Belt Drive Design

Problem Statement: Design a belt drive to transmit 2000 watts of power over a distance of 3 meters. The driver pulley is connected to a motor running at 1500 RPM, and the belt material is made of rubber. The coefficient of friction between the belt and pulley is assumed to be 0.3. Calculate the appropriate dimensions for the belt and pulleys.

Solution:

Step 1: Power to be transmitted:

The power transmitted by the belt drive is given as:

P=2000W

Step 2: Driver Pulley Speed and Pulley Diameters:

Assuming the driver pulley speed N1=1500RPM, we can choose the driver and driven pulley diameters as:

The velocity ratio VR is given by:

VR=D2D1=400200=2

Step 3: Belt Speed:

The belt speed v can be calculated using the formula:

v=πD1N160

Substituting the known values:

v=π×0.2×150060=15.7m/s

Step 4: Power Transmission Capacity of the Belt:

The power transmitted by the belt can be expressed as:

P=(T1T2)v

Where T1 is the tension on the tight side and T2 is the tension on the slack side. Using the equation for the ratio of tensions:

T1T2=eμθ

Assuming μ=0.3 (coefficient of friction) and θ=πradians (angle of contact between belt and pulley):

T1T2=e0.3×π2.57

Step 5: Tensions in the Belt:

The power equation becomes:

2000=(T1T2)×15.7

Substitute T1=2.57T2 into the equation:

2000=(2.57T2T2)×15.7

Solving for T2:

T2=20001.57×15.781.27N

Now, calculate T1:

T1=2.57×81.27208.87N

Step 6: Belt Dimensions:

The cross-sectional area A of the belt is related to the tension by the formula:

T1=σA

For rubber, the allowable tensile stress σ is around 3 MPa. Therefore:

A=T1σ=208.873×106=69.62mm2

Assuming a rectangular cross-section with width-to-thickness ratio of 2:1:

2xx=69.62x2=34.81x5.9mm

Thus, the belt dimensions are:

Summary of Belt Drive Design: