Need help understanding eigenvalues and eigenvectors

  • #1
Cisneros778
48
0

Homework Statement


Find the principal stresses and the orientation for the principal axis of stress for the following cases of plane stress.

σx = 4,000 psi
σy = 0 psi
τxy = 8,000 psi


Homework Equations


See picture.

The Attempt at a Solution


https://mail.google.com/mail/u/0/?ui=2&ik=bc68d58ae7&view=att&th=139dbef260c42514&attid=0.1&disp=inline&realattid=f_h79p3pz70&safe=1&zw

I solved this problem using Mohr's Circle. However, the solution to the problem is different and I would like to understand it.
I do not know what the steps mean. Why does the determinate of that function must equal zero?
And what is the n1 and n2 about?
Finally, the 2x2 matrices adds two shear stresses where I was only given one τxy where does this other value come from?
 

Answers and Replies

  • #2
Cisneros778
48
0
Any help please?
 
  • #3
cmmcnamara
122
1
I'm not sure if anyone else is having trouble here, but your picture doesn't seem to be loading?
 
  • #4
22,328
5,204

Homework Statement


Find the principal stresses and the orientation for the principal axis of stress for the following cases of plane stress.

σx = 4,000 psi
σy = 0 psi
τxy = 8,000 psi


Homework Equations


See picture.

The Attempt at a Solution


https://mail.google.com/mail/u/0/?ui=2&ik=bc68d58ae7&view=att&th=139dbef260c42514&attid=0.1&disp=inline&realattid=f_h79p3pz70&safe=1&zw

I solved this problem using Mohr's Circle. However, the solution to the problem is different and I would like to understand it.
I do not know what the steps mean. Why does the determinate of that function must equal zero?
And what is the n1 and n2 about?
Finally, the 2x2 matrices adds two shear stresses where I was only given one τxy where does this other value come from?

In terms of components and unit vectors, the stress tensor can be written as a sum of terms (similar to a vector) as follows:

σ = σxx ixix + τxy ixiy+ τyx iyix+ σyy iyiy

Since the stress tensor is symmetric, τyx = τxy

Therefore, the stress tensor is given by:

σ = σxx ixix + τxy ixiy+ τxy iyix+ σyy iyiy

This is where the other τxy you were asking about comes from.


If n is a unit vector oriented in some arbitrary horizontal direction within your material, then, in terms of its components in the x and y directions, n can be written as the following sum of terms:

n = nx ix + ny iy

According to the so-called Cauchy stress relationship, the traction (force per unit area) acting on a plane perpendicular to the unit normal n is obtained by dotting the stress tensor σ with the unit normal n:

[itex]\Sigma[/itex]=(σxx nx + τxy ny) ix + (τxy nx + σyy ny) iy

If n corresponds to one of the principal direction of stress, then the traction on the plane normal to n is perpendicular to the plane, and parallel to n:


[itex]\Sigma[/itex]=(σxx nx + τxy ny) ix + (τxy nx + σyy ny) iy = λ (nx ix + ny iy)

where λ is the principal stress.

From the above equation, we get:

σxx nx + τxy ny = λ nx

τxy nx + σyy ny) = λ ny

This defines your eigenvalue problem. The components of the normal define the eigenvector, and the principal stress defines the eigenvalue.
 

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