Principal Strains

Having learned how to find the deformation of various material vectors emanating from a material point $P$, we now investigate which of these lines is stretched the most. Let

$$\mathbf{P}\mathbf{Q} = 10^{-3}(N_1,N_2,N_3)$$ (3.9.1)

where $(N_1,N_2,N_3)$ are components of a unit vector along $\mathbf{P}\mathbf{Q}$. That is, $\mathbf{N} = (N_1,N_2,N_3)$ is a unit vector along $\mathbf{P}\mathbf{Q}$. Our aim is to find $\mathbf{N}$ such that the stretch at $P$ along $\mathbf{N}$ is maximum or minimum. The vector $\mathbf{P}^\prime \mathbf{Q}^\prime$ into which $\mathbf{P}\mathbf{Q}$ is deformed is given by

$$(P^\prime Q^\prime )_j = F_{jA}(PQ)_A = 10^{-3}F_{jA}N_A.$$ (3.9.2)

Therefore,

$$\frac{\vert\mathbf{P}^\prime\mathbf{Q}^\prime\vert^2}{\vert\mathbf{P}\mathbf{Q}\vert^2} = F_{jA}N_AF_{jB}N_B = C_{AB}N_AN_B.$$ (3.9.3)

Thus the problem reduces to finding a unit vector $\mathbf{N}$ such that $C_{AB}N_AN_B$ is maximum. By using the method of Lagrange multipliers we need to find $\mathbf{N}$ such that

$$C_{AB}N_AN_B - \lambda (N_AN_A-1)$$ (3.9.4)

takes on extreme values for all $\mathbf{N}$. In (3.9.4) $\lambda$ is a Lagrange multiplier. Such an $\mathbf{N}$ is given by

$$\frac{\partial}{\partial N_I}[C_{AB}N_AN_B - \lambda (N_AN_A-1)] = 0$$

which is equivalent to

$$C_{AB}N_B - \lambda N_A = 0,$$

or

$$(C_{AB}- \lambda \delta_{AB})N_B = 0.$$ (3.9.5)

These are three linear homogeneous equations in $N_1,N_2,N_3$. A non-trivial solution of these equations exists if and only if

$$\det[C_{AB} - \lambda\delta_{AB}] = 0\ ,$$

or

$$\lambda^3 - I\lambda^2 + II\lambda - III = 0\ ,$$ (3.9.6)

where

$$I=\ C_{AA},$$

$$II =\ 1/2[-C_{AB}C_{BA} + C_{AA}C_{BB}] = (\det[\mathbf{C}]) (C^{-1})_{AA},$$

$$III =\ \det[\mathbf{C}].$$

$I,\ II,\ III$ are called principal invariants of $[\mathbf{C}]$. Equation (3.9.6) is cubic in $\lambda$ and will have three roots. Since the matrix $[\mathbf{C}]$ is symmetric and positive definite, all three roots of (3.9.6) are positive. Let us denote the three roots of (3.9.6) by $\lambda^2_1,\ \lambda^2_2,\ \lambda^2_3$.²For each one of these roots of (3.9.6) we find the corresponding $\mathbf{N}$ from (3.9.5). That is, once the three roots of (3.9.6) have been found, we use (3.9.5) to find the directions $\mathbf{N}^{(1)},\ \mathbf{N}^{(2)},\ \mathbf{N}^{(3)}$ along which stretches assume extreme values. For example, $\mathbf{N}^{(1)}$ is obtained by solving

$$(C_{AB} - \lambda^2_1\delta_{AB})N^{(1)}_B = 0,\ N^{(1)}_1N^{(1)}_1 + N^{(1)}_2N^{(1)}_2 + N^{(1)}_3N^{(1)}_3 = 1.$$ (3.9.7)

Having found $\mathbf{N}^{(1)}$, we use (3.9.3) to find the stretch in this direction.

$$\ \mathbf{N}^{(1)} =\ \sqrt{C_{AB}N^{(1)}_AN^{(1)}_B}\ ,$$

$$=\ \sqrt{\lambda^2_1\delta_{AB}N^{(1)}_AN^{(1)}_B} = \sqrt{\lambda^2_1N^{(1)}_AN^{(1)}_A}\ ,$$

$$=\ \lambda_1.$$

Thus the three roots of (3.9.6) are squares of extreme values of stretches at the point $P$. Let us now assume that $\lambda^2_1\ne \lambda^2_2\ne \lambda^2_3$ and find the angle between directions $\mathbf{N}^{(1)}$ and $\mathbf{N}^{(2)}$. Rewriting (3.9.7) as

$$C_{AB}N^{(1)}_B = \lambda^2_1\delta_{AB}N^{(1)}_B =\lambda^2_1N_A^{(1)}\ ,$$

we obtain

$$N^{(2)}_AC_{AB}N^{(1)}_B = \lambda^2_1N^{(2)}_AN^{(1)}_A\ .$$ (3.9.8)

Similarly

$$N^{(1)}_AC_{AB}N^{(2)}_B = \lambda^2_2N^{(1)}_AN^{(2)}_A\ .$$ (3.9.9)

Since $C_{AB}= C_{BA}$, therefore, the left-hand sides of (3.9.8) and (3.9.9) are equal. Thus

$$(\lambda^2_1 - \lambda^2_2) N^{(1)}_AN^{(2)}_A = 0.$$

Since $\lambda^2_1 \ne \lambda^2_2$ by assumption,

$$N^{(1)}_AN^{(2)}_A = 0.$$ (3.9.10)

Thus $\mathbf{N}^{(1)}$ and $\mathbf{N}^{(2)}$ are perpendicular to each other. By using the same argument for $\mathbf{N}^{(2)}$ and $\mathbf{N}^{(3)}$, we conclude that whenever $\lambda^2_1\ne \lambda^2_2\ne \lambda^2_3$, $\mathbf{N}^{(1)},\ \mathbf{N}^{(2)}$ and $\mathbf{N}^{(3)}$ are mutually perpendicular to each other.

Let us now find the change in the angle between the lines $\mathbf{N}^{(1)}$ and $\mathbf{N}^{(2)}$ during the deformation. If $\theta$ is the angle between the deformed positions of $\mathbf{N}^{(1)}$ and $\mathbf{N}^{(2)}$, then by (3.8.1) and (3.8.4)

$$\cos\theta = \ \frac{C_{FG}N^{(1)}_FN^{(2)}_G}{(C_{AB}N^{(1)}_AN^{(1)}_B)^{1/2}(C_{DE}N^{(2)}_ DN^{(2)}_E)^{1/2}}\ ,$$

$$\noalign{\vskip .1in} = \ \frac{\lambda^2_1\delta_{AB}N^{(1)}_BN^{(2)}_A}{\lambda_1\lambda_2} = \frac{\lambda_1N^{(1)}_AN^{(2)}_A}{\lambda_2} = 0.$$

$$\therefore\hspace{.2in} \theta =\ 90^\circ .$$ (3.9.11)

That is, the angle between directions $\mathbf{N}^{(1)}$ and $\mathbf{N}^{(2)}$ does not change during the deformation. The same holds true for directions $\mathbf{N}^{(2)}$ and $\mathbf{N}^{(3)}$. Because of this property, the directions $\mathbf{N}^{(1)},\ \mathbf{N}^{(2)}$ and $\mathbf{N}^{(3)}$ are called principal axes of stretch and $\lambda_1,\lambda_2,\lambda_3$ are called principal stretches.

Whenever any two or all roots of eqn. (3.9.6) are equal, equations (3.9.10) and (3.9.11) hold for suitable choices of $\mathbf{N}^{(1)},\ \mathbf{N}^{(2)}$ and $\mathbf{N}^{(3)}$. If $\lambda^2_1 = \lambda^2_2 \ne \lambda^2_3$, then $\mathbf{N}^{(3)}$ is uniquely determined but $\mathbf{N}^{(1)}$ and $\mathbf{N}^{(2)}$ can be taken as any two directions in a plane perpendicular to $\mathbf{N}^{(3)}$. When $\lambda^2_1 = \lambda^2_2 = \lambda^2_3$, then any three linearly independent directions which need not be mutually perpendicular to each other can be taken as $\mathbf{N}^{(1)},\ \mathbf{N}^{(2)}$, and $\mathbf{N}^{(3)}$.

Recalling that

$$2E_{AB} = C_{AB} - \delta_{AB},$$

we obtain

$$2E_{AB}N^{(1)}_B =\ C_{AB}N^{(1)}_B - \delta_{AB}N^{(1)}_B\ ,$$

$$=\ (\lambda^2_1 - 1)N^{(1)}_A.$$ (3.9.12)

To arrive at this equation, we used (3.9.7). Thus $\mathbf{N}^{(1)},\ \mathbf{N}^{(2)}$ and $\mathbf{N}^{(3)}$ are the directions along which $E_{AB}N_AN_B$ takes on extreme values

$$(\lambda^2_1 - 1)/2,\ (\lambda^2_2 - 1)/2,\ (\lambda^2_3 - 1)/2$$ (3.9.13)

respectively. If we define strain in the direction $\mathbf{P}\mathbf{Q}$ as

$$(\vert\mathbf{P}^\prime\mathbf{Q}^\prime\vert^2 - \vert\mathbf{P}\mathbf{Q}\vert^2)/\vert\mathbf{P}\mathbf{Q}\vert^2$$ (3.9.14)

instead of the engineering strain

$$(\vert\mathbf{P}^\prime\mathbf{Q}^\prime\vert - \vert\mathbf{P}\mathbf{Q}\vert)/\vert\mathbf{P}\mathbf{Q}\vert,$$ (3.9.15)

we see from (3.8.7) that $E_{AB}N_AN_B$ gives strain in the direction $\mathbf{N}$. Thus the three numbers given by (3.9.13) are the extreme values of strains and since the angle between the mutually orthogonal directions $\mathbf{N}^{(1)},\ \mathbf{N}^{(2)},\ \mathbf{N}^{(3)}$ does not change, (3.9.13) are the principal strains and $\mathbf{N}^{(1)},\ \mathbf{N}^{(2)},\ \mathbf{N}^{(3)}$ are the axes of principal strain. Note that there is no shear strain between any two of the three directions $\mathbf{N}^{(1)},\ \mathbf{N}^{(2)}$ and $\mathbf{N}^{(3)}$.

APPENDIX: SOLUTION OF A CUBIC EQUATION

The solution of a general cubic equation, which may have imaginary roots, in a closed form involves the use of hyperbolic functions. However, the cubic equation obtained from

$$\det[C_{AB} - \lambda \delta_{AB}] = 0$$ (A1)

in which $C_{AB}= C_{BA}$ has only real roots, and can be solved as follows.

The cubic equation

$$y^3 + py^2 + qy + s = 0,$$ (A2)

can be reduced to the form

$$x^3 + ax + b = 0$$ (A3)

by substituting

$$y = x + p/3.$$ (A4)

In (A3)

$$a = (3q-p^2)/3 \ \ b = (2p^3 - 9pq + 27s)/27\ .$$

The reduced cubic equation (A3) can be solved by transforming it to the trignometric identity

$$4\cos^3\theta - 3\cos\theta - \cos 3\theta = 0.$$

(See pages 93-95, CRC Standard Mathematical Tables). The solution of the equation

$$\lambda^3 - I\lambda^2 + II \lambda - III = 0$$

obtained by this method is

$$\lambda = r\cos\theta + I/3,$$

where

$$\cos 3\theta =\ (2I^3 - 9(I)(II) + 27 III)/2(I^2 - 3II)^{3/2}\ ,$$

$$r=\ 2(I^2 - 3II)^{1/2}/3.$$

Example: The deformation of a body is given by

$$u_1 = 3X^2_1 + X_2,\hspace{.2in} u_2 = 2X^2_2 + X_3,\hspace{.2in} u_3 = 4X^2_3 + X_1.$$

a)

Find the principal strains at the material point (1,1,1) in the reference configuration.

b)

Find the axis of the maximum strain through the material point (1,1,1) in the reference configuration. Also find the direction cosines of the line into which this axis is deformed.

Solution:

(a)

At any point

$$[F_{iA}] = \left[\begin{array}{ccc}1 + 6X_1 & 1 & 0\\ 0 & 1 + 4X_2 & 1\\ 1 & 0 & 1 + 8X_3\end{array}\right]\ .$$

Thus

$$[F_{iA}\vert _{(1,1,1)}] = \left[\begin{array}{ccc} 7 & 1 & 0\\ 0 & 5 & 1\\ 1 & 0 & 9\end{array}\right]\ ,$$

$$[C_{AB}] = [\mathbf{F}]^T[\mathbf{F}] =\ \left[\begin{array}{ccc} 7 & 0 & 1\\ 1 & 5 & 0\\ 0 & 1 & 9\end{ar... ...\left[\begin{array}{ccc} 7 & 1 & 0\\ 0 & 5 & 1\\ 1 & 0 & 9\end{array}\right]\ ,$$

$$=\ \left[\begin{array}{ccc}50 & 7 & 9\\ 7 & 26 & 5\\ 9 & 5 & 82\end{array}\right]\ .$$

The principal invariants of $[\mathbf{C}]$ are given by

$$I=\ C_{AA} = 50 + 26 + 82 = 158\ ,$$

$$III =\ \det[\mathbf{C}] = 50[26(82) - 25] - 7[7(82)-45] + 9[35 - 234]= 99856\ ,$$

$$II =\ \det [\mathbf{C}] (C^{-1})_{AA} = \left\vert\begin{array}{rr}26 &... ...t + \left\vert\begin{array}{rr} 50 & 7\\ 7 & 26\end{array}\right\vert = 7377\ ,$$

$$r=\ 2(I^2-3II)^{1/2}/3 = 35.484\ ,$$

$$\cos 3\theta =\ (2I^3 - 9I(II) + 27 III)/2(I^2 - 3II)^{3/2} = 0.31382\ ,$$

$$3\theta =\ 71.7^\circ,\ 360-71.7^\circ,\ 360 + 71.7^\circ \ ,$$

$$\theta =\ 23.9^\circ,\ 96.1^\circ,\ 143.9^\circ\ ,$$

$$\lambda^2_1 =\ 85.11,\ \lambda^2_2 = 48.9,\ \lambda^2_3 = 24\ .$$

$\therefore$ Principal strains at the material point $(1,1,1)$ are 42.06, 23.95, 11.5.

(b)

To obtain $\mathbf{N}^{(1)}$ we need to solve

$$(C_{AB} - \lambda^2_1\delta_{AB})N^{(1)}_B = 0,$$

$$N^{(1)}_BN^{(1)}_B = 1.$$

$$\begin{array}{ll} -35.11 N^{(1)}_1 + 7N^{(1)}_2 + 9N^{(1)}_3 ... ...1)}_2 N^{(1)}_2 + N^{(1)}_3 N^{(1)}_3 = 1\ . & (iv) \end{array}$$

$7(i) + 35.11(ii)$ gives

$$-2026.35N^{(1)}_2 + 238.55N^{(1)}_3 = 0\ ,$$

$$N^{(1)}_3 = 8.494N^{(1)}_2.$$

$5(i)-9(ii)$ gives

$$-238.55N^{(1)}_1 + 567N^{(1)}_2 = 0\ ,$$

$$N^{(1)}_1 = 2.377N^{(1)}_2\ .$$

Substituting for $N^{(1)}_3$ and $N^{(1)}_1$ into (iv), we get

$$(2.377)^2N^{(1)}_2 N^{(1)}_2 + N^{(1)}_2 N^{(1)}_2 + (8.494)^2N^{(1)}_2 N^{(1)}_2 = 1.$$

$$N^{(1)}_2 = 0.1126$$

$$\therefore\hspace{.2in} N^{(1)}_3 = 0.957,\ N^{(1)}_1 = 0.268\ .$$

$\therefore$ The axis of the maximum strain at the material point $P\ (1,1,1)$ is $(0.268, 0.1126, 0.957)$.

To find the direction cosines of the line into which this is deformed, let

$$\mathbf{P}\mathbf{Q} = ds (0.268,\ 0.1126,\ 0.957).$$

Then $(P^\prime Q^\prime )_j = F_{jA}(PQ)_A$,

$$[\mathbf{P}^\prime\mathbf{Q}^\prime] = ds\left[\begin{array}{ccc}... ...ray} \right] = ds\left[\begin{array}{l}1.989\\ 1.52\\ 8.881\end{array}\right]$$

$\therefore$ Direction cosines of $\mathbf{P}^\prime \mathbf{Q}^\prime$ are $(0.216,\ 0.164,\ 0.962)$.

Exercise: Given the displacement components

$$u_1 = 0.1X^2_2,\ u_2 = u_3 = 0.$$

a)

Find the principal strains at the material point $(1,1,0)$ in the reference configuration.

b)

Find the axis of the maximum strain at the point $(1,1,0)$ in the reference configuration.

Exercise: Given the following displacement components

$$u_1 = 2X^2_1 + X_1X_2,\ u_2 = X^2_2,\ u_3 = 0.$$

Find the principal strains and their axes at the material point $(1/2,0,0)$ in the reference configuration.