Calculating torque from yield stress understanding

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SUMMARY

The discussion centers on calculating torque for a torque arm design, specifically using yield stress values. The user inquired whether to use approximately 90% of the yield stress (185 MPa) of the torque rod material in their torque calculations. It was clarified that yield stress, measured in Pascals (Pa), cannot be directly substituted into the torque formula, which requires torque in Newton-meters (N·m). The correct approach involves calculating torque using the formula torque = force x distance, ensuring that all units are consistent.

PREREQUISITES
  • Understanding of torque calculations and their units (N·m)
  • Knowledge of yield stress and its measurement (Pa or N/mm²)
  • Familiarity with basic mechanics, specifically force and distance relationships
  • Ability to convert units appropriately for engineering calculations
NEXT STEPS
  • Research the relationship between yield stress and torque calculations in mechanical design
  • Learn about unit conversions between Pascals and Newton-meters
  • Explore torque arm design principles and best practices
  • Study examples of torque calculations in engineering applications
USEFUL FOR

Mechanical engineers, product designers, and anyone involved in torque arm design or mechanical systems requiring precise torque calculations.

vitaljunk
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Hello,
I'm new here and so if I've put this in the wrong section or anything, please let me know and I'll fix/change whatever needs to be done.

I just wanted to see if someone could help me understand something.

When choosing a torque value, is it just ~90% of the yield stress of the material of the torque rod? I have to design a torque arm that needs to withstand a certain volume before hitting a switch and need to decide where to put the force/hang the volume along the arm. I already know the force, and the material of the torque arm and rod, so need to chose a torque value so I can figure out the distance along the arm for the force. Sorry if I'm being a bit vague, I want to solve it myself, but am just having some trouble with understanding things. If you need more info, please let me know.

Anyway, I found the yield stress of the material of the torque rod, it was around 185MPA. I just plugged that straight into torque = force x distance. Is that right? Should I have converted that to Pa? Or because torque is in Newton meters and stress is in pascals and they're not the same right, so I need to do something else to it? I'm not sure what to do. By doing that (putting 185 into the equation as torque) I got a really small value, so I think I've done it wrong and am currently lost.

Could someone please point me in the right direction?

Thank you for your help in advance, and sorry about the rambling.
 
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Vitaljunk;
You mast definitely can not input the yield stress as torque! Torque is measured in n.m whereas the yield stress is measured in Pa or N/mm^2.

If you could attach a picture of what you want to achieve I could try to help you a bit more, also include some of your equations and what you try to achieve.

As I understand you are trying to have a weight hang on a rod and let the rod release once a certain weight is reached? is this correct, from the weight and the distance to the centre of the arm around which it swings will give you a torque figure that you want.

For me to help you I would have to understand what you are on about a bit better

Try to explain...
 
I have encountered a vertically oriented hydraulic cylinder that is designed to actuate and slice heavy cabling into sections with a blade. The cylinder is quite small (around 1.5 inches in diameter) and has an equally small stroke. The cylinder is single acting (i.e. it is pressurized from the bottom, and vented to atmosphere with a spring return, roughly 200lbs of force on the spring). The system operates at roughly 2500 psi. Interestingly, the cylinder has a pin that passes through its...

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