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Chapter 7: Problem 31

(a) What is the approximate ductility (\% EL) of a brass that has a yield strength of \(345 \mathrm{MPa}\) \((50,000 \mathrm{psi}) ?\) (b) What is the approximate Brinell hardness of a 1040 steel having a yield strength of \(620 \mathrm{MPa}(90,000 \mathrm{psi}) ?\)

Short Answer

Expert verified
Question: Calculate the approximate ductility of brass with a yield strength of 275 MPa and the approximate Brinell hardness of a 1040 steel material with a yield strength of 690 MPa. Answer: The approximate ductility of brass is 106.25% and the approximate Brinell hardness of the 1040 steel is 345.

Step by step solution

01

Recall the formula for ductility

To calculate the percentage elongation (%EL), we can use the formula: \%EL \(\approx\) \%RA, where \%RA is the percentage reduction in area.
02

Determine the relationship between yield strength and \%RA for brass

For brass, the relationship between yield strength (YS) in MPa and the percentage reduction in area (\%RA) can be found using the equation: \%RA \(\approx\) 1000 - 3.25(YS)
03

Substitute the given yield strength into the equation

Given the yield strength of 275 MPa, substitute the value into the equation to find the \%RA: \%RA \(\approx\) 1000 - 3.25(275)
04

Calculate \%RA and \%EL

Solve for \%RA and \%EL: \%RA \(\approx\) 1000 - 3.25(275) \(\approx\) 1000 - 893.75 \(\approx\) 106.25 Since \%EL \(\approx\) \%RA, the approximate ductility of brass is 106.25\%. Part (b): Calculate approximate Brinell hardness of 1040 steel with given yield strength
05

Recall the formula for Brinell hardness

The relationship between Brinell hardness (HB) and yield strength (YS) can be described by the formula: HB \(\approx\) 0.5(YS)
06

Substitute the given yield strength into the equation

Given the yield strength of 690 MPa, we can substitute the value into the equation to find the Brinell hardness: HB \(\approx\) 0.5(690)
07

Calculate Brinell hardness

Solve for HB: HB \(\approx\) 0.5(690) \(\approx\) 345 Thus, the approximate Brinell hardness of the 1040 steel is 345.

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Most popular questions from this chapter

List four major differences between deformation by twinning and deformation by slip relative to mechanism, conditions of occurrence, and final result.

An undeformed specimen of some alloy has an average grain diameter of \(0.050 \mathrm{mm} .\) You are asked to reduce its average grain diameter to \(0.020 \mathrm{mm}\). Is this possible? If so, explain the procedures you would use and name the processes involved. If it is not possible, explain why.

The lower yield point for an iron that has an average grain diameter of \(1 \times 10^{-2} \mathrm{mm}\) is \(230 \mathrm{MPa}(33,000 \mathrm{psi}) .\) At a grain diameter of \(6 \times 10^{-3} \mathrm{mm},\) the yield point increases to 275 MPa \((40,000\) psi ) . At what grain diameter will the lower yield point be \(310 \mathrm{MPa}\) \((45,000 \mathrm{psi}) ?\)

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Consider a single crystal of nickel oriented such that a tensile stress is applied along a [001] direction. If slip occurs on a (111) plane and in a \([\overline{1} 01]\) direction, and is initiated at an applied tensile stress of \(13.9 \mathrm{MPa}(2020 \mathrm{psi})\) compute the critical resolved shear stress.
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