i- Modulus of elasticity 2- Yield strength 3- Tensile strength
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Q: Define the term Absolute Maximum Shear Strain?
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Q: 25 20 15 10 5 0.3 0.6 0.9 1.2 1.5 Elongation (mm) Force (kN)
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- In a forming process the length of material changes from 50 mm to 60 mm. Show the maximum true stress value on the graph using the flow stress equation. o = Ke" Where the parameters are o: True Stress in MPa K: Strength Coefficient, K = 200 ɛ : True Strain, s = In () n: Strain Hardening Exponent, n = 0.27Name: 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 0.0009Q1- 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))?
- 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.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 strengtha . 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)
- STATICS AND STENGITH OF MATERIALS Question: Load (kip) Elongation (in.) 0 0 A tension test was performed on a steel specimen having an original diameter of 0.503 in. and gage length of 2.00 in. The data is listed in the table beside. 1.50 0.0005 4.60 0.0015 8.00 0.0025 11.00 0.0035 11.80 0.0050 1. Plot the stress-strain diagram (better to use a software like MS Excel) and determine approximately 11.80 0.0080 12.00 0.0200 16.60 0.0400 a. the modulus of elasticity, 20.00 0.1000 21.50 0.2800 b. the yield stress, 19.50 0.4000 18.50 0.4600 c. the ultimate stress, d. the fracture stress. Use a scale of 1 in. = 20 ksi and 1 in. = 0.05 in.>in. 2. Redraw the elastic region (on the same graph), using the same stress scale but a strain scale of 1 in. = 0.001 in.>in.30. In a typical tension test a dog-bone shaped specimen is pulled in a machine. During the test, the force F needed to pull the specimen and the length L of a gauge section are measured. This data is used for plotting a stress-strain diagram of the material. Two definitions, engineering and true, exist for stress and strain. The engineering stress F de and strain e are defined by oe F %3D and A0 L-Lo where Lo and Ao are the initial gauge 6, = Lo length and the initial cross-sectional area of the specimen, respectively. The true stress o, and strain e are defined by o, 는는 and q %3D = In A0 Lo The following are measurements of force and gauge length from a ten- sion test with an aluminum specimen. The specimen has a round cross sec- tion with a radius of 0.25 in. (before the test). The initial gauge length is 0.5 in. Use the data to calculate and generate the engineering and true stress- strain curves, both on the same plot. Label the axes and use a legend to identify the curves. Units:…Data taken from a stress-strain test for a ceramic are given in the table. The curve is linear between the origin and the first point. Figure σ (ksi) 0 33.2 45.5 49.4 51.5 53.4 € (in./in.) 0 0.0006 0.0010 0.0014 0.0018 0.0022 < 1 of 1 Part A Plot the stress-strain diagram. (Figure 1) +SGA0 No elements selected σ(ksi) Submit Request Answer 50 40 30+ 20 10 0.5 1.0 1.5 2.0 2.5 €x 10-³(in./in.) ? Select the elements from the list and add them to the canvas setting the appropriate attributes. Press [CTRL+M) to get to the main menu.
- 650 600 550 500 450 400 350 300 250 200 150 100 50 0.002 0.004 0.006 0.008 0.01 0.012 0.014 0.016 0.018 0.02 Strain mm/mm The graph above shows the stress-strain relationship of a steel bar under tension test. The steel bar diameter = 8 mm and the bar gauge length = 100 mm. Determine the following: If the specimen is loaded to 550 MPa and then unloaded. What would be the modulus of resilience of the sample after reloading? Stress (MPa)640 480 160 0.002 0,004 0.006 0.008 0.010 Strain (mm/mm) A tensile test specimen of Aluminum alloy having a gauge length of 280 mm was tested to fracture. Stress and strain data obtained during the test are shown in the attached Figure. **Answers are approximate - do your best to read the graph. Determine: (a) the modulus of elasticity in GPa (4sigs). Enter this value into D2L answer box. (b) the proportional limit in MPa (3sigs). (c) the ultimate strength in MPa (3sigs). (d) Ductility of material based on percent elongation. (3sigs) Your Answer: Stress (MPa)Question: In a forming process the length of material changes from 50 mm to 60 mm. Show the maximum true stress value on the graph (True Stress vs. True Strain) plotted in Excel Program or any plotting program not by hand using the flow stress equation given below. o = Kɛ" Where the parameters are o: True Stress in MPa K: Strength Coefficient, K = (200 + N) in MPa, N is Your Rank Number in the Class List ɛ : True Strain, s = In ('/1,) n: Strain Hardening Exponent, n = 0.27 Notes: - Never change the template and only write your own identity data. - Your N value is 1 - Take at least 3 digits after the decimal point. - Specify the units. - You can use Excel or any plotting program not by hand then paste your graph into then solution section.