EN0175-11

EN0175-11 - EN0175 10 / 10 / 06 Strain in a solid 2 v dx2 v...

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EN0175 10 / 10 / 06 Strain in a solid x v y v u v 1 d y v 2 d y v 1 d x v 2 d x v 1 3 2 Consider an arbitrary fiber within the elastic body, In the undeformed configuration, we can represent the fiber as a small vetor: where is the length and is the unit vector along the fiber direction (orientation of the fiber). 0 d d l m x v v = 0 d l m v In the deformed configuration, the same fiber is represented as l n y d d v v = . Write the deformed position of a particle as ) , , ( ) , , ( 3 2 1 3 2 1 x x x u x x x x y y r r v v + = = where is clearly the displacement vector. We can write a differential segment ) , , ( 3 2 1 x x x u r y v d as x F y v v d d = where j i ij x y F = is called the deform gradient tensor. This suggests that 0 d d l m F l n v v = The ratio between the deformed length to undeformed length: ε λ + = = 1 d d 0 l l ( : strain) is defined as stretch. Therefore, n m F v v = 1
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EN0175 10 / 10 / 06 F contains information about both stretch and rotation (change of orientation from m v to n v ). In order to separate stretch from rigid body rotation, consider the dot product of two fibers, 2 1 2 1 2 1 d d d d d d x F F x x F x F y y T v v v v v v = = where F F C T = is the right Cauchy-Green strain. Since the dot product only depends on the relative angle between the two vectors, rigid body rotation has been effectively “filtered” out of F . We can easily see that
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EN0175-11 - EN0175 10 / 10 / 06 Strain in a solid 2 v dx2 v...

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