Compression members - The Behaviour and Design of Steel Structures to EC3 - page 86

θ 1 , θ 2 asshown.Becauseoftheseactions,thereareendshears ( M 1 + M 2 + N ∆)/ L
and end moments M 1 , M 2 .The equilibrium equation for the member is

EI d 2 v

d x 2 =− Nv − M 1 + ( M 1 + M 2 ) x L + N ∆ x L ,

and the solution of this is

M 1 cos π/ k cr + M 2
N cr sin π/ k cr

sin π x

k cr L + M 1

cos π x v =−

k cr L − 1 N cr

M 1 + M 2
N cr

x

L + x

+

L ,

where N cr is the elastic buckling load transmitted by the member (see
equation 3.23) and k cr = L cr / L . By using this to obtain expressions for the end
rotations θ 1 , θ 2 ,andbysubstitutingequation3.33,thefollowingequationscanbe
obtained:

N cr L
α 1

+ 1 − π

k cr cot π

1 − π

k cr cosec π

M 1 + M 2 + N cr ∆ = 0

k cr k cr

N cr L
α 2

.

1 − π

k cr cosec π

+ 1 − π

k cr cot π

M 1 + M 2 + N cr ∆ = 0

k cr k cr

(3.69)

If the compression member is braced so that joint translation is effectively pre-
vented, the sway terms N cr ∆ disappear from equations 3.69. The moments M 1 ,
M 2 can then be eliminated, whence

π
k cr

2 1

1 − π

( EI / L ) 2
α 1 α 2

+ EI
L α 1 + 1 k cr cot π

+ tan π/ 2 k cr
π/ 2 k cr

= 1.

α 2 k cr

(3.70)

By writing the relative stiffness of the braced member at the end 1 as

γ 1 = 2 EI / L
α 1

= ( 2 EI / L ) 12 , (3.43)

α

1

with a similar definition for γ 2 , equation 3.70 becomes

π
k cr

2 γ 1 + γ 2

2

1 − π

γ 1 γ 2
4 k cr cot π

+ tan (π/ 2 k cr )
π/ 2 k cr

+ = 1. (3.42)

k cr

3.10.5Buckling of an unbraced member

If there are no reaction points or braces to supply the member end shears ( M 1 +
M 2 + N ∆)/ L , the member will sway as shown in Figure 3.17b, and the sway