AE321_HW6

62% D 18:54 From the shear stress - shear strain diagram shown below, all the following are true except: Shear Stress (MPa) 450 400 350 300 250 200 150 100 50 0.2 04 1.2 1.4 1.6 1.8 Shear Strain (rad) Shear Stress-Strain Curve for Brass Select one: a. The ultimate Shear stress and Shear strain are 430 MPa and 1.72 rad. b. Shear stress and Shear strain are 250 MPa and 0.05 rad at Yielding. c. Shear stress and Shear strain are 430 MPa and 1.72 rad at fracture. d. The elastic zone ends at a shear stress of 400 MPa and shear strain of 1 rad II

2) Calculate the following: A) Estimate the modulus of resilience. It uses the expanded scale. B) Estimate the modulus of toughness. It uses the regular scale with each block being 0.04 in width. 8 σ (ksi) 80 70 60 50 40 30 20 10 0 0 0 Stress vs. Strain Uses Regular Scale Uses Expanded Scale 0.04 0.08 0.12 0.16 0.20 0.24 0.28 0.0005 0.001 0.0015 0.002 0.0025 0.003 0.0035 € (in./in.)

Problem 2 The fatigue data for a steel alloy are given as follows. Stress amp, 470 440 390 350 300 290 290 290 S(MPa) Cycle to failure (N) 1.0x104 3.0x10* 1.0x105 3.0x105 1.0x106 3.0x10 1.0x107 1.0x10 f) Make an S-N plot of the data. g) Select S-N curve, select add trend line and in trend line options click power and display equation on the chart. h) Determine the material constants A and B by fitting this plot to the equation S=AN°. i) Is A and B the same as in problem 1? Why or why not? j) Determine the Fatigue limit or Fatigue strength whichever is applicable for this metal. k) Estimate fatigue strength at 2.0x 10* and 6.0x105 cycles.

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)

What value can not be found from the stress-strain graph? Please select one:a. Modulus of elasticityb. Disgracec. Toughnessd. Poisson ratioe. Yield strength

The data recorded during a tensile test on a metal alloy bar is shown in table below. Test piece dimensions: diameter 12.0 mm, gauge length 60mm. Gauge Length, mm Load, KN 0 10 30 50 70 75 80 81 81 76 60.000,+ 60.030 d) Fracture strength e) Modulus of resilience 60.105 60.195 60.270 61.125 63.500 64.500 66.000 67.695 Convert the load-length data to engineering stress and strain. Plot the stress-strain curve and determine the following: a) Modulus of Elasticity b) Yield stress for an offset 0.2% c) Ultimate strength

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In the graph, the module of toughness can be shown to be represented by: A) The area associated with necking B) The rectangular area label yielding C) The triangular area in the elastic region D) The sum of the entire shaded area under the curve

What is the materials ultimate strength in MPa

1. The data for a tensile test is given in table 1. The sample has a diameter of 4.5 mm and a length of 250mm. Mass Force Stress Elongation Strain kg N MPа mm 344 0.84 405 1.12 458 1.62 469 2.42 458 2.98 421 3.65 Table 1 Maximum elastic strain occurs when a force of 3.37 kN is applied. a) Calculate the yield strength of the material. b) What is the value of Young's modulus of this material c) Calculate the true ultimate tensile strength of the material. d) What will the uniform strain be? e) Compare the resilience of this material with spring made of material that has a resilience of 300KJ/m³. Would you use the material of the tensile test for a spring and explain why.

1 - A tensile test is carried out on a specimen of mild steel of gauge length 40 mm and diameter 7.42 mm. The results are: Load (kN) 10 17 25 30 34 37.5 38.5 36 Extension (mm) 0 0.05 0.08 0.11 0.14 0.20 0.40 0.60 0.90 At fracture the final length of the specimen is 40.90 mm. Plot the load/ extension graph and determine (a) The modulus of elasticity for mild steel, (b) The yield stress, (c) The ultimate tensile strength, (d) The percentage elongation 2 - An aluminum alloy specimen of gauge length 75 mm and of diameter 11.28 mm was subjected to a tensile test, with these results: Load (kN) 2.0 6.5 11.5 13.6 16.0 18.0 19.0 20.5 19.0 Extension (mm) 0 0.012 0.039 0.069 0.080 0.107 0.133 0.158 0.225 0.310 The specimen fractured at a load of 19.0 kN. Determine (a) the modulus of elasticity of the alloy, (b) The percentage elongation. (c) The ductility in term of reduction of area (d) Fracture stress (e) Tensile strength

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.

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

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. Material Yield Strength Tensile Strain at Fracture Elastic Strength Fracture Strength (MPa) Modulus (MPa) (МPa) (GPa) A 410 1440 0.63 265 410 B 200 220 0.40 105 250 815 950 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?

A cylindrical specimen of brass that has a diameter of 20 mm, a tensile modulus of 110 GPa, and a Poisson's ratio of 0.35 is pulled in tension with a force of 40, 000 N. If the deformation is totally elastic and the original length is 120 mm, what is the shear modulus. G. of the specimen in GPa? A 20.74 40.74 50.74 30.74

Q1- The following data was obtained tensile test ɔn a specimen of 10 mm diameter and gauge length Lo=50 mm: 0 38 | 76.2 92.7 107 149 160.4 | 155 142 124.7 load(KN) | elongation(mm) 0 0.02 0.12 0.25 0.50 2.03 3.55 4.06 5.1 5.84 Using the graphic paper supplied, Plot engineering stress-strain curves, then determine the (ay, ey), (ou , eu), (of,, ef), modulus of elasticity, then detected in the draw ((stroin hardening (m)), (Necking), (uniform plastic region and non-uniform plastic region) and (elastic & plastic) region))?

Ultimate strength, s, Yield strength, s, Percent Material SAE no. Condition ksi MPa ksi MPa elongation 36 43 48 64 393 57 65 75 296 331 Annealed 1020 1020 448 517 36 20 Hot rolled 1020 Cold drawn 441 Annealed Hot rolled Cold drawn 75 90 97 517 621 669 51 60 352 414 565 30 25 16 1040 1040 1040 82 19 22 24 641 WQT 700 WQT 900 WQT 1100 WQT 1300 127 118 107 93 90 1040 876 1040 814 621 552 434 1040 738 80 1040 87 600 63 32 372 25 54 141 129 103 70 1080 Annealed 89 614 OQT 700 OQT 900 OQT 1100 OQT 1300 1303 1234 1000 972 889 710 483 12 13 17 189 179 1080 1080 1080 145 1080 117 807 23 352 26 Annealed Cold drawn 600 772 51 95 1141 87 14 655 1186 1141 112 172 129 97 1331 οQT 700 OQT 900 OQT 1100 OQT 1300 193 146 1141 1007 800 648 889 669 15 20 1141 1141 116 1141 94 68 469 28 Annealed 60 414 26 655 1593 1289 1014 4140 4140 4140 95 231 1462 1193 903 12 15 212 OQT 700 OQT 900 OQT 1100 OQT 1300 173 187 147 4140 131 18 4140 118 814 101 696 23 724 40 238 Annealed 105 276 17 5160 5160 1641 9. OQT 700…

Name: Surname: Student ID: ME 216 MECHANICS OF MATERIALS QUIZ #2 6) QUESTION σ (MPa) 25.03.2019 490 420 350 280 210 140 70 (mm/mm) 0 0.002 0.004 0.006 0.008 0.010 A tension test was performed on a 7075-T6 aluminum tensile test specimen. The results are given in the figure above. According to this data, find: a. Young modulus of the material, b. Yield stress of the material (be careful, there is no any sharp yield point), c. Modulus of the resilience, d. If the material is stressed up to 420 MPa stress, what is the permanent elongation remained in the material if its length is 90 mm. e. For a 20 mm diameter bar made up of from the same material, what is the maximum load that can be applied without plastic deformation. ANSWER E=A ٤ E = 70.106 σ = 70 MPa at ε=0,0009 = 77,776 Pa 0.0009

- Lo° do A solid cylindrical bar is subjected to tensile test. The initial diameter of the bar is 60 mm. The change in diameter (Ad) after the test is 0,01438 mn The applied load is 130 kN. The inital length of the bar is 240 mm. The change in length (Al) after the test is 0,1987 mm. Evaluate the axial strain. O A) 0,1987 O B) 0,01438 C) 0,000239 O D) 0,000275 E) 0.000828

stress 0.02 0.018 0.016 0.014 0.012 0.01 0.008 0.006 0.004 0.002 0.05 0.1 0.15 0.2 0.25 0.3 0.35 Determine : 1- Stress, Stratn care 2. elastic Zone 3. yield sbresscoys 4- plastic Zone 5- UItimate Stress Cous 6- Elongation 7- propotionaL stress cops 8-fructurc stress coF)

Question Comp If the element is rotated by 60 degree clockwise, find ? 20 MPa 50 MPa 150 00 -20.00) 157 07 50.00 40 00 -30.00 -20.00 100 10.00 20.000V -7.016 -20.0 0 000 20 0CDO.CO 40.00 1 a. 19.3 MPa Ob. 38.7 MPa Oc-19.3 MPa O d.-38.7 MPa P •A:.0 E (20.18.-31.65) -32.02 7500, 00 30 02 (29 02 0105) سماء صافية 37°C - 0

Calculate: 1. stress 2. strain 3. diameter of wire corrected to zero error

The choice of the blank: 607080901051151301401802002202402552800.110.130.150.17

Q4: A- The following data was obtained from a tensile test on a specimen of 10mm diameter and gauge length Lo=50mm. Load (KN) 0. 38 76.2 92.7 107 149 160.4 155 142 124.7 Elongation (mm) 0.025 0.152 0.254 0.508 2.032 3.556 4.064 5.1 5.842 Using the graphic paper supplied, plot the load-elongation diagram and determine: 1-0.2% proof stress. 2- The Young's modulus. 3- Ultimate stress. 4- Fracture stress.

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

Graph the following data on a stress-strain curve. Identify/Determine the following: Elastic Limit, Plastic Limit, Proportional Limit, Upper Yield Point, Lower Yield Point, Ultimate Strength, Rupture Strength, and Modulus of Elasticity. Engr. True Stress Strain o, MPa 102.3 0.00050 204.7 0.00102 293.5 0.00146 272.9 0.00230 264.0 0.00310 268.9 0.00500 267.6 0.00700 264.0 0.01000 318.6 0.0490 372.0 0.1250 394.4 0.2180 395.0 0.2340 384.2 0.3060 360.4 0.3300 327.5 0.3480 290.0 0.3600 266.8 0.3660

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From the data what is the materials ultimate strength in MPa

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