Find modulus of elasticity Stress=30 Pa Strain=0.5
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A: As per given question α=11.7 m•c E=200gpa ΔT=50°c We have to find load carried by each bar
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- A 12 mm thick steel tire has a width of 110 mm has an internal diameter of 800 mm. Tire is heated and shrunk to a steel wheel 800.5 mm in diameter. E=200 MPa. Find the tensile stress in the tire. Look for the compressive pressure between the tire and wheel.Find the stresses in each direction, also find the change in volume of the block of dimension 120 mm x 60 mm x 43 mm, subjected to 3 mutually perpendicular loads. The load along length, breadth and depth directions are 10 kN (tensile), 25 kN (tensile), 13 kN (compressive) respectively, Take E as 180 GPa, Poisson's ratio as 0.3. The stress along length direction (Unit in kN/m2)= _____________ The stress along width direction (Unit in kN/m2)= _____________ The compressive stress along depth direction (Unit in kN/m2)= _____________ The change in volume of the block is (unit in mm3) = ______________Find the stress if strain 0.55 and modulus of elasticity is 50 MPa
- Find the stresses in each direction, also find the change in volume of the block of dimension 110 mm x 55 mm x 40 mm, subjected to 3 mutually perpendicular loads. The load along length, breadth and depth directions are 14 kN (tensile), 20 kN (compressive), 10 kN (compressive) respectively. Take E as 160 GPa, Poisson's ratio as 0.3 (ENTER ONLY THE VALUES IN THE BOXES BY REFERRING THE UNIT GIVEN IN BRACKET & UPLOAD YOUR HAND WRITTEN ANSWERS IN THE LINK PROVIDED) The stress along length direction (Unit in MN/m?)=. The compressive stress along width direction (Unit in MN/m2)=. The compressive stress along depth direction (Unit in MN/m2)=. The change in volume of the block is (unit in mm3) =.Find the stresses in each direction, also find the change in volume of the block of dimension 110 mm x 40 mm x 43 mm, subjected to 3 mutually perpendicular loads. The load along length, breadth and depth directions are 10 kN (tensile), 25 kN (compressive), 22 kN (compressive) respectively. Take E as 120 GPa, Poisson's ratio as 0.3 (figure - 0.5 mark, calculation - 1.5 marks, solution - 2 marks) (ENTER ONLY THE VALUES IN THE BOXES BY REFERRING THE UNIT GIVEN IN BRACKET & UPLOAD YOUR HAND WRITTEN ANSWERS IN THE LINK PROVIDED) The stress along length direction (Unit in MN/m2)= The compressive stress along width direction (Unit in MN/m2?)=, The compressive stress along depth direction (Unit in MN/m?)= The change in volume of the block is (unit in mm)Compute strain Stress-40 MPa Modulus of elasticity=100 MPa
- 10/10FIND STRESS IN STEEL MATERIALS IN COMPOSITE COLUMIN SHOWN BELOW. IF E st= 200 kN/MM2, Ealu.=70 kN/MM2 100 kN RALU=30 MM Aluminum Steel 1m. R ST = 20 MMFind Stress Force=2 N Cross sectional area=1 m2A rubber ball is inflated to a pressure of 60KPa. At that pressure, the diameter of the ball is 230mm and the wall thickness is 1.2mm. The rubber has a modulus of elasticity of E = 3.5MPa and Poisson’s ratio of 0.45. (a)Compute the maximum tensile stress. (b)If the allowable tensile stress of the rubber ball wall is 2.5MPa, compute the internal pressure the rubber ball could be inflated.
- Find the initial length of rod if strain value is 0.0003 and change in length value 0.3 mmProblem 1 of 2: Nodal displacements for the elements 1-3 and 2-3 of a plane truss were determined to be u = 1.5 mm, uz = 0.2 mm, uz = -2.7 mm, v = 3.1 mm, v2 = 3.3 mm, v, =-0.8 mm L. %3D What are their axial stresses, if E = 2.0-101 Pa and L = 2 m for both elements? Solution:A steel rod is stretched between two walls. At 20C, the tensile force in the rod is 5000N. If the stress is not to exceed 130MPa at -20C, find the minimum allowable diameter of the rod. Use a=11.17x10^-6C and E=200GPa. -Draw and label the diagram correctly, No diagram in the solution will be marked wrong. -Shortcut solution will be marked wrong.- Direction of the assumption of the equilibrium equation must be shown, no direction will be marked wrong. -Show complete solution and explanation