The AM1004-T61 magnesium alloy tube AB is capped with a rigid plate E. The gap between E and end C of the 6061-T6 aluminum alloy solid circular rod CD is 0.2 mm when the temperature is at 30 degree C.

To solve this problem, we need to calculate the change in length of both the magnesium alloy tube and the aluminum alloy rod due to the increase in temperature from 30°C to 80°C, and then use this information to calculate the normal stress developed in each component. First, we use the formula for thermal expansion to calculate the change in length of each component: ?L = L0 ? ?T where ?L is the change in length, L0 is the original length, ? is the coefficient of thermal expansion, and ?T is the change in temperature. For the magnesium alloy tube AB: ?L_AB = L0_AB ?_mg ?T = (0.6 m) (26 ? 10^-6 / °C) (80°C - 30°C) = 0.078 m For the aluminum alloy rod CD: ?L_CD = L0_CD ?_Al ?T = (0.3 m) (24 ? 10^-6 / °C) (80°C - 30°C) = 0.034 m Next, we use the changes in length to calculate the normal stress developed in each component at the new temperature using the formula: ? = F / A where ? is the normal stress, F is the force applied, and A is the cross-sectional area of the component. For the magnesium alloy tube AB, the force applied is due to the compression of the rigid cap E against the end C of the aluminum alloy rod CD, so the force can be calculated using Hooke's law: F = - k ?x where k is the spring constant of the rigid cap and ?x is the change in length of the aluminum alloy rod. Assuming the spring constant is the same in both compression and tension, we can use the formula for the spring constant of a rod to calculate k: k = EA / L where E is the modulus of elasticity, A is the cross-sectional area, and L is the original length. For the aluminum alloy rod CD, the cross-sectional area is given by: A = ?/4 D? = ?/4 (0.03 m)? = 7.07 ? 10^-4 m? The original length is: L = L0_CD = 0.3 m So the spring constant is: k = EA / L = (68.9 ? 10^9 N/m?) (7.07 ? 10^-4 m?) / (0.3 m) = 1.606 ? 10^6 N/m Therefore, the force applied by the rigid cap is: F = - k ?x = -(1.606 ? 10^6 N/m) (0.034 m) = -54.524 N The cross-sectional area of the magnesium alloy tube AB is given by: A = ?/4 (D? - d?) = ?/4 [(0.04 m)? - (0.036 m)?] = 1.185 ? 10^-3 m? So the normal stress developed in the magnesium alloy tube at the new temperature is: ?_AB = F / A = (-54.524 N) / (1.185 ? 10^-3 m?) = -45.96 ? 10^6 N/m? = -45.96 MPa (compressive) For the aluminum alloy rod CD, the cross-sectional area is the same as before, so the normal stress developed at the new temperature is: ?_CD = F / A = (-54.524 N) / (7.07 ? 10^-4 m?) = -77.03 ? 10^6 N/m? = -77.03 MPa (compressive) Therefore, the normal stresses developed in the magnesium alloy tube AB and the aluminum alloy rod CD at the new temperature of 80°C are -45.96 MPa and -77.03 MPa, respectively (both compressive).