A brass alloy is known to have a yield strength of 275 MPa, a tensile strength of 380 MPa, and an elastic modulus of 103 GPa. A cylindrical specimen of this alloy 12.7 mm in diameter and 250 mm long is stressed in tension and found to elongate 7.6 mm. On the basis of the information given, is it possible to compute the magnitude of the load that is necessary to produce this change in length? If so, calculate the load. If not, explain why.

Computation of the load is not possible since e (test) > e (yield).

Explanation:

Given

E = 103 GPa = 103*10⁹ Pa

σy = 275 MPa = 275*10⁶ Pa

L₀ = 250 mm = 0.25 m

ΔL = 7.6 mm = 0.0076 m

D = 12.7 mm = 0.0127 m

We are asked to ascertain whether or not it is possible to compute, for brass, the magnitude of the load necessary to produce an elongation of 7.6 mm (0.30 in).

It is first necessary to compute the strain at yielding from the yield strength and the elastic modulus, and then the strain experienced by the test specimen. Then, if

e (test) < e (yield)

deformation is elastic, and the load may be computed using Equation

P = ΔL*A*E/L₀

However, if

e (test) > e (yield)

computation of the load is not possible inasmuch as deformation is plastic and we have neither a stress-strain plot nor a mathematical expression relating plastic stress and strain. We compute these two strain values as

ε (test) = ΔL / L₀ = 7.6 mm / 250 mm = 0.0304

and

ε (yield) = σy / E = (275*10⁶ Pa) / (103*10⁹ Pa) = 0.00267

Therefore, computation of the load is not possible since e (test) > e (yield).

0.0304 > 0.00267