Integration of partial derivative

  • Thread starter acg8934
  • Start date
  • #1
4
0

Homework Statement


Given a body in a state of plane stress with no body forces where
[tex]\sigma[/tex]x=x2y
[tex]\sigma[/tex]y=(y3-3y)/3

Find [tex]\tau[/tex]xy


Homework Equations


For plane stress
[tex]\partial[/tex][tex]\sigma[/tex]x/[tex]\partial[/tex]x + [tex]\partial[/tex][tex]\tau[/tex]xy/[tex]\partial[/tex]y + X = 0

[tex]\partial[/tex][tex]\sigma[/tex]y/[tex]\partial[/tex]y + [tex]\partial[/tex][tex]\tau[/tex]xy/[tex]\partial[/tex]x + Y = 0

No body forces: X=0 Y=0

Simplified:
[tex]\partial[/tex][tex]\sigma[/tex]x/[tex]\partial[/tex]x = -[tex]\partial[/tex][tex]\tau[/tex]xy/[tex]\partial[/tex]y

[tex]\partial[/tex][tex]\sigma[/tex]y/[tex]\partial[/tex]y = -[tex]\partial[/tex][tex]\tau[/tex]xy/[tex]\partial[/tex]x

The Attempt at a Solution



[tex]\partial[/tex][tex]\sigma[/tex]x/[tex]\partial[/tex]x = 2xy

[tex]\partial[/tex][tex]\sigma[/tex]y/[tex]\partial[/tex]y = y2-1

Putting this into above equations gives:
[tex]\partial[/tex][tex]\tau[/tex]xy/[tex]\partial[/tex]y = -2xy
[tex]\partial[/tex][tex]\tau[/tex]xy/[tex]\partial[/tex]x = -y2+1

[tex]\int[/tex][tex]\partial[/tex][tex]\tau[/tex]xy = [tex]\int[/tex]-2xy[tex]\partial[/tex]y
[tex]\tau[/tex]xy = -y2x

[tex]\int[/tex][tex]\partial[/tex][tex]\tau[/tex]xy = [tex]\int[/tex]-y2+1[tex]\partial[/tex]x
[tex]\tau[/tex]xy = -y2x + x

These should both be equal, but I'm not sure what I'm doing wrong. I think its my integration.
Please help.
 

Answers and Replies

  • #2
AEM
360
0

Homework Statement


Given a body in a state of plane stress with no body forces where
[tex]\sigma[/tex]x=x2y
[tex]\sigma[/tex]y=(y3-3y)/3

Find [tex]\tau[/tex]xy


Homework Equations


For plane stress
[tex]\partial[/tex][tex]\sigma[/tex]x/[tex]\partial[/tex]x + [tex]\partial[/tex][tex]\tau[/tex]xy/[tex]\partial[/tex]y + X = 0

[tex]\partial[/tex][tex]\sigma[/tex]y/[tex]\partial[/tex]y + [tex]\partial[/tex][tex]\tau[/tex]xy/[tex]\partial[/tex]x + Y = 0

No body forces: X=0 Y=0

Simplified:
[tex]\partial[/tex][tex]\sigma[/tex]x/[tex]\partial[/tex]x = -[tex]\partial[/tex][tex]\tau[/tex]xy/[tex]\partial[/tex]y

[tex]\partial[/tex][tex]\sigma[/tex]y/[tex]\partial[/tex]y = -[tex]\partial[/tex][tex]\tau[/tex]xy/[tex]\partial[/tex]x

The Attempt at a Solution



[tex]\partial[/tex][tex]\sigma[/tex]x/[tex]\partial[/tex]x = 2xy

[tex]\partial[/tex][tex]\sigma[/tex]y/[tex]\partial[/tex]y = y2-1

Putting this into above equations gives:
[tex]\partial[/tex][tex]\tau[/tex]xy/[tex]\partial[/tex]y = -2xy
[tex]\partial[/tex][tex]\tau[/tex]xy/[tex]\partial[/tex]x = -y2+1

[tex]\int[/tex][tex]\partial[/tex][tex]\tau[/tex]xy = [tex]\int[/tex]-2xy[tex]\partial[/tex]y
[tex]\tau[/tex]xy = -y2x

[tex]\int[/tex][tex]\partial[/tex][tex]\tau[/tex]xy = [tex]\int[/tex]-y2+1[tex]\partial[/tex]x
[tex]\tau[/tex]xy = -y2x + x

These should both be equal, but I'm not sure what I'm doing wrong. I think its my integration.
Please help.
You forgot that when you do partial integration, say with respect to y, you have to add an arbitrary function of x. Similarly when you integrate partially with respect to x you have to add an arbitrary function of y. This will give you the necessary freedom to determine your unknown function.
 
  • #3
4
0
Ok, but doing that gives me

[tex]\tau[/tex]xy=-y2x + f(x) for the first

and

[tex]\tau[/tex]xy=-y2x + x + f(y)

Is that correct? If so how do I figure out what the actual [tex]\tau[/tex]xy is?

Thanks
 
  • #4
34,553
6,266
Don't use f as both a function of x alone and another function of y alone.

So you have T(x, y) = -y^2x + f(x), and
T(x, y) = -y^2x + x + g(y).
Edit: added x in the line above.

Does that help?

My LaTeX doesn't seem to be rendering correctly, so I switched tau to T.
 
Last edited:
  • #5
4
0
How do I find f(x) and g(y)?

I know the tau's should be equal.
 
  • #6
34,553
6,266
I omitted x from the 2nd representation by accident. Both forms of T(x, y) have -y2x. The first representation doesn't have any functions of y alone, so g(y) = ? The 2nd representation has x, so f(x) = ?
 
  • #7
4
0
So does that mean f(x)=x and g(y)=0? So Txy=-y^2x+x?
 
  • #8
34,553
6,266
Almost. g(y) = C, a constant. So T(x, y) = -xy^2 + x + C.
 

Related Threads on Integration of partial derivative

  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
5
Views
4K
  • Last Post
Replies
6
Views
3K
  • Last Post
2
Replies
26
Views
3K
  • Last Post
Replies
0
Views
976
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
5
Views
16K
Replies
2
Views
5K
Replies
3
Views
6K
Replies
0
Views
875
Top