Example 5.2 A cylindrical test sample of diameter 12.8 mm is tested in tension up to fracture andfound to have engineering fracture strength o, = 450 MPa. If the diameter of cross-sectional area atfracture is 10.7 mm, determine(a) The ductility in terms of per cent reduction in the area(b) True stress at fracture(c) True fracture strain

Answered: Image /qna-images/answer/9a5331f2-5e08-43a1-b7f6-319441bfe7fb.jpg

Answered: Example 5.2 A cylindrical test sample…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

See similar textbooks

Similar questions

A ceramic part for a jet engine has a yield strength of 75,000 psi and a plane strain fracture toughness of 5,000 psiVin. To be sure that the part does not fail, we plan to ensure that the maximum applied stress is only one-third of the yield strength. We use a nondestructive test that will detect any internal faws greater than 0.05 in. long. Assuming that f = 1.4, does our nondestructive test have the required sensitivity? Explain. A manufacturine proorssthat unintention
A ceramic part for a jet engine has a yield strength of 66,215 psi and a plane strain fracture toughness of 5000 psi (in) ¾. To be sure that the part does not fail, we plan to assure that the maximum applied stress is only one-third the yield strength. We use a nondestructive test that will detect any internal flaws greater than 0.05 in. Iong. Assuming that f = 1.2 , does our non- destructive test have the required sensitivity? Type 1 for yes and O for no.
1. For the stress-strain curve shown below, please estimate the properties indicated. (a) Fracture Strain Please do your work on a separate sheet of paper, and put your answers in the boxes on the right. Be sure to include the proper symbol and units. Stress Strain 70 60 50 Stress (ksi) 240 30 20 10 70 0 0.000 60 50 Stress (ksi) 40 20 10 KULL 0 0.000 0.010 0.050 0.100 Strain (in/in) Stress Strain 0.020 0.030 Strain (in/in) 0.040 0.150 0.050 (b) Ultimate Tensile Stress (c) Fracture Stress (d) Proportional Limit (e) Elastic Modulus (1) Yield Stress (g) Tensile Toughness (Modulus of Toughness) (h) Modulus of Resilience
Consider a cylindrical specimen of a steel alloy with 8.5 mm diameter and 80 mm long that is pulled in tension. Estimate the following mechanical properties using Fig. 1: a. Modulus of Elasticity and Resilience in MPa and psi b. Ultimate Tensile Strength in MPa and psi c. Fracture Strength in MPa and psi d. Ductility or % elongation at fracture in MPa and psi 2000 10³ psi MPa 300 2000 200 1000 100 0 0.000 0.005 0.010 0.015 Strain 0.020 0.040 0.060 Strain Fig. 1 Engineering Stress-Strain Curve Stress (MPa) 1000 0 0.000 Stress 0.080 300 200 100 0 Stress (10³ psi)
Problem: A tensile test specimen of stainless steel alloy having a with a diameter of 12.6 mm and a 50 mm gage length was tested to fracture. Load and deformation data obtained during the test are given: Determine: (a) the modulus of elasticity. (b) the proportional limit. Change in length (mm) 0.00 0.00 0.05 Load (c) the ultimate strength. (kN) 0.00 0.00 25.79 (d) the yield strength (0.20% offset). (e) the fracture stress. 51.59 0.10 (f) the true fracture stress if the final diameter 66.55 0.20 of the specimen at the location of the fracture 73.36 0.30 80.18 0.45 was 8.89 mm 87.77 0.63 0.81 1.01 1.18 95.35 101.42 105.98 112.05 118.13 123.46 1.41 1.66 1.94 2.27 128.03 2.60 2.82 131.85 133.39 134.95 135.73 3.15 3.35 3.68 3.93 4.26 136.53 136.56 135.84 135.14 133.66 4.66 4.94 5.22 5.47 5.70 5.82 130.67 128.44 124.69 120.92 117.91 116.52 5.95 fracture
2.33. AISI 403 stainless steel has a fracture toughness K1c=77 MPa /m and a uni- axial yield stress Sy = 690 MPa. If a component contains a random distribution of through cracks up to 1 mm in length, determine whether it will exhibit ductile or brittle behaviour in tension.
Problem: A tensile test specimen of stainless steel alloy having a with a diameter of 12.6 mm and a 50 mm gage length was tested to fracture. Load and deformation data obtained during the test are given: Determine: (a) the modulus of elasticity. (b) the proportional limit. Load (kN) 0.00 Change in length (mm) (c) the ultimate strength. 0.00 (d) the yield strength (0.20% offset). 0.00 25.79 0.00 (e) the fracture stress. 0.05 51.59 0.10 (f) the true fracture stress if the final diameter 66.55 73.36 80.18 87.77 0.20 of the specimen at the location of the fracture 0.30 0.45 was 8.89 mm 0.63 95.35 0.81 101.42 105.98 112.05 1.01 1,18 1.41 only solve d, e,f 118.13 123.46 128.03 131.85 133.39 134.95 135.73 1.66 1.94 2.27 2.60 2,82 3.15 3.35 3.68 136.53 136.56 3.93 135.84 4.26 135.14 4,66 133.66 4.94 130.67 5.22 5.47 128.44 124.69 5.70 120.92 5.82 117.91 5.95 fracture 116.52
Mild Steel is tested in the laboratory. The data obtained are:Length of Specimen = 320 mmDiameter of Specimen = 30 mmElongation under a load of 20 kN = 0.055 mmLoad at Yield Point = 130 kNMaximum Load = 210 kNLoad at fracture = 115 kNLength of Specimen at failure = 380 mmDiameter of Specimen at failure = 17.5 mm Calculate: (1).Young’s Modulus (2). Yield Stress (3). Ultimate Stress
A tensile test for a copper specimen has been performed and the following data are obtained. - Percentage of Elongation = 69 % - Percentage of Reduction in Area = 39 % - Final length after fracture = 35.5 mm - Final Diameter after fracture = 3.6 mm & - Ultimate stress = 396 MPa SOLUTION: iv) Initial Diameter (in mm) = v) Ultimate Load (in N) =
A Copper specimen of circular cross-section is subjected to a tensile test. The data obtained are: Length of Specimen = 329 mm; Diameter of Specimen = 39 mm; Load at Yield Point = 130 kN; Maximum Load = 219 kN; Load at fracture = 118kN Determine the following: 1) Find initial area of the test specimen 2) Yield Stress 3) Ultimate Stress 4) Fracture stress 5) Find the strain for 0.058mm elongation.
A tensile test for a copper specimen has been performed and the following data are obtained. - Percentage of Elongation = 69 % - Percentage of Reduction in Area = 39 % - Final length after fracture = 35.5 mm - Final Diameter after fracture = 3.6 mm & - Ultimate stress = 396 MPa SOLUTION: i) Initial Length (in mm) = ii) Final Area (in mm2) = iii) Initial Area (in mm2) =
A specimen of steel 20 mm diameter with a gauge length of 200 mm is tested to destruction. It has an extension of 0.25 mm under a load of 80 kN and the load at elastic limit is 102 kN. The maximum load is 130 kN. The total extension at fracture is 56 mm and diameter at neck is 15 mm. Find (i)The stress at elastic limit.(ii) Young’s modulus.(iii) Percentage elongation. (iv)Percentage reduction in area. (v)Ultimate tensile stress.

SEE MORE QUESTIONS

Recommended textbooks for you

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

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

SEE MORE TEXTBOOKS