Compression members - The Behaviour and Design of Steel Structures to EC3 - page 87

moment N ∆ will be completely resisted by the end moments M 1 and M 2 , so that

N ∆ =− ( M 1 + M 2 ) .

If this is substituted into equation 3.69 and if the moments M 1 and M 2 are
eliminated, then

π
k cr

2 1 π

( EI / L ) 2
α 1 α 2

− 1 = EI
L α 1 + 1 k cr cot π

(3.71)

α 2 k cr

where k cr = L cr / L . By writing the relative stiffnesses of the unbraced member at
end1as

γ 1 = 6 EI / L
α 1

= ( 6 EI / L ) 12 (3.47)

α

1

with a similar definition of γ 2 , equation 3.71 becomes

γ 1 γ 2 (π/ k cr ) 2 − 36
6 (γ 1 + γ 2 )

= π

k cr cot π

k cr .

(3.46)

3.10.6 Stiffness of bracing for a rigid frame

IfthesimplysupportedmembershowninFigure3.22hasasufficientlystiffsway
brace acting at one end, then it will buckle as if rigidly braced, as shown in
Figure3.22a, ataload π 2 EI / L 2 . Ifthestiffness α ofthebraceisreducedbelowa
minimum value α L , the member will buckle in the rigid body sway mode shown
inFigure3.22b.Theelasticbucklingload N cr forthismodecanbeobtainedfrom
the equilibrium condition that

N cr ∆ = α∆ L ,

where α∆ is the restraining force in the brace, so that

N cr = α L .

The variation of N cr with α is compared in Figure 3.22c with the braced mode
buckling load of π 2 EI / L 2 . It can be seen that when the brace stiffness α is
less than

α L = π 2 EI / L 3 ,

(3.50)

the member buckles in the sway mode, and that when α is greater than α L ,it
buckles in the braced mode.