CONTINUUM MECHANICS for ENGINEERS

()
(9.7-9a)
(9.7-9b)
(9.7-9c)
From these equations it is again apparent that ε 22 = ε 33, and that in order to
develop the solution details for a particular material we need the expression
for J S, the shear compliance of that material as shown by the example that
follows.
Example 9.7-2
Develop the solution for the problem of Example 9.7-1 using the hereditary
integral form of constitutive equations for a bar that is Kelvin in distortion,
elastic in dilatation.
Solution
For a Kelvin material, the shear (creep) compliance, Eq 9.4-6, is
J s = (1 – e –t/τ )/G so that Eq 9.7-9a becomes for σ 11 = σ oU(t)
which may be integrated directly using Eq 9.4-12 to yield
or by a simple rearrangement
t
⎡ σ oUt⎤
⎛ σ o ⎞
2⎢ε11
−
σ δ
⎣ 9
⎥ = ⎜ − ⎟ ( ′ ) − ′
⎦ ∫ o t JSt t dt
K 0 ⎝ 3 ⎠
⎡
2⎢ε
⎣
⎡
2⎢ε
⎣
22
33
()
( ) ′
σ ⎤ σ
−
δ
9
⎥ = −
⎦ 0 3
⎛
t
oUt o⎞
⎜ ⎟ ( t′ ) J − ′ ∫
S t t dt
K ⎝ ⎠
()
( ) ′
σ ⎤ σ
−
δ
9
⎥ = −
⎦ 0 3
⎛
t
oUt o⎞
⎜ ⎟ ( t′ ) J − ′ ∫
S t t dt
K ⎝ ⎠
()
( ) ′
⎡ σ ⎤ 2σ
δ
2 ε11
1 τ
⎢ −
⎣ 9
⎥
⎦ 0 3
=
t
t− t′
′ ⎛ − ⎞
oUt o t
⎜ − e ⎟ dt′
K ∫ G ⎝ ⎠
(9.7-10)
(9.7-11)
(9.7-12)
in agreement with Eq 9.7-6. Likewise, from Eq 9.7-9b we obtain for this loading
⎡
2⎢ε
⎣
22
( )
t ⎡ − ⎤
1 e τ 1
ε11()= t σoU()
t
⎢ −
+
⎥
⎢ 3G
9K⎥
⎣⎢
⎦⎥
t
⎡
− ⎤
3K
G e τ
ε11()= t σoU()
t
⎢ +
−
⎥
⎢ 9KG 3G⎥
⎣⎢
⎦⎥
()
( )
t
t− t′
σ ⎤ σ δ ′ ⎛ − ⎞
oUt o t
−
1 τ
9
⎥ = − −
⎦ 0 3
⎜ e ⎟ dt′
K ∫ G ⎝ ⎠