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 elastic 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.

Question

Engineering

Cameron Collins

28 July, 07:04

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 elastic 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.

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Answers (2)

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Ellie Kent · Answer 1

It's not possible to compute the magnitude of the load

Explanation:

Given dа ta:

Yield strength = 275Mpa

Tensile strength = 380Mpa

Elastic modulus = 103Gpa = 103 * 1000MPa

Diameter = 5.5 mm

Length = 267 mm

Elongation = 7.0 mm

Calculating the strain test value using the formula'

ε (test) = ΔL/Lo

Substituting, we have

ε = 7.0/267

= 0.0262

Calculating the yield strain using the formula

ε (yield) = бy/E

Substituting, we have

ε (yield) = 275 / 103 * 1000

= 275/103000

= 0.00267

From the calculation above, the strain test is greater than the yield. Therefore, it's not possible to compute the magnitude of the load.

ε (test) ⊃ ε (yield)

Rishi Stuart · Answer 2

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)