This article provides a detailed worked example of designing angle steel sections subjected to pure tension and compression in accordance with Eurocode 3. These steel sections serve as internal web members within a 30-meter span Pratt truss roof structure. They are securely welded to the truss’s top and bottom chords, ensuring structural integrity
Analysis of Truss
The 30m Pratt truss has been analysed in the article “Design of Trusses”, and the axial force diagram is presented below:
The table of the maximum force in members is also published below:
From the above table, the maximum tension and compressive force on web members are 45.93KN and 69.24 respectively.
The web member under maximum compressive force has a length of 2.62m. However, since the failure under compressive force can be due to buckling which is length-dependent, we shall conservatively design the member with highest length under compression for the maximum compressive force. This saves design time and also helps in fabrication.
The compression web member with highest length is the kingpost which has a length of 4.3m. It shall be designed for maximum compressive force of 45.93KN
Member Properties
Angle section: UKA 200 x 200 x 24
The properties of the steel section listed below are excerpted from SCI P363
h = 200mm b = 200mm t = 24mm, r1 = 18mm, r2 = 9mm, A = 90.6cm2, c = 5.84, Iy = 3330cm4, Iz = 3330cm4, Iu = 5280cm4, Iv = 1380cm4, iy = 6.06cm, iz = 6.06cm, iu = 7.64cm, iv = 3.90cm, Wply = 235cm3, Wplz = 235cm3, fy = 265MPa fu = 430MPa
Verification of Angle Section under Pure Tensile Force
For tension members the expression below must be satisfied
$
\frac{N_{ED}}{N_{tRd}}\,\,\leqslant \,\,1
$
Since the joint is without holes (i.e.; welded), then only the gross cross-sectional area is considered. The tensile resistance is:
$
N_{tRd}\,\,=\frac{A\,\,f_y}{\gamma _{mo}}\,\,
$
$
N_{tRd}\,\,=\frac{90.6 x 10^2 x 265}{1 x 10^3}\,\,
$ = 2400.9KN
45.93 KN ≤ 2400.9KN, the section is adequate in tension
Verification of Angle Section under Pure Compressive Force
Cross-section Classification
We shall classify the cross-section of the angle section under pure compression (you should click here to study a detailed article on cross-section classification of steel cross-sections).
Influence of material strength
The UK National Annex recommends that EN 10025-2:2004 replaces Table 3 of EN 1993-1-1:2005 to determine the yield strength (fy) and ultimate strength (fu) of steel sections
Hence, for thickness of 24mm for grade S275 | fy = 265MPa, fu = 430MPa (Table 7, EN 10025-2)
$$
\varepsilon \,\,=\,\,\sqrt{\frac{235}{265}}\,\,=\,\,0.940
$$
Section Classification
Since the dimension of both flanges are equal (equal angle) then section classification is only carried out once.
To classify an angle section, two conditions must be satisfied
- Condition 1: h/t ≤ 15 ℇ
(Using Table 5.2 (Sheet 3 of 3) of EN 1993-1-1-2005)
h/t = 200/24 = 8.33
15 ℇ = 15 x 0.94 = 14.1
Since h/t ≤ 15 ℇ first condition satisfied
- Condition 2: (b + h)/2t ≤ 5 ℇ
(b + h)/2t = (200 + 200)/(2 × 24) = 8,33
11.5 ℇ = 11.5 x 0.94 = 10.81
Since (b + h)/2t ≤ 11.5 ℇ, second condition satisfied
Since both conditions are satisfied, the angle is class 3
Compression Resistance
The expression below has to be satisfied for the section to be adequate in compression
$
\frac{N_{ED}}{N_{cRd}}\,\,\leqslant \,\,1
$
$
N_{cRd}\,\,=\frac{A\,\,f_y}{\gamma _{m}}\,\,
$ for class 1, 2, & 3 sections
$
N_{cRd}\,\,=\frac{90.6 x 10^2\,\,265}{\ 1 x 10^3}\,\,
$ = 2400.9KN
Since 69.24 KN ≤ 2400.9, section is adequate
Flexural Buckling Resistance
The flexural buckling is determined for y-axis, z-axis, and v-axis. About any of the axis, the section is adequate for flexural buckling if the expression below is satisfied
$
\frac{N_{ED}}{N_{bRd}}\,\,\leqslant \,\,1
$
Flexural Buckling Resistance about y-axis
$
N_{bRd}\,\,=\chi \frac{A\,\,f_y}{\gamma _m}
$ for class 1, 2, & 3 sections
$
\,\,\chi \,\,=\,\,\frac{1}{\phi \,\,+\,\,\sqrt{\phi ^2\,\,-\,\,\lambda ^2}}\,\,\leqslant \,\,1
$
Φ = 0.5 (1 + α(λ – 0.2) + λ²)
$
\,\,\lambda \,\,=\,\,\sqrt{\frac{A\,\,f_y}{N_{cr}}}\,\,=\,\,\frac{L_{cr}}{i}\,\,\frac{1}{\lambda _1}
$ for class 1, 2 & 3 cross-sections
Determine the Non-dimension Slenderness
iy =6.06cm
λ1 = 93.9ℇ = 93.9 x 0.940 = 88.4
Lcr = 4300mm
$
\,\,\lambda \,\,=\,\,\frac{4300}{6.06 x 10}\,\,\frac{1}{88.4}
$= 0.8
since the web members are welded to the top and bottom chords, they are assumed to have adequate restraint like in the case of two bolts as fasteners. In this case of adequate restraint, the effective slenderness (λeff) in Annex BB shall be adopted and whatever eccentricity in the joints neglected.
(NB: The angle sections would have been designed as section under combined bending and compression if no adequate restraint such as in the case of a single bolt to cater for its eccentricity.)
λeff,y = 0.5 + 0.7λy
λeff,y = 0.5 + 0.7 x 0.8
λeff,y = 1.06
Determine the Reduction factor
For angle sections, buckling curve b is to be used (Table 6.2 EN 1993-1-1)
α = 0.34 (Table 6.1 EN 1993-1-1)
Φ = 0.5 (1 + 0.34(1.06 – 0.2) + 1.06²) = 0.9
$
\,\,\chi \,\,=\,\,\frac{1}{0.9 \,\,+\,\,\sqrt{0.9^2\,\,-\,\,1.06 ^2}}
$ = 0.44
Since 0.44 ≤ 1, χ = 0.44
Nb,Rd = 0.44 x 2400.9
Nb,Rd = 1044.5KN
Check
$
\,\,\frac{N_{Ed}}{N_{bRd}}
$ = $
\,\,\frac{69.24}{1044.5}
$ = 0.066
Since 0.066 ≤ 1, the section is adequate
Flexural Buckling Resistance about z-axis
Since the angle section is equal angle which has the same section properties in z-axis as on the y-axis, the buckling resistance on z-axis is the same as on the y-axis
Flexural Buckling Resistance about v-axis
$
N_{bRd}\,\,=\chi \frac{A\,\,f_y}{\gamma _m}
$ for class 1, 2, & 3 sections
$
\,\,\chi \,\,=\,\,\frac{1}{\phi \,\,+\,\,\sqrt{\phi ^2\,\,-\,\,\lambda ^2}}\,\,\leqslant \,\,1
$
Φ = 0.5 (1 + α(λ – 0.2) + λ²)
$
\,\,\lambda \,\,=\,\,\sqrt{\frac{A\,\,f_y}{N_{cr}}}\,\,=\,\,\frac{L_{cr}}{i}\,\,\frac{1}{\lambda _1}
$ for class 1, 2 & 3 cross-sections
Determine the Non-dimension Slenderness
iv =3.9cm
λ1 = 93.9ℇ = 93.9 x 0.940 = 88.4
Lcr = 4300mm
$
\,\,\lambda \,\,=\,\,\frac{4300}{3.9 x 10}\,\,\frac{1}{88.4}
$= 1.25
since the web members are welded to the top and bottom chords, they are assumed to have adequate restraint like in the case of two bolts as fasteners. In this case of adequate restraint, the effective slenderness (λeff) in Annex BB shall be adopted and whatever eccentricity in the joints neglected.
(NB: The angle sections would have been designed as section under combined bending and compression if no adequate restraint such as in the case of a single bolt to cater for its eccentricity.)
λeff,z = 0.35 + 0.7λz
λeff,z = 0.35 + 0.7 x 1.25
λeff,y = 1.22
Determine the Reduction factor
For angle sections, buckling curve b is to be used (Table 6.2 EN 1993-1-1)
α = 0.34 (Table 6.1 EN 1993-1-1)
Φ = 0.5 (1 + 0.34(1.22 – 0.2) + 1.22²) = 1.12
$
\,\,\chi \,\,=\,\,\frac{1}{1.12 \,\,+\,\,\sqrt{1.12^2\,\,-\,\,1.22 ^2}}
$ = 0.36
Since 0.36 ≤ 1, χ = 0.36
Nb,Rd = 0.36 x 2400.9
Nb,Rd = 866.52KN
Check
$
\,\,\frac{N_{Ed}}{N_{bRd}}
$ = $
\,\,\frac{69.24}{866.52}
$ = 0.08
Since 0.08 ≤ 1, the section is adequate
