5: An aluminum rod (E-70 GPa and v=0.25) is inserted into a hollow steel tube (E=210 Gra and v-0.28). The aluminum rod projects (015 mm) as shown and has a diameter of (40 mm). The steel tube has an inside diameter of (50 mm) and is (10 mm) thick. If the applied load (P= 200 kN), determine: 1) The axial stresses in the aluminum rod and the steel tube. 2) Load carried by each bar. 3) The change in diameter of the aluminum. P. 0.15mm 250mm Aluminum

Rigid bar ABC is supported by a pin at bracket A and by tie rod (1). Tie rod (1) has a diameter of 5 mm, and it is supported by double- shear pin connections at B and D. The pin at bracket A is a single-shear connection. The pin at B is 7 mm in diameter. Assume a = 560 mm, b = 280 mm, h = 450 mm, and 0= 70%. If the normal stress in tie rod (1) cannot exceed 200 MPa and the shear stress in pin B cannot exceed 140 MPa, determine the maximum load Pmax that can be supported by the structure. D (1) a Answer: Pmax= i B KN

Determine the nodal displacements and the element stresses. E = 105 GPa, v = 0.3, and t = 5 mm. You don't have to calculate everything manually, but can write a code to get some input from you and find the stiffness matrices for instance. 400 mm 3 5 400 mm 30 kN 2

Part 1. A composite assembly of length, L-375 mm, consisting of a steel (G-77 GPa) core (2) connected by rigid plates at the ends of an aluminum [G-35 GPa) tube (1) is shown. The aluminum tube has outside diameter D₁ - 50 mm and inside diameter d₁-30 mm. = 100 MPa, and the The solid steel core has diameter d₂-25 mm. The allowable shear stress of aluminum tube (1) is (rallow) allowable shear stress of steel core (2) is (Tallow)2= 135 MPa. Determine (a) the allowable torque T that can be applied to the composite shaft. (b) the corresponding torques produced in tube (1) and core (2). (c) the angle of twist produced by the allowable torque T. T D₁ L Calculate the polar moment of inertia, J₁, of the aluminum tube. Answer: J₁ (106) mm².