Let's address each question step by step, starting with the given data and moving through each calculation. --- ### **Given Data** #### **Member AD (Composite: Aluminum Rod inside Steel Hollow Tube)** - **Radius of aluminum rod, \( r_a \) = 10 mm** - **Outer radius of steel tube, \( r_o \) = 25 mm** - **Thickness of steel tube, \( t \) = 5 mm** - **Length, \( L \) = 1 m = 1000 mm** - **\( E_{\text{steel}} \) = 200 GPa = 200,000 MPa** - **\( E_{\text{aluminum}} \) = 70 GPa = 70,000 MPa** - **\( \sigma_{\text{allow, aluminum}} \) = 20 MPa** - **\( \alpha_{\text{aluminum}} \) = \( 23 \times 10^{-6} \)/°C** - **\( \alpha_{\text{steel}} \) = \( 12 \times 10^{-6} \)/°C** #### **Member BE (Steel Rod)** - **Radius, \( r \) = 10 mm** - **\( E_{\text{steel}} \) = 200 GPa** - **Length, \( L \) = 1 m** #### **Member CF (Tension Cable Wire)** - **\( E_w \) = 100 GPa** - **Area, \( A_w \) = 30 mm²** - **Length, \( L \) = 0.5 m** --- ## **1. Maximum Axial Force on the Aluminum Rod in AD** - **Allowable stress, \( \sigma_{\text{allow, aluminum}} = 20\, \text{MPa} \)** - **Area of aluminum rod, \( A_a = \pi r_a^2 = \pi (10^2) = 314.16\, \text{mm}^2 \)** \[ F_{\text{max, al}} = \sigma_{\text{allow, aluminum}} \cdot A_a \] \[ = 20\, \text{MPa} \times 314.16\, \text{mm}^2 \] \[ = 6283.19\, \text{N} = 6.28\, \text{kN} \] --- ## **2. Express Total Force on AD in terms of Aluminum Rod Force** Let \( F_{AD} \) = total force carried by member AD (composite), Let \( F_a \) = force carried by aluminum rod, - The total force in AD is carried by both the aluminum rod and the steel tube: \[ F_{AD} = F_a + F_s \] If only the aluminum rod is loaded to its max: \[ F_{AD, \text{max}} = F_{a, \text{max}} + F_s \] But if steel is not yet stressed or only aluminum is loaded: \[ F_{AD} = F_a \] --- ## **3. Maximum Displacement at Point A (from Aluminum Rod Max Stress)** - **Displacement, \( \delta = \frac{F L}{A E} \)** - Use max load for aluminum rod: - \( F = F_{a, \text{max}} = 6283.19\, \text{N} \) - \( L = 1000\, \text{mm} \) - \( A = 314.16\, \text{mm}^2 \) - \( E = 70,000\, \text{MPa} \) \[ \delta_A = \frac{6283.19 \times 1000}{314.16 \times 70,000} \] \[ = \frac{6,283,190}{21,991,200} \] \[ = 0.286\, \text{mm} \] --- ## **4. Express Vertical Displacement at Point C in Terms of A and B** Let: - \( \delta_A \) = vertical displacement at A - \( \delta_B \) = vertical displacement at B - \( \delta_C \) = vertical displacement at C Assuming rigid bar and supports, with distances from A to B = 1 m, B to C = 2 m: - By similar triangles (rotation about E): \[ \frac{\delta_B - \delta_A}{1} = \frac{\delta_C - \delta_A}{3} \] \[ \delta_C = 3(\delta_B - \delta_A) + \delta_A = 3\delta_B - 2\delta_A \] --- ## **5. Maximum Load \( P \) Assembly Can Carry (Aluminum Rod Max Stress)** - From equilibrium: - Set up moment about A or B - Let forces in AD, BE, CF be \( F_{AD} \), \( F_{BE} \), \( F_{CF} \) - \( F_{AD, \text{max}} = F_{a, \text{max}} \) - Write moment about A for equilibrium: - \( P \) acts at B, \( F_{CF} \) at C, \( F_{BE} \) at E (vertical), \( F_{AD} \) at D (vertical) Assuming all vertical: \[ \sum M_A = 0: \quad P \times 1 + F_{CF} \times 3 - F_{BE} \times 1 - F_{AD} \times 1 = 0 \] But more info needed about internal force distribution (usually, solve using compatibility and equilibrium). For this problem, if only AD's max is considered, and assuming other members are much stronger: \[ P_{max} = F_{AD, \text{max}} = 6.28\, \text{kN} \] --- ## **6. Temperature Change for Zero Displacement at Point A (Upward \( P \))** For zero displacement at A, thermal expansion must offset mechanical contraction: \[ \delta_{thermal} = \delta_{mechanical} \] \[ \alpha_{al} \Delta T \cdot L = \frac{F_{a, \text{max}} L}{A_a E_{al}} \] \[ \Delta T = \frac{F_{a, \text{max}}}{A_a E_{al} \alpha_{al}} \] \[ = \frac{6283.19}{314.16 \times 70,000 \times 23 \times 10^{-6}} \] \[ = \frac{6283.19}{507.98} \] \[ = 12.37^\circ\text{C} \] --- ## **Summary Table** | **Step** | **Value** | |----------|-----------| | 1. Max axial force (aluminum rod) | **6.28 kN** | | 2. Total force on AD | \( F_{AD} = F_a + F_s \) | | 3. Max disp. at A | **0.286 mm** | | 4. Disp. at C | \( \delta_C = 3\delta_B - 2\delta_A \) | | 5. Max load P | **6.28 kN** (if only AD governs) | | 6. Required ΔT for zero disp. at A | **12.37°C** | --- **If you need full algebraic development or further steps (like equilibrium with all members), let me know!**

**Final Answers:** 1. **Maximum axial force in aluminum rod (Member AD):** \(\boxed{6.28\, \text{kN}}\) 2. **Total force on member AD in terms of aluminum force:** \(\boxed{F_{AD} = F_a + F_s}\) 3. **Maximum displacement at Point A (due to aluminum max stress):** \(\boxed{0.286\, \text{mm}}\) 4. **Vertical displacement at Point C in terms of displacements at A and B:** \(\boxed{\delta_C = 3\, \delta_B - 2\, \delta_A}\) 5. **Maximum force \( P \) the assembly can carry (assuming aluminum governs):** \(\boxed{6.28\, \text{kN}}\) 6. **Temperature change \(\Delta T\) needed for zero displacement at Point A (with upward \( P \)):** \(\boxed{12.37^\circ \text{C}}\)