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Monday, April 11, 2016

Comparative Example on Design of Axially Loaded Steel Column Using EC3 and BS 5950-1:2000


In structural members subjected to compressive forces (e.g. columns and struts), secondary bending caused by, for example, imperfections within materials during fabrication processes, inaccurate positioning of loads or asymmetry of the cross-section etc, can induce premature failure either in a part of the cross-section, such as the outstand flange of an I section, or on the element as a whole. In such cases the failure mode is predominantly buckling.

The design of most compressive members is governed by their overall buckling capacity, i.e. the maximum compressive load which can be carried before failure occurs by excessive deflection in the plane of greatest slenderness.

Compression members (i.e. struts) should be checked for;
(1) resistance to compression
(2) resistance to buckling

The design of struts for buckling is well covered in clause 4.7 of BS 5950-1:2000 while the buckling resistance of members is covered in clause 5.5 of ENV 1993-1-1:1992 (EC3). The checks for uniform compression in EC3 is found in clause 5.4.4.

In the solved example which is downloadable, the capacity of a 4m universal column (UC 305 × 305 × 158) pinned at both ends was investigated to carry an axial load of 3556 KN using Eurocode3 and BS 5950.

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SOLUTION
The first step is to outline the properties of the section we wish to investigate.

Properties of UC 305 × 305 × 158

D = 327.1 mm; B = 311.2 mm; tf = 25mm; tw = 15.8mm; r = 15.2mm; d = 246.7mm, b/tf = 6.22; d/tw = 15.6; Izz = 38800 cm4; Iyy = 12600 cm4; iy-y = 13.9cm; iz-z = 7.9 cm, A = 201 cm2


DESIGN BY EC3



Thickness of flange tf = 25mm. Since tf > 16mm, Design yield strength Fy = 265 N/mm2 (Table 3.1 EC3)

Section classification
ε = √(235/Fy ) = √(235/265) = 0.942

We can calculate the outstand of the flange (flange under compression)
C = (b - tw - 2r) / (2 ) = (327.1 - 15.8 -2(15.2)) / (2 ) = 140.45mm.
We can then verify that C/tf = 140.45/25 = 5.618
5.618 < 9ε i.e. 5.618 < 8.478.

Therefore the flange is class 1 plastic

Web (Internal compression)
d/tw = 15.6 < 33ε so that 15.6 < 31.088. Therefore the web is also class 1 plastic

Resistance of the member to uniform compression
NC,Rd = (A.Fy)/γmo = (201 × 102 × 265) / 1.0 = 5326500 N = 5326.5 KN

NEd/NC,Rd = 3556/5326.5 = 0.6676 < 1

Therefore section is ok for uniform compression.

Buckling resistance of member (clause 5.5 ENV 1993-1-1:1992)

Since member is pinned at both ends, critical buckling length is the same for all axis Lcr = 4000mm

Slenderness ratio ¯λ = Lcr/(ri λ1)
λ1 = 93.9ε = 93.9 × 0.942 = 88.454

In the major axis
(¯λy ) = 4000/(139 × 88.454) = 0.3253
In the minor axis
(¯λz ) = 4000/(79 × 88.454) = 0.5724

Check D/b ratio = 327.1/311.2 = 1.0510 < 1.2, and tf < 100 mm (Table 5.5.3 ENV 1993-1-1:1992)

Therefore buckling curve b is appropriate for y-y axis, and buckling curve c for z-z axis. The imperfection factor for buckling curve b α = 0.34 and curve c = 0.49 (Table 5.5.1)

Φ = 0.5 [1 + α(¯λ - 0.2) + ¯λ2]

Φz = 0.5 [1+ 0.49 (0.5724 - 0.2) + 0.57242 ] = 0.7551
Φy = 0.5 [1+ 0.34 (0.3253 - 0.2) + 0.32532 ] = 0.5742

X = 1/(Φ+ √(Φ2 -¯λ2))
Xz = 1/( 0.7551 + √(0.75512 - 0.57242)) = 0.8015 < 1 ok
Xy = 1/( 0.5742+ √(0.57242 - 0.32532)) = 0.9548 < 1 ok

In this case, the lesser holds for the calculation of the buckling load.

Therefore Nb,Rd = (Xz A.Fy)/γm1 = (0.8015 × 201 × 102 × 265) / (1.0 ) = 4269189.75 N = 4269.189 KN

NEd/Nb,Rd = 3556/4269.189 = 0.8329 < 1

Therefore section is ok for buckling.

Summarily, the section is ok to resist axial load on it.




DESIGN BY BS 5950-1:2000

From Table 9 BS 5950-1:2000
Since the thickness of the flange is > 16mm but < 40mm, Py = 265 N/mm2

Obviously, the column will buckle about the z-z axis, which is the weaker axis in terms of buckling. It is however important to realise that axes are not labelled in the same way using the two codes. The z-z axis in EC3 is the y-y axis in BS 5950, while the y-y axis in EC3 is called the x-x axis in BS 5950.

ε = √(275/Fy) = √(275/265) = 1.0186
b/tf = 6.22 < 15ε = 15.28; d/tw = 15.6 < 40ε = 40.74

Hence the UC section is not slender.
Therefore, let us focus on the weaker axis (z-z) in order to verify the buckling load.
Slenderness ratio = λ = Lcr/rz = 4000/79 = 50.633

From Table 23 of BS 5950, for H sections of thickness < 40mm, strut curve c is appropriate for calculating our compressive strength PC (N/mm2). We can go through the stress of calculating the compressive strength PC using the formulars outlined in ANNEX C of BS 5950 or simply read them from Table 24 of the code. We will use the two methods in this example.

Calculating using formular
Limiting slenderness λ0 = 0.2 [(π2E)/Py ]0.5 = 0.2 [(π2 × 210000) / 265]0.5 = 17.6875

Perry factor for flexural buckling under axial load η = [a(λ - λ0)]/1000 where a = 5.5 for strut curve c
η = [5.5(50.633 – 17.6875)] / 1000 = 0.1812
Euler load PE = (π2E)/λ2 = (π2 × 210000) / 50.6332 = 808.447 N/mm2
Φ = (Py + (η+1)PE) / 2 = [265 + (0.1812 + 1) × 808.447] / 2 = 609.9688 N/mm2

Therefore, the compressive strength Pc = (PE Py) / (Φ + (Φ2 - PE Py)0.5)

Pc = (808.447 × 265) / (609.9688 + [609.96882 - (808.447 × 265)]0.5) = 212.699 N/mm2

If we had decided to read from chart, strut curve C, λ < 110, Py = 265 N/mm2

Knowing that λ = 50.633. Interpolate between λ = 50 and 52 to obtain Pc = 212.0505 N/mm2
So let us use PC = 212.699 N/mm2
Hence Buckling load, PX = AgPC = 201 × 102 × 212.699 = 4275249.9 N = 4275.2499 KN

Therefore, N/PX = 3556/4275.2499 = 0.8317 < 1.0. Hence the section is ok for buckling.

The difference in result between EC2 buckling load (Nb,Rd =4269.189 KN) and BS 5950 buckling load (PX = 4275.2499 KN) for axially loaded columns disregarding load factor is just about 0.144%.

You can download the paper HERE and compare the results from the two codes.

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