Determine normal stresses for pure bending moments

AI Thread Summary
To determine normal stresses σx, σy, and σz using the flexure formula, it is crucial to identify the correct moments associated with each stress type. The flexure formula σ = Mc/I + P/A is essential for calculating these stresses, but confusion often arises regarding which moment corresponds to which stress. Visualizing the behavior of the material under bending, such as compression on one side and tension on the other, can clarify the stress types involved. An additional approach involves applying forces to create a turning moment, allowing for the analysis of combined effects on the stresses. Understanding these principles is key to solving complex bending problems effectively.
sgvaibhav
Messages
65
Reaction score
0

Homework Statement


Hi

I am having tough time figuring out how to determine normal stresses σx, σy, σz using the flexure formula.

All that confuses me is which moment to take for which σ. My for σx or Mz for σx and so on..

I will post an example question.

In the question below, find the stresses σx, σy, σz at point H and point K

Homework Equations



σ=Mc/I + P/A

The Attempt at a Solution


The actual question is very big and requires determining principles planes, principle stress and stresses in different orientation to be found out which i know how to do as well.
I can proceed with the question only if i can determine the stresses σx, σy, σz, which is were i am getting stuck in all such types of questions.
 

Attachments

  • question2.png
    question2.png
    16.6 KB · Views: 520
Physics news on Phys.org
Imagine CD being made of cheese or soft rubber. Can you see that the point C will initially deflect parallel to the x axis, and be bending in the xz plane, (raised to the level that contains C and D) What normal stresses due to bending in CD will that give rise to? Can you see that the thinner edge CD closest to the viewer will go into compression, and the far side into tension. Which of the the three choices of stress do you think that causes?
 
My first answer was to check that you can visualise the behaviour and correctly identify the stress type in a simple case. H and K are more complicated. One approach is to add another 10 kN force at A in the +x direction (in other words, parallel to the existing force) and then cancel its effect by adding a 10 kN force at A in the -x direction. You then have three forces one of which is the 10 at A in the +x direction and the other two forming a turning moment about the y axis. You can then add the effects of these separate actions. Does that help?
 
Back
Top