ultimate strength in MPa
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- 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.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.
- 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 . 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)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 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.You have been given the following set of data from a soil-geotextile friction test: Normal stress (kPa) Peak shear strength (kPa) Residual shear strength (kPa) 17 20.2 13 35 31 22 70 52 39.5 140 94 74.5 (a) Plot the Mohr-Coulomb failure envelope (x-axis=normal stress; y-axis=shear stress) (b) Determine the peak friction angle, and the residual friction angle, . (c) Determine the peak adhesion and the residual adhesion. (d) Calculate the fabric efficiency based on a soil friction angle of = 35-degrees using the peak friction angle. **HINT: the equation for efficiency of soil friction angle mobilization is:
- Civil engineers often use the straight-line equation, = b 0 + b 1x, to model the relationship between the shear strength y of masonry joints and precompression stress, x. To test this theory, a series of stress tests were performed on solid bricks arranged in triplets and joined with mortar. The precompression stress was varied for each triplet and the ultimate shear load just before failure (called the shear strength) was recorded. The stress results for n = 7 resulted in a Coefficient of Determination of 0.8436. Given that r 2= 0.8436, give a practical interpretation of r 2, the coefficient of determination for the least squares model. a. We expect to predict the shear strength of a triplet test to within about 0.8436 tons of its true value. b. About 84.36% of the total variation in the sample of y-values can be explained by (or attributed to) the linear relationship between shear strength and precompression stress. c. In repeated sampling, approximately 84.36% of…Specimen (Rebar) Original Length Length when it was clamped Diameter (mm) Internal Diameter (mm) External 1 300 mm 163mm 10 15.030 2 300 mm 180 mm 10.020 14.386 3 300 mm 197mm 9.40 15.028 Dimension of Specimen Specimen A0 P max Tensile strength Modulus of elasaticity Percent Elongation 1 474 N 179.381 2.776 MPa 19.76 MPa 13.4% 2 475 N 159.070 2.895 MPa 18.74 MPa 16.67% 3 485 N 179.381 2.706 MPa 20.058 Mpa 13.4% Result of testing Observation What is the possible discussion of these tablesP.S THIS IS NOT GRADED, ITS JUST EXAMPLESCalculate 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) 0.342