**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) |
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A |
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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. |

Step-by-Step

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)

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 |
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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. |

Question: 14.8

Equation (14.43) may be rearranged into the form
[...

Question: 14.7

By Equation (14.12) and Table 14.2,
S_{u s...

Question: 14.5

The mean and alternating loads are
P_m=\fr...

Question: 14.4

Refer to the numerical values given in Example 14....

Question: 14.3

The direct shear factor, from Equation (14.7), is ...

Question: 14.2

The spring index is C=D / d=0.5 / 0.0625=8...

Question: 14.1

The mean diameter of the spring is D [/lat...

Question: 14.9

From Table 7.3, C_r =1. The modifi...