The torque required to rotate 675 lbs mass?

AI Thread Summary
To determine the torque required to rotate a 675 lbs mass, the design must account for the equipment's center of gravity and its non-symmetrical mass distribution. Key considerations include the mounting of the fixture to ensure the center of gravity aligns with the axis of rotation and calculating the worst-case torque based on maximum displacement from this axis. The force exerted by a person on a handwheel and the necessary handwheel diameter must also be evaluated. Additionally, the speed at which the operator needs to turn the equipment influences whether acceleration torque calculations are necessary. Understanding the moment of inertia and the axis of rotation is crucial for accurate torque assessment.
shawnycoconut
Messages
2
Reaction score
0
1651161749058.png


Hello everyone,
I am trying to design a fixture that can clamp onto and turn an equipment 180 degrees for maintenance.
I am trying to figure out what torque will be required at the hand wheel to turn it.
The piece of equipment will be clamped at its center of gravity. However, its mass is not symmetrical.

Pleases guide me along this problem. Thank you
 
Engineering news on Phys.org
Welcome to PF.

What is the MOI of the object?
 
Here are the mass properties of the equipment that needed to be turned.
source: Solidworks Mass = 673.280 pounds

Volume = 3065.995 cubic inches

Surface area = 9939.267 square inches

Center of mass: ( inches )
X = 0.012
Y = 7.023
Z = 0.002

Principal axes of inertia and principal moments of inertia: ( pounds * square inches )
Taken at the center of mass.
Ix = ( 0.000, 1.000, 0.001) Px = 23871.534
Iy = (-0.997, 0.000, 0.072) Py = 37167.538
Iz = ( 0.072, -0.001, 0.997) Pz = 37745.029

Moments of inertia: ( pounds * square inches )
Taken at the center of mass and aligned with the output coordinate system.
Lxx = 37170.503 Lxy = 1.103 Lxz = -41.273
Lyx = 1.103 Lyy = 23871.541 Lyz = 9.484
Lzx = -41.273 Lzy = 9.484 Lzz = 37742.057

Moments of inertia: ( pounds * square inches )
Taken at the output coordinate system.
Ixx = 70379.321 Ixy = 55.828 Ixz = -41.258
Iyx = 55.828 Iyy = 23871.633 Iyz = 18.187
Izx = -41.258 Izy = 18.187 Izz = 70950.963
 
shawnycoconut said:
Pleases guide me along this problem.
1) How will you get it mounted so that the CG will be on the axis of rotation? What is your location tolerance?

2) Given the worst case location tolerance (CG farthest from axis of rotation), what is the worst case torque (CG is horizontally displaced from axis of rotation)?

3) How much force can a normal person exert on a handwheel handle? From that, calculate the minimum handwheel diameter.

4) Is there a reason why the operator cannot turn it by grabbing the object directly? No handwheel.

5) How fast does the operator need to turn it 180 degrees? If more than about 10 seconds, then acceleration/deceleration torque can be ignored. I faster than 4 or 5 seconds, then you need to calculate acceleration/deceleration torque about the axis of rotation.

The above is enough for you to get started. Work through the above carefully, then we can help you with the next steps.
 
  • Like
Likes Lnewqban
jrmichler said:
4) Is there a reason why the operator cannot turn it by grabbing the object directly? No handwheel.
Excellent question!
 
$$ \sum T = I \frac{d^2 \theta}{dt^2}$$

What is the moment of inertia of the body about the axis you wish to rotate it about? Your coordinates of the mass center and MOI calculated by the program are RELATIVE to the CM coordinate system in your 3D model. Which axis are you rotating about. Is that axis coincident with a principal axis? I simple diagram of the part ( indicating the axis of rotation ) would greatly help.Then you are going to have some opposing torque that is a function of the bearing load ( and the quality, type of bearing), and a torque applied by the operator ( through the wheel )
 
Last edited:
Back
Top