Frames - The Behaviour and Design of Steel Structures to EC3 - page 375

8.5.2 Example 2 - buckling of a braced frame

Problem . Determine the effective length factors of the members of the braced
frame shown in Figure 8.3a, and estimate the frame buckling load factor α cr .
Solution.
For the vertical member 13, using equation 8.5,

k 1 = 1.0 (theoretical value for a frictionless hinge).
Using the form of equation 3.45 and Figure 3.19,

k 3 = 2 × 2210 × 10 4 / 10000

0.5 × 4 × 8503 × 10 4 / 12000 + 2.0 × 2210 × 10 4 / 10000 = 0.238
k cr 13 = 0.74 (using Figure 3.21a).

( k cr 13 = 0.73 if a base stiffness ratio of 0.1 is used so that

k 1 = ( 2 × 2210 × 10 4 / 10000 )/( 2 × 1.1 × 2210 × 10 4 / 10000 ) = 0.909 )

N cr 13 = π 2 × 210000 × 2210 × 10 4 /( 0.74 × 10000 ) 2 N = 836.5 kN

α cr 13 = 836.5 / 71.6 = 11.7

For the horizontal member 35, using equation 8.5,

k 3 = 2 × 8503 × 10 4 / 12000

0.5 × 3 × 2210 × 10 4 / 10000 + 2 × 8503 × 10 4 / 12000 = 0.810.
k 5 = 0 (theoretical value for a fixed end).

k cr 35 = 0.66 (using Figure 3.21a).

( k cr 35 = 0.77 if a base stiffness ratio of 1.0 is used so that k 5 = 0.5 )

N cr 35 = π 2 × 210000 × 8503 × 10 4 N = 2809.6 kN

( 0.66 × 12000 ) 2

α cr 35 = 2809.6 / 25.3 = 111 > 11.7 = α cr 13

∴ α cr = 11.7 (compare with the computer analysis value of α cr = 11.32 given
in Figure 8.3d).
Aslightlymoreaccurateestimatemightbeobtainedbyusingequation8.5with
α r obtained from Figure 3.19 as

25.3 × 10 3
( 2 × π 2 × 210000 × 8503 × 10 4 / 12000 2 ) = 0.990

α r = 1 −