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A solid cylinder of diameter 50 mm and height 100 mm is made of a material with Young's modulus of 200 GPa. Calculate the elongation of the cylinder when a tensile load of 100 kN is applied to it.
The formula for elongation in a cylinder due to tension is given by δ = PL/EA, where P is the applied load, L is the length of the cylinder, E is the Young's modulus, and A is the cross-sectional area.
The cross-sectional area of the cylinder is A = πd^2/4 = π(50mm)^2/4 = 1963.5 mm^2.
The length of the cylinder is L = 100 mm.
Substituting these values along with E = 200 GPa and P = 100 kN, we get: δ = (100 kN x 100 mm)/(200 GPa x 1963.5 mm^2) = 0.0255 mm Therefore, the elongation of the cylinder is 0.0255 mm.
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A cantilever beam of length 3 m and rectangular cross-section of width 200 mm and height 400 mm is made of a material with Young's modulus of 200 GPa and Poisson's ratio of 0.3. If the beam is subjected to a point load of 50 kN at its free end, calculate the maximum bending stress in the beam.
The formula for maximum bending stress in a cantilever beam due to a point load at its free end is given by σ = (My)/I, where M is the bending moment, y is the distance from the neutral axis to the outermost fiber, and I is the moment of inertia.
The bending moment is given by M = PL, where P is the point load and L is the length of the beam. Substituting the values, we get: M = (50 kN x 3 m) = 150 kNm
The moment of inertia of a rectangular cross-section is given by I = (bh^3)/12, where b is the width and h is the height. Substituting the values, we get: I = (200 mm x 400 mm^3)/12 = 10.67x10^6 mm^4
The distance from the neutral axis to the outermost fiber is half the height, y = h/2 = 200 mm.
Substituting these values, we get: σ = (150 kNm x 200 mm)/(10.67x10^6 mm^4) = 2.81 MPa Therefore, the maximum bending stress in the beam is 2.81 MPa.
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A machine element with a cross-sectional area of 50 mm^2 is subjected to a tensile load of 100 kN. If the yield strength of the material is 200 MPa, determine if the element will yield or not.
The formula for stress is given by σ = P/A, where P is the load and A is the crosssectional area.
Substituting the values, we get: σ = (100 kN)/(50 mm^2) = 2000 MPa
The stress is greater than the yield strength of the material, which is 200 MPa. Therefore, the element will yield.
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A pump is designed to lift water from a depth of 30 m with a flow rate of 0.1 m^3/s. If the efficiency of the pump is 70%, calculate the power required to run the pump.
The formula for power required by the pump is given by P = ρghQ/η, where ρ is the density of water, g is the acceleration due to gravity, h is the height of lift, Q is the flow rate, and η is the efficiency of the pump.
Substituting the values, we get: P = (1000 kg/m^3 x 9.81 m/s^2 x 30 m x 0.1 m^3/s)/(0.7) = 4178.3 W
Therefore, the power required to run the pump is 4178.3 W.
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A car engine has a displacement of 2.0 L and a maximum power output of 150 hp at 6000 rpm. If the engine efficiency is 25%, calculate the torque output of the engine at its maximum power.
The formula for torque is given by T = P/(2πN/60), where P is the power, N is the engine speed in rpm, and T is the torque.
Substituting the values, we get: T = (150 hp x 746 W/hp x 0.25)/(2π x 6000/60) =
180.9 Nm
Therefore, the torque output of the engine at its maximum power is 180.9 Nm.
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