A vertical two start square threaded screw of a 120 mm nominal (Outside) diameter and 25 mm pitch supports a vertical load of 25 kN. The axial thrust on the screw is taken by a collar bearing of 200 mm outside diameter and 80 mm inside diameter. Find the force required at the end of a lever which is 300 mm long in order to lift and lower the load. The coefficient of friction for the vertical screw and nut is 0.15 and that for collar bearing is 0.20

To solve this problem, we need to calculate the force required at the end of a lever to lift and lower a load using a vertical two-start square threaded screw. We’ll break this down step by step, considering the screw, collar bearing, and lever.

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Given Data:

1. Screw Details:

   • Nominal diameter ($d_o$) = 120 mm

   • Pitch ($p$) = 25 mm

   • Number of starts ($n$) = 2

   • Load ($W$) = 25 kN = 25,000 N

   • Coefficient of friction for screw and nut ($\mu$) = 0.15

2. Collar Bearing Details:

   • Outside diameter ($D_o$) = 200 mm

   • Inside diameter ($D_i$) = 80 mm

   • Coefficient of friction for collar (${\mu }_c$) = 0.20

3. Lever Details:

   • Length of lever ($L$) = 300 mm

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Step 1: Calculate Mean Diameter of Screw ($d_m$)

The mean diameter is the average of the nominal diameter and the core diameter. For a square thread:

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

  $$d_c=120-25=95\text{mm}$$

• Mean diameter ($d_m$):

  $$d_m=\frac{d_o+d_c}{2}=\frac{120+95}{2}=107.5\text{mm}$$

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Step 2: Calculate Lead ($l$)

For a multi-start thread, the lead is the distance the nut moves in one full rotation of the screw. For a two-start thread:

$$l=n\times p=2\times 25=50\text{mm}$$

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Step 3: Calculate Helix Angle ($\alpha$)

The helix angle is the angle between the thread helix and a plane perpendicular to the screw axis. It is given by:

$$\tan (\alpha )=\frac{l}{\pi d_m}$$$$\tan (\alpha )=\frac{50}{\pi \times 107.5}=0.148$$$$\alpha ={\tan }^{-1}(0.148)=8.43\mathrm{°}$$

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Step 4: Calculate Friction Angle ($\varphi$)

The friction angle is related to the coefficient of friction ($\mu$):

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

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Step 5: 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:

   $$T_{screw}=W\cdot \frac{d_m}{2}\cdot \tan (\alpha +\varphi )$$$$T_{screw}=25,000\cdot \frac{107.5}{2}\cdot \tan (8.43\mathrm{°}+8.53\mathrm{°})$$$$T_{screw}=25,000\cdot 53.75\cdot \tan (16.96\mathrm{°})$$$$T_{screw}=25,000\cdot 53.75\cdot 0.305$$$$T_{screw}=409,531.25\text{Nmm}=409.53\text{Nm}$$

2. Torque to Overcome Collar Friction:
   The collar friction torque depends on the mean radius of the collar ($r_c$):

   $$r_c=\frac{D_o+D_i}{4}=\frac{200+80}{4}=70\text{mm}$$$$T_{collar}={\mu }_c\cdot W\cdot r_c$$$$T_{collar}=0.20\cdot 25,000\cdot 70$$$$T_{collar}=350,000\text{Nmm}=350\text{Nm}$$

3. Total Torque to Lift the Load:

   $$T_{lift}=T_{screw}+T_{collar}$$$$T_{lift}=409.53+350=759.53\text{Nm}$$

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Step 6: Torque Required to Lower the Load ($T_{lower}$)

The torque required to lower the load is calculated similarly, but the friction angle is subtracted from the helix angle:

$$T_{screw\mathrm{\_}lower}=W\cdot \frac{d_m}{2}\cdot \tan (\alpha -\varphi )$$$$T_{screw\mathrm{\_}lower}=25,000\cdot 53.75\cdot \tan (8.43\mathrm{°}-8.53\mathrm{°})$$$$T_{screw\mathrm{\_}lower}=25,000\cdot 53.75\cdot \tan (-0.1\mathrm{°})$$

Since $\tan (-0.1\mathrm{°})$ is very small, $T_{screw\mathrm{\_}lower}\approx 0$.

The total torque to lower the load is primarily due to collar friction:

$$T_{lower}=T_{collar}=350\text{Nm}$$

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Step 7: Force Required at the End of the Lever

The force ($F$) required at the end of the lever is calculated using the torque equation:

$$T=F\cdot L$$$$F=\frac{T}{L}$$

1. Force to Lift the Load:

   $$F_{lift}=\frac{T_{lift}}{L}=\frac{759.53}{0.3}=2531.77\text{N}$$

2. Force to Lower the Load:

   $$F_{lower}=\frac{T_{lower}}{L}=\frac{350}{0.3}=1166.67\text{N}$$

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Final Results:

• Force to Lift the Load: $2531.77\text{N}$

• Force to Lower the Load: $1166.67\text{N}$

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Conclusion

To lift the 25 kN load, a force of approximately 2531.77 N is required at the end of the 300 mm lever. To lower the load, a force of approximately 1166.67 N is required. These calculations account for the friction in both the screw threads and the collar bearing.