Calculation Inputs Output Average Normal Stresses Shear Stresses Explanation The 3D stresses are usually given by six stress components, the normal stresses at the x- y- and z- directions and the shear stresses at the xy , yz, and xz planes. These stresses are then put in a 3 x 3 symmetric matrix shown below.
T 3 = [ σ x τ x y τ x z τ x y σ y τ y z τ x z τ y z σ z ] T_3 = \begin{bmatrix} \sigma_x & \tau_{xy} & \tau_{xz} \\ \tau_{xy} & \sigma_y & \tau_{yz} \\ \tau_{xz} & \tau_{yz} & \sigma_z \\\end{bmatrix} T 3 = σ x τ x y τ x z τ x y σ y τ yz τ x z τ yz σ z In order to get the maximum normal and shear stresses, we get the eigenvalues of the matrix above. Say, that the cut plane has the following directional cosine:
v = ( v x , v y , v z ) \mathbf{v} = (v_x,\ v_y,\ v_z) v = ( v x , v y , v z ) We can find the normal stress on this plane by,
σ v = σ x v x 2 + σ y v y 2 + σ z v z 2 + 2 τ x y v x v y + 2 τ y z v y v z + 2 τ x z v x v z \sigma_v = \sigma_x v_{x}^{2} + \sigma_y v_{y}^{2} + \sigma_z v_{z}^{2} + 2 \tau_{xy} v_{x} v_{y} + 2 \tau_{yz} v_{y} v_{z} + 2 \tau_{xz} v_{x} v_{z} σ v = σ x v x 2 + σ y v y 2 + σ z v z 2 + 2 τ x y v x v y + 2 τ yz v y v z + 2 τ x z v x v z To find the eigenvalues and eigenvectors, we use the following characteristic polynomial equation:
d e t ( σ I 3 − T 3 ) = σ 3 − A σ 3 − B σ − C = 0 det(\sigma I_3-T_3)\ =\ \sigma^3- A\sigma^3 - B\sigma-C=0 d e t ( σ I 3 − T 3 ) = σ 3 − A σ 3 − B σ − C = 0
A = σ x + σ y + σ z B = σ x σ y + σ y σ z + σ x σ z − τ x y 2 − τ y z 2 − τ x z 2 C = σ x σ y σ z + 2 τ x y τ y z τ x z − σ x τ y z 2 − σ y τ x z 2 − σ z τ x y 2 A = \sigma_x + \sigma_y + \sigma_z \\B = \sigma_x \sigma_y + \sigma_y \sigma_z + \sigma_x \sigma_z - \tau_{xy}^2 - \tau_{yz}^2 - \tau_{xz}^2 \\C = \sigma_x \sigma_y \sigma_z + 2 \tau_{xy} \tau_{yz} \tau_{xz} - \sigma_x \tau_{yz}^2 - \sigma_y \tau_{xz}^2 - \sigma_z \tau_{xy}^2 A = σ x + σ y + σ z B = σ x σ y + σ y σ z + σ x σ z − τ x y 2 − τ yz 2 − τ x z 2 C = σ x σ y σ z + 2 τ x y τ yz τ x z − σ x τ yz 2 − σ y τ x z 2 − σ z τ x y 2 We can then find the normal stresses using the following system of equations:
A = σ 1 + σ 2 + σ 3 B = σ 1 σ 2 + σ 2 σ 3 + σ 1 σ 3 C = σ 1 σ 2 σ 3 A = \sigma_1 + \sigma_2 + \sigma_3 \\B = \sigma_1 \sigma_2 + \sigma_2 \sigma_3 + \sigma_1 \sigma_3 \\C = \sigma_1 \sigma_2 \sigma_3 A = σ 1 + σ 2 + σ 3 B = σ 1 σ 2 + σ 2 σ 3 + σ 1 σ 3 C = σ 1 σ 2 σ 3 For a more graphical approach, we can use the following approach:
σ a v e 1 = σ x + σ y 2 σ a v e 2 = σ y + σ z 2 σ a v e 3 = σ x + σ z 2 \sigma_{ave1}\ =\ \frac{\sigma_x\ +\ \sigma_y}{2}\\\sigma_{ave2}\ =\ \frac{\sigma_y\ +\ \sigma_z}{2}\\\sigma_{ave3}\ =\ \frac{\sigma_x\ +\ \sigma_z}{2} σ a v e 1 = 2 σ x + σ y σ a v e 2 = 2 σ y + σ z σ a v e 3 = 2 σ x + σ z The obtained values will serve as the centers of our circles. The radii of the circles can be obtained by:
r a d i u s 1 = ∣ σ x − σ a v e 1 ∣ r a d i u s 2 = ∣ σ y − σ a v e 2 ∣ r a d i u s 3 = ∣ σ z − σ a v e 3 ∣ radius_1\ =\ |\sigma_x\ -\sigma_{ave1}|\\radius_2\ =\ |\sigma_y\ -\sigma_{ave2}|\\radius_3\ =\ |\sigma_z\ -\sigma_{ave3}| r a d i u s 1 = ∣ σ x − σ a v e 1 ∣ r a d i u s 2 = ∣ σ y − σ a v e 2 ∣ r a d i u s 3 = ∣ σ z − σ a v e 3 ∣ References Related Resources Check out our full library of CalcTree templates here ! Mohr's Circle for 2D Stresses Weight to Volume relationships in soils