsi}}{1-0.3359}\\ =52.70\text{kpsi}\end{array}$

Substitute $163.7\text{kpsi}$ for $S_{su}$, $52.70\text{kpsi}$ for $S_{se}$, $\text{7}\text{lbf}$ for $F_a$ and $11\text{lbf}$ for $F_m$ in Equation (VIII).

$\begin{array}{c}{S}_{sa}=\frac{{\left(\frac{\text{7}\text{lbf}}{11\text{lbf}}\right)}^2{\left(163.7\text{kpsi}\right)}^2}{2\left(52.70\text{kpsi}\right)}\left\{-1+\sqrt{1+{\left[\frac{2\left(52.70\text{kpsi}\right)}{\left(\frac{\text{7}\text{lbf}}{11\text{lbf}}\right)\left(163.7\text{kpsi}\right)}\right]}^2}\right\}\\ =\frac{0.404\left(26797.69\right)\text{kpsi}}{105.4}\left\{-1+\sqrt{1+{\left[\frac{105.4}{104.172}\right]}^2}\right\}\\ =102.71\text{kpsi}\left\{-1+1.422\right\}\\ \approx 43.40\text{kpsi}\end{array}$

Substitute $43.40\text{kpsi}$ for $S_{sa}$ and $1.5$ for $n_f$ in Equation (IX).

$\begin{array}{c}\alpha =\frac{43.40\text{kpsi}}{1.5}\\ =28.93\text{kpsi}\\ \approx 29\text{kpsi}\end{array}$

Substitute $7\text{lbf}$ for $F_a$ and $\text{0}\text{.080}\text{in}$ for $d$ in Equation (X).

$\begin{array}{c}\beta =\frac{8\left(7\text{lbf}\right)}{\pi {\left(\text{0}\text{.080}\text{in}\right)}^2}\\ =\frac{56\text{lbf}}{0.020{\text{in}}^2}\left(\frac{\text{1}\text{kpsi}}{{\text{10}}^3\text{lbf}/{\text{in}}^{\text{2}}}\right)\\ =2.785\text{kpsi}\end{array}$

Substitute $2.785\text{kpsi}$ for $\beta$ and $29\text{kpsi}$ for $\alpha$ in Equation (XI).

$\begin{array}{c}C=\frac{2\left(29\text{kpsi}\right)-2.785\text{kpsi}}{4\left(2.785\text{kpsi}\right)}+\sqrt{{\left[\frac{2\left(29\text{kpsi}\right)-2.785\text{kpsi}}{4\left(2.785\text{kpsi}\right)}\right]}^2-\frac{3\left(29\text{kpsi}\right)}{4\left(2.785\text{kpsi}\right)}}\\ =\frac{55.21}{11.14}+\sqrt{\left(24.56\right)-7.809}\\ =4.95+4.09\\ =9.046\end{array}$

Substitute $6.95$ for $C$ and $0.0\text{80}\text{in}$ for $d$ in Equation (XII).

$\begin{array}{c}D=9.046\left(0.0\text{80}\text{in}\right)\\ =0.723\text{in}\end{array}$

Substitute $9.046$ for $C$ in Equation (XIII).

$\begin{array}{c}{K}_B=\frac{4\times 9.046+2}{4\times 9.046-3}\\ =\frac{38.18}{33.18}\\ =1.15\end{array}$

Substitute $\text{7}\text{lbf}$ for $F_a$, $0.723\text{in}$ for $D$, $1.15$ for $K_B$ and $0.0\text{80}\text{in}$ for $d$ in Equation (XIV).

$\begin{array}{c}{\tau }_a=1.15\left(\frac{8\left(\text{7}\text{lbf}\right)\left(0.723\text{in}\right)}{\pi {\left(0.080\text{in}\right)}^3}\right)\\ =1.15\frac{40.48}{0.001608}\text{lbf}/{\text{in}}^{\text{2}}\left(\frac{1\text{kpsi}}{{\text{10}}^3\text{lbf}/{\text{in}}^{\text{2}}}\right)\\ =28.95\text{kpsi}\\ \approx 29\text{kpsi}\end{array}$

Substitute $29\text{kpsi}$ for ${\tau }_a$ and $43.40\text{kpsi}$ for $S_{sa}$ in Equation (XV).

$\begin{array}{c}{n}_f=\frac{43.40\text{kpsi}}{29\text{kpsi}}\\ =1.49\\ \approx 1.50\end{array}$

Refer to table 10-5 “Mechanical properties of some spring wires” to obtain the modulus of rigidity for A313 stainless wire as ${\text{10}}^{\text{7}}\text{psi}$.

Substitute ${\text{10}}^{\text{7}}\text{psi}$ for $G$, $\text{0}\text{.08}\text{in}$ for $d$, $\text{0}\text{.723}\text{in}$ for $D$ and $\text{9}\text{.5}\text{lbf}/\text{in}$ for $k$ in Equation (XVI).

$\begin{array}{c}{N}_a=\frac{{\text{10}}^{\text{7}}\text{psi}{\left(\text{0}\text{.08}\text{in}\right)}^4}{8\left(\text{9}\text{.5}\text{lbf}/\text{in}\right){\left(\text{0}\text{.723}\text{in}\right)}^3}\\ =\frac{409.6\text{psi}\cdot \text{in}}{28.72\text{lbf}/\text{in}}\left(\frac{1\text{lbf}/{\text{in}}^2}{1\text{psi}}\right)\\ =14.26\text{coils}\end{array}$

Substitute $14.26\text{coils}$ for $N_a$ in Equation (XVII).

$\begin{array}{c}{N}_t=14.26\text{coils}+2\\ =16.26\text{coils}\end{array}$

Substitute $\text{18}\text{lbf}$ for $F_{\max }$ and $\text{9}\text{.5}\text{lbf}/\text{in}$ for $k$ in Equation (XVIII).

$\begin{array}{c}{y}_{\max }=\frac{\text{18}\text{lbf}}{\text{9}\text{.5}\text{lbf}/\text{in}}\\ =1.895\text{in}\end{array}$

Substitute $0.15$ for $\xi$ and $1.\text{895}\text{in}$ for $y_{\max }$ in Equation (XIX).

$\begin{array}{c}{y}_s=\left(1+0.15\right)1.\text{895}\text{in}\\ =2.179\text{in}\end{array}$

Substitute $16.26\text{coils}$ for $N_t$ and $0.08\text{in}$ for $d$ in Equation (XX).

$\begin{array}{c}{L}_s=0.08\text{in}\left(16.26\text{coils}\right)\\ =1.30\text{in}\end{array}$

Substitute $2.179\text{in}$ for $y_s$ and $1.30\text{in}$ for $L_s$ in Equation (XXI).

$\begin{array}{c}{L}_o=1.30\text{in}+2.179\text{in}\\ =3.47\text{in}\end{array}$

Since, the free length is lesser than the $5.26$ times mean coil diameter. Hence the end condition constant for A313 stainless wire is $0.5$.

Substitute $\text{0}\text{.723}\text{in}$ for $D$ and $0.5$ for $\alpha$ in Equation (XXII).

$\begin{array}{c}{\left(L_{{}_o}\right)}_{cr}=2.63\left(\frac{\text{0}\text{.723}\text{in}}{0.5}\right)\\ =3.802\text{in}\\ \approx 3.8\text{in}\end{array}$

Substitute $\text{18}\text{lbf}$ for $F_{\max }$, $7\text{lbf}$ for $F_a$ and $29\text{kpsi}$ for ${\tau }_a$ in Equation (XXIII).

$\begin{array}{c}{\tau }_s=1.15\left(\frac{\text{18}\text{lbf}}{7\text{lbf}}\right)\text{29}\text{kpsi}\\ \text{=1}\text{.15}\left(2.571\right)\text{29}\text{kpsi}\\ =85.74\text{kpsi}\end{array}$

Substitute $85.74\text{kpsi}$ for ${\tau }_s$ and $85.5\text{kpsi}$ for $S_{sy}$ in Equation (XXIV).

$\begin{array}{c}{n}_s=\frac{85.5\text{kpsi}}{85.74\text{kpsi}}\\ =0.997\end{array}$

Since the steel is A313 stainless wire. Hence specific weight is $0.283\text{lb}/{\text{ft}}^3$ and the acceleration due to gravity is $\text{386}\text{in}/{\text{s}}^{\text{2}}$.

Substitute $\text{386}\text{in}/{\text{s}}^{\text{2}}$ for $g$, $0.283\text{lb}/{\text{ft}}^3$ for $\gamma$, $14.26$ for $N_a$, $\text{0}\text{.08}\text{in}$ for $d$, $\text{0}\text{.723}\text{in}$ for $D$ and $\text{9}\text{.5}\text{lbf}/\text{in}$ for $k$ in Equation (XXV).

$\begin{array}{c}f=\sqrt{\frac{386\text{in}/{\text{s}}^{\text{2}}\left(\text{9}\text{.5}\text{lbf}/\text{in}\right)}{{\pi }^2{\left(\text{0}\text{.08}\text{in}\right)}^2\left(\text{0}\text{.723}\text{in}\right)\left(14.26\right)0.283\text{lb}/{\text{ft}}^3}}\\ =\sqrt{\frac{3667\text{lbf}/{\text{s}}^2}{0.1842\text{lb}\cdot {\text{in}}^3/{\text{ft}}^3}}\\ =\sqrt{19896.96}\text{Hz}\\ \approx 141.05\text{Hz}\end{array}$

Repeat all the steps for other values of the wire diameter. All the calculated values for other values of wire diameter are shown in below table.

The following table shows the first iteration.

		$d_1$	$d_2$	$d_3$	$d_4$
1	$d$	0.080	0.0915	0.1055	0.1205
2	$m$	0.146	0.146	0.263	0.263
3	$A$	169	169	128	128
4	$S_{ut}$	244.363	239.618	231.257	223.311
5	$S_{su}$	163.723	160.544	154.942	149.618
6	$S_{sy}$	85.5	83.86	80.94	78.15
7	$S_{se}$	52.70	53.23	54.26	55.34
8	$S_{sa}$	43.40	43.56	43.63	43.69
9	$\alpha$	29	29.04	29.09	29.12
10	$\beta$	2.75	2.12	1.60	1.22
11	$C$	9.046	12.30	16.85	22.43
12	$D$	0.723	1.126	1.778	2.703
13	$K_B$	1.15	1.10	1.07	1.05
14	${\tau }_a$	29	29.04	29.09	29.12
15	$n_f$	1.5	1.5	1.5	1.5
16	$N_a$	14.26	6.45	2.89	1.40
17	$N_t$	16.26	8.45	4.89	3.40
18	$L_s$	1.3	0.774	0.51	0.41
19	$y_s$	2.17	2.17	2.17	2.17
20	$L_o$	4.38	3.64	3.39	3.28
21	${\left(L_o\right)}_{cr}$	3.802	5.924	9.35	14.21
22	${\tau }_s$	85.74	85.87	86.02	86.13
23	$n_s$	0.997	0.977	0.941	0.907
24	$f$	141.05	145.55	149.93	152.96

Since, the factor of safety lesser than one, Hence the design is not suitable.

Repeat all the steps for second iteration.

The following table shows the second iteration.

		$d_1$	$d_2$	$d_3$	$d_4$
1	$d$	0.080	0.0915	0.1055	0.1205
2	$m$	0.146	0.146	0.263	0.263
3	$A$	169	169	128	128
4	$S_{ut}$	244.363	239.618	231.257	223.311
5	$S_{su}$	163.723	160.544	154.942	149.618
6	$S_{sy}$	85.5	83.86	80.94	78.15
7	$S_{se}$	52.70	53.23	54.26	55.34
8	$S_{sa}$	43.40	43.56	43.63	43.69
9	$\alpha$	21.75	21.78	21.81	21.84
10	$\beta$	2.75	2.12	1.60	1.22
11	$C$	6.995	8.86	12.29	16.48
12	$D$	0.512	0.811	1.29	1.98
13	$K_B$	1.22	1.15	1.10	1.07
14	${\tau }_a$	21.756	21.78	21.81	21.84
15	$n_f$	2	2	2	2
16	$N_a$	40.24	17.28	7.47	3.53
17	$N_t$	42.24	19.28	9.47	5.53
18	$L_s$	3.37	1.76	1.00	0.667
19	$y_s$	2.17	2.17	2.17	2.17
20	$L_o$	6.25	4.64	3.87	3.54
21	${\left(L_o\right)}_{cr}$	2.69	4.26	6.82	10.44
22	${\tau }_s$	64.33	64.40	64.51	64.60
23	$n_s$	1.32	1.30	1.25	1.21
24	$f$	98.93	104.82	109.340	112.409

Substitute $\text{0}\text{.811}\text{in}$ for $D$ and $\text{0}\text{.092}\text{in}$ for $d$ in Equation (XXVI).

$\begin{array}{c}{D}_o=\text{0}\text{.811}\text{in}+\text{0}\text{.092}\text{in}\\ =\text{0}\text{.903}\text{in}\end{array}$

Thus, the outer diameter for the spring is $\text{0}\text{.903}\text{in}$.

Thus, the specifications of the spring are A313 stainless steel wire.

The wire diameter for the spring is $\text{0}\text{.0915}\text{in}$.

The free length for the spring is $4.64\text{in}$.

The total number of the coils for the spring is $19.3$.