## Problem Restatement Given: - A steel shaft with three segments of different diameters and subjected to axial loads. - **Diameters:** - \( d_{AB} = .4 \) in (length = 80 in) - \( d_{BC} = .8 \) in (length = 150 in) - \( d_{CD} = .3 \) in (length = 100 in) - **Axial Loads:** - \( AB \): 5 kip (tension) - \( BC \): 2 kip (tension) - \( CD \): 6 kip (tension) - **Stress-strain diagram**: Provided in figure (not shown here, but assume elastic-perfectly plastic behavior for calculation). **Tasks:** 1. Determine the **approximate elongation** of the shaft when the load is applied. 2. If the load is removed, what is the **permanent elongation** of the shaft? 3. Assume same stress-strain behavior in tension and compression. --- ## Step 1: Find Internal Forces in Each Segment From the free-body diagram: - **Segment AB**: Force = 5 kip - **Segment BC**: Force = 2 kip - **Segment CD**: Force = 6 kip --- ## Step 2: Calculate Elongation (Assume Elastic Region) ### Formula for Elongation \[ \delta = \frac{PL}{AE} \] Where: - \(P\) = Axial force (lb) - \(L\) = Length (in) - \(A\) = Cross-sectional area (in²) - \(E\) = Young's modulus (psi) ### Areas \[ A = \frac{\pi}{4} d^2 \] - \(A_{AB} = \frac{\pi}{4} (.4)^2 = .126\) in² - \(A_{BC} = \frac{\pi}{4} (.8)^2 = .503\) in² - \(A_{CD} = \frac{\pi}{4} (.3)^2 = .071\) in² ### Forces (convert kip to lb) - \(P_{AB} = 5 \times 100 = 500\) lb - \(P_{BC} = 2 \times 100 = 200\) lb - \(P_{CD} = 6 \times 100 = 600\) lb ### Young's Modulus (Steel): Assume \(E = 30 \times 10^6\) psi --- ### Elongations #### Segment AB \[ \delta_{AB} = \frac{500 \times 80}{.126 \times 30 \times 10^6} = \frac{400,000}{3.78 \times 10^6} = .106 \text{ in} \] #### Segment BC \[ \delta_{BC} = \frac{200 \times 150}{.503 \times 30 \times 10^6} = \frac{300,000}{15.09 \times 10^6} = .0199 \text{ in} \] #### Segment CD \[ \delta_{CD} = \frac{600 \times 100}{.071 \times 30 \times 10^6} = \frac{600,000}{2.13 \times 10^6} = .282 \text{ in} \] --- ### **Total Elongation** \[ \delta_{total} = .106 + .0199 + .282 \approx .408 \text{ in} \] --- ## Step 3: Permanent Elongation (If Yielding Occurs) If any segment exceeds yield stress, permanent elongation occurs. You need the yield stress from the stress-strain diagram (not provided here). If all stresses are below yield, permanent elongation = . --- ## Step 4: Stress in Each Segment \[ \sigma = \frac{P}{A} \] - \( \sigma_{AB} = \frac{500}{.126} = 39,683 \) psi - \( \sigma_{BC} = \frac{200}{.503} = 3,978 \) psi - \( \sigma_{CD} = \frac{600}{.071} = 84,507 \) psi If yield strength of steel (typically 36,000 psi) is exceeded, CD will yield. --- ## **Summary Table** | Segment | Diameter (in) | Length (in) | Force (lb) | Area (in²) | Stress (psi) | Elongation (in) | |---------|---------------|-------------|------------|------------|--------------|-----------------| | AB | .4 | 80 | 500 | .126 | 39,683 | .106 | | BC | .8 | 150 | 200 | .503 | 3,978 | .0199 | | CD | .3 | 100 | 600 | .071 | 84,507 | .282 | --- ## **Final Answers** ### 1. **Approximate elongation when load is applied:** \[ \boxed{.408 \text{ in}} \] ### 2. **Permanent elongation (if CD yields):** - If yield stress = 36,000 psi, CD will yield. - Permanent elongation = plastic strain in CD \(\times\) length CD. ### 3. **Assumptions:** - Same stress-strain in tension & compression. - Only segment CD yields (if yield = 36,000 psi). --- **Note:** For more exact values, refer to the provided stress-strain diagram to check the amount of plastic (permanent) strain for the stress in CD. If you have that figure, you can compute permanent elongation as: \[ \delta_{perm, CD} = (\epsilon_{CD, final} - \epsilon_{CD, elastic}) \times L_{CD} \] Plug in the values from your diagram.