5.6.1 Design considerations
The horizontal wind bracing system consists of lattice girders (of warren type) formed by the bottom chord members of the vertical roof girders and the other chord members at the bot- tom chord level of the roof trusses. Two sets of wind girders, one along column line A and another one along column line E, are to be provided (see Fig. 1.2). The chord members are 6 m apart, and bracings connect them to form a horizontal roof wind girder (see Fig. 5.8).
The girder spans 24 m between the main roof trusses. The depth of the girder is 2.25 m.
5.6.2 Functions
The wind girder on each column line, either A or E, is assumed to take the full horizontal wind reactions from the intermediate columns as point loads, and forces are induced in the individual members owing to the loadings.
5.6.3 Loadings (wind loads) From previous calculations,
effective external wind pressure qp(z) = 1.4 kN/m2.
External wind pressure/m height of intermediate column = wk = 1.4 × 6 = 8.4 kN.
The intermediate column is assumed to be fi xed at the base and simply supported at the top. Therefore the reaction at the top, Rt = Wk at a node, is equal to
(3/8)wkH = 3/8 × 8.4 × 33 = 104 kN (unfactored).
5.6.4 Forces in members of braced girder
The braced girder (1-6-10-5) shown in Fig. 5.8 is loaded with point wind loads of Wk = 104 kN (unfactored) at the nodal points 2, 3 and 4. It is assumed that all three point loads are taken by the braced girder. To calculate the forces in the members, we fi rst fi nd the reactions at the support:
Rl = 1.5 × 104 kN = 156 kN = Rr We then calculate the forces as follows.
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5.6.4.1 Forces in members (unfactored) Consider node 1.
Length of diagonal (1-7) = (2.252 + 62)0.5 = 6.41 m.
∑V = 0.
Therefore
force in member (1-7) × 2.25/6.41 = member (1-6) = Rl = 156;
force in member (1-7) = 156 × 6.41/2.25 = 444.4 kN (tension).
∑H = 0.
Therefore
(1-2) = (1-7) × 6/6.41 = 444.4 × 6/6.41 = 417 kN (compression).
Next, consider node 7.
∑V = 0.
Therefore
(2-7) + (3-7) × 2.25/6.41 − (1-7) × 2.25/6.41) = 0;
(3-7) = [(444.4 × 2.25/6.41 − 104)] × 6.41/2.25 = 148.8 kN (compression).
∑H = 0.
Therefore
(7-8) = (1-7) × 6/6.41 + (3-7) × 6/6.41 = (444.4 + 148.8) × 6/6.41) = 555.3 kN (tension).
5.6.4.2 Ultimate design forces in members
With a partial safety factor γwk = 1.5 (due to wind), we have the following ultimate design forces in the members.
In chord member (1-2), force = 1.5 × 416 = 624 kN (compression).
In chord member (2-3), force = 624 kN (compression).
In chord member (6-7), force = 0.0.
In chord member (7-8), force = 1.5 × 555.3 = 833 kN (tension).
In diagonal member (1-7), force = 1.5 × 444.4 = 667 kN (tension).
In diagonal member (3-7), force = 1.5 × 148.8 = 223 kN (compression).
In addition, the chord members (1-2), (2-3), (3-4) and (4-5), acting as bottom chord members of the vertical roof girder, have ultimate tension forces due to the vertical loads (DL + LL) (see Fig. 5.11).
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In case 1, when WL is acting with DL only, the load combination to be considered is 1.5 × WL + 1.0 × DL. Therefore, in chord member (1-2), the ultimate design force is 624 − 437 = 187 kN (compression). In chord member (2-3), the ultimate design force is 624 − 583 = 41 (compression).
In case 2, when WL is acting simultaneously with (DL + LL), the partial safety factor is taken equal to 0.9 × 1.5 = 0.9. Therefore, in chord member (1-2), the ultimate design force is 0.9 × 624 − 878 = −316 kN (tension), and in chord member (2-3), 0.9 × 624 − 1171 = − 609 kN (tension) (see Fig. 5.11).
5.6.5 Design of section of members 5.6.5.1 Chord member (7-8)
Ultimate design force = −833 kN (tension).
For the initial sizing of the section, we try two angles 120 × 120 × 12: Ag = 55 cm2. Using two 26 mm holes,
Anet = 55 − 2 × 2.6 × 1.2 = 48.76 cm2. For the design of the section,
Npl,Rd = Anetfy/γMo = 48.76 × 100 × 275/103 = 1340 kN > NEd (833 kN) Satisfactory Therefore we adopt two angles 120 × 120 × 12 back to back with a 12 mm gap between the vertical faces.
5.6.5.2 Chord member (2-3)
Ultimate design force = 187 kN (compression) when WL and DL are acting, no LL.
Ultimate design force = 1171 kN (tension) when DL + LL are acting, no WL.
Firstly, we design the member as a compression member.
For the initial sizing of the section, we try two angles 200 × 200 × 20: Ag = 153 cm2, iy = 6.11 cm, l = 600 cm.
For the section classifi cation, h/t = 200/20 = 10,
15ε = 15 × 0.92 = 13.8, (h + b)/2t = 400/40 = 10, 11.5ε = 11.5 × 0.92 = 10.6.
Referring to Table 5.2 (sheet 3) of Eurocode 3, Part 1-1, for class 3 classifi cation, h/t ≤ 15ε and (h + b)/2t ≤ 11.5ε. In our case, the section satisfi es both conditions.
For the buckling resistance in compression,
Nb,Rd = χAfy/γM1 (6.47)
λ¯ = Lcr/(iyλ1) = 600/(6.11 × 93.9 × 0.92) = 1.14
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where λ1 = 93.9ε. Referring to Table 6.2 of Eurocode 3, Part 1-1, for angles, we follow the buckling curve “b” in Fig. 6.4 of that Eurocode, and fi nd χ = 0.5. Therefore
Nb,Rd = 0.5 × 153 × 100 × 275/103 = 2100 kN >> NEd (187 kN). OK Now, we design the member as a tension member.
Anet = 153 − 2 × 2.6 × 2 = 142.6 cm2. Therefore
Npl,Rd = Anetfy/γMo = 142.6 × 275/10 = 3922 kN >> NEd (1171 kN) OK Therefore we adopt two angles 200 × 2000 × 20 back to back with a 15 mm gap between the vertical faces.
5.6.5.3 Diagonal members (3-7) and (3-9) Ultimate design force = 667 kN (tension).
Ultimate design force = 223 kN (compression).
Firstly, we design the member as a compression member.
For the initial sizing of the section, we try two angles 120 × 120 × 12 A = 55 cm2: Anet
= 55 − 2 × 26 × 1.2 = 48.8 cm2, iy = 3.65 cm, Lcr = 641 cm.
For the section classifi cation,
h/t = 120/12 = 10, 15ε = 15 × 0.92 = 13.8,
(h + b)/2t = 240/24 = 10, 11.5 × ε =11.5 × 0.92 = 10.58.
Referring to Table 5.2 of Eurocode 3, Part 1-1, to satisfy the conditions for class 3 section classifi cation, h/t ≤ 15ε and (h + b)/2t ≤ 11.5t, and we have 10 ≤ 13.5 and 10 ≤ 10.58. So, the section satisfi es the section classifi cation.
For the buckling resistance in compression, Nb,Rd = χAfy/γM1
λ¯ = Lcr/(iyλ1) = 641/(3.65 × 93.9 × 0.92) = 2.03.
Referring to Table 6.2 of Eurocode 3, Part 1-1, for angles, we follow the buckling curve
“b” in Fig. 6.4 of that Eurocode. With λ¯ = 2.03, χ = 0.2. Therefore
Nb,Rd = 0.2 × 55 × 275/10 = 303 kN > 223 kN OK
Now, we design the member as a tension member.
Npl,Rd = Anetfy/γMo = 48.8 × 275/10 = 1342 kN > 667 kN OK
Therefore we adopt two angles 120 × 120 × 12 back to back with a 12 mm gap between the vertical faces.
For the diagonals (1-7) and (5-9), we adopt two angles 120 × 120 × 10 back to back.
See Fig. 5.13, showing the forces and member sizes for the horizontal wind girder.
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main roof truss
main roof truss intermediate trusses
4 spaces @ 6.0 m = 24.0 m Wk= 104 kN Wk= 104 kN Wk= 104 kN
loads are unfactored
RL = 156 kN RR = 156 kN
HORIZONTAL LATTICED WIND GIRDER
1 2 3
4 5
0 1 9
8 7
6
Member Location
Ultimate force compression
(kN)
Ultimate force tension
(kN)
Member size 2 angles b×× h × t (1–2) chord 187 1171 200× 200 × 20 (2–3) chord 41 1171 200× 200 × 20 (3–4) chord 41 1171 200× 200 × 20 (4–5) chord 87 1171 200× 200 × 20
(6–7) chord 0 0 120× 120 × 12
(9–10) chord 0 0 120× 120 × 12
(7–8) chord 0 833 120× 120 × 12 (8–9) chord 0 833 120× 120 × 12 (1–7) diagonal 0 667 120× 120 × 12 (5–9) diagonal 0 667 120× 120 × 12 (3–7) diagonal 223 0 120× 120 × 12 (3–9) diagonal 223 0 120× 120 × 12
Fig. 5.13. Wind girder: loadings and ultimate forces in members, and sizes of members
References
Eurocode, 2002a. BS EN 1991-1-1: 2002, Actions on structures. General actions. Densities, self-weight, imposed loads for buildings.
Eurocode, 2002b. BS EN 1990: 2002(E), Basis of structural design.
Eurocode, 2005a. BS EN 1991-1-4: 2005, Actions on structures. General actions. Wind actions.
Eurocode, 2005b. BS EN 1993-1-1: 2005, Eurocode 3. Design of steel structures. General rules and rules for buildings.
Reynolds, C.E. and Steedman, J.C., 1995. Reinforced Concrete Designer’s Handbook, Spon, London.
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Case Study I: Analysis and Design of Structure of Melting Shop and Finishing Mill Building