Sure! Let's break down the problem step-by-step: --- ## **Problem Overview** - **Rods**: Two cylindrical rods, **AB** **BC**, at **B. - **Diameters: AB: \( d_1 1.9 \) in - BC: \( d_2 = 2.8 \) in - **Forces**: - Two **compressive forces** of **26 kips** each at point B (applied to BC). - **Tensile force \( P \)** applied at end A (applied to AB). **Goal:** Find the value of \( P \) (in kips) so that the **tensile stress in AB is twice the compressive stress in BC**. --- ## **Step 1: Draw the Force Diagram** - At **point A**: Tensile force \( P \) (pulling to the right). - At **point B**: Two compressive forces, each 26 kips (pushing to the left), applied to BC. --- ## **Step 2: Internal Axial Forces** ### **Rod AB** - Internal force in AB = \( P \) (tension). ### **Rod BC** - At B, BC receives: - \( P \) from the left (from AB, if you cut just right of B) - \( 2 \times 26 = 52 \) kips compressive force from the right - **Net internal force in BC**: The internal axial force in BC (right of B) must balance the external forces: \[ \text{Internal force in BC} = P - 52 \text{ (compression if negative)} \] --- ## **Step 3: Stresses in Each Rod** - **Stress** \( \sigma = \dfrac{\text{Force}}{\text{Area}} \) ### **Cross-sectional Areas** - \( A_{AB} = \frac{\pi}{4} d_1^2 \) - \( A_{BC} = \frac{\pi}{4} d_2^2 \) ### **Stresses** - \( \sigma_{AB} = \dfrac{P}{A_{AB}} \) (tension) - \( \sigma_{BC} = \dfrac{P - 52}{A_{BC}} \) (compression, negative if \( P < 52 \)) --- ## **Step 4: Set Up the Relationship** We want: \[ \sigma_{AB} = 2 |\sigma_{BC}| \] Since \( \sigma_{BC} \) is compressive (negative), \( |\sigma_{BC}| = -\sigma_{BC} \): \[ \dfrac{P}{A_{AB}} = 2 \left| \dfrac{P - 52}{A_{BC}} \right| \] Assuming \( P < 52 \) (so BC is in compression): \[ \dfrac{P}{A_{AB}} = 2 \cdot \dfrac{52 - P}{A_{BC}} \] --- ## **Step 5: Substitute Values** - \( d_1 = 1.9 \) in, \( d_2 = 2.8 \) in - \( A_{AB} = \frac{\pi}{4} (1.9)^2 = 2.835 \) in² - \( A_{BC} = \frac{\pi}{4} (2.8)^2 = 6.157 \) in² Plug in: \[ \dfrac{P}{2.835} = 2 \cdot \dfrac{52 - P}{6.157} \] --- ## **Step 6: Solve for \( P \)** Multiply both sides by 2.835 and 6.157 to clear denominators: \[ P \cdot 6.157 = 2 \cdot (52 - P) \cdot 2.835 \] \[ 6.157P = 2 \cdot 2.835 \cdot (52 - P) \] \[ 6.157P = 5.67 \cdot (52 - P) \] \[ 6.157P = 5.67 \cdot 52 - 5.67P \] \[ 6.157P + 5.67P = 5.67 \cdot 52 \] \[ 11.827P = 294.84 \] \[ P = \frac{294.84}{11.827} \approx 24.9 \text{ kips} \] --- ## **Final Answer** \[ \boxed{P = 24.9 \text{ kips}} \] --- ## **Summary Table** | Segment | Diameter (in) | Area (in²) | Internal Force | Stress Equation | Stress | |---------|---------------|------------|---------------|----------------|--------| | AB | 1.9 | 2.835 | \(P\) | \( P/A_{AB} \) | Tension| | BC | 2.8 | 6.157 | \(P - 52\) | \( (P-52)/A_{BC} \) | Compression| --- **Let me know if you need more details on any step!**