Compression members - The Behaviour and Design of Steel Structures to EC3 - page 89

the internal resisting torque

d x − EI w d 3 φ

M x = GI t d φ

d x 3
in which GI t and EI w are the torsional and warping rigidities (see Chapter 10).
Thus, at torsional buckling

d x − EI w d 3 φ

N

A i 0 d φ d x = GI t d φ d x 3 .

(3.72)

The solution of this which satisfies the boundary conditions of end twisting
prevented ((φ) 0, L = 0 ) andendsfreetowarp (( d 2 φ/ d x 2 ) 0, L = 0 ) (seeChapter10)
is φ = (φ) L / 2 sin π x / L in which (φ) L / 2 is the undetermined magnitude of the
angle of twist rotation at the centre of the member, and the buckling load N cr , T is

GI t + π 2 EI w
L 2

N cr , T = 1
i 0

.

This solution may be generalised for compression members with other end
conditions by writing it in the form

GI t + π 2 EI w
L cr , T

N cr , T = 1
i 0

(3.54)

in which the torsional buckling effective length L cr , T is the distance between
inflexion points in the twisted shape.

3.12 Worked examples

3.12.1 Example 1 - checking a UB compression member

Problem. The457 × 191UB82compressionmemberofS275steelofFigure3.28a
is simply supported about both principal axes at each end ( L cr , y = 12.0 m), and
has a central brace which prevents lateral deflections in the minor principal plane
( L cr , z = 6.0 m). Check the adequacy of the member for a factored axial compres-
siveloadcorrespondingtoanominaldeadloadof160kNandanominalimposed
load of 230 kN.

Factored axial load. N Ed = ( 1.35 × 160 ) + ( 1.5 × 230 ) = 561 kN
Classifying the section.
For S275 steel with t f = 16 mm, f y = 275 N/mm 2

EN 10025-2

ε = ( 235 / 275 ) 0.5 = 0.924
c f /( t f ε) =[ ( 191.3 − 9.9 − 2 × 10.2 )/ 2 ] /( 16.0 × 0.924 ) = 5.44 < 14

T5.2