2- Ctr 11119 - Notes - Structural Steelwork - Lattice Girder Design I - EC3

M Sc Advanced Structural Engineering M.Sc. Lecture 2 Lattice Girder Design I

Dr. W.M.C. McKenzie

1

Lattice Girders Lattice girders are plane frames with an open-web construction usually having parallel chords and internal web bracing members. They are very useful in long-span long span construction with typical depth:span ratios of about 1/10 to 1/14. The framing of lattice girders should be triangulated with panel lengths equal and positioned to correspond to applied load positions where possible. When loading occurs between panel points secondary framing can be introduced to prevent bending effects in the chord members.

Dr. W.M.C. McKenzie

2

Typical Lattice Girder with Fire-proofing

Figure 2.1 Dr. W.M.C. McKenzie

3

Principal Types and Components of a Lattice Girder Point load between the nodes compression chord

N-Type Girder Figure 2.2 diagonal bracing

tension chord

Uniformly Distributed Load between the nodes compression chord diagonal bracing t i chord h d tension

Warren Girder Figure 2.3

Dr. W.M.C. McKenzie

4

Lattice Girders: Principal Types and Components If there are no applied loads between the nodes, the members of the girder are designed as simple ties or struts as for a typical pin-jointed frame. When applied loads do occur between the nodes there are two techniques which can be used to allow for the additional secondary effects, effects they are: ¡

design for additional secondary axial forces, or

¡

design for additional secondary bending forces.

Dr. W.M.C. McKenzie

5

Lattice Girders: Analysis Consider the two-bay lattice girder supporting a series of purlins as shown.

Figure 2.4 For analysis purposes the lattice girder is considered to be a pin-jointed, plane-frame with a series of point loads. Dr. W.M.C. McKenzie

6

Lattice Girders: Analysis (cont.) P/2

P

P

A

P

B

P/2 C

Secondary Bracing Members AB and BC are subjected to a combined effect caused by the f frame action ti i d i inducing b th primary both i

D

E

F

and secondary axial loads

Figure 2.5 25 P/2

P

P

A

P

B

P/2 C

Secondary Bending M b Members AB and d BC are subjected bj t d to t a combined effect caused by the primary axial loads and a secondary

D

E


F

bending moment due to the mid-span point loads. 7

Lattice Girders: Analysis: Secondary bracing P/2

P P/2

P P/2

A

Figure 2.7

P P/2

P/2 P/2

B

D

C

E

F

1st Stage: Determine the Primary axial loads 2P

P

FBC

FAB FAD

Figure g 2.8 Dr. W.M.C. McKenzie

FAE FDE

P

FBE

FEC

FCF

FEF 8

Lattice Girders: Analysis: Secondary bracing (cont.) P/2 A

P

P

P

P/2 C

B

2nd Stage: Determine the Secondary axial loads Secondary y bracing g to

D

E

transfer the mid-span point

F

loads to the nodes

Figure 2.9 Consider panel length AB P F'AB A

F'AB

B

Forces induced in the members by the

F'AE

F'1

F'2


secondary bracing F'AB , F'AE , F'1 etc. 9

Lattice Girders: Secondary bracing (cont.) 3nd Stage: Combine the results from stages 1 and 2 to obtain the final axial loads in the members A

B

FAB + F'AB F'1 FAE + F'AE

F'1 FCE + F'CE

FBE

FAD

D

F'2

F'2

FAE FDE

C

FBC + F'BC

FCE E

FEF

FCF

F

Figure 2.11

Design Forces = (Primary Axial Loads + Secondary Axial Loads) Dr. W.M.C. McKenzie

10

Lattice Girders: Secondary bracing – Example 2.1

2,0 m

The lattice girder shown supports a series of joists as indicated. Using the data given design suitable section for: (i) the th ttop and db bottom tt chords h d (ii) the main diagonals and uprights, (iii) the secondary bracing. Figure 2.12 2 12

4 Bays @ 55,0 0 m each

Figure 2.13

Design Data: Characteristic permanent load (including self-weight) gk Characteristic variable load qk Spacing of lattice girders Dr. W.M.C. McKenzie

= = =

0,6 kN/m2 0,75 kN/m2 3,5 m 11

Lattice Girders: Example 2.1 - Solution The lattice girder has secondary bracing and the top chord members will be subject j to both primary p y and secondary y axial loads. P

A

F

P

P

B K

P

P

P

C

G

L

P

P/2

D H

M

I

E N

J

2,0 m

P/2

4 Bays @ 5,0 m each VF

VJ

Figure 2.14 Spacing of lattice girders

=

3,5 m

Spacing of joists

=

2,5 m

Dr. W.M.C. McKenzie

12

Lattice Girders: Example 2.1 - Solution (cont.) Determined combination factors: EC 0

UK National Annex Tables NA.A1 and NA.A.1.2(B)

ψ0,imposed = 0,7

(EC: 0,7)

γG,1 = 1,35

(EC: 1,35)

γQ,1 = 1,5

(EC: 1,5)

EC 0

UK National Annex Clause NA.2.2.3.2

ξG,1 = 0,925 0 925 Dr. W.M.C. McKenzie

(EC: 0,85) 0 85) 13

Lattice Girders: Example 2.1 - Solution (cont.) Ultimate limit state: Fundamental Combination - Clause 6.4.3.2 (i.e. excluding accidental and seismic effects) Use the less favourable of Equations (6.10a) & (6.10b) EC 0

UK National Annex

Equation (6.10a)

∑γ j ≥1

G,jGk,j

+ γ Q,1ψ 0,1Qk,1 +

∑γ

Q,iψ 0,iQk,i

i >1

Self-weight of girder Fd = (1,35 × 0,6) = 0,81 kN/m2 Variable action

Fd = (1,5 × 0,7 × 0,75) = 0,79 kN/m2

Total Design load

Fd = (0,81 + 0,79) = 1,6 kN/m2

Dr. W.M.C. McKenzie

14

Lattice Girders: Example 2.1 - Solution (cont.) Equation (6.10b)

∑ξ

G , j γ G,jGk,j

+ γ Q,1Qk,1 +

j ≥1

∑γ

Q,iψ 0,iQk,i

i >1

Self-weight g of ggirder Fd = ((0,925 , × x 1,35 , × 0,6) , ) = 0,75 , kN/m2 Variable action

Fd = (1,5 × 0,75) = 1,13 kN/m2

Total design load

Fd = (0,75 + 1,13) = 1,88 kN/m2 Adopt Equation (6.10b) value: 1,88 kN/m2

Area supported/joist

= (2,5 × 3,5)

Design Load/joist P = (1,88 × 8,75) Total load/girder Dr. W.M.C. McKenzie

= (8 × 16,45) 16 45) = 131,6 131 6 kN

VF = VJ

=

8,75 m2

=

16,45 kN

=

65 8 kN 65,8 15

Lattice Girders: Example 2.1 - Solution (cont.) Distribute the internal point loads to the adjacent nodes using simple statics. 16,45 kN

P/2

P/2

A

F

16,45 kN

16,45 kN

P/2

P/2

B K

16,45 kN

16,45 kN

P/2

16,45 kN

P/2

P/2

C

G

L

16,45 kN

P/2

D H

M

E N

I

8.23 kN

J

2,0 m

8.23 kN

4 Bays @ 5,0 m each

Figure 2.15

16,45 kN

32.9 kN

A

F

32.9 kN

B K

68,5 kN 32.9 kN

C

G

L

16,45 kN

D H

M

I

E N

J

2,00 m

65,8 kN

4 Bays @ 5,0 m each 65,8 kN Dr. W.M.C. McKenzie

Figure 2.16

65,8 kN 16

Lattice Girders: Example 2.1 - Solution (cont.) Determine the Primary Axial Loads in the members – use Method of Sections.

2

A

32 9 kN 32,9

3

B

F

C

2

3

I

Section 1-1

Dr. W.M.C.

32 9 kN 32,9

Section 2-2 2,0 m 65 8 kN 65,8

McKenzie

65,8 kN

16 45 kN 16,45

A

65 8 kN 65,8

J

4 Bays @ 5,0 m each

16 45 kN 16,45

F

E

Figure 2.17

65,8 kN

FFA

16 45 kN 16,45

D H

G 1

32 9 kN 32,9

F

2,0 m

1

32 9 kN 32,9

Section 3-3 A

FAG 65 8 kN 65,8

Figure 2.18

B

F

FBC G

2,0 m

16 45 kN 16,45

FGH

50m 5,0 17

Lattice Girders: Example 2.1 - Solution (cont.) Consider the vertical equilibrium at joint F Σ Fy = 0 F FA

Section 1-1

+ 65,8 + FFA = 0

F

FFA = − 65,8 kN

65 8 kN 65,8

Length of diagonal = 5,385 m Sinθ = (2,0/5,385) = 0,371 Cosθ = (5,0/5,385) = 0,929

Figure 2.19

Section 2-2

16,45 kN

2,,0 m

A

F

65,8 kN

Strut

Use the three equations of statics Σ Fx = 0

θ FAG

Σ Fy = 0

ΣM=0

Σ Fy = 0 + 65,8 – 16,45 – FAG Sinθ = 0 FAG = + (49,35/0,371)

= + 133,0 kN

Tie


18

Lattice Girders: Example 2.1 - Solution (cont.) 32,9 kN

A

B

FBC

Section 3-3 F

Figure 2.21

65,8 kN

H

G

2,0 m

16,45 kN

FGH 5,0 m

5,0 m

Σ M about node B = 0 [+ (65,8 × 5,0) – (16,45 × 5,0) – (FGH × 2,0)] = 0 Σ M about node H = 0

FGH = + 123,4 kN

Tie

[+ (65,8 × 10,0) – (16,45 × 10,0) – (32,9 × 5,0) + (FBC × 2,0)] = 0 FBC = − 164,5 , kN Dr. W.M.C. McKenzie

Strut 19

Lattice Girders: Example 2.1 - Solution (cont.) Determine the Secondary Axial Loads in the members – use Joint Resolution. 16,45 kN B 8 23 kN 8,23

F'BC

F'BH

F'BC F''1

F'2

C 8 23 kN 8,23

Forces induced in the members by the secondary bracing

Figure 2.22 Consider the vertical equilibrium at joint B Σ Fy = 0 + 8,23 − F'BH Sinθ = 0 Σ Fx = 0 − F 'BC + F'BH Cosθ = 0

F'BH = + (8,23/0,371) = 22,2 kN F '2 = 22,2 22 2 kN Tension F'BC = (22,2 0,929) = 20,6 kN F '1 = 16,45 kN

Dr. W.M.C. McKenzie

Tension

Compression

Compression 20

Lattice Girders: Example 2.1 - Solution (cont.) Maximum Primary Axial Loads 16,45 kN

A 65 8 kN 65,8

32,9 kN

32,9 kN

164,5 kN

B

32,9 kN

C

16,45 kN

D

E

133,0 kN F

G

123,4 kN

65,8 kN

H

65,8 kN

16,45 kN

B

Dr. W.M.C. McKenzie

J

Figure 2.23

Secondary Axial Loads

Figure g 2.24

I

20,6 kN

22,2 kN

16,45kN

20,6 kN

C

22,2 kN

21

Lattice Girders: Example 2.1 - Solution (cont.) Design Forces Total Design Force = Primary Force + Secondary Force Top Chord: Total design force = − (164,5 + 20,6) = − 185,1 kN

Compression

Bottom Chord: Total design force = + (123,4 + 0)

= + 123,4 kN

Tension

Diagonal: Total design force = + (133,0 + 22,2) = + 155,2 kN

Tension

Upright: U i h Total design force = − (65,8 + 0)

= − 65,8 kN

Compression

Secondary Strut: Total design force = − (16,45)

= − 16,45 kN

Compression

Secondary Tie: Total design force = + (22,2)

= + 22,2 kN

Tension

Dr. W.M.C. McKenzie

22

Lattice Girders: Example 2.1 - Solution (cont.) Design the members in accordance with EC3 (All S 275 steel) Top Chord: Effective buckling lengths lzz = lyy = 2,5 m 2/80 x 60 x 7 unequal angles, long legs back-to-back to both sides of a gusset plate with an 8 mm space. Two or more bolts in line at each end. Bottom otto C Chord: o d: 2/50 x 50 x 6 equal angles back-to-back to both sides of a gusset plate with an 8 mm space. One 14 mm diameter hole deducted each angle. Diagonal: 1/75 x 75 x 8 equal angle connected to a gusset plate. 18 mm diameter holes with at least three bolts in line. Upright: Effecti e buckling Effective b ckling lengths lzz = lyy = 2,0 20m 1/80 x 60 x 7 unequal angle, long leg connected to a gusset plate with two or more bolts in line at each end. Secondary Strut: Effective ff i buckling b kli lengths l h lzz = lyy = 1,0 m 50 x 50 x 6 equal angle. Two or more bolts in line at each end. Secondary y Tie: 50 x 50 x 6 equal angle. One 14 mm diameter hole deducted from each angle. Dr. W.M.C. McKenzie

23

Lattice Girders: Example 2.1 - Solution (cont.) Material properties and partial factors γM Assume all material to be non-alloy y structural steel ggrade EN 10024-2 S 275 EC3-1-1

Table 3.1

t = 6 mm < 40 mm

fy = 275 MPa fu = 430 MPa

(Note: In UK National Annex use EN 10025: Table 7 i.e. t ≤ 16 mm) EC3-1-1

Section Properties

EC3-1-1

γM0 = 1,0 γM1 = 1,0 γM2 = 1,25 1 25

Clause 6.1(1)

SPT

Clause 6.2.2.(2) ( )

Dr. W.M.C. McKenzie

Section Property Tables Anet = Agross – deductions for bolt holes 24

Lattice Girders: Example 2.1 - Solution (cont.) Bottom Chord: 2/50 x 50 x 6 equal angles, legs back-to-back to both sides of a gusset plate with an 8 mm space. M16 non-preloaded bolts in line at each end. (123,4 kN - Tension) Gross area – Ag = 11,4 cm2 SPT EC3-1-1

EC3-1-1

EC3-1-1

EC3-1-1

N u,Rd =

Clause 6.2.3 Equation 6.5 Equation 6.6

N Ed ≤ 1,0 N t,Rd ,

N pl,Rd =

Af y

γ M0

=

1140 × 275 1,0 × 10

3

= 313,5 kN

Equation 6.7 0,9 Anet f u

γ M2

Nt,Rd = 286,1 kN Dr. W.M.C. McKenzie

=

0,9 × [1140 − (2 × 18 × 6)] × 430 3

1,25 × 10 123 4 123,4 = 0,43 ≤ 1,0 286,1

= 286,1 kN

Bottom chord is adequate 25

Lattice Girders: Example 2.1 - Solution (cont.) Diagonals: 1/75 x 75 x 8 equal angle, connected to a gusset plate with M16 non preloaded bolts in line at each end. non-preloaded end (155,2 (155 2 kN - Tension) Gross area: Ag = 11,4 cm2

SPT

EC3-1-1

EC3-1-8

Equation 6.6

EC3-1-1

N pl,Rd =

Af y

γ M00

Equation 6.5

=

1140 × 275 1,0 , × 10

3

N Ed ≤ 1,0 N t,Rd

= 313,5 kN

Table 3.3/Figure g 3.9

Edge distance e2 ≥ 1,2d0 = (1,2 × 16) = 19,2 mm Dr. W.M.C. McKenzie

Assume e2 = 20,0 mm 26

Lattice Girders: Example 2.1 - Solution (cont.)

EC3-1-8

EC3-1-8

N u,Rd

Clause 3 3.10.3/Equation 10 3/Equation 3.13 3 13

Table 3.8

N u,Rd u Rd =

Assume the pitch = 2,5d0

β 3 Anet f u γ M2

β3 = 0,5cm2

β 3 Anet f u 0,5 × ⎡⎣1140 − (18 × 6 )⎤⎦ × 430 = = = 177,5 177 5 kN 3 γ M2 1,25 × 10

Nt,Rd = 177,5 kN

Dr. W.M.C. McKenzie

155,22 155 = 0,87 ≤ 1,0 177,5

Diagonals are adequate

27

Lattice Girders: Example 2.1 - Solution (cont.) Secondary ties: 1/50 x 50 x 6 equal angle, connected to a gusset plate with one M16 non-preloaded non preloaded bolt at each end. end (22,2 (22 2 kN - Tension) Gross area: Ag = 5,69 cm2

SPT

EC3-1-1

EC3-1-8

Equation 6.6

EC3 1 1 EC3-1-1

N pl,Rd =

Af y

γ M0

E Equation ti 6.5 65

=

569 × 275 , × 10 1,0

3

N Ed ≤ 1,0 10 N t,Rd

= 156,5 kN

Table 3.3/Figure 3 3/Figure 3.9 39

Edge g distance e2 ≥ 1,2d , 0 = ((1,2 , × 16)) = 19,2 , mm Dr. W.M.C. McKenzie

Assume e2 = 20,0 mm 28

Lattice Girders: Example 2.1 - Solution (cont.) EC3-1-8

N u,Rd =

Clause 3.10.3/Equation q 3.13

2,0 ( e2 − 0,5d 0 ) tf u

Nt,Rd 53 7 kN t Rd = 53,7

Dr. W.M.C. McKenzie

γ M2

=

N u,Rd =

2,0 ( e2 − 0,5d 0 ) tf u

2,0 ⎡⎣ 20,0 − ( 0,5 × 14 )⎤⎦ × 6 × 430 3

1,25 × 10

22,2 = 0,41 ≤ 1,0 53,7

γ M2

= 53,7 kN

Secondary ties are adequate

29

Lattice Girders: Example 2.1 - Solution (cont.) Upright: 1/80 x 60 x 7 unequal angle with long-leg connected to a gusset plate with two or more bolts in line at each end. (65,8 kN - Compression) Gross area: Ag = 9,38 cm2; i = 1,28 cm

SPT

EC3-1-1

Clause 3.2.6

E = 210000 MPa

Check Section Classification EC3-1-1 EC3 11

EC3-1-1

EC3-1-1

Clause 5.5.2

Table 3.1 31

ε = (235/275)0,5, = 0,92 0 92 mm

Table 5.2

h 80 = = 11,4 11 4 t 7

b + h 80 + 60 = = 10,0 10 0 2t 2×7 Dr. W.M.C. McKenzie

≤ 15ε = (15 × 0,92 0 92 ) = 13,8 13 8

≤ 11,5 11 5ε = (11,5 11 5 × 0,92 0 92 ) = 10,58 10 58

The Section is Class 3 30

Lattice Girders: Example 2.1 - Solution (cont.) Compression Resistance of Cross-section – Nc,Rd EC3-1-1

EC3-1-1

N Ed ≤ 1,0 N c,Rd

Clause 6.2.4/Equation 6.9

Equation 6.10

N c,Rd =

Af y

γ M0

=

938 × 275 1,0 × 10

3

= 257,9 kN

Resistance to Flexural Buckling – Nb,Rd EC3-1-1

EC3-1-1

Clause 6.3.1.1/Equation 6.46

Equation 6.47

Dr. W.M.C. McKenzie

N b,Rd =

N Ed ≤ 1,0 N b,Rd

χ Af y γ M1 31

Lattice Girders: Example 2.1 - Solution (cont.) Buckling Curves

χ=

1 2

Φ+ Φ −λ

EC3-1-1

2


but χ ≤ 1,0

(

where λv is as defined in Clause 6.3.1.2/3

Clause 6.3.1.2/3

λv =

For Class 1,2 and 3 cross-sections λv =

λ1 = π

210000 = 86,81 275

Dr. W.M.C. McKenzie

)

where Φ = 0,5 ⎡1+α λ − 0,2 + λ 2 ⎤ ⎣ ⎦

Annex BB/Clause BB.1.2/Equation BB.1

λeff = 0,35 + 0,7λv EC3-1-1

EC3-1-1

λv =

Af y N cr

Af y N cr

where N cr =

=

π 2 EI L2cr

Lcr where λ1 = π iλ1

E fy

Lcr 2000 = = 1,8 18 iλ1 12,8 × 86,81 32



λeff = 0,35 + 0,7λv = [0,35 + (0,7 × 1,8)] = 1,61 Select the appropriate pp p buckling g curve from Table 6.2 EC3-1-1

Table 6.2

Use buckling curve b

Determine the imperfection factor from Table 6.1 EC3-1-1

Table 6.1

(

)

α = 0,34

Φ = 0,5 ⎡1+α λ − 0,2 + λ 2 ⎤ = 0,5 ⎡1 + 0,34 (1,61 − 0,2 ) + 1,612 ⎤ ⎣ ⎦ ⎣ ⎦ 1 1 = χ= = 0,3 03 2 2 2 2 2,04 + 2,04 − 1,61 Φ+ Φ −λ Dr. W.M.C. McKenzie

= 2,04

33



EC3-1-1

Equation 6.47

N b,Rd =

χ Af y γ M1

=

0,3 × 938 × 275 3

1 0 × 10 1,0

N Ed 65,8 = = 0,85 ≤ 1,0 N b,Rd 77 77,39 39

N Ed ≤ 1,0 N b,Rd ,

= 77,39 kN

Uprights are adequate

Note: The value for χ could have been determined from Figure 6.4 using the value , as indicated in Clause 6.3.1.2(3). ( ) of the non-dimensional slenderness = 1,61 Dr. W.M.C. McKenzie

34

Lattice Girders: Example 2.1 - Solution (cont.) Secondary strut: 1/50 x 50 x 6 equal angle, connected to a gusset plate with two or more bolts in line at each end. (16,45 kN - Compression) Gross area: Ag = 5,69 cm2; i = 0,968 cm

SPT

EC3-1-1

Clause 3.2.6

E = 210,000 MPa

Check Section Classification EC3-1-1 EC3 11

Table 3.1 31

EC3 1 1 EC3-1-1

Table 5.2

b + h 50 + 50 = = 8,3 83 2t 2×6 Dr. W.M.C. McKenzie

ε = (235/275)0,5, = 0,92 0 92 mm h 50 = = 8,3 83 t 6

≤ 15ε = (15 × 0,92 0 92 ) = 13,8 13 8

≤ 11,5 11 5ε = (11,5 11 5 × 0,92 0 92 ) = 10,58 10 58

The Section is Class 3 35


EC3-1-1

N Ed ≤ 1,0 N c,Rd


Equation 6.10

N c,Rd =

Af y

γ M0

=

569 × 275 1,0 × 10

3

= 156,5 kN


EC3-1-1


Equation 6.47

Dr. W.M.C. McKenzie

N b,Rd =

N Ed ≤ 1,0 N b,Rd

χ Af y γ M1 36


χ=

1 2

Φ+ Φ −λ

EC3-1-1

but χ ≤ 1,0

(

)

where Φ = 0,5 ⎡1+α λ − 0,2 + λ 2 ⎤ ⎣ ⎦

where λv is as defined in Clause 6.3.1.2

Clause 6.3.1.2

λv =

For Class 1,2 and 3 cross-sections

λ1 = π



λeff = 0,35 + 0,7λv EC3-1-1

2

EC3-1-1

210000 = 86,81 86 81 275

Dr. W.M.C. McKenzie

Af y N cr

λv =

λv =

where N cr =

Af y N cr

=

π 2 EI L2cr

Lcr where λ1 = π iλ1

E fy

Lcr 1000 = = 1,2 12 iλ1 9,68 × 86,81 37



λeff = 0,35 + 0,7λv = [0,35 + (0,7 × 1,2)] = 1,19 Select the appropriate buckling curve from Table 6.2 62 EC3-1-1

Table 6.2

Use buckling curve b


Table 6.1

(

)

α = 0,34

Φ = 0,5 ⎡1+α λ − 0,2 + λ 2 ⎤ = 0,5 ⎡1 + 0,34 (1,19 − 0,2 ) + 1,192 ⎤ ⎣ ⎦ ⎣ ⎦ 1 1 χ= = = 0,48 0 48 2 2 2 2 Φ+ Φ −λ 1,38 + 1,38 − 1,19 Dr. W.M.C. McKenzie

= 1,38

38



EC3-1-1

Equation 6.47

N b,Rd =

χ Af y γ M1

=

0,48 × 569 × 275 3

1 0 × 10 1,0

N Ed 22,2 = = 0,3 ≤ 1,0 N b,Rd 75,1 75 1 b Rd

N Ed ≤ 1,0 N b,Rd ,

= 75,1 kN

Secondary y struts are adequate q

Note: As in the case of the uprights, the value for χ could have been determined from Figure 6.4 using the value of the non-dimensional slenderness = 1,61 as indicated in Clause 6.3.1.2(3). Dr. W.M.C. McKenzie

39

Lattice Girders: Example 2.1 - Solution (cont.) Top chord: 2/80 x 60 x 7 unequal angle with long-leg connected to a gusset plate with two or more bolts in line at each end. (185,1 kN - Compression) z Packing plates 8 mm thick

y

y

Figure 2.25 z Section properties: SPT

Ag = 18,8

iyy,two angles = 2,51 cm ivv, one angle = 1,28 1 28 cm Dr. W.M.C.

z

u

cm2;

izz,two angles = 2,59 2 59 cm

McKenzie

v

y

y u

z

v 40

Lattice Girders: Example 2.1 - Solution (cont.) Check Section Classification

EC3-1-1

Clause 3.2.6

E = 210,000 MPa

EC3-1-1

Table 3.1

ε = (235/275)0,5 = 0,92 mm

EC3-1-1

Table 5.2 52

h 80 = = 11,4 11 4 t 7

b + h 80 + 60 = = 10,0 10 0 2t 2×7

≤ 15ε = (15 × 0,92 0 92 ) = 13,8 13 8

≤ 11,5 11 5ε = (11,5 11 5 × 0,92 0 92 ) = 10,58 10 58

The Section is Class 3 Dr. W.M.C. McKenzie

41


EC3-1-1

N Ed ≤ 1,0 N c,Rd


Equation 6.10

N c,Rd =

Af y

γ M0

=

1880 × 275 1 0 × 10 1,0

3

= 517,0 kN


EC3-1-1


Equation q 6.47

Dr. W.M.C. McKenzie

N b,Rd =

N Ed ≤ 1,0 N b,Rd

χ Af y γ M1 42

Lattice Girders: Example 2.1 - Solution (cont.) Closely Spaced Built-up Members EC3-1-1

Clause 6.4.4/Figure 6.12/Table 6.9

M i Maximum spacing i between b t interconnections i t ti is i given i by b in i Table T bl 6.9 69

EC3-1-1

Table 6.9

Maximum spacing = 15imin,single angle

15imin = (15 ×12,8) = 192,0 mm ? mm ? mm ? mm ? mm

Figure 2.26

? mm ? mm

Dr. W.M.C. McKenzie

43


χ=

1 2

Φ+ Φ −λ

EC3-1-1

2

EC3-1-1

but χ ≤ 1,0


(

)

where Φ = 0,5 ⎡1+α λ − 0,2 + λ 2 ⎤ ⎣ ⎦

Annex BB/Clause BB.1.1

Out-of-plane Out of plane buckling length Lcr is normally taken as the system length. length In this case assume restraint is provided by joists, i.e. Lcr = 2.5 m EC3-1-1

Clause 6.3.1.2

λv =

For Class 1,2 1 2 and 3 cross cross-sections sections

λv =

210000 λ1 = π = 86,81 86 81 275

λv =

Dr. W.M.C. McKenzie

Af y N cr Af y N cr

where N cr = =

π 2 EI L2cr

Lcr where h λ1 = π iλ1

E fy

Lcr 2500 = = 1,15 1 15 iλ1 25,1 × 86,81 44

Lattice Girders: Example 2.1 - Solution (cont.) Select the appropriate buckling curve from Table 6.2 EC3-1-1

Use buckling curve c for a T-section

Table 6.2


α = 0,49

Table 6.1

(

)

Φ = 0,5 ⎡1+α λ − 0,2 + λ 2 ⎤ = 0,5 ⎡1 + 0,49 (1,15 − 0,2 ) + 1,152 ⎤ ⎣ ⎦ ⎣ ⎦

χ=

1 2

Φ+ Φ −λ

Dr. W.M.C. McKenzie

2

=

1 2

2

1 39 + 11,39 1,39 39 − 11,15 15

= 1,39

= 0,46

45



EC3-1-1

Equation 6.47 6 47

N b,Rd b Rd =

χ Af y γ M1

=

0,46 × 1880 × 275 3

1,0 × 10

N Ed 185,1 = = 0,78 0 78 ≤ 1,0 10 N b,Rd 237,8

N Ed ≤ 1,0 N b,Rd b Rd

= 237,8 , kN

Top Chord is adequate

Note: The value for χ could have been determined from Figure 6.4 using the value of , as indicated in Clause 6.3.1.2(3). ( ) the non-dimensional slenderness = 1,15 Dr. W.M.C. McKenzie

46

THE END FIN DAS ENDE Tο Tέλοs KONIEC

Dr. W.M.C. McKenzie

完

SAMAPT

47

2- Ctr 11119 - Notes - Structural Steelwork - Lattice Girder Design I - EC3

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