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# Small Planar Motions of Beams

Module by: Ronald Nordgren. E-mail the author

Summary: The equations for large planar motion of beams in a previous module are reduced to the case of small planar motions of nearly straight beams and beam-columns. These equations will be used to analyze the stability of columns in a subsequent module.

## Small Planar Motions of Beams

If a rod is nearly straight it is called a beam or a beam-column if it is vertical. When the beam is oriented along the x axis, the slope angle is small θ01θ01 For small deflections the slope angle of the deformed beam also is small θ1θ1 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 e1,e1, 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 right-hand-sides 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 + s q x d s . T s = T 0 - 0 s q x d s = T + s 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 ( ) = - 1 2 0 y s 2 d s , x ( s ) = s - 1 2 0 s y s 2 d s , x ( ) = - 1 2 0 y s 2 d s ,
(11)

where 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(),x(),y()y() 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.

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