Chapter 10, Problem 22P

Solve Prob. 10–21 by iterating with an initial value of C = 10. If you have already solved Prob. 10–21, compare the steps and the results.

To determine

The design parameters for the spring over a rod.

Answer to Problem 22P

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 $\xi$.

Write the expression for the wire diameter.

$d=\frac{D_i}{C-1}$ (XVIII)

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 $0.800\text{in}$ for $D_i$ and $10$ for $C$ in Equation (XVIII).

$\begin{array}{c}d=\frac{0.800\text{in}}{10-1}\\ =0.08888\text{in}\\ \approx 0.089\text{in}\end{array}$

Refer to table A-28 “decimal equivalent of wire and sheet-metal gauges” to obtain the wire diameter for square and ground ends as $0.090\text{in}$.

Since the calculated wire diameter is very close to the wire diameter taken from the table, so it is acceptable for the given spring index.

Substitute $0.090\text{in}$ for $d$ and $0.890\text{in}$ for $D$ in Equation (II).

$\begin{array}{c}C=\frac{0.890\text{in}}{0.090\text{in}}\\ =9.88\end{array}$

Substitute $9.88$ for $C$ in Equation (III).

$\begin{array}{c}{K}_B=\frac{\left(4\times 9.88\right)+2}{\left(4\times 9.88\right)-3}\\ =\frac{41.52}{36.52}\\ =1.136\\ \approx 1.136\end{array}$

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 $\text{0}\text{.090}\text{in}$.

Substitute $0.090\text{in}$ for $d$, $201\text{kpsi}\cdot \text{in}$ for $A$ and $0.145$ for $m$ in Equation (IV).

$\begin{array}{c}{S}_{ut}=\frac{201\text{kpsi}\cdot \text{in}}{{\left(0.090\text{in}\right)}^{0.145}}\\ =\frac{201\text{kpsi}}{0.7052}\\ =284.99\text{kpsi}\end{array}$

Substitute $284.99\text{kpsi}$ for $S_{ut}$ in Equation (V).

$\begin{array}{c}{S}_{sy}=0.45\times 284.99\text{kpsi}\\ =128.24\text{kpsi}\end{array}$

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 $d=0.090\text{in}$ as $\text{11}\text{.75}\times {\text{10}}^6\text{psi}$.

Substitute $\text{11}\text{.75}\times {\text{10}}^6\text{psi}$ for $G$, $\text{0}\text{.090}\text{in}$ for $d$, $0.890\text{in}$ for $D$ and $10\text{lbf}/\text{in}$ for $k$ in Equation (VI).

$\begin{array}{c}{N}_a=\frac{11.75\times {\text{10}}^6\text{psi}{\left(\text{0}\text{.090}\text{in}\right)}^4}{8\left(10\text{lbf}/\text{in}\right){\left(0.890\text{in}\right)}^3}\\ =\frac{770.91\text{psi}\cdot \text{in}}{56.39\text{lbf}/\text{in}}\left(\frac{1\text{lbf}/{\text{in}}^2}{1\text{psi}}\right)\\ =13.66\text{coils}\end{array}$

Substitute $13.66\text{coils}$ for $N_a$ in Equation (VII).

$\begin{array}{c}{N}_t=13.66\text{coils}+2\\ =15.66\text{coils}\end{array}$

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 $15.66\text{coils}$ for $N_t$ and $0.090\text{in}$ for $d$ in Equation (IX).

$\begin{array}{c}{L}_s=0.090\text{in}\left(15.66\text{coils}\right)\\ =1.4094\text{in}\\ \approx 1.41\text{in}\end{array}$

Substitute $2.3\text{in}$ for $y_s$ and $1.41\text{in}$ for $L_s$ in Equation (X).

$\begin{array}{c}{L}_o=2.3\text{in}+1.41\text{in}\\ =3.71\text{in}\end{array}$

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 $0.890\text{in}$ for $D$ and $0.5$ for $\alpha$ in Equation (XI).

$\begin{array}{c}{\left(L_{{}_o}\right)}_{cr}=2.63\left(\frac{0.890\text{in}}{0.5}\right)\\ =4.681\text{in}\end{array}$

Substitute $23\text{lbf}$ for $F_s$, $0.890\text{in}$ for $D$, $1.135$ for $K_B$ and $0.090\text{in}$ for $d$ in Equation (XII).

$\begin{array}{c}{\tau }_a=1.135\left(\frac{8\left(23\text{lbf}\right)\left(0.890\text{in}\right)}{\pi {\left(0.090\text{in}\right)}^3}\right)\\ =1.135\left(\frac{163.76}{2.290\times {10}^{-3}}\right)\text{lbf}/{\text{in}}^{\text{2}}\left(\frac{1\text{kpsi}}{{\text{10}}^3\text{lbf}/{\text{in}}^{\text{2}}}\right)\\ =81.15\text{kpsi}\\ \approx 81.15\text{kpsi}\end{array}$

Substitute $20\text{lbf}$ for $F_{\max }$, $23\text{lbf}$ for $F_s$ and $81.15\text{kpsi}$ for ${\tau }_a$ in Equation (XIII).

$\begin{array}{c}{\tau }_s=1.15\left(\frac{20\text{lbf}}{23\text{lbf}}\right)81.15\text{kpsi}\\ \text{=1}\text{.15}\left(0.8695\right)81.15\text{kpsi}\\ =81.143\text{kpsi}\end{array}$

Substitute $81.143\text{kpsi}$ for ${\tau }_s$ and $128.24\text{kpsi}$ for $S_{sy}$ in Equation (XIV).

$\begin{array}{c}{n}_s=\frac{128.24\text{kpsi}}{81.143\text{kpsi}}\\ =1.58\end{array}$

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

Substitute $2.6$ for relative cost material, $15.66$ for $N_t$, $\text{0}\text{.090}\text{in}$ for $d$ and $0.890\text{in}$ for $D$ in Equation (XV).

$\begin{array}{c}\text{fom}=\frac{-2.6{\pi }^2{\left(\text{0}\text{.090}\text{in}\right)}^2\left(0.890\text{in}\right)\left(15.66\right)}{4}\\ =-\frac{2.8969}{4}\\ =-0.7242\\ \approx -0.725\end{array}$

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	$d_1$	$d_2$
	$C$	$10$	$10.5$
1	$d$	$0.089$	$0.084$
2	$d$ (from table)	$0.090$	$0.085$
3	$D_i$	$0.800$	$0.800$
4	$D$	$0.890$	$0.885$
5	$C$	$9.88$	$10.41$
6	$N_a$	$13.66$	$11.061$
7	$N_t$	$15.66$	$13.061$
8	$L_s$	$1.41$	$1.110$
9	$L_o$	$3.71$	$3.410$
10	${\left(L_o\right)}_{cr}$	$4.68$	$4.65$
11	$A$	$201.00$	$201.00$
12	$m$	$0.145$	$0.145$
13	$S_{ut}$	$284.99$	$287.36$
14	$S_{sy}$	$128.24$	$129.31$
15	$K_B$	$1.135$	$1.128$
16	${\tau }_s$	$81.14$	$95.22$
17	$n_s$	$1.58$	$1.35$
18	$\text{fom}$	$-0.725$	$-0.536$

Since the factor of safety is greater than the given factor of safety, so the design is suitable for the helical spring $\left(1.358>1.2\right)$.

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 22P

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 $0.090\text{in}$ for $d$ and $0.890\text{in}$ for $D$ in Equation (XVIII).

$\begin{array}{c}{D}_o=0.890\text{in}+0.090\text{in}\\ =0.980\text{in}\end{array}$

Substitute $0.090\text{in}$ for $d$ and $0.890\text{in}$ for $D$ in Equation (II).

$\begin{array}{c}C=\frac{0.890\text{in}}{0.090\text{in}}\\ =9.88\end{array}$

Substitute $9.88$ for $C$ in Equation (III).

$\begin{array}{c}{K}_B=\frac{\left(4\times 9.88\right)+2}{\left(4\times 9.88\right)-3}\\ =\frac{41.52}{36.52}\\ =1.136\\ \approx 1.136\end{array}$

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 $\text{0}\text{.090}\text{in}$.

Substitute $0.090\text{in}$ for $d$, $201\text{kpsi}\cdot \text{in}$ for $A$ and $0.145$ for $m$ in Equation (IV).

$\begin{array}{c}{S}_{ut}=\frac{201\text{kpsi}\cdot \text{in}}{{\left(0.090\text{in}\right)}^{0.145}}\\ =\frac{201\text{kpsi}}{0.7052}\\ =284.99\text{kpsi}\end{array}$

Substitute $284.99\text{kpsi}$ for $S_{ut}$ in Equation (V).

$\begin{array}{c}{S}_{sy}=0.45\times 284.99\text{kpsi}\\ =128.24\text{kpsi}\end{array}$

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 $d=0.090\text{in}$ as $\text{11}\text{.75}\times {\text{10}}^6\text{psi}$.

Substitute $\text{11}\text{.75}\times {\text{10}}^6\text{psi}$ for $G$, $\text{0}\text{.090}\text{in}$ for $d$, $0.890\text{in}$ for $D$ and $10\text{lbf}/\text{in}$ for $k$ in Equation (VI).

$\begin{array}{c}{N}_a=\frac{11.75\times {\text{10}}^6\text{psi}{\left(\text{0}\text{.090}\text{in}\right)}^4}{8\left(10\text{lbf}/\text{in}\right){\left(0.890\text{in}\right)}^3}\\ =\frac{770.91\text{psi}\cdot \text{in}}{56.39\text{lbf}/\text{in}}\left(\frac{1\text{lbf}/{\text{in}}^2}{1\text{psi}}\right)\\ =13.66\text{coils}\end{array}$

Substitute $13.66\text{coils}$ for $N_a$ in Equation (VII).

$\begin{array}{c}{N}_t=13.66\text{coils}+2\\ =15.66\text{coils}\end{array}$

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 $15.66\text{coils}$ for $N_t$ and $0.090\text{in}$ for $d$ in Equation (IX).

$\begin{array}{c}{L}_s=0.090\text{in}\left(15.66\text{coils}\right)\\ =1.4094\text{in}\\ \approx 1.41\text{in}\end{array}$

Substitute $2.3\text{in}$ for $y_s$ and $1.41\text{in}$ for $L_s$ in Equation (X).

$\begin{array}{c}{L}_o=2.3\text{in}+1.41\text{in}\\ =3.71\text{in}\end{array}$

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 $0.890\text{in}$ for $D$ and $0.5$ for $\alpha$ in Equation (XI).

$\begin{array}{c}{\left(L_{{}_o}\right)}_{cr}=2.63\left(\frac{0.890\text{in}}{0.5}\right)\\ =4.681\text{in}\end{array}$

Substitute $23\text{lbf}$ for $F_s$, $0.890\text{in}$ for $D$, $1.135$ for $K_B$ and $0.090\text{in}$ for $d$ in Equation (XII).

$\begin{array}{c}{\tau }_a=1.135\left(\frac{8\left(23\text{lbf}\right)\left(0.890\text{in}\right)}{\pi {\left(0.090\text{in}\right)}^3}\right)\\ =1.135\left(\frac{163.76}{2.290\times {10}^{-3}}\right)\text{lbf}/{\text{in}}^{\text{2}}\left(\frac{1\text{kpsi}}{{\text{10}}^3\text{lbf}/{\text{in}}^{\text{2}}}\right)\\ =81.15\text{kpsi}\\ \approx 81.15\text{kpsi}\end{array}$

Substitute $20\text{lbf}$ for $F_{\max }$, $23\text{lbf}$ for $F_s$ and $81.15\text{kpsi}$ for ${\tau }_a$ in Equation (XIII).

$\begin{array}{c}{\tau }_s=1.15\left(\frac{20\text{lbf}}{23\text{lbf}}\right)81.15\text{kpsi}\\ \text{=1}\text{.15}\left(0.8695\right)81.15\text{kpsi}\\ =81.143\text{kpsi}\end{array}$

Substitute $81.143\text{kpsi}$ for ${\tau }_s$ and $128.24\text{kpsi}$ for $S_{sy}$ in Equation (XIV).

$\begin{array}{c}{n}_s=\frac{128.24\text{kpsi}}{81.143\text{kpsi}}\\ =1.58\end{array}$

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

Substitute $2.6$ for relative cost material, $15.66$ for $N_t$, $\text{0}\text{.090}\text{in}$ for $d$ and $0.890\text{in}$ for $D$ in Equation (XV).

$\begin{array}{c}\text{fom}=\frac{-2.6{\pi }^2{\left(\text{0}\text{.090}\text{in}\right)}^2\left(0.890\text{in}\right)\left(15.66\right)}{4}\\ =-\frac{2.8969}{4}\\ =-0.7242\\ \approx -0.725\end{array}$

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	$d_1$	$d_2$
	$C$	$10$	$10$
1	$d$	$0.089$	$0.086$
2	$d$ (from table)	$0.090$	$0.085$
3	$D_o$	$0.980$	$0.980$
4	$D$	$0.890$	$0.865$
5	$C$	$9.88$	$10.17$
6	$N_a$	$13.66$	$11.84$
7	$N_t$	$15.66$	$13.84$
8	$L_s$	$1.41$	$1.177$
9	$L_o$	$3.71$	$3.477$
10	${\left(L_o\right)}_{cr}$	$4.68$	$4.55$
11	$A$	$201.00$	$201.00$
12	$m$	$0.145$	$0.145$
13	$S_{ut}$	$284.99$	$287.36$
14	$S_{sy}$	$128.24$	$129.31$
15	$K_B$	$1.135$	$1.135$
16	${\tau }_s$	$81.14$	$93.64$
17	$n_s$	$1.58$	$1.38$
18	$fom$	$-0.725$	$-0.555$

Since the factor of safety is greater than the given factor of safety, so the 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 $13.84$.