TOPIC : SHAFT

AREAS OF DESCRIPTION
CONCERN
TERMINOLOGY 1. Shaft – a rotating member transmitting power
2. Axle – a stationary member carrying rotating wheels, pulleys, etc.
3. Spindle – a short shaft or axle on a machines
4. Machine Shaft – shaft which is an integral part of the machine
5. Transmission Shaft – shaft is used to transmit power between the source and the
machine absorbing power.
6. Line shaft or main shaft – transmission shaft drive by a prime mover.
7. Countershaft, Jackshaft, short shaft – transmission shaft intermediate between the line
shaft and the drive machine.
MATERIALS & SIZES Materials: Cold Rolled, Hot Rolled, Forged Carbon Steel;
Commercially available sizes: refer to Machine Design by Faires p. 269
& 181 for Doughtie
DESIGN FORMULA 1. Relation of Power, Torque & Speed
a. Bending Force produced by belts on the shaft:
F1 + F2 = C (F1 – F2)
Where:
C = 2 for flat belt & 1.5 for V-belt
F1 = tension in the belt on the tight side
F2 = tension on the belt on the slack
Side
F1 + F2 = Fb, bending force on the shaft
F1 - F2 = F, net driving force

b. Bending Force produces by the belt:

Fb = 2 (F1 – F2 ) ------------------------- Flat belt
= 2 T/r
Fb = 1.5 T/r ------------------------------- V-belts

c. Power
P = 2 ii TN
Where:
P = power transmitted, kw
T = torque of torsional moment, N-m
N = Speed, rpm
F = Transmitted load or Tangential force
r = radius in meter
or HP = Tn / 63000
Where:
T = Ft – lb
n = rpm

2. Stresses on Shafts
a. Shaft subjected to torsional only
a1 for solid shaft.

Tc
d Ss =
J

20d
 TL
θ=¿ ¿
JG
Simplifying: d =
√
3 16 T
π Ss

a2 for hollow shaft ;

Simplifying the equation:

where:
d0 d i
do=
√
3 16 T
Π Ss ¿¿
¿ k = d i / d0

a3 θ=TL/JG
using allowable torsional deflection: 1 degree per length
of 20 d ; L = 20d & θ=π /180
Solving for d based from allowable deflection:

d= √640 T (180)/π G
2

b. Shaft subject to pure bending

b1 for solid shaft St = 32m/ π d 3

3 4
b2 for hollow shaft, St = 32m/ π d 0 (1−k )

c. Shaft subject to torsion and bending

c1 for solid circular shaft:

16
Ssmax ¿ 3
√ M 2+ T 2
πd
16
Stmax ¿ 3 [M + √ M + T ]
2 2

πd

Where:
M= bending moment
T = torsional moment
Ssmax = max. shear stress
Stmax = max, tensile or compressive stress.

b2 for hollow shaft:

16 1
[M + √ M + T ]
2 2
Stmax ¿ 3
πd 1−k 4
16 1
Ssmax ¿ 3 [ √ M +T ]
2 2
4
πd 1−k

When shaft is subjected to an axial load Fa:

4 L fa
Faverage ¿ 3 3
π d 0 −d 0

3. Strength of shaft with assumed allowable stresses (PSME Code p. 18)

Main Power transmitting shaft:

P= D3N/80
 Line shaft carrying pulley;

P= D3N/53.5

Small short shaft;

P= D3N/38

DESIGN From Machinist Handbook and other technical/ design resources, the major formula &
CONSIDERATIONS design considerations are as follows:
1. Allowable twisting moment for a shaft of any cross-section such as circular, square ,
etc.
T = Ss x Zp

2. For a shaft delivering P Hp at N rpm, the twisting moment t T being transmitted is

6300 P
T=
N

3. The diameter of a solid circular shaft required to transmit a given torque:

D= √ 5.1 T /Ssor D= √ 321000 /NSs
3

Note:
Ss allowable used in practice = 4000 psi for main power transmitting shafts.
Ss = 6000 psi for line shafts carrying pulley
Ss = 8500 psi for small, short shafts counter shaft
Thus: we have the following formula similar to PSME Code:

4. For main power transmitting shafts:

P = D3N/80 or P¿ √
3
80 P /N

5. For lineshaft carrying pulleys:

P = D3N/535 or P¿ √
3
53.5 P /N

6. For small, shorts shafts:

P = D3N/38 or P¿ √
3
38 P /N

Metric :

a. Torsional strength of shafting

T = Ss x Zp
Where:
T = twisting moment in N-mn

Ss = allowable torsional shearing stress in N/mn 2

Zp = polar section modulus in mm 3

b. For a shaft delivering power of P Kw at N rpm,

5 6
9.55 x 10 P 10 P
T¿ ∨T =
N W

Where:

T=N-mn & W = angular velocity in rad/sec.

c. The diameter D of a solid circular shaft required to transmit a given torque T:

D¿ √
3
5.1T /Ss or D= √ 48.7 x 106 P/ N

D¿ √
3
5.1 x 106 P /W

Where:
D=mm p=KW Ss = torsional shearing stress, N/mn 2
T= N-mm N=rpm

For main power transmitting shafts:

D3 N
P¿ or D¿ √
3
1.77 x 106 P/ N
1.77 x 106

For small, short shafts:

D3 N
D¿ √ 0.83 x 106 P/ N
3
P¿ 6 or
0.63 x 10

Where:

P = KW N=rpm
D=mm

Problems:

1. What should be the diameter of a line shaft to transmit 10 hp if the shaft makes 150
rpm.

D¿
√
3 53.5 x 10
150
=1.53, 19/16 in.

2. What horsepower would a short shaft, 2 in. θ , carrying but two pulleys close to the
bearings transmits if the shaft make 300 rpm?

3 3
D N 2 x 300
P¿ = = 63 hp.
38 38

3. What would be the dia. Of a power transmitting shaft to transmit 150 KW at 500
rpm.

D¿ √ 1.77 x 106 (150)/500=81mm .

3

4. What power would be a short shaft, 50 mm ∅ , transmit at 400 rpm.

3
50 400
P¿ 6 = 60 KW
0.83 x 10

Are you ready and confident you have internalized these shaft concepts? If so, you may
evaluate yourself. Proceed to the practice tasks. . . .
TOPIC : SHAFTS

DIRECTION : Select the best answer for each of the following question. Mark only one for each item by shading the
box corresponding to the letter of your choice on the answer sheet provided. Strictly no
erasures allowed. Use pencil no.1

1. Compute the lineshaft diameter to transmit 12 hp at 180 rpm with torsional deflection of 0.08 deg. per ft.
length.
A. 3 inches B. 5 cm C. 2.35 inches D. 62mm

2. The allowable stresses that are generally used in practice are 4 ksi for main power transmitting shaft, 6 ksi for
line shaft carrying pulleys and 8.5 ksi for small short shaft/countershaft, etc. With these allowable stresses, what
will be the simplified formula to determine transmitted hp, diameter of the shaft or rpm.
A. HP = D3N/80 B. HP = D3N/53.5 C. All of these D. HP = D3N/38
Board Exam April 1997

3. A line shaft runs at 360 rpm. An 18” pulley on the same shaft is belt connected to a 12 inches pulley on the
countershaft. From 15” pulley on the countershaft, motion is transmitted to the machine. Compute the required
diameter of the pulley on the machine to give a spindle speed of 660 rpm.
A. 16 inches B. 12.25 inches C. 10.5 inches D. 8.5 inches

4. Two parallel shafts connected by pure rolling turn in the same direction having a speed ratio of 2.75. What is the
distance of two shaft if smaller D is 22 cm in diameter.
A. 16.6 cm B. 30.25 cm C. 25.25 cm D. 19.25 cm
Board Exam Oct. 1996

5. Determine the torque received by the motor shaft running at 4250 rpm transmitting 11 hp through 10”
diameter, 20 deg. involute gears. The shaft is supported by ball bearings at both ends and the gear is fixed at the
middle 8” shaft length.
A. 163 in-lb B. 167 in-lb C. 132 in-lb D. 138 in-lb
Board Exam April 1998

6. Compute the diameter of the solid shaft transmitting 75 hp at 1800 rpm. The nature of the load & the type of
service is such that the allowable Ss based on pure torsion is 6000 psi.
A. 1 7/8” B. 2 1/16” C. 1 6/16” D. 3 1/8”

7. A small countershaft is 1 ½” in diameter and has Ss all of 8500 psi. Find the hp delivered by the shaft at a speed
of 15.7 rad. Per second.
A. 7.20 B. 1.4 C. 13.31 D. 14.72

8. The shaft is subjected to a steady load of 36, 000 in-lb at a shear stress of 10,000 psi. Compute the diameter of
the said shaft in inches.
A. 1 7/8” B. 2 1/4” C. 3” D. 2 ¾”

9. A 16 ft steel line shaft has no bending action except its own weight. What power in hp can the shaft deliver at a
speed of 200 rpm. Consider that the torsional deflection will not exceed 0.008 per ft length.
A. 13.2 B. 15.8 C. 24.4 D. 14.6
10. The torsional deformation of SAE 1040 steel shaft is 0.8 deg. in a length of ½”. The shear stress is 69 Mpa.
Compute the diameter of the shaft in mm. Steel Modulus of Elasticity is 79300 mPa or N/sq. mm
A. 51 B. 50 C. 75 D. 62