Frames - The Behaviour and Design of Steel Structures to EC3 - page 376

so that
k 3 = 2 × 2210 × 10 4 / 10000
0.5 × 4 × 0.990 × 8503 × 10 4 / 12000 + 2 × 2210 × 10 4 / 10000
= 0.240

but k cr 13 is virtually unchanged, and so therefore is α cr .

8.5.3 Example 3 - buckling of an unbraced two-storey frame

Problem . Determine the storey buckling load factors of the unbraced two-storey
frame shown in Figure 8.4a, and estimate the frame buckling load factor, by
using

a. the storey deflection method of equation 8.8, and
b. the member buckling load method of equation 8.7.

(a) Solution using equation 8.8
For the upper storey, using the first-order analysis results shown in Figure 8.4,

10
( 20 + 40 + 20 ) ×

5000
( 121.4 − 85.5 ) = 17.41

α cr , su =

For the lower storey, using the first-order analysis results shown in Figure 8.4,

( 10 + 20 )

( 20 + 40 + 20 + 40 + 80 + 40 ) × 5000

α cr , sl =

85.5 = 7.31 < 17.41 = α cr , su

∴ α cr = 7.31 (compare with the computer analysis value of α cr = 6.905 given in
Figure 8.4d).
(b) Solution using equation 8.7
For the upper-storey columns, using equation 8.4,

2210 × 10 4 / 5000
2210 × 10 4 / 5000 + 1.5 × 3415 × 10 4 / 12000 = 0.508

k T =

2210 × 10 4 / 5000 + 5259 × 10 4 / 5000
2210 × 10 4 / 5000 + 5259 × 10 4 / 5000 + 1.5 × 14136 × 10 4 / 12000

k B =

= 0.458

k cr , u = 1.44 ( using Figure 3.21b )

N cr , u = π 2 × 210000 × 2210 × 10 4 /( 1.44 × 5000 ) 2 N = 883.6 kN