Computational Physics

Exercises 445

For a pair potential, this can be worked out further to yield

C ̃αβ γ δ=^1

2 

∑

i=j

{

1

rij^2

[

V′′(rij)−

1

rij

V′(rij)

]

aαijaβijaγijaδij

}

. (13.63)

We have used the tilde (∼) for the elasticity matrix because it is given in terms of
thexycomponents of the strain. The relation with theCmatrix given above for two
dimensions, which used(∂ux/∂y+∂uy/∂x)/2 as the third component, is given by

C 11 =C ̃xxxx, C 22 =C ̃yyyy (13.64a)

C 12 =C ̃xxyy, C 21 =C ̃yyxx (13.64b)

C 13 =

1

2

(C ̃xxxy+C ̃xxyx), C 23 =

1

2

(C ̃yyxy+C ̃yyyx) (13.64c)

C 33 =

1

4

(C ̃xyxy+C ̃xyyx +C ̃yxxy+C ̃yxyx) (13.64d)

For a Lennard–Jones potential we find, in reduced units:

C=





76.8 25.5 0

25.6 76.8 0

0 0 25.6



. (13.65)

From this we find, for the case of plane stress:ν= 1 /3 andE=68. The fact that
ν= 1 /3 shows an important shortcoming of a pair potential: irrespective of the
specific form of the potential, a pair potential always leads toν= 1 /3.

Exercises

13.1 In this problem, we study the natural coordinates for triangles. We consider an
‘archetypical’ triangle as shown in Figure 13.9. Now consider a mapping of this
triangle to some other triangle, also shown in Figure 13.9. This can be obtained from
the archetypical one by a translation over the vectorraa′, followed by a linear
transformation. The matrixUof this linear transformation can be found as
U=

(

x′b−x′a x′c−xa′

y′b−y′a y′c−y′a

)

, (13.66)

where(xa′,y′a)are the Cartesian coordinates of the vectorra′etc. We have

(

x′

y′

)

=U

(

x

y

)

+

(

xa

ya

)

(13.67)

Now we take for the natural coordinates in the archetypical trianglex,yand

1 −(x+y). It is clear that these coordinates assume the value 1 ona,bandc

respectively and vanish at the other points. We want the linear transformation of

these coordinates to have the same property. We therefore consider the function

g(x′,y′)=f(x,y)