Information Technology Reference
In-Depth Information
that is,
x δφ
x δφ
∂τ
dA d x
y + ∂τ
xy
xz
ω(τ
xz a z
+ τ
xy a y
)
ds dx
+
A ω
z
x φ
∂τ
dA
y + ∂τ
xy
xz
+
δ
A ω
δ
ω(τ
xz a z
+ τ
xy a y
)
ds
z
xz dA
dA dx
1
G δτ
1
G δτ
τ
+ τ
+
A ( δτ
xy z
+ δτ
xz y
)
xy
xy
xz
A
+
[
(
M
M t
)δφ
] L
[
φ)δ
M t ] 0
=
0
(15)
From Eq. (4), and Chapter 1, Eqs. (1.142) and (1.143),
δφ ω(τ xz a z + τ xy a y )
ds dx
x
x δφ
dA d x
xy ∂ω
xz ∂ω
=−
τ
y + τ
z
A
x δφ
∂τ
dA d x
y + ∂τ
xy
xz
A ω
z
xy ∂ω
z
xz ∂ω
y dA d x
x δφ
=−
τ
y +
+ τ
z
A
∂τ
xy
dA d x
x δφ
x δφ
y + ∂τ
xz
+
A
xy z
τ
xz y
)
dA d x
A ω
z
∂ω
z 2
∂ω
y 2 dA d x
G
x δφ M t dx
x δφ
=
y +
+
z
A
∂τ xy
dA d x
x δφ
y + ∂τ xz
+
A ω
z
∂τ xy
dA d x
x δφ
y + ∂τ xz
G J
x δφ
φ dx
L
0
=
δφ
M t |
A ω
(16)
z
A
z 2
y 2 dA
with J
=
∂ω/∂
y
+
+
∂ω/∂
z
.
Substitution of (16) into (15) gives
x δφ
Applicable to the
direction of
the bar axis.
φ dx
GJ
[ M t
δφ
] L
[
φδ
M t ] 0
+
[
δ(
M t
φ)
] 0
x φ
∂τ
dA
y + ∂τ
xy
xz
+
δ
A ω
δ
ω(τ
xz a z
+ τ
xy a y
)
ds
Applicable to
the bar's
cross-section.
z
xz dA
dA dx
1
G δτ
1
G δτ
τ
+ τ
+
A ( δτ
xy z
+ δτ
xz y
)
=
0
xy
xy
xz
A
(17)
The axial and cross-sectional parts must be equal to zero separately. For the cross-sectional
terms, using
τ
τ
xy
xz
τ xy =
τ xz =
and
φ )
φ )
(
G
(
G
Search WWH ::




Custom Search