Engineerin Engineerin Engineerin EngineeringgggstrainStressstressstrain(in./in.)(in./in.)(ksi)(ksi)00760.08300.001750.10550.002730.12600.005690.14680.01650.16720.02560.18740.0451(Fracture)0.19750.06Plot the data as engineering stress (MPa) versus engineering strain (mm/mm) anddetermine the ultimate strength of the steel

Step 1: Analysis of the Data Table introduces the engineering stress in MPa and strain in mm/mm. The…

Answered: Engineerin Engineerin Engineerin…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

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hings The following results were obtained during a tensile test to destruction on a mild steel test piece of diameter 15.96mm and 80mm gauge length Extension for a load of 40KN = 0.08mm Maximum load applied during test = 93kN Diameter at fracture = 12.85mm %3D Final length between gauge points 106mm From these results, determine: i) The modulus of elasticity for mild steel ii) The tensile strength iii) Percentage of elongation iv) Percentage of reduction in area
Ql- mm ,length 40 mm. The final diameter 18 mm and length 45 mm. Draw the force-extension diagram and determine the following : a) Ultimate tensile strength, c) Modulus of elasticity, d) Percent elongation and e) Percent reduction in area. The following data collected from tensile test of a metal bar diameter 20 b) The fracture strength, Force - kN 75 90 | 105 120 131 125 110 25 50 4.7 Extension-mm 0.02 0.04 0.06 0.20 - 0.60 1.60 3.00 4.0 Fracture
6. The following data were collected from a 12-mm diameter test specimen of magnesium (lo = 30.00 mm). After fracture, the total length was 32.61 mm and the diameter was 11.74 mm. Load (N) Al (mm) 0.0000 5,000 10,000 15,000 20,000 25,000 26,500 0.0296 0.0592 0.0888 0.15 0.51 0.90 1.50 (maximum load) 2.10 2.79 (fracture) 27,000 26,500 25,000 a) Plot the data as engineering stress versus engineering strain. b) Compute the modulus of elasticity. c) Determine the yield strength at a strain offset of 0.002. d) Determine the tensile strength of this alloy. e) What is the approximate ductility, in percent elongation? f) Compute the modulus of resilience. g) Compute from the data and plot true stress versus true strain diagram.
I want to solve the required squares and their units of measurement in regularity, please
Q2c) Listed in the table below is the tensile stress-strain data for different grades of steels. Utilizing the data given answer the three queries given below. Tensile Strain at Fracture Strength Strength Fracture Strength Modulus Material Yield Elastic (MPa) (MPa) (MPa) (GPa) А 410 1440 0.63 265 410 200 220 0.40 105 250 C 815 950 0.25 500 610 800 650 0.14 720 210 E Fractures before yielding 650 550 1) Which will experience the greatest percent reduction in area? Why? 2) Which is the strongest? Why? 3) Which is the stiffest? Why? B.
Engineering materials
Which of these cracks will lead to failure first from the table in the picture? And present your findings in a table format and comment briefly.
Cyclic Stresses (Fatigue); The S-N Curve 4. The fatigue data for a ductile cast iron are given as follows: Cycles to Failure Stress Amplitude [MPa (ksi)] 248 (36.0) 236 (34.2) 224 (32.5) 213 (30.9) 201 (29.1) 193 (28.0) 193 (28.0) 193 (28.0) 1 × 105 3 x 105 1 x 10⁰ 3 x 106 1 x 107 3 x 107 1 108 3 × 108 (a) Make an S-N plot (stress amplitude versus logarithm cycles to failure) using these data. (b) What is the fatigue limit for this alloy? (c) Determine fatigue lifetimes at stress amplitudes of 230 MPa (33,500 psi) and 175 MPa (25,000 psi). (d) Estimate fatigue strengths at 2 × 105 and 6 × 106 cycles.
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The following curve is a result of a tensile test. The gauge length was 2 in, and the specimen diameter was 0.505 in The fracture area was round with a diameter of 0.391 in. Remember to com 20000 Force (pounds force) 15000 10000 5000 0 16. The yield strength is 0 0.2% O 80,000 pl O 12.000 p 060,000 pl 85,000 p 0.1 0.2 Extension (inch) 0.3 0.4
7. The following data were collected from a 20-mm diameter test specimen of a ductile cast iron (lo = 40.00 mm). After fracture, the total length was 47.42 mm and the diameter was 18.35 mm. Load Al (N) (mm) 0.0000 25,000 0.0185 50,000 75,000 90,000 105,000 120,000 131,000 0.0370 0.0555 0.20 0.60 1.56 4.00 (maximum load) 7.52 (fracture) 125,000 a) Plot the data as engineering stress versus engineering strain. b) Compute the modulus of elasticity. c) Determine the yield strength at a strain offset of 0.002. d) Determine the tensile strength of this alloy. e) What is the approximate ductility, in percent elongation? f) Compute the modulus of resilience. g) Compute from the data and plot true stress versus true strain diagram.
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