Q19 and Q20 Q19-Derive an expression for true strain as a function of D and Do for a tensile testspecimen of round cross-section.Q20-The following data are collected during a tensile test in which the starting gagelength = 125.0 mm and the cross- sectional area = 62.5 mm2:23,042131.25Load (N)27,578153.0017,79327,579140.0528,91320,4620 125.23Length (mm)The maximum load is 28,913 N and the final data point occurred immediately prior tofailure.(a)Plot the engineering stress strain curve. Determine: (b) yield strength Y,(c) modulus of elasticity E,(d) tensile strength UTS (e). determine the strength coefficient and the strainhardening exponent. Be sure not to use data after the point at which necking147.01160.10occurred...Ans b-Y = 310.27 MPa , c- e=168,625 MPa. d.UTS=426.6 MPa e- n = 0.283,K = 910.2 MPa0.283The flow curve equation is: o = 910.2 e

as per our guidelines, i solved 1st question only

Answered: Q19-Derive an expression for true…

Engineering

Mechanical Engineering

Q19-Derive an expression for true strain as a function of D and Do for a tensile test specimen of round cross-section.

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

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.
Annealed low-carbon steel has a flow curve with strength coefficient = 80,000 lb/in2 and strain-hardening exponent = 0.25. A tensile test specimen with gage length = 2.0 in is stretched to a length = 3.5 in. Determine the flow stress and average flow stress that the metal experienced during this deformation. use correct stratgy and correct E 0.55962
How is the nominal or engineering strain found directly from the strain gage?
The 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)
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 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 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)
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)
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 %. THE QUESTION IS : FIND THE 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 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
Was there a discrepancy between the values of Young's Modulus found in the tensile test and the bending test? If so, why does this occur?
10. A torsion test shows that the shear modulus of an aluminumspecimen is 4.6 x 106psi. When the same specimen is used in atensile test, the modulus of elasticity is found to be 12.2 x 106psi. Find the Poisson’s ratio for the specimen.