Explain in detail the effect of Temperature on the Elastic Modulus and Yield Strength ofEngineering materials.
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Explain in detail the effect of Temperature on the Elastic Modulus and Yield Strength of
Engineering materials.
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- strength of material subject columns chapter about reduction factor, in working stress1. The tensile strength vs. temperature curves of aluminum are plotted below. Explain the meaning of these two curves. 400 Ultimate strength 300 Yield strength 200 100 -200 -150 -100 -50 50 100 150 200 250 Temperature, °C Strength, kPa3. Discuss how to determine the following material properties of structural steel using an engineering stress-strain curve: a. Modulus of elasticity b. Yield point and yield strength c. Ultimate strength 4. Discuss the effect of high temperature on material properties of structural steel.
- Use the engineering stress strain diagram provided below to answer parts (A) to (H) below (the stress-strain diagram has already been drawn for you): stress strain diagram 400 350 0 300 0 250 0 200 0 150.0 100.0 50.0 O 05 0 1 0.15 strain A. Determine the tensile strength of this alloy. B. show the elastic, plastic and total strain on the diagram stress (Mpa)Discuss the concept of strain hardening in materials?The following figure shows the tensile stress-strain curve for a plain-carbon steel. 600 80 500 MPa 600 10 psi 60 80 400 400 60- 300 40 40 200 200 20 20 100 0.005 0.05 0.10 0.15 Strain (a) What is this alloy's tensile strength? MPа (b) What is its modulus of elasticity? GPa (c) What is the yield strength? i MPa Stress (MPa) Stress (10 psi)
- Two bars of different materials and same size are subjected to the same tensile force. If the bars have unit elongation in the ratio 2: 5, then the ratio of modulus of elasticity of the two materials will be (a) 2:5 (c) 4 :3 (b) 5:2 (d) 3:4Question 5 A steel component is subjected to a biaxial state of stress in which both stresses are tensile in nature, and having magnitudes 100 and 60 MPa. The plane at which the resultant stress has maximum obliquity with the normal (in degrees) isExplain how dislocations have paradoxically rationalized why metals can be bothmechanically weak and strong. How have dislocations helped to explain work hardening of metals.
- Please annotate the attached image to highlight the different phases of a polymer stress/strain graph. From the attached graph, calculate the following: 1-modulus of elasticity 2-tensile strength 3-the ductility in % of elongation 4- yield strength at a strain offset of 0.002Q2c) Listed in the table below is the tensile stress-strain data for different grades of steels. Utilizing the data given answer the three queries given below. Material Yield Tensile Strain at Fracture Elastic StrengthStrengthFractureStrengthModulus (MPa) (MPa) (MPa) (GPa) A 410 1440 0.63 265 410 В 200 220 0.40 105 250 C 815 950 0.25 500 610 D 800 650 0.14 720 210 E Fractures before yielding 650 550 1) Which will experience the greatest percent reduction in area? Why? 2) Which is the strongest? Why? 3) Which is the stiffest? Why?Testing a round steel alloy bar with a diameter of 15 mm and a gauge length of 250 mm produced the stress–strain relationship shown in Figure Determinea. the elastic modulusb. the proportional limitc. the yield strength at a strain offset of 0.002d. the tensile strengthe. the magnitude of the load required to produce an increase in length of 0.38 mmf. the final deformation, if the specimen is unloaded after being strained by the amount specified in (e)g. In designing a typical structure made of this material, would you expect the stress applied in (e) reasonable? Why?