---------------- is a relatively new technology that enables the complicated shaping of ultra high strength steels which is not achievable with standard cold stamping procedures.
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---------------- is a relatively new technology that enables the complicated shaping of ultra high strength steels which is not achievable with standard cold stamping procedures.
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- The most common method for production of steel is Group of answer choices Slip casting Continuous casting Hot pressing Investment castingi need quickly please please please Which of these cast iron would allow the greatest amount of plastic deformation before fracture . Nodular cast iron Grey cast iron White cast iron None of them can be worked. If the diameter of steel specimen stretched by tension load is doubled , then it’s tensile strength will be Halved Remain unaffected Doubled Become four times Temperature is proportional with Percent of elongation Tensile strength Modulus of elasticity All above The tensile strength of a steel specimen that have a 182 HB , equals to 490 MPa 910 MPa 627 MPa Not given…Is high hardenability desirable in steels? Explain your answer briefly [do not use chatgpt]
- A batch of casted mild steel has a modulus of elasticity of 200 GPa and a yield strength of 250 MPa. Calculate for its modulus of resilience. After cold working the steel, the yield strength increases to 310 MPa. Calculate for the percent reduction in the average grain diameter given σo = 70 MPa and k = 0.74.. Properties of steel can be altered by applying a variety of heat treatments. Pleaseexplain full annealing for steel.Decode the characteristics of a 6063 T831 aluminum.
- d) No increase Which of the following is not a property of steel materials? C a) Homogeneous and isotropic b) Linearly elastic stress-strain behavior c) Recyclable d) Fire-resistantSteel, Brass, and Copper rods are connected as shown in the figure. Initially, the temperature was 15 degrees Celsius and the stress on the bars is zero. Eventually, the temperature increased to 25 degrees Celsius. Determine the total deformation on the brass. Steel Brass Copper Est = 200 GPa 12(10-)/°C apr Ebr Ecu 17(10-)/°C 120 GPa 100 GPa %3D ast = 21(10-6)/°C acu Acu = 515 mm? |Ast 200 mm2 Abr = 450 mm2 300 mm -200 mm 100 mm O -0.0109mm O 0.0241mm O -0.0241mm O 0.0109mm oooOWhich one of the following processes causes a dramatic increase in the hardness of a steel part? A) quenching B) tempering (c) normalizing D full annealing
- The aluminum (E=15x10^10psi, α=11.6x10^-6/°F) shell is fully bonded to the brass (E=10.6x10^6psi, α=12.9x10^-6/°F) sore, and the assembly is unstressed at a temperature of 78°F. Considering only axial deformations, determine the stress when the temperature reaches 180°F (a) in the brass core (b) in the aluminum shellThe assembly is composed of a steel shell and an aluminum core that has been welded to a rigid plate. The gap between the plate and the aluminum is initially 1- mm. If the assembly's temperature is reduced by 180°C, determine (a) the final axial stresses in each material and (b) the deflection of the rigid bar. To support your response, draw a deformation diagram with appropriate labels. Use the following properties: Aluminum core Steel shell Diameters (mm) d = 15 mm do = 30 mm d₁ = 20 mm E (GPa) 70 200 2 m a (/°C) 22 x 10-6 12 x 10-6Task (3) you are asked to perform tensile test on specimens of two different materials (A and B) and you obtained the stress-strain diagram of the two specimens as shown in Figure 2: 400 350 300 250 Material A 200 150 100 50 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 Strain (mm/mm) 50 Material B 30 20 10 0.02 0.04 0.06 0.08 0.1 Strain (mm/mm) Figure 2: Tensile Test Analysis Stress (MPa) Stress (MPa)