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This variational principle is useful in problems in which the strains are of immediate con-
cern, such as in numerical comparisons with experimental results in which strains are
measured directly. Also, this form is useful in the numerical analysis of large structures in
which continuity requirements at member intersections are given in terms of strains rather
than stresses.
EXAMPLE 2.11 Beam Theory
Use the Hellinger-Reissner functional to derive the governing equations for a beam. Include
the effects of shear deformation.
We will employ
R of Eq. (2.96). There are only two stresses or strains of significance
in engineering beam theory (see Chapter 1, Section 1.8). These are
σ x ,
τ xz and
x ,
γ xz .
From
Eq. (2.15)
2
x
2 E +
2
x
2
xz
2 G
U 0 (σ ) = σ
2
(
1
+ v)
2 E
= σ
2 E + τ
2
xz
τ
(1)
The integrand of the second term of
R [Eq. (2.96)] is
σ ij u i, j = σ 11 u 1 , 1 + σ 13 u 1 , 3 + σ 31 u 3 , 1
xz
x
z + ∂w
u
u
= σ
x + τ
= σ
+ τ
γ
= σ
(2)
x
x
xz
xz
ij
ij
x
Suppose the only applied loading is the distributed force p z .
Then
R becomes
x + τ xz
dV
L
x
xz
σ
2 E τ
2 G + σ x
u
u
z + ∂w
R =
p z w
dx
(3)
x
V
0
To proceed, introduce the kinematic relations for engineering beam theory into (3). From
Chapter 1, Eq. (1.98), if the extension of the centerline is ignored,
x = x u
+
z
x θ.
From
Eq. (1.99),
γ xz = x u
+ x w = θ + x w.
The stresses and net resultant forces are related by
V
A
σ x =
Mz
/
I,
τ xz = τ average =
(see Chapter 1, Section 1.8). Then
L
Mz
I
2
V
A
2
dA d x
L
Mz 2
1
2 E
1
2 G
θ
V
+
w)
x
x
=
+
+
p z w
dx
R
I
A
0
A
0
L
dx
L
M 2
2 EI
V 2
2 Gk s A +
=
M
θ +
V
+
w)
p z w
dx
(4)
x
x
0
0
where the area A has been modified by the shear factor k s . Now, set
δ
=
0 , where
R
variations are taken independently with respect to the forces
(
M, V
)
and displacements
(w
,
θ).
Thus,
L
M
EI δ
V
Gk s A δ
δ
=
M
V
+
θδ
M
+
M
δ∂
θ + +
x
w)δ
V
R
x
x
0
dx
+
V
(δθ + δ∂ x w)
p z δw
=
0
(5)
Integrate by parts the integrals containing
δ∂
x
θ
and
δ∂
x
w.
Then (5) becomes
L
δ R = M
δw L
M
EI + x θ
V
Gk s A + θ + x w
δθ +
V
0 +
δ
M
+
δ
V
0
dx
+ ( x M
+
V
)δθ + ( x V
p z )δw
=
0
(6)
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