Page |63.1) Calculate stress and strain for the data in Table 1. Please show 1 calculation of stressand 1 of strain. The rest you must tabulate.(12) A 1045 hot-rolled steel tension test specimen has the original diameter and length of 6 mmand 25 mm, respectively. The load and change in length data were recorded as shown inTable 1 below:Table 1. Load and change in lengthLoad (KN)Change in Length Load (KN)Change in Length(mm)|(mm)20.562.262.940.0120.723.365.580.0220.613.838.520.0319.974.0011.160.0418.72Fracture.12.630.0513.020.0613.160.0813.220.1016.150.6118.501.0420.271.80

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Answered: A 1045 hot-rolled steel tension test…

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

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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.
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…
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.
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)
A high-impulse universal testing machine is being used to determine the Young's Modulus of a rectangular solid specimen (l = 20 cm, t = 1.5 cm, w = 2 cm). The machine produces 35,000 Ns that causes the impact force of compression resulted to a 1 mm deformation in the length of the specimen in 0.05 s. Determine the Young's modulus of the specimen expressed in MPa. Round off your answer to the nearest whole number
The following data were obtained from a standard tensile stress test on a mild steel specimen. Graph the result and determine proportional limit stress and Young’s Modulus.
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)
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 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 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)????

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A 1045 hot-rolled steel tension test specimen has the original diameter and length of 6 mn and 25 mm, respectively. The load and change in length data were recorded as shown n Table 1 below: