Chapter 10, Problem 21P

A static service music wire helical compression spring is needed to support a 20-lbf load after being compressed 2 in. The solid height of the spring cannot exceed $1\frac{1}{2}$ in. The free length must not exceed 4 in. The static factor of safely must equal or exceed 1.2. For robust linearity use a fractional overrun to closure ξ of 0.15. There are two springs to be designed. Start with a wire diameter of 0.075 in.

1. (a) The spring must operate over a $\frac{3}{4}-$in rod. A 0.050-in diametral clearance allowance should be adequate to avoid interference between the rod and the spring due to out-of-round coils. Design the spring.
2. (b) The spring must operate in a 1-in-diameier hole. A 0.050-in diametral clearance allowance should be adequate to avoid interference between the spring and the hole due to swelling of the spring diameter as the spring is compressed and out-of-round coils. Design the spring.

To determine

The design parameters for the spring over a rod.

Answer to Problem 21P

The design parameters for the spring over a rod are:

• The wire diameter for the spring is $\text{0}\text{.085}\text{in}$.
• The free length for the spring is $3.410\text{in}$.
• The total number of the coils for the spring is $13.06$.

Explanation of Solution

Write the expression for the inner diameter of the spring.

$D_i=D-d$ . (I)

Here, the inner diameter of the spring is $D_i$.

Write the expression for the mean coil diameter.

$C=\frac{D}{d}$ (II)

Here, the mean coil diameter is $D$.

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

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

Here, the Bergstrasser factor is $K_B$.

Write the expression for ultimate tensile strength.

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

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.

$S_{sy}=0.45S_{ut}$ . (V)

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

Write the expression for the number of the active coils.

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

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.

$N_t=N_a+2$ (VII)

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

Write the expression for the deflection under the steady load.

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

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.

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

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

Write the expression for the free length of the spring.

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

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

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

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

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

Write the expression for the alternating shear stress component.

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

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

Write the expression for the shear force of the spring.

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

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

Write the expression for the factor of the safety.

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

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

Write the expression for the relative cost material.

$\text{fom}=\frac{-\left(\text{relativematerialcost}\right){\pi }^2{d}^2{N}_{t}D}{4}$ (XV)

Write the expression for the spring rate.

$k=\frac{F_{\max }}{y}$ (XVI)

Here, the deflection in the spring is $y$, the maximum force on the spring is $F_{\max }$ and the spring rate is $k$.

Write the expression for the spring load.

$F_s=F_{\max }\left(1+\xi \right)$ (XVII)

Here, the force on the solid spring is $F_s$ and the robust linearity is $\xi$.

Conclusion:

Substitute $20\text{lbf}$ for $F_{\max }$ and $2\text{in}$ for $y$ in Equation (XVI).

$\begin{array}{c}k=\frac{20\text{lbf}}{2\text{in}}\\ =10\text{lbf}/\text{in}\end{array}$

Substitute $20\text{lbf}$ for $F_{\max }$ and $0.15$ for $\xi$ in Equation (XVII).

$\begin{array}{c}{F}_s=20\text{lbf}\left(1+0.15\right)\\ =23\text{lbf}\end{array}$

Substitute $0.075\text{in}$ for $d$ and $0.875\text{in}$ for $D$ in Equation (I).

$\begin{array}{c}{D}_i=0.875\text{in}-0.075\text{in}\\ =0.800\text{in}\end{array}$

Substitute $0.075\text{in}$ for $d$ and $0.875\text{in}$ for $D$ in Equation (II).

$\begin{array}{c}C=\frac{0.875\text{in}}{0.075\text{in}}\\ =11.66\end{array}$

Substitute $11.66$ for $C$ in Equation (III).

$\begin{array}{c}{K}_B=\frac{\left(4\times 11.66\right)+2}{\left(4\times 11.66\right)-3}\\ =\frac{48.64}{43.64}\\ =1.1145\\ \approx 1.115\end{array}$

Refer to table 10-4 “for estimating minimum tensile strength of the spring wires” to obtain the intercept and slope constant $201\text{kpsi}\cdot \text{in}$ for $A$ and $0.145$ for $m$ at the wire diameter $\text{0}\text{.075}\text{in}$.

Substitute $0.075\text{in}$ for $d$, $201\text{kpsi}\cdot \text{in}$ for $A$ and $0.145$ for $m$ in Equation (IV).

$\begin{array}{c}{S}_{ut}=\frac{201\text{kpsi}\cdot \text{in}}{{\left(0.075\text{in}\right)}^{0.145}}\\ =\frac{201\text{kpsi}}{0.6868}\\ =292.62\text{kpsi}\end{array}$

Substitute $292.62\text{kpsi}$ for $S_{ut}$ in Equation (V).

$\begin{array}{c}{S}_{sy}=0.45\times 292.62\text{kpsi}\\ =131.679\text{kpsi}\\ \approx \text{131}\text{.68}\text{kpsi}\end{array}$

Refer to table 10-5 “mechanical properties of some spring wires.” to obtain the modulus of rigidity for square and ground ends at wire diameter $d=0.075\text{in}$ as $\text{11}\text{.75}\times {\text{10}}^6\text{psi}$.

Substitute $\text{11}\text{.75}\times {\text{10}}^6\text{psi}$ for $G$, $\text{0}\text{.075}\text{in}$ for $d$, $0.875\text{in}$ for $D$ and $10\text{lbf}/\text{in}$ for $k$ in Equation (VI).

$\begin{array}{c}{N}_a=\frac{11.75\times {\text{10}}^6\text{psi}{\left(\text{0}\text{.075}\text{in}\right)}^4}{8\left(10\text{lbf}/\text{in}\right){\left(0.875\text{in}\right)}^3}\\ =\frac{371.77\text{psi}\cdot \text{in}}{53.593\text{lbf}/\text{in}}\left(\frac{1\text{lbf}/{\text{in}}^2}{1\text{psi}}\right)\\ =6.936\text{coils}\end{array}$

Substitute $6.936\text{coils}$ for $N_a$ in Equation (VII).

$\begin{array}{c}{N}_t=6.936\text{coils}+2\\ =8.936\text{coils}\end{array}$

Substitute $0.15$ for $\xi$ and $2\text{in}$ for $y_{\max }$ in Equation (VIII).

$\begin{array}{c}{y}_s=\left(1+0.15\right)2\text{in}\\ =2.3\text{in}\end{array}$

Substitute $8.936\text{coils}$ for $N_t$ and $0.075\text{in}$ for $d$ in Equation (IX).

$\begin{array}{c}{L}_s=0.075\text{in}\left(8.936\text{coils}\right)\\ =0.670\text{in}\end{array}$

Substitute $2.3\text{in}$ for $y_s$ and $0.670\text{in}$ for $L_s$ in Equation (X).

$\begin{array}{c}{L}_o=2.3\text{in}+0.670\text{in}\\ =2.970\text{in}\end{array}$

Since, the free length is lesser than the $5.26$ times mean coil diameter. Hence the end condition constant for music wire is $0.5$ $\left(\alpha =0.5\right)$.

Substitute $0.875\text{in}$ for $D$ and $0.5$ for $\alpha$ in Equation (XI).

$\begin{array}{c}{\left(L_{{}_o}\right)}_{cr}=2.63\left(\frac{0.875\text{in}}{0.5}\right)\\ =4.6025\text{in}\\ \approx 4.603\text{in}\end{array}$

Substitute $23\text{lbf}$ for $F_s$, $0.875\text{in}$ for $D$, $1.115$ for $K_B$ and $0.075\text{in}$ for $d$ in Equation (XII).

$\begin{array}{c}{\tau }_a=1.115\left(\frac{8\left(23\text{lbf}\right)\left(0.875\text{in}\right)}{\pi {\left(0.075\text{in}\right)}^3}\right)\\ =1.115\left(\frac{161}{1.3253\times {10}^{-3}}\right)\text{lbf}/{\text{in}}^{\text{2}}\left(\frac{1\text{kpsi}}{{\text{10}}^3\text{lbf}/{\text{in}}^{\text{2}}}\right)\\ =135.45\text{kpsi}\\ \approx 135.45\text{kpsi}\end{array}$

Substitute $20\text{lbf}$ for $F_{\max }$, $23\text{lbf}$ for $F_s$ and $135.45\text{kpsi}$ for ${\tau }_a$ in Equation (XIII).

$\begin{array}{c}{\tau }_s=1.115\left(\frac{20\text{lbf}}{23\text{lbf}}\right)135.45\text{kpsi}\\ \text{=1}\text{.115}\left(0.8695\right)135.45\text{kpsi}\\ =131.32\text{kpsi}\end{array}$

Substitute $131.32\text{kpsi}$ for ${\tau }_s$ and $131.68\text{kpsi}$ for $S_{sy}$ in Equation (XIV).

$\begin{array}{c}{n}_s=\frac{131.68\text{kpsi}}{131.32\text{kpsi}}\\ =1.002\end{array}$

Assume the relative cost material for the spring as $2.6$.

Substitute $2.6$ for relative cost material, $8.936$ for $N_t$, $\text{0}\text{.075}\text{in}$ for $d$ and $0.875\text{in}$ for $D$ in Equation (XV).

$\begin{array}{c}\text{fom}=\frac{-2.6{\pi }^2{\left(\text{0}\text{.075}\text{in}\right)}^2\left(0.875\text{in}\right)\left(8.936\right)}{4}\\ =-\frac{1.1286}{4}\\ =-0.282\end{array}$

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)

	Parameter	$d_1$	$d_2$	$d_3$
1	$d$	$0.075$	$0.080$	$0.085$
2	$D_i$	$0.800$	$0.800$	$0.800$
3	$D$	$0.875$	$0.880$	$0.885$
4	$C$	$11.66$	$11.00$	$10.41$
5	$N_a$	$6.936$	$8.828$	$11.061$
6	$N_t$	$8.936$	$10.828$	$13.061$
7	$L_s$	$0.670$	$0.866$	$1.110$
8	$L_o$	$2.970$	$3.166$	$3.410$
9	${\left(L_o\right)}_{cr}$	$4.603$	$4.629$	$4.655$
10	$A$	$201.00$	$201.00$	$201.00$
11	$m$	$0.145$	$0.145$	$0.145$
12	$S_{ut}$	$292.62$	$289.90$	$287.36$
13	$S_{sy}$	$131.68$	$130.45$	$129.313$
14	$K_B$	$1.115$	$1.122$	$1.129$
15	${\tau }_s$	$131.32$	$112.94$	$95.29$
16	$n_s$	$1.002$	$1.155$	$1.357$
17	$\text{fom}$	$-0.282$	$-0.391$	$-0.536$

Since the factor of safety is greater than the given factor of safety hence design is suitable for the helical spring $\left(1.357>1.2\right)$.

Thus, the dimensions of the spring are:

• The wire diameter for the spring is $\text{0}\text{.085}\text{in}$.
• The free length for the spring is $3.410\text{in}$.
• The total number of the coils for the spring is $13.06$.

To determine

The design parameters for the spring in a hole.

Answer to Problem 21P

The design parameters for the spring in a hole are:

• The wire diameter for the spring is $\text{0}\text{.085}\text{in}$.
• The free length for the spring is $3.477\text{in}$.
• The total number of the coils for the spring is $13.84$.

Explanation of Solution

Write the expression for the outer diameter of the spring.

$D_o=D+d$ . (XVIII)

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

Conclusion:

Substitute $20\text{lbf}$ for $F_{\max }$ and $2\text{in}$ for $y$ in Equation (XVI).

$\begin{array}{c}k=\frac{20\text{lbf}}{2\text{in}}\\ =10\text{lbf}/\text{in}\end{array}$

Substitute $20\text{lbf}$ for $F_{\max }$ and $0.15$ for $\xi$ in Equation (XVII).

$\begin{array}{c}{F}_s=20\text{lbf}\left(1+0.15\right)\\ =23\text{lbf}\end{array}$

Substitute $0.075\text{in}$ for $d$ and $0.875\text{in}$ for $D$ in Equation (XVIII).

$\begin{array}{c}{D}_o=0.875\text{in}+0.075\text{in}\\ =0.950\text{in}\end{array}$

Substitute $0.075\text{in}$ for $d$ and $0.875\text{in}$ for $D$ in Equation (II).

$\begin{array}{c}C=\frac{0.875\text{in}}{0.075\text{in}}\\ =11.66\end{array}$

Substitute $11.66$ for $C$ in Equation (III).

$\begin{array}{c}{K}_B=\frac{\left(4\times 11.66\right)+2}{\left(4\times 11.66\right)-3}\\ =\frac{48.64}{43.64}\\ =1.1145\\ \approx 1.115\end{array}$

Refer to table 10-4 “for estimating minimum tensile strength of the spring wires” to obtain the intercept and slope constant $201\text{kpsi}\cdot \text{in}$ for $A$ and $0.145$ for $m$ at the wire diameter $\text{0}\text{.075}\text{in}$.

Substitute $0.075\text{in}$ for $d$, $201\text{kpsi}\cdot \text{in}$ for $A$ and $0.145$ for $m$ in Equation (IV).

$\begin{array}{c}{S}_{ut}=\frac{201\text{kpsi}\cdot \text{in}}{{\left(0.075\text{in}\right)}^{0.145}}\\ =\frac{201\text{kpsi}}{0.6868}\\ =292.62\text{kpsi}\end{array}$

Substitute $292.62\text{kpsi}$ for $S_{ut}$ in Equation (V).

$\begin{array}{c}{S}_{sy}=0.45\times 292.62\text{kpsi}\\ =131.679\text{kpsi}\\ \approx \text{131}\text{.68}\text{kpsi}\end{array}$

Refer to table 10-5 “mechanical properties of some spring wires.” to obtain the modulus of rigidity for square and ground ends at wire diameter $d=0.075\text{in}$ as $\text{11}\text{.75}\times {\text{10}}^6\text{psi}$.

Substitute $\text{11}\text{.75}\times {\text{10}}^6\text{psi}$ for $G$, $\text{0}\text{.075}\text{in}$ for $d$, $0.875\text{in}$ for $D$ and $10\text{lbf}/\text{in}$ for $k$ in Equation (VI).

$\begin{array}{c}{N}_a=\frac{11.75\times {\text{10}}^6\text{psi}{\left(\text{0}\text{.075}\text{in}\right)}^4}{8\left(10\text{lbf}/\text{in}\right){\left(0.875\text{in}\right)}^3}\\ =\frac{371.77\text{psi}\cdot \text{in}}{53.593\text{lbf}/\text{in}}\left(\frac{1\text{lbf}/{\text{in}}^2}{1\text{psi}}\right)\\ =6.936\text{coils}\end{array}$

Substitute $6.936\text{coils}$ for $N_a$ in Equation (VII).

$\begin{array}{c}{N}_t=6.936\text{coils}+2\\ =8.936\text{coils}\end{array}$

Substitute $0.15$ for $\xi$ and $2\text{in}$ for $y_{\max }$ in Equation (VIII).

$\begin{array}{c}{y}_s=\left(1+0.15\right)2\text{in}\\ =2.3\text{in}\end{array}$

Substitute $8.936\text{coils}$ for $N_t$ and $0.075\text{in}$ for $d$ in Equation (IX).

$\begin{array}{c}{L}_s=0.075\text{in}\left(8.936\text{coils}\right)\\ =0.670\text{in}\end{array}$

Substitute $2.3\text{in}$ for $y_s$ and $0.670\text{in}$ for $L_s$ in Equation (X).

$\begin{array}{c}{L}_o=2.3\text{in}+0.670\text{in}\\ =2.970\text{in}\end{array}$

Since, the free length is lesser than the $5.26$ times mean coil diameter. Hence the end condition constant for music wire is $0.5$ $\left(\alpha =0.5\right)$.

Substitute $0.875\text{in}$ for $D$ and $0.5$ for $\alpha$ in Equation (XI).

$\begin{array}{c}{\left(L_{{}_o}\right)}_{cr}=2.63\left(\frac{0.875\text{in}}{0.5}\right)\\ =4.6025\text{in}\\ \approx 4.603\text{in}\end{array}$

Substitute $23\text{lbf}$ for $F_s$, $0.875\text{in}$ for $D$, $1.115$ for $K_B$ and $0.075\text{in}$ for $d$ in Equation (XII).

$\begin{array}{c}{\tau }_a=1.115\left(\frac{8\left(23\text{lbf}\right)\left(0.875\text{in}\right)}{\pi {\left(0.075\text{in}\right)}^3}\right)\\ =1.115\left(\frac{161}{1.3253\times {10}^{-3}}\right)\text{lbf}/{\text{in}}^{\text{2}}\left(\frac{1\text{kpsi}}{{\text{10}}^3\text{lbf}/{\text{in}}^{\text{2}}}\right)\\ =135.45\text{kpsi}\\ \approx 135.45\text{kpsi}\end{array}$

Substitute $20\text{lbf}$ for $F_{\max }$, $23\text{lbf}$ for $F_s$ and $135.45\text{kpsi}$ for ${\tau }_a$ in Equation (XIII).

$\begin{array}{c}{\tau }_s=1.115\left(\frac{20\text{lbf}}{23\text{lbf}}\right)135.45\text{kpsi}\\ \text{=1}\text{.115}\left(0.8695\right)135.45\text{kpsi}\\ =131.32\text{kpsi}\end{array}$

Substitute $131.32\text{kpsi}$ for ${\tau }_s$ and $131.68\text{kpsi}$ for $S_{sy}$ in Equation (XIV).

$\begin{array}{c}{n}_s=\frac{131.68\text{kpsi}}{131.32\text{kpsi}}\\ =1.002\end{array}$

Assume the relative cost material for the spring as $2.6$.

Substitute $2.6$ for relative cost material, $8.936$ for $N_t$, $\text{0}\text{.075}\text{in}$ for $d$ and $0.875\text{in}$ for $D$ in Equation (XV).

$\begin{array}{c}\text{fom}=\frac{-2.6{\pi }^2{\left(\text{0}\text{.075}\text{in}\right)}^2\left(0.875\text{in}\right)\left(8.936\right)}{4}\\ =-\frac{1.1286}{4}\\ =-0.282\end{array}$

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-(2)

	Parameter	$d_1$	$d_2$	$d_3$
1	$d$	$0.075$	$0.080$	$0.085$
2	$D_o$	$0.950$	$0.950$	$0.950$
3	$D$	$0.875$	$0.870$	$0.865$
4	$C$	$11.66$	$10.87$	$10.17$
5	$N_a$	$6.936$	$9.136$	$11.84$
6	$N_t$	$8.936$	$11.136$	$13.84$
7	$L_s$	$0.670$	$0.891$	$1.177$
8	$L_o$	$2.970$	$3.191$	$3.477$
9	${\left(L_o\right)}_{cr}$	$4.603$	$4.576$	$4.550$
10	$A$	$201.00$	$201.00$	$201.00$
11	$m$	$0.145$	$0.145$	$0.145$
12	$S_{ut}$	$292.62$	$289.90$	$287.36$
13	$S_{sy}$	$131.68$	$130.45$	$129.313$
14	$K_B$	$1.115$	$1.123$	$1.133$
15	${\tau }_s$	$131.32$	$111.78$	$93.43$
16	$n_s$	$1.002$	$1.167$	$1.384$
17	$fom$	$-0.282$	$-0.398$	$-0.555$

Since the factor of safety is greater than the given factor of safety hence design is suitable for the helical spring $\left(1.384>1.2\right)$.

Thus, the dimensions of the spring are:

• The wire diameter for the spring is $\text{0}\text{.085}\text{in}$.
• The free length for the spring is $3.477\text{in}$.
• The total number of the coils for the spring is $13.84$.