the modulus of elasticity the percentage elongation the percentage reduction in area
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During a tensile test on a specimen the following results were obtained
load(KN) 15 30 40 50 55 60 65 70 75 80 82 80 70
extension(mm) 0.05 0.094 0.127 0.157 1.778 2.79 3.81 5.08 7.62 12.7 16 19.05 22.9
diameter of gauge length=19mm
diameter at fracture=16.49mm
gauge length=100mm
gauge lengthat fracture=121mm
plot the complete load extension graph and the straight line portion to an enlarged scale .hence determine
the modulus of elasticity
the percentage elongation
the percentage reduction in area
the nominal stress at fracture
the actual stress at fracture
the tensile strength
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- The results of a tensile test are shown in Table 1.5.2. The test was performed on a metal specimen with a circular cross section. The diameter was 3 8 inch and the gage length (The length over which the elongation is measured) was 2 inches. a. Use the data in Table 1.5.2 to produce a table of stress and strain values. b. Plot the stress-strain data and draw a best-fit curve. c. Compute the, modulus of elasticity from the initial slope of the curve. d. Estimate the yield stress.A tensile test was performed on a metal specimen having a circular cross section with a diameter 0. 510 inch. For each increment of load applied, the strain was directly determined by means of a strain gage attached to the specimen. The results are, shown in Table: 1.5.1. a. Prepare a table of stress and strain. b. Plot these data to obtain a stress-strain curve. Do not connect the data points; draw a best-fit straight line through them. c. Determine the modulus of elasticity as the slope of the best-fit line.A tensile test was performed on a metal specimen with a diameter of 1 2 inch and a gage length (the length over which the elongation is measured) of 4 inches. The dam were plotted on a load-displacement graph. P vs. L. A best-fit line was drawn through the points, and the slope of the straight-line portion was calculated to be P/L =1392 kips/in. What is the modulus of elasticity?
- A tensile test was performed on a metal specimen having a circular cross section with a diameter of 1 2 inch. The gage length (the length over which the elongation is measured) is 2 inches. For a load 13.5 kips, the elongation was 4.6610 3 inches. If the load is assumed to be within the linear elastic rang: of the material, determine the modulus of elasticity.2.7 The following readings were recorded during a tensile test of a mild steel specimen 24 mm wide by 10 mm thick and with a gauge length of 200 mm. Force (kN) 16 32 48 64 68 72 76 79 Elongation (mm) 0,066 0,133 0,198 0,264 0,281 0,304 0,355 1,125 Force (kN) 76,8 83,7 103,8 111 112,8 108 96 Elongation (mm) 3,75 6,66 15,00 25,00 36,5 45,0 50,0 The test piece fractured at a 50-mm elongation. Plot the load-extension dia- gram using the following scales: 10 divisions on the x-axis = 5-mm exten- sion; 10 divisions on the y-axis = 10 kN. On the same graph paper and using the scale 10 divisions on x-axis = 0,5-mm extension, and 10 divisions on y-axis = 10 kN, replot the elastic portion of the graph. Using the graphs, determine the (a) modulus of elasticity for mild steel; (b) yield stress; (c) ultimate tensile stress; and (d) percentage elongation.During a tensile test on a specimen the following results were obtained: Load (kN) 15 30 40 50 55 60 65 Extension (mm) 0.05 0.094 0.127 0.157 1.778 2.79 3.81 Load (kN) 70 75 80 82 80 70 Extension (mm) 5.08 7.62 12.7 16.0 19.05 22.9 Diameter of gauge length - 19 mm Gauge length = 100 mm Diameter at fracture = 16.49 mm Gauge length at fracture = 121 mm Plot the complete load extension graph and the straight line portion to an enlarged scale. Hence determine: (a) the modulus of elasticity; (d) the nominal stress at fracture; (b) the percentage elongation; (e) the actual stress at fracture; (c) the percentage reduction in area; (f) the tensile strength. !116 GN/m2; 21%; 24.7%; 247 MN/m2; 328 MN/m2; 289 MN/m2.] 1.10 Figure 1.24 shows a special spanner used to tighten screwed components. A torque is applied at the tommybar and is transmitted to the pins which engage into holes located into the end of a screwed component. (a) Using the data given in Fig. 1.24 calculate: (i) the diameter D of…
- The following data was obtained from a tensile test on Load (kN) 16 32 56 72 95 110 132 142 140 135 Extension (mm) 0.2 0.4 0.7 0.9 1.5 2.5 5.0 8.5 10.0 12.0 a specimen of 10mm diameter and gauge length 60mm. Using the graph paper supplied, plot the load-extension diagram and determine: (1) The tensile strength (2) 0.1% Proof Stress (3) Young's Modulus for the specimen1 a. During a tensile test on a specimen the following results were obtained: Load (kN) 15 |30 40 50 55 60 Extension 1.03 2.00 2.73 3.42 3.80 4.10 (mm) Diameter of gauge length =19mm Gauge length = 150mm Load at fracture = 80kN Diameter at fracture =16.5mm Gauge length at fracture = 170mm Ultimate load = 120kN Plot the load/extension graph to an appropriate scale. Hence determine the Modulus of elasticity ii. Percentage elongation iii. Percentage reduction in area iv. The nominal stress at fracture v. The actual stress at fracture vi. Ultimate tensile strength i.Q1- The followving data was obtained tensile test on a specimen of 6 mm diameter and gauge length Lo=30 mm: 92.7 160.4 155 4.06 142 124.7 5.1 5.84 0| 38 load(KN) elongation(mm) 0 0.02 Using the graphic paper supplied, Plot engineering stress-strain curves, then determine the 76.2 107 149 0.12 0.25 0.50 2.03 3.55 (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))?
- Sample: Malleable Steel (AISI 4145) Original Diameter: 6.14mm Gauge Length: 55 mm Final Length: 68.12 mm Final Diameter: 3.54mm Load (kN) Deformation (mm) Stress (MPa) 12.95 Strain (%) 0.3838 0.54 0.7841 0.95 26.48 1.3899 1.63 1.9485 2.07 3.3090 3.10 4.5821 4.22 5.9359 5.03 7.0340 5.51 8.2413 6.22 10.9446 7.22 13.1951 8.18 12.8228 8.77 12.2583 9.11 12.5915 9.61 13.2536 10.33 13.6636 10.71 13.9772 11.35 14.5433 12.63 15.1155 13.88 15.4970 15.10 15.6484 17.09 15.4031 17.79 14.7655 18.73 13.6721 19.28 10.4617 19.88ngineering / Civil Engineering / Q&A Library/1 a. a. During a tensile test on a speci... 1 a. During a tensile test on a specimen the following results were obtaineo Load (kN) 15 30 40 50 55 60 Extension 1.03 2.00 2.73 3.42 3.80 4.10 (mm) Diameter at fracture -16.5mm Diameter of gauge length =19mm Gauge length = 150mm Load at fracture - R0KN Gauge length at fracture = 170mm Ultimate load-120kN Plot the load/extension graph to an appropriate scale. Hence determine the i. Modulus of elasticity ii. Percentage elongation iii. Percentage reduction in area iv. The nominal stress at fractureThe (G-E) diagram obtained in the tensile test performed on a metal sample with a diameter of 16 mm is as follows. The loads at points A, B and C and the elongation measured on l. 16 cm gauge length were determined as follows: B A B C Load (kgf) 4800 8400 7200 Elongation (mm) 0.192 28.8 38.4 a) Calculate the proportionality limit, modulus of elasticity, tensile strength, maximum uniform elongation, and contraction-elongation ratio of the metal. b) Since the measured diameter of the metal at break is 12 mm, find the constriction ratio and the actual stress at break.