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The deformation of an object is defined by a tensor field, i.e., this strain tensor is defined for every point of the object. This field is linked to the field of the stress tensor by the generalized Hooke's law.
In case of small deformations, the strain tensor is the Green tensor or Cauchy's infinitesimal strain tensor, defined by the equation:
Where u represents the displacement field of the object's configuration (i.e., the difference between the object's configuration and its natural state). This is the 'symmetric part' of the Jacobian matrix. The 'antisymmetric part' is called the small rotation tensor.
For large (finite) deformations see Finite Deformation Tensors.
Additional recommended knowledge
Demonstration in simple cases
When the [AB] segment, parallel to the x1-axis, is deformed to become the [A'B' ] segment, the deformation being also parallel to x1
the ε11 strain is (expressed in algebraic length):
the strain is
The series expansion of u1 is
And in general
Pure shear strain
Let us now consider a pure shear strain. An ABCD square, where [AB] is parallel to x1 and [AD] is parallel to x2, is transformed into an AB'C'D' rhombus, symmetric to the first bisecting line.
The tangent of the γ angle is:
for small deformations,
and u2(A) = 0. Thus,
Considering now the [AD] segment:
where γ12 is the engineering strain, which is equal to 2γ.
It is interesting to use the average because the formula is still valid when the rhombus rotates; in such a case, there are two different angles and and the formula allows for neglecting the variation of angle due to rigid-body motion (which gives no contribution to the strain).
Relative variation of the volume
The dilatation (the relative variation of the volume) δ = ΔV/V0, is the trace of the tensor:
Actually, if we consider a cube with an edge length a, it is a quasi-cube after the deformation (the variations of the angles do not change the volume) with the dimensions and V0 = a3, thus
as we consider small deformations,
therefore the formula.
In case of pure shear, we can see that there is no change of the volume.
Derivation of the strain tensor
Let the position of a point in a material be specified by a vector with components xi. Let the point then move a small distance to a new position specified by a vector with components
where ui is a vector function of . Let xi + dxi be a point close to xi. After the motion, it will be in a new position given by:
Since the ui are small, we may approximate them by the first two terms in their Taylor series
where we have used to represent and we have used Einstein notation in which repeated indices in a product are assumed to be summed (i.e. index j in this case). is the Jacobian matrix of the ui function. If we represent the unit matrix by δij then the above equation may be written:
It is seen that the final term (the displacement matrix) specifies the infinitesimal change in the position (dx'i) of the nearby particle. If the ui are constants, the displacement matrix will be the unit matrix, and the resulting displacement will simply be a rigid translation. Any matrix may be written as the sum of an antisymmetric matrix and a symmetric matrix. Writing the diplacement matrix (in parentheses in the above equation) in this manner yields:
The first two terms are the unit matrix and the antisymmetric part of the displacement matrix. These are the first two terms in the Taylor series of a rigid rotation about the translated point x'i. They constitute an infinitesimal rotation and therefore do not represent a deformation of the material. It is the second, symmetric matrix which represents the deformation of the material and this is just the strain tensor :
|This article is licensed under the GNU Free Documentation License. It uses material from the Wikipedia article "Strain_tensor". A list of authors is available in Wikipedia.|