# Elasticity problem

1. Jan 7, 2013

### physicist10

Hello, I am struggling with this problem. It is probably the easiest problem ever...

What I did: The plane has 2 stress components. σn and σs.

σn is a multiple of (l, m, k) vector. For σs, I made up a vector (a, b, c) which is orthogonal to (l, m, k). And I equated all vectors.

I'm probably doing something wrong. Any help is appreciated!

2. Jan 7, 2013

### physicist10

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

2. Relevant equations

General plane formulas.

3. The attempt at a solution

I thought that the plane has 2 stress components. σn and σs.

σn is a multiple of (l, m, k) vector. For σs, I made up a vector (a, b, c) which is orthogonal to (l, m, k). And I equated all vectors.

I'm probably doing something wrong. Any help is appreciated!

3. Jan 7, 2013

### Zirkus

Hello, I am not an expert on elasticity but this really looks quite straightforward. Let's first find the stress vector T (I'm using T instead of σ to avoid confusion with the stress tensor). You will get it by multiplying the (diagonal) stress tensor by your normal vector as T=(σ1l, σ2m, σ3n). It has two components as you wrote, Tn and Ts. The magnitude of Tn is simply the dot product of T and n and its direction is along n as you wrote. Vector Ts has to be the complement to the total stress vector.
And for the second part - the shear stress will be maximum if vector T lies in your plane, e.g. the dot product of T and n is zero.

Last edited: Jan 7, 2013
4. Jan 7, 2013

### Studiot

Whilst this thread properly belongs in the homework section, this needs comment.

What is a stress vector?

5. Jan 7, 2013

### physicist10

Aha, he's probably talking about the traction vector.

However, this was very helpful. Thanks

6. Jan 7, 2013

### Zirkus

We defined it similarly like the article on wiki does, so I won't rewrite it...Stress on Wikipedia
Most likely there are other methods or other terminology, I'm not a native speaker so i can't tell the subtle differences that well, sorry about that.

7. Jan 7, 2013

### physicist10

Thanks Zirkus!