Infinitesimal Deformations

When the displacements and displacement gradients are small, we can neglect second order terms in (3.8.6) and approximate the strain tensor by

$$e_{AB} = 1/2 (u_{A,B} + u_{B,A}).$$ (3.14.1)

Since

$$\frac{\partial u_A}{\partial X_B} =\ \frac{\partial u_A}{\partial x_j}\frac{\partial x_j}{\partial X_B},$$

$$=\ \frac{\partial u_A}{\partial x_j}\left(\delta_{jB} + \frac{\partial u_j}{\partial X_B}\right),$$

$$=\ \frac{\partial u_A}{\partial x_j}\delta_{jB} + \frac{\partial u_A}{\partial x_j}\frac{\partial u_j}{\partial X_B},$$

$$\simeq\ \frac{\partial u_A}{\partial x_j}\delta_{jB},$$ (3.14.2)

where we have neglected the second order term in the displacement gradients; for infinitesimal deformations,

$$e_{AB} = \frac{1}{2}\left(\frac{\partial u_A}{\partial X_B} + \fr... ...partial x_j} + \frac{\partial u_j}{\partial x_i}\right) \delta_{iA}\delta_{jB}.$$ (3.14.3)

Thus one can differentiate displacements with respect to the current coordinates or the referential coordinates to evaluate the infinitesimal strain tensor.

Since $e_{AB}$ is symmetric, therefore, it has at least one orthogonal triad of eigenvectors. The eigenvectors of $e_{AB}$ are the principal axes of engineering strain and its eigenvalues are principal infinitesimal strains. From equations (3.8.3) and (3.8.6) we obtain

$$\vert\mathbf{P}^\prime \mathbf{Q}^\prime \vert/\vert\mathbf{PQ}\vert = \sqrt{1 + 2E_{11}}\simeq \sqrt{1 + 2e_{11}} \simeq 1 + e_{11}.$$ (3.14.4)

Recalling that $\mathbf{PQ} = ds\ (1,0,0)$ we see that $e_{11}$ equals the change in length per unit length of an infinitesimal line element parallel to $X_1$-axis. Similar interpretation holds for $e_{22}$ and $e_{33}$. From equations (3.8.4) and (3.8.6) we conclude that

$$\cos\theta = \frac{2E_{12}}{\sqrt{1 + 2E_{11}}\sqrt{1 + 2E_{22}}}\simeq \frac{2e_{12}}{(1 + e_{11}) (1 + e_{22})} \simeq 2e_{12}\ .$$ (3.14.5)

Here $\theta$ is the angle between the deformed positions of lines initially parallel to $X_1$ and $X_2$ axes. If the change $\gamma$ in the angle is small so that $\theta = \frac{\pi}{2} - \gamma$, then eqn. (3.14.5) becomes

$$\sin\gamma\simeq \gamma = 2e_{12}\ .$$ (3.14.6)

Thus $2e_{12}$ equals the infinitesimal change in the angle between two lines originally parallel to $X_1$ and $X_2$ axes. This change in the angle is called the shearing strain.

From equations (3.9.3) and (3.8.6) we obtain

$$\frac{\vert\mathbf{P}^\prime \mathbf{Q}^\prime\vert}{\vert\mathbf{PQ}\vert} = 1 + e_{AB} N_AN_B\ .$$ (3.14.7)

Thus the engineering strain in any direction $\mathbf{N}$ equals $N_Ae_{AB}N_B$. One can similarly show that the shearing strain between two orthogonal directions $\mathbf{M}$ and $\mathbf{N}$ equals $2M_Ae_{AB}N_B$. Recalling that

$$F_{iA} = \delta_{iA} + u_{i,A} = \delta_{iA} + H_{iA}\ ,$$

and that

$$C_{AB} = F_{iA}F_{iB}\ ,$$

we obtain

$$[\mathbf{C}]=\ [\mathbf{F}]^T[\mathbf{F}]\simeq [\mathbf{1}] + [\mathbf{H}] + [\mathbf{H}]^T\ ,$$ (3.14.8)

$$[\mathbf{U}] =\ [\mathbf{C}]^{1/2} \simeq [\mathbf{1}] + \frac{1}{2}([\mathbf{H}] + [\mathbf{H}]^T)\ ,$$

$$=\ [\mathbf{1}] + [\mathbf{e}]\ ,$$ (3.14.9)

$$[\mathbf{R}] =\ [\mathbf{F}][\mathbf{U}]^{-1}$$

$$=\ ([\mathbf{1}] + [\mathbf{H}])([\mathbf{1}] + [\mathbf{e}])^{-1}$$

$$\simeq\ [\mathbf{1}] + \frac{1}{2}([\mathbf{H}] - [\mathbf{H}]^T)\ .$$ (3.14.10)

The skew symmetric tensor

$$\omega_{AB} = \frac{1}{2}(u_{A,B} - u_{B,A})$$ (3.14.11)

gives the infinitesimal rotation. For any infinitesimal vector $\mathbf{PQ}$ we have

$$\{\mathbf{P}^\prime \mathbf{Q}^\prime\} = [\mathbf{F}]\{\mathbf{P... ...([\mathbf{1}] + [\mathbf{e}] + [{\mbox{\boldmath {$\omega$}}} ])\{\mathbf{PQ}\}$$

and therefore

$$\{\mathbf{P}^\prime \mathbf{Q}^\prime\} - \{\mathbf{PQ}\} = [\mathbf{e}]\{\mathbf{PQ}\} + [{\mbox{\boldmath {$\omega$}}}]\{\mathbf{PQ}\}\ .$$ (3.14.12)

Thus the deformation of a line element $\mathbf{PQ}$ equals the sum of the deformations caused by the infinitesimal strain tensor and the infinitesimal rotation tensor. Similarly, if

$$\mathbf{u} = \mathbf{u}^1 + \mathbf{u}^2\ ,$$ (3.14.13)

then

$$[\mathbf{e}] = [\mathbf{e}^1] + [\mathbf{e}^2]\ ,$$ (3.14.14)

$$[{\mbox{\boldmath {$\omega$}}} ] = [{\mbox{\boldmath {$\omega$}}}^1] + [{\mbox{\boldmath {$\omega$}}}^2]\ .$$ (3.14.15)

Thus the infinitesimal strains and rotations caused by a given displacement equal the sum of the infinitesimal strains and rotations caused by the components of the given displacement.

Exercise: Prove that for small deformations

$$J = 1 + e_{AA},$$ (3.14.16)

$$\rho = \rho_0 (1 - e_{AA}),$$ (3.14.17)

and for isochoric deformations

$$e_{AA} = 0.$$ (3.14.18)

Exercise: Consider the displacement field

$$u_A = k[(2X^2_1 + X_1X_2)\delta_{A1} + X^2_2\delta_{A2}].$$

Using both the infinitesimal strain theory and the finite strain theory, find the change in length per unit length for the material line element $\mathbf{PQ} = ds\ (1,1,0)$ that emanates from the material particle $P\ (1,1,1)$ in the reference configuration for $k = 10^{-4},10^{-3},10^{-2},10^{-1},1,10$. Plot these changes as a function of $k$.

Exercise: With reference to a rectangular Cartesian coordinate system, the state of strain at a point is given by the matrix.

$$[\mathbf{e}] = 10^{-4}\left[\begin{array}{crr} 5 & 3 & 0\\ 3 & 4 & -1\\ 0 & -1 & 2\end{array}\right].$$

a)

What is the engineering strain in the direction $2\mathbf{e}_1 + 2\mathbf{e}_2 + \mathbf{e}_3$ at the point $(1,1,1)$ in the reference configuration?

b)

What is the shearing strain between two perpendicular lines (in the reference configuration) emanating from the point $(1,1,1)$ in the directions of $2\mathbf{e}_1 + 2\mathbf{e}_2 + \mathbf{e}_3$ and $-3\mathbf{e}_1 + 6\mathbf{e}_3$?

For an infinitesimal rigid body motion, the displacement vector $\mathbf{u}$ is given by

$$u_A = c_A + b_{AB} X_B$$ (3.14.19)

in which $c_A$ is a constant and $b_{AB} = -b_{BA}$ is a skew-symmetric tensor. For the displacement given by (3.14.19), $e_{AB} = 0$. Naturally the following question arises: Is the rigid body motion the only deformation for which the infinitesimal strain tensor vanishes identically? The answer, as proved in the following example, is yes.

Example: Regarding $e_{AB} = 0$ as six partial differential equations, solve for $\mathbf{u}$.

Solution: $e_{AB} = 0$ corresponds to

$$\frac{\partial u_1}{\partial X_1} = \frac{\partial u_2}{\partial X_2} = \frac{\partial u_3}{\partial X_3} = 0,$$ (3.14.20)

and

$$\frac{\partial u_1}{\partial X_2} + \frac{\partial u_2}{\partial ... ...2} = \frac{\partial u_3}{\partial X_1} + \frac{\partial u_1}{\partial X_3} = 0.$$ (3.14.21)

By differentiating (3.14.21)$_1$ with respect to $X_2$ and (3.14.21)$_3$ with respect to $X_3$ and making use of (3.14.20)$_{2,3}$ we obtain

$$\frac{\partial^2u_1}{\partial X^2_2} = \frac{\partial^2u_1}{\partial X^2_3} = 0\ .$$

This when combined with (3.14.20)$_1$ yields

$$u_1 = c_1 + b_{12}X_2 + b_{13}X_3 + a_1X_2X_3,$$ (3.14.22)

in which $c_1,\ b_{12},\ b_{13}$ and $a_1$ are constants. By following a procedure similar to that used to obtain (3.14.22) for $u_1$, we obtain for $u_2$ and $u_3$ the following.

$$u_2 = c_2 + b_{21}X_1 + b_{23}X_3 + a_2X_1X_3,$$ (3.14.23)

$$u_3 = c_3 + b_{31} X_1 + b_{32}X_2 + a_3X_1X_2.$$ (3.14.24)

Substituting from (3.14.22), (3.14.23) and (3.14.24) into (3.14.21) we get

$$b_{12} + b_{21} + (a_1 + a_2)X_3 = 0,$$

$$b_{23} + b_{32} + (a_2 + a_3)X_1 = 0,$$

$$b_{31} + b_{13} + (a_3 + a_1)X_2 = 0.$$

Since these equations hold for all values of $X_1,\ X_2,\ X_3$ which correspond to various points in the body, therefore,

$$b_{12} + b_{21} = b_{23} + b_{32} = b_{31} + b_{13} = 0,$$

$$a_1 + a_2 = a_2 + a_3 = a_3 + a_1 = 0.$$

The last set of equations implies that $a_1 = a_2 = a_3 = 0$. From the other set of equations, it follows that $b_{AB} = -b_{BA}$. Hence equations (3.14.22), (3.14.23) and (3.14.24) which are a solution of $e_{AB} = 0$ reduce to

$$u_A = c_A + b_{AB} X_B.$$

Recalling equations (3.14.13) and (3.14.14) we see that the strain caused by displacements $\mathbf{u}^1$ and $\mathbf{u}^1 + \mathbf{u}^2$ is the same if $\mathbf{u}^2$ is a rigid body motion. To prevent the rigid motion of a body, one needs to fix three noncolinear points of the body.

The strain field given by eqns. (3.14.20) and (3.14.21) is a very special one in that it is identically zero throughout the body. What if we were given

$$\begin{split}&e_{11} = f(X_1,X_2),\\ &e_{22} = g(X_1,X_2), \\ &e_{12} = h(X_1,X_2), \end{split}$$ (3.14.25)

and asked to find the corresponding two-dimensional displacement field? Can we always find a displacement field that will produce the strains specified by (3.14.25)? The answer is of course no. Since

$$e_{11} = \frac{\partial u_1}{\partial X_1},\ e_{22} = \frac{\part... ...e_{12} = \frac{\partial u_1}{\partial X_2} + \frac{\partial u_2}{\partial X_1},$$

therefore,

$$2\frac{\partial^2e_{12}}{\partial X_1\partial X_2} = \frac{\parti... ...ac{\partial^2e_{11}}{\partial X^2_2} + \frac{\partial^2e_{22}}{\partial X^2_1}.$$

If we substitute for $e_{11},\ e_{22}$ and $e_{12}$ from (3.14.25) we obtain

$$2\frac{\partial^2h}{\partial X_1\partial X_2} = \frac{\partial^2f}{\partial X^2_2} + \frac{\partial^2g}{\partial X^2_1}.$$ (3.14.26)

Thus unless the given functions $f,\ g$, and $h$ satisfy (3.14.26) we will not be able to find a displacement field that will produce the desired strain field.⁴ Another way of saying essentially the same thing is that we have three equations (3.14.25)$_{1,2,3}$ for two unknowns $u_1$ and $u_2$. Unless the given expressions for $e_{11}$, $e_{22}$ and $e_{12}$ are related somehow, we will not, in general, be able to find $u_1$ and $u_2$. That relation is the equation (3.14.26) which is known as a compatability condition. In the three dimensional case we can derive compatability conditions like (3.14.26) in a similar way. However, another, perhaps neater, approach to the problem is the following.

Given the displacement $\mathbf{u}^P$ of a point $P$ in the body and the strain field $e_{AB}$ in the neighborhood of $P$, we would like to find the displacement $\mathbf{u}^Q$ of a neighboring point $Q$. Now

$$u^Q_A - u^P_A = \ \int^Q_Pdu_A = \int^Q_Pu_{A,B} dX_B ,$$

$$=\ \ \int^Q_P(e_{AB} + \omega_{AB})dX_B.$$ (3.14.27)

Note that

$$\int^Q_P\omega_{AB} dX_B =\ \int^Q_P\omega_{AB} d(X_B - X^Q_B),$$

$$=\ \omega_{AB}(X_B - X^Q_B)\bigg\vert^Q_P - \int_P(X_B - X^Q_B)\omega_{AB,C}dX_C,$$ (3.14.28)

where we have integrated by parts. Combining (3.14.27) and (3.14.28) we obtain

$$u^Q_A = u^P_A - \omega^P_{AB} (X^P_B - X^Q_B) + \int^Q_P R_{AC} dX_C\ ,$$ (3.14.29)

where

$$R_{AC} = e_{AC} - (X_B - X^Q_B)\omega_{AB,C}\ .$$ (3.14.30)

Since the path of integration in (3.14.29) from point $P$ to point $Q$ is arbitrary, in order to obtain a unique value of $\mathbf{u}^Q$, the integral in eqn. (3.14.29) must be path independent. The necessary and sufficient condition for this is

$$R_{AC,D} = R_{AD,C}.$$ (3.14.31)

Substituting from (3.14.30) into (3.14.31) and noting that

$$\omega_{AB,C} = \frac{1}{2}(u_{A,BC} - u_{B,AC}) + \frac{1}{2}(u_{C,AB} - u_{C,AB}) = e_{AC,B} - e_{BC,A},$$

we arrive at

$$e_{AB,CD} + e_{CD,AB} - e_{AD,BC} - e_{BC,AD} = 0.$$ (3.14.32)

Even though there are 81 equations given by (3.14.32) only the following six are non-trivial.

$$\begin{split}&e_{11,22} + e_{22,11} - 2e_{12,12} = 0,\\ &e_{2... ...&e_{33,12} + e_{12,33} - e_{32,31} - e_{31,32} = 0. \end{split}$$ (3.14.33)

Out of these six equations only three are linearly independent. However, we will not prove that here.

Example: For the two-dimensional small strain theory, the strains for a cantilever beam are given by

$$e_{11} = AX_1X_2,\ e_{22} = - \nu AX_1X_2,\ 2e_{12} = A(1 + \nu )(a^2-X^2_2),$$ (3.14.31)

where $A$, $a$, $\nu$ are positive constants and $\nu \le \frac{1}{2}$. Assume that the displacements $u_1$, $u_2$ relative to axes $X_1,\ X_2$ are functions of $X_1,\ X_2$.

a)

Show that continuous single-valued displacements $(u_1,u_2)$ are possible.

b)

Hence derive formulas for $(u_1,u_2)$ as explicit functions of $(X_1,X_2)$ with the conditions $u_1 = u_2 = u_{1,2} = 0$ for $X_1 = L,\ X_2 = 0$.

Solution: a) From the given expressions for $e_{11},\ e_{22}$ and $e_{12}$ we obtain

$$e_{11,22} = e_{22,11} = e_{12,12} = 0$$

so that the only non-trivial compatability equation (3.14.33)$_1$ is satisfied. $J = 1 + e_{11} + e_{22} = 1 + A(1 - \nu )X_1X_2$. For the given constraints on $A,\ a,\ \nu,\ X_1$ and $X_2$, $J > 0$. Thus continuous single-valued displacements are possible.

b) Integrating

$$\frac{\partial u_1}{\partial X_1} = AX_1X_2,\ \frac{\partial u_2}{\partial X_2} = - \nu AX_1X_2,$$

we obtain

$$u_1 = \frac{A}{2}X^2_1 X_2 + f(X_2),\ u_2 = -\frac{\nu}{2} AX_1X^2_2 + g(X_1).$$

Substituting for $u_1$ and $u_2$ into

$$\frac{\partial u_1}{\partial X_2}+ \frac{\partial u_2}{\partial X_1} = A(1 + \nu )(a^2 - X^2_2),$$

we obtain

$$\frac{df}{dX_2} - \frac{\nu}{2}AX^2_2 + A(1 + \nu ) X^2_2 = A(1 + \nu )a^2 - \frac{dg}{dX_1} - \frac{A}{2}X^2_1.$$

Since the left-hand side of this equation is a function of $X_2$ and the right-hand side a function of $X_1$, for the two sides to be always equal, each must equal a constant say $c$. Thus

$$\frac{df}{dX_2} - \frac{\nu}{2}AX^2_2 + A(1 + \nu )X^2_2 = c$$

$$A(1 + \nu )a^2 - \frac{dg}{dX_1} - \frac{A}{2} X^2_1 = c.$$

Therefore

$$f = \frac{\nu}{6} AX^3_2 - \frac{A}{3} (1 + \nu )X^3_2 + cX_2 + d,$$

$$g = A(1 + \nu )a^2 X_1 - \frac{A}{6}X^3_1 - cX_1 + e,$$

where $d$ and $e$ are constants of integration. Thus

$$\begin{split}&u_1 = \frac{A}{2}X^2_1X_2 + \frac{\nu}{6} AX^3_... ...+ A(1 + \nu ) a^2X_1 - \frac{A}{6} X^3_1 - cX_1 + e.\end{split}$$ (3.14.32)

In order that $u_1 = u_2 = u_{1,2} = 0$ for $X_1 = L,\ X_2 = 0$ we must have

$$d = 0,\ c = -\frac{A}{2}L^2,\ e = -A(1 + \nu )a^2L + \frac{A}{6}L^3 - \frac{A}{2}L^3.$$

Substituting for $d,\ c$ and $e$ into (3.14.32) we arrive at the following.

$$\begin{split}&u_1 = \frac{A}{2}X^2_1X_2 - \frac{A}{3} \left(1... ...ac{A}{6} (L^3 - X^3_1 ) + \frac{PAL^2}{2} (X_1 - L).\end{split}$$ (3.14.34)

For points on the plane $X_2 = 0$,

$$u_1 = 0,\ u_2 = A(1 + \nu )a^2 (X_1 - L) + \frac{A}{6}(L^3 - X^3_1) + \frac{AL^2}{2} (X_1-L)\ .$$ (3.14.33)

Thus longitudinal lines on the plane $X_2 = 0$ which is the neutral surface are not stretched. However, points on the longitudinal line do move vertically. Equation (3.14.34) corresponds to the deflection equation usually studied in the first Strength of Materials or Mechanics of Deforms course. However, in that course the following boundary conditions are used. For

$$X_1 = L,\ X_2 = 0,\ u_1 = u_2 = u_{2,1} = 0.$$

With these, eqn. (3.14.32) gives

$$u_1 = \frac{A}{2} X^2_1 X_2 + A(1 + \nu )a^2X_2 - \frac{A}{3} \left(1 + \frac{\nu}{2}\right) X^3_2 - \frac{AL^2}{2} X_2\ ,$$

$$u_2 = - \frac{\nu}{2} AX_1X^2_2 - \frac{A}{6} X^3_1 + \frac{AL^2}{2} X_1 - \frac{AL^3}{3}\ .$$

Thus

$$u_2 (0,0) = - \frac{AL^3}{3}$$

which agrees with the strength of materials solution.

Exercise: Check whether or not the following distribution of the state of infinitesimal strain satisfies the compatability conditions:

$$[\mathbf{e}] = \left[\begin{array}{ccc} X^2_1 & X^2_2 + X^2_3 & X_1X_3\\ X^2_2 + X^2_3 & 0 & X_1\\ X_1X_3 & X_1 & X^2_2\end{array}\right].$$

Exercise: Given the strain field

$$e_{12} = e_{21} = X_1X_2,$$

and all other $e_{AB} = 0$,

a)

Does it satisfy the equations of compatability?

b)

By attempting to integrate the strain field, show that it cannot correspond to a displacement field.