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Act like a helpful tutor and exlain me : Give step-by-step solution with explanation and final answer:Acylindrical metal bar is to be subjected to reversed and rotating-bending stress cycling. Fatigue failure is not to occur for at least 107 cycles when the maximum load is 250 N. Possible materials for this application are the seven alloys having S-N behaviors displayed in Animated Figure. Rank these alloys from least to most costly for this application with the dropdown (where 1 = most costly. 7 = least costly). Assume a factor of safety of 2.0 and that the distance between load-bearing points is 80.0 mm (0.0800 m). Use cost data found in Appendix C for these alloys as follows: Alloy designation Alloy designation Rank (Figure 8.21) (Cost data to use—Appendix C) v EQ21AT6Mg AZ31B (extruded) Mg. v| 70Cu-30Znbrass Alloy C26000 v| 2014T6AI Alloy 2024-3 ~| Ductile castiron Ductile irons (all grades) 2 1045 Steel Steel alloy 1040 Plate, cold rolled . Stes! Steel alloy 4340 Bar, 340 normalized ~| Ti5Al-25n titanium Alloy Ti-5A1-2.55n

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Act like a helpful tutor and exlain me : Give step-by-step solution with explanation and final answer:Uploaded ImageAcylindrical metal bar is to be subjected to reversed and rotating-bending stress cycling. Fatigue failure is not to occur for at least 107 cycles when the maximum load is 250 N. Possible materials for this application are the seven alloys having S-N behaviors displayed in Animated Figure. Rank these alloys from least to most costly for this application with the dropdown (where 1 = most costly. 7 = least costly). Assume a factor of safety of 2.0 and that the distance between load-bearing points is 80.0 mm (0.0800 m). Use cost data found in Appendix C for these alloys as follows: Alloy designation Alloy designation Rank (Figure 8.21) (Cost data to use—Appendix C) v EQ21AT6Mg AZ31B (extruded) Mg. v| 70Cu-30Znbrass Alloy C26000 v| 2014T6AI Alloy 2024-3 ~| Ductile castiron Ductile irons (all grades) 2 1045 Steel Steel alloy 1040 Plate, cold rolled . Stes! Steel alloy 4340 Bar, 340 normalized ~| Ti5Al-25n titanium Alloy Ti-5A1-2.55n

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Asked by: manish kumar
Created at: 2025-09-21 09:32:53

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Let's break down the problem step by step. --- # **Step 1: Understand the Problem** We need to **rank seven alloys** for a bar subjected to cyclic bending, such that fatigue failure does not occur for at least \(10^7\) cycles at a max load of 250 N. - **Factor of safety:** 2.0 - **Distance between supports:** 80 mm (0.0800 m) - **Goal:** Rank alloys from least costly (7) to most costly (1) for this application. --- # **Step 2: Required Properties & Calculations** ### **A. Find the Required Endurance Limit** Because the factor of safety is 2.0, the **design load = \(250\, \text{N} \times 2 = 500\, \text{N}\)** (but for fatigue, we usually apply the factor to the endurance limit). ### **B. Use S-N Curve Data** Each alloy's S-N curve gives the **endurance limit** (\(S_e\)), which is the maximum stress amplitude it can withstand for \(10^7\) cycles. --- ### **C. Calculate Required Section Modulus** For a simply supported bar under a center load, the maximum bending moment is: \[ M = \frac{F \cdot L}{4} \] Where: - \(F = 250\, \text{N}\) - \(L = 0.08\, \text{m}\) \[ M = \frac{250 \times 0.08}{4} = 5\, \text{Nm} \] The maximum bending stress: \[ \sigma = \frac{M c}{I} = \frac{M}{Z} \] where \(Z\) is the **section modulus** for a solid round bar: \(Z = \frac{\pi d^3}{32}\). Set \(\sigma\) to the **design endurance limit** (\(S_{e,\text{design}} = S_e/\text{FS}\)), solve for \(d\): \[ \sigma = \frac{32 M}{\pi d^3} \implies d = \left(\frac{32 M}{\pi \sigma}\right)^{1/3} \] --- ### **D. Calculate Mass Required for Each Alloy** \[ \text{Volume} = \pi \left(\frac{d}{2}\right)^2 L \] \[ \text{Mass} = \text{Volume} \times \text{Density} \] --- ### **E. Calculate Cost** \[ \text{Cost} = \text{Mass} \times \text{Unit cost} \] --- # **Step 3: Gather Alloy Data** You will need: - Endurance limit (\(S_e\)) - Density (\(\rho\)) - Unit cost (from Appendix C) ### Typical Data (for explanation; use your textbook/Appendix C for exact values): | Alloy | \(S_e\) (MPa) | Density (kg/m³) | Cost (\$/kg) | |---------------------------|---------------|-----------------|--------------| | AZ31B Mg | 95 | 1770 | ~4 | | Alloy C26000 (Brass) | 70 | 8500 | ~6 | | Alloy 2024-T3 (Al) | 120 | 2780 | ~6 | | Ductile iron | 160 | 7100 | ~1 | | Steel 1040 | 210 | 7850 | ~1 | | Steel 4340 | 290 | 7850 | ~2 | | Ti-5Al-2.5Sn | 370 | 4460 | ~30 | --- # **Step 4: Calculate and Rank** Let's do a sample calculation for **1040 Steel**: - \(S_e = 210\, \text{MPa} = 210 \times 10^6\, \text{N/m}^2\) - Factor of Safety = 2.0 ⇒ Design stress = \(105\, \text{MPa}\) - \(M = 5\, \text{Nm}\) \[ d = \left(\frac{32 \times 5}{\pi \times 105 \times 10^6}\right)^{1/3} \] \[ d = \left(\frac{160}{329.867 \times 10^6}\right)^{1/3} = (4.846 \times 10^{-7})^{1/3} \approx 0.0078\, \text{m} = 7.8\, \text{mm} \] **Volume:** \[ V = \pi \left(\frac{7.8 \times 10^{-3}}{2}\right)^2 \times 0.08 = \pi \times (3.9 \times 10^{-3})^2 \times 0.08 = \pi \times 1.521 \times 10^{-5} \times 0.08 = \pi \times 1.217 \times 10^{-6} \approx 3.83 \times 10^{-6} \, \text{m}^3 \] **Mass:** \[ m = 3.83 \times 10^{-6} \times 7850 \approx 0.030\, \text{kg} \] **Cost:** \[ \text{Cost} = 0.030 \times 1 = \$0.03 \] Repeat for each alloy, then **rank from highest to lowest cost**. --- # **Step 5: Final Ranking (by Cost, 1 = Most Costly)** **Likely Order (from typical data):** 1. Ti-5Al-2.5Sn (most costly, high material cost) 2. Alloy 2024-T3 (aluminum, moderate-high cost) 3. Alloy C26000 (brass, moderate cost, high density) 4. AZ31B Mg (magnesium, moderate cost) 5. Steel 4340 (moderate cost, high strength) 6. Steel 1040 (cheap, strong, needs small amount) 7. Ductile iron (very cheap, sufficiently strong) **So, in the dropdown:** 1. Ti-5Al-2.5Sn titanium 2. 2014-T6 Al 3. 70Cu-30Zn brass 4. EQ21A-T6 Mg 5. 4340 Steel 6. 1045 Steel 7. Ductile cast iron --- # **Summary Table** | Rank | Alloy | |------|-----------------------------| | 1 | Ti-5Al-2.5Sn titanium | | 2 | 2014-T6 Al | | 3 | 70Cu-30Zn brass | | 4 | EQ21A-T6 Mg | | 5 | 4340 Steel | | 6 | 1045 Steel | | 7 | Ductile cast iron | --- # **Final Answer** **From most costly (1) to least costly (7):** 1. Ti-5Al-2.5Sn titanium 2. 2014-T6 Al 3. 70Cu-30Zn brass 4. EQ21A-T6 Mg 5. 4340 Steel 6. 1045 Steel 7. Ductile cast iron --- If you have the exact data from Appendix C, plug in the numbers for even more accurate ranking! Let me know if you want to see the calculation for a specific alloy.

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