Plot the True stress strain graph, pointing each point on the graph, and calculate graphically 1- Modulus of elasticity 2 The values of K and n 3 True tensile strength 4- True fracture stress
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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. 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.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.
- Question 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 yield strength (in MPa) of the material. - Calculate the percent elongation of the specimen at yield point. (Use at least five decimal units) - Calculate the stiffness (in MPa) of the specimen material. - Calculate the ultimate strength (in MPa) of the material. - Calculate the percent elongation of the specimen at point of ultimate strength.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.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 results were obtained in a tensile test on a mild steel specimen of original diameter 28 mm and gauge length 56 mm. Load at limit of proportionality : 96 kN Extension at 96 kN load : 0.068 mm Load at yield point : 105 kN Maximum load : 170 kN When the two parts were fitted together after being broken, the length between gauge length was found to be 65.8 mm and the diameter at the neck was 18.8 mm. Calculate: (a). E, (b).σy, (c).σu, (d). PRIA and PECalculate the values of Poisson’s ratio of a rod under uniaxial tensile test. Initially it has 15.5 mm diameter and 62 mm gauge length, then it changes to 15.415 mm diameter and 62.102 mm gauge length under a load of 20 kN.a) 0.186 b) 0.242 c) 0.259 d) 0.34238) A tensile test specimen has a gauge length =2in. and a diameter= 0.875in. Yielding occurs at a load of 35,500lbs. The corresponding gauge length= 2.0113in (neglect the 0.2% yield point. The maximum load of 45,000lb is reached at a gauge length=2.543in. Determine the modulus of elasticity (neglect the 0.2% offset, and round to the nearest whole Msi)
- When a bar of 22 mm diameter is subjected to an axial pull of 60 kN the extension on the 49 mm gauge length is 0.1 mm and there is a decrease in diameter of 0.013 mm. Calculate the Young's Modulus. Provide your answer in GN/m2 to the nearest whole number.The following data was obtained as a result of tensile testing of a standard 0.505 inch diameter test specimen of magnesium. After fracture, the gage length is 2.245 inch and the diameter is 0.466 inch. a). Calculate the engineering stress and strain values to fill in the blank boxes and plot the data. Load(lb) Gage Length (in) Stress (kpsi) Strain 0 2 1000 2.00154 2000 2.00308 3000 2.00462 4000 2.00615 5000 2.00769 5500 2.014 6000 2.05 6200 (max) 2.13 6000 (fracture) 2.255 b). Calculate the modulus of elasticity c). If another identical sample of the same material is pulled only to 6000 pounds and is unloaded from there, determine the gage length of the sample after unloading.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 decimal