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Chapter 14: Problem 40

Design a straight-ended helical torsion spring for a static load of 430 in-lb at a deflection of \(55^{\circ}\) with a safety factor of 2 . Specify all parameters necessary to manufacture the spring. State all assumptions.

Short Answer

Expert verified
The spring can be made of ASTM A228 steel, with wire diameter 0.179 in, mean coil diameter 2.321 in, allowing for a maximum shear stress of 110,000 psi, and having a spring constant of 448 in-lb/rad. Its coil pitch should be about 0.358 in.

Step by step solution

01

- Assume a spring material

First, assume a spring material. For this exercise, the commonly used spring steel, such as ASTM A228, is used as the spring material, with a shear strength \( S_{s} = 220,000 psi \). This assumption of the material is important as it influences the other aspects of the spring design.
02

- Calculate Maximum Shear Stress

The maximum allowable shear stress, \( \tau_{a} \), is determined by taking the shear strength of the material and dividing it by the safety factor of 2: \( \tau_{a} = S_{s} / N = 220,000 psi / 2 = 110,000 psi \).
03

- Calculate Wire Diameter

One could use the formula for the shear stress in the statically loaded springs: \( \tau = (16*P*D)/(pi*d^3) \). Here, P=430 in-lb is the load, D is the spring diameter which can be reasonably assumed - let's say D = 2.5 in. Placing these values, and solving for wire diameter d could obtain: \( d = ((16*P*D)/(pi*\tau))^{1/3} = 0.179 in \).
04

- Calculate Mean Coil Diameter

The mean coil diameter Dm = D - d = 2.5 in - 0.179 in = 2.321 in.
05

- Calculate Coil Pitch

The angle of twist can be calculated using the definition of radians (\(55^{\circ}\) = 0.96 rad). Then we can calculate spring constant \( k = P / \theta = 430 in-lb / 0.96 rad = 448 in-lb/rad \). Finally, to manufacture the spring we also need a coil pitch. It should be somewhat larger than wire's diameter, but its exact value mostly depends on desired spring flexibility. Let's assume one coil length to be twice the diameter, that is approx. 0.358 in.

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Most popular questions from this chapter

Design a straight-ended helical torsion spring for a dynamic load of \(150-350\) in-lb over a deflection of \(50^{\circ}\) with a safety factor of \(1.4\). Specify all parameters necessary to manufacture the spring. State all assumptions.

Given the following data for a helical torsion spring, loaded in fatigue, find the spring index, unloaded coil diameter, minimum loaded coil diameter, and safety factor in fatigue. State all assumptions and sources of empirical data used. Deflection at assembly \(=0.15 \mathrm{rev}\), working deflection \(=0.35 \mathrm{rev}, k=10 \mathrm{~N}-\mathrm{m} / \mathrm{rev}, N_{a}=25\), \(4.50 \mathrm{~mm}\) oil-tempered wire, unpeened.

A spring made from ASTM A228 wire with ends squared and ground, wire diameter \(d=3 \mathrm{~mm}\), outside diameter \(D_{o}=27 \mathrm{~mm}, 14\) total coils, and free length \(L_{f}=80 \mathrm{~mm}\) has been chosen for an application. Determine the static safety factor if the spring is subjected to a static load of \(175 \mathrm{~N}\).

Design a helical compression spring for a static load of \(60 \mathrm{lb}\) at a deflection of \(1.50\) in with a safety factor of \(2.0\) to work in a \(1.06\)-in- dia hole. Specify all parameters necessary to manufacture the spring.

A helical compression coil spring is needed to provide a time-varying force that ranges from a minimum of \(100 \mathrm{lb}\) to a maximum of \(300 \mathrm{lb}\) over a deflection of 1 in. It needs to work free over a shaft of \(1.25\)-in dia. Use a cold-drawn carbon steel wire having an \(S_{u t}=250000\) psi. A spring index of 6 , a clash allowance of \(15 \%\), and squared and ground ends are desired.
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