steel tension test specimen has a diameter of 6.00 mm and a gage length of 25 mm. The following data are of id test: Load, KN 0 2.94 Deformation, mm 0 0.01 Tensile Test Result Load, KN 13.22 13.25 Deformation, mm 0.10 0.14 Load, KN 20.72 20.61
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Explain in your own words how are these properties determined.
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- S1:Q1 a) Figure 1a provides dimensions of a test sample that is being subjected to a tensile stress. If the stress at section Y-Y is 188N/mm2 and the force is 220 kN, calculate the dimension X. Y 1.8cm- Y Not to scale Figure 1aA results below were recorded during tensile test for a steel specimen to be used to manufacture water reticulation steel pipes. The diameter of the specimen measured by an electronic vernier was found to be 0.597cm. Other measurements noted were the gauge length of 5.010cm,elongation 0.0635mm at load of 7.11kn, 7.33kn load at the yield point and 8.15kn load at the point of failure. The reduction of the cross-sectional at failure is estimated to be 35.06%. Calculate the yield stress in N/mm^22.7 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.
- a. Assuming the computer and the printer are turned on, what will be the next step to do when trying to perform a test?b. What part of the UTM will you use to remove the tensile test specimen from the grips?c. In what direction does the lower crosshead move during a tensile test?d. In what direction does the lower crosshead move during a compression test?e. In what direction does the lower crosshead move during a bending test?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 data 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 y∆L 5 1392 kips yin. What is the modulus of elasticity?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…
- Procedure: 1. Compression tests on specimens shall be made as soon as practicable after removal from the moist storage. A 28-day test shall be performed within +-20 hours of the 28thday. Test specimens shall be kept moist by any convenient method during the period between removals from moist storage and testing. They shall be tested in moist condition. 2. All test specimens for a given test age shall be broken within the permissible time tolerance prescribed below. TEST AGE PERMISSIBLE TOLERANCE 24 HOURS +-0.5 HOURS OR 2.1% 3 DAYS 2 HOURS OR 2.8% 7 DAYS 6 HOURS OR 3.6% 28 DAYS 20 HOURS OR 3.0% 90 DAYS 2 DAYS OR 2.2 % 3. With a clean rag or rush clean the bearing faces of the bearing blocks, test the specimens and exclusion controller (elastomeric cps). 4. Rest the specimen on the lower extrusion controller, place the top extrusion controller on the specimen, and check the spacing between the sides of the specimen and the extrusion controllers to ensure no contact between the cylinder…During a tensile test on a specimen the following results were obtainedload(KN) 15 30 40 50 55 60 65 70 75 80 82 80 70extension(mm) 0.05 0.094 0.127 0.157 1.778 2.79 3.81 5.08 7.62 12.7 16 19.05 22.9diameter of gauge length=19mmdiameter at fracture=16.49mmgauge length=100mmgauge lengthat fracture=121mmplot the complete load extension graph and the straight line portion to an enlarged scale .hence determinethe modulus of elasticitythe percentage elongationthe percentage reduction in areathe nominal stress at fracturethe actual stress at fracturethe tensile strengthThe nominal stress-strain diagram for a specimen having an original diameter of 10 mm and a gauge length of 50 mm is shown below. Using the given nominal stress-nominal strain diagram, answer this question. The inset shows a detailed view of the curve at low strains. 400 350 300 250 Stress (MPa) 200 150 100 50 0 0.00 Stress (MPa) 350 kN 300 250 200 150 100 50 0 0.000 0.002 0.004 0.006 0.008 0.010 0.012 Strain 0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16 Strain 0.18 What is the maximum force that could be applied to the specimen before it plastically deforms?
- Given the torque-angle deformation curve below, estimate the shear modulus in GPa?. Where the length of the specimen is 273mm and the diameter is 19.3mm (Round your answer to nearest integer, e.g. 100, do not add units, use exact value of pi not 3.14) SC.O 45.0 35.0 25.0 20.0 5.0 0.20 040 050 0.60 0.70 0.80 0.90 Angle (") A S en eta ine f Your Answer: Answer Torque (N-m)The (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.A steel specimen of 300 mm length and 30 mm diameter is subjected to tensile test in a computerizedUTM. Under 54 kN of tensile load, the final length and diameter are observed as 300.112 mm and29.99634 mm respectively. Calculate (a) the Poission’s ratio and (b) the values of three moduli.