Computational Physics

13.3 Linear elasticity 433

Theξidepend onxandy– the relation is given inEq. (13.11). From this, and from
(13.17), we find

εεεe=





yb−yy 0 yc−ya 0 ya−yb 0

0 xc−xb 0 xa−xb 0 xb−xa

xc−xb yb−yc xa−xb yc−ya xb−xa ya−yb

















vax

vay

vbx

vby

vcx

vcy













.

(13.38)

We call the 3×6 matrix on the right hand sideB. Using the relation

σσσ=Cεεε, (13.39)

we can rewrite the element integral of the left hand side ofEq. (13.35)as
∫

e

(δv)TBTCBvde, (13.40)

wherevis a six-dimensional vector,Bisa3×6 matrix andCa3×3 matrix.
Note that there is no dependence on the coordinatesxandyin this expression.
This can be traced back to the fact that we can express the integrand in terms of
the strain, which contains derivatives of the deformationuwhich in turn is alinear
function within the element. The integral is obtained by multiplying the constant
integrand by the surface areaAof the integrand. The stiffness matrixkis therefore
given by

k=ABTCB. (13.41)

This is a 6×6 matrix which connects the six-dimensional vectorsv.
The right hand side ofEq. (13.35)also involves an integral expression. This
contains the external force. Taking this to be gravity, it is constant. We must evaluate
the integral

fe·

∫

e

[(δv)aξa+(δv)bξb+(δv)cξc]de. (13.42)

This can be written in the form

fe·Gv, (13.43)

whereGis the 2×6 matrix:

A

3

(

101010

010101

)

. (13.44)

We have now reworked (13.35) to the form

δvTKv=δvTGf, (13.45)