Angular speed that breaks a spinning body apart with inertial stress

Click For Summary
SUMMARY

The discussion focuses on determining the angular speed at which a spinning hollow cylindrical body fails due to inertial stress. Two sources provide conflicting formulas for calculating maximum radial stress, with one stating it occurs at \( r=\sqrt{r_1*r_2} \) and the other at \( r=0 \). The criteria for failure include whether the tangential or radial stress exceeds the ultimate stress, or if the combined stresses exceed the ultimate stress squared. Understanding principal stresses and applying appropriate failure theories are essential for accurately assessing material failure under stress conditions.

PREREQUISITES
  • Understanding of stress tensor concepts
  • Familiarity with failure theories in materials science
  • Knowledge of mechanical properties of materials (e.g., brittle vs. ductile)
  • Basic principles of rotational dynamics and angular velocity
NEXT STEPS
  • Research the application of principal stress theory in material failure analysis
  • Study different failure theories such as von Mises and Tresca criteria
  • Explore the effects of material properties on stress tolerance, focusing on various states (annealed, hardened, etc.)
  • Learn about the calculation of tangential and radial stresses in rotating bodies using formulas from engineering mechanics
USEFUL FOR

Mechanical engineers, materials scientists, and students studying mechanics of materials who are interested in understanding the failure mechanisms of rotating bodies under stress.

olgerm
Gold Member
Messages
532
Reaction score
35
How to find the angular speed, on which a spinning hollow cylindrical body breaks due to inertial stress(force)?
I found 2 sources(http://www.roymech.co.uk/Useful_Tables/Cams_Springs/Flywheels.html (last 2 equations) , https://www.engineersedge.com/mechanics_machines/solid_disk_flywheel_design_14642.htm (eq. 4) ) that offer different formulas to calculate tangential and radial stresses. One of them claims that maximum radial stress is at ##r=\sqrt{r_1*r_2}## and other one that it occurs at ## r=0##. Which one of these is correct?
Is it that the body breaks if
##\sigma_{tangential}>\sigma_{ultimate\ stress}\ or\ \sigma_{radial}>\sigma_{ultimate\ stress}##
or if
##(\sigma_{tangential}^2+\sigma_{radial}^2)>\sigma_{ultimate\ stress}^2##
 
Last edited:
Physics news on Phys.org
More generally how determine if a body breaks if I know stress tensor appllyied on that body and mechanical propetis of that body?
 
I heard that it has something to do with principal stresses. Can someone explain that to me with more details how to determine whether body breaks based on principal stresses applied on the body?
 
olgerm said:
More generally how determine if a body breaks if I know stress tensor appllyied on that body and mechanical propetis of that body?
What about simple case when it pushed from 2 sides so that the stress tensor is:
##\begin{bmatrix}
\sigma_{xx}&0 &0 \\
0&\sigma_{yy}&0\\
0&0&0
\end{bmatrix}##
How to know if a body breaks under this stress?
 
You will need to know something about the material properties (is it brittle or will it deform plastically?) and you will need a failure theory. The inputs to most failure theories are the principal stresses, so you are part way there.
 
Dr.D said:
you will need a failure theory.
Which criterion should In use to determine whether a metal body breaks into pieces or not?

Is one of these good approximation?
olgerm said:
body breaks if
##\sigma_{tangential}>\sigma_{ultimate\ stress}\ or\ \sigma_{radial}>\sigma_{ultimate\ stress}##
or if
##(\sigma_{tangential}^2+\sigma_{radial}^2)>\sigma_{ultimate\ stress}^2##
 
The question is too broad. You have to specify what the metal is and what its state is (annealed, hardened, drawn, etc.). This is not a simple question, and not the sort of thing that can be adequately answered on PF.
 

Similar threads

  • · Replies 8 ·
Replies
8
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
6K
  • · Replies 21 ·
Replies
21
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
13
Views
2K
  • · Replies 66 ·
3
Replies
66
Views
6K
  • · Replies 13 ·
Replies
13
Views
2K