how to design transversal bracing?...upper and bottom chord h3 and s3 transversal bracing chords...
TRANSCRIPT
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