Mathematical Physics - Volume II - Numerical Methods

6.4 Lattices

253

The deformation energy of the equivalent micropolar continuum is

U continuum = (6.52) where V = ( √ 3 / 2 ) L 2 t denotes the volume of the continuum corresponding to the unit cell (Figure 6.16b), while t is the thickness (which, in the general case, should be distinguished from the thickness of the half-beam t ( b ) perpendicular to the lattice plane, shown in Figure 6.16). The condition of equivalence (6.28) of deformation energies (6.51) and (6.52), leads to the following expressions γ αβ C αβ γδ γ γδ + κ α D αβ κ β V 1 2

( b ) γ h n

δ ( b ) ˜ R ( b ) i , D

6 ∑ b = 1

6 ∑ b = 1

n ( b )

( b ) β n

( b ) δ R

n ( b )

( b ) α S

( b ) + ˜ n β

( b ) ˜ n

( b )

C αβ γδ =

α n

α n

(6.53)

αβ =

where

E ( b ) A ( b ) √ 3 Lt

12 E ( b ) I ( b ) √ 3 L 3 t

E ( b ) I ( b ) √ 3 Lt

˜ R ( b ) =

R ( b ) =

S ( b ) =

, (6.54) It should be noted that, given (6.54) 1 and (6.54) 2 and the above-mentioned definitions of the cross-sectional properties, it is straightforward to show that , .

h ( b ) L ( b ) !

2

= ¯ h ( b )

˜ R ( b ) R ( b )

2

(6.55)

=

where the height of the beam cross section normalized with its length (Figure 6.16), ¯ h ( b ) = h ( b ) / L ( b ) , represents a geometric parameter inverse of the slenderness of the beam element. Referring to the properties of the unit cell illustrated in Figure 6.16b θ ( b ) = ( b − 1 ) π / 3 , n ( b ) = n ( b ) 1 , n ( b ) 2 = ( cos θ ( b ) , sin θ ( b ) ) , ˜ n ( b ) = ˜ n ( b ) 1 , ˜ n ( b ) 2 = ( − sin θ ( b ) , cos θ ( b ) ) , b = 1 , 2 , 3 , 4 , 5 , 6 and assuming equality of beams ( R ( b ) = R , . . . ) , the non-zero stiffness components are obtained from (6.53) in the following form

R 3 +

˜ R R

R 1 −

˜ R R

C 1111 = C 2222 = 3 4

3 4

, C 1122 = C 2211 =

R 1 − R 1 + 3

˜ R R

R 1 + 3

˜ R R

3 4 3 4

3 4

(6.56)

C 1221 = C 2112 =

, C 1212 =

˜ R R

C 2121 =

D 1 = D 22 = 2 S .

,

Corresponding Lame coefficients are

√ 3 4

√ 3 4

R 1 +

˜ R R

R 1 −

˜ R R

t ( b ) t

t ( b ) t

3 4

¯ h ( 1 + ¯ h 2 ) E ( b ) , λ = 3 4

¯ h ( 1 − ¯ h 2 ) E ( b ) (6.57)

µ =

=

=