6.3 Modified Couple Stress Theory of Thermoelastic Curvilinear Panels
139
σ 13 = G
1 − 3α z
2
ϕ +
∂ w
∂ x
− βku
, ε 22 = ε 33 = −ν ε 11 + (1 + ν) γ T,
(6.23)
m 12 =
1
2
Gl
2
2
∂ϕ
∂ x
− (1 + 3αz
2
)
∂ϕ
∂ x
+
∂
2 w
∂ x 2 − βk
∂u
∂ x
,
(6.24)
m 23 =
1
2
Gl
2
(k − 6αz)
ϕ +
∂w
∂ x
− βku
− (1 − β)kϕ
.
(6.25)
General energy of an elastic beam deformation is as follows:
U =
1
2
V
(σ i j δε i j + m i j δχ i j )dw =
=
1
2
L
0
A
(σ 11 ε 11 + 2σ 13 ε 13 + 2m 12 χ 12 + 2m 23 χ 23 )d Adx.
(6.26)
Variation of elastic deformations energy for the kinematically allowed displacement fields takes the following form:
δU =
L
0
A
σ 11 δ ε 11 + σ 13 δ ε 13 + m 12 δχ 12 + m 23 δχ 23 )d Adx,
(6.27)
where A is the transversal beam cross section.
Substituting the variations δ ε 11 , δ ε 13 , δχ 12 , δχ 23 from (6.13), (6.14), (6.16),
(6.17) into (6.26) yields
δU =
L
0
A
σ 11 δ
∂u
∂ x
+ kw +
1
2
∂w
∂ x
− βku
2
+ z
∂ϕ
∂ x
− αz
3 ∂ϕ
∂ x
− αz
3 ∂
2 w
∂ x 2
+
+ σ 13 δ
(1 − 3αz
2
)
ϕ +
∂w
∂ x
− βku
+
(6.28)
+
m 12
2
δ
2
∂ϕ
∂ x
− (1 + 3αz
2
)
∂ϕ
∂ x
+
∂
2 w
∂ x 2 − βk
∂u
∂ x
+
+
m 23
2
δ
(k − 6αz)
ϕ +
∂w
∂ x
− βku
− (1 − β)kϕ
.
We define classical versus higher order resultants with regard to the transversal
beam cross section as follows:
139
σ 13 = G
1 − 3α z
2
ϕ +
∂ w
∂ x
− βku
, ε 22 = ε 33 = −ν ε 11 + (1 + ν) γ T,
(6.23)
m 12 =
1
2
Gl
2
2
∂ϕ
∂ x
− (1 + 3αz
2
)
∂ϕ
∂ x
+
∂
2 w
∂ x 2 − βk
∂u
∂ x
,
(6.24)
m 23 =
1
2
Gl
2
(k − 6αz)
ϕ +
∂w
∂ x
− βku
− (1 − β)kϕ
.
(6.25)
General energy of an elastic beam deformation is as follows:
U =
1
2
V
(σ i j δε i j + m i j δχ i j )dw =
=
1
2
L
0
A
(σ 11 ε 11 + 2σ 13 ε 13 + 2m 12 χ 12 + 2m 23 χ 23 )d Adx.
(6.26)
Variation of elastic deformations energy for the kinematically allowed displacement fields takes the following form:
δU =
L
0
A
σ 11 δ ε 11 + σ 13 δ ε 13 + m 12 δχ 12 + m 23 δχ 23 )d Adx,
(6.27)
where A is the transversal beam cross section.
Substituting the variations δ ε 11 , δ ε 13 , δχ 12 , δχ 23 from (6.13), (6.14), (6.16),
(6.17) into (6.26) yields
δU =
L
0
A
σ 11 δ
∂u
∂ x
+ kw +
1
2
∂w
∂ x
− βku
2
+ z
∂ϕ
∂ x
− αz
3 ∂ϕ
∂ x
− αz
3 ∂
2 w
∂ x 2
+
+ σ 13 δ
(1 − 3αz
2
)
ϕ +
∂w
∂ x
− βku
+
(6.28)
+
m 12
2
δ
2
∂ϕ
∂ x
− (1 + 3αz
2
)
∂ϕ
∂ x
+
∂
2 w
∂ x 2 − βk
∂u
∂ x
+
+
m 23
2
δ
(k − 6αz)
ϕ +
∂w
∂ x
− βku
− (1 − β)kϕ
.
We define classical versus higher order resultants with regard to the transversal
beam cross section as follows:
