Design a screw jack of lift 270 mm to carry a axial load of 250 KN

 Designing a screw jack involves determining the dimensions of the screw, nut, and other components to safely carry the specified axial load and lift height. Below is a step-by-step design procedure for a screw jack to lift a load of 250 kN with a lift of 270 mm.

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Design Requirements:

1. Load to be lifted ($W$): 250 kN = 250,000 N

2. Lift height ($L$): 270 mm

3. Material for Screw: High-strength steel (e.g., C40)

4. Material for Nut: Bronze or phosphor bronze (softer than the screw)

5. Factor of Safety (FOS): 3 to 5 (assume 4 for this design)

6. Thread Type: Square thread (for high efficiency and load capacity)

7. Coefficient of Friction ($\mu$): 0.15 (for steel screw and bronze nut)

8. Collar Bearing: Assume a thrust collar with a coefficient of friction (${\mu }_c$) = 0.20

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Step 1: Select Screw Material and Allowable Stress

• Material for Screw: C40 steel

• Yield strength (${\sigma }_y$): 330 MPa (for C40 steel)

• Allowable stress (${\sigma }_{all}$):

  $${\sigma }_{all}=\frac{{\sigma }_y}{\text{FOS}}=\frac{330}{4}=82.5\text{MPa}$$

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Step 2: Determine Core Diameter of Screw ($d_c$)

The core diameter is calculated based on the compressive stress acting on the screw:

$${\sigma }_c=\frac{W}{\frac{\pi }{4}{d}_c^2}$$

Rearranging for $d_c$:

$$d_c=\sqrt{\frac{4W}{\pi {\sigma }_{all}}}$$$$d_c=\sqrt{\frac{4\times 250,000}{\pi \times 82.5}}$$$$d_c=\sqrt{\frac{1,000,000}{259.2}}=\sqrt{3858.3}=62.12\text{mm}$$

Round up to the nearest standard size: $d_c=65\text{mm}$.

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Step 3: Determine Nominal Diameter ($d_o$) and Pitch ($p$)

For a square thread:

• Nominal diameter ($d_o$) = Core diameter ($d_c$) + Pitch ($p$)

• Assume a pitch ($p$) = 10 mm (standard for heavy-duty screws):

$$d_o=d_c+p=65+10=75\text{mm}$$

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Step 4: Check for Buckling

The screw is subjected to compressive load and must be checked for buckling. The critical buckling load is given by:

$$P_{cr}=\frac{{\pi }^{2}EI}{(KL{)}^2}$$

where:

• $E$ = Modulus of elasticity for steel = 210 GPa = $210\times 10^3\text{MPa}$

• $I$ = Moment of inertia of the screw = $\frac{\pi }{64}{d}_c^4$

• $K$ = End condition factor (assume $K=2$ for one end fixed and one end free)

• $L$ = Lift height = 270 mm

Calculate $I$:

$$I=\frac{\pi }{64}(65{)}^4=8.76\times 10^6{\text{mm}}^4$$

Calculate $P_{cr}$:

$$P_{cr}=\frac{{\pi }^2\times 210\times 10^3\times 8.76\times 10^6}{(2\times 270{)}^2}$$$$P_{cr}=\frac{1.82\times 10^{13}}{291,600}=62,400\text{kN}$$

Since $P_{cr}\gg W$, the screw is safe against buckling.

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Step 5: Determine Helix Angle ($\alpha$) and Friction Angle ($\varphi$)

• Lead ($l$): For a single-start thread, $l=p=10\text{mm}$.

• Mean diameter ($d_m$):

  $$d_m=\frac{d_o+d_c}{2}=\frac{75+65}{2}=70\text{mm}$$

• Helix angle ($\alpha$):

  $$\tan (\alpha )=\frac{l}{\pi d_m}=\frac{10}{\pi \times 70}=0.0455$$$$\alpha ={\tan }^{-1}(0.0455)=2.61\mathrm{°}$$

• Friction angle ($\varphi$):

  $$\varphi ={\tan }^{-1}(\mu )={\tan }^{-1}(0.15)=8.53\mathrm{°}$$

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Step 6: Calculate Torque Required to Lift the Load ($T_{lift}$)

The torque required to lift the load is the sum of the torque to overcome screw friction and the torque to overcome collar friction.

1. Torque to Overcome Screw Friction:

   $$T_{screw}=W\cdot \frac{d_m}{2}\cdot \tan (\alpha +\varphi )$$$$T_{screw}=250,000\cdot \frac{70}{2}\cdot \tan (2.61\mathrm{°}+8.53\mathrm{°})$$$$T_{screw}=250,000\cdot 35\cdot \tan (11.14\mathrm{°})$$$$T_{screw}=250,000\cdot 35\cdot 0.196=1,715,000\text{Nmm}=1,715\text{Nm}$$

2. Torque to Overcome Collar Friction:
   Assume a collar with an outside diameter of 100 mm and inside diameter of 50 mm:

   $$r_c=\frac{100+50}{4}=37.5\text{mm}$$$$T_{collar}={\mu }_c\cdot W\cdot r_c$$$$T_{collar}=0.20\cdot 250,000\cdot 37.5=1,875,000\text{Nmm}=1,875\text{Nm}$$

3. Total Torque to Lift the Load:

   $$T_{lift}=T_{screw}+T_{collar}=1,715+1,875=3,590\text{Nm}$$

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Step 7: Design of Nut

• Height of Nut ($H$):

  $$H=n\cdot p$$

  where $n$ is the number of threads in contact. Assume $n=4$:

  $$H=4\cdot 10=40\text{mm}$$

• Shear Stress in Nut:

  $${\tau }_{nut}=\frac{W}{\pi d_{o}H}$$$${\tau }_{nut}=\frac{250,000}{\pi \times 75\times 40}=26.5\text{MPa}$$

  This is within the allowable stress for bronze.

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Step 8: Power Required

Assume the screw is rotated at 10 rpm:

$$\omega =\frac{2\pi N}{60}=\frac{2\pi \times 10}{60}=1.047\text{rad/s}$$$$P=T_{lift}\cdot \omega =3,590\cdot 1.047=3,760\text{W}=3.76\text{kW}$$

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Final Design Specifications:

1. Screw:

   • Core diameter ($d_c$) = 65 mm

   • Nominal diameter ($d_o$) = 75 mm

   • Pitch ($p$) = 10 mm

   • Material: C40 steel

2. Nut:

   • Height ($H$) = 40 mm

   • Material: Bronze

3. Motor Power: 3.76 kW

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This design ensures the screw jack can safely lift a load of 250 kN with a lift of 270 mm.