BELT AND CHAIN DRIVES II
Lecture #8
Course Name : DESIGN OF MACHINE ELEMENTS
Course Number: MET 214
To identify the relationships existing between belt forces (tension in the belt) and the
torques applied to the pulleys of a belt drive system operating in the steady state,
consider initially the situation shown below.
Q: How can the tension existing in the belt shown above be increased?
Q: Why is the amount of belt tension important in pulley applications?
Q: When the pulleys are stationary, is the belt tension the same throughout the entire
belt?
In order to systematize the issues involved in the use of a belt drive system, the figure
shown below will serve as a point of reference for the ensuing discussion.
Using the figure in the previous slide as a reference, note that the motor is the power
source for driving the load labeled as load A. To transfer power to the load A, the
motor must apply power to shaft #1. Since the motor supplies power to shaft #1,
power from the motor is positive and the direction of rotation of the motor is in the
same direction as the torque that the motor applies to shaft #1 for reasons discussed
in a previous lecture.
Note: Shaft #1 is rigid, and pulley #1 is affixed to shaft #1, (not shown in detail). Since
shaft #1 is rigid, shaft #1 will transfer the torque that the motor applies to shaft #1 to
pulley #1 causing pulley #1 to rotate in the same direction as the motor. Since pulley
#1 and the motor are all connected by shaft #1, pulley #1 and shaft #1 rotate with the
same speed as the motor and in the same direction as the motor.
Assuming the system is operating in the steady state, the net torque acting on pulley
#1 must be zero. Consequently the torque applied to pulley #1 by the belt, as shown in
the free body diagram below must offset the torque applied to the pulley #1 by shaft
#1. Accordingly, the following relationship can be written for the torque applied to
pulley #1 by the shaft and the tensions existing in the belt.
T A   F1  F 2 
`
DA
2
Where 𝑇𝐴 = Torque applied to pulley A by shaft
𝐷𝐴 = Diameter of pitch circle of pulley A
𝐹1 , 𝐹2 = forces applied to pulley A by belt
As is evident from the equation on the previous slide, in order for the pulleys to rotate
at a constant rate in the steady state, an imbalance in the belt tension must be
developed. Furthermore, in order to avoid slippage, both sides of the belt must be
maintained in tension. Consequently, to create and imbalance that permits rotation
while avoiding slippage, one side of the belt must be increased in tension beyond the
level existing on the other side. The side of the belt that experiences the greater
amount of tension is referred to as the tight side. The other side of the belt is referred
to as the slack side. In the figure shown on the previous slide, 𝐹1 is the tight side and 𝐹2
is the slack side.
According to Newton’s 3rd law, pulley #1 will apply a torque to shaft #1 that is equal and
opposite to the torque that shaft #1 applies to pulley #1. As a consequence, the torque
applied to shaft #1 by the motor and the torque applied to shaft #1 by pulley #1 will be
equal and opposite as shown in the free body diagram for shaft #1 presented below.
For reason explained in a previous lecture, since the torque that pulley #1 applies to
shaft #1 is in the direction opposite of the direction of rotation of shaft #1, the algebraic
sign for the power associated with pulley #1 is negative and therefore pulley #1 removes
power from shaft #1 . The resulting power balance for shaft #1 is as follows.
𝑃𝑚 = 𝑃𝑝1 → 𝑇𝑚 𝑛𝑚 = 𝑇𝑝1 𝑛𝑚 → 𝑇𝑚 = 𝑇𝑝1
Since the power associated with Pulley #1 is negative, pulley #1 will remove power from
shaft #1 and will transfer the power to pulley #2 via the belt shown in the figure. Pulley
#2 will receive power from pulley #1 and will transfer (supply) the power to shaft #2.
Shaft #2 will transfer the power from pulley #2 to the load A. Consequently, shaft #2
supplies power to load A or in other words, load A removes power from the shaft or
alternatively, shaft #2 transfers power to load A.
A free body diagram for pulley #2 is shown in the figure below. Since the belt transfers
the tensions established by pulley #1 to pulley #2, so that power can be transferred from
pulley #1 to pulley #2, the belt applies a torque to pulley #2. Since the system is
operating in the steady state, the torque applied to pulley #2 by shaft #2 must offset the
torque applied to pulley #2 by the belt. Consequently, the torque that shaft #2 applies to
the pulley #2 must be in a direction opposite to the torque the belt applies to pulley #2.
The torque that shaft #2 applies to pulley 2 is shown as 𝑇𝐵 in the figure. Due to Newton’s
third law, the direction of the torque the pulley applies to the shaft will be in a direction
opposite to the direction of the torque that shaft #2 applies to pulley #2. This is a
confusing point and the student is advised to proceed with caution.
It must be noted that pulley #2 will rotate in the same direction as pulley #1 due to the belt
transferring forces and/or power from pulley #1 to pulley #2. With the above considerations in mind,
an equation relating belt tensions to the torque that shaft #2 applies to pulley #2 can be formulated.
Where 𝑇𝐵 = Torque applied to pulley B by shaft
`
𝐷𝐵 = Diameter of pitch circle of pulley B
DB
T B   F1  F 2 
𝐹1 , 𝐹2 = forces applied to pulley B by belt
2
Due to Newton’s 3rd law, the torque that pulley #2 applies to shaft #2 must be equal and opposite to
the torque that shaft #2 applies to pulley #2. Accordingly, the direction of the torque that pulley #2
applies to shaft #2 must be in the same direction as the direction of rotation of pulley #2. In order to
maintain equilibrium with the external torques applied to shaft #2, the torque that load A applies to
shaft #2 must be equal and opposite to the torque that pulley #2 applies to shaft #2. A free body
diagram of shaft #2 is presented below.
It must be noted that the direction of rotation of shaft #2 is in the same direction as the
torque that pulley #2 applies to shaft #2 Accordingly, pulley #2 transfers positive power
to shaft #2. In essence, pulley #2 is functioning as a power source for shaft #2. Since the
load A applies a torque to shaft #2 that is in a direction opposite to the direction of
rotation of shaft #2, in order to maintain a balance of torques acting on shaft #2,load A
has a negative algebraic sign associated with power and therefore load A is removing
power from shaft #2.
In order to simplify the process of determining the direction of torque that the pulleys
and/or loads apply to the shafts in a belt drive systems, a few terms will be defined, and
a brief discussion of how to apply the principles associated with the new terms will be
provided.
When a pair of pulleys are connected by a belt, one of the pulleys is to be considered the
driver pulley and the other pulley is to be considered the driven pulley. The driver pulley
of a pair of pulleys is the pulley that is “closest” to the motor. The pulley “closest” to the
motor is determined by counting the number of components that are involved in
transferring back to the motor through the system from the pulley in question. The
pulley of the pair that requires the least number of component transfers in order to
transfer through the system back to the motor is the pulley that is “closest” to the motor
and therefore is the driver pulley of the pair. In the system shown previously, pulley #1 is
the driver pulley. The driven pulley of a pair is the pulley that is not the driver. In the
system shown previously, pulley #2 was the driven pulley.
It must be noted that the direction of rotation of shaft #2 is in the same direction as the
torque that pulley #2 applies to shaft #2 Accordingly, pulley #2 transfers positive power
to shaft #2. In essence, pulley #2 is functioning as a power source for shaft #2. Since the
load A applies a torque to shaft #2 that is in a direction opposite to the direction of
rotation of shaft #2, in order to maintain a balance of torques acting on shaft #2,load A
has a negative algebraic sign associated with power and therefore load A is removing
power from shaft #2. A power balance can be written for shaft #2. Since the power
balance for shaft #2 is similar to the balance for shaft #1, the balance for shaft #2 will
not be presented, but is left as an exercise.
Determination of tight side and slack side of a belt:
The direction of rotation of the driver pulley can be used to identify the tight side and
slack side of a belt.
The direction of rotation of the driver pulley “pulls” on the tight side and pushes on the
slack side. Assuming the driver pulley is labeled with an arrow that represents the
direction of rotation of the driver pulley, the tail of the arrow will align with the tight side
and the tip of the arrow will point in the direction of the slack side of the belt assuming
the arrow overlaps the portion of the belt that is in contact with the pulley.
After establishing which pulleys are driver pulleys and which pulleys are driven pulleys in
a power transmission system by using the rules provided on a previous slide, the
direction of rotation of all pulleys in the system can be determined. When considering a
pair of pulleys, the driven pulley will rotate in the same direction as the driver pulley.
The driver pulley of a pair of pulleys will rotate in the same direction as the torque that is
applied to the shaft of the driver pulley by the component supplying power to the shaft.
Recall that pulley #1 rotates in the same direction as the motor rotates, which is the
same direction of the torque that the motor applies to shaft #1 in the system shown
previously. After determining the direction of rotation of all pulleys, apply the following
rules.
If a pulley is a driver pulley, the direction of the torque that the driver pulley will apply
to the shaft is in a direction that is opposite to the direction of rotation of the shaft.
If a pulley is a driven pulley, the direction of the torque that the driven pulley will apply
to the shaft is in the same direction as the direction of rotation of the driven pulley.
As is readily apparent, using the rules associated with driver and driven pulleys is much
simpler than using Newton’s 3rd law to determine the direction of torques that pulleys
apply to a shaft.
Reinforcement: The driver pulley of a pair of pulleys is the pulley that removes (receives)
power from the shaft that it is connected to. The driver pulley transfers the power
received from the shaft to another pulley, referred to as the driven pulley, by a belt
connecting the driver pulley to the driven pulley. The driven pulley receives power from
the driver pulley through the belt and transfers (supplies) the power to the shaft
connected to the driven pulley. In the process of transferring power from one pulley to
another pulley of a pulley pair, the driver pulley receives (removes) power from the shaft
connected to the driver pulley. The driven pulley supplies power to the shaft connected to
the driven pulley. Since the driven pulley supplies power to the shaft connected to the
driven pulley, the driven pulley is functioning in the capacity of the power source for shaft
connected to the driven pulley. In essence, the driven pulley becomes the source of power
for the shaft connected to the driven pulley.
Direction of Rotation:
The driver pulley rotates in the same direction as the torque that is applied to the shaft of
the driver pulley by the component that is supplying power to the shaft.
The driven pulley rotates in the same direction as the driver pulley.
Accordingly, the following relationship exists between torque T and the difference F T - FS
T A   FT  F S 
T B   FT  FS

DA
where
TA 
2
FT 
DB
FS 
2
DA 
Torque being applied to shaft A to establish pulley A as drive pulley
Tension in belt on tight side
Tension in belt on slack side
Diameter of pulley A
Reaction torque on shafts due to pulleys
A) Reaction torque on a shaft due to the driver pulley is in the direction opposite to the direction of
rotation of the driver pulley.
Reaction torque on shaft connected to driven pulley is in the same direction as the direction of
rotation of the driven pulley.
As established as part of the discussion concerning belt drives, the torque applied to a
pulley and the pulling force exerted by a belt on the pulley are related as follows
TA = (FT – FS )
DA
2
TA =Fp
DA
2
Where
Fp  pulling force on belt
Fp = FT – FS
FT  tight side tension
FS  slack side tension
DA  diameter of pulley attached to shaft
Since the tight side tension and the slack side tension exist in combination, it is
necessary to consider the sum of the tight side tension and the slack side tension since
this combination of tensions produces a bending force on the shaft connected to the
pulley. The bending force resulting from the combination of tight side and slack side
tensions leads to shaft deflection and/or bending stresses in the shaft. The bending
force due to belt tensions in a belt drive system is calculated as follows
Fb = FT + FS
Fb = bending force for use in shear and moment diagrams
In order to calculate bending force Fb , additional information concerning the force
system in the belt is needed since there are two unknowns and only one equation.
For the purposes of this class, the approach for determining bending forces will follow
the procedure used in the book by Mott.
To determine the bending force Fb a second equation involving FT and FS is needed so
that the two unknowns, FT and FS can be solved by using two equations.
The ratio of the tight side tension to the slack side tension will be assumed to be a
known value stated in any problem. The ratio will be the second equation need to solve
for both unknowns.
FT
C
Where C = 5 is standard, however some sources report that C can vary
from 2 to 5.
FS
The ratio of the tight side tension to the slack side tension can be used with the pulley
force Fp to solve for FT and FS as shown below
Fp = FT - FS =
(CFS -FS ) =
FS =
T
1
r C 1
T
FT
r
FS
C
 FT = CFS
T
r
 FT = CFS =
T
C
r C 1
Fb = FS + FT =
Fb =
T
C
r C 1
+
T
1
r C 1
T ( C  1)
r ( C  1)
Fb = Fp
( C  1)
( C  1)
> 1 for any C from 2 to 5, so Fb > Fp
( C  1)
( C  1)
Fb will be used to create shear and moment diagrams. Shear and moment diagrams will
be used in conjunction with torque diagrams to determine shaft size using the 1927
standard.
Example: The shaft shown below is being driven by a 5 H.P. electric motor (not shown) at a speed of
n = 175 rpm. Assuming the tight side to slack side ratio is C = 4, and the radius of the pulley is 5
inches, and the system rotates with the load applied suddenly, with minor shocks, and the shaft
material has a yield strength in tension of 𝑆𝑌 = 51,000 𝑝𝑠𝑖 and a factor of safety N= 3 is to be
used in the design, calculate the quantities requested.
1) label direction of rotation of pulley on the pulley
2) calculate the pulling force of the pulley
3) calculate the bending force due to the pulley
4) identify the tight side and slack side of the belt by labeling the belt
5) draw a torque diagram for the shaft
6) calculate bearing reactions
7) draw a shear and moment diagrams
8) calculate the diameter of the shaft required for the application using the 1927
standard as presented in the class note.
9) provide an expression for the angle of twist 𝜑 as a function of G, the shear modulus.