1. Introduction

The Purpose of this document is to give a practical example of how to calculate the number of brake rollers that need to be installed on a specific configuration of conveyor belt. In order to successfully understand the terminology used in this document the following basic concepts regarding conveyor belt layouts and friction need to be understood.

1.1 Conveyor Belt Angle

The correct angle at which a conveyor belt operates is important in determining the correct number of BELT-BRAKETM to install. Both the height of the conveyor as well as the height difference between the head pulley and the tail pulley is required as seen in Figure 1 to determine the conveyor belt angle. The angle can be calculated from normal trigonometric relationships.

Figure 1: Conveyor Angle

Sinθ = (H/L)     therefore     θ = Sin-1(H/L)                                                         (1)

1.2 Normal Force

A normal force is a force that acts perpendicularly to a surface. When the angle at which a surface is orientated is at a zero angle (θ=0°) the normal force is equal to the weight of the object that is lying on the surface. However as soon as the surface is raised at a certain angle the normal force is fractionally less depending on the angle at which the surface is inclined. The normal force can be calculated for the following four vector force combinations and layouts as seen in Figure 2.

Figure 2: Normal Forces [1]

The configuration that is applicable to the mass being distributed on a conveyor belt at a certain angle theta (θ) is

N = mgCosθ     [N]                                                                                                (2)

Hence only a portion of the mass acts perpendicular to the surface of the conveyor belt and the other portion want to roll back down the incline.

1.3 Coefficient of Friction

Friction is the phenomena of the interlocking forces between the irregularities between two surfaces. These forces increase when trying to move the surfaces relative to each other until just prior to motion occurring when the forces are at there greatest. After there is relative motion the friction forces are found to be slightly less than the amount just prior to the relevant movement between the two surfaces.

The experimental Static coefficient of friction can be determined by placing a mass at an incline and determining the tangent of the threshold angle at which the mass starts sliding down the incline.

The experimental Kinetic coefficient of friction is determined by using the tangent of the minimum angle at which the mass will maintain a constant velocity down the incline when initially assisted to overcome the static friction threshold and set into motion. The Kinetic coefficient of friction is found to be constant for various velocities and is always less that the Static coefficient of friction.

Figure 3: Determining Friction Forces [1]

It is important to note that coefficient of kinetic and static friction is a function of the two materials which are being used and hence there is not a coefficient of friction for say wood or for glass. There is however a unique coefficient between say tar and a tyre. It is also important to note that the surface finish plays a big role in the coefficient.

1.4 Weight distribution

When there is certain average mass distributed over a load-carrying member, which in this case is the conveyor belt we need to know what the weight distribution in N/m is. The symbol q is used for the weight distribution. Both the mass being transported per hour and the weight of the conveyor belt in kg/m needs to be used in order to accurately determine the weight distribution.

To give an example of how we calculate the weight distribution we use the following example.

Example 1:

1) Velocity of Conveyor Belt (V) = 2 m/s

2) Tonnage per hour (T) = 1000 tons per hour

3) Mass of belt per meter (mB) = 30 kg/m

The mass distribution is calculated as follows:

1 Ton = 1000 kg therefore 1000 ton = 1.0x106 kg (1000 000 kg)

Mass transported per second    = (1.103)/3600 = 277.78 kg/s

Mass distribution = Mass per second/ belt velocity (V) + Belt mass (mB)

= 277.78/2 + 30kg/m   = 169 kg/m>

Weight distribution = Mass Distribution* Gravitational Constant

Therefore                     q = 1657N/m (g = 9.81 m/s2)

Mass Distribution (q) =g((T.1000/3600.V) + MB))          [N/m]                               (3)

This value is needed to determine the normal and horizontal force components that react through the BELT-BRAKETM rollers and hence its structure.

1.5 Configuration Factor

A trough will typically be a three or a five roll configuration and therefore a different percentage of the mass that is distributed on a section of the conveyor will act on the idler or on the installed BELT-BRAKETM. On average the percentage of mass that acts on the centre idler of a three roll configuration is 64% and on a 5 roll configuration approximately 38 %. The remaining percentage on a three roll configuration is then 18% per wing idler.

Figure 4: Roller Configuration

Figure 4 illustrates the approximate factor for both the three and five roller through. Therefore we need to incorporate a factor which will be used to determine correctly the normal force that will act on the BELT-BRAKETM. We will call this factor the configuration factor CF. The configuration is designate the subscript 3 to the three roll and 5 to the 5 roll configuration.

Configuration factor:  CF3 = 0.64; 0.18      or  CF5 = 0.38; 0.20; 0.11             (4)

2. Calculating the number of BELT-BRAKETM rollers to be installed.

The number of BELT-BRAKETM rollers that needs to be installed can be calculated as follows.

Firstly the amount of weight that needs to be stopped will be formulated as a function of the angel of the conveyor. This amount will be multiplied by a safety factor. The amount of braking force that can be realised by the BELT-BRAKETM will then be formulated as a function of the angle of the conveyor, the friction coefficient and the Configuration Factor. These two can then be equated to determine how many BELT-BRAKETM rollers need to be installed for a specific configuration.

2.1 Total mass that needs to be stopped.

The amount of friction force that is required to stop the mass that is lying on the conveyor belt is the component of mass that want to roll back down the specific incline. In the same way that the component of the mass that is acting normal to the conveyor belt is given by m.Cosθ so the component that wants to roll back is given by m.Sinθ

This total mass that want to roll down the conveyor is given by the total length of the belt (L) multiplied by the mass distribution (q). The total mass is also multiplied by the safety factor (SF). This total quantity is called the force required to stop (FRS) and is given in Equation 5.

FRS =(SF)LqSinθ       [N]                                                                                         (5)

2.2 Friction force that can be realised by a BELT-BRAKETM

To differentiate between the friction force that is required and the friction force that can be realised we allocate FFB to the friction force that a BELT-BRAKETM is capable of producing. We assume that the total weight that acts on a trough is equal to the weight distribution that lies between two troughs. This distance will be designated the symbol LR. Figure 5 indicates LR more clearly.

Therefore the friction force that can be realised per BELT-BRAKETM is the normal force m.Cosθ multiplied by the friction coefficient  µ  multiplied by the Configuration factor given Equation 4 (CF). The mass that acts on any roller is the mass distribution q multiplied by LR.

FFB = CF.µ.q.LR.cosθ     [N]                                                                                   (6)


Figure 5: Mass distribution between two troughs

The total amount of friction force that is required is the total number of BELT-BRAKETM rollers that will be installed (NBR) multiplied by the amount that each can stop or the friction force that can be realised by each BELT-BRAKETM (FFB). Therefore the friction force required is given the designation FFR.

FFR = NBR FFB                      [N]                                                                             (7)

All that is still required is to determine the number of rollers to install. This is done by noting that the friction force that is required FFR in Equation 7 should at least be greater or equal to the Force that we require to stop FRS in Equation 5.

FFR ≥ FRS                                                                                                               (8)


CF.NBRLR.µCosθ ≥ q.L.(SF).Sinθ

NBR ≥ L.Tanθ.SF / LR.µ.CF                                                                                     (9)

3. Symbols

Symbol   Description
µ [Dimensionless]> Static Friction Coefficient
θ [Degrees] Angle of Conveyor belt
N [Newton] Normal Force
NBR [Dimensionless] Number BELT-BRAKETM Rollers
L [m] Length of Conveyor Belt
g [m.s-2] Gravitational Constant
q [N.m-1] Weight Distribution
V [m.s-1] Velocity
NBR [Dimensionless] No of Brake Rollers to be Installed
FFR [N] Stopping force realised by all BELT-BRAKETM installed
FRS [N] Stopping Friction force required to stop
FFB [N] Stopping Force realised by a BELT-BRAKETM

LR [m] Distance between brake rollers
T Kg/hour Tonnage per hour
SF [Dimensionless] Safety Factor normally 2
LR [m] Distance between Idler Rollers
mB [kg/m] Belt Mass

4. References


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