13.4 Stability of Continuous Beams and Frames 477
3.Assumel 2 D2l 1 andEI 1 DEI 2 DEI. In this case parameter ̨D
l 2
l 1
r
EI 1
EI 2
D
2 and critical load equation becomes' 1 . /C0:5' 1 .2 /D 0 ; parameter of
critical loadD1:967. So the critical loadPcrD
3:869EI
l 12
. It can be seen that
increasing of the one span by two times has profound effect on the critical load
(coefficient 3.869 instead of 2 for two-span beam with equal spans).
Let us illustrate the displacement methodin canonical form for stability analysis of
the multistory frame.
Example 13.4.The two-story frame in Fig.13.15a is subjected to compressed axial
forcesPand ̨P. Geometrical parameters of the frame arehandl D ˇh,the
bending stiffness of the members areEIfor members of the second level andkEIfor
column of the first level. Derive the stability equation in terms of arbitrary positive
numbers ̨; ˇ,andk.
h
h
P
l=bh
EI
EI
EI
kEI
aP i
i/b
ki
i/b
1
2
r 11
Z 1 = 1
r 21
M 1
b
(^4) i
b
4 i
Z 2 = 1
r 22
r 12
M 2
4 kij 2 (u 2 )
2 ij 3 (u)
2 ij 3 (u)
4 ij 2 (u) 4 ij 2 (u)
abcd
Fig. 13.15 Two-story frame
Solution.The primary system is presented in Fig.13.15b. Let us assign member
1–2 as basic one; its flexural stiffness per unit length equalsiDEI=h. The flexural
stiffness per unit length for each member are shown in Fig.13.15b. Parameters of
critical force for member 1–2 and 2-Aare determined as follows:
1 Dh
r
P
EI
D; 2 Dh
r
PC ̨P
kEI
D
r
1 C ̨
k
:
Bending moment diagrams caused byZ 1 D 1 andZ 2 D 1 are shown in Fig.13.15c, d,
respectively.
Unit reactions are
r 11 D4i' 2 . /C
4i
ˇ
I r 12 Dr 21 D2i' 3 . /I
r 22 D4i' 2 . /C4ki' 2 . 2 /C
4i
ˇ
: