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Chapter 7: Problem 30

A cylindrical specimen of stainless steel having a diameter of \(12.8 \mathrm{~mm}(0.505 \mathrm{in}\).) and a gauge length of \(50.800 \mathrm{~mm}(2.000 \mathrm{in}\) ) is pulled in tension. Use he load-elongation characteristics shown in the ollowing table to complete parts (a) through (f). (a) Plot the data as engineering stress versus engineering strain. (b) Compute the modulus of elasticity. (c) Determine the yield strength at a strain offset of \(0.002\). (d) Determine the tensile strength of this alloy. (e) What is the approximate ductility, in percent elongation? (f) Compute the modulus of resilience.

Short Answer

Expert verified
Question: Given the dimensions of a cylindrical stainless steel specimen and a table of load-elongation characteristics, calculate the following (Show your workings): (a) Modulus of elasticity (b) Yield strength at a strain offset of 0.002 (c) Percent elongation as an approximation of ductility (d) Modulus of resilience

Step by step solution

01

Calculate cross-sectional area of the specimen

First, find the cross-sectional area (A) of the cylindrical specimen: \(A = \pi (\frac{d}{2})^2\), where \(d\) is the diameter of the specimen. For the given diameter of d = 12.8 mm, the cross-sectional area A is: \(A = \pi (\frac{12.8}{2})^2\) Now, calculate the cross-sectional area using the diameter given.
02

Calculate engineering stress and engineering strain

Using the table of load-elongation data, we can calculate the engineering stress (\(\sigma\)) and engineering strain (\(\epsilon\)) as follows: Engineering stress (\(\sigma\)) is defined as the applied load (P) divided by the original cross-sectional area (A): \(\sigma = \frac{P}{A}\) Engineering strain (\(\epsilon\)) is the change in length (ΔL) divided by the original length (L): \(\epsilon = \frac{\Delta L}{L}\) Replace the load, elongation, and gauge length values from the table and the cross-sectional area calculated in Step 1 to compute the engineering stress and engineering strain for each data point.
03

Determine the modulus of elasticity

The modulus of elasticity (E) is the slope of the initial linear portion of the engineering stress vs. engineering strain curve: \(E = \frac{\sigma}{\epsilon}\) From the calculated engineering stress and strain data, find two points within the linear elastic region, and use them to calculate the slope (modulus of elasticity).
04

Determine the yield strength at a strain offset of 0.002

To determine the yield strength at a strain offset of 0.002, first draw a line parallel to the initial linear portion of the stress-strain curve that intersects the curve at a strain of 0.002. The stress value at the intersection point is the yield strength. Find the intersection point on the graph and determine the corresponding stress value.
05

Calculate ductility as percent elongation

Ductility can be approximated as the percent elongation of the specimen, which is the ratio of the change in length (ΔL) to the original length (L) multiplied by 100: \(\text{Percent elongation} = \frac{\Delta L}{L} \times 100\) Look at the final elongation value in the table, divide it by the original length (50.8 mm) and multiply the result by 100 to find the approximate percent elongation.
06

Compute the modulus of resilience

The modulus of resilience (U) is the area under the stress-strain curve up to the yield point, which represents the amount of energy per unit volume that a material can absorb before yielding: \(U = \frac{1}{2} \sigma_y \epsilon_y\), where \(\sigma_y\) and \(\epsilon_y\) are the yield strength and the corresponding strain, respectively. Find the value for yield strength and its corresponding strain, and then calculate the modulus of resilience using the formula above.

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Most popular questions from this chapter

A single crystal of a metal that has the FCC crystal structure is oriented such that a tensile stress is applied parallel to the [100] direction. If the critical resolved shear stress for this material is \(0.5 \mathrm{MPa}\), calculate the magnitude(s) of applied stress(es) necessary to cause slip to occur on the (111) plane in each of the [101], [10\overline{1} ] \text { , and } [ 0 \overline { 1 1 } ] \text { } directions.

A cylindrical rod of steel \(\left(E=207 \mathrm{GPa}, 30 \times 10^{\circ}\right.\) psi) having a yield strength of \(310 \mathrm{MPa}(45,000\) psi) is to be subjected to a load of \(11,100 \mathrm{~N}\) (2500 \(\left.\mathrm{Ib}_{i}\right)\). If the length of the rod is \(500 \mathrm{~mm}(20.0 \mathrm{in}\).), what must be the diameter to allow an elongation of \(0.38 \mathrm{~mm}(0.015\) in.)?

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

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