Power transfer through shaft

In summary, a shaft with a diameter of 50mm between A and B and a diameter of 40mm between B and C, running at 300 rev/min, transmits power from pulley A to pulleys B and C, which each drive a machine in a workshop. The distance between A and B is 3m and that between B and C is 2.4m, giving a total shaft length of 5.4m. The maximum permissible shear stress in the shaft is 80MPa. The maximum power that can be supplied from each of the pulleys B and C is 30.2kW and 31.4kW respectively. The total angle of twist of one end of the
  • #1
MMCS
151
0
A shaft runs at 300 rev/min and transmits power from a pulley A at one end to two
pulleys, B and C, which each drive a machine in a workshop. The distance between
pulleys A and B is 3 m and that between B and C is 2.4 m (i.e. the total length of the
shaft is 5.4 m). The shaft has a diameter of 50 mm between A and B and a diameter
of 40 mm between B and C. If the maximum permissible shear stress in the shaft is 80
MPa, calculate the maximum power which may be supplied from each of the pulleys B
an C, assuming that both machines would be in operation at the same time. Also
calculate the total angle of twist of one en
d of the shaft relative to the other when
running on full load. G = 80 (GPa)

Answers:
(30.2 kW; 31.4 kW; 13.7°)


Attached is my working, i can arrive at the answer of 31.4KW for the pulley at C at the smaller diameter step, however when i try the same calculate with the second moment of area (J) for the larger diameter i get an answer in excess of 60000KW. How would i calculate the torque for the larger diameter section?

as you can see from my working I've used the standard formula rearranged for torque, then used the standard power equation.

Thanks
 

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  • #2
Here is a diagram of the set up..
Pulleys.jpg


Relevant Equations:
Power = T * ω .....(1)
where
T = Torque in Nm
ω = Angular velocity in Rads/S

The Polar Second Moment of Area (J) for a solid round shaft is
J = πd4/32......(2)
where d is the diameter in meters.

Max Stress τmax = Td/2J....(3)

Substitute (2) into (3) and rearrange to give an equation for the Max Torque in terms of the Max Stress..

T = (τmax * π * d3)/16...(4)

Shaft from B to C

Now we can calculate the max torque and power allowed in the section of shaft from B to C...

Using (4) and...

τmax = 80 * 106Pa
d = 0.04m

T = (80*106 * π *0.043)/16
= 1005Nm

Then using (1) we can calculate the max power

PowerC = 1005 * 300 * 2π/60
= 31.6kW
This is slightly higher than the book answer of 31.4kW. Looks like they rounded the Torque down to 1000Nm.

Shaft from A to B

Now we can calculate the max torque in the section of shaft from A to B and the power available at B...

Using (4) and...

τmax = 80 * 106Pa
d = 0.05m

T = (80*106 * π *0.053)/16
= 1963Nm

However this time to calculate the power available at B we must subtract the torque required at C ..

T = 1963 - 1005 = 958NM

Then using (1) we can calculate the max power that can be delivered to machine B..

PowerB = 958 * 300 * 2π/60
= 30.1kW
This is fractionally lower than the book answer of 30.2kW. I haven't tried to account for the difference.
 
  • #3
Angle of Twist

Relevant Equations

The angle of twist in each section is given by
θ = TL/GJ......(5)

where
T = Torque
L = Length
J = Polar Second Moment of Area of that section

From my post above..
J = πd4/32......(2)

Substituting for J in (5)...

θ = 32TL/Gπd4

For the section AB

T = 1963Nm
L = 3m
d = 0.05m
G = 80 * 109Pa

so

θAB = (32 * 1963 * 3) / (80 * 109 π * 0.054)

= 0.12 Rads
= 6.8 degrees

For the section BC

T = 1000Nm
L = 2.4m
d = 0.04m
G = 80 * 109Pa

so

θAB = (32 * 1000 * 2.4) / (80 * 109 π * 0.044)

= 76800 / 643398
= 0.12 rADS
= 6.8 degrees same as section AB

Total Twist = 6.8 + 6.8 = 13.6 degrees
 

1. How does power transfer through a shaft work?

Power transfer through a shaft involves the use of mechanical force to rotate a shaft, which then transmits power from one machine or component to another. This can be achieved through various mechanisms, such as gears, pulleys, or belts.

2. What is the purpose of a shaft in power transfer?

The main purpose of a shaft in power transfer is to transmit power from a source, such as an engine or motor, to a load, such as a machine or tool. It serves as a link between the power source and the load, allowing for the efficient transfer of energy.

3. How is the power transferred through a shaft measured?

The power transferred through a shaft is typically measured in terms of torque (rotational force) and rotational speed. Torque is usually measured in units such as Newton-meters (Nm) or foot-pounds (ft-lb), while rotational speed is measured in revolutions per minute (RPM) or radians per second (rad/s).

4. What are some common types of shafts used for power transfer?

There are several types of shafts that are commonly used for power transfer, including straight shafts, splined shafts, keyed shafts, and tapered shafts. The type of shaft used will depend on the specific application and the amount of power that needs to be transmitted.

5. What factors should be considered when selecting a shaft for power transfer?

When selecting a shaft for power transfer, it is important to consider factors such as the amount of power that needs to be transmitted, the speed and torque requirements, the type of load, and the environmental conditions. It is also important to ensure that the shaft is properly sized and designed for the specific application to ensure efficient and reliable power transfer.

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