We refer to the design of the purlins. The analytical method will be followed.
5.2.2 Design considerations
The side rails should be continuous over two spans of 6 m, as the restrictions on transport do not allow lengths of more than 12 m. The bending of the side rails due to vertical dead loads occurs in the minor axis of the member, and the bending due to wind loads occurs in the major axis.
5.2.3 Design data Vertical spacing = 2 m.
Span = 6 m continuous over two spans.
5.2.4 Loadings 5.2.4.1 Dead loads
As calculated for purlins (assuming UB305 × 165 × 40 kg/m) = 0.154 kN/m.
Total dead load = Gk = 0.154 × 6 = 3.08 kN.
5.2.4.2 Wind loads
Effective external pressure on vertical wall face = pe = qpcpe = 1.27 × 1.1 = 1.4 kN/m2, where qp is the peak velocity pressure and cpe is the resultant pressure coeffi cient, both already calculated.
Spacing of side rail = 2 m.
So, wind pressure/m run = 1.4 × 2 = 2.8 kN/m
× total wind pressure on side rail = Wk = 2.8 × 6 = 16.8 kN.
5.2.5 Characteristic moments
5.2.5.1 Due to vertical dead loads (moments along minor axis) Over the central support = Mgk,v = 0.125GkL = 0.125 × 3.08 × 6 = 2.31 kN m.
At midspan = Mgk,y = 0.07GkL = 0.07 × 3.08 × 6 = 1.3 kN m.
5.2.5.2 Due to horizontal wind loads (moments along major axis) Over the central support = Mwk,z = 0.125WkL = 0.125 × 16.8 × 6 = 12.6 kN m.
At midspan = Mwk,z = 0.096WkL = 0.096 × 16.8 × 6 = 9.7 kN m.
5.2.6 Ultimate design moments 5.2.6.1 Due to vertical dead loads
The ultimate design moments are calculated with a safety factor for permanent actions (dead loads) γGj = 1.35.
Steel.indb 106
Steel.indb 106 5/21/2010 6:54:21 PM5/21/2010 6:54:21 PM
Ultimate design moment over the central support = Mugk,y
= γGjMgk,y = 1.35 × 2.31 = 3.1 kN m.
At midspan = Mugk,y = γGjMgk,y = 1.35 × 1.3 = 1.8 kN m.
5.2.6.2 Due to horizontal wind loads
The ultimate design moments are calculated with a safety factor for variable actions γwk = 1.5.
Ultimate design moment over the central support = Muwk,z = 1.5Mwk,z = 1.5 × 12.6
= 18.9 kN m.
At midspan = Muwk,z = 1.5Mwk,z = 1.5 × 9.7 = 14.6 kN m.
5.2.7 Ultimate shear at support This is due to horizontal wind loads.
Ultimate shear at central support = Ved = 1.5 × 16.8 × 0.625 = 15.8 kN m.
5.2.8 Design of section
The member will be designed in accordance with Eurocode 3, Part 1-1.
5.2.8.1 Design strength
The design yield strength fy is that of grade S 275 steel, fy = 275 N/mm2. 5.2.8.2 Initial sizing of section
In an elastic analysis, to limit the defl ection of the side rail to L/200, the minimum depth of the member should be as follows:
For a simply supported member: depth = h = L/20, where L is the span.
For a continuous member: depth = h = L/25.
In our case, L = 6000 mm, and the side rail is assumed to be continuous. So, minimum depth of side rail = 6000/25 = 240 mm.
Try a section UB 305 × 165 × 40 kg/m (see Fig. 5.1).
Depth of section = h = 303.4 mm.
Depth of straight portion of web = d = 265.2 mm.
Width of fl ange = b = 165 mm.
Thickness of fl ange = tf = 10.2 mm.
Thickness of web = tw = 6.0 mm.
Root thickness = r = 8.9 mm.
Radius of gyration about major axis = iy = 12.9 cm.
Radius of gyration about minor axis = iz = 3.96 cm.
Plastic modulus about major axis = Wpl,y = 623 cm3.
•
•
Steel.indb 107
Steel.indb 107 5/21/2010 6:54:22 PM5/21/2010 6:54:22 PM
Plastic modulus about minor axis = Wpl,z = 142 cm3. Elastic modulus about major axis = Wy = 560 cm3. Elastic modulus about minor axis = Wz = 92.6 cm3. Cross-sectional area = A = 51.3 cm2.
5.2.8.3 Section classifi cation Flange
Stress factor ε = √(235/275) = 0.92.
Outstand of fl ange c = (b − tw − 2r)/2 = (165 − 6.0 − 2 × 8.9)/2 = 70.4 mm.
Ratio c/tf = 70.4/10.2 = 6.9; 9ε = 8.28.
Referring to Table 5.2 (sheet 1) of Eurocode 3, Part 1-1, the limiting value c/tf ≤ 9ε. So, the fl ange satisfi es the conditions for class 1 classifi cation.
Web
Ratio d/tw = 44.2.
72ε = 66.2.
Referring to Table 5.2 (sheet 2) of Eurocode 3, Part 1-1, the limiting value d/tw≤ 72ε. So, the web satisfi es the conditions for class 1 classifi cation.
5.2.8.4 Moment capacity
We consider the moments due to wind.
Maximum horizontal design moment along major axis over support = Muy
= 18.9 kN m.
Maximum ultimate shear = Ved = 15.8 kN (the above values have already been calculated).
The moment capacity should be calculated in the following ways:
When the web is not susceptible to buckling. When the web depth-to-thickness ratio = d/tw ≤ 72ε for class 1 classifi cation, it should be assumed that the web is not susceptible to buckling, and the moment capacity should be calculated from the equa- tion Myrd = fyWpl,y, provided that the shear force Ved < 0.5Vpl,Rd. In our case, we have assumed a section in which d/tw < 72ε. So, the web is not susceptible to buckling.
When the ultimate shear force Ved ≤ 0.5Vpl,Rd. In our case, maximum ultimate shear at central support Ved = 15.8 kN, and, referring to equation (6.18) of Eurocode 3, Part 1-1,
design plastic shear capacity of section in major axis = Vpl,Rd = Avfy/√3y/γM0, where
Av = A − 2btf + 2(tw + 2r)tf = 5130 − 2 × 116.5 × 10.2 + 2 × (6 + 2 × 8.9) × 10.2
= 3239 mm2.
•
•
Steel.indb 108
Steel.indb 108 5/21/2010 6:54:22 PM5/21/2010 6:54:22 PM
But Av should not be less than hwtw = (303.4 − 2 × 10.2) × 6 = 1698 mm2 < Av. OK Therefore
Vpl,Rd = (3239 × 275/(3)0.5)/103 = 514 kN
Thus, Ved < 0.5 × 514 satisfi es the condition. So, no reduction of moment capacity needs to be considered. Therefore
plastic moment capacity MyRd = fyWpl,y = 275 × 623 × 103/106
= 171 kN m < MyEd (Muy = 18.9 kN m). OK 5.2.8.5 Shear buckling resistance
The shear buckling resistance need not be checked if the ratio d/tw ≤ 72ε. In our case, d/
tw = 265.2/6 = 44.2 and 72ε = 66.2. So, d/tw < 72ε
Therefore, the shear buckling resistance need not be checked.
5.2.8.6 Buckling resistance moment
Along the major axis of the member, when a horizontal wind pressure is acting, Muy = 18.9 kN m
As explained before in the purlin calculations, the compression fl ange is connected to side sheeting at midspan and is restrained against torsional buckling. But the compression fl ange is subject to torsional buckling over the support owing to its unrestrained condition, and thus the buckling resistance moment is reduced. The length of the unrestrained portion may be considered as one-quarter of the span on either side of the support, i.e. the point of contrafl exure in the bending moment diagram. So, let us consider the case when wind pressure acts on the member along its major axis. The buckling resistance moment may be calculated in the following way. The length between the restraints of the compression fl ange over the support is
Le = 1/4 × 6000 = 1500 mm
Referring to the equation in Clause 6.3.2.4,
λ¯f = kcLc/(if,zλ1) ≤ λc0(Mc,Rd/My,Ed) (6.59) As already calculated in the design of the purlin, Mc,Rd = 171 kN m > MEd (18.9 kN m).
So, OK.
Along the minor axis of the member, when a vertical dead load is acting, Muz = 3.1 kN m
Buckling moment of resistance = Mb,Rd(z) = Wzfy/γM1 = 142 × 103 × 275/106 = 39.1 kN m > Muz (3.1 kN m)
Therefore
Muy/Mb,Rd( y) + Muz/Mb,Rd(z) = 18.9/171 + 3.1/39.1 = 0.11 + 0.08 = 0.19 < 1 OK
Steel.indb 109
Steel.indb 109 5/21/2010 6:54:22 PM5/21/2010 6:54:22 PM
Therefore, we adopt UB305 × 165 × 40 kg/m for the side rail.