Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 7: Problem 31

(a) What is the approximate ductility (\%EL) of a brass that has a yield strength of \(275 \mathrm{MPa}\) \((40,000 \mathrm{psi}) ?\) (b) What is the approximate Brinell hardness of a 1040 steel having a yield strength of 690 \(\mathrm{MPa}(100,000 \mathrm{psi})\) ?

Short Answer

Expert verified
Question: Calculate the ductility and Brinell hardness for the following materials: (a) brass with a yield strength of 275 MPa and (b) 1040 steel with a yield strength of 690 MPa. Answer: (a) The approximate ductility of brass with a yield strength of 275 MPa is 5% EL. (b) The approximate Brinell hardness of 1040 steel with a yield strength of 690 MPa is 310.5.

Step by step solution

01

a) Ductility Calculation (Brass)

To find the approximate ductility of brass with a yield strength of \(275\,\text{MPa}\), we'll use a percentage elongation formula: $$ \% EL = 60 - 0.2 \times \sigma_y $$ where \(\% EL\) is the percentage elongation and \(\sigma_y\) is the yield strength in MPa. Plugging the given yield strength value in the formula for ductility: $$ \% EL = 60 - 0.2 \times (275\,\text{MPa}) $$
02

a) Ductility Calculation Result

Now we can calculate the ductility: $$ \% EL = 60 - 0.2 \times (275\,\text{MPa}) = 60 - 55 = 5 $$ The approximate ductility of the brass with a yield strength of \(275\,\text{MPa}\) is 5\% EL.
03

b) Brinell Hardness Calculation (1040 Steel)

To find the approximate Brinell hardness of a 1040 steel with a yield strength of \(690\,\text{MPa}\), we'll use the following formula: $$ HB = 0.45 \times \sigma_y $$ where \(HB\) is the Brinell hardness and \(\sigma_y\) is the yield strength in MPa. Plugging the given yield strength value in the formula for Brinell hardness: $$ HB = 0.45 \times (690\,\text{MPa}) $$
04

b) Brinell Hardness Calculation Result

Now we can calculate the Brinell hardness: $$ HB = 0.45 \times (690\,\text{MPa}) = 310.5 $$ The approximate Brinell hardness of the 1040 steel with a yield strength of \(690\,\text{MPa}\) is 310.5.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

(a) Show, for a tensile test, that $$ \% \mathrm{CW}=\left(\frac{\epsilon}{\epsilon+1}\right) \times 100 $$ if there is no change in specimen volume during the deformation process (i.e., \(A_{0} l_{0}=A_{d} l_{d}\) ). (b) Using the result of part (a), compute the percent cold work experienced by naval brass (the stress-strain behavior of which is shown in Figure 6.12) when a stress of 400 MPa \((58,000\) psi) is applied.

Is it possible for two screw dislocations of opposite sign to annihilate each other? Explain your answer.

An uncold-worked brass specimen of average grain size \(0.008 \mathrm{~mm}\) has a yield strength of 160 MPa \((23,500\) psi). Estimate the yield strength of this alloy after it has been heated to \(600^{\circ} \mathrm{C}\) for \(1000 \mathrm{~s}\), if it is known that the value of \(k_{y}\) is \(12.0 \mathrm{MPa} \cdot \mathrm{mm}^{1 / 2}\left(1740 \mathrm{psi} \cdot \mathrm{mm}^{1 / 2}\right)\)

(a) Compare planar densities (Section \(3.11\) and Problem 3.54) for the (100), (110), and (111) planes for FCC. (b) Compare planar densities (Problem 3.55) for the (100), (110), and (111) planes for BCC.

Two previously undeformed specimens of the same metal are to be plastically deformed by reducing their cross-sectional areas. One has a circular cross section, and the other is rectangular; during deformation the circular cross section is to remain circular, and the rectangular is to remain as such. Their original and deformed dimensions are as follows: $$ \begin{array}{lcr} \hline & \begin{array}{c} \text { Circular } \\ \text { (diameter, } \mathbf{m m}) \end{array} & \begin{array}{c} \text { Rectangular } \\ (\mathbf{m m}) \end{array} \\ \hline \text { Original dimensions } & 15.2 & 125 \times 175 \\ \text { Deformed dimensions } & 11.4 & 75 \times 200 \\ \hline \end{array} $$ Which of these specimens will be the hardest after plastic deformation, and why?
See all solutions

Recommended explanations on Physics Textbooks

Space Physics

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Kinematics Physics

Read Explanation

Linear Momentum

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Famous Physicists

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free