Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 7: Problem 28

Two previously undeformed cylindrical specimens of an alloy are to be strain hardened by reducing their cross-sectional areas (while maintaining their circular cross sections). For one specimen, the initial and deformed radii are \(16 \mathrm{~mm}\) and \(11 \mathrm{~mm}\), respectively. The second specimen, with an initial radius of \(12 \mathrm{~mm}\), must have the same deformed hardness as the first specimen; compute the second specimen's radius after deformation.

Short Answer

Expert verified
Answer: The deformed radius of the second specimen is approximately \(\sqrt{\frac{\pi (12)^2 - \left(\frac{\pi (16)^2 - \pi (11)^2}{\pi (16)^2} \times \pi (12)^2\right)}{\pi}}\) mm.

Step by step solution

01

Calculate the cross-sectional area of the first specimen (before and after deformation)

We need to calculate the cross-sectional area of the cylindrical specimen, which is given by the formula \(A = \pi r^2\). So, let's calculate the areas before (A_initial) and after (A_final) deformation. Initial radius (r_initial): 16 mm Deformed radius (r_final): 11 mm \(A_{initial} = \pi ({r_{initial}})^2 = \pi (16)^2 \) \(A_{final} = \pi ({r_{final}})^2 = \pi (11)^2 \)
02

Calculate the reduction in the cross-sectional area of the first specimen

Now that we have the initial and final cross-sectional areas of the specimen, let's calculate the reduction percentage as follows: \(Reduction_{percentage} = \frac{A_{initial} - A_{final}}{A_{initial}} \times 100\) Plug in the values we calculated in Step 1: \(Reduction_{percentage} = \frac{\pi (16)^2 - \pi (11)^2}{\pi (16)^2} \times 100\)
03

Calculate the initial cross-sectional area of the second specimen

Given that the initial radius of the second specimen is 12 mm, we can calculate its initial cross-sectional area using the same formula as in Step 1: Initial radius (r_initial_2): 12 mm \(A_{initial\_2} = \pi ({r_{initial\_2}})^2 = \pi (12)^2 \)
04

Calculate the final cross-sectional area of the second specimen

Now we know the initial cross-sectional area and the reduction percentage, we can calculate the final cross-sectional area of the second specimen: \(A_{final\_2} = A_{initial\_2} - (Reduction_{percentage} \times A_{initial\_2})\) Plug in the values: \(A_{final\_2} = \pi (12)^2 - \left(\frac{\pi (16)^2 - \pi (11)^2}{\pi (16)^2} \times \pi (12)^2\right)\)
05

Calculate the deformed radius of the second specimen

Using the final cross-sectional area, we can find the deformed radius by rearranging the area formula, and solving for the radius: \(r_{final\_2} = \sqrt{\frac{A_{final\_2}}{\pi}}\) Plug in the value: \(r_{final\_2} = \sqrt{\frac{\pi (12)^2 - \left(\frac{\pi (16)^2 - \pi (11)^2}{\pi (16)^2} \times \pi (12)^2\right)}{\pi}}\) After calculating, the deformed radius of the second specimen is obtained.

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

Consider a metal single crystal oriented such that the normal to the slip plane and the slip direction are at angles of \(43.1^{\circ}\) and \(47.9^{\circ}\), respectively, with the tensile axis. If the critical resolved shear stress is \(20.7\) MPa (3000 psi), will an applied stress of 45 MPa (6500 psi) cause the single crystal to yield? If not, what stress will be necessary?

(a) Define a slip system. (b) Do all metals have the same slip system? Why or why not?

The critical resolved shear stress for iron is 27 MPa (4000 psi). Determine the maximum possible yield strength for a single crystal of Fe pulled in tension.

(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.

(a) What is the driving force for recrystallization? (b) For grain growth?
See all solutions

Recommended explanations on Physics Textbooks

Electricity and Magnetism

Read Explanation

Solid State Physics

Read Explanation

Work Energy and Power

Read Explanation

Translational Dynamics

Read Explanation

Physics of Motion

Read Explanation

Rotational Dynamics

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