HOW TO DESIGN TRANSVERSAL BRACING?

…worked examples for BO003 and BO006

WORKED EXAMPLE 5: TRANSVERSAL BRACING OF TRUSS GIRDER

ROOF STRUCTURE

Make assessment of transversal bracing of truss girder roof structure. Bracing members are made of solid

timber C20. Diagonals of K-shape bracing are designed as solid sections. Service class no 2 is considered.

Axonometry

Spatial structure members:

blue – truss girder + columns

orange – purlins

red – longitudinal vertical bracing

green – transversal bracing (roof

part + wall part)

Building geometry

L = 12 m truss girder span = building width

d = 48 m building length

h = 8 m building high

α = 12° pitch of the roof (slope)

Material

Solid timber C20 according to the EN 338

fc,0,k = 19 MPa characteristic compressive strength along the grain

MPa15,133,1

199,0

M

kc,0,

moddc,0,

fkf design compressive strength along the grain

Load states

LS1 Wind pressure and wind suction on gable walls

2

p kN/m864,0zq (for CZ wind area II, terrain category II and building high 8 m above terrain)

pressure coefficients for external pressure for areas D and E for geometrical ratio 166,0488 dh

cpe,D = +0,7 cpe,E = -0,3

wind pressure on windward wall and wind suction on leeward wall

2

Dpe,pDe, kN/m60,07,0864,0 czqw 2

Epe,pEe, kN/m26,03,0864,0 czqw

actions on transversal roof bracing

kN/m40,22

860,0

2De,Dw,

hwf kN/m04,1

2

826,0

2Ee,Ew,

hwf

LS2 Wind friction on the roof cladding

Friction length

m24234;2min488;4212min48;42min-fr hbdd

Friction area

2

frfr m29412cos

1224

cos

LdA

Friction force

kN08,5864,029402,0pfrfrfr zqAcF

where cfr = 0,02 is friction coefficient for asphalt shingle

Transfer to uniformly distributed load

kN/m41,012cos12

08,5

cos

frfr

L

Ff

LS3 Stabilizing load (forces)

kN/m52,012cos1230

23,480,41

cosf,3

dc,

st

Lk

Nnkf L

where

length coefficient

111,1

1min

12cos12

15

1

min

cos

15

1

min

L

kL

factor of manufacturing quality

30f,3 k (for bad manufacturing quality)

number of roof girders (upper chords) stabilized by one transversal bracing

44

16

bracingltransversaofnumber

girdersroofofnumbern

weighted average of design normal forces in upper chords where lengths of upper chord li is weight

i

ii

l

lNN

dc,

Upper chord

Load state Load combination

LS1 permanent

LS2 snow

LC1=1,35×LS1+1,5×LS2

[kN] [kN]

H1 -5,50 -13,25 -27,30

H2 -8,51 -20,48 -42,20

H3 -10,60 -25,53 -52,60

H4 -10,60 -25,53 -52,60

H5 -10,82 -26,05 -53,67

H6 -10,82 -26,05 -53,67

upper chords notation and lengths

kN23,48124,1124,1124,1124,1818,0818,0

124,167,53124,167,53124,160,52124,160,52818,020,42818,030,27dc,

N

Load combinations

There are special load combination for each transversal bracing (on the safe side)

kN/m12,452,05,140,2stQDw,1 fff

kN/m52,0st2 ff

kN/m52,0st3 ff

kN/m70,252,05,141,05,104,1stQfrQEw,4 ffff

All of the transversal bracing are the same geometry and members cross sections. The most loaded of them

will be analysed with load f

kN/m12,470,2;52,0;52,0;12,4max;;;max 4321 fffff

Internal forces

Uniformly distributed load f is transformed on forces Fi acting in joints. The purlins which are not joint to the

bracing (dashed verticals) are not considered in the calculation of internal forces.

kN37,32

163612,4

2

11

lfF

kN00,82

250,2636,112,4

2

212

llfF

kN27,92

250,2250,212,4

2

323

llfF

kN27,92

250,2250,212,4

2

434

llfF

Reactions

kN74,24

2

07,907,983,730,32

2ba

iF

RR

Normal force in diagonals D1

kN16,1554cos2

30,374,24

cos2 1

1aD1

FRN

Normal force in diagonals D2

kN58,1145cos2

83,730,374,24

cos2 2

21aD2

FFRN

Normal force in diagonals D3

kN86,345cos2

07,983,730,374,24

cos2 3

321aD3

FFFRN

In one diagonal is tension force and in the second there is compression force of the same value. The

compression is more unfavourable situation thus the diagonals will be check on flexural buckling.

Diagonal D1

b = 40 mm section depth

h = 160 mm section high

NEd = -15,16 kN

Buckling to the z-axis (in y direction) is decisive.

Design compression stress along the grain

MPa37,26400

10.16,15 3

Eddc,0,

A

N

Critical length (half of the actual length of diagonal due to joining to the neglected verticals = purlins)

mm1157zcr, L

Moment of inertia

4633

z mm10.853,01604012

1

12

1 hbI

Radius of gyration

mm5,116400

10.853,0 6

zz

A

Ii

Slenderness

1005,11

1157

z

zcr,

z i

L

Relative slenderness

74,16400

19

π

100

π 05,0

kc,0,zzrel,

E

f

Factor for reduction factor calculation

16,274,13,074,12,015,03,015,0 22

zrel,zrel,cz k

Reduction factor

29,074,116,216,2

11222

zrel,

2

zz

zc,

kk

k

Reliability criterion

0,162,015,310,29

2,37

dc,0,zc,

dc,0,

fk

=> condition is satisfied

Diagonal D2

b = 40 mm section depth

h = 160 mm section high

NEd = -11,58 kN

Buckling to the z-axis (in y direction) is decisive.

Design compression stress along the grain

MPa81,16400

10.58,11 3

Eddc,0,

A

N

Critical length (half of the actual length of diagonal due to joining to the neglected verticals = purlins)

mm1360zcr, L

Moment of inertia

4633

z mm10.853,01604012

1

12

1 hbI

Radius of gyration

mm5,116400

10.853,0 6

zz

A

Ii

Slenderness

1185,11

1360

z

zcr,

z i

L

Relative slenderness

05,26400

19

π

118

π 05,0

kc,0,zzrel,

E

f

Factor for reduction factor calculation

78,205,23,005,22,015,03,015,0 22

zrel,zrel,cz k

Reduction factor

21,005,278,278,2

11222

zrel,

2

zz

zc,

kk

k

Reliability criterion

0,166,015,310,21

1,81

dc,0,zc,

dc,0,

fk

=> condition is satisfied

Diagonal D3

Has the same geometry as diagonal D3 bur smaller normal force.

Vertical V1

Vertical V1 is actually roof purlin

b = 100 mm section depth

h = 160 mm section high

Normal force in vertical V1 can be calculate from static condition

of vertical forces

0y F => 0sin V1D1a NNR

kN02,1445sin16,1574,24sinD1aV1 NRN

Design compression stress along the grain

MPa876,016000

10.02,14 3

Eddc,0,

A

N

Because of reliability criterions of member stressed by normal force and biaxial bending reduction factors for

both axis have to be calculated

Critical length

mm3200zcr,ycr, LL

Moment of inertia

4633

y mm10.1,3416,010,012

1

12

1 hbI

4633

z mm10.3,1316,010,012

1

12

1 hbI

Radius of gyration

mm2,4616000

10.1,34 6y

y A

Ii

mm9,2816000

10.3,13 6

zz

A

Ii

Slenderness

3,692,46

3200

y

ycr,

y i

L

1119,28

3200

z

zcr,

z i

L

Relative slenderness

20,16400

19

π

3,69

π 05,0

kc,0,y

yrel, E

f

92,16400

19

π

111

π 05,0

kc,0,zzrel,

E

f

Factor for reduction factor calculation

31,120,13,020,12,015,03,015,0 22

yrel,yrel,cy k

51,292,13,092,12,015,03,015,0 22

zrel,zrel,cz k

Reduction factor

54,020,131,131,1

11

222

yrel,

2

yy

yc,

kk

k

24,092,151,251,2

11

222

zrel,

2

zz

zc,

kk

k

Reliability criterions

0,184,08,13

66,27,0

8,13

13,8

15,1354,0

876,0

dz,m,

zm,

m

dy,m,

ym,

dc,0,yc,

dc,0,

fk

ffk

=> condition is satisfied

0,187,08,13

66,2

8,13

13,87,0

15,1324,0

876,0

dz,m,

zm,

dy,m,

ym,

m

dc,0,zc,

dc,0,

ffk

fk

=> condition is satisfied

it is known from roof purlin assessment

Upper and bottom chord H3 and S3

Transversal bracing chords (both upper and bottom) are actually upper chord of truss girder

b = 50+40+50=140 mm section depth

h = 180 mm section high

Normal force in chords H3 and S3 can be simplified calculate from bending

moment on simple beam

kNm51,7712cos1212,48

1cos

8

1 22 LfM

kN22,242,3

51,77H3S3

h

MNN

Design compression stress along the grain

MPa33,3

23400

10.22,2467,53 3

tot

Eddc,0,

A

N

Reliability criterions

0,126,015,310,986

3,33

dc,0,zc,

dc,0,

fk

=> condition is satisfied

0,142,115,310,178

3,33

dc,0,zc,

dc,0,

fk

=> condition is not satisfied

it is known from truss girder assessment