# Solution ## Given Data - **Beam BCD**: Pinned at B, supported by cable AD. - **Truss C-D-E-F**: Pinned to beam, carries load \( P \) at E. - **Beam cross-section**: - Thickness, \( t = 25 \) mm (into page) - Width at D & B, \( w_{B,D} = 200 \) mm - Width at C, \( w_C = 350 \) mm - **Cable DB**: - Diameter, \( d_{DB} = 50 \) mm - Elongation required, \( \Delta L = .04 \) m - **Material Properties**: - Normal yield stress, \( \sigma_{y} = 450 \) MPa - Shear yield stress, \( \tau_{y} = 350 \) MPa - Young's Modulus, \( E = 200 \) GPa - **Pin at B**: - Double shear - Diameter, \( d_{pin} = 40 \) mm - Allowable shear stress, \( \tau_{allow} = 250 \) MPa --- ## (a) Load \( P \) to Elongate Cable DB by \( .04 \) m ### Step 1: Cable Cross-Section Area \[ A_{DB} = \frac{\pi}{4} d_{DB}^2 = \frac{\pi}{4} (.05)^2 = .001963 \text{ m}^2 \] ### Step 2: Relationship Between Force and Elongation \[ \Delta L = \frac{F_{DB} L_{DB}}{A_{DB} E} \] But \( L_{DB} \) (length of cable) is not given. **Assume** the cable runs directly from D to B (if not, use given geometry). **If \( L_{DB} \) is not specified, express \( P \) in terms of \( L_{DB} \):** \[ F_{DB} = \frac{\Delta L \cdot A_{DB} \cdot E}{L_{DB}} \] Substitute values: \[ F_{DB} = \frac{.04 \times .001963 \times 200 \times 10^9}{L_{DB}} \] \[ F_{DB} = \frac{.04 \times .001963 \times 200 \times 10^9}{L_{DB}} \] \[ F_{DB} = \frac{.015704 \times 200 \times 10^9}{L_{DB}} \] \[ F_{DB} = \frac{3.1408 \times 10^9}{L_{DB}} \] ### Step 3: Relate \( F_{DB} \) to Load \( P \) - The load \( P \) at E causes force in the cable via the truss and beam. - The precise geometry is needed to relate \( P \) to \( F_{DB} \). - **If we assume \( F_{DB} = P \)** (worst-case, vertical cable), then: \[ P = F_{DB} = \frac{3.1408 \times 10^9}{L_{DB}} \] **If geometry is given, use correct truss/beam analysis to relate \( P \) to \( F_{DB} \).** --- ## (b) Normal Stress in Beam Halfway Between C and D ### Step 1: Find Bending Moment at Midpoint - The force \( F_{DB} \) will create a reaction at B and a moment in the beam. - The maximum moment between C and D is at midpoint. - Let \( L_{CD} \) be the distance between C and D. \[ M_{mid} = \text{(depends on loading and support, assume simply supported or cantilevered as per diagram; specify if known)} \] ### Step 2: Section Properties at Midpoint - At halfway between C and D, width \( w_{mid} \): \[ w_{mid} = \frac{w_C + w_D}{2} = \frac{350 + 200}{2} = 275 \text{ mm} = .275 \text{ m} \] - Thickness: \( t = 25 \) mm = .025 m - Section modulus \( S \): \[ I = \frac{1}{12} w_{mid} t^3 \] \[ c = \frac{t}{2} \] \[ S = \frac{I}{c} = \frac{w_{mid} t^3 / 12}{t/2} = \frac{w_{mid} t^2}{6} \] ### Step 3: Normal (Bending) Stress \[ \sigma_{max} = \frac{M_{mid}}{S} \] Plug in \( M_{mid} \) and \( S \). ### Step 4: Factor of Safety \[ FS = \frac{\sigma_y}{\sigma_{max}} \] - If \( FS > 1 \), the beam is safe. --- ## (c) Factor of Safety for 40 mm Pin in Double Shear ### Step 1: Shear Area for Double Shear \[ A_{shear} = 2 \left( \frac{\pi}{4} d_{pin}^2 \right) = 2 \left( \frac{\pi}{4} (.04)^2 \right) = 2 \times .001257 = .002513 \text{ m}^2 \] ### Step 2: Shear Stress in Pin \[ \tau_{pin} = \frac{F_{DB}}{A_{shear}} \] ### Step 3: Factor of Safety \[ FS_{pin} = \frac{\tau_{allow}}{\tau_{pin}} \] - If \( FS_{pin} > 1 \), the pin is safe. --- ## Summary Table | Part | Formula | Result (in terms of \( L_{DB} \)) | |------|---------|-----------------------------------| | (a) | \( F_{DB} = \frac{3.14 \times 10^9}{L_{DB}} \) | \( P = F_{DB} \) | | (b) | \( \sigma_{max} = \frac{M_{mid}}{S} \) | Calculate using above | | (c) | \( FS_{pin} = \frac{250 \text{ MPa}}{\frac{F_{DB}}{A_{shear}}} \) | Calculate using above | --- ## Notes - The exact numerical answers require the distances \( L_{DB} \) (cable length) and \( L_{CD} \) (beam segment), and the loading geometry. Plug in these values as appropriate for your actual problem configuration. --- ### [No image provided, so none included.]