d.Modulus of resiliencein N-mm/mm3 e.Modulus of toughnessin N-mm/mm3
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d.Modulus of resiliencein N-mm/mm3
e.Modulus of toughnessin N-mm/mm3
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- The stress-strain diagram of a reinforcement steel having a cross-sectional diameter of 12 mmdiameter and 100 mm gage length is determined after its tensile strength test as follows. Based on the stressstrain diagram determine the followings properties of the material (Poisson’s ratio of the material is 0.32) : a) Modulus of elasticityb) Yield strengthc) Toughnessd) Resiliencee) Shear modulusf) Bulk modulusg) Ductility as described bypercent change in lengthA tension test was performed on a steel specimen having an original diameter of 12.5 mm and gauge length of 50 mm. The data is listed in the table. Plot the stress–strain diagram and determine approximately the modulus of elasticity, the yield stress, the ultimate stress, and the rupture stress. Use a scale of 25 mm = 140 MPa and 25 mm = 0.05 mm/mm. Redraw the elastic region, using the same stress scale but a strain scale of 25 mm = 0.001 mm/mm.The strain rosettes shown in Figures P13.50 were used to obtain normal strain data at a point on the freesurface of a machine component.(a) determine the strain components e x, e y, and γ xy at the point.(b) determine the principal strains and the maximum in-planeshear strain at the point.
- The following data were obtained from the tensile test of Aluminum alloy. The initial diameter of testspecimen was 0.505 inch and gauge length was 2.0 inch. Plot the stress strain diagram and determine(a) Proportional Limit (b) Modulus of Elasticity (c) Yield Stress at 0.2% offset (d) Ultimate Stress and(e) Nominal Rupture Stress.Please annotate the attached stress strain graph of 0.9% carbon steel to show the different phases. From the graph work out the following: 1) modulus of elasticity 2) tensile strength 3) the ductility in % elongation 4) yield strength at a strain offset of 0.002The following data are obtained from a tensile test of a copper specimen. - The load at the yield point is 150 kN. - Length of the specimen is 24 mm. - The yield strength is 78 kN/mm2. - The percentage of elongation is 42 %. Determine the following iii) Stress under an elastic load of 14 kN, iv) Young's Modulus if the elongation is 1.7 mm at 14 kN and (v) Final diameter if the percentage of reduction in area is 26 %. I wont answer for this three questions only: 1)Final Area of the Specimen at Fracture (in mm) 2)Final Diameter of the Specimen after Fracture (in mm) 3)Initial Cross-sectional Area (in mm2)
- 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.A mild steel bar went through a tensile test conducted at the University of San Carlos Mechanical Engineering Laboratory. The following data was obtained from the test: i. Diameter of the steel bar = 30 mm ii. Gauge length of the bar = 200 mm iii. Load at the elastic limit = 250 kN iv. Extension at a load of 150 kN = 0.21 mm v. Maximum load = 380 kN vi. Total extension = 60 mm vii. Diameter of the rod at the failure = 22.5 mm Determine the Young’s modulus, the stress at the elastic limit, the percentage elongation, and the percentage decrease in area.Draw the tensile curves of the following materials, taking into account the appropriate elongation at break, modulus of elasticity and tensile stress, on the side stress-strain graph. a)Low carbon steel (Ecelik=200Gpa),(Syield=300MPa),(Stensile=400MPa) b)High carbon steel (Esteel=200GPa),(Stensile=700MPa) c)Aluminum (Ealuminum=70GPa), (Syield=200MPa),(Stensile=300MPa) d)Cast iron (Ecast iron=120GPa),(Stensile=200MPa)
- The strain rosettes shown in Figures P13.50 were used to obtain normal strain data at a point on the free surface of a machine component. In each of Problems P13.50(a) determine the strain components e x, e y, and γ xy at the point. b) determine the principal strains and the maximum in-planeshear strain at the point.The following data were obtained from a standard tensile stress test on a mild steel specimen. Graph the result and determine proportional limit stress and Young’s Modulus.The data shown in the table below were obtained from a tensile test of high-strength steel. The test specimen had a diameter of 13mm and a gage length of 50mm. At fracture, the elongation between the gage marks was 3.0mm and the minimum diameter was 10.7mm. Plot the conventional stress-strain curve for the steel and determine the propotional limit, modulus of elasticity (i.e the slope of the initial part of the stress-strain curve), yield stress at 0.1% offset, ultimate stress, percent elongation in 50mm, and percent reduction area. TENSILE-TEST DATA Load(kN) Elongation(mm) 5 0.005 10 0.015 30 0.048 50 0.084 60 0.099 64.5 0.109 67.0 0.119 68.0 0.137 69.0 0.160 70.0 0.229 72.0 0.259 76.0 0.330 84.0 0.584 92.0 0.853 100.0 1.288 112.0 2.814 113.0 Fracture