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or in matrix form
d
z
dx
=
+
Az
P
(1.134)
where
1
k
s
GA
w
V
M
0
−
1
0
0
0
−
1
EI
00
0
z
=
A
=
P
=
(1.135)
00
0
0
p
z
0
00
1
0
Equations (1.133) or (1.134) can be solved using the methodology developed in Chapters
4 and 5. It is of interest to note that as in the case of the mixed method equations for the
general elastic solids, the mixed method governing equations for the beam of Eq. (1.133)
do not involve derivatives of geometrical or material parameters and all derivatives are of
the first order. This is not the case for the displacement governing equations of Eqs. (1.127)
and (1.128). In many instances, these characteristics are highly advantageous when solving
the equations. For example, many numerical integration schemes operate with first order
derivatives only, and equations with higher order derivatives must first be reduced to this
form.
If the axial terms are included, Eq. (1.134) would be defined using
1
EA
u
0
w
N
V
M
00 0
0
0
0
0
0
−
1
k
s
GA
00
−
10
0
1
EI
00 0
0
0
z
=
A
=
P
=
(1.136)
p
x
00 0
0
0
0
00 0
0
0
0
−
p
z
0
00 0
0
1
0
1.8.7
Stress Formulas
The solution of the governing equations of this section provides the state variables
,V,
and
M
along the beam. Given the moment
M
and shear force
V
, the major normal and
shear stresses in the beam can be computed. Recall from Eq. (1.106a)
du
0
w
,
θ
/
dx
=
N
/(
EA
)
and from Eqs. (1.97) and (1.106b)
d
θ/
dx
=
M
/(
EI
)
. Substitution of these into
σ
=
E
=
x
x
E
(
du
0
/
dx
+
zd
θ/
dx
)
gives
N
A
+
M
I
σ
x
=
z
(1.137)
When the bending moment is zero,
A
, which indicates that the stress due to the
axial force
N
is equal to the average normal stress on the cross-section. That is, this stress is
uniformly distributed over the cross-section. The other term,
σ
x
=
N
/
I
, is referred to as
the
flexure formula
. This bending stress is a linearly distributed stress equal to zero at
z
σ
=
Mz
/
x
0,
the location of the centroid. It assumes its maximum value when
z
reaches its maximum
value at an outer edge of the cross section.
To find the shear stress
=
τ
xz
,
substitute
σ
=
Mz
/
I
in the first term of Eq. (1.120a) and
x
use Eq. (1.121c). Then
Vz
I
. Assume the cross-section
is rectangular of height
t
and suppose the shear stresses are distributed uniformly across
the width. Integrate
/
I
+
∂τ
/∂
z
=
0or
−
∂τ
/∂
z
=
Vz
/
xz
xz
−
∂τ
/∂
z
=
Vz
/
I
with respect to
z
from the
z
position where
τ
xz
is to
xz
be evaluated, to the level of
z
=
t
/
2, where
t
/
2defines the top (or bottom) surface of the
I
t
/
2
V
−
τ
|
t
/
2
=
beam. Then
zdz
. Since there are no loads in the
x
direction on the upper
xz
z
z
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