Applied Statics and Strength of Materials (6th Edition)
6th Edition
ISBN: 9780133840544
Author: George F. Limbrunner, Craig D'Allaird, Leonard Spiegel
Publisher: PEARSON
expand_more
expand_more
format_list_bulleted
Textbook Question
Chapter 10, Problem 10.20SP
During a tensile test of a steel specimen, the strain at a stress of 35 MPa was calculated to be 0.000 170 (point A). The strain at a stress of 134 MPa was calculated to be 0.000 630 (point B). Determine the modulus of elasticity for this material using the slope between these two points. Calculate the expected stress that would correspond to a strain of 0.000 250. The proportional limit is 200 MPa.
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solutionStudents have asked these similar questions
Tensile test specimens are extracted from the "X" and "y" directions of a rolled sheet of metal. "x" is the rolling direction, "y" is
transverse to the rolling direction, and "z" is in the thickness direction. Both specimens were pulled to a longitudinal strain =
0.15 strain. For the sample in the x-direction, the width strain was measured to be ew= -0.0923 at that instant. For the sample in
the y-direction, the width strain was measured to be gw=-0.1000 at that instant.
The yield strength of the x-direction specimen was 50 kpsi and the yield strength of the y-direction specimen was 52 kpsi.
Determine the strain ratio for the x direction tensile test specimen. Determine the strain ratio for the y-direction tensile test
specimen. Determine the expected yield strength in the z-direction. Give your answer in units of kpsi (just the number). If the
sheet is plastically deformed in equal biaxial tension (a, = 0, to the point where & = 0.15, calculate the strain, 6, that would be
expected.
Solve it quickly please
Problem F2
stress (ksi)
130
120
110
100
90
80
70
60
50
40
30
20
10
0
0.000 0.025 0.050
0.075 0.100 0.125 0.150 0.175
strain (in/in)
Above you will find the experimental
stress-strain diagram of 1045 steel.
Calculate the permanent set if a
cylindrical specimen with a diameter of
2 in is loaded to the ultimate stress and
then unloaded. Provide your answer in
units of in/in with 3 significant figures
after the decimal. The elastic modulus
of 1045 steel is 29,000 ksi.
Chapter 10 Solutions
Applied Statics and Strength of Materials (6th Edition)
Ch. 10 - A 916 - in. - diameter steel rod is tested in...Ch. 10 - A concrete cylinder 150 mm in diameter was tested...Ch. 10 - Prob. 10.3PCh. 10 - The data from the tension test of a steel specimen...Ch. 10 - An 18-in.-long titanium alloy rod is subjected to...Ch. 10 - ASTM A36 steel rods are used to support a balcony....Ch. 10 - A 450-mm-long AISI 1020 steel rod is subjected to...Ch. 10 - A tension member in a roof truss is composed of...Ch. 10 - A short, solid, compression member of circular...Ch. 10 - A main cable in a large bridge is designed for a...
Ch. 10 - Test results of a steel specimen indicated an...Ch. 10 - A concrete canoe in storage is supported by two...Ch. 10 - A load is applied to a rigid bar that is...Ch. 10 - Prob. 10.14CPCh. 10 - Write a program that will allow a user to input...Ch. 10 - A 12 - in. - diaiíct.cr structural nickel steel...Ch. 10 - Compute the modulus of elasticity of a copper...Ch. 10 - A concrete cylinder 6 in. in diameter was tested...Ch. 10 - An aluminum bar 2 in. by 12 - in. in cross section...Ch. 10 - During a tensile test of a steel specimen, the...Ch. 10 - A 12.5-mm-diameter steel rod was subjected to a...Ch. 10 - Prob. 10.22SPCh. 10 - A standard steel specimen having a diameter of...Ch. 10 - 10.24 A tension member in a structure is composed...Ch. 10 - A pair of wire cutters is designed to operate...Ch. 10 - Calculate the end bearing length required for a...Ch. 10 - Design a 3-m-long rod subjected to a tensile load...Ch. 10 - The collar bearing shown is subjected to a...Ch. 10 - A 10-ft-long steel member is subjected to a...Ch. 10 - Two steel bars A and B support a load P, as shown....Ch. 10 - Prob. 10.31SP
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.Similar questions
- O 0.6% 19% If an isotropic material has a Young's modulus of 85 Gpa and a Poisson's ratio of 0.25, calculate its shear modulus. Select one: O G = 29 Gpa O G = 34 Gpa G = 25 Gpa O G= 77 Gpa O G= 46 Gpa If a rubber material is deformed as shown in the following figure, determine the normal strain along diagonal BD. C 2 mm 4 mm EN DO Oarrow_forwardIf the engineering strain in a tensile bar is 0.0025 and Poisson’s ratio is 0.33, find the original length and the original diameter if the length and diameter under load are 2.333 ft. and 1.005 in. respectively.arrow_forwardStress strain behavior of a cast iron with a diameter of 15.8mm and gauge length of 60.80mm is shown in the graphs. Find the ductility in percent elongationarrow_forward
- uniform diameter d= 1.75 in. Segment (1) is an aluminum Ę₁ = 10000 ksi L₁ = 90 in. Segment (2) is a copper Ę₂ = 17000 ksi L₂ = 126 in. When Pis applied, a strain gage attached to copper segment (2) measures a normal strain of &2 = 1700 uin./in. in the longitudinal direction. What is the elongation of segment (1)? Answer in inch rounded-off to 3 decimal places L₁ A (1) Aluminum B L2 (2) Copperarrow_forwardA 5 mm diameter aluminium alloy test bar is subjected to a load of 500 N. If the diameter of the bar at this load is 4 mm, the true strain is 1. 0.56 2. 0.22 3. 0.25 4. 0.45arrow_forwardQuestion 3: Please refer to the metal bar and its cross-section shown below. Two strain gages a and and b are attached to the bar as shown; note that 3 is the angle of their orientation with the x-axis. Assume: the Young's modulus of the metal = 16, 500 ksi; the Poisson's ratio of the metal = 0.33; P, = 13 kips; P, = 20 kips; d = 4.0 in.; t = 0.75 in.; L = 18 in.; and B = 40°. Determine the strains expected in the strain gages. 가 2 B b a H Ps N H y darrow_forward
- solve first one only and use b h and x to find the strainarrow_forwardA sample is deformed until the uniaxial strain is 0.251. Calculate the corresponding true strain. • Express your answer in scientific notation using 3 significant figures, e.g., 1.23*10^(-4). Use the pop-up MathType window to enter your answer. Checkarrow_forwardDetermine the elastic tensile load "F" that acts on mild steel specimen of Diameter 9 mm & Modulus of Elasticity 201 GPa. It is found that the specimen has undergone an extension of 1 mm due to the elastic load "F". Also, determine the length "L" of the specimen, if the strain- induced due to the load "F" on the specimen is 3.6 x 10-³. Solution Cross-sectional Area, A (in mm²) = 63.617 One possible correct answer is: 63.617251235193 Tensile Stress, (in N/mm²) 723.6 One possible correct answer is: 723.6 Tensile Load, (in N) = = 46033.261 One possible correct answer is: 46033.442993786 Length of the Specimen (in mm) 4.366 X One possible correct answer is: 277.77777777778arrow_forward
- The question is related to Modulus of rigidity and is attached as an image.arrow_forward8arrow_forwardDetermine the elastic tensile load "F" that acts on mild steel specimen of Diameter 5 mm & Modulus of Elasticity 207 GPa. It is found that the specimen has undergone an extension of 1.2 mm due to the elastic load "F". Also, determine the length "L" of the specimen, if the strain-induced due to the load "F" on the specimen is 3 x 10-³. Solution Cross-sectional Area, A (in mm²) = Tensile Stress, (in N/mm²) = Tensile Load, (in N) = Length of the Specimen (in mm)arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
- Mechanics of Materials (MindTap Course List)Mechanical EngineeringISBN:9781337093347Author:Barry J. Goodno, James M. GerePublisher:Cengage Learning
Mechanics of Materials (MindTap Course List)
Mechanical Engineering
ISBN:9781337093347
Author:Barry J. Goodno, James M. Gere
Publisher:Cengage Learning
An Introduction to Stress and Strain; Author: The Efficient Engineer;https://www.youtube.com/watch?v=aQf6Q8t1FQE;License: Standard YouTube License, CC-BY