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Question 14.6: Load-Carrying Capacity of a Helical Extension Spring Hook A ......

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.
F14.12
Step-by-Step
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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.

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