Holooly Plus Logo

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


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)
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.
The 'Blue Check Mark' means that this solution was answered by an expert.
Learn more on how do we answer questions.

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)


(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.

Related Answered Questions