CONTINUUM MECHANICS for ENGINEERS

6.16 For an elastic body whose x 3 axis is an axis of elastic symmetry of
order N = 6, show that the nonzero elastic constants are C 11 = C 22, C 33,
C 55 = C 44, C 66 = C , and C 13 = C 2 11 − C12
23.
6.17 Develop a formula in terms of the strain components for the strain
energy density W for the case of an orthotropic elastic medium.
Answer:
1
( )
1
( ) + ( + )
1
2 11 1 12 2 13 3 1
W = C ε + 2C ε + 2C ε ε C ε 2C
ε ε
( 44 4 55 5 66 6 )
1
2 33 3
2 2 2 2
+ C ε + C ε + C ε + C ε
6.18 Show that, for an elastic continuum having x1 as an axis of elastic
symmetry of order N = 2, the strain energy density has the same form
as for a continuum which has an x2x3 plane of elastic symmetry.
Answer:
6.19 Let the stress field for a continuum be given by
2 22 2 23 3 2
2W = ε1( C11ε1+ 2C12ε2+ 2C13ε3+ 2C14ε4)+ ε2C22ε2+ 2C23ε3+ 2C24ε4
( )
2
2
+ ε3( C33ε3 + 2C34ε4)+
C 44ε4 + ε5( C55ε5 + 2C56ε6)+
C66ε6
⎡x1+
x2
σ 12
⎢
[ σ ij]=
⎢
σ x − x
⎣
⎢ 0 0
12 1 2
where σ12 is a function of x1 and x2. If the equilibrium equations are
satisfied in the absence of body forces and if the stress vector on the
plane x1 = 1 is given by t , determine σ12 as a
function of x1 and x2. Answer: σ12 = x1 – x2 + 5
6.20 Invert Eq 6.4-3b to obtain Hooke’s law in the form
ˆ ( e1)
= ( 1+ x eˆ 6 eˆ
2) 1+ ( −x2)
2
( )
σij ε ε δ
= 2G
ij + v kk ij
1−2v 0 ⎤
⎥
0
⎥
x ⎦
⎥
2
which, upon combination with Eqs 6.4-2 and 6.4-1, leads to the Navier
equation
G
( ui,jj + uj,ij)+ ρbi=
0
1−2v