Civil Engineering Reference
In-Depth Information
8.4.2.1.3 Axial Compression and Bending from End Moments
M
A
+
M
B
L
z
.
M
p
(z)
=
M
MA
+
M
MB
=
M
A
−
(8.55)
Substitution of Equation 8.55 into Equation 8.33 yields
M
A
+
z
.
d
2
y
d
z
2
+
M
A
EI
+
M
B
EIL
k
2
y
=−
(8.56)
The general solution to Equation 8.56 is
M
A
+
M
B
EILk
2
M
A
EIk
2
y
=
A
sin
kz
+
B
cos
kz
+
z
−
(8.57)
Considering boundary conditions of
y(
0
)
=
y(L)
=
0
(M
A
cos
kL
M
B
)
EIk
2
sin
kL
+
M
A
EIk
2
cos
kz
M
A
+
M
B
EILk
2
M
A
EIk
2
.
y
=−
sin
kz
+
+
z
−
(8.58)
Differentiation of Equation 8.58 yields
d
y
d
z
=−
(M
A
cos
kL
+
M
B
)
EIk
sin
kL
M
A
EIk
sin
kz
+
M
A
+
M
B
EILk
2
cos
kz
−
,
(8.59)
d
2
y
d
z
2
(M
A
cos
kL
+
M
B
)
EI
sin
kL
M
A
EI
cos
kz
,
=
sin
kz
−
(8.60)
d
3
y
d
z
3
k(M
A
cos
kL
+
M
B
)
kM
A
EI
=
cos
kz
+
sin
kz
.
(8.61)
EI
sin
kL
The bending moment and shear forces are
EI
d
2
y
d
z
2
(M
A
cos
kL
+
M
B
)
M
Z
=−
=−
sin
kz
+
M
A
cos
kz
,
(8.62)
sin
kL
EI
d
3
y
d
z
3
k(M
A
cos
kL
+
M
B
)
V
z
=−
=−
cos
kz
−
kM
A
sin
kz
.
(8.63)
sin
kL
The maximum moment occurs where the shear force is zero. If Equation 8.63 is
equated to zero, we obtain
M
A
cos
kL
M
B
M
A
sin
kL
+
tan
kz
M
=−
,
(8.64)
where
z
M
is the location of the maximum moment along the
z
axis. Expressions
for sin(
kz
M
)
and cos(
kz
M
)
may be obtained from Equation 8.64 for substitution into
Equation 8.62 to obtain the maximum bending moment,
M
max
,as
⎛
(M
A
/M
B
)
2
⎞
−
2
(M
A
/M
B
)
cos
kL
+
1
⎝
⎠
.
M
max
=−
M
B
(8.65)
sin
2
kL
