EBK MANUFACTURING PROCESSES FOR ENGINEE
6th Edition
ISBN: 9780134425115
Author: Schmid
Publisher: YUZU
expand_more
expand_more
format_list_bulleted
Concept explainers
Videos
Question
Chapter 2, Problem 2.9Q
To determine
Whether the specific work divide by volume of the specimen under the stress-strain curve and load elongation curve represents same or not also explain whether it is same for all the value of strain or not.
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
The following data are obtained from a tensile test of a copper specimen.
- The load at the yield point is 143 kN.
- Length of the specimen is 29 mm.
- The yield strength is 71 kN/mm2.
- The percentage of elongation is 48 %.
Determine the following
Diameter of the specimen,
Final length of the specimen,
Stress under an elastic load of 18 kN,
Young's Modulus if the elongation is 1 mm at 18 kN and
Final diameter if the percentage of reduction in area is 29 %.
Initial Cross-sectional Area 2.01 mm2.
The Diameter of the Specimen 1.59 mm.
Final Length of the Specimen 42.92 mm.
Stress at the elastic load 8955.22 N/mm2.
Find:
Young's Modulus of the Specimen (in N/mm2)
Final Area of the Specimen at Fracture (in mm)
Final Diameter of the Specimen after Fracture (in mm)
The following data are obtained from a tensile test of a copper specimen.
- The load at the yield point is 157 kN.
- Length of the specimen is 23 mm.
- The yield strength is 89 kN/mm2.
- The percentage of elongation is 45 %.
Determine the following
Diameter of the specimen,
Final length of the specimen,
Stress under an elastic load of 18 kN,
Young's Modulus if the elongation is 1.3 mm at 18 kN
and
Final diameter if the percentage of reduction in area is
25 %.
Fine this ans
1-Initial Cross-sectional Area (in mm2)
2-The Diameter of the Specimen (in mm)
3-Final Length of the Specimen (in mm)
4-Stress at the elastic load (in N/mm2)
5-Young's Modulus of the Specimen (in N/mm2)
6-Final Area of the Specimen at Fracture (in mm)
7-Final Diameter of the Specimen after Fracture (in mm)
As Fast As you can
Please mak sure the answer is correct 100%
Please match the following to the appropriate areas or sublocations illustrated on the steel stress-strain curve shown below:
(ultimate tensile stress- yield stress - repture stress)
(4) The maximum stress point on the stress strain curve.
(2) The point where the proportional limit ends and the moment the elastic limit of the specimen is reached, the specimen will return to its original state after the loading is removed. Typically occurs before the steel specimen starts to plastically yield.
(5) The point at which the steel specimens has underwent necking and breaks. This typically occurs after the maximum stress is reached during the experiment.
Chapter 2 Solutions
EBK MANUFACTURING PROCESSES FOR ENGINEE
Ch. 2 - Prob. 2.1QCh. 2 - Prob. 2.2QCh. 2 - Prob. 2.3QCh. 2 - Prob. 2.4QCh. 2 - Prob. 2.5QCh. 2 - Prob. 2.6QCh. 2 - Prob. 2.7QCh. 2 - Prob. 2.8QCh. 2 - Prob. 2.9QCh. 2 - Prob. 2.10Q
Ch. 2 - Prob. 2.11QCh. 2 - Prob. 2.12QCh. 2 - Prob. 2.13QCh. 2 - Prob. 2.14QCh. 2 - Prob. 2.15QCh. 2 - Prob. 2.16QCh. 2 - Prob. 2.17QCh. 2 - Prob. 2.18QCh. 2 - Prob. 2.19QCh. 2 - Prob. 2.20QCh. 2 - Prob. 2.21QCh. 2 - Prob. 2.22QCh. 2 - Prob. 2.23QCh. 2 - Prob. 2.24QCh. 2 - Prob. 2.25QCh. 2 - Prob. 2.26QCh. 2 - Prob. 2.27QCh. 2 - Prob. 2.28QCh. 2 - Prob. 2.29QCh. 2 - Prob. 2.30QCh. 2 - Prob. 2.31QCh. 2 - Prob. 2.32QCh. 2 - Prob. 2.33QCh. 2 - Prob. 2.34QCh. 2 - Prob. 2.35QCh. 2 - Prob. 2.36QCh. 2 - Prob. 2.37QCh. 2 - Prob. 2.38QCh. 2 - Prob. 2.39QCh. 2 - Prob. 2.40QCh. 2 - Prob. 2.41QCh. 2 - Prob. 2.42QCh. 2 - Prob. 2.43QCh. 2 - Prob. 2.44QCh. 2 - Prob. 2.45QCh. 2 - Prob. 2.46QCh. 2 - Prob. 2.47QCh. 2 - Prob. 2.48QCh. 2 - Prob. 2.49PCh. 2 - Prob. 2.50PCh. 2 - Prob. 2.51PCh. 2 - Prob. 2.52PCh. 2 - Prob. 2.53PCh. 2 - Prob. 2.54PCh. 2 - Prob. 2.55PCh. 2 - Prob. 2.56PCh. 2 - Prob. 2.57PCh. 2 - Prob. 2.58PCh. 2 - Prob. 2.59PCh. 2 - Prob. 2.60PCh. 2 - Prob. 2.61PCh. 2 - Prob. 2.62PCh. 2 - Prob. 2.63PCh. 2 - Prob. 2.64PCh. 2 - Prob. 2.65PCh. 2 - Prob. 2.66PCh. 2 - Prob. 2.67PCh. 2 - Prob. 2.68PCh. 2 - Prob. 2.69PCh. 2 - Prob. 2.70PCh. 2 - Prob. 2.71PCh. 2 - Prob. 2.72PCh. 2 - Prob. 2.73PCh. 2 - Prob. 2.74PCh. 2 - Prob. 2.75PCh. 2 - Prob. 2.76PCh. 2 - Prob. 2.78PCh. 2 - Prob. 2.79PCh. 2 - Prob. 2.80PCh. 2 - Prob. 2.81PCh. 2 - Prob. 2.82PCh. 2 - Prob. 2.83PCh. 2 - Prob. 2.84PCh. 2 - Prob. 2.85PCh. 2 - Prob. 2.86PCh. 2 - Prob. 2.87PCh. 2 - Prob. 2.88PCh. 2 - Prob. 2.89PCh. 2 - Prob. 2.90PCh. 2 - Prob. 2.91PCh. 2 - Prob. 2.92PCh. 2 - Prob. 2.93PCh. 2 - Prob. 2.94PCh. 2 - Prob. 2.95PCh. 2 - Prob. 2.96PCh. 2 - Prob. 2.97PCh. 2 - Prob. 2.98PCh. 2 - Prob. 2.99PCh. 2 - Prob. 2.100PCh. 2 - Prob. 2.101P
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
- Question 4 You do a series of tensile tests on plates of a magnesium alloy that have been subjected to prior cold rolling to true plastic strains of 0.1, 0.2 and 0.3. The resulting true stress-true strain curves are shown below (including a zoomed in version expanding on the small strain region). It is reasonable to approximate the the 0.2% offset yield strength of the magnesium as ✓ MPa for 0.1 plastic strain, ✓ MPa for 0.2 and ✓ MPa for 0.3. Assuming the yield strength is proportional to the square root of the prior true plastic strain results in a hardening coefficient of approximately k= ✓ MPa. Hence, we can predict that we need a prior plastic strain of approximately ✓to obtain a hgth of 120 MPa True Stress (MPa) 250 200 100 85 95 50 105 115 125 135 145 300 350 0.12 0.14 150 0.16 0.18 155 165 175 200 250 0.1 True Stress (MPa) Ep = 0.1 Zoomed version of left plotarrow_forwardQuestion 4 Save Answer You do a series of tensile tests on plates of a magnesium alloy that have been subjected to prior cold rolling to true plastic strains of 0.1, 0.2 and 0.3. The resulting true stress-true strain curves are shown below (including a zoomed in version expanding on the small strain region). It is reasonable to approximate the the 0.2% offset yield strength of the magnesium as ✓ MPa for 0.2 and ✓ MPa for 0.3. Assuming the yield strength is ✓ MPa for 0.1 plastic strain, proportional to the square root of the prior true plastic strain results in a hardening coefficient of approximately k= MPa. Hence, we can predict that we need a prior plastic strain of approximately True Stress (MPa) True Stress (MPa) True Stress (MPa) 250 200 150 100 50 0 0.00 250 200 150 100 50 0 0.00 250 200 150 100 50 0 0.00 Ep = 0.1 0.01 Ep=0.2 0.01 Ep=0.3 0.01 0.02 True Strain 0.02 0.03 0.02 0.03 True Strain 0.03 0.04 0.04 0.04 True Stress (MPa) 0.05 0.000 0.001 True Stress (MPa) Ep=0.1 True…arrow_forwardDraw a typical stress vs strain tensile test curve for the following materials (two seperate graphs) and label the axis. A ductile metallic test specimen that is stretched to failure displaying a characteristic yield point and show the following parts on the curve. 1- Yield point 2- Ultimate Tensile Strength 3- Breaking point 4- Elastic Region 5- Plastic Region 6- Necking regionarrow_forward
- 1. A tensile test was conducted on a metal "505" specimen and the following stress-strain curves were generated, both curves generated from the same set of data. Use the graphs to fill in the mechanical properties of the material tested in the box below. Don't forget units! Stress vs Strain Stress, psi Stress, psi 80000 70000 60000 50000 40000 30000 20000 10000 0 0.00 80000 70000 60000 50000 40000 30000 20000 10000 0.02 0 0.000 0.002 0.04 0.004 0.06 0.006 0.08 0.10 Strain Stress vs Strain 0.008 0.12 Elastic Modulus, E: 0.2% Offset Yield Strength, oo: Tensile Strength, ou: Breaking Strength, of: % Elongation: 0.14 0.010 0.012 0.014 Strain 0.16 0.18 0.016 0.018 0.20 0.020arrow_forwardThe following data are obtained from a tensile test of a copper specimen. - The load at the yield point is 158 kN. - Length of the specimen is 26 mm. The yield strength is 75 kN/mm2. - The percentage of elongation is 40 %. Determine the following (i) Diameter of the specimen, ii) Final length of the specimen, iii) Stress under an elastic load of 15 kN, iv) Young's Modulus if the elongation is 1 mm at 15 kN and (v) Final diameter if the percentage of reduction in area is 21 %. Solution Initial Cross-sectional Area (in mm-) The Diameter of the Specimen (in mm) Final Length of the Specimen (in mm) Stress at the elastic load (in N/mm-) Young's Modulus of the Specimen (in N/mm-) Final Area of the Specimen at Fracture (in mm)arrow_forwardThe following data are obtained from a tensile test of a copper specimen. - The load at the yield point is 159 kN. - Length of the specimen is 29 mm. - The yield strength is 82 kN/mm2. - The percentage of elongation is 47 %. Determine the following (i) Diameter of the specimen, ii) Final length of the specimen, iii) Stress under an elastic load of 19 kN, iv) Young's Modulus if the elongation is 2.8 mm at 19 kN and (v) Final diameter if the percentage of reduction in area is 21 %. Solution Initial Cross-sectional Area (in mm2) Answer for part 1 The Diameter of the Specimen (in mm) Answer for part 2 Final Length of the Specimen (in mm) Answer for part 3 Stress at the elastic load (in N/mm2) Answer for part 4 Young's Modulus of the Specimen (in N/mm2) Answer for part 5 Final Area of the Specimen at Fracture (in mm) Answer for part 6 Final Diameter of the Specimen after Fracture (in mm)arrow_forward
- The following data are obtained from a tensile test of a copper specimen. - The load at the yield point is 146 kN. - Length of the specimen is 25 mm. - The yield strength is 83 kN/mm2. - The percentage of elongation is 40 %. Determine the following (i) Diameter of the specimen, ii) Final length of the specimen, iii) Stress under an elastic load of 14 kN, iv) Young's Modulus if the elongation is 1.1 mm at 14 kN and (v) Final diameter if the percentage of reduction in area is 25 %. Solution Initial Cross-sectional Area (in mm2) Answer for part 1 The Diameter of the Specimen (in mm) Answer for part 2 Final Length of the Specimen (in mm) Answer for part 3 Stress at the elastic load (in N/mm2) Answer for part 4 Young's Modulus of the Specimen (in N/mm2) Answer for part 5 Final Area of the Specimen at Fracture (in mm) Answer for part 6 Final Diameter of the Specimen after Fracture (in mm) Answer for part 7arrow_forwardThe following data are obtained from a tensile test of an aluminium specimen. - The load at the yield point is 147 kN - Length of the specimen is 29 mm. - The yield strength is 79 kN/mm2. - The percentage of elongation is 49 %. Determine the following (i) Diameter of the specimen, ii) Final length of the specimen, iii) Stress under an elastic load of 15 kN, iv) Young's Modulus if the elongation is 2.8 mm at 15 kN (1.5 Marks) and (v) Final diameter if the percentage of reduction in area is 20 %. (1.5 Initial Cross-sectional Area (in mm2) The Diameter of the Specimen (in mm) Final Length of the Specimen (in mm) Stress at the elastic load (in N/mm2) Young's Modulus of the Specimen (in N/mm2) Final Area of the Specimen at Fracture (in mm) Final Diameter of the Specimen after Fracture (in mm)arrow_forwardThe following data are obtained from a tensile test of a copper specimen. - The load at the yield point is 147 kN. - Length of the specimen is 28 mm. - The yield strength is 75 kN/mm2. - The percentage of elongation is 49 %. Determine the following (i) Diameter of the specimen, ii) Final length of the specimen, iii) Stress under an elastic load of 16 kN, iv) Young's Modulus if the elongation is 1 mm at 16 kN and (v) Final diameter if the percentage of reduction in area is 22 Final Area of the Specimen at Fracture (in mm) Final Diameter of the Specimen after Fracture (in mm) Initial Cross-sectional Area (in mm2)arrow_forward
- The following data are obtained from a tensile test of a copper specimen. - The load at the yield point is 157 kN. - Length of the specimen is 23 mm. - The yield strength is 89 kN/mm2. - The percentage of elongation is 45 %. Determine the following (i) Diameter of the specimen, ii) Final length of the specimen, iii) Stress under an elastic load of 18 kN, iv) Young's Modulus if the elongation is 1.3 mm at 18 kN and (v) Final diameter if the percentage of reduction in area is 25 %. Solution: Initial Cross-sectional Area (in mm2) The Diameter of the Specimen (in mm) Final Length of the Specimen (in mm) Stress at the elastic load (in N/mm2) Young's Modulus of the Specimen (in N/mm2) Final Area of the Specimen at Fracture (in mm) Final Diameter of the Specimen after Fracture (in mm)arrow_forwardThe following data are obtained from a tensile test of a copper specimen. - The load at the yield point is 151 kN. - Length of the specimen is 23 mm. - The yield strength is 79 kN/mm2. - The percentage of elongation is 45 %. Determine the following (i) Diameter of the specimen, ii) Final length of the specimen, iii) Stress under an elastic load of 17 kN, iv) Young's Modulus if the elongation is 2.4 mm at 17 kN and (v) Final diameter if the percentage of reduction in area is 24 %. Find: 1)Stress at the elastic load (in N/mm2) 2)Young's Modulus of the Specimen (in N/mm2) 3)Final Area of the Specimen at Fracture (in mm) 4)Final Diameter of the Specimen after Fracture (in mm)arrow_forwardThe following data are obtained from a tensile test of a copper specimen. - The load at the yield point is 151 kN. - Length of the specimen is 23 mm. - The yield strength is 79 kN/mm2. - The percentage of elongation is 45 %. Determine the following (i) Diameter of the specimen, ii) Final length of the specimen, iii) Stress under an elastic load of 17 kN, iv) Young's Modulus if the elongation is 2.4 mm at 17 kN and (v) Final diameter if the percentage of reduction in area is 24 %. THE QUESTION IS : FIND THE Final Diameter of the Specimen after Fracture (in mm)????arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
Elements Of ElectromagneticsMechanical EngineeringISBN:9780190698614Author:Sadiku, Matthew N. O.Publisher:Oxford University Press
Mechanics of Materials (10th Edition)Mechanical EngineeringISBN:9780134319650Author:Russell C. HibbelerPublisher:PEARSON
Thermodynamics: An Engineering ApproachMechanical EngineeringISBN:9781259822674Author:Yunus A. Cengel Dr., Michael A. BolesPublisher:McGraw-Hill Education
Control Systems EngineeringMechanical EngineeringISBN:9781118170519Author:Norman S. NisePublisher:WILEY
Mechanics of Materials (MindTap Course List)Mechanical EngineeringISBN:9781337093347Author:Barry J. Goodno, James M. GerePublisher:Cengage Learning
Engineering Mechanics: StaticsMechanical EngineeringISBN:9781118807330Author:James L. Meriam, L. G. Kraige, J. N. BoltonPublisher:WILEY
Elements Of Electromagnetics
Mechanical Engineering
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Oxford University Press
Mechanics of Materials (10th Edition)
Mechanical Engineering
ISBN:9780134319650
Author:Russell C. Hibbeler
Publisher:PEARSON
Thermodynamics: An Engineering Approach
Mechanical Engineering
ISBN:9781259822674
Author:Yunus A. Cengel Dr., Michael A. Boles
Publisher:McGraw-Hill Education
Control Systems Engineering
Mechanical Engineering
ISBN:9781118170519
Author:Norman S. Nise
Publisher:WILEY
Mechanics of Materials (MindTap Course List)
Mechanical Engineering
ISBN:9781337093347
Author:Barry J. Goodno, James M. Gere
Publisher:Cengage Learning
Engineering Mechanics: Statics
Mechanical Engineering
ISBN:9781118807330
Author:James L. Meriam, L. G. Kraige, J. N. Bolton
Publisher:WILEY
Material Properties 101; Author: Real Engineering;https://www.youtube.com/watch?v=BHZALtqAjeM;License: Standard YouTube License, CC-BY