i- Modulus of elasticity 2- Yield strength 3- Tensile strength
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Q: Define the term Absolute Maximum Shear Strain?
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- Following experimental data are obtained from tensile test of a rectangular test specimen with original thickness of 2,5 mm, gauge width of 24 mm and gauge length of 101 mm: Load (N) Elongation (mm) 0 0 24372 0,183 23008 0,315 28357 5,777 35517 12,315 27555 17,978 23750 23,865 Based on the information above; draw stress-strain diagram of the material and answer the following questions. - Determine the true stress (in MPa) at yield point. - Determine the true stress (in MPa) at point of ultimate strength. - Determine the true stress (in MPa) at fracture point. - Determine the true strain (in mm/mm) at yield point. (Use at least five decimal units) - Determine the true strain (in mm/mm) at point of ultimate strength. (Use at least five decimal units) - Determine the true strain (in mm/mm) at fracture point. (Use at least five decimal units)Following experimental data are obtained from tensile test of a rectangular test specimen with original thickness of 2,5 mm, gauge width of 24 mm and gauge length of 101 mm: Load (N) Elongation (mm) 0 0 24372 0,183 23008 0,315 28357 5,777 35517 12,315 27555 17,978 23750 23,865 Based on the information above; draw stress-strain diagram of the material and answer the following questions. Question 1 ;Determine the elastic energy absorption capacity (in N.mm) of that specimen. Question 2; Determine the plastic energy absorption capacity (in N.mm) of that specimen.I want answers to all four questions if possible. Thanks for help :) Following experimental data are obtained from tensile test of a rectangular test specimen with original thickness of 2,5 mm, gauge width of 24 mm and gauge length of 101 mm: Load (N) Elongation (mm) 0 0 24372 0,183 23008 0,315 28357 5,777 35517 12,315 27555 17,978 23750 23,865 Based on the information above; draw stress-strain diagram of the material and answer the following questions. - Calculate the fracture strength (in MPa) of the material. - Calculate the percent elongation of the specimen at fracture point. - Determine the modulus of resilience (in N.mm/mm3) of the material. (Use at least five decimal units) - Determine the toughness index number (in N.mm/mm3) of the material.
- 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.In a forming process the diameter of the specimen is 1 cm and its length changes from 50 mm to 60 mm. Show the maximum force value on the graph of Force vs. True Strain plotted in Excel Program or any plotting program not by hand using the flow stress equation given below. σ= Kεn and σ=F/A Where the parameters are σ : True Stress in MPa K: Strength Coefficient, K=210 in MPa ε : True Strain, ε=ln (l/l0) n: Strain Hardening Exponent, n=0.27 F: Force in 10 A : Area in mm2 Take at least 3 digits after the decimal point. Specify the units. You can use Excel or any plotting program not by hand then paste your graph into the solution section.In a tensile test on a specimen of black mild steel of 12 mm diameter, the following results were obtained for a gauge length of 60 mm. Load W(kN) 5 10 15 20 25 30 35 40 Extension x (10-3 mm) 14 27.2 41 54 67.6 81.2 96 112 When tested to destruction. Maximum load = 65 kN; load at fracture = 50 kN, diameter at fracture = 7.5 mm, total extension on gauge length = 17 mm. Find young's modulus, specific modulus, ultimate tensile stress, breaking stress, true stress at fracture, limit of proportionality, percentage elongation, percentage reduction in area. The relative density of the steel is 7.8. Draw the straight line graph.
- In a tensile test on a specimen of black mild steel of 12 mm diameter, the following results were obtained for a gauge length of 60 mm. Load W(kN) 5 10 15 20 25 30 35 40 Extension x (10-3 mm) 14 27.2 41 54 67.6 81.2 96 112 When tested to destruction. Maximum load = 65 kN; load at fracture = 50 kN, diameter at fracture = 7.5 mm, total extension on gauge length = 17 mm. Find young's modulus, specific modulus, ultimate tensile stress, breaking stress, true stress at fracture, limit of proportionality, percentage elongation, percentage reduction in area. The relative density of the steel is 7.8. Draw the straight line graph. Answer: breaking stress,true stress at fracture and limit of proportionality.The following data were obtained during a tensile test of an alluminium alloy specimen. Initial diameter = 18mm Original gauge length = 45mm Final gauge length = 53mm Load at elastic limit = 82kN Yield load = 93kN Maximum load = 168kN Determine the Yield stress - answer to be provided in MPa correct to 3 decimalThe 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 7
- The following data are obtained from a tensile test of a copper specimen. - The load at the yield point is 142 kN. - Length of the specimen is 23 mm. - The yield strength is 82 kN/mm2. - The percentage of elongation is 48 %. 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 1.6 mm at 15 kN and (v) Final diameter if the percentage of reduction in area is 20 %. 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)The following data were obtained from the tensile test of Aluminum alloy. The initial diameter of testspecimen was 0.505 inch and gauge length was 2.0 inch. Plot the stress strain diagram and determine(a) Proportional Limit (b) Modulus of Elasticity (c) Yield Stress at 0.2% offset (d) Ultimate Stress and(e) Nominal Rupture Stress.The 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)
