3.3 Analysis of Deformation
85
flipped) at these particular values of θ . In other words, only one orientation exists
where the shear strain is zero, and at this orientation the normal strains take their
maximum and minimum values. The axes that define this special orientation are
called principal axes of strain, and the normal strain components relative to these
axes are principal strains. Since the element does not undergo shear, it remains
rectangular after deformation, although its dimensions generally change.
To extend this analysis to 3D, consider an arbitrary rectangular element that
undergoes deformation defined by the strain tensor E. Before deformation, the
orientation of the element is defined by three mutually orthogonal unit vectors N i
(i = 1, 2, 3) that lie normal to the element faces but are not necessarily parallel
to the coordinate axes (Fig. 3.9b). Our objectives are to find the orientation of this
element for which the shear strains vanish and to compute the corresponding normal
(principal) strains.
For a given deformation, only one set of principal axes for strain exists at each
point. The principal directions N i must satisfy the relation
N i · E · N j = E (i) δ ij ,
(3.62)
where the E (i) are principal strains. The left-hand side of this equation extracts the
components of E; the shear strains (i = j ) are set to zero, while the normal strains
(i = j ) are set equal to E (i) . An equivalent relation is given by
E · N j = E (j ) N j ,
(3.63)
as can be seen by dotting both sides with N i to get (3.62). This equation can be
written in the form of the eigenvalue problem
(E − E I) · N = 0,
(3.64)
where the subscript j has been dropped with the understanding that, in three dimensions, solving this equation yields three eigenvalues (principal strains E (i) ) and three
corresponding eigenvectors (principal directions N i ). Since E is symmetric, it can
be shown that the eigenvalues are real and the eigenvectors are mutually orthogonal.
As discussed in Sect. 2.4.3, the eigenvalues are obtained by solving the polynomial equation
det(E − EI) = −E
3
+ ¯
I 1 E
2
− ¯
I 2 E + ¯
I 3 = 0,
(3.65)
where
¯
I 1 = tr E
¯
I 2 =
1
2 [(tr E)
2
− tr(E · E)] =
1
2 ( ¯
I
2
1 − tr E
2 )
¯
I 3 = det E
(3.66)
85
flipped) at these particular values of θ . In other words, only one orientation exists
where the shear strain is zero, and at this orientation the normal strains take their
maximum and minimum values. The axes that define this special orientation are
called principal axes of strain, and the normal strain components relative to these
axes are principal strains. Since the element does not undergo shear, it remains
rectangular after deformation, although its dimensions generally change.
To extend this analysis to 3D, consider an arbitrary rectangular element that
undergoes deformation defined by the strain tensor E. Before deformation, the
orientation of the element is defined by three mutually orthogonal unit vectors N i
(i = 1, 2, 3) that lie normal to the element faces but are not necessarily parallel
to the coordinate axes (Fig. 3.9b). Our objectives are to find the orientation of this
element for which the shear strains vanish and to compute the corresponding normal
(principal) strains.
For a given deformation, only one set of principal axes for strain exists at each
point. The principal directions N i must satisfy the relation
N i · E · N j = E (i) δ ij ,
(3.62)
where the E (i) are principal strains. The left-hand side of this equation extracts the
components of E; the shear strains (i = j ) are set to zero, while the normal strains
(i = j ) are set equal to E (i) . An equivalent relation is given by
E · N j = E (j ) N j ,
(3.63)
as can be seen by dotting both sides with N i to get (3.62). This equation can be
written in the form of the eigenvalue problem
(E − E I) · N = 0,
(3.64)
where the subscript j has been dropped with the understanding that, in three dimensions, solving this equation yields three eigenvalues (principal strains E (i) ) and three
corresponding eigenvectors (principal directions N i ). Since E is symmetric, it can
be shown that the eigenvalues are real and the eigenvectors are mutually orthogonal.
As discussed in Sect. 2.4.3, the eigenvalues are obtained by solving the polynomial equation
det(E − EI) = −E
3
+ ¯
I 1 E
2
− ¯
I 2 E + ¯
I 3 = 0,
(3.65)
where
¯
I 1 = tr E
¯
I 2 =
1
2 [(tr E)
2
− tr(E · E)] =
1
2 ( ¯
I
2
1 − tr E
2 )
¯
I 3 = det E
(3.66)