applet-magic.com
Thayer Watkins
Silicon Valley
USA

 The Analysis of Belt Drives

## Flat Belts

The standard analysis of the forces on a small segment of a flat belt on a sheave is as follows. Let θ be the angle for one edge of the segment and θ+Δθ at the other edge. The tension T in the belt is a function of θ. At steady state operation there must be an equilibrium of the forces on the belt segment.

In the diagram the sheave is turning in a counterclockwise direction and driving the belt in the same direction. For the belt segment shown a balance of forces in the tangential (in this case,horizontal) direction requires that the tension force in the belt in the direction of travel balance with the force at the other edge plus the friction force. The friction force is equal to a friction coefficient fc times the perpendicular pressure P(θ) exerted by the sheave on the belt times the area of the segment against the sheave. If the belt width is w and the length of the segment is r, the sheave radius, times the angle increment Δθ then the area is wrΔθ. The friction force on the belt acts in the direction of travel if the sheave is driving the belt. Thus,

#### T(θ+Δθ)cos(Δθ/2) + fcP(θ)wrΔθ = T(θ)cos(Δθ/2)

Since
T(θ+Δθ) is, to the first approximation, T(θ) + T'(θ)Δθ and cos(Δθ/2) goes to unity as Δθ goes to zero, the above equation reduces to:

#### T'(θ)Δθ + fcPwrΔθ = 0 or T'(θ) = -fcP(θ)wr

This says that the tension decreases at a rate of fcPwr per unit angle in the direction of travel of the belt.

In the radial (in this case, vertical) direction the balance of forces requires that the net effect of the tension force and the pressure of the sheave against the belt balance the "centrifugal force" on the belt segment moving around the perimeter of the sheave. This centrifugal force is equal to the mass of the belt segment, say dm, times the velocity of the belt squared divided by the radius of the sheave; i.e., dm(v2/r). The mass of the segment is equal to the belt's linear mass density ρ times the length of the belt segment rΔθ. Thus,

#### T(θ)sin(Δθ/2) + T(θ+Δθ)sin(Δθ/2) - P(θ)wrΔθ = ρrΔθ(v2/r)

Since to the first approximation sin(Δθ/2) = Δθ/2 the last equation reduces to:

#### T(θ) - P(θ)wr = ρv2

These differential equations can be solved for directly T(θ) but it is more convenient and enlightening to first solve for the pressure P(θ). If ρ and v are constant then T'(θ) = P'(θ)wr. Thus

#### P'(θ)wr = -fcP(θ)wr and hence canceling the wr on both sides and dividing by P gives P'/P = -fc which is equivalent to d(ln(P))/dθ = -fc

and thus integration from the initial angle of contact θ0 to an arbitrary angle θ gives the solution

#### P(θ) = P(θ0)exp(-fcθ).

Since T(θ) = P(θ)wr + ρv2 the solution for T is

#### T(θ) = wrP(θ0)exp(-fcθ) + ρv2

The power transmitted to the belt by the sheave is equal to vT(θ1) - vT(θ0) where θ0 is the angle where the belt make contact with the sheave and θ1 is the angle at which the belt separates from the sheave.

(To be continued.)