FLAT BELT FORMULAS BELT TENSION RATIO
OPEN BELT FORMULAS F 1−F c fθ
=e
F 2−F c
CENTRIFUGAL FORCE
ρb t v 2
F c=
Length of Belt g
2
π ( D2−D1 )
L= ( D1 + D2 ) +2 C+ NET BELT PULL/ EFFECTIVE BELT
2 4C
Angle of Wrap Or Contact
PULL/ NET BELT TENSION
θ1=180−2 sin−1 ( D 2C−D )
2 1
θ2 +θ1=360
F 1−F2= ( F1−F c ) ( e fθ −1
e fθ )
D −D
θ2=180+ 2sin−1 ( 2C )
2 1
(
F 1−F2=bt S d−
ρ v2
g )( efθ −1
e fθ )
CROSS BELT FORMULAS
DESIGN STRESS
F1
Sd =
bt
SU
Sd = X Joint Factor
Length of Belt
FS
2
π ( D2 + D1 ) Basic assumptions if the following are not given in
L= ( D1 + D2 ) +2 C+
2 4C the problem
Angle of Wrap Or Contact
D2−D1 LEATHER BELT
θ1=θ 2=180+2 sin−1
ARC OF CONTACT
( 2C ) Sd =400 X Joint Factor , psi
s1=r 1 θ2 s2=r 2 θ2 RUBBER BELT
Sd =350 X Joint Factor , psi
Where,
D1 – diameter of the driving pulley JOINT FACTOR
D2 – diameter of the driven pulley
Cemented Joint: JF = 1
r1 – radius of the driving pulley
r2 – radius of the driven pulley Wire Laced By Machine: JF = 0.88
C – center distance
θ1 – angle of contact of the driving pulley Metal Hooks: JF = 0.35
θ2 – angle of contact of the driven pulley
POWER TRANSMITTED
P=( F 1−F 2 ) v
mm, V = 18.3 m/s Note: the density of leather belt is
BELT SPEED
approximately 0.035 lb/in3 or 974 kg/m3.
a. 205N b. 209N c. 305N d.215N
NO SLIP MENTIONED
4. Two pulleys which are 24 inches in diameter and
v=π D1 N 1=π D2 N 2 which run at 370 rpm are connected by a leather belt
3/8 inch thick. If the belt transmits 30 Hp and the
WITH SLIP MENTIONED distance between shafts is 10 ft, compute the width of
the belt required considering the allowable design
v1 =π D1 N 1 ( 1−%slip ) stress of 400 psi, the coefficient of friction is 0.45 with
contact angle of 180 deg.
a. 2 in b. 4 in c. 5 in d.1.5 in
v 2=π D 2 N 2 ( 1+%slip ) 5. What is the approximate length of the belt required
a. 26.3 ft b. 34.5 ft c. 28.9 ft d. 31.4 ft
6. An electric motor running at 1200 rpm drives a punch
INSTALLATION OF IDLER PULLEY press shaft at 200 rpm by means of a 130 mm wide
For old transmission capacity and 8 mm thick belt. When the clutch is engaged the
belt slips. To correct this condition, an idler pulley was
e fθ −1
( F 1−F 2) O=( F 1−F c ) ( e fθ ) installed to increase the angle of contact but the same
belt and pulley were used. The original contact angle
on the 200 mm motor pulley is 160 degrees. The
original tension ratio is 2.4 and the net tension is 12
For new transmission capacity N/mm of the belt width. If an increase in transmission
' capacity of 20% will prevent slippage, determine the
e f θ −1
( F 1−F 2) N =( F1 −Fc ) f θ
e ( '
) new angle of contact.
a. 320o b. 220o c. 310o d. 120o
Hence; 7. A 914 mm driving pulley and a 1219 mm driven pulley
'
e f θ −1 are arranged on a 3 meter centers. The output of the
( F1−F 2 ) NEW
=
(
ef θ
'
)
=( 1+%increase )
driven shaft is 84 Hp. The belt speed is assumed to be
1280 m/min with a coefficient of friction of 0.30 and a
slip of 1.5% each pulley. Determine the rpm of the
( F 1−F2 )OLD e fθ −1
( e fθ ) 8.
driven shaft.
a. 329 b. 452.9 c. 454
Determine the rpm of the driving shaft.
d. 543
a. 329 b. 452.9 c. 454 d. 543
Where,
ρ – mass density of belt 9. Determine the power, in kW, input at the driver
For leather: ρ = 0.035 lb/in3 considering that the friction loss is 5% on each shaft.
a. 45.5 b. 32.5 c. 73.4 d. 69.6
For rubber: ρ = 0.0435 lb/in3
f – coefficient of friction 10. Determine the net belt tension considering that the
For leather on iron or steel: f = 0.3 friction loss is 5% on each shaft.
a. 32.1 kN b. 3.21 kN c. 2.31 kN d. 2.41 kN
For leather on paper pulley: f = 0.5
b – width of the belt 11. Find the angle of contact on the small pulley for an
open belt drive with a 72 in center distance. The
t – thickness of the belt pulley diameters are 6 in and 12 in.
g – gravitational constant a. 175.22o b. 170.34o
F1 – angle of contact of the driving pulley c. 185.34o d. 165.34o
12. A 3/8 in flat belt is 12 in wide and is used on a 24 in
F2 – angle of contact of the driven pulley diameter pulley rotating 600 rpm. The specific weight
of the belt is 0.035 lb/in 3. The angle of contact is 150 o.
Sample Problems If the coefficient of friction is 0.3 and the allowable
stress is 3000 psi how much hp can it transmit?
1. Determine the length of belt, in cm, needed for a two a. 34.5 hp b. 69.5 hp
flat belt pulleys having a diameter of 70 cm and 30 c. 78.5 hp d. 54.5 hp
cm. The center distance is 400 cm and pulleys rotate 13. A 5 mm round belt connects a 20 mm pulley with a 40
in same direction. Assume power transmitted is 10 hp mm pulley. The center distance is 150 mm. The 20
at 200 rpm. mm pulley rotates at 100 rpm and the coefficient of
a. 958 b. 859 c. 589 d. 895 friction of the belt is 0.25. Find the horsepower
3. Compute the centrifugal force: for a leather belt with capacity for this arrangement if the allowable belt
the following given: w = 100 mm, thickness = 6.4 stress is 2.6 N/mm2.
a. 0.12 kW b. 0.23 kW
c. 0.00282 kW d. 0.054 kW
