Introduction to Elasticity/Stress example 3

Example 3

Given:

A stress field whose components in the basis ${\displaystyle (\mathbf {e} _{1},\mathbf {e} _{2},\mathbf {e} _{3})\,}$ are given by the matrix

${\displaystyle [{\boldsymbol {\sigma }}]={\begin{bmatrix}6~x_{1}~x_{3}^{2}&0&-2~x_{3}^{3}\\0&1&2\\-2~x_{3}^{3}&2&3~x_{1}^{2}\end{bmatrix}}}$

Find:

1. Assuming negligible body forces, show whether this field satisfies equilibrium.
2. Find the traction acting at a point P whose position vector is ${\displaystyle \mathbf {x} =2~\mathbf {e} _{1}+3~\mathbf {e} _{2}+2~\mathbf {e} _{3}\,}$ acting on a plane ${\displaystyle 2~x_{1}+x_{2}-x_{3}=5\,}$.
3. Determine the normal traction at this point on the plane.
4. Determine the projected shear traction at this point on the plane.
5. Determine the principal stresses at point P.
6. Determine the principal directions of stress at point P.

Solution

This problem is very similar to example 2.

The stress is a function of ${\displaystyle x_{1}\,}$ and ${\displaystyle x_{3}\,}$. We first plug the stress into the equilibrium equations (Cauchy's first law) and find that the sum is zero. Hence the stress satifies equilibrium.

Next, we find the stress at the given point byplugging in the values of ${\displaystyle x_{1}\,}$ and ${\displaystyle x_{3}\,}$ into the given stress.

Then we find the normal to the given plane ${\displaystyle a_{i}~x_{i}=b\,}$ using the relation

${\displaystyle {\widehat {\mathbf {n} }}=\pm {\cfrac {a_{i}~\mathbf {e} _{i}}{\sqrt {a_{j}~a_{j}}}}}$

and find the traction vector and thenormal and projected shear tractions.

We find the eigenvalues (principal stresses) by first forming the cubic characteristic equation and then solving for it. Then we plug the values of the principal stresses into the eigenfunction and solve for the direction of the principal strains.

Since the stress tensor is symmetric, we need an extra relation between ${\displaystyle n_{1}\,}$, ${\displaystyle n_{2}\,}$, and ${\displaystyle n_{3}\,}$ to reduce the system down to two equations and two unknowns. This relation is ${\displaystyle n_{1}^{2}+n_{2}^{2}+n_{3}^{2}=1\,}$, which means that each eigenvector is a unit vector.

The Maple output is shown below.

> with(linalg):
> sigma :=
> linalg[matrix](3,3,[6*x1*x3^2,0,-2*x3^3,0,1,2,-2*x3^3,2,3*x1^2]);

[       2              3]
[6 x1 x3     0    -2 x3 ]
[                       ]
sigma := [   0        1      2   ]
[                       ]
[      3              2 ]
[ -2 x3      2    3 x1  ]

> e1 := linalg[matrix](3,1,[1,0,0]):
> e2 := linalg[matrix](3,1,[0,1,0]):
> e3 := linalg[matrix](3,1,[0,0,1]):
> sigi1i := diff(sigma[1,1],x1)+diff(sigma[2,1],x2)+diff(sigma[3,1],x3);

sigi1i := 0

> sigi2i := diff(sigma[1,2],x1)+diff(sigma[2,2],x2)+diff(sigma[3,2],x3);

sigi2i := 0

> sigi3i := diff(sigma[1,3],x1)+diff(sigma[2,3],x2)+diff(sigma[3,3],x3);

sigi3i := 0

> X := evalm(2*e1 + 3*e2 + 2*e3):
> sig := linalg[matrix](3,3):
> for i from 1 to 3 do
>  for j from 1 to 3 do
>   sig[i,j] := eval(eval(eval(sigma[i,j], x1 = X[1,1]), x2 =X[2,1]), x3
> = X[3,1]);
>  end do;
> end do;
> evalm(sig);

[ 48    0    -16]
[               ]
[  0    1      2]
[               ]
[-16    2     12]

> a1 := 2: a2 := 1: a3 := -1: ajaj := sqrt(a1*a1 + a2*a2 + a3*a3):
> n := evalm(a1*e1/ajaj + a2*e2/ajaj + a3*e3/ajaj);

[  1/2 ]
[ 6    ]
[ ---- ]
[  3   ]
[      ]
[  1/2 ]
n := [ 6    ]
[ ---- ]
[  6   ]
[      ]
[   1/2]
[  6   ]
[- ----]
[   6  ]

> sigT := transpose(sig):
> t := evalm(sigT&*n);

[    1/2]
[56 6   ]
[-------]
[   3   ]
[       ]
t := [   1/2 ]
[  6    ]
[- ---- ]
[   6   ]
[       ]
[    1/2]
[-7 6   ]

> tT := transpose(t):
> N := evalm(tT&*n): Nt := evalf(N[1,1]);

Nt := 44.16666667

> tdott := evalm(tT&*t):
> St := evalf(sqrt(tdott[1,1] - N[1,1]^2));

St := 20.83599984

> I1 := sig[1,1]+sig[2,2]+sig[3,3];

I1 := 61

> I2 := sig[1,1]*sig[2,2] + sig[2,2]*sig[3,3] + sig[3,3]*sig[1,1] -
> sig[1,2]*sig[2,1] - sig[2,3]*sig[3,2] - sig[3,1]*sig[1,3];

I2 := 376

> I3 := det(sig);

I3 := 128

> prinSig := solve(x^3 - I1*x^2 + I2*x - I3 = 0, x):
> s1 := evalf(prinSig[1]); s2 := evalf(prinSig[2]); s3 :=
> evalf(prinSig[3]);

-8
s1 := 54.09271724 - 0.2 10   I

-8
s2 := 0.361501126 - 0.8160254040 10   I

-8
s3 := 6.545781610 + 0.9160254040 10   I

> sig1 := 54.09271724; sig2 := 6.545781610; sig3 := .361501126;

sig1 := 54.09271724

sig2 := 6.545781610

sig3 := 0.361501126

> with(LinearAlgebra): Id := IdentityMatrix(3):
> Left1 := evalm(sig - sig1*Id):
> n3 := sqrt(1 - n1^2 - n2^2):
> n := linalg[matrix](3,1,[n1,n2,n3]):
> Ax1 := evalm(Left1 &* n):
> sols1 := solve({Ax1[1,1] = 0, Ax1[2,1] = 0});

sols1 := {n2 = 0.01340427809, n1 = -0.9344527487}

> N1_1 := sols1[2]; N1_2 := sols1[1]; N1_3 := sqrt(1 -
> (.1340427809e-1)^2 - (-.9344527487)^2);
>

N1_1 := n1 = -0.9344527487

N1_2 := n2 = 0.01340427809

N1_3 := 0.3558347730

> Left2 := evalm(sig - sig2*Id):
> Ax2 := evalm(Left2 &* n):
> sols2 := solve({Ax2[1,1] = 0, Ax2[2,1] = 0});

sols2 := {n1 = 0.3412802400, n2 = 0.3188798847}

> N2_1 := sols2[1]; N2_2 := sols1[2]; N2_3 := sqrt(1 - (.3412802400)^2 -
> (.3188798847)^2);

N2_1 := n1 = 0.3412802400

N2_2 := n1 = -0.9344527487

N2_3 := 0.8842191001

> Left3 := evalm(sig - sig3*Id):
> Ax3 := evalm(Left3 &* n):
> sols3 := solve({Ax3[1,1] = 0, Ax3[2,1] = 0});

sols3 := {n1 = 0.1016162335, n2 = -0.9477003447}

> N3_1 := sols3[1]; N3_2 := sols3[2]; N3_3 := sqrt(1 - (.1016162335)^2 -
> (-.9477003447)^2);

N3_1 := n1 = 0.1016162335

N3_2 := n2 = -0.9477003447

N3_3 := 0.3025528017