Advanced Methods of Structural Analysis

13.3 Stability of Columns with Rigid andElastic Supports 469

For the second portion of the column, the origin 02 is placed at the point where

forceP 2 is applied. Initial parameters for this portion coincide with corresponding

parameters at the end of the first portion (atx Dl 1 ); they are' 1 D y 10 ¤ 0;

M 1 ¤ 0 IQ 1 D 0. The slope at the end of the second portion (at thexDl 2 )

according second equation (13.10) can be presented as

' 2 .xDl 2 /D' 1 cosn 2 l 2 M 1

n 2 sinn 2 l 2

P 1 CP 2

;n 2 D

s

P 1 CP 2

EI 2

(b)

In this equations' 1 andM 1 are initial parameters for portion 2. Substitution of (a)

in (b) yields

' 2 .xDl 2 /D' 0



cosn 1 l 1 cosn 2 l 2 n 1 EI 1 sinn 1 l 1

n 2 sinn 2 l 2

P 1 CP 2



:

For a clamped support the slope' 2 .x Dl 2 /D 0 .Since' 0 ¤ 0 , then stability

equation becomes

cosn 1 l 1 cosn 2 l 2 n 1 EI 1 sinn 1 l 1

n 2 sinn 2 l 2

P 1 CP 2

D0:

After a simple rearrengement, this equation may be presented as

tann 1 l 1 tann 2 l 2 

n 1

n 2

.1Cˇ/D0:

This equation may be presented in other form. Let the total length of column

l 1 Cl 2 Dlandl 2 D ̨l,where ̨is any positive number. In this casel 1 D.1 ̨/l

and stability equations becomes

tanŒn 1 .1 ̨/ ltann 2 ̨l

n 1

n 2

.1Cˇ/D0: (c)

Limiting cases

1.Let ̨D 0. This case corresponds to uniform column of lengthl, stiffnessEI 1
and loaded byP. Stability equation becomes tann 1 lD1. The root of equation
is.n 1 l/minD=2, so the critical loadPcr1D

2 EI 1

4l^2

.

2.Let ̨D 1. This case corresponds to the uniform column of lengthl, stiffnessEI 2
and loaded by forceP. Stability equation becomes tann 2 lD1, so the critical
loadPcr 2 D

2 EI 2

4l^2

.

In general case, the equation (c) should be solved numerically. Let ̨ D 0:5,

EI 2 D 2 EI 1 ,andˇ D 3. For these parameters, the stability equation becomes

tan'tan

p

2'D

p

2 ,where'D0:5n 1 l. The root of this equation is 0.719, thus

the critical load