- #1
Dustinsfl
- 2,281
- 5
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}}##?
$$
[t_{ij}] = \begin{bmatrix}
9 & 3 & 0\\
3 & 9 & 0\\
0 & 0 & 18
\end{bmatrix}.
$$
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}}##?