Physics Asked on December 11, 2020
There are relations for the line and volume elements in continuum mechanics. For example:
begin{align}
frac{V}{V_0}&={rm det}(F)tag{1}
lambda^2&=(F^TFe_1cdot e_1)tag{2}
end{align}
with $F$ being the deformation gradient,
$$lambda=delta x/delta X tag{3}$$
is the stretch and $e_1$ is the unit vector in direction where stretch is to be found.
Is there a similar relation between infinitesimal areas (for ratio of deformed and undeformed areas)?
Your first two equations can be written in other forms, so how about this? In 1-D:
$$lambda=frac{L}{^0L}$$
(I use a pre-superscript for initial values and post-subscripts for spatial directions). In 2-D:
$$lambda_1lambda_2=frac{L_1}{^0L_1}frac{L_2}{^0L_2}=frac{A}{^0A}$$
In 3-D:
$$lambda_1lambda_2lambda_3=frac{L_1}{^0L_1}frac{L_2}{^0L_2}frac{L_3}{^0L_3}=frac{V}{^0V}$$
Can't recall ever using the 2-D case but it might arise in a plane stress or plane strain calculation.
Answered by rdt2 on December 11, 2020
There is, and it is usually referred to as Nanson's formula/equation.
Consider an area element with initial size $da$ and normal orientation vector $vec{n}$ that is deformed into an area element of size $dA$ with normal orientation vector $vec{N}$.
The relationship describing these oriented area elements before/after deformation is:
$$vec{n} da = (det{pmb{F}})(pmb{F}^{-T}cdot vec{N})dA$$
where $pmb{F}$ is the deformation gradient tensor. See A.J.M. Spencer's book on continuum mechanics or this page for a derivation of it.
Answered by aghostinthefigures on December 11, 2020
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