A Stress tensor and pressure relationship

[Seyn]

I'm studying with Currie's Fundamental Mechanics of Fluids, Fourth edition.
I'm confusing with the above statement; the force acting in the x2 direction is P2=s12n1 which is n1 direction.
Is there anybody help me to understand this subject?

 

[Ibix]

the force acting in the x2 direction is P2=s12n1 which is n1 direction.

No, $P_2=\sigma_{12}n_1$ is the second component of a vector. That vector is the result of multiplying your original normal vector by a matrix, so can be in an arbitrary direction. Its second component is the component in the direction of your second basis vector.

Note that I think the author has tried to simplify the notation slightly, and I think he's done it in an unhelpful way. The point is that there's a normal vector $\vec n$ which he has chosen to be parallel to the first basis vector, so in component form it is $(n_1,0,0)^T$. So I'd say that "the unit normal vector ... is $n_1$" is possibly confusing the issue - $n_1$ is a component of a vector (in this case, the only non-zero component, but still only a component), not a vector itself.

 

[Seyn]

No, $P_2=\sigma_{12}n_1$ is the second component of a vector. That vector is the result of multiplying your original normal vector by a matrix, so can be in an arbitrary direction. Its second component is the component in the direction of your second basis vector.

Note that I think the author has tried to simplify the notation slightly, and I think he's done it in an unhelpful way. The point is that there's a normal vector $\vec n$ which he has chosen to be parallel to the first basis vector, so in component form it is $(n_1,0,0)^T$. So I'd say that "the unit normal vector ... is $n_1$" is possibly confusing the issue - $n_1$ is a component of a vector, not a vector itself.

Do you mean the point is,
$\vec{P}=(P_1,P_2,P_3)$ and,
$P_j=\sigma_{ij}n_i$
so
$\vec{P}=\sigma_{ij}n_i \hat{e_j}$??

 

[Ibix]

Yes.

 

[Seyn]

Yes.

Thank you for your kind explanation!

 

[Ibix]

No problem! You're welcome.

 

[Chestermiller]

You are confused because the result the author gives is wrong. A unit vector in the 1 direction is (1,0,0). The components of the stress vector on the surface of constant x (i.e., 1) are $P_1=\sigma_{ 11}$, $P_2=\sigma_{ 12}$, and $P_3=\sigma_{ 13}$. So the stress vector is $(P_1,P_2,P_3)=(\sigma_{11},\sigma_{12},\sigma_{13})$

 

[Seyn]

You are confused because the result the author gives is wrong. A unit vector in the 1 direction is (1,0,0). The components of the stress vector on the surface of constant x (i.e., 1) are $P_1=\sigma_{ 11}$, $P_2=\sigma_{ 12}$, and $P_3=\sigma_{ 13}$. So the stress vector is $(P_1,P_2,P_3)=(\sigma_{11},\sigma_{12},\sigma_{13})$

Thanks for the reply. Do you mean the term $n_i$s are the component of arbitrary surface vector$\vec{n}$, and the author chose an unit vector in the 1 direction as the surface vector for the first example which is $(1,0,0)$? As a result, because $n_1=1$, $n_2=0$, $n_3=0$, the surface force vector is $P_j=\sigma_{ij} n_i=(1 \cdot \sigma_{11}+0 \cdot \sigma_{21}+0 \cdot \sigma_{31},1 \cdot \sigma_{12}+0 \cdot \sigma_{22}+0 \cdot \sigma_{23},1 \cdot \sigma_{13}+0 \cdot \sigma_{23}+0 \cdot \sigma_{33})=(\sigma_{11},\sigma_{12},\sigma_{13})$??

 

[Chestermiller]

Thanks for the reply. Do you mean the term $n_i$s are the component of arbitrary surface vector$\vec{n}$, and the author chose an unit vector in the 1 direction as the surface vector for the first example which is $(1,0,0)$? As a result, because $n_1=1$, $n_2=0$, $n_3=0$, the surface force vector is $P_j=\sigma_{ij} n_i=(1 \cdot \sigma_{11}+0 \cdot \sigma_{21}+0 \cdot \sigma_{31},1 \cdot \sigma_{12}+0 \cdot \sigma_{22}+0 \cdot \sigma_{23},1 \cdot \sigma_{13}+0 \cdot \sigma_{23}+0 \cdot \sigma_{33})=(\sigma_{11},\sigma_{12},\sigma_{13})$??

yes

 

[Chestermiller]

See post #6 of this thread: https://www.physicsforums.com/threa...faces-equal-and-opposite.635431/#post-4071668

Also see post #12 of this thread: https://www.physicsforums.com/threads/so-tension-is-not-a-force.960204/#post-6091082

 

[Seyn]

See post #6 of this thread: https://www.physicsforums.com/threa...faces-equal-and-opposite.635431/#post-4071668

Also see post #12 of this thread: https://www.physicsforums.com/threads/so-tension-is-not-a-force.960204/#post-6091082

These are amazing. Thank you very much!