1. A steel rod 30mm & 800mm long has an allowance elongation not to exceed 1.5 mm, find the allowable load in KN?
Solution:
Properties of steel;
E = 30,000,000 psi or 20,758,714 kPa
Formula;
Y= FL/AE
where:
Y=elongation
F=load
A= cross-sectional area
thus,
F = 274 kN answer
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