A machine slide weighing 4 kN is elevated by a two-start square threaded screw with the following specifications: Screw diameter: 40 mm Pitch: 6 mm Rate of elevation: 0.6 m/min The coefficient of friction at the thread and the thrust collar is 0.12. The thrust collar has an inside diameter of 30 mm and an outside diameter of 50 mm. Calculate the power of the motor required to drive the slide.

Calculating the Power Required to Drive a Machine Slide Using a Square Threaded Screw

In mechanical systems, screws are often used to convert rotational motion into linear motion. A common application is in machine slides, where a screw mechanism is used to elevate or move heavy loads. In this blog, we will calculate the power required to drive a machine slide using a square threaded screw, considering friction at the threads and the thrust collar.

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Problem Statement

A machine slide weighing 4 kN is elevated by a two-start square threaded screw with the following specifications:

• Screw diameter: 40 mm

• Pitch: 6 mm

• Rate of elevation: 0.6 m/min

The coefficient of friction at the thread and the thrust collar is 0.12. The thrust collar has an inside diameter of 30 mm and an outside diameter of 50 mm. Calculate the power of the motor required to drive the slide.

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Step-by-Step Solution

1. Determine the Lead of the Screw:
   Since the screw is two-start, the lead (linear distance traveled per revolution) is twice the pitch:

   $$\text{Lead}=2\times \text{Pitch}=2\times 6=12\text{mm}=0.012\text{m}$$

2. Calculate the Speed of Rotation:
   The slide is elevated at a rate of 0.6 m/min. The rotational speed (in rpm) is:

   $$\text{Rotational speed}=\frac{\text{Rate of elevation}}{\text{Lead}}=\frac{0.6}{0.012}=50\text{rpm}$$

3. Calculate the Mean Diameter of the Screw:
   The mean diameter ($d_m$) of the screw is:

   $$d_m=\frac{\text{Outer diameter}+\text{Inner diameter}}{2}=\frac{40+30}{2}=35\text{mm}=0.035\text{m}$$

4. Calculate the Helix Angle ($\alpha$):
   The helix angle is given by:

   $$\tan \alpha =\frac{\text{Lead}}{\pi \cdot d_m}=\frac{0.012}{\pi \cdot 0.035}\approx 0.109$$$$\alpha ={\tan }^{-1}(0.109)\approx 6.22^{\circ }$$

5. Calculate the Frictional Torque at the Thread:
   The frictional torque ($T_{\text{thread}}$) is:

   $$T_{\text{thread}}=\frac{W\cdot d_m}{2}\cdot \frac{\mu +\tan \alpha }{1-\mu \tan \alpha }$$

   Substituting the values:

   $$T_{\text{thread}}=\frac{4000\cdot 0.035}{2}\cdot \frac{0.12+0.109}{1-0.12\cdot 0.109}\approx 70\cdot 0.229\approx 16.03\text{Nm}$$

6. Calculate the Frictional Torque at the Collar:
   The mean radius of the collar ($r_c$) is:

   $$r_c=\frac{50+30}{4}=20\text{mm}=0.02\text{m}$$

   The frictional torque ($T_{\text{collar}}$) is:

   $$T_{\text{collar}}=\mu \cdot W\cdot r_c=0.12\cdot 4000\cdot 0.02=9.6\text{Nm}$$

7. Total Torque Required:

   $$T_{\text{total}}=T_{\text{thread}}+T_{\text{collar}}=16.03+9.6=25.63\text{Nm}$$

8. Calculate the Power Required:
   The power ($P$) is given by:

   $$P=$$

   Substituting the values:

   $$P=\frac{2\pi \cdot 25.63\cdot 50}{60}\approx 134.5\text{W}$$

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Final Answer

The power required to drive the machine slide is approximately 134.5 W.

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