Chapter 10, Problem 37P

Design an infinite-life helical coil extension spring with full end loops and generous loop-bend radii for a minimum load of 9 lbf and a maximum load of 18 lbf, with an accompanying stretch of $\frac{1}{4}$ in. The spring is for food-service equipment and must be stainless steel. The outside diameter of the coil cannot exceed 1 in, and the free length cannot exceed $2\frac{1}{2}$ in. Using a fatigue design factor of n_f = 2, complete the design. Use the Gerber criterion with Table 10-8.

To determine

The design parameter for the infinite life helical spring.

Answer to Problem 37P

The dimensions of the spring are:

• The wire diameter for the spring is $\text{0}\text{.081}\text{in}$.
• The outer diameter for the spring is $0.4536\text{in}$.
• The free length for the spring is $2.9751\text{in}$.
• The number of active coils for the spring is $28.89$.

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 ultimate tensile strength is $S_{ut}$, 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 yield strength.

$S_y=0.55S_{ut}$ . (VI)

Here, the yield strength is $S_y$.

Write the expression for the fatigue strength.

$S_r=0.45S_{ut}$ . (VII)

Here, the fatigue strength of the spring is $S_r$.

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

$r=\frac{F_a}{F_m}$ . (VIII)

Here, the load line slope is $r$, the alternating shear force component is $F_a$ and the midrange component is $F_m$.

Write the expression for the Gerber ordinate intercept.

$S_e=\frac{S_r/2}{1-{\left(\frac{S_r}{2S_{ut}}\right)}^2}$ (IX)

Here, the ordinate intercept for shear is $S_e$.

Write the expression for the amplitude component of the strength.

$s_a=\frac{r^2{S}^2{}_{ut}}{2s_e}\left[-1+\sqrt{1+{\left(\frac{2S_e}{r{S}_{ut}}\right)}^2}\right]$ (X)

Here, the load line factor is $r$.

Write the expression for bending stress for hook.

${\left({\sigma }_a\right)}_A=\frac{4F_a}{\pi d^2}\left(\frac{4C^2-C-1}{C-1}+1\right)$ (XI)

Here, the bending stress for hook is ${\left({\sigma }_a\right)}_A$ and the spring index is $C$.

Write the expression for the spring index.

$C=0.5\left[\frac{\pi d^2{S}_a}{144}+\sqrt{\left(\frac{\pi d^2{S}_a}{144}\right)-\frac{\pi d^2{S}_a}{36}+2}\right]$ (XII)

Here, the wire diameter is $d$.

Write the expression for the mean coil diameter.

$D=Cd$ (XIII)

Here, the mean coil diameter is $D$.

Write the expression for the lowest spring force.

$F_i=\frac{\pi d^3{\tau }_i}{8D}$ (XIV)

Here, the initial tension in the spring is $F_i$ and the initial uncorrected stress factor is ${\tau }_i$.

Write the expression for the spring rate of the spring.

$k=\frac{F_{\max }-F_{\min }}{y_{\max }}$ (XV)

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

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 modulus of rigidity is $G$.

Write the expression for the number of coil of body.

$N_b=N_a-\frac{G}{E}$ (XVII)

Here, the number of coil of body is $N_b$ and the modulus of elasticity is $E$.

Write the expression for the free length of the spring.

$L_o=\left(2C-1+N_b\right)d$ (XVIII)

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

Write the expression for the maximum length of the spring.

$L_{\max }=L_o+\frac{\left(F_{\max }-F_i\right)}{k}$ (XIX)

Here, the maximum length of the spring is $L_{\max }$.

Write the expression for the fatigue factor of the safety.

${\left(n_f\right)}_A=\frac{S_a}{{\left({\sigma }_a\right)}_A}$ (XX)

Here, the fatigue factor of the safety is ${\left(n_f\right)}_A$.

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

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

Here, the Bergstrasser factor is $K_B$.

Write the expression for the alternating shear stress component.

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

Here, the alternating shear stress component is ${\tau }_a$.

Write the expression for the mid range shear stress component.

${\tau }_m=\frac{F_m}{F_a}{\tau }_a$ (XXIII)

Here, the mid range shear stress component is ${\tau }_m$.

Write the expression for the radius of the wire.

$r_2=2d$ . (XXIV)

Here, the radius of the wire is $r_2$.

Write the expression for the spring index at hook point $B$.

$C_2=\frac{2r_2}{d}$ (XXV)

Here, the spring index at hook point $B$ is $C_2$.

Write the expression for the Bergstrasser factor at hook point $B$.

${\left(K_B\right)}_B=\frac{4C-1}{4C-4}$ (XXVI)

Here, the Bergstrasser factor at hook point $B$ is ${\left(K_B\right)}_B$.

Write the expression for the alternating shear stress component at hook point $B$.

${\left({\tau }_a\right)}_B=\frac{{\left(K_B\right)}_B}{K_B}{\tau }_m$ (XXVII)

Here, the alternating shear stress component at hook point $B$ is ${\left({\tau }_a\right)}_B$.

Write the expression for the mid range shear stress component at the hook point $B$.

${\left({\tau }_m\right)}_B=\frac{{\left(K_B\right)}_B}{K_B}{\tau }_m$ (XXVIII)

Here, the mid range shear stress component at the hook point $B$ is ${\left({\tau }_m\right)}_B$.

Write the expression for bending stress at hook point $B$.

${\left({\sigma }_a\right)}_{\max }=\frac{4F_{\max }}{\pi d^2}\left(\frac{4C^2-C-1}{C-1}+1\right)$ (XXIX)

Write the expression for the shear stress for body.

${\tau }_i=\frac{F_i}{F_a}{\tau }_a$ (XXX)

Write the expression for the load line factor for body.

$r=\frac{{\tau }_a}{\left({\tau }_m-{\tau }_i\right)}$ (XXXI)

Write the expression for the yield strength for the body.

${\left(S_{sa}\right)}_y=\frac{r}{r+1}\left(S_{sy}-{\tau }_i\right)$ (XXXII)

Here, the yield strength for the body is ${\left(S_{sa}\right)}_y$.

Write the expression for the fatigue factor of the safety for body.

${\left(n_f\right)}_b={\left(n_f\right)}_b=\frac{1}{2}\frac{\left(S_{sa}\right)}{\left({\tau }_{\mathrm{m}}\right)}\left(\frac{\left({\tau }_a\right)}{\left(S_{se}\right)}\right)\left\{-1+\sqrt{1+{\left[\frac{2\left({\tau }_{\mathrm{m}}\right)\left(S_{se}\right)}{\left(S_{sa}\right)\left({\tau }_a\right)}\right]}^2}\right\}$ (XXXIII)

Here, the fatigue factor of the safety for body is ${\left(n_f\right)}_b$.

Write the expression for the maximum shear stress for hook shear.

${\tau }_{\max }={\left({\tau }_a\right)}_B+{\left({\tau }_{\mathrm{m}}\right)}_B$ (XXXIV)

Write the expression for the fatigue factor of the safety for hook shear.

${\left(n_f\right)}_B=\frac{1}{2}\frac{{\left(S_{sa}\right)}_B}{{\left({\tau }_{\mathrm{m}}\right)}_B}\left(\frac{{\left({\tau }_a\right)}_B}{{\left(S_{se}\right)}_B}\right)\left\{-1+\sqrt{1+{\left[\frac{2{\left({\tau }_{\mathrm{m}}\right)}_B{\left(S_{se}\right)}_B}{{\left(S_{sa}\right)}_B{\left({\tau }_a\right)}_B}\right]}^2}\right\}$ (XXXV)

Here, the fatigue factor of the safety for hook shear is ${\left(n_f\right)}_B$.

Write the expression for the relative cost material.

$\text{fom}=\frac{-\left(7.6\right){\pi }^2{d}^2\left(N_b+2\right)D}{4}$ (XXXVI)

Write the expression for the repeating allowable stress.

$S_{sr}=0.30S_{ut}$ (XXXVII)

Here, the repeating allowable stress is $S_{sr}$.

Write the expression for the repeating allowable stress at body.

${\left(S_{sr}\right)}_B=0.30S_{ut}$ (XXXVIII)

Here, the repeating allowable stress at body is ${\left(S_{sr}\right)}_B$.

Write the expression for the fatigue factor of the safety.

${\left(n_y\right)}_b=\frac{{\left(S_{sa}\right)}_y}{\left({\tau }_a\right)}$ (XXXIX)

Here, the fatigue factor of the safety for body is ${\left(n_y\right)}_b$.

Write the expression for the maximum allowable stresses for helical springs at hook shear.

$S_{sy}=0.30S_{ut}$ . (XL)

Here, the allowable yield stress for helical springs at hook shear is $S_{sy}$.

Write the expression for the uncorrected torsional stress.

${\tau }_i=\frac{33500}{\exp \left(0.105C\right)}\pm 1000\left(4-\frac{C-3}{6.5}\right)$ . (XLI)

Here, the uncorrected torsional stress is ${\tau }_i$.

Write the expression for the outer diameter.

$D_o=D+d$ (XLII)

Here, the outer diameter is $D_o$.

Conclusion:

Substitute $18\text{lbf}$ for $F_{\max }$ and $9\text{lbf}$ for $F_{\min }$ in Equation (I).

$\begin{array}{c}{F}_a=\frac{18\text{lbf}-9\text{lbf}}{2}\\ =\frac{9\text{lbf}}{2}\\ =4.5\text{lbf}\end{array}$

Substitute $18\text{lbf}$ for $F_{\max }$ and $9\text{lbf}$ for $F_{\min }$ in Equation (II).

$\begin{array}{c}{F}_m=\frac{18\text{lbf}+9\text{lbf}}{2}\\ =\frac{27\text{lbf}}{2}\\ =13.5\text{lbf}\end{array}$

Refer to table 10-4 “for estimating minimum tensile strength of the spring wires” to obtain the intercept and slope constant $169\text{kpsi}\cdot \text{in}$ for $A$ and $0.146$ for $m$ at the wire diameter $\text{0}\text{.081}\text{in}$.

Substitute $0.081\text{in}$ for $d$, $169\text{kpsi}\cdot \text{in}$ for $A$ and $0.146$ for $m$ in Equation (III).

$\begin{array}{c}{S}_{ut}=\frac{169\text{kpsi}\cdot \text{in}}{{\left(0.081\text{in}\right)}^{0.146}}\\ =\frac{169\text{kpsi}}{0.6928}\\ =243.9\text{kpsi}\end{array}$

Substitute $243.9\text{kpsi}$ for $S_{ut}$ in Equation (IV).

$\begin{array}{c}{S}_{sy}=0.35\times 243.9\text{kpsi}\\ =85.36\text{kpsi}\\ \approx 85.4\text{kpsi}\end{array}$

Substitute $243.9\text{kpsi}$ for $S_{ut}$ in Equation (V).

$\begin{array}{c}{S}_{su}=0.67\times 243.9\text{kpsi}\\ =163.4\text{kpsi}\end{array}$

Substitute $243.9\text{kpsi}$ for $S_{ut}$ in Equation (VI).

$\begin{array}{c}{S}_y=0.55\times 243.9\text{kpsi}\\ =134.14\text{kpsi}\\ \approx 134.2\text{kpsi}\end{array}$

Substitute $243.9\text{kpsi}$ for $S_{ut}$ in Equation (VII).

$\begin{array}{c}{S}_r=0.45\times 243.9\text{kpsi}\\ =109.75\text{kpsi}\\ \approx 109.8\text{kpsi}\end{array}$

Substitute $4.5\text{lbf}$ for $F_a$ and $\text{13}\text{.5}\text{lbf}$ for $F_m$ in Equation (VIII).

$\begin{array}{c}r=\frac{\text{4}\text{.5}\text{lbf}}{\text{13}\text{.5}\text{lbf}}\\ =0.333\end{array}$

Substitute $109.8\text{kpsi}$ for $S_r$ and $243.9\text{kpsi}$ for $S_{ut}$ in Equation (IX).

$\begin{array}{c}{S}_e=\frac{\frac{109.8\text{kpsi}}{2}}{1-{\left(\frac{109.8\text{kpsi}}{2\times 243.9\text{kpsi}}\right)}^2}\\ =\frac{54.9\text{kpsi}}{1-0.0506}\\ =57.8\text{kpsi}\end{array}$

Substitute $243.9\text{kpsi}$ for $S_{su}$, $57.8\text{kpsi}$ for $S_e$, $0.333$ for $r$ in Equation (X).

$\begin{array}{c}{S}_{sa}=\frac{{\left(0.33\text{in}\right)}^2{\left(243.9\text{kpsi}\right)}^2}{2\left(57.8\text{kpsi}\right)}\left\{-1+\sqrt{1+{\left[\frac{2\left(57.8\text{kpsi}\right)}{\left(0.33\text{in}\right)\left(243.9\text{kpsi}\right)}\right]}^2}\right\}\\ =\frac{6478.15\text{kpsi}}{115.6}\left\{-1+\sqrt{1+{\left[\frac{115.6}{80.48}\right]}^2}\right\}\\ =56.03\text{kpsi}\left\{-1+1.750\right\}\\ \approx 42.02\text{kpsi}\end{array}$

For hook bending:

Substitute $\text{0}\text{.081}\text{in}$ for $d$ and $42.02\text{kpsi}$ for $S_{sa}$ in Equation (XII).

$\begin{array}{l}C=0.5\left[\frac{\pi {\left(\text{0}\text{.081}\text{in}\right)}^{2}42.02\text{kpsi}}{144}+\sqrt{\begin{array}{l}{\left(\frac{\pi {\left(\text{0}\text{.081}\text{in}\right)}^{2}42.02\text{kpsi}}{144}\right)}^2\\ -\pi \frac{{\left(\text{0}\text{.081}\text{in}\right)}^{2}42.02\text{kpsi}}{36}+2\end{array}}\right]\\ =0.5\left[0.006014\text{kpsi}\cdot {\text{in}}^2\left(\frac{{10}^3\text{psi}}{1\text{kpsi}}\right)+\sqrt{\begin{array}{c}\left(3.616\times {10}^{-5}\right){\text{kpsi}}^2\cdot {\text{in}}^4\left(\frac{{10}^6{\text{psi}}^2}{1{\text{kpsi}}^2}\right)\\ -0.0240\text{kpsi}\cdot {\text{in}}^2\left(\frac{{10}^3\text{psi}}{1\text{kpsi}}\right)+2\end{array}}\right]\\ =4.6\end{array}$

Substitute $4.6$ for $C$ and $\text{0}\text{.081}\text{in}$ for $d$ in Equation (XIII).

$\begin{array}{c}D=4.6\times \text{0}\text{.081}\text{in}\\ =\text{0}\text{.3726}\text{in}\end{array}$

Substitute $4.6$ for $C$ in Equation (XLI).

$\begin{array}{c}{\tau }_i=\frac{33500}{\exp \left(0.105\left(4.6\right)\right)}-1000\left(4-\frac{4.6-3}{6.5}\right)\text{psi}\\ =\text{20667}\text{.14}-1000\left(3.75\right)\text{psi}\\ \text{=16}\text{.91329}\text{kpsi}\end{array}$

Substitute $\text{16}\text{.91329}\text{kpsi}$ for ${\tau }_i$, $\text{0}\text{.3726}\text{in}$ for $D$ and $\text{0}\text{.081}\text{in}$ for $d$ in Equation (XIV).

$\begin{array}{c}{F}_i=\frac{\pi {\left(\text{0}\text{.081}\text{in}\right)}^3\left(\text{16}\text{.91329}\text{kpsi}\right)}{8\left(\text{0}\text{.3726}\text{in}\right)}\\ =\frac{0.0282}{2.9808}\text{kpsi}\cdot {\text{in}}^2\left(\frac{{10}^3\text{lbf}/{\text{in}}^{\text{2}}}{1\text{kpsi}}\right)\\ =9.46\text{lbf}\end{array}$

Substitute $\text{0}\text{.25}\text{in}$ for $k$, $18\text{lbf}$ for $F_{\max }$ and $9\text{lbf}$ for $F_{\min }$ in Equation (XV).

$\begin{array}{c}k=\frac{18\text{lbf}-9\text{lbf}}{0.25\text{in}}\\ =\frac{9\text{lbf}}{0.25\text{in}}\\ =36\text{lbf}/\text{in}\end{array}$

Refer to table 10-5 “mechanical properties of some spring wires.” obtain the modulus of rigidity and modulus of elasticity for A313 stainless steel wore as ${\text{10}}^7\text{psi}$ for $G$ and $28\times {10}^6\text{psi}$ for $E$

Substitute ${\text{10}}^7\text{psi}$ for $G$, $36\text{lbf}/\text{in}$ for $k$, $\text{0}\text{.3726}\text{in}$ for $D$ and $\text{0}\text{.081}\text{in}$ for $d$ in Equation (XVI).

$\begin{array}{c}{N}_a=\frac{{\left(\text{0}\text{.081}\text{in}\right)}^4\left({\text{10}}^7\text{psi}\right)}{8\times \left(36\text{lbf}/\text{in}\right){\left(\text{0}\text{.3726}\text{in}\right)}^3}\\ =\frac{430.46}{14.8977}\\ =\text{28}\text{.89}\text{turns}\end{array}$

Substitute $\text{28}\text{.89}\text{turns}$ for $N_a$, $28\times {10}^6\text{psi}$ for $E$ and ${10}^7\text{psi}$ for $G$ in Equation (XVII).

$\begin{array}{c}{N}_b=\text{28}\text{.89}\text{turns}-\frac{{10}^7\text{psi}}{28\times {10}^6\text{psi}}\\ =\text{28}\text{.53}\text{turns}\end{array}$

Substitute $4.6$ for $C$, $\text{28}\text{.53}\text{turns}$ for $N_b$ and $\text{0}\text{.081}\text{in}$ for $d$ in Equation (XVIII).

$\begin{array}{c}{L}_o=\left(2\left(4.6\right)-1+\text{28}\text{.53}\text{turns}\right)0.08\text{1}\text{in}\\ =\left(36.73\right)\left(0.08\text{1}\text{in}\right)\\ =\text{2}\text{.9751}\text{in}\end{array}$

Substitute $\text{2}\text{.9751}\text{in}$ for $L_o$, $\text{18}\text{lbf}$ for $F_{\max }$, $9.46\text{lbf}$ for $F_i$ and $36\text{lbf}/\text{in}$ for $k$ in Equation (XIX).

$\begin{array}{c}{L}_{\max }=\text{2}\text{.9751}\text{in}+\frac{\left(\text{18}\text{lbf}-9.46\text{lbf}\right)}{36\text{lbf}/\text{in}}\\ =\text{2}\text{.9751}\text{in}+0.2569\text{in}\\ =3.212\text{in}\end{array}$

Substitute $0.081\text{in}$ for $d$, $4.6$ for $C$ and $\text{4}\text{.5}\text{lbf}$ for $F_a$ in Equation (XI).

$\begin{array}{c}{\left({\sigma }_a\right)}_A=\frac{4\left(\text{4}\text{.5}\text{lbf}\right)}{\pi {\left(0.081\text{in}\right)}^2}\left(\frac{4{\left(4.6\right)}^2-4.6-1}{4.6-1}+1\right)\\ =873.27\left(22.95\right){\text{lbf}/\text{in}}^2\left(\frac{{10}^{-3}\text{kpsi}}{{1\text{lbf}/\text{in}}^2}\right)\\ =20.046\text{kpsi}\end{array}$

Substitute $20.046\text{kpsi}$ for ${\left({\sigma }_a\right)}_A$ and $42.02\text{kpsi}$ for $S_a$ in Equation (XX).

$\begin{array}{c}{\left(n_f\right)}_A=\frac{42.02\text{kpsi}}{20.046\text{kpsi}}\\ =2.096\end{array}$

For body:

Substitute $4.6$ for $C$ in Equation (XXI).

$\begin{array}{c}{K}_B=\frac{\left(4\times 4.6\right)+2}{\left(4\times 4.6\right)-3}\\ =\frac{20.4}{15.4}\\ =1.3246\end{array}$

Substitute $\text{4}\text{.5}\text{lbf}$ for $F_a$, $1.3246$ for $K_B$, $\text{0}\text{.3726}\text{in}$ for $D$ and $\text{0}\text{.081}\text{in}$ for $d$ in Equation (XXII).

$\begin{array}{c}{\tau }_a=\frac{8\left(1.3246\right)\left(\text{4}\text{.5}\text{lbf}\right)\left(\text{0}\text{.3726}\text{in}\right)}{\pi {\left(\text{0}\text{.081}\text{in}\right)}^3}\\ =\frac{17.767}{\left(0.0016695\right)}{\text{lbf}/\text{in}}^2\left(\frac{{10}^{-3}\text{kpsi}}{{1\text{lbf}/\text{in}}^2}\right)\\ =10641.65\times {10}^{-3}\text{kpsi}\\ =10.641\text{kpsi}\end{array}$

Substitute $10.641\text{kpsi}$ for ${\tau }_a$, $\text{13}\text{.5}\text{lbf}$ for $F_m$ and $\text{4}\text{.5}\text{lbf}$ for $F_a$ in Equation (XXIII).

$\begin{array}{c}{\tau }_m=\left(\frac{\text{13}\text{.5}\text{lbf}}{\text{4}\text{.5}\text{lbf}}\right)\left(10.641\text{kpsi}\right)\\ =3\left(10.641\text{kpsi}\right)\\ =31.923\text{kpsi}\end{array}$

Substitute $243.9\text{kpsi}$ for $S_{ut}$ in Equation (XXXVII).

$\begin{array}{c}{S}_{sr}=0.30\left(243.9\text{kpsi}\right)\\ =73.17\text{kpsi}\end{array}$

Substitute $163.4\text{kpsi}$ for $S_{sa}$ and $73.17\text{kpsi}$ for $S_{sr}$ in Equation (IX).

$\begin{array}{c}{S}_{se}=\frac{73.17\text{kpsi}/2}{1-{\left(\frac{73.17\text{kpsi}}{2\left(163.4\text{kpsi}\right)}\right)}^2}\\ =\frac{36.58}{0.9498}\text{kpsi}\\ =38.85\text{kpsi}\end{array}$

Substitute $10.641\text{kpsi}$ for ${\tau }_a$, $31.923\text{kpsi}$ for ${\tau }_m$, $163.4\text{kpsi}$ for $S_{sa}$ and $38.5\text{kpsi}$ for $S_{se}$ in Equation (XXXIII).

$\begin{array}{c}{\left(n_f\right)}_b=\left[\begin{array}{l}\frac{1}{2}\left({\left(\frac{163.4\text{kpsi}}{31.923\text{kpsi}}\right)}^2\frac{\left(10.641\text{kpsi}\right)}{\left(38.5\text{kpsi}\right)}\right)\\ \left\{-1+\sqrt{1+{\left[\frac{2\left(31.923\text{kpsi}\right)\left(38.5\text{kpsi}\right)}{\left(163.4\text{kpsi}\right)\left(10.641\text{kpsi}\right)}\right]}^2}\right\}\end{array}\right]\\ =\frac{1}{2}\left(26.199\right)\left(0.2763\right)\left[-1+1.7316\right]\\ =2.64\end{array}$

Substitute $\text{0}\text{.081}\text{in}$ for $d$ in Equation (XXIV).

$\begin{array}{c}{r}_2=2\left(\text{0}\text{.081}\text{in}\right)\\ =0.162\text{in}\end{array}$

Substitute $\text{0}\text{.162}\text{in}$ for $r_2$ in Equation (XXV).

$\begin{array}{c}{C}_2=\frac{2\left(\text{0}\text{.162}\text{in}\right)}{\text{0}\text{.081}\text{in}}\\ =4\end{array}$

Substitute $4$ for $C$ in Equation (XXVI).

$\begin{array}{c}{\left(K\right)}_B=\frac{\left(4\times 4\right)-1}{\left(4\times 4\right)-4}\\ =\frac{15}{12}\\ =1.25\end{array}$

Substitute $1.25$ for ${\left(K\right)}_B$, $1.30$ for $K_B$ and $\text{11}\text{.16}\text{kpsi}$ for ${\tau }_a$ in Equation (XXVII).

$\begin{array}{c}{\left({\tau }_a\right)}_B=\frac{1.25}{1.30}\left(\text{11}\text{.16}\text{kpsi}\right)\\ =10.73\text{kpsi}\end{array}$

Substitute $1.25$ for ${\left(K\right)}_B$, $1.30$ for $K_B$ and $33.47\text{kpsi}$ for ${\tau }_m$ in Equation (XXVIII).

$\begin{array}{c}{\left({\tau }_m\right)}_B=\frac{1.25}{1.30}\left(33.47\text{kpsi}\right)\\ =32.1826\text{kpsi}\end{array}$

Substitute $243.9\text{kpsi}$ for $S_{ut}$ in Equation (XXXVIII).

$\begin{array}{c}{\left(S_{sr}\right)}_B=0.28\left(243.9\text{kpsi}\right)\\ =68.29\text{kpsi}\\ \approx 68.3\text{kpsi}\end{array}$

Substitute $163.4\text{kpsi}$ for $S_{sa}$ and $68.3\text{kpsi}$ for ${\left(S_{sr}\right)}_B$ in Equation (IX).

$\begin{array}{c}{\left(S_{se}\right)}_B=\frac{68.3\text{kpsi}/2}{1-{\left(\frac{68.3\text{kpsi}}{2\left(163.4\text{kpsi}\right)}\right)}^2}\\ =\frac{34.15}{0.9563}\text{kpsi}\\ =35.70\text{kpsi}\end{array}$

Substitute $10.73\text{kpsi}$ for ${\left({\tau }_a\right)}_B$, $32.1826\text{kpsi}$ for ${\left({\tau }_m\right)}_B$, $35.7\text{kpsi}$ for ${\left(S_{se}\right)}_B$ and $38.5\text{kpsi}$ for $S_{se}$ in Equation (XXXV).

$\begin{array}{c}{\left(n_f\right)}_B=\left[\begin{array}{l}\frac{1}{2}\left({\left(\frac{163.4\text{kpsi}}{32.18\text{kpsi}}\right)}^2\frac{\left(10.73\text{kpsi}\right)}{\left(35.7\text{kpsi}\right)}\right)\\ \left\{-1+\sqrt{1+{\left[\frac{2\left(32.18\text{kpsi}\right)\left(35.7\text{kpsi}\right)}{\left(163.4\text{kpsi}\right)\left(10.73\text{kpsi}\right)}\right]}^2}\right\}\end{array}\right]\\ =\frac{1}{2}\left(25.78\right)\left(0.3005\right)\left[-1+1.6484\right]\\ =2.51\end{array}$

For yield bending:

Substitute $0.081\text{in}$ for $d$, $4.6$ for $C$ and $18\text{lbf}$ for $F_{\max }$ in Equation (XXIX).

$\begin{array}{c}{\left({\sigma }_a\right)}_{\max }=\frac{4\left(18\text{lbf}\right)}{\pi {\left(0.081\text{in}\right)}^2}\left(\frac{4{\left(4.6\right)}^2-4.6-1}{4.6-1}+1\right)\\ =3493.11\left(22.95\right){\text{lbf}/\text{in}}^2\left(\frac{{10}^{-3}\text{kpsi}}{{1\text{lbf}/\text{in}}^2}\right)\\ =80.166\text{kpsi}\end{array}$

Substitute $80.166\text{kpsi}$ for ${\left({\sigma }_a\right)}_{\max }$ and $134.2\text{kpsi}$ for $S_a$ in Equation (XX).

$\begin{array}{c}{\left(n_f\right)}_A=\frac{134.2\text{kpsi}}{80.166\text{kpsi}}\\ =1.67\end{array}$

Substitute $\text{8}\text{.75}\text{lbf}$ for $F_i$, $\text{4}\text{.5}\text{lbf}$ for $F_a$ and $10.641\text{kpsi}$ for ${\tau }_a$ in Equation (XXX).

$\begin{array}{c}{\tau }_i=\left(\frac{\text{8}\text{.75}\text{lbf}}{\text{4}\text{.5}\text{lbf}}\right)10.641\text{kpsi}\\ =1.944\left(10.641\text{kpsi}\right)\\ =20.69\text{kpsi}\end{array}$

Substitute $20.69\text{kpsi}$ for ${\tau }_i$, $31.923\text{kpsi}$ for ${\tau }_m$ and $10.641\text{kpsi}$ for ${\tau }_a$ in Equation (XXXI).

$\begin{array}{c}r=\frac{10.641\text{kpsi}}{\left(31.923\text{kpsi}-20.69\text{kpsi}\right)}\\ =\frac{10.641\text{kpsi}}{11.233\text{kpsi}}\\ =0.927\end{array}$

Substitute $20.69\text{kpsi}$ for ${\tau }_i$, $0.947$ for $r$ and $85.4\text{kpsi}$ for $S_{sy}$ in Equation (XXXII).

$\begin{array}{c}{\left(S_{sa}\right)}_y=\frac{0.947}{\left(0.947+1\right)}\left(85.4\text{kpsi}-20.69\text{kpsi}\right)\\ =0.4863\left(64.71\text{kpsi}\right)\\ =31.46\text{kpsi}\end{array}$

Substitute $31.46\text{kpsi}$ for ${\left(S_{sa}\right)}_y$ and $10.641\text{kpsi}$ for ${\tau }_a$ in Equation (XXXI).

$\begin{array}{c}{\left(n_y\right)}_b=\frac{31.46\text{kpsi}}{10.641\text{kpsi}}\\ =2.9564\end{array}$

For hook shear:

Substitute $243.9\text{kpsi}$ for $S_{ut}$ in Equation (XXXX).

$\begin{array}{c}{\left(S_{sy}\right)}_B=0.30\left(243.9\text{kpsi}\right)\\ =73.17\text{kpsi}\end{array}$

Substitute $10.73\text{kpsi}$ for ${\left({\tau }_a\right)}_B$ and $32.1826\text{kpsi}$ for ${\left({\tau }_m\right)}_B$ in Equation (XXXIV).

$\begin{array}{c}{\tau }_{\max }=10.73\text{kpsi}+32.18\text{kpsi}\\ =\text{42}\text{.91}\text{kpsi}\end{array}$

Substitute $73.17\text{kpsi}$ for $S_{sy}$ and $42.91\text{kpsi}$ for ${\tau }_{\max }$ in Equation (XXXI).

$\begin{array}{c}{\left(n_y\right)}_B=\frac{73.17\text{kpsi}}{42.9\text{kpsi}}\\ =1.705\end{array}$

Substitute $\text{28}\text{.53}\text{turns}$ for $N_b$, $\text{0}\text{.3726}\text{in}$ for $D$ and $\text{0}\text{.081}\text{in}$ for $d$ in Equation (XXXVI).

$\begin{array}{c}fom=-\frac{7.6\left({\pi }^2\right){\left(\text{0}\text{.081}\text{in}\right)}^2\left(\text{28}\text{.53}\text{turns}+2\right)\text{0}\text{.3726}\text{in}}{4}\\ =-\frac{5.5982}{4}\\ =-1.39\end{array}$

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.

Table-(1)

S/No	Parameters	$d_1$	$d_2$	$d_3$
1	$d$	$0.081$	$0.085$	$0.092$
2	$S_{ut}$	$243.9$	$242.21$	$239.42$
3	$S_{su}$	$163.4$	$162.28$	$160.41$
4	$S_r$	$109.8$	$108.99$	$107.74$
5	$S_e$	$57.80$	$57.40$	$56.74$
6	$S_a$	$42.02$	$41.86$	$41.36$
7	$C$	$4.6$	$5.48$	$6.54$
8	$D$	$0.3726$	$0.466$	$0.602$
9	$D_o$	$0.4536$	$0.551$	$0.694$
10	$F_i$ (cal.)	$8.57$	$7.87$	$6.79$
11	$F_i$(rd)	$9.46$	$9.75$	$10.75$
12	$k$	$36$	$36$	$36$
13	$N_a$	$28.89$	$17.90$	$11.38$
14	$N_b$	$28.53$	$17.54$	$11.02$
15	$L_o$	$2.9751$	$2.33$	$2.12$
16	$L_{\max }$	$3.212$	$2.56$	$2.32$
17	${\left({\sigma }_a\right)}_A$	$20.046$	$20.92$	$20.68$
18	${\left(n_f\right)}_A$	$2$	$2$	$2$
19	$K_B$	$1.32$	$1.26$	$1.21$
20	$\left({\tau }_a\right)$	$10.34$	$10.99$	$10.77$
21	$\left({\tau }_m\right)$	$31.92$	$32.98$	$32.32$
22	$S_{sr}$	$73.17$	$72.66$	$71.82$
23	$S_{se}$	$38.85$	$38.24$	$37.80$
24	${\left(n_f\right)}_b$	$2.64$	$2.54$	$2.56$
25	${\left(K\right)}_B$	$1.25$	$1.25$	$1.25$
26	${\left({\tau }_a\right)}_B$	$10.70$	$10.87$	$11.08$
27	${\left({\tau }_m\right)}_B$	$32.11$	$32.61$	$33.24$
28	${\left(S_{sr}\right)}_B$	$68.29$	$67.81$	$67.04$
29	${\left(S_{se}\right)}_B$	$35.70$	$35.45$	$35.05$
30	${\left(n_f\right)}_B$	$2.51$	$2.46$	$2.38$
31	$S_y$	$134.2$	$133.21$	$131.68$
32	${\left({\sigma }_A\right)}_{\max }$	$80.166$	$83.68$	$82.72$
33	${\left(n_y\right)}_A$	$1.59$	$1.59$	$1.44$
34	${\tau }_i$	$20.69$	$23.82$	$25.74$
35	$r$	$0.927$	$1.157$	$1.44$
36	${\left(S_{sy}\right)}_b$	$85.4$	$84.77$	$83.80$
37	${\left(S_{sa}\right)}_y$	$31.46$	$32.68$	$34.30$
38	${\left(n_y\right)}_b$	$2.95$	$2.97$	$3.18$
39	${\left(S_{sy}\right)}_B$	$73.17$	$72.66$	$71.82$
40	${\left({\tau }_B\right)}_{\max }$	$42.91$	$43.48$	$44.32$
41	$\text{fom}$	$-1.39$	$-1.234$	$-1.245$

Substitute $\text{0}\text{.3726}\text{in}$ for $D$ and $\text{0}\text{.081}\text{in}$ for $d$ in Equation (XLII).

$\begin{array}{c}{D}_o=\text{0}\text{.3726}\text{in}+\text{0}\text{.081}\text{in}\\ =0.4536\text{in}\end{array}$

Thus, the dimensions of the spring are:

• The wire diameter for the spring is $\text{0}\text{.081}\text{in}$.
• The outer diameter for the spring is $0.4536\text{in}$.
• The free length for the spring is $2.9751\text{in}$.
• The number of active coils for the spring is $28.89$.