States of Stress at P: Principal Stress Values & Normals

[Dustinsfl]

The state of stress at $\mathbf{P}$, when referred to axes $P_{x_1x_2x_3}$ is given in ksi unites by the matrix
$$
[t_{ij}] = \begin{bmatrix}
9 & 3 & 0\\
3 & 9 & 0\\
0 & 0 & 18
\end{bmatrix}.
$$
Determine
(1)the principal stress values at $\mathbf{P}$ and

The trace of $t_{ij}$ is 36, $t_{ij}^2$ is
$$
\begin{bmatrix}
90 & 54 & 0\\
54 & 90 & 0\\
0 & 0 & 324
\end{bmatrix},
$$
and the determinant is 1296.
So the characteristic polynomial is $p(\sigma) = \sigma^3 - 36\sigma^2 + 396\sigma - 1296 = (\sigma - 6)(\sigma - 12)(\sigma - 18)$.
So the principal stress values are $\sigma_{\text{\MakeUppercase{\romannumeral 1}}} = 6$, $\sigma_{\text{\MakeUppercase{\romannumeral 3}}} = 12$, and $\sigma_{\text{\MakeUppercase{\romannumeral 3}}} = 18$.

(2)the unit normal $\hat{\mathbf{n}}^* = n_i\hat{\mathbf{e}}^*_i$ of the plane on which $\sigma_{\text{N}} = 12$ ksi and $\sigma_{\text{S}} = 3$ ksi.

How do I find $\sigma_{\text{N}}$?

 

[Chestermiller]

The stress vector on a plane is obtained by dotting the stress tensor with a unit vector normal to the plane. Don't forget that the unit normal is a unit vector, so the sum of the squares of its three components equals 1. To get the normal component of the stress vector on the plane, dot the stress vector with the unit normal again. The tangential component of the stress vector is equal to the stress vector minus the normal component times the unit normal. This should give you enough equations to solve for the components of the unit normal, given the stress tensor and the normal- and tangential components of the stress vector.

 

[Dustinsfl]

The stress vector on a plane is obtained by dotting the stress tensor with a unit vector normal to the plane. Don't forget that the unit normal is a unit vector, so the sum of the squares of its three components equals 1. To get the normal component of the stress vector on the plane, dot the stress vector with the unit normal again. The tangential component of the stress vector is equal to the stress vector minus the normal component times the unit normal. This should give you enough equations to solve for the components of the unit normal, given the stress tensor and the normal- and tangential components of the stress vector.

So take $t_{ij}\cdot n$ where $n = \langle a,b,c\rangle$ and then take $t_{ij}\cdot n\cdot n$ and set it equal to 12?
That gives $a(9a+3b)+b(3a+9b)+18c^2 = 12$. How do I find a,b,c or do I not need them?

 

[Chestermiller]

You need another equation, namely the shear stress on the plane being equal to 3 ksi. I told you how to handle this. You then have a second equation involving a, b, and c. The third equation is the sum of the squares of a, b, and c is equal to 1.

 

[Dustinsfl]

You need another equation, namely the shear stress on the plane being equal to 3 ksi. I told you how to handle this. You then have a second equation involving a, b, and c. The third equation is the sum of the squares of a, b, and c is equal to 1.

What is my stress vector?
$\sigma_{\text{S}}^{\max}$ can be two combination since
$$
\frac{1}{2}\lvert\sigma_i - \sigma_j\rvert = 3
$$
for $i\neq j$ and $i,j = 1,2$ or $i,j = 2,3$ by using the maximum shear stress equation.

 

[Chestermiller]

What is my stress vector?

You've already calculated your stress vector: (9a+3b), (9b+3a), 18c
The normal stress times the unit normal is: 12a + 12b + 12c
The tangential (shear) component of the stress vector is the total stress vector minus the normal stress times the unit normal:

(-3a+3b), (-3b+3a), 6c = 3(b-a), 3(a-b), 6c

The magnitude of this shear stress vector must be equal to 3ksi:

9(a-b)^2+9(a-b)^2+36c^2=9

or 2(a-b)^2+4c^2=1

If you make use of the condition that a^2+b^2+c^2 = 1

the above equation reduces to 4ab-2c^2=1

From your normal stress equation, you got:
9a^2+9b^2+18c^2+6ab=12
Again making use of the condition that a^2+b^2+c^2 = 1
you get:
6ab+9c^2=3
or 2ab+3c^2=1
Now we have enough to solve for c²:

c^2=\frac{1}{8}
From this we get:
c=\frac{\sqrt{2}}{4}
and
2ab=\frac{5}{8}
and a^2+b^2=\frac{7}{8}
So,
(a+b)^2=\frac{3}{2}
(b-a)^2=\frac{1}{4}
So
(b+a)=\frac{\sqrt{6}}{2}
(b-a)=\frac{1}{2}
So
b=\frac{\sqrt{6}+1}{4}
a=\frac{\sqrt{6}-1}{4}

I think I did the "arithmetic" correctly. If not, I'm sure you will find any mistake I might have made. But, in any event, the methodology is correct.

Chet

 

[Dustinsfl]

Thanks that was helpful but the book has the solution as
$$
\frac{1}{2\sqrt{2}}\langle 1, \sqrt{6}, 1\rangle.
$$

 

[Chestermiller]

I don't know what to say. I checked the solution for the components of the unit normal vector, and they seemed to check. In the book solution, the only component that matches mine is c. Any chance there is a typo in the book solution? Try my solution, and see what it gives for the normal stress vector and the tangential stress vector. See for yourself if, from my solution, the magnitudes of these vectors are 12 and 3.

Chet

 

[Chestermiller]

I don't know what to say. I checked the solution for the components of the unit normal vector, and they seemed to check. In the book solution, the only component that matches mine is c. Any chance there is a typo in the book solution? Try my solution, and see what it gives for the normal stress vector and the tangential stress vector. See for yourself if, from my solution, the magnitudes of these vectors are 12 and 3.

Chet

I checked my solution, and it is correct. The results in the book must be wrong.

 

[Dustinsfl]

I am not surprised the book is wrong. It seems to be consistently wrong.

Thanks.