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dusya [7]
4 years ago
11

A helical compression spring is wound using 3 mm diameter music wire. The spring has an outside diameter of 35 mm with plain gro

und ends and 16 coils If the spring does not yield when the spring is fully compressed, what is curvature correction factors Kc and the maximum shear that will be in the spring?

Engineering
2 answers:
Lerok [7]4 years ago
8 0

Answer:

Kc = 1.0844

Maximum shear = 848.43 Mpa

Explanation:

Find attached of the calculations

Elina [12.6K]4 years ago
5 0

Answer:

Kc = 1.0844

Smax = 848.4352 MPa

Explanation:

Given data:

d = diameter = 3 mm

D = outside diameter = 35 mm

N = coils = 16

Spring fully compressed

Question: What is the curvature correction factors, Kc = ? and the maximum shear, Smax = ?

First, Dm needs to be calculated:

Dm = D - d = 35 - 3 = 32 mm

The spring index:

C=\frac{D_{m} }{d} =\frac{32}{3} =10.6667

K_{c} =\frac{(\frac{4C-1}{4C-4})+\frac{0.615}{C}  }{1+\frac{0.5}{C} } =\frac{\frac{4*10.6667-1}{4*10.6667-4} +\frac{0.615}{10.6667} }{1+\frac{0.5}{10.6667} } =1.0844

Properties at a diameter of 3 mm:

Area = A = 2211 MPa mm^m

m = 0.145

The shear stress:

S=\frac{A}{d^{m} } =\frac{2211}{3^{0.145} } =1885.4115MPa

According the distortion theory, the maximum shear:

S_{max} =0.45S=0.45*1885.4115=848.4352MPa

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