(gy. sy), (gu, cu), (of , sf), modulus of elasticity, then detected in the draw (strain hardening (n)), (Necking), (uniform plastic region and non-uniform plastic region) and (elastic & plastic) region))?
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- Q1- The following data was obtained tensile test ɔn a specimen of 10 mm diameter and gauge length Lo=50 mm: 0 38 | 76.2 92.7 107 149 160.4 | 155 142 124.7 load(KN) | elongation(mm) 0 0.02 0.12 0.25 0.50 2.03 3.55 4.06 5.1 5.84 Using the graphic paper supplied, Plot engineering stress-strain curves, then determine the (ay, ey), (ou , eu), (of,, ef), modulus of elasticity, then detected in the draw ((stroin hardening (m)), (Necking), (uniform plastic region and non-uniform plastic region) and (elastic & plastic) region))?Load (N) Elongation (mm) In a tensile test of stainless steel, the following data were obtained: 7,200 11,250 13,500 16,200 18,900 20,200 20,700 13,200 24.90 0.0 0.505 2.03 5.08 10.16 15.24 21.80 The dimensions of the cylindrical specimen were initial length L₁= 50.8 mm and initial area A₁ = 35.8 mm² and the area at fracture was Ar = 10.0 mm². Plot the stress-strain diagram and determine the following: a) Modulus of elasticity b) The yield strength at a strain offset of 0.002 c) Ultimate strength d) Fracture strengthIn a tensile test of stainless steel, the following data were obtained: Load (N) 7,200 11,250 13.500 16,200 18.900 Elongation (mm) 0.0 0.505 2.03 5.08 20,200 15.24 20,700 13,200 24.90 10.16 21.80 The dimensions of the cylindrical specimen were initial length L₁ = 50.8 mm and initial arca A₁ - 35.8 mm² and the area at fracture was Ar 10.0 mm². Plot the stress-strain diagram and determine the following: a) Modulus of elasticity b) The yield strength at a strain offset of 0.002 c) Ultimate strength d) Fracture strength
- 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.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)The stress-strain data from a tensile test on a cast-iron specimen are € (10-3) 0 0.20 0.44 0.80 1.0 1.5 2.0 26 32 40 46 49 54 NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. 2.8 3.4 4.0 5.0 a (kpsi) 0 5 10 16 19 Determine the tangent modulus, E, at a value of a=0 psi. 106 psl. The tangent modulus, Eis X
- Problem :- The following data was obtained tensile test on a specimen of 6 mm diameter and gauge length Lo=30 mm: load(KN) 0 40 80 95 90 135 150.2 145 130 120 elongation(mm) 0 0.022 0.13 0.25 0.55 2.01 3.88 4.12 5.22 6 Plot engineering and True (stress-strain) curves, then determine the (oy, ɛy), (ou, ɛu), (of, , ef), ductility, modulus of elasticity. Then select on the drawing ((strain hardening (n)) , (Necking), (uniform plastic region and non-uniform plastic region) and (elastic & plastic) region)?a . Sketch stress strain curve if the result shown in table represent the force and extension happened in steel, and show Mechanical properties that we get from tensile test on curve? (10p)Note : Lo=80 mm , Do=10 mm , use excel to plot the curve ExtensionLoad(mm) (N)0 0.900.83 4694.341.67 4831.412.50 4781.083.33 4918.834.17 4926.585.00 5257.075.83 5437.016.66 5575.888.33 5775.189.16 5847.5210.83 5965.4111.67 6010.5312.50 6042.5713.33 6072.2614.16 6092.9315.00 6113.2416.67 6140.3617.50 6146.3718.33 6148.1419.16 6149.1725.00 5940.2125.83 5675.3326.67 4725.52b. What is meant by modulus of rigidity? if it increases what does happen to material? (2p)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.
- A 1045 hot-rolled steel tension test specimen has the original diameter and length of 6 mm and 25 mm, respectively. The load and change in length data were recorded as shown in Table 1 below: Table 1. Load and change in length Load (KN) Change in Length Load (KN) Change in Length (mm) |(mm) 20.56 2.26 2.94 0.01 20.72 3.36 5.58 0.02 20.61 3.83 8.52 0.03 19.97 4.00 11.16 0.04 18.72 Fracture. 12.63 0.05 13.02 0.06 13.16 0.08 13.22 0.10 16.15 0.61 18.50 1.04 20.27 1.80I 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.1 - A tensile test is carried out on a specimen of mild steel of gauge length 40 mm and diameter 7.42 mm. The results are: Load (kN) 10 17 25 30 34 37.5 38.5 36 Extension (mm) 0 0.05 0.08 0.11 0.14 0.20 0.40 0.60 0.90 At fracture the final length of the specimen is 40.90 mm. Plot the load/ extension graph and determine (a) The modulus of elasticity for mild steel, (b) The yield stress, (c) The ultimate tensile strength, (d) The percentage elongation 2 - An aluminum alloy specimen of gauge length 75 mm and of diameter 11.28 mm was subjected to a tensile test, with these results: Load (kN) 2.0 6.5 11.5 13.6 16.0 18.0 19.0 20.5 19.0 Extension (mm) 0 0.012 0.039 0.069 0.080 0.107 0.133 0.158 0.225 0.310 The specimen fractured at a load of 19.0 kN. Determine (a) the modulus of elasticity of the alloy, (b) The percentage elongation. (c) The ductility in term of reduction of area (d) Fracture stress (e) Tensile strength