Mechanics of Materials (10th Edition)
10th Edition
ISBN: 9780134319650
Author: Russell C. Hibbeler
Publisher: PEARSON
expand_more
expand_more
format_list_bulleted
Concept explainers
Textbook Question
thumb_up100%
Chapter 3.4, Problem 3.11P
The stress-strain diagram for an aluminum alloy specimen having an original diameter of 0.5 in. and a gage length of 2 in. is given in the figure. If the specimen is loaded until it is stressed to 60 ksi, determine the approximate amount of elastic recovery and the increase in the gage length after it is unloaded.
Expert Solution & Answer
Trending nowThis is a popular solution!
Students have asked these similar questions
3.5 please need assistance
The stress strain curve determined from a sample of material
is shown in the figure. The original diameter was 14.3 mm
and it changed to 14.28858 when a 43.5 kN load was applied.
Calculate Poisson's ratio for the sample.
o (MPa)
550
400
0.00275
Poisson's ratio v =
0.0
A tensile test specimen has a gage length = 3.0 in and
diameter 0.75 in. Yielding occurs at a load of 38,000 lb. The
corresponding gage length = 3.0103 in (neglect the 0.2 percent
yield point). The maximum load of 54,000 lb is reached at a gage
length = 3.453 in. Determine the modulus of elasticity (neglect
the 0.2% offset, and round to the nearest whole Msi).
37
Chapter 3 Solutions
Mechanics of Materials (10th Edition)
Ch. 3.4 - Define a homogeneous material.Ch. 3.4 - Indicate the points on the stress-strain diagram...Ch. 3.4 - Define the modulus of elasticity E.Ch. 3.4 - At room temperature, mild steel is a ductile...Ch. 3.4 - Engineering stress and strain are calculated using...Ch. 3.4 - As the temperature increases the modulus of...Ch. 3.4 - A 100-mm-long rod has a diameter of 15 mm. If an...Ch. 3.4 - A bar has a length of 8 in. and cross-sectional...Ch. 3.4 - A 10-mm-diameter rod has a modulus of elasticity...Ch. 3.4 - The material for the 50-mm-long specimen has the...
Ch. 3.4 - The material for the 50-mm-long specimen has the...Ch. 3.4 - If the elongation of wire BC is 0.2 mm after the...Ch. 3.4 - A tension test was performed on a steel specimen...Ch. 3.4 - Data taken from a stress-strain test for a ceramic...Ch. 3.4 - Data taken from a stress-strain test for a ceramic...Ch. 3.4 - The stress-strain diagram for a steel alloy having...Ch. 3.4 - The stress-strain diagram for a steel alloy having...Ch. 3.4 - The stress-strain diagram for a steel alloy having...Ch. 3.4 - The rigid beam is supported by a pin at C and an...Ch. 3.4 - The rigid beam is supported by a pin at C and an...Ch. 3.4 - Acetal plastic has a stress-strain diagram as...Ch. 3.4 - The stress-strain diagram for an aluminum alloy...Ch. 3.4 - The stress-strain diagram for an aluminum alloy...Ch. 3.4 - The stress-strain diagram for an aluminum alloy...Ch. 3.4 - A bar having a length of 5 in. and cross-sectional...Ch. 3.4 - The rigid pipe is supported by a pin at A and an...Ch. 3.4 - The rigid pipe is supported by a pin at A and an...Ch. 3.4 - Direct tension indicators are sometimes used...Ch. 3.4 - The rigid beam is supported by a pin at C and an...Ch. 3.4 - The rigid beam is supported by a pin at C and an...Ch. 3.4 - The stress-strain diagram for a bone is shown, and...Ch. 3.4 - The stress-strain diagram for a bone is shown and...Ch. 3.4 - The two bars are made of a material that has the...Ch. 3.4 - The two bars are made of a material that has the...Ch. 3.4 - The pole is supported by a pin at C and an A-36...Ch. 3.4 - The bar DA is rigid and is originally held in the...Ch. 3.7 - A 100-mm-long rod has a diameter of 15 mm. If an...Ch. 3.7 - A solid circular rod that is 600 mm long and 20 mm...Ch. 3.7 - A 20-mm-wide block is firmly bonded to rigid...Ch. 3.7 - A 20-mm-wide block is bonded to rigid plates at...Ch. 3.7 - The acrylic plastic rod is 200 mm long and 15 mm...Ch. 3.7 - The plug has a diameter of 30 mm and fits within a...Ch. 3.7 - The elastic portion of the stress-strain diagram...Ch. 3.7 - The elastic portion of the stress-strain diagram...Ch. 3.7 - The brake pads for a bicycle tire are made of...Ch. 3.7 - The lap joint is connected together using a 1.25...Ch. 3.7 - The lap joint is connected together using a 1.25...Ch. 3.7 - The rubber block is subjected to an elongation of...Ch. 3.7 - The shear stress-strain diagram for an alloy is...Ch. 3.7 - A shear spring is made from two blocks of rubber,...Ch. 3 - The elastic portion of the tension stress-strain...Ch. 3 - The elastic portion of the tension stress-strain...Ch. 3 - The rigid beam rests in the horizontal position on...Ch. 3 - The wires each have a diameter of 12 in., length...Ch. 3 - The wires each have a diameter of 12 in., length...Ch. 3 - diameter steel bolts. If the clamping force in...Ch. 3 - The stress-strain diagram for polyethylene, which...Ch. 3 - The pipe with two rigid caps attached to its ends...Ch. 3 - The 8-mm-diameter bolt is made of an aluminum...Ch. 3 - An acetal polymer block is fixed to the rigid...
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
- An element of material in plain strain has the following strains: x = 0.001 and y = 0.0015. (a) Determine the strains for an element oriented at an angle = 250. (b) Find the principal strains of the element. Confirm the solution using Mohr’s circle for plane strain.arrow_forwardA solid spherical ball of magnesium alloy (E = 6.5 × l0-6 psi, v = 0.35) is lowered into the ocean to a depth of 8000 ft. The diameter of the ball is 9.0 in. (a) Determine the decrease ?d in diameter, the decrease, ?V in volume, and the strain energy U of the ball. (b) At what depth will the volume change be equal to 0.0324% of the original volume?arrow_forwardAn element of material in plain strain is subjected to strains x = 0.0015, , y . = -0.0002, and xy = 0.0003. (a) Determine the strains for an element oriented at an angle = 20°. (b) Determine the principal strains of the element. Confirm the solution using Mohr’s circle for plane strain.arrow_forward
- The elastic portion of the stress–strain diagram for an aluminum alloy is shown in the figure. The specimen from which it was obtained has an original diameter of 12.7 mm andagagelengthof50.8mm.IfaloadofP=60kNisapplied to the specimen, determine its new diameter and length. Take n = 0.35.arrow_forwardTensile 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.arrow_forwardA tension test was performed on a steel specimen having an original diameter of 12.5 mm and gauge length of 50 mm. The data is listed in the table. Plot the stress–strain diagram and determine approximately the modulus of elasticity, the yield stress, the ultimate stress, and the rupture stress. Use a scale of 25 mm = 140 MPa and 25 mm = 0.05 mm/mm. Redraw the elastic region, using the same stress scale but a strain scale of 25 mm = 0.001 mm/mm.arrow_forward
- The stress-strain diagram for a steel alloy having an original diameter of 0.5 in. and a gage length of 2 in. is given in the figure. Determine approximately the modulus of elasticity for the material, the load on the specimen that causes yielding, and the ultimate load the specimen will support. o (ksi) 80 70 60 50 40 30 20 10 e (in./in.) 0 0.04 0.08 0.12 0.16 0.20 0.24 0.28 0 .0.0005 0.0010.0015 0.002 0.0025 0.0030.0035arrow_forwardplease help answer and explainarrow_forwardNylon strips are fused to glass plates. When moderately heated the nylon will become soft while the glass stays approximately rigid. Determine the average shear strain in the nylon due to the load P when the assembly deforms as indicated. 2 mm P. 3 mm 5 mm 3 mm 5 mm 3 mmarrow_forward
- The stress–strain diagram for a steel alloy having an original diameter of 0.5 in. and a gage length of 2 in. is given in the figure. If the specimen is loaded until it is stressed to 70 ksi, determine the approximate amount of elasticrecovery and the increase in the gage length after it is unloaded.arrow_forwardThe stress-strain diagram for an aluminum alloy specimen having an original diameter of 0.5 in. and a gauge length of 2 in. is given in the figure. If the specimen is loaded until it is stressed to 60 ksi, determine the approximate amount of elastic recovery and the increase in the gage length after it is unloaded.arrow_forwardThe elastic portion of the tension stress–strain diagram for an aluminum alloy is shown in the figure. The specimen used for the test has a gage length of 2 in. and a diameter of 0.5 in. When the applied load is 9 kip, the new diameter of the specimen is 0.49935 in. Calculate the shear modulus Gal for the aluminum.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
Strain energy and strain energy density introduced; Author: Engineer4Free;https://www.youtube.com/watch?v=m14sqLGg4BQ;License: Standard youtube license