# Digital Solution Format --- ## Step 1: Problem Statement and Structure **Given:** - Truss structure with joints A, B, C, D, E, F. - Load of \(8\,\text{kN}\) at joint E. - Dimensions and geometry as shown. --- ## Step 2: Force Method Analysis **Redundant Member:** EF Remove EF to get the primary structure and calculate member forces due to the external load. --- ## Step 3: P-Forces (Forces due to External Load) ### Free Body Diagram (FBD) at Joint E \[ \theta = \tan^{-1}\left(\frac{3}{4}\right) = 36.87^\circ \] #### At Joint E: \[ \sum_x = : \quad F_{ED}\cos\theta + 8 = \] \[ F_{ED} = \frac{-8}{\cos 36.87} = -10\, \text{kN} \quad \text{(Compression)} \] \[ \sum F_y = : \quad F_{EC} + F_{ED}\sin 36.87 = \] \[ F_{EC} = -F_{ED}\sin 36.87 = 6\,\text{kN} \quad \text{(Tension)} \] --- ### At Joint F \[ \sum F_y = : \quad F_{AF} = \] --- ## Step 4: K-Forces (Unit Load in Direction of Redundant) ### Apply Unit Load in EF #### At Joint E: \[ \sum F_x = : \quad F_{ED}\cos 36.87 = 1 \implies F_{ED} = 1.25\,\text{kN} \, (\text{Tension}) \] \[ \sum F_y = : \quad F_{EC} + F_{ED}\sin 36.87 = \implies F_{EC} = -.75\,\text{kN} \, (\text{Compression}) \] #### At Joint F: \[ \sum F_x = : \quad F_{AF}\cos 36.87 = 1 \implies F_{AF} = 1.25\,\text{kN} \, (\text{Tension}) \] \[ \sum F_y = : \quad F_{BF} + F_{AF}\sin 36.87 = \implies F_{BF} = -.75\,\text{kN} \, (\text{Compression}) \] --- ## Step 5: Tabulate Results | Member | P-force (kN) | K-force (kN) | Length (m) | \(P \cdot K \cdot L\) | \(K^2 L\) | |--------|--------------|--------------|------------|-----------------------|-----------| | AF | | 1.25 | 5 | | 7.8125 | | BF | | -.75 | 3 | | 1.6875 | | CE | 6 | -.75 | 3 | -13.5 | 1.6875 | | DE | -10 | 1.25 | 5 | -62.5 | 7.8125 | | EF | -1 | 1 | 4 | 4 | 4 | \[ \sum P \cdot K \cdot L = -76,\quad \sum K^2 L = 23 \] --- ## Step 6: Compatibility Equation (Force Method) \[ R = -\frac{\sum P \cdot K \cdot L}{\sum K^2 L} = -\frac{-76}{23} = 3.304\,\text{kN} \quad (\text{Tension}) \] --- ## Step 7: Final Forces in Members \[ \text{Force} = P + R \cdot K \] | Member | Final Force (kN) | |--------|------------------| | AF | \( + 3.304 \times 1.25 = 4.13\) (T) | | BF | \( + 3.304 \times -.75 = -2.478\) (C) | | CE | \(6 + 3.304 \times -.75 = 3.522\) (T) | | DE | \(-10 + 3.304 \times 1.25 = -5.87\) (C) | | EF | \( + 3.304 \times 1 = 3.304\) (T) | --- ## Step 8: Final Truss with Forces ``` [Structure Diagram with labeled member forces] ``` - AF: \(4.13\,\text{kN}\) (T) - BF: \(2.478\,\text{kN}\) (C) - CE: \(3.522\,\text{kN}\) (T) - DE: \(5.87\,\text{kN}\) (C) - EF: \(3.304\,\text{kN}\) (T) --- **Note:** (T) = Tension, (C) = Compression --- **End of Solution (as per images, digitally formatted).**