## Introduction to Ratio of Driving Tension for Flat Belts

The belts are flexible machine elements that are used to transmit power. The transmission of power between the two pulleys that are mounted to shafts that may be parallel or perpendicular or inclined at certain angle. The diameter of the pulleys may same or different depending on the requirement of the speed. The belt drives are always preferred when the shock loads are involved. The factors like linear velocity of the belt, angle of lap and belt tension plays a major role in determining the power capacity of the belt drive.

## Types of Belts

The following are the important types of belts:

• Flat belts

• Trapezoidal belts or V-belts

Circular belts or rope

## Ratio of Driving Tensions for Flat Belt Drives

Let us assume the pulley shown in figure is a driven pulley and is driven by a driving pulley or driver. Also we assume that the direction of rotation of the pulleys are in the clockwise.

Let, T1 and T2 be the tight side tension and slack side tension or loose side tension respectively.

µ be the coefficient of friction between the belt and the pulley

θ be the angle of lap or angle of contact or angle of embracement in radians.

Now, consider a small element PQ of the belt with an included angle dq at the centre of the pulley. Considering the equilibrium the following forces are present on the belt element PQ:

• Tension T in the belt at the point P

• Tension (T + δT) in the  belt at the point Q

• Normal reaction RN

• Force of friction or frictional force F

The magnitude of F will be equal to µ × RN

Resolving all the forces in a horizontal direction,

RN = (T + δT) × sin(δθ/2) + (T × sin(δθ/2))

Since, (δθ/2) is very small angle, the value of sin(δθ/2) can be written as (δθ/2) in the above equation, we get

RN = (T + δT) × (δθ/2) + (T × (δθ/2))

= T × (δθ/2) + (δT × (δθ/2)) + T × (δθ/2)

Since the value of δT×(δθ/2) is too small and hence neglecting the term, we get

RN = T × δθ

Now, resolving all the forces in a vertical direction, we get

µ × RN = (T + δT) × cos(δθ/2) - T × cos(δθ/2)

Since the angle (δθ/2) is very small, substituting the value of cos(δθ/2) as 1 in the above equation, we get

µ × RN = (T + δT) - T = δT

RN = (δT / µ)

Now equating the both the values of RN, we get

(δθ/µ) = T × δθ

(δT / T) = µ × δθ

Integrating the above equation and applying the limits for the tension from T1 to T2 and the angle of lap from 0 to θ respectively, we get

This above equation is known as the Ratio of Driving Tensions for the Flat Belts.

This equation gives the relationship between the tensions on the tight side and the slack or loose side in terms of the angle of lap and the frictional coefficient between the belt and the pulley.

## Angle of Lap or Angle of Contact

The angle of lap for the open belt drives can be determined from the relation

Where,

Where, r1 and r2 are the radius of smaller and larger pulleys respectively and x is the centre distance between the driving and driven pulleys.

Example

A flat belt runs over a driving pulley of diameter 500 mm at 180 rpm. The maximum tension in the belt is limited to 2000 N. Taking the coefficient of friction between the pulley and the belt as 0.25 and the angle of lap as 160o, determine the power transmitted by the belt.

Given Data:

Diameter of the pulley,                   D = 500 mm = 0.5 m

Rotational speed of the pulley,        N = 180 rpm

Maximum tension in the belt,            T1 = 2000 N

Coefficient of friction,                      µ = 0.25

Angle of contact,                             θ = 160 × ( p / 180) = 2.793 rad

To determine the power transmission, we need to find out the velocity of the belt.

The linear velocity of the belt,

Let, T2 be the tension on the slack side of the belt

We know that the ratio of driving tensions for the flat belt is given by

Power transmitted by the belt, P = (T1 – T2)×v

= (2000 – 995) × 4.712

P = 4735.56 W

P = 4.74 kW

Ratio of Driving Tension for Flat Belts

## Related Topics

All Engineering Topics

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## Introduction to Ratio of Driving Tension for Flat Belts

The belts are flexible machine elements that are used to transmit power. The transmission of power between the two pulleys that are mounted to shafts that may be parallel or perpendicular or inclined at certain angle. The diameter of the pulleys may same or different depending on the requirement of the speed. The belt drives are always preferred when the shock loads are involved. The factors like linear velocity of the belt, angle of lap and belt tension plays a major role in determining the power capacity of the belt drive.

## Types of Belts

The following are the important types of belts:

• Flat belts

• Trapezoidal belts or V-belts

Circular belts or rope

## Ratio of Driving Tensions for Flat Belt Drives

Let us assume the pulley shown in figure is a driven pulley and is driven by a driving pulley or driver. Also we assume that the direction of rotation of the pulleys are in the clockwise.

Let, T1 and T2 be the tight side tension and slack side tension or loose side tension respectively.

µ be the coefficient of friction between the belt and the pulley

θ be the angle of lap or angle of contact or angle of embracement in radians.

Now, consider a small element PQ of the belt with an included angle dq at the centre of the pulley. Considering the equilibrium the following forces are present on the belt element PQ:

• Tension T in the belt at the point P

• Tension (T + δT) in the  belt at the point Q

• Normal reaction RN

• Force of friction or frictional force F

The magnitude of F will be equal to µ × RN

Resolving all the forces in a horizontal direction,

RN = (T + δT) × sin(δθ/2) + (T × sin(δθ/2))

Since, (δθ/2) is very small angle, the value of sin(δθ/2) can be written as (δθ/2) in the above equation, we get

RN = (T + δT) × (δθ/2) + (T × (δθ/2))

= T × (δθ/2) + (δT × (δθ/2)) + T × (δθ/2)

Since the value of δT×(δθ/2) is too small and hence neglecting the term, we get

RN = T × δθ

Now, resolving all the forces in a vertical direction, we get

µ × RN = (T + δT) × cos(δθ/2) - T × cos(δθ/2)

Since the angle (δθ/2) is very small, substituting the value of cos(δθ/2) as 1 in the above equation, we get

µ × RN = (T + δT) - T = δT

RN = (δT / µ)

Now equating the both the values of RN, we get

(δθ/µ) = T × δθ

(δT / T) = µ × δθ

Integrating the above equation and applying the limits for the tension from T1 to T2 and the angle of lap from 0 to θ respectively, we get

This above equation is known as the Ratio of Driving Tensions for the Flat Belts.

This equation gives the relationship between the tensions on the tight side and the slack or loose side in terms of the angle of lap and the frictional coefficient between the belt and the pulley.

## Angle of Lap or Angle of Contact

The angle of lap for the open belt drives can be determined from the relation

Where,

Where, r1 and r2 are the radius of smaller and larger pulleys respectively and x is the centre distance between the driving and driven pulleys.

Example

A flat belt runs over a driving pulley of diameter 500 mm at 180 rpm. The maximum tension in the belt is limited to 2000 N. Taking the coefficient of friction between the pulley and the belt as 0.25 and the angle of lap as 160o, determine the power transmitted by the belt.

Given Data:

Diameter of the pulley,                   D = 500 mm = 0.5 m

Rotational speed of the pulley,        N = 180 rpm

Maximum tension in the belt,            T1 = 2000 N

Coefficient of friction,                      µ = 0.25

Angle of contact,                             θ = 160 × ( p / 180) = 2.793 rad

To determine the power transmission, we need to find out the velocity of the belt.

The linear velocity of the belt,

Let, T2 be the tension on the slack side of the belt

We know that the ratio of driving tensions for the flat belt is given by

Power transmitted by the belt, P = (T1 – T2)×v

= (2000 – 995) × 4.712

P = 4735.56 W

P = 4.74 kW

Ratio of Driving Tension for Flat Belts

## Related Topics

All Engineering Topics

More Q&A
user
The text passage provides a detailed explanation of the ratio of driving tensions for flat belt drives, including the derivation of the relevant equation and an example problem with a solution.

TYPE: SINGLE_QUESTION_WITH_ANSWER

The passage contains a single mathematical question in the form of an example problem, and the answer to that question is provided within the text.