This 3-page document provides an example calculation for determining the axial force resistance of a laced column. It includes dimensions, material properties, and calculations of various parameters needed for the analysis. Key steps and results are shown, such as determining the effective length, flexural buckling resistance, and checking that the lacing can resist the required shear force. In the end, it is determined that the lacing can adequately resist the applied axial load of 270 kN.

1. TALAT Lecture 2301
Design of Members
Axial Force
Example 5.5 : Axial force resistance of laced column
3 pages
Advanced Level
prepared by Torsten Höglund, Royal Institute of Technology, Stockholm
Date of Issue: 1999
 EAA - European Aluminium Association
TALAT 2301 – Example 5.5 1

2. Example 5.5. Axial force resistance of laced column
Width b 400 . mm fo 200 . MPa kN 1000 . newton
30 . mm 20 . mm E 70000 . MPa MPa 10 . Pa
6
Longitudinal members d o di
γ M1 1.0
Lacing bd 40 . mm td 10 . mm
500 . mm
2 2
a d a b d = 640.3 mm
lc
Effective length lc 5 .m = 10
a
Axial force N Ed 270 . kN
5.4.3 Slenderness parameters
2 2
do di
π.
2
Single main Al A l = 392.699 mm
member 4
4 4
do di
π. I l = 3.191 . 10 mm
4 4
Il
64
Il
il i l = 9.014 mm
Al
Effective 2 .a . 1 . f o
λ l λ = 1.888
length 2 . a il π E l
4 .A l A = 1.571 . 10 mm
3 2
Four main A
member b
2
4 .I l 4 .A l. I = 6.296 . 10 mm
7 4
I
2
I
i i = 200.2 mm
A
Effective lc 1 fo
λ o . . λ o 0.425
=
length l c i π E
b d .t d
2
Lacing Ad A d = 400 mm
b d .t d
3
I d = 3.333 . 10 mm
3 4
Id
12
Effective Id d . 1 . fo
id i d = 2.9 mm λ d λ d 3.774
=
length d Ad id π E
f o .A .d
3
Composite 2
λ c λ o λ c 0.466
=
member E .A d .a .b
2
TALAT 2301 – Example 5.5 2

3. 5.8.4 Flexural buckling, buttoned strut
Table 5.5 α 0.2 λ o 0.1 k1 1 k2 1 λ λ c λ c = 0.466
and 5.6
1
0.5 . 1 α . λ
2
(5.33) φ λ o λ χ φ = 0.645 χ = 0.916
2 2
φ φ λ
fo
5.8.3 (1) N b.Rd χ .k 1 .k 2 . .A N b.Rd = 287.8 kN
γ M1
5.8.4 Flexural buckling, single longitudinal member
Table 5.5 α 0.2 λ o 0.1 k1 1 k2 1 λ λ l
and 5.6
1
0.5 . 1 α . λ
2
(5.33) φ λ o λ χ φ = 2.46 χ = 0.248
2 2
fo φ φ λ
5.8.3 (1) N l.Rd χ .k 1 .k 2 . .A
l N l.Rd = 19.4 kN
γ M1
5.8.4 Flexural buckling, lacing
0.015 . N Ed q χ .l c
q χ = 1.022 kN . m
1
TALAT qχ Vχ V χ = 2.555 kN
(5.22) N Ed .λ 2 2
c
l c. 1 d
0.9 . A . f o Nl V χ. 0.02 . N b.Rd = 5.757 kN N l = 4.091 kN
b
λ λ d
1
0.5 . 1 α . λ
2
(5.33) φ λ o λ χ φ = 7.989 χ = 0.067
2 2
φ φ λ
fo
5.8.3 (1) N l.Rd χ .k 1 .k 2 . .A
d N l.Rd = 5.3 kN < N l OK !
γ M1
TALAT 2301 – Example 5.5 3