Mechanical properties of materials H.W.: The following data where obtained during the tensile test of mild steel circular bar 12.75 mm diameter and 203.2 mm gauge length. Determine the following:
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A: solutionArea=π4×D2=π4×142=49π mm2length=L=50 mmstrain=elongationlengthstress=Load area
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- 1.5-6 The data shown in the table were obtained from a tensile test of a metal specimen with a diameter of 0.500 inch and a gage length (the length over which the elongation is measured) of 2.00 inches. The specimen was not loaded to failure. a. Generate a table of stress and strain values. b. Plot these values and draw a best-fit line to obtain a stress-strain curve. c. Use the slope of the best-fit line to estimate the modulus of elasticity. Load (kips) PI223 SIN 0 2.5 3.5 10 11.5 12 Elongation (in.) 0 0.0010 0.0014 0.0020 0.0024 0.0036 0.0044 0.0050 0.0060 0.0070 0.0080 0.0120 0.01802.Discuss elastic region in a tensile test and shear test. It is always linear. What is engineering stress, strain ? Give Hooke's law, for tensile and shear deformation with corresponding modulus.A steel bar, whose cross section is 0.60 inch by 4.10 inches, was tested in tension. An axial load of P = 31,025 lb. produced a deformation of 0.115 inch over a gauge length of 2.10 inches and a decrease of 0.0080 inch in the 0.60-inch thickness of the bar. a. Determine the lateral strain. b. Determine the axial strain. c. Determine the Poisson’s ratio v. d. Determine the decrease in the 4.05-in. cross-sectional dimension (in inches).
- Q3 / An element in plane stress at the surface of a large machine is subjected to stresses shown below. Using Mohr's circle determine the following quantities (a) the principal stresses and their directions and (b) the maximum shearing stresses and the directions of the planes on which they occur. 100 MPa 50 MPa 75 MPa 75 MPa 50 MPa 100 MPa2. For the state of stress shown, determine the range of value of for which the normal is equal to or less than 100 MPa and 50 MPa. stress x' + 90 MPa 60 MPa =The state of stresses in the x and y directions at a point in a given material is shown in the figure below. If the value of the shear stress in the x-y axes Ty is 19N/mm2, find the maximum direct stress that can be found at this point in any possible direction. Round off your answer to the nearest N/mm2. 10 N/mm2 Txy 20 N/mm2 20 N/mm2 Txy 10 N/mm2
- The components of plane stress at a critical point on a steel shell are shown in the figure below. The yield stress (Gy) of the material in simple tension is 250 MPa. 60 MPa 40 MPa 70 MPa (a) Determine the magnitude of the principal stresses and maximum shear stress. Sketch these on appropriately orientated elements. (b) Calculate the factor of safety against yield on the basis of i. Tresca failure criterion ii. Von Mises failure criterionQ: A piece of material is subjected to tensile stress of 70 N/mm² in one direction and a compressive stress of 50 N/mm² in a direction at right angles to the previous one. Find fully the stresses on a plane the normal of which makes an angle of 40 degree with the 70 N/mm2 stress.The state of stresses in the x and y directions at a point in a given material is shown in the figure below. If the value of the shear stress in the x-y axes Ty is 11 N/mm², find the minimum direct stress that can be found at this point in any possible direction. Round off your answer to the nearest N/mm?. If the answer is negative do insert the "-" sign. 10 N/mm2 Txy 20 N/mm2 20 N/mm2 Ixy 10 N/mm2
- The stress concentration occurs whenever there is an abrupt change in the cross-section of a component or there is any discontinuity in the material. The figure given below shows a flat plate with a hole of diameter d. The plate is fixed at one end and the other end is subjected to a tensile load of P = 44 kN due to which there is a change in length of 0.4 mm. The thickness of the plate is 11.5 mm. The maximum stress developed in the flat plate is 228 MPa. Take Young's modulus(E) = 210 GPa and theoretical stress concentration factor =2, Calculate the following values: i) Width of the plate (W2) in mm ( ii) Nominal Stress in MPa ( iii) Diameter of the hole (d) in mm30. An aluminum alloy rod has a circular cross section with a diameter of 8mm. The rod is subjected to a tensile load of 5kN. Assume that the material is in the elastic region and E = 69 GPa. If Poisson's Ratio is 0.33, what will be the lateral strain? E= 6/8 v = -E(lateral)/(axial)The stress concentration occurs whenever there is an abrupt change in the cross-section of a component or there is any discontinuity in the material. The figure given below shows a flat plate with a hole of diameter d. The plate is fixed at one end and the other end is subjected to a tensile load of P = 41 kN due to which there is a change in length of 0.4 mm. The thickness of the plate is 11 mm. The maximum stress developed in the flat plate is 235 MPa. Take Young's modulus(E) = 210 GPa and theoretical stress concentration factor =2, Calculate the following values: i) Width of the plate (W2) in mm ii) Nominal Stress in MPa iii) Diameter of the hole (d) in mm Hole with stress concentration factor 2 35 mm W2 35 mm P 300 mm 350 mm- 250 mm