the value of estimated shear yield stress,
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- (d) Figure Q2 (d) shows a typical aluminum stress-strain curve. A mechanical engineer has to carry out a tensile test for a rod of aluminum that can withstand an applied force of 20kN. To ensure safety, the maximum allowable stress on the rod is limited to 400 MPa, which is below the yield strength of the aluminum. The rod must be at least 380 cm long and must deform elastically not more than 0.8 cm when the force is applied. Design the appropriate rod.Uniaxial yield stress of a material is 260 MPa Determine shear yield stress using Von Mises criteria.A Copper specimen of circular cross-section is subjected to a tensile test. The data obtainedare:Length of Specimen = 329 mm;Diameter of Specimen = 39 mm;Load at Yield Point = 130 kN;Maximum Load = 219 kN;Load at fracture = 118kNDetermine the following:1) Find initial area of the test specimen2) Yield Stress3) Ultimate Stress4) Fracture stress5) Find the strain for 0.058mm elongation.6) Mark Yield stress, Ultimate stress and Fracture stress on a Stress-Strain diagram.
- Draw a typical stress vs strain tensile test curve for the following materials (two seperate graphs) and label the axis. A ductile metallic test specimen that is stretched to failure displaying a characteristic yield point and show the following parts on the curve. 1- Yield point 2- Ultimate Tensile Strength 3- Breaking point 4- Elastic Region 5- Plastic Region 6- Necking regionThe following stress-strain curve was prepared based on a tensile test of a specimen that had a circular cross-section. The gage diameter of the specimen was 0.25 inches and the gage length was 4 inches. The stress scale of the stress-strain diagram is given with the factor a = 10 ksi. Estimate: (a) The modulus of elasticity. (b) The ultimate strength. (c) The yield strength (0.2% offset). (d) The percent elongation at fracture. 2013 Michael Swanbom STRESS VS. STRAIN BY NC SA 7a bat Sat 2at at 0.05 STRAIN 0.01 0.04 0.06 0.08 0.02 0.03 0.07 0.09 STRESSFrom the shear stress - shear strain diagram shown below, all the following are true except: Shear Stress (MPa) 500 450 400 350 300 250 200 150 100 50 0. 0.2 0.4 0.6 0.8 1.2 1.4 1.6 1.8 Shear Strain (rad) Shear Stress-Strain Curve for Brass Select one: O a. Shear stress and Shear strain are 430 MPa and 1.72 rad at fracture. O b. Shear stress and Shear strain are 250 MPa and 0.05 rad at Yielding. Oc. The elastic zone ends at a shear stress of 400 MPa and shear strain of 1 rad Od. The ultimate Shear stress and Shear strain are 430 MPa and 1.72 rad.
- 1) Fic 7.11 ypical engineering strain behavior to ture, potnt F. The sille strength 75 s licated at point M. The circular insets et the geometry of the deformed 75 pecimen at various ts along the curve. Please indicate in the stress-strain diagram given below the stress levels that can fail the material due to creep rather than instantaneously. (Please note that this is NOT the same question as the one in the final homework you were given)The figure below shows the tensile engineering stress-strain behavior for a steel alloy. (a) What is the modulus of elasticity? (b) What is the yield strength at a strain offset of 0.002? (c) What is the tensile strength? Stress (MPa) 600 500 400 300 200 100 T I 0.00 Stress (MPa) 0.04 500 400 300 200 100 0.000 0.08 0.002 Strain 0.004 Strain 0.12 I 0.006 0.16 0.20 In the previous problem, A load of 85,000 N (19,100 lbf) is applied to a cylindrical specimen of the steel alloy that has a cross-sectional diameter of 15 mm (0.59 in.). (a) Will the specimen experience elastic and/or plastic deformation? Why? (b) If the original specimen length is 250 mm (10 in.), how much will it increase in length when this load is applied?1. What are the elastic modulus (E) and the Poisson's ratio () used to indicate? 2. Illustrate the differences between actual stress and engineered stress with strain, and also describe their underlying physical concepts. 3. If the engineering strain is 2% for a specific state of uniaxial stress, what is the real strain? Please solve for all in full detail and step by step
- Consider a cylindrical specimen of a steel alloy with 8.5 mm diameter and 80 mm long that is pulled in tension. Estimate the following mechanical properties using Fig. 1: a. Modulus of Elasticity and Resilience in MPa and psi b. Ultimate Tensile Strength in MPa and psi c. Fracture Strength in MPa and psi d. Ductility or % elongation at fracture in MPa and psi 2000 10³ psi MPa 300 2000 200 1000 100 0 0.000 0.005 0.010 0.015 Strain 0.020 0.040 0.060 Strain Fig. 1 Engineering Stress-Strain Curve Stress (MPa) 1000 0 0.000 Stress 0.080 300 200 100 0 Stress (10³ psi)1. For the stress-strain curve shown below, please estimate the properties indicated. (a) Fracture Strain Please do your work on a separate sheet of paper, and put your answers in the boxes on the right. Be sure to include the proper symbol and units. Stress Strain 70 60 50 Stress (ksi) 240 30 20 10 70 0 0.000 60 50 Stress (ksi) 40 20 10 KULL 0 0.000 0.010 0.050 0.100 Strain (in/in) Stress Strain 0.020 0.030 Strain (in/in) 0.040 0.150 0.050 (b) Ultimate Tensile Stress (c) Fracture Stress (d) Proportional Limit (e) Elastic Modulus (1) Yield Stress (g) Tensile Toughness (Modulus of Toughness) (h) Modulus of ResilienceBy application of tensile force, the cross-sectional area of bar Pis first reduced by 30% and then by an additional 20% Another bar Q of the same material is reduced in cross-sectional area by 50% in a single step by applying tensile force After deformation, the true strain in bar Pand bar Q will, respectively be (a) 0.5 and 05 (b) 0.58 and 0.69 (c) 0.69 and 0.69 (d) 0.78 and 1.00