Recording HelpStress (MPa)600500-400-300200100-00.00Tell me what youExercise 2Stress (MPa)0.04500400300200-1000.000 0.002 0.004 0.006Strain0.08Strain0.120.160.20Consider a cylindrical specimen of a steel alloy 10.0mmintension. Determine, its elongation when a load of20,000 N is applied.

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Answered: Consider a cylindrical specimen of a…

Consider a cylindrical specimen of a steel alloy 10.0 mm in diameter 75 mm long that is pulled tension. Determine, its elongation when a load of 20,000 N is applied. and in

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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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)
Desuon T EHide blocks Determine the Young's modulus of a metal alloy having the following tensile stress-strain diagram. ot yet nswered Stress-Strain Diagram Marked out of 5.00 1200 P Flag question 1000 800 1000 500 800 600 400 400 200 0.005 0.01 0.015 002 0.025 0.03 0.05 0.1 0.15 Strain (mm/mm) 0.2 0.25 Select one: O E = 1000 Mpa O E= 250 Gpa OE = 120 Gpa E = 50 Gpa E = 200 Gpa Stress (Mpa)
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
The modulus of rigidity (shear modulus) of a circular Cast Iron specimen subjected to a torsion test is: During Elastic Zone Angle of twist (°) TD Specimen Specimen Torque 2J Radius Length (N.m) 2.5 15 rp Y = L 3 mm 50 mm 10.8 20 Select one: O a. 88.39 GPa. O b. 13.586 GPa. O c. 236.73 MPa. O d. 236.73 GPa.
Q1- Answer only one a or b a- Listed with the drawing (only) the types of phase diagrams and pointed all phases. b - Listed the types of nondestructive test method and explains one of methods which suitable for internal defects and give the advantage and dis advantage of these methods. Q2- The following data was obtained from a tensile test on Load (kN ) Extension (mm) 0.2 16 32 0.4 56 72 95 110 132 142 140 135 10.0 12.0 0.7 0.9 1.5 2.5 5.0 8.5 A specimen of 10mm diameter and gauge length 60mm. Determine the following:- 1-Draw the stress-strain curve. 2- Pointed on the curve only proof stress, ultimate tensile strength. Q3- Only Answer two branches. a- calculate the atomic backing factor for Bec. b- 10 mm diameter of Brinall hardness ball when the load 500 kgr and HB= 300 determined the diameter of indenter. C- The diminutions of the low alloy steel impact test sample as shown in fig below and the notched depth is 1.5 mm determine the impact toughness if the impact energy absorption at…
The modulus of rigidity (shear modulus) of a circular Cast Iron specimen subjected to a torsion test is: During Elastic Zone Angle of twist (°) TD Specimen Torque (N.m) Specimen T = 2) Radius Length 2.5 15 rp 3 mm 50 mm 10.8 20 Select one: O a. 13.586 GPa. O b. 236.73 GPa. O c. 236.73 MPa. O d. 88.39 GPa. web.whatsapp.com. now +961 76 301 678 balorfo 9-20 DM
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
The following data were collected from a 12 mm diameter test specimen of Magnesium. LOAD (N) 0 5000 10000 15000 20000 25000 26500 27000 26500 25000 GAUEGE LENGTH (mm) 30.000 30.0296 30.0592 30.0888 30.15 30.51 30.90 31.50 (maximum load) 32.10 32.79 (fracture) After the fracture, the gauge length is 32.61 mm and the diameter is 11.74 mm. a) What is the elastic modulus? b) Percent elongation at fracture? c) Percent elongation after fracture? d) What is the Poisson's ratio? e)Draw the engineering stress-strain diagram corresponding to the values in the table. Call this plot I. Now consider this experiment is repeated at a higher temperature with an identical sample. Draw the new engineering stress-strain diagram, call it plot Il and highlight the differences (on the same graph) between I and II.
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
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.
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 EXAMPLES
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.