Design rod to support load of 175,000 pounds. What diameter rod is appropriate?   Safety Factor, SF= Strength of Material (S)/ Applied Stress (sigma) = 1 Applied stress (sigma) = Force/Area Yield strength of Steel = 100,000 psi (yield means will undergo plastic deformation)

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QI/ An aluminum bar carries the axial loads at the positions shown in Fig. 1, below. The diameter for part AB, BC and CD is 35 mm, 40 mm and 30 mm, respectively. The elastic modulus for aluminum is 72.4 x 10 N/mm. Determine the internal stresses at section (A-B), (B-C) and (C- D), and the elongation (AL) in length at each part of aluminum due to loading. в D 37 kN 10 kN 45 kN 18 kN 1000 mm 800 mm I 600 mm Fig. 1

What maximum load P can be applied without exceeding an allowable stress of 70 MPa for aluminum or 120 MPa for steel? Can a larger load P be carried if the length of the aluminum rod is changed. the length of the steel portion being kept the same? If so, determine this length

The solid rod has a diameter of 20mm. Find the following: Angle of twist at C with respect to A (in radians) Angle of twist at B with respect to A (in radians) Maximum shear stress at A (in MPa) (Shear stress at support, at a point 7mm from center. (in MPa) A-36 steel with the following properties: Modulus of Elasticity, E = 29x10^3 ksi or 200 GPa Shear Modulus/Modulus of Rigidity, G = 11.5x10^3 ksi or 79.3 GPa Poisson's Ratio, nu = 0.260 Temperature Coefficient for thermal expansion, a = 6.6x10^-6/°F or 11.7x10^-6/°C