If a rod is nearly straight it is called a beam or a
beamcolumn if it is vertical. When the beam is oriented along the
x axis, the slope angle is small θ0≪1θ0≪1 For
small deflections the slope angle of the deformed beam also is small θ≪1θ≪1 and equations (Reference) become
∂
x
0
∂
s
0
=
cos
θ
0
≃
1

1
2
θ
0
2
≃
1
,
∂
y
0
∂
s
0
=
sin
θ
0
≃
θ
0

1
6
θ
0
3
≃
θ
0
,
∂
x
∂
s
=
cos
θ
≃
1

1
2
θ
2
≃
1
,
∂
y
∂
s
=
sin
θ
≃
θ

1
6
θ
3
≃
θ
.
∂
x
0
∂
s
0
=
cos
θ
0
≃
1

1
2
θ
0
2
≃
1
,
∂
y
0
∂
s
0
=
sin
θ
0
≃
θ
0

1
6
θ
0
3
≃
θ
0
,
∂
x
∂
s
=
cos
θ
≃
1

1
2
θ
2
≃
1
,
∂
y
∂
s
=
sin
θ
≃
θ

1
6
θ
3
≃
θ
.
(1)
Thus we may take
θ
0
=
∂
y
0
∂
s
0
,
θ
=
∂
y
∂
s
.
θ
0
=
∂
y
0
∂
s
0
,
θ
=
∂
y
∂
s
.
(2)Then, by (Reference) and Equation 2, we have
M
=
B
∂
2
y
∂
s
2

κ
0
,
κ
0
=
∂
2
y
0
∂
s
0
2
M
=
B
∂
2
y
∂
s
2

κ
0
,
κ
0
=
∂
2
y
0
∂
s
0
2
(3)and, by (Reference),
V
=

∂
M
∂
s
=

∂
∂
s
B
∂
2
y
∂
s
2

κ
0
.
V
=

∂
M
∂
s
=

∂
∂
s
B
∂
2
y
∂
s
2

κ
0
.
(4)where, by (Reference)
∂
κ
0
∂
s
=
∂
3
y
0
∂
s
0
3
∂
s
0
∂
s
=
∂
3
y
0
∂
s
0
3
∂
s
0
∂
s
=
1
1
+
e
∂
3
y
0
∂
s
0
3
∴
∂
κ
0
∂
s
≃
∂
3
y
0
∂
s
0
3
,
for
e
≪
1
.
∂
κ
0
∂
s
=
∂
3
y
0
∂
s
0
3
∂
s
0
∂
s
=
∂
3
y
0
∂
s
0
3
∂
s
0
∂
s
=
1
1
+
e
∂
3
y
0
∂
s
0
3
∴
∂
κ
0
∂
s
≃
∂
3
y
0
∂
s
0
3
,
for
e
≪
1
.
(5)
Since VsinθVsinθ is negligible for small θ, (Reference), for
e≪1,e≪1, become
∂
T
∂
s
+
q
x
=
m
∂
2
x
∂
t
2
,

∂
2
∂
s
2
B
∂
2
y
∂
s
2

κ
0
+
∂
∂
s
T
∂
y
∂
s
+
q
y
=
m
∂
2
y
∂
t
2
.
∂
T
∂
s
+
q
x
=
m
∂
2
x
∂
t
2
,

∂
2
∂
s
2
B
∂
2
y
∂
s
2

κ
0
+
∂
∂
s
T
∂
y
∂
s
+
q
y
=
m
∂
2
y
∂
t
2
.
(6)
For statics problems the righthandsides of these last two equations are zero
and the first equation can be integrated to give
T
s
=
T
0

∫
0
s
q
x
d
s
=
T
ℓ
+
∫
s
ℓ
q
x
d
s
.
T
s
=
T
0

∫
0
s
q
x
d
s
=
T
ℓ
+
∫
s
ℓ
q
x
d
s
.
(7)Then, with TsTs known, the second equation applies to the
stability of columns as will be considered in the next.section.
For static cases when T=0T=0 or when the T term is negligible,
Equation 622 reads
∂
2
∂
s
2
B
∂
2
y
∂
s
2

κ
0
=
q
y
,
∂
2
∂
s
2
B
∂
2
y
∂
s
2

κ
0
=
q
y
,
(8)and if B=B= const,,s=x,s=x, and κ0=0,κ0=0, this becomes
B
∂
4
y
∂
x
4
=
q
y
,
B
∂
4
y
∂
x
4
=
q
y
,
(9)which is the familiar basic equation of elementary beam theory. As will be
seen, for stability analysis it is necessary to include the T term in
Equation 622 .
In certain applications it is important not to identify s with x.x. By
Equation 1 and Equation 2, we have
∂
x
∂
s
=
1

1
2
θ
2
=
1

1
2
∂
y
∂
s
2
.
∂
x
∂
s
=
1

1
2
θ
2
=
1

1
2
∂
y
∂
s
2
.
(10)On integration, with x0=0,x0=0, there results
x
(
s
)
=
s

1
2
∫
0
s
∂
y
∂
s
2
d
s
,
x
(
ℓ
)
=
ℓ

1
2
∫
0
ℓ
∂
y
∂
s
2
d
s
,
x
(
s
)
=
s

1
2
∫
0
s
∂
y
∂
s
2
d
s
,
x
(
ℓ
)
=
ℓ

1
2
∫
0
ℓ
∂
y
∂
s
2
d
s
,
(11)where ℓ is the length of the beam. Thus, xsxs can be
evaluated once y(s)y(s) is determined. In particular, the position of the end of
the beam x(ℓ),x(ℓ),y(ℓ)y(ℓ) can be determined with the effect of lateral
deflection included. In some applications, the ends of the beam are held fixed
and applied lateral forces cause axial force TsTs and axial
stress to develope in the beam. An example of this is a pipe crossing a river
where the ends are anchored or their are intermediate anchors and a current
applies a lateral force. We ewll return to this problem later.