A brass alloy is known to have a yield strength of 275 MPa (40,000 psi), a tensile strength of 380 MPa (55,000 psi), and an elas

Question

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A brass alloy is known to have a yield strength of 275 MPa (40,000 psi), a tensile strength of 380 MPa (55,000 psi), and an elas

tic modulus of 103 GPa (15.0×106 psi). A cylindrical specimen of this alloy 5.5 mm (0.22 in.) in diameter and 267 mm (10.52 in.) long is stressed in tension and found to elongate 7.0 mm (0.28 in.). 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 not, explain why.

Engineering

2 answers:

Ugo [173] · Answer 1 · 2022-03-29T08:31:05+03:00

Answer:

Computation of the load is not possible because E(test) >E(yield)

Explanation:

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 7.0 mm (0.28 in.). It is first necessary/ important 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) is less than E(yield), deformation is elastic and the load may be computed. However is E(test) is greater than E(yield) computation/determination of the load is not possible even though defamation is plastic and we have neither a stress-strain plot or a mathematical relating plastic stress and strain. Therefore, we can compute these two values as:

Calculation of E(test is as follows)

E(test) = change in l/lo= Elongation produced/stressed tension= 7.0mm/267mm

=0.0262

Computation of E(yield) is given below:

E(yield) = σy/E=275Mpa/103 ×10^6Mpa= 0.0027

Therefore, we won't be able to compute the load because for computation to take place, E(test) <E(yield). In this case, E(test) is greater than E(yield).