## Q. 14.6

Load-Carrying Capacity of a Helical Extension Spring Hook

A helical extension spring with hook ends is made of a music wire of mean coil radius $D$, wire diameter $d$, mean hook radius $r_m$, and inner hook radius $r_i$ (Figure 14.12). The preload is $P_i$ and the free end is [$h_f$

Find:

a. The material properties and initial torsional stress in the wire using Equation (a).

$\tau_i=0.7 \frac{S_u}{C}$    (a)

b. Maximum load when yielding in tension impends at section $A$.

c. Distance between the hook ends.

Given: $d=2.5 mm , D=12.5 mm ,\left(r_m\right)_A=6.25 mm ,\left(r_m\right)_B=3.75 mm$

$N_a=150, \quad P=50 N , \quad h_f=290 mm$

$A=2060 \quad \text { and } \quad b=-0.163$      (from Table 14.2)

Assumption: Modulus of rigidity will be $G$=79 GPa.

 TABLE 14.2 Coefficients and Exponents for Equation (14.12) $A$ Material ASTM No. $b$ MPa ksi Hard-drawn wire A227 −0.201 1510 237 Music wire A228 −0.163 2060 186 Oil-tempered wire A229 −0 193 1610 146 Chrome-vanadium wire A232 −0.155 1790 173 Chrome-silicon wire A401 −0 091 1960 218 Source: Associated Spring-Barnes Group, Design Handbook, Associated Spring-Barnes Group, Bristol, CN, 1987. ## Verified Solution

a. Ultimate tensile strength, estimated from Equation (14.12),

$S_{u s}=A d^b$      (14.12)

$S_u A d^b=2060(2.5)^{-0.163}=1774 MPa$

By Equation (7.5b) and Table 14.3, we obtain $S_y= S _{y s} / 0.577=(0.40 / 0.577) S_y=0.693 S_u$. The yield strength is thus

$S_{y s}=0.577 S_y$     (7.5b)

$S_y=0.693\left(1774 \times 10^6\right)=1229.4\left(10^6\right)$

The spring index equals $C=D/d$ =12.5/2.5=5. Equation (a) results in then

$\tau_i=0.7 \frac{S_u}{C}$    (a)

$\tau_i=0.7 \frac{S_u}{C}=0.7 \frac{1774}{5}=248.4 MPa$

b. Combined normal stress at section A in the hook is obtained by superimposing bending and axial stresses. The former is defined by Equation (14.34a), and the latter equals $P /(\pi d 2 / 4)$. At the onset of yield, we therefore have

$\sigma_A=K \frac{16 P D}{\pi d^3}$      (14.34a)

$\sigma_A=K \frac{16 P D}{\pi d^3}+\frac{4 P}{\pi d^2}=S_y$      (14.35)

where $K=r_m / r_i \text { with } r_m=6.25 mm \text { and } r_i=6.25-2.5 / 2=5 mm$. Introducing the given data, Equation (14.34a) leads to

$\sigma_A=\left(\frac{6.25}{5}\right)\left[\frac{16 P(12.5)}{\pi(2.5)^3\left(10^{-6}\right)}\right]+\frac{4 P}{\pi(2.5)^2\left(10^{-6}\right)}=1229.4\left(10^6\right)$

or

$(5.09296 P)+(0.2037 P) 10^6=1229.4\left(10^6\right)$

Solving the maximum load when yielding begins in the hook gives $P$=232.1 N

c. Inserting the given data into Equation (14.11), we obtain the spring rate as

$k=\frac{P}{\delta}=\frac{G d^4}{8 D^3 N_s}=\frac{d G}{8 C^3 N_a}$    (14.11)

$k=\frac{d G}{8 N_a C^3}=\frac{\left(2.5 \times 10^{-3}\right)\left(79 \times 10^9\right)}{8(150)(5)^3}=1317 N / m$

The deflection from Equation (14.32) is then

$k=\frac{P-P_i}{\delta}=\frac{d G}{8 N_a C^3}$        (14.32)

$\delta=\frac{P-P_i}{k}=\frac{232.1-50}{1317}=01383 m =1383 mm$

The distance between hook ends equals

$h_f+\delta=290+138.3=428.3 mm$

Comments: Force $P$ required to cause the torsional stress at section $B$ in the hook may also readily be determined, using Equation (14.34b) and Table 14.3. In so doing, a smaller load is obtained (see Problem 14.27), which shows that failure by yielding first takes place by shear stress in the hook.

$\tau_B=K \frac{8 P D}{\pi d^3}$    (14.34b)

 TABLE 14.3 Approximate Strength Ratios of Some Common Spring Materials Material $S_{y s} / S_u$ $S_{e s}^{\prime} / S_u$ Hard-drawn wire 0.42 0.21 Music wire 0.40 0.23 Oil-tempered wire 0.45 0.22 Chrome-vanadium wire 0.52 0.20 Chrome-silicon wire 0.52 0.20 Source: Associated Spring-Barnes Group, Design Handbook, Associated Spring-Barnes Group, Bristol, CN, 1987. Notes: $S_{y s}$, yield strength in shear; $S_u,$ ultimate strength in tension; $S_{e s}^{\prime}$, endurance limit (or strength) in shear.