Engineering Mechanics

Chapter 33 : Transmission of Power by Belts and Ropes „„„„„ 679

Note: 1. We see that the above expression is a function of (r 1 + r 2 ). It is thus obvious, that if sum of
the radii of the two pulleys be constant, length of the belt required will also remain
constant ; provided the distance between the centres of the pulleys remain unchanged.

2. If width of the belt is given then the effective corresponding diameters (and then radii) will
   be given by the sum of actual diameter of the pulleys and width of the belt.
   Example 33.4. Find the length of belt necessary to drive a pulley of 500 mm diameter
   running parallel at a distance of 12 meters from the driving pulley of diameter 1600 mm.
   Solution. Given: Diameter of the driven pulley (d 2 ) = 500 mm = 0·5 m or radius (r 2 ) = 0·25 m;
   Distance between the centres of the two pulleys (l) = 12 m and diameter of the driving pulley
   (d 1 ) = 1600 mm = 1·6 m or radius (r 1 ) = 0·8 m.
   In this example, no mention has been made whether the belt is open or crossed. Therefore we
   shall find out the value of length of the belt in both the cases.

Length of the belt if it is open
We know that the length of the belt if it is open,

L=

2

12

12

(– )

()2

rr

rr l

l

π+ + +

=

(0·8 – 0·25)^2

(0·8 0·25) (2 12)

12

π+ +×+ m

= 27·32 m Ans.
Length of the belt if it is crossed

We know that length of the belt if it is crossed,

L=

2

12

12

()

()2

rr

rr l

l

+

π+ + +

=

(0·8 0·25)^2

(0·8 0·25) (2 12)

12

• π+ +×+
  = 27·39 m Ans.
  Example 33.5. Find the length of the belt required for driving two pulleys in a cross belt
  drive of 600 mm and 300 mm diameter when 3·5 m apart. Take thickness of the belt as 5 mm.

Solution. Given: Diameter of driving pulley (d 1 ) = 600 + 5 = 605 mm = 0·605 m or radius
(r 1 ) = 0·3025m; Diameter of driven pulley (d 2 ) = 300 + 5 = 305 mm = 0·305 m or radius (r 2 ) = 0·1525
m and distance between the centres of the pulleys (l) = 3·5 m

We know that length of the cross belt drive

L=

2

12

12

()

()2

rr

rr l

l

+

π+ + +

=

(0·3025 0·1525)^2

(0·3025 0·1525) (2 3·5) m

3·5

+

π+ +×+

= 8.488 m Ans.

EXERCISE 33.1

1. A diesel engine shaft having a speed of 180 r.p.m. is required to drive a machine shaft with the
   help of a belt. Find the speed of the machine shaft, if the diameters of the engine shaft and
   machine shaft are 300 mm and 200 mm respectively. [Ans. 270 r.p.m.]


2. In a workshop, a machine shaft is driven by an electric motor with the help of belts across a
   main shaft and a counter-shaft. The diameters of the driving pulleys are 500 mm, 400 mm and
   300 mm respectively whereas the diameters of the driven pulleys are 250 mm, 200 mm and 150
   mm respectively. Find the speed of the machine shaft when the electric motor runs at 150 r.p.m.
   [Ans. 1200 r.p.m.]