Name: AuiShon R Student ID: Question #3 (30 pts) A helical compression spring is

Question

Name: AuiShon R Student ID: Question #3 (30 pts) A helical compression spring is made of hard-drawn spring steel wire 0.080-in in diameter and mean coil diameter of 0.8 in. The ends are plain and ground, and there are 8 total coils. Show all your work. (Other parameters: N,-7, Sut-226.23 kpsi, S,y-101.80 kpsi, G=11.5 Mps) 1.What force is needed to compress the spring to its solid length (10 pts)? (Hint: Replace with Ssy in the stress eq following page to find the force.) uation in the 2.Estimate the spring constant (10 pts). 3.Estimate the displacement of the spring when the force calculated in 1 is applied to the spring (10 pts).

Explanation / Answer

Given:

Material = HD steel wire;  d = 0.080 in.; D = 0.8 in.; Nt = 8; Na = 7; Sut = 226.23 kpsi;

Ssy = 101.80 kpsi = 101.80 * 103 psi; G = 11.5 Mpsi = 11.5*106 psi;

Solution:

1. Force needed (Fs) = (d3Ssy/ns) / 8KBD

where ns = 1.2 for solid-safe.

KB = (4C + 2) / (4C -3)

C = D/d = 0.8/0.080 = 10

KB = ((4*10) + 2) / ((4*10) -3) = 1.1351

Fs = (*0.0803*101.80*103 /1.2) / (8*1.1351*0.8) = 136.4540/7.2646 = 18.7834

Fs = 18.7834 lbf

2. Spring constant (k) = d4G / 8D3Na = 0.0804*11.5*106 / (8*0.83*7) = 471.04/28.672 = 16.4286

k = 16.4286 lbf/in

3. Displacement (ys) = Fs/k = 18.7834 / 16.4286 = 1.1433

ys = 1.1433 in

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