# Screw Jack Design Problem: Step-by-Step Solution ## **Given Data** - **Load to lift (\( W \))**: \( 80 \) kN \( = 80,000 \) N - **Material properties (Screw):** - Elastic strength (tension/compression): \( 200 \) MPa - Shear strength: \( 120 \) MPa - **Nut material (Phosphor-bronze):** - Elastic limit (tension): \( 100 \) MPa - Elastic limit (compression): \( 90 \) MPa - Elastic limit (shear): \( 80 \) MPa - **Maximum bearing pressure (nut-screw):** \( 18 \) MPa - **Factor of Safety (FOS):** \( 2 \) --- ## **1. Design of Screw** ### **Step 1: Calculation of Design Stresses** Divide each ultimate stress by FOS: - **Screw (Steel):** - Allowable tension/compression: \( \frac{200}{2} = 100 \) MPa - Allowable shear: \( \frac{120}{2} = 60 \) MPa - **Nut (Phosphor-bronze):** - Allowable tension: \( \frac{100}{2} = 50 \) MPa - Allowable compression: \( \frac{90}{2} = 45 \) MPa - Allowable shear: \( \frac{80}{2} = 40 \) MPa - **Bearing pressure:** \( \frac{18}{2} = 9 \) MPa --- ### **Step 2: Core Diameter of Screw (based on tension/compression)** The screw is subjected to axial load: \[ \text{Axial stress} = \frac{W}{A_{\text{core}}} \] \[ A_{\text{core}} = \frac{W}{\text{allowable tensile stress}} \] \[ A_{\text{core}} = \frac{80,000}{100 \times 10^6} = 8 \times 10^{-4} \text{ m}^2 = 800 \text{ mm}^2 \] \[ d_{\text{core}} = \sqrt{\frac{4A}{\pi}} = \sqrt{\frac{4 \times 800}{\pi}} \approx 31.9 \text{ mm} \] --- ### **Step 3: Shear Area and Check** Shear area is also at root (core). Shear stress: \[ \text{Shear stress} = \frac{W}{A_{\text{core}}} \] \[ \text{Allowable shear stress} = 60 \text{ MPa} \] \[ A_{\text{core (shear)}} = \frac{80,000}{60 \times 10^6} = 1.33 \times 10^{-3} \text{ m}^2 = 1333 \text{ mm}^2 \] \[ d_{\text{core (shear)}} = \sqrt{\frac{4 \times 1333}{\pi}} \approx 41.2 \text{ mm} \] **Use the higher diameter for safety.** **Take:** \[ d_{\text{core}} = 42 \text{ mm (rounded up to next standard size)} \] --- ### **Step 4: Major Diameter of Screw** Let’s assume a square thread, common for screw jacks. - **Depth of thread (\( h \))**: Usually about \(\frac{p}{2}\) - **Pitch (\( p \))**: Assume \( p = 7 \) mm (can be adjusted) \[ d_{\text{core}} = d - p \] \[ d = d_{\text{core}} + p = 42 + 7 = 49 \text{ mm} \] So, - **Core diameter:** \( 42 \) mm - **Major diameter:** \( 49 \) mm - **Pitch:** \( 7 \) mm --- ## **2. Design of Nut** The nut must be checked for: - **Bearing pressure** - **Shear strength** - **Tensile/compressive strength** --- ### **Step 1: Number of Threads in Engagement (Based on Bearing Pressure)** \[ \text{Bearing Area} = \pi \times d_{\text{mean}} \times l \] Where: - \( d_{\text{mean}} = \frac{d + d_{\text{core}}}{2} = \frac{49 + 42}{2} = 45.5 \) mm Bearing pressure limit (allowable): \( 9 \) MPa Set up equation: \[ W = \text{bearing pressure} \times \text{bearing area} \] \[ 80,000 = 9 \times 10^6 \times \pi \times 45.5 \times l \times 10^{-3} \] \[ l = \frac{80,000}{9 \times 10^6 \times \pi \times .0455} \] \[ l = \frac{80,000}{1,286,000} \approx .0622 \text{ m} = 62.2 \text{ mm} \] Number of threads engaged: \[ n = \frac{l}{p} = \frac{62.2}{7} \approx 8.9 \approx 9 \] --- ### **Step 2: Nut Failure in Shear** The threads can shear off. Shear area per thread: \( A_{\text{shear per thread}} = \pi \times d_{\text{core}} \times t \), where \( t \) is thread thickness (approx \( \frac{p}{2} = 3.5 \) mm) Total shear area: \[ A_{\text{shear}} = \pi \times d_{\text{core}} \times n \times t \] \[ A_{\text{shear}} = \pi \times 42 \times 9 \times 3.5 = \pi \times 42 \times 31.5 \approx 4,153 \text{ mm}^2 \] Shear force allowed: \[ F_{\text{shear}} = A_{\text{shear}} \times \text{allowable shear stress} = 4,153 \times 40 = 166,120 \text{ N} > 80,000 \text{ N} \] **Safe in shear.** --- ### **Step 3: Nut (Body) Tensile/Compressive Strength** - **Outer diameter of nut (\( D \)):** Assume minimum wall thickness = \( d \), so \( D = d + 2d = 3d = 147 \) mm (over-conservative, can be optimized in practice) \[ A_{\text{nut}} = \frac{\pi}{4}(D^2 - d^2) \] \[ A_{\text{nut}} = \frac{\pi}{4}(147^2 - 49^2) = \frac{\pi}{4}(21,609 - 2,401) = \frac{\pi}{4}(19,208) \approx 15,090 \text{ mm}^2 \] Tensile strength: \[ F_{\text{tensile}} = A_{\text{nut}} \times \text{allowable tensile stress} = 15,090 \times 50 = 754,500 \text{ N} > 80,000 \text{ N} \] **Safe in tension/compression.** --- ## **Summary of Final Design** ### **Screw:** - **Core diameter:** \( 42 \) mm - **Major diameter:** \( 49 \) mm - **Pitch:** \( 7 \) mm (square thread) - **Material:** Steel (as specified) ### **Nut:** - **Material:** Phosphor-bronze - **Number of threads engaged:** \( 9 \) - **Length of nut:** \( 62 \) mm - **Outer diameter (recommended):** \( 147 \) mm --- ## **Conclusion** - The screw core diameter is governed by shear, so **use at least \( 42 \) mm core diameter**. - **Engage at least 9 threads** in the nut, giving a nut length of about **62 mm**. - The outer diameter of the nut can be **147 mm** for strength, but may be optimized further. - All stresses are within safe limits with the given factor of safety. --- **If you need design drawings or further optimization based on thread standards or manufacturability, let me know!** --- ### **Diagram (for reference)**  *Alt text: Schematic of screw thread showing major diameter, core diameter, and pitch*