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where x d is a derivative, with respect to x , that operates on the preceding variables u 0 ,
w
,
and
. This derivation shows that, for one-dimensional problems, the adjoint relationship
takes the form
θ
b
b
u T u D T s dx
u T A T s
b
a
u T D s s dx
=
|
(1.119)
a
a
The equilibrium equations of Eqs. (1.112) and (1.113) for the beam element of Fig. 1.16
can be derived from the theory of elasticity. For our beam, which lies in and is loaded in the
xz plane, we can safely consider the y direction stresses to be zero, i.e.,
0.
From the theory of elasticity, the equilibrium equations of Eq. (1.51) for non-zero stresses
σ x ,
σ y = τ xy = τ yz =
σ z , and
τ xz are
∂σ x
x + ∂τ xz
=
0
(1.120a)
z
∂τ xz
x + ∂σ z
z =
0
(1.120b)
where the body forces are taken to be zero. We wish to integrate these relations over the
cross-sectional area. Suppose the cross-section is rectangular of width b . T he boundary
conditions for the upper and lower surfaces of the beam element are
σ z =
p z (
x
)/
b,
τ xz =
0
at the upper surface and
σ z does
not vary through the width ( y direction) for a prescribed z value. Then, using these
boundary conditions, integration over the cross-sectional area A of the above differential
equilibrium equations gives
σ z =
0 ,
τ xz =
0 at the lower surface. Suppose the stress
dN
dx =
0
(1.121a)
dV
dx =−
p z (
x
)
(1.121b)
where N , the axial force, and V , the shear force, are stress resultants defined as
N
=
A σ x dA,
V
=
A τ xz dA
Multiply Eq. (1.120a) by z , integrate over A , apply integration by parts, and use the
prescribed surface conditions to find
dM
dx =
V
(1.121c)
= A z
where M
σ x dA . Thus, the equilibrium conditions of Eqs. (1.112) and (1.113) have
been derived directly from the elasticity equations.
1.8.4
Boundary Conditions
The boundary conditions for a beam are referred to the ends of the beam where either
conditions on forces
can be imposed. For example, if an end of
a beam, for which axial effects are ignored, is fixed, then
(
S p
)
or displacements
(
S u
)
w = θ =
0on S u . If the end of
the beam is free, then M
=
V
=
0on S p . And for a simply supported end,
w =
M
=
0on
S u
+
S p
=
S . In summary, with
V
M
w
θ
s
=
u
=
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