# Stress Tensor - simple problem (I think)

1. Sep 13, 2009

### rarkup

1. The problem statement, all variables and given/known data

Stress tensor at a point Q in a body has components:

pij:
| 1 -1 0 |
|-1 2 1 |
| 0 1 3 |

(i) Calculate components of the stress force f across a small area of surface at Q normal to n = (2,1,-1).
(ii) The component of f in the direction of n is called the normal stress while the component of f tangential to the surface is called the shear stress. Find the normal stress and the shear stress at Q.

2. Relevant equations

f = fi = pijnj

3. The attempt at a solution

(i) I take the stress force at Q across the surface normal to n to be the product of the stress tensor pij and the normal vector n and arrive at:

f = (1x2 + -1x1 + 0x-1)i + (-1x2 + 2x1 + 1x-1)j + (0x2 + 1x1 +3x-1)k
= ( 1, -1, -2 )

(ii) I think this is a simple algebraic exercise - ie what components of ( 1, -1, -2) project onto n to give the normal stress, and which components project onto the plane normal to n to give the shear stress.

Am I reading this correctly, and what is the most efficient way of computing these components? I can't help but think the answer might be staring at me from within the tensor pij.

2. Sep 13, 2009

### tiny-tim

Welcome to PF!

Hi rarkup! Welcome to PF!
Yup, that's fine!
Just find the component of ( 1, -1, -2) in the direction of ( 2, 1, -1) in the usual way (and then subtract that from the whole stress to get the shear stress) …

this has nothing to do with the tensor any more.

3. Sep 13, 2009

### rarkup

Thanks for the welcome and your response tiny-tim.

Now it's time to dust off the Algebra books and relearn how to compute components of a vector in the direction of another.