In reference to the Tensile Test result for a steel sample shown in the figure below. If the original sample diameter was 12 mm and gauge length was 200 mm F 8아 70 6아 É 60- 30 2아 10 AL 12 14 20 16 18 elongation AL (mm) 9. 10 Hardness of this steel is approximately: force F (kN)
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- The rails of a railroad track are welded together at their ends (to form continuous rails and thus eliminate the clacking sound of the wheels) when the temperature is 60°F. What compressive stress ?? =6.5×10-6 /? is produced in the rails when they are heated by the sun to 120"F if the coefficient of thermal expansion a = the modulus of elasticity E = 30 × 106 psi?A copper specimen subjected to the Brinell Hardness Test using hardened steel ball indenter of diameter 12 mm and the indentation diameter 3.87 mm is measured using an optical magnifying lens with a ruler. Draw the Brinell Hardness Test setup neatly and determine the force applied on the specimen. Take Brinell Hardness Number for copper as 807. Calculate: 1-Surface Area of Indentation (in mm2) 2-Applied Force (in N)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 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 lengthThe 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 FractureConsider the tensile stress-strain diagram below. If samples 1 and 2 are different materials, which sample is tougher? A. Both samples are equally tough/ Albei monsters is ewe taai B. Sample 2 C. Sample 1
- – The weight is suspended from structural A–36 steel alloy and 2014-T6 aluminum alloy wires, eachhaving the same initial length of 10 ft and cross-sectional area of 7×10–3 in2 (Figure Q1). If the weight is increasedgradually, determine which wire yields first? Explain which material model and yield criterion you will use? Justifyyour choices3. Two brass rods (one square, one circular) are extruded; initial and final dimensions are provided below. Which part is expected to have the greater tensile strength? Which part is expected to have the greater ductility? Part Initial Dimensions Final Dimensions Cylindrical D=2.5 in D = 2.35 in. Square 4 × 4 in. 3.9 × 3.9 in.A standard Vickers Hardness test was conducted on a metal specimen.Determine the Vicker’s hardness number and the tensile stress of thespecimen. The applied load was 10 kgf, and the standard indentor left anindentation with diagonal d = 0.217 mm. Assume that the metal is hard.
- Mild steel 1 Young;s modulus 1219.5 Yield strain and stress (0.4101,500.08) Failure stress and strain :not able to find because the given data shows the experiment did not reach the failure point. if the material stress and strain does not reach a failure point ,what dose it means , does it means that the material is more stronger?Determine the tensile yield strength (0.2% offset) and the maximum strength of a metal alloy having the following tensile stress-strain diagram. Select one: The tensile yield strength Sy = 100 Mpa and the maximum strength Smax = 250 Mpa. The tensile yield strength Sy = 170 Mpa and the maximum strength Smax = 250 Mpa. The tensile yield strength Sy = 150 Mpa and the maximum strength Smax = 200 Mpa. The tensile yield strength Sy = 240 Mpa and the maximum strength Smax = 250 Mpa. The tensile yield strength Sy = 80 Mpa and the maximum strength Smax = 250 Mpa.A tensile test for a copper specimen has been performed and the following data are obtained. - Percentage of Elongation = 69 % - Percentage of Reduction in Area = 39 % - Final length after fracture = 35.5 mm - Final Diameter after fracture = 3.6 mm & - Ultimate stress = 396 MPa SOLUTION: iv) Initial Diameter (in mm) = v) Ultimate Load (in N) =