2- ctr 11119 - notes - structural steelwork - lattice girder design i - ec3

8/12/2019 2- Ctr 11119 - Notes - Structural Steelwork - Lattice Girder Design I - EC3

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. .

Lecture 2

Lattice Girder Design I

Dr. W.M.C. McKenzie 1


2/47

Lattice Girders

Lattice girders are plane frames with an open-web construction usually having

parallel chords and internal web bracing members.

-

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.

prevent bending effects in the chord members.

2Dr. W.M.C. McKenzie


3/47

Typical Lattice Girder with Fire-proofing


Figure 2.1


4/47

Principal Types and Components of a Lattice Girder

compression chordPoint load between the nodes

N-Type Girder

tension chorddiagonal bracing

Figure 2.2

Uniformly Distributed Load between the nodes

compression chord

diagonal bracing


tens on c or

Figure 2.3


5/47

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

,

design for additional secondary axial forces,

or

design for additional secondary bending forces.



6/47

Lattice Girders: Analysis

Consider the two-bay lattice girder supporting a series of purlins as shown.

For analysis purposes the lattice girder is considered to be a pin-jointed, plane-frame

Figure 2.4


with a series of point loads.


7/47

Lattice Girders: Analysis (cont.)

A B C

P/2 P P P P/2Secondary Bracing

a combined effect caused by therame act on n uc ng ot pr mary

and secondary axial loadsD E F

P/2 P P P P/2 Secondary Bending

.

em ers an are su ec e o

a combined effect caused by the

primary axial loads and a secondary

bending moment due to the mid-spanD E F


point loads.Figure 2.6


8/47

Lattice Girders: Analysis: Secondary bracing

P/2 P P P P/2

P/2 P/2 P/2 P/2

A B C

D E FFigure 2.7

P 2P P

1st Stage: Determine the Primary axial loads

FAB FBC

FAEFBE

FECFCFFAD

Fi ure 2.8


FDE

FEF


9/47

Lattice Girders: Analysis: Secondary bracing (cont.)

A B C

P/2 P P P P/22nd Stage:

Secondar bracin to

axial loads

transfer the mid-span point

loads to the nodesD E F

Consider panel length ABFigure 2.9

A BA BForces induced in the

members by the

F'AB F'AB

secondary bracing

F' , F' , F' etc.

F'AE F'2F'1


Figure 2.10


10/47

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 CFAB +F'AB FBC +F'BC

F'1 F'1F' F'2

FAD FCF

FAE +F'AE FCE +F'CEFBE

D E FDE EF

Figure 2.11



11/47

Lattice Girders: Secondary bracing Example 2.1

The lattice girder shown supports a series of joists as indicated. Using the data

given design suitable section for:

e op an o om c or s

(ii) the main diagonals and uprights,

m

.

2,0

,

Desi n Data: Figure 2.13

Characteristic permanent load (including self-weight)gk = 0,6 kN/m2

Characteristic variable load qk = 0,75 kN/m2


Spacing of lattice girders = 3,5 m


12/47

Lattice Girders: Example 2.1 - Solution

The lattice girder has secondary bracing and the top chord members will be

sub ect to both rimar and secondar axial loads.

P/2 P P P P P P P P/2

K L M N 2,0m

F G I JH

4 Bays @ 5,0 m each

F J

Figure 2.14

Spacing of lattice girders = 3,5 m

Spacing of joists = 2,5 m



13/47

Lattice Girders: Example 2.1 - Solution (cont.)

Determined combination factors:

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

EC 0

0,imposed = 0,7 (EC: 0,7)

G,1 = 1,35 (EC: 1,35)

Q,1 = 1,5 (EC: 1,5)

UK National Annex Clause NA.2.2.3.2EC 0


G,1 = , ,


14/47


Ultimate limit state: Fundamental Combination - Clause 6.4.3.2

i.e. excludin accidental and seismic effects

Use the less favourable of Equations (6.10a) & (6.10b)

UK National AnnexEC 0

G,j k,j Q,1 0,1 k,1 Q,i 0,i k,i1 1j i

G Q Q >

+ +Equation (6.10a)

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

Variable action d = (1,5 0,7 0,75) = 0,79 kN/m2

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



15/47


Equation (6.10b) , G,j k,j Q,1 k,1 Q,i 0,i k,iG j G Q Q + + 1 1j i >

Self-wei ht of irderF = 0 925 x 1 35 0 6 = 0 75 kN/m2Variable action Fd = (1,5 0,75) = 1,13 kN/m2

2

Total design load Fd

= (0,75 + 1,13) = 1,88 kN/m2

Area supported/joist = (2,5 3,5) = 8,75 m2Design Load/joist P = (1,88 8,75) = 16,45 kN

= = = =15Dr. W.M.C. McKenzie

, , F J ,


16/47


Distribute the internal point loads to the adjacent nodes using simple statics.

8.23 kN 16,45 kN 16,45 kN 16,45 kN 16,45 kN 16,45 kN 16,45 kN 16,45 kN 8.23 kN

m

A B C D E

P/2 P/2P/2 P/2P/2 P/2P/2 P/2

F G I JK L M N 2

,0

4 Bays @ 5,0 m each

H

65,8 kN 68,5 kN

16,45 kN 32.9 kN 32.9 kN 32.9 kN 16,45 kN

Figure 2.15

m

A B C D E

K L M N 2,

4 Bays @ 5,0 m each

F G I JH


65,8 kN 65,8 kNFigure 2.16


17/47


Determine thePrimary Axial Loads in the members use Method of Sections.

, , , , ,

A B C D E2 3

1

2,0

4 Ba s 5 0 m each

F G I JH

2 31

65,8 kN 65,8 kNFigure 2.17

, ,

A BFBC

Section 1-1 Section 3-3

,

A

Section 2-2

F

2,0m

F GFGH

2,0m

FFAG

F


, m,,,

Figure 2.18


18/47


Consider the vertical equilibrium at joint F

F Fy = 0

Section 1-1F

+ 65,8 +FFA = 0

FFA = 65,8 kN Strut

Length of diagonal = 5,385 m

Sin = (2,0/5,385) = 0,37165,8 kN

Figure 2.19

Use the three equations of statics16,45 kN

Cos = (5,0/5,385) = 0,929Section 2-2

Fx = 0 Fy = 0 M= 0,0

m

A Fy = 02 F

65,8 kN

+ 65,8 16,45 FAG Sin = 0FAG = + (49,35/0,371) = + 133,0 kN Tie

18Dr. W.M.C. McKenzieFigure 2.20


19/47


16,45 kN 32,9 kN

Section 3-3

2,

0m

F G

BC

H

5,0 m65,8 kN

FGH5,0 mFigure 2.21

about node B =[+ (65,8 5,0) (16,45 5,0) (FGH 2,0)] = 0

FGH = + 123,4 kN Tie

Mabout node H

= 0

[+ (65,8 10,0) (16,45 10,0) (32,9 5,0) + (FBC 2,0)] = 0F = 164 5 kN Strut



20/47


Determine the Secondary Axial Loads in the members use Joint Resolution.

16,45 kN

Forces induced in

the members b the

B CF'BC F'BC

secondary bracing, ,

F'BH F'2F'1

Figure 2.22

F = 0Consider the vertical equilibrium at joint B

+ 8,23 F'BH Sin = 0 F'BH = + (8,23/0,371) = 22,2 kN Tension'Fx = 0

F 'BC +F'BH Cos = 0 F'BC = (22,2 0,929) = 20,6 kN Compression 2 ,

20Dr. W.M.C. McKenzieF '1 = 16,45 kN Compression


21/47


16,45 kN 32,9 kN 32,9 kN 32,9 kN 16,45 kN

Maximum Primary Axial Loads

A B C D E164,5 kN

F G H I J123,4 kN

,133,0 kN

65,8 kN 65,8 kN

Secondar Axial Loads

Figure 2.23

B C

16,45 kN

20,6 kN 20,6 kN

22,2 kN 22,2 kN16,45kN

Fi ure 2.24



22/47


Design Forces

Top Chord:

Total desi n force = 164 5 + 20 6 = 185 1 kN Com ressionBottom Chord:

Total design force = + (123,4 + 0) = + 123,4 kN Tension

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

Upright:

Total design force = (65,8 + 0) = 65,8 kN Compression

Total design force = (16,45) = 16,45 kN Compression

Secondary Tie:


Total design force = + (22,2) = + 22,2 kN Tension


23/47


24/47


Material properties and partial factors M

Assume all material to be non-alloy structural steel grade EN 10024-2 S 275

Table 3.1EC3-1-1 t= 6 mm < 40 mm

fy = 275 MPa

=u

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

Clause 6.1(1)EC3-1-1

M0

= 1,0

M1 = 1,0

=

Section Properties Section Property TablesS P T

M2 ,

Clause 6.2.2.(2)EC3-1-1 Anet =Agross deductions for bolt holes



25/47


Bottom Chord: 2/50 x 50 x 6 equal angles, legs back-to-back to both sides ofa gusset plate with an 8 mm space. M16 non-preloaded bolts in line at each

-S P T Gross area Ag = 11,4 cm

2

Clause 6.2.3EC3-1-1

Equation 6.5EC3-1-1Ed

t Rd

1,0N

N

Equation 6.6EC3-1-1y

pl,Rd 3M0

1140 275 313,5 kN1,0 10

AfN

= = =

Equation 6.7EC3-1-1

net uu,Rd 3

M2

0,9 0,9 [1140 (2 18 6)] 430286,1 kN

1,25 10

A fN

= = =


Nt,Rd = 286,1 kN Bottom chord is adequate,

0,43 1,0286,1

=


26/47


Diagonals: 1/75 x 75 x 8 equal angle, connected to a gusset plate with M16

- . , -

Gross area:

Ag = 11,4 cm2 Equation 6.5EC3-1-1

t,Rd

1,0N


pl,Rd 3M0

1140 275313,5 kN

1,0 10

AfN

= = =

Table 3.3/Figure 3.9EC3-1-8

Edge distance e2 1,2d0 = (1,2 16) = 19,2 mm Assumee = 20,0 mm



27/47


EC3-1-8 3 net uf

N

=. . . ,M2

Table 3.8EC3-1-8 Assume the pitch= 2,5d0 3 = 0,5cm2

( )0,5 1140 18 6 430 u,Rd 3

M2

,1,25 10

= = =

Nt,Rd = 177,5 kN Diagonals are adequate,

0,87 1,0177,5

=



28/47


Secondary ties: 1/50 x 50 x 6 equal angle, connected to a gusset plate with

- . , -

Gross area: Ng = 5,69 cm

2 qua on .- -t,Rd

,N


pl,Rd 3

569 275156,5 kN

1 0 10

AfN

= = =

- - . .

Ed e distance e 1 2d = 1 2 16 = 19 2 mm Assumee = 20,0 mm



29/47


30/47


Upright: 1/80 x 60 x 7 unequal angle with long-leg connected to a gusset platewith two or more bolts in line at each end. (65,8 kN - Compression)

S P T Gross area:Ag = 9,38 cm2; i = 1,28 cm

Clause 3.2.6EC3-1-1 E= 210000 MPa

Check Section Classification

EC3-1-1 = 0 5 =

Clause 5.5.2EC3-1-1

. ,

80hTable 5.2EC3-1-1 11,4

7t= = , , = =

80 60b h+ +

30Dr. W.M.C. McKenzieThe Section is Class 3

,2 2 7t

= =

, , , ,= =


31/47


Compression Resistance of Cross-section Nc,Rd

Clause 6.2.4/Equation 6.9EC3-1-1Ed

c,Rd

1,0N

yc,Rd 3

M0

938 275257,9 kN

1,0 10

AfN

= = =

Equation 6.10EC3-1-1

Resistance to Flexural Buckling Nb,Rd

Clause 6.3.1.1/Equation 6.46EC3-1-1Ed

b,Rd

1,0N

N


b,Rd

fN

=



32/47


Buckling Curves Clause 6.3.1.2/Equation 6.49EC3-1-1

( ) 22 2

1 but 1,0 where = 0,5 1+ 0,2

= +

+

Annex BB/Clause BB.1.2/Equation BB.1EC3-1-1

eff v v0,35 0,7 where is as defined in Clause 6.3.1.2/3 = +

2

Clause 6.3.1.2/3EC3-1-1y

v cr 2cr cr

where =NN L

=

y crv 1

cr 1 y

whereN i f

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

210000= = 2000L


,275 v

1

,12,8 86,81i

= = =


33/47


Annex BB/Clause BB.1.2/Equation BB.1EC3-1-1

[ ]eff v0,35 0,7 = 0,35 + (0,7 1,8) 1,61 = + =

Select the appropriate buckling curve from Table 6.2

Table 6.2EC3-1-1 Use buckling curve b

Determine the imperfection factor from Table 6.1

Table 6.1EC3-1-1 = 0,34

( ) ( )2 2= 0,5 1+ 0,2 0,5 1 0,34 1,61 0,2 1,61 + = + + = 2,041 1

==


2 2 2 2

2,04 2,04 1,61 + + = ,


34/47



b,Rd

1,0N

N


yb,Rd 3

0,3 938 275AfN

= = = 77,39 kN

,

Ed 65,8 0,85 1,0N

= = Uprights are adequateb,Rd ,

Note:The value for could have been determined from Figure 6.4 using the valueof the non-dimensional slenderness = 1,61 as indicated in Clause 6.3.1.2 3 .



35/47


Secondary strut: 1/50 x 50 x 6 equal angle, connected to a gusset plate withtwo or more bolts in line at each end. (16,45 kN - Compression)

S P T Gross area:Ag = 5,69 cm2; i = 0,968 cm

Clause 3.2.6EC3-1-1 E= 210,000 MPa


EC3-1-1 = 0,5 =. ,

50hTable 5.2- - ,6t

= = , , = =

50 50b h+ +

35Dr. W.M.C. McKenzieThe Section is Class 3

,2 2 6t

= =

, , , ,


36/47




c,Rd

1,0N

yc,Rd 3

M0

569 275156,5 kN

1,0 10

fN

= = =




b,Rd

1,0N

N


b,Rd

fN

=


M1


37/47


38/47

L tti Gi d E l 2 1 S l ti ( t )


39/47



b,Rd

1,0N

N


yb,Rd 3

0,48 569 275AfN

= = = 75,1 kN

,

Ed 22,2 0,3 1,0N

= = Secondary struts are adequate, ,

Note:

As in the case of the uprights, the value for could have beendetermined from Figure 6.4 using the value of the non-dimensional


slenderness = 1,61 as indicated in Clause 6.3.1.2(3).



40/47


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)

Packing plates 8 mm thick

z

zFigure 2.25

S P T

Section properties:

Ag = 18,8 cm2;

zvu

izz,two angles = , cm

iyy,two angles = 2,51 cmyy


vv, one angle ,

zvu



41/47



Clause 3.2.6EC3-1-1 E= 210,000 MPa

Table 3.1EC3-1-1 = (235/275)0,5 = 0,92 mm

EC3-1-180h

= = = =. ,7t

, ,

80 60b h+ +

The Section is Class 3

,2 2 7t

= =

, , , ,= =


Lattice Girders: Example 2 1 Solution (cont )


42/47




c,Rd

1,0N

N

yc,Rd 3

1880 275517,0 kN

AfN

= = =Equation 6.10EC3-1-1

,


Clause 6.3.1.1/Equation 6.46EC3-1-1Ed 1,0

N

N

,

E uation 6.47EC3-1-1y

b Rd

fN

=


M1



43/47


Closely Spaced Built-up Members

Clause 6.4.4/Figure 6.12/Table 6.9EC3-1-1

ax mum spac ng e ween n erconnec ons s g ven y n a e .

Table 6.9EC3-1-1

Maximum spacing = 15imin,single angle

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

? mm

? mm

? mmFigure 2.26


? mm



44/47


Buckling Curves Clause 6.3.1.2/Equation 6.46EC3-1-1

( ) 22 2

but 1,0 where = 0,5 1+ 0,2

= +

+

Annex BB/Clause BB.1.1EC3-1-1

- -

2

cr .

this case assume restraint is provided by joists, i.e.Lcr= 2.5 m

Clause 6.3.1.2EC3-1-1y

v cr 2cr cr

where =NN L

=

Af Lv 1cr 1 y

w ereN i f

= = = , -

210000 cr 2500L= = =


1 ,275

v1

,25,1 86,81i



45/47


Select the appropriate buckling curve from Table 6.2

Table 6.2EC3-1-1 Use buckling curve c for a T-section

Determine the imperfection factor from Table 6.1

Table 6.1EC3-1-1 = 0,49

( ) ( )2 2= 0,5 1+ 0,2 0,5 1 0,49 1,15 0,2 1,15 + = + + = 1,39

2 2 2 2

1 1 ==

= 0,46


, , ,

Lattice Girders: Example 2 1 - Solution (cont )


46/47


Clause 6.3.1.1/Equation 6.46EC3-1-1Ed 1,0

N

N

,

EC3-1-1 .

y 0,46 1880 275AfN

= = = 237 8 kN,

M1 1,0 10

Ed 185,1N

b,Rd

, ,237,8N

= =

Note:

The value for could have been determined from Figure 6.4 using the value ofthe non-dimensional slenderness = 1 15 as indicated in Clause 6.3.1.2 3 .



47/47

Dr. W.M.C. McKenzie 47

2- ctr 11119 - notes - structural steelwork - lattice girder design i - ec3

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