# Q3 Solution: Cable Force with Temperature Drop Let's solve **step-by-step**: --- ## **Step 1: Structure Analysis Without Temperature Effect** ### **1.1: System Overview** - **AB** is a horizontal beam (fixed at **A**). - **BC** is a cantilevered section (total beam length = 8 m). - **Cable DB** supports point **B**. - **Point Load:** \( P = 280 \) kN at \( C \). - **Cable Length \( DB \):** Use Pythagoras: \( DB = \sqrt{(6\,m)^2 + (6\,m + 2\,m)^2} = \sqrt{36 + 64} = \sqrt{100} = 10\,m \) *(Correction: DB connects D(,6) to B(6,), so DB = sqrt(6²+6²) = sqrt(72) = 8.485 m)* ### **1.2: Redundant Force Method (Using \( M_A \) as Redundant)** - Treat the structure as **statically indeterminate** (one redundant: \( M_A \)). #### **Free Body Diagram** - Draw the beam with point load at **C** (8 m from **A**) and cable at **B**. #### **Vertical Force and Moment Equilibrium:** - Sum of moments about **A** (without cable force): \[ \sum M_A = : \quad 280 \times 8 - R_B \times 6 = \implies R_B = \frac{280 \times 8}{6} = 373.33\,\text{kN} \] This is the reaction at **B** if no cable, but the cable reduces this. #### **Let Cable Force = \( T \) (acts along DB)** - Vertical component of cable at **B**: \( T_v = T \sin \theta \) - \( \sin \theta = \frac{6}{\sqrt{6^2 + 6^2}} = \frac{6}{8.485} = .707 \) - **Vertical Equilibrium:** \[ R_A + T_v = 280 \] (assuming reactions at A plus cable support the load). #### **Moment about A:** \[ \sum M_A = : \quad 280 \times 8 - T_v \times 6 = \] \[ T_v = \frac{280 \times 8}{6} = 373.33\,\text{kN} \] \[ T = \frac{T_v}{\sin \theta} = \frac{373.33}{.707} = 528\,\text{kN} \] --- ## **Step 2: Include Temperature Drop Effect** ### **Cable Contracts with Temperature Drop:** - \( \Delta T = -20^\circ C \) - \( \alpha = 1.2 \times 10^{-5}/^\circ C \) - **Free Contraction:** \( \delta_{T,free} = L \alpha \Delta T \) \[ \delta_{T,free} = 8.485\,m \times 1.2 \times 10^{-5} \times (-20) \] \[ = 8.485 \times (-2.4 \times 10^{-4}) = -.00204\,m = -2.04\,mm \] ### **Compatibility:** - The structure prevents cable from shortening, so it develops extra tension. - **Increase in force to prevent contraction:** \[ \delta_T = \frac{T_{add} L}{AE} \] Set \( \delta_T = 2.04\,mm \) (to counteract contraction): \[ T_{add} = \frac{\delta_T AE}{L} \] Where, - \( A = 600\,mm^2 = 6 \times 10^{-3}\,m^2 \) - \( E = 150\,GPa = 150 \times 10^9\,N/m^2 \) - \( L = 8.485\,m \) \[ T_{add} = \frac{.00204 \times 6 \times 10^{-3} \times 150 \times 10^9}{8.485} \] \[ = \frac{.00204 \times 900 \times 10^6}{8.485} = \frac{1.836 \times 10^6}{8.485} = 216,300\,N = 216.3\,kN \] ### **Final Cable Force:** \[ T_{total} = T_{load} + T_{temp} = 528\,\text{kN} + 216.3\,\text{kN} = \boxed{744.3\,\text{kN}} \] --- ## **Summary Table** | Step | Description | Value | |------|-----------------------------|------------| | 1 | Cable force (load only) | 528 kN | | 2 | Additional force (temp drop)| 216.3 kN | | | **Total Cable Force** | **744.3 kN**| --- ## **Graphical Representation**  *Alt text: Diagram of beam ABC with cable DB, showing point load at C and cable force at B.* --- ## **Final Answer** > **The cable force under the given load and a temperature drop of \(20^\circ C\) is \(\boxed{744.3\,\text{kN}}\).** --- ### **Example Calculation Recap** - **Step 1:** Find cable force from equilibrium: \( T = 528\,\text{kN} \) - **Step 2:** Add force required to prevent thermal contraction: \( T_{add} = 216.3\,\text{kN} \) - **Total:** \( 528 + 216.3 = 744.3\,\text{kN} \) If you need a hand-drawn-style graph or FBD, let me know!