Civil Engineering Reference
In-Depth Information
The moment,
M
z
,at
z
is then
M
Z
=
M
p
(z)
+
Py
.
(8.32)
EI
d
2
y/
d
z
2
into Equation 8.32 yields
Substitution of
M
Z
=−
d
2
y
d
z
2
+
k
2
y
=−
M
p
(z)
EI
,
(8.33)
where
k
2
P
EI
.
Differentiating twice yields
=
d
2
M
p
(z)
d
z
2
.
d
4
y
d
z
4
+
k
2
d
2
y
d
z
2
1
EI
=−
(8.34)
Equation 8.34 is the differential equation for axial compression and bending.
8.4.2.1.1 Axial Compression and Bending from a Uniformly Distributed
Transverse Load
−
wz(L
z)
=
M
w(z)
=
M
p
(z)
,
(8.35)
2
d
2
M
p
(z)
d
z
2
=−
w
,
(8.36)
and the differential equation for axial compression and flexure (Equation 8.34) is
d
4
y
d
z
4
+
k
2
d
2
y
d
z
2
w
EI
.
=
(8.37)
The solution of Equation 8.37 is (Chen and Lui, 1987)
tan
kL
1
w
EIk
4
w
2
EIk
2
z(L
y(z)
=
2
sin
kz
+
cos
kz
−
−
−
z)
,
(8.38)
tan
kL
1
,
d
2
y(z)
d
z
2
w
EIk
2
=−
2
sin
kz
+
cos
kz
−
(8.39)
tan
kL
1
.
EI
d
2
y
d
z
2
w
EIk
2
M
Z
=−
=
2
sin
kz
+
cos
kz
−
(8.40)
The maximum moment at the center span is
sec
kL
1
8
(
sec
(kL/
2
)
,
wL
2
8
w
k
2
−
1
)
M
z
=
L
/
2
=
2
−
=
(8.41)
k
2
L
2
