Advanced Methods of Structural Analysis

13.4 Stability of Continuous Beams and Frames 475

Note that subscript 2 at function'is related to the clamped–clamped member

subjected to angular displacement of theone support (Table A.22), while the sub-

script 1 at the parameteris related to the compressed-bent member 1. Canonical

equation of the displacement method isr 11 . 1 /ZD 0. Nontrivial solution of this

equation leads to equation of stabilityr 11 D 0 or in expanded form

r 11 D

4 EI 1

l 1

' 2 . 1 /C

4 EI 2

l 2

D0:

Special cases:

1.Assume thatl 2! 0. In this case, the second term

4 EI 2

l 2

!1, rigid joint

is transformed to clamped support and the initial frame in whole is transformed

into the vertical clamped–clamped column. Stability equation becomes' 2 . 1 /D

1. Root of this equation (Table A.25) is 1 D 2 and critical force becomes

PcrD

 12 EI

l^21

D

42 EI

l^21

D

2 EI

.0:5l 1 /^2

;

where D0:5is effective-length factor for clamped–clamped column.
2.AssumeEI 2! 0. In this case, the rigid joint is transformed to hinge and the
initial frame is transformed into the vertical clamped-pinned column. Stability
equation becomes' 2 . 1 /D 0. Root of this equation is 1 D4:488(Table A.25)
and critical forcePcrD

 12 EI

l 12

D

4:488^2 EI

l 12

D

2 EI

.0:7l 1 /^2

;D0:7

3.Ifl 1 Dl 2 ;EI 1 DEI 2 , then stability equation becomes' 2 . 1 /C 1 D 0.

The root of this equation is 1 D5:3269and critical load equals

PcrD

 12 EI

l^21

D

28:397EI

l 12

:

Now let us consider a nonuniform two-span continuous beam shown in Fig.13.14a.

We need to derive a stability equation and determine the critical load.

P

l 1 l 2

EI 1 EI 2 Elastic curve

r 11

Z= 1

3 EI (^1) j
l^1 (u^1 )
1
3 EI (^2) j
l^1 (u^2 )
2
ab
Fig. 13.14 Continuous compressed beam: (a) Design diagram; (b) Primary system and unit
bending moment diagram