Let's solve the problem step-by-step as per the given questions. --- ## **Given Data** - **Diameter of rod BC, \( d \) = 40 mm = .04 m** - **Length BC:** \[ \text{Horizontal component} = .75~\text{m} \\ \text{Vertical component} = 1.~\text{m} \\ \text{Length } (L) = \sqrt{(.75^2 + 1^2)} = 1.25~\text{m} \] - **\( E = 200 \) GPa = \( 200 \times 10^9 \) Pa** - **\( \nu = .32 \)** - **Load \( P = 150 \) kN = \( 150,000 \) N** --- ## **(a) Write Equations of Equilibrium** Let: - \( A_x \) and \( A_y \) = reactions at A (horizontal and vertical) - \( F_{BC} \) = force in rod BC (direction along BC) - Point B has no vertical/horizontal support, only internal force from BC. ### **Free-Body Diagram Analysis** **Sum of Forces:** \[ \sum F_x = : \quad A_x - F_{BC} \cdot \cos\theta = \] \[ \sum F_y = : \quad A_y - 2P + F_{BC} \cdot \sin\theta = \] Where \( \theta \) is the angle of rod BC from the horizontal: \[ \theta = \tan^{-1}\left(\frac{1.}{.75}\right) = 53.13^\circ \] **Sum of Moments about A:** - The forces P act at 1 m and 2 m from A. - The vertical component of BC acts at B (2.75 m from A). \[ \sum M_A = : \quad -P \cdot 1 - P \cdot 2 + (F_{BC} \cdot \sin\theta) \cdot 2.75 = \] --- ## **(b) Determine Force in BC and Reactions at A** ### **Step 1: Calculate \( F_{BC} \)** From moment equilibrium: \[ -150,000 \cdot 1 - 150,000 \cdot 2 + (F_{BC} \cdot \sin 53.13^\circ) \cdot 2.75 = \] \[ -150,000 - 300,000 + (F_{BC} \cdot .8) \cdot 2.75 = \] \[ -450,000 + 2.2 F_{BC} = \] \[ 2.2 F_{BC} = 450,000 \] \[ F_{BC} = \frac{450,000}{2.2} = 204,545.45~\text{N} \] ### **Step 2: Reactions at A** \[ A_x = F_{BC} \cdot \cos\theta = 204,545.45 \times .6 = 122,727.27~\text{N} \] \[ A_y = 2P - F_{BC} \cdot \sin\theta = 300,000 - 204,545.45 \times .8 = 135,454.55~\text{N} \] --- ## **(c) Average Normal Stress in BC** \[ \sigma = \frac{P}{A} \] Where \( A = \frac{\pi}{4} d^2 = \frac{\pi}{4} (.04)^2 = .001257~\text{m}^2 \) \[ \sigma = \frac{204,545.45}{.001257} = 162,767,998~\text{Pa} = 162.77~\text{MPa} \] --- ## **(d) Strain in BC** \[ \epsilon = \frac{\sigma}{E} \] \[ \epsilon = \frac{162,767,998}{200 \times 10^9} \] \[ \epsilon = .000814 \] --- ## **(e) Elastic Elongation and Decrease in Diameter** ### **Elastic Elongation** \[ \delta = \frac{PL}{EA} \] \[ \delta = \frac{204,545.45 \times 1.25}{200 \times 10^9 \times .001257} = \frac{255,681.81}{251,400,000} = .001017~\text{m} = 1.02~\text{mm} \] ### **Decrease in Diameter** \[ \text{Lateral strain:} \quad \epsilon_{lat} = -\nu \epsilon_{long} \] \[ \Delta d = d \cdot \epsilon_{lat} = .04 \times (-.32 \times .000814) = .04 \times -.00026048 = -.00001042~\text{m} = -.0104~\text{mm} \] --- ## **Summary Table** | Step | Quantity | Value | |------|---------------------------------------|--------------:| | (b) | Force in BC, \( F_{BC} \) | 204,545 N | | | \( A_x \) | 122,727 N | | | \( A_y \) | 135,455 N | | (c) | Normal stress, \( \sigma \) | 162.77 MPa | | (d) | Strain, \( \epsilon \) | .000814 | | (e) | Elongation, \( \delta \) | 1.02 mm | | | Decrease in diameter, \( \Delta d \) | .0104 mm | --- **Let me know if you'd like the step-by-step calculations typed differently or want explanations for specific steps!**