) The data below represent the stress-strain relationship of a material; use Newton's divided difference method to estimate the stress corresponding to a strain of (290 10 m/m) a, 10°N/m 87.5 96.6 176 263 351 e, 10m/m 153 204 255 306 357
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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)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 numberIn a forming process the diameter of the specimen is 1 cm and its length changes from 50 mm to 60 mm. Show the maximum force value on the graph of Force vs. True Strain plotted in Excel Program or any plotting program not by hand using the flow stress equation given below. σ= Kεn and σ=F/A Where the parameters are σ : True Stress in MPa K: Strength Coefficient, K=210 in MPa ε : True Strain, ε=ln (l/l0) n: Strain Hardening Exponent, n=0.27 F: Force in 10 A : Area in mm2 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 the solution section.
- The following data was obtained as a result of tensile testing of a standard 0.505 inch diameter test specimen of magnesium. After fracture, the gage length is 2.245 inch and the diameter is 0.466 inch. a). Calculate the engineering stress and strain values to fill in the blank boxes and plot the data. Load(lb) Gage Length (in) Stress (kpsi) Strain 0 2 1000 2.00154 2000 2.00308 3000 2.00462 4000 2.00615 5000 2.00769 5500 2.014 6000 2.05 6200 (max) 2.13 6000 (fracture) 2.255 b). Calculate the modulus of elasticity c). If another identical sample of the same material is pulled only to 6000 pounds and is unloaded from there, determine the gage length of the sample after unloading.An element of material in plane strain is subjected to strains εxx=480×10–6 , εyy=70×10–6 and τxy=470×10–6 . Determine the strains for an element rotated through an angle θ=80⁰, the principal strains and the maximum shear strains using (a) transformation relations, (b) using Mohr’s circle of strains.Based in the table 1. (need to answer question 4,5,6)1) Develop a best-fit equation for the relationship between stress and strain. Employ Naïve–Gauss elimination method whenever necessary.S=___28.65___+___294.43___e + ___235.22____e2 2) Determine the coefficient of determination for the equation. R2 =___0.636____ 3) Calculate the stress value to the most accurate value at strain value 0.53.s = ___250.76____Pa4) The yield point is the point on a stress–strain curve that indicates the limit of elastic behaviour and the beginning of plastic behavior. In this case, the yield point occurs at a stress value of 80. Determine the corresponding strain value at the yield point. In any relevant method, use a stopping criterion of 0.05% e =_______ 5) The ultimate strength is the maximum point on the stress–strain curve. This corresponds to the maximum stress that can be sustained by a structure in tension. Compute the ultimate strength point of the polymeric material (strain value that gives…
- Using the equations of strain in isotropic elasticity, find the stress tensor that cause the following strain state determined from an experiment (material has Young’s modulus of 190 GPa): ?? = 0.001 0.001 0 0.001 −0.002 0.0005 0 0.0005 0shear moduli is 70GPa.A tensile test was conducted on a mild steel& the following data was obtained as follows. Diameter of the steel bar = 3 cm,Gauge length of the bar = 20 cm. Load at elastic limit = 250 kN . Extension at a load of 150kN = 0.21 mm . Maximum load = 380 kN . Total extension = 60 mm . Diameter of the rod at failure = 2. 25 cm Determine: (a) Young’s modulus (b) stress at elastic limit (c)the percentage of elongation (d) Percentage decrease in area.In our previous topic regarding strain, repeat the problem on the strain using thefollowing parameters:F = 150000 +/- 150 lbfSide length = 1.2 +/- 0.025 inModulus of Elasticity = 25000 +/- 11.5 ksiDetermine the relative error. Is it above or below the accepted 5% error?
- 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. Question 1 ;Determine the elastic energy absorption capacity (in N.mm) of that specimen. Question 2; Determine the plastic energy absorption capacity (in N.mm) of that specimen.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. - Determine the true stress (in MPa) at yield point. - Determine the true stress (in MPa) at point of ultimate strength. - Determine the true stress (in MPa) at fracture point. - Determine the true strain (in mm/mm) at yield point. (Use at least five decimal units) - Determine the true strain (in mm/mm) at point of ultimate strength. (Use at least five decimal units) - Determine the true strain (in mm/mm) at fracture point. (Use at least five decimal units)A 316 steel plate sample is under the following loading condition:?(sigma) = 90 40 0 40 −50 0 0 0 0 MPaa. Plot the shear and normal stresses on a given oblique plane (rotated plane) on a graph (stress vs. angle) and show the max. shear plane and the principal planes.b. Draw the Mohr circle for the loading state above to scale (using a proper computer program or using a compass with correct scales on the axes) and show the angles you found in (a).c. If the steel sample has a Young’s modulus of 193 GPa, compute the dimensional changes of the plate under the above loading condition