The maximum load that may be applied to a specimen with a cross-sectional area of 130 mm² is; 35535 N
<h3>How to find Elastic Modulus?</h3>
We are told that for an alloy, the yield stress is 345-MPa and the elastic modulus (E) is 103-GPa.
Now, we want to find the maximum load that may be applied to a specimen with a cross-sectional area of 130-mm² without plastic deformation. Thus;
We are given the parameters;
Yield Stress; σ = 345 Mpa = 345 * 10⁶ Pa
Elastic Modulus; E = 103 GPa = 103 * 10⁹ Pa
Cross sectional Area; A = 130 mm² = 103 * 10⁻⁶ m²
Formula for stress without Plastic deformation is;
σ = F_max/Area
where;
σ is stress
F_max is maximum force
Area is Area
Thus making maximum force the subject of the formula gives;
F_max = σ * A
Plugging in the relevant values for stress and area gives us;
F_max = 345 * 10⁶ * 103 * 10⁻⁶
F_max = 35535 N
The maximum load that may be applied to a specimen with a cross-sectional area of 130 mm² is gotten to be 35535 N
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