analysis of portal frame building
TRANSCRIPT
Analysis of portal frame building
In accordance to EN 1993-1-1(2005)
1 description
A/ The portal frame is the main structural element of the
building.
The frame is designed for the following loads
Wind loads can be positive as on AB or negative (suction)as on BC,CD and DE. Roof loads are
positive and up to down direction
B/ If The joints at B,C and D are not rigid,they will open up and the frame will be unstable
Roof loads such as workmen, snow or hail
Wind loads
C/ 1) Vertical loading on the frame results in A and E tending to be pushed outwards.if the foundation
cannot resist this horizontal push,outward movement will occur,and the frame will loose structural
strength
2) Wind subjects the portal frame to uplift forces(the roof tends to fly-off)like an plane wing,to
overturning forces on the sides and ends of the building,
These destabilizing forces are resisted essentially by the weight of the building,and in this regard,the
foundations contribute significantly to this weight. Generally speaking it is a fact that portal frame
buildings of this kind are light weight structures, and as such they tend to collapse “sideward” and
“upwards” rather than downwards”. The effect of wind on a light building cannot be overemphasized.
The destabilization it causes is a major design consideration, and in this context, foundations can be
regarded as the building’s “anchors
D/ the rafter of the portal frame is a slender structural element,and it is restrained it will buckled when
loaded.
In a braced roof this restraint is provided by the purlins acting together with a braced bay.The purlins
provide the restraining force for the rafters,and the braced bay acts as a “buttress” wich absorbs these
purlin restraining forces.
While this system is effective in restraining the top flange of the rafter I -beam,the bottom flange
remains relatively unrestrained, and to achieve the requisite restraint,short lengths of angle iron are
connected at intervals between the bottom flange of the I-beam and the purlins.This simple and
necessary anti-buckling feature is sometimes neglected in the design of the portal frames.
E/
A building frame subjected to wind forces along its length will tend to collapse as shown above ,while a
building with a braced side bay as shown below will be stable,since the braced bay will functions as a
“buttress” to resist the wind forces, and transform them to the foundations
2 portal frame design
2.1 Basic data
Total length b= 70 m
Bay width d= 25 m
Spacing s= 7 m
Height h= 7.5 m
Roof slope α= 5°
Purlin spacing
sp=1.5 m
Cladding rail
spacing Sp’=2.0m
E=210000N/mm2 G=80770N/mm2 Steel:S235
Articulated purlin purlin
cladding rail
column Internal portal frame
door 4*5m
α= 5°
7.5m
6.41m
25.0m
Internal portal frame
2.2 Loads
2.2.1 Permanent loads
Self-weight of the beam
Roofing with purlins G= 0.35 KN/m2
For an internal frame G=0.35*7=2.45 KN/m
G=-2.45 KN/m
α=5°
6.406 m 7.5 m
25.0 m
2.2.2 Construction loads
Q=0.5 KN/m2 EN 1991-1-6 clause 4.11(2005)
For an internal portal frame Q=0.5*7=3.5 KN/m
2.2.3 Wind loads
Take from the document treated “ wind actions to EN 1991-1-4(2005) as a values described below
2.2.4 Approximation calculations
1/ wind forces applied to duopitch roofs and partial variables live loads
These actions are very small in ccomparison with the wind actions on vertical walls(0.5% to 1.3%). In
this case they will be neglected for calculations.
1.5
w1
-6.55 (G)
-8.88(F)
92.0 (F1 etF2)
11 11 1.5
-2.34 (E)
-3.74 (I)
-3.74 (H)
+2.34(D)
-3.74 (J)
2/ wind forces (up-to fly)
The actions applied to duopitch roofs are oriented as described above (perpendicular to rafters).For
simplifications we admit that these forces will be oriented vertically as gravity forces.
3/ Forces transmitted by purlins
The forces transmitted to rafters by purlins, (are ponctual forces and must be applied in calculations
of rafters),will be converted to linear forces.The error caused by this simplification is ≈0.5%,and
conduct to increase the moments at B and D
4/ Stiffness at B and D
To conduct manually calculations we consider that the inertia of the column and the rafter are equals
Ic=IR
The coefficient of stiffness R
C
I hk
S I will be
hk
s
This simplification ,justified by the presence of the haunchs ,conduct to increase the moment at C and
decrease moments at B and D.It will be compensated by the simplification applied to purlin
calculations,wich act in opposite sens.
2.3 Simple cases
1/Case 1 Vertical actions
dead (G) variable (Q) loads
Y
C
s IR f=1.09
- + + -
B D
- h=6.41 -
HA A E HE x
VA VE
Rigidity coefficient at B and D
. .
. .
R
C
Istiffness of rafter hK
stiffness of comumn S I
In application of Castigliano théorem and with the structure symetry
0ABCDE
M dMds
EI dH where H is the horizontal force
Displacement 1 in AB column.
In any ordinate point ,y, of the column AB , the moment is .M H y ,then
dMy
dH and
32
1
0 0
1 1. .
3
h h
R R R
Hy Hhy dy Hy dy
EI EI EI
Displasment 2 in BC rafter
The moment expression at abscissa ,x,is:
2 2cos
sin cos2
xM H h x q Vx
sindM
h xdH
and
2 2
2
0
cossin cos sin
2
sx
H h x q Vx h x dx
We have cos2
l
s and sin
f
s then
22 2 2
2
1 5 1. . . . . . .
3 96 12R
f sH h s h f s q l f s hl s
EI
Then the equation 1 2 0 give the result
2
3 2 2
2
5 8
323 3R
C R
ql s f hH
I h h s f f
I I h h
and may be reduced if we use the rigidity
hk
s in place of the real expression
R
C
I hk
s I
We obtain the simplified expression
2
2
5 8
32 3 3
ql f hH
h k f h f
Conclusion
B DM M Hh
2
2
5 8
32 3 3A E
ql f hH H H
h k f h f
2
8C
qlM H h f
2A E
qlV V
2/Case 2 Vertical actions( wind up to fly)
B DM M Hh
2
2
5 8
32 3 3A E
ql f hH H H
h k f h f
2
8C
qlM H h f
2A E
qlV V
Y
q
C
- -
+ B + D
HA A E HE x
VA VE
3/Case 3 Horizontal actions( wind 1 pressure)
2
.2
B E
qhM H h
.D EM H h
2
2
5 6 2
16 3 3E
kh h fqhH
h k f h f
.A EH q h H
2
4C E
qhM H h f
2
2E A
qhV V
l
Y
C
- -
+
+ B + D
q -
HA A E HE x
VA VE
4/ Case 4 Horizontal actions( wind 1 succions)
2
.2
D A
qhM H h
.B AM H h
2
2
5 6 2
16 3 3A
kh h fqhH
h k f h f
.E AH q h H
2
4C A
qhM H h f
2
2E A
qhV V
l
Y
C
+ -
+ B D
+ q
-
HA A E HE x
VA VE
Calculation of the rafter in bending
Dead loads G=2.45KN/m
W1 wind in long span (internal surpressure)
Wc,3
Wc,1 Wc,2
,1 2.34 /cw KN m see fig above
,2 2.34 /cw KN m
,3 3.74 /cw KN m
W2 wind in long span (internal depressure)
Wc,3
Wc,1 Wc,2
Take , 0.3p ic EN 1991-1-4(2005) (7.2.9 (6)note 2)
,1 , , 0.7 0.3 0.668*7 4.68 /c p e p i pw c c q s KN m
,2 0 /cw KN m
We have choose the max value of G zone for wind calculation but not the better
,3 , , 1.2 0.3 0.668*7 4.21 /c p e p i pw c c q s KN m
And For the zones H;I,and J this value is
,3 , , 0.6 0.3 0.668*7 1.4 /c p e p i pw c c q s KN m
W3 wind gear ( with internal surpressure)
Wc,3
Wc,1 W3 Wc,2
We take a middle value of the zones G,H and I as described in wind actions to
EN 1991-1-4(2005)
We take also a middle value of the zones A,B and C then we will have
,1 ,2 0.668*7 4.676 4.68 /c cw w KN m
,3 1.0*7 7 /cw KN m
Calculus actions
It is to determinate:
--the support reactions HA ; HE ;VA and VE
-- the max bending moments MB ;MC and MD
These forces are obtained from the actions mentioned in tables above
Values for calculations: S=12.55; f=1.09 ; h=6.41; k=0.511
2
2
2.45*25 5*1.09 8*6.4116.494
32 6.41 0.511 3 1.09 3*6.41 1.09A EH H KN
2.45*2530.625
2A EV V KN
22.45*25
16.494 6.41 1.09 67.78
CM KNm
16.494*6.41 105.73B DM M KNm
actions case q(KN/m) HA(KN) HE(KN) VA(KN) VE(KN) MB(KNm) MC(KNm) MD(KNm)
G 1 2.45 16.494 16.494 30.625 30.625 105.73 67.7 105.73
Q 1 3.5 23.31 23.31 43.75 43.75 149.42 98.62 149.42
W1 ;Wc,1 3 2.34 11.39 3.61 1.92 1.92 24.933 3.04 23.14
W1; Wc,2 4 2.34 3.61 11.39 1.92 1.92 23.14 3.04 24.933
W1 Wc,3 2 3.74 24.9 24.9 46.75 46.75 159.61 105.44 159.61
Total 39.90 17.12 50.59 42.91 207.68 105.44 111.54
W2 ;Wc,1 3 4.68 22.69 7.21 3.85 3.85 49.93 6.0 46.22
W2 ;Wc,2 4 0 0 0 0 0 0 0 0
W2 ;Wc,3 2 4.21 28.03 28.03 52.63 52.63
179.67 118.68 179.67
Total 50.72 35.24 56.48 48.78
229.6 124.68 133.45
W3 ;Wc,1 4 4.68 22.79 7.21 3.85 3.85 49.93 6.0 46.22
W3 ;Wc,2 4 4.68 7.21 22.79 3.85 3.85 46.22 6.0 49.93
W3 ;Wc,3 2 7 46.61 46.61 87.5 87.5 298.77 197.3 298.77
Total 31.03 31.03 87.5 87.5 295.06 185.3 295.06
3 Load combinations
Partial factor
max 1.35G permanent loads
min 1.0G permanent loads
1.50Q variable loads
When there is more then one variable action acting,requiring the actions to be combined, the
expression is
ULS : , , , ,
1
0.9g j K j Q i K i
j i
G Q
SLS , ,
1
0.9K j K i
j i
G Q
These combinations are obtained from the NADF2 (French,national annex )
the coefficient 1.2 applied for wind will be omitted if we use combinations above
ULS combination
combination Reactions (KN) Bending moments (KNm)
HA HE VA VE MB MC MD
1011.35 1.5G Q 57.23
57.23
106.97
106.97
366.87
239.33
366.87
10211.35 1.5G W
useless
10321.35 1.5G W
53.81
30.59
43.38
31.83
201.66
95.63
57.44
10431.35 1.5G W
24.28
24.28
89.91
89.91
299.85
186.56
299.85
10511.5G W
43.36
9.19
45.26
33.74
205.79
90.46
61.58
10621.5G W
59.59
36.37
54.09
42.55
238.67
119.32
94.45
10731.5G W
30.05
30.05
100.63
100.63
336.86
210.25
336.86
10811.35 1.8G W
49.55
8.55
49.72
35.89
231.09
98.4
58.04
10921.35 1.8G W
69.03
41.17
60.32
46.46
270.54
133.03
97.47
11031.35 1.8G W
33.59
33.59
116.17
116.17
388.37
242.15
388.37
The maximum values are collected in the table
Reactions (KN) Bending moments (KNm)
HA HE VA VE MB MC MD
57.23
57.23
106.97
106.97
388.37
239.33
388.37
69.03
41.17
116.17
116.17
366.87
242.15
366.87
4/ Rafter
4.1/Resistance
The maximum moment in:
- Apex connection : MB= MD=-366.87 KNm
- Eave connection : MC=+239.33 KNm
The expression RdM M must be verified for bending
With 0
.pl y
Rd
M
W fM
.We have 0
.pl y
M
W fM
then
0. Mpl
y
MW
f
For apex connection 366.87
235000plW
eave connection 239.33
235000plW
- In apex connection 31561.1plW cm IPE 360+(1/2) IPE 360
- In eave connection 31018.4plW cm IPE 360
The 1.5 IPE360 section is considered as welded beam . the table below show it’s
characteristics
Caractéristiques du profil P.R.S.
Caractéristiques
géométriques
Caractéristiques
mécaniques
Axe neutre
élastique
Axe neutre
plastique
h = 540 mm
hw = 514,6 mm
tw = 8 mm
bf = 170 mm
tf = 12,7 mm
g = 66,21 kg/m
A = 84,35 cm2
Iy = 39105,7 cm4
Wel.y = 1448,4 cm3
Wpl.y = 1668,1 cm3
iy = 21,53 cm
Iz = 1042,1 cm4
Wel.z = 122,6 cm3
Wpl.z = 191,7 cm3
iz = 3,52 cm
It = 32 cm4
Iw = 722861 cm6
Zane = 270 mm
Yane = 85 mm
Zanp = 270 mm
Yanp = 85 mm
This choice is preliminary and will be completed by others
4.2/ Vertical deflection
Vertical deflection of the rafter
The vertical deflection will be calculated under G Q
The moment in a section is
2
2 2x B
ql qM M x x
By integration of the equation
2
2
d y M
dx EI
We have
2 22
0 0
1
2 2
l l
B
dy M ql qdx M x x dx
dx EI EI
For 2
lx
we have 0
dy
dx
then
322 3
0
1. .
4 6 2 24
l
B B
ql q l qly M x x x M dx
EI
For x=0 we have y=0 then
4 2
max
15 48 .
384By ql M l
EI
E=210000 MPa=210000N/mm2=2.1x108 KN/m2
I=16270 cm4
q= G Q
=2.45+3.5=5.95KN/m
L=25.1m
MB=105.73+149.42=255.15 KNm
For IPE 360 the vertical deflection is
4 2
max 8 8
5*5.95*25.1 48*255.15*25.10.3119 31.2
384*2.1*10 *16270*10y m cm
In this case we must upgrade to IPE 500 and we obtain a limit value but less
because we haven’t consider the presence of apex
4 2
max 8 8
5*5.95*25.1 48*255.15*25.10.1052 10.52
384*2.1*10 *48200*10y m cm
251012.55
200 200adm
lf cm
CARACTERISTIQUES GEOMETRIQUES IPE 500
CARACTERISTIQUES MECANIQUES
h = 500 mm
b = 200 mm
tw = 10,2 mm
tf = 16 mm
r = 21 mm
d = 426 mm
g = 90,70 kg/m
A = 116,00 cm2
Iy = 48 200,00 cm4
Wel.y = 1 928,00 cm3
Wpl.y = 2 194,00 cm3
iy = 20,43 cm
Avz = 59,87 cm2
Iz = 2 142,00 cm4
Wel.z = 214,20 cm3
Wpl.z = 335,90 cm3
iz = 4,31 cm
It = 89,29 cm4
We remark that IPE 500 is very suffisant to resist under positif and negative bending moment
4.3/Classification
The section is class 1 as a similar (but not the same) verification for the column (see§5)
4.4/Buckling resistance
This figure shows different Sections categories and buckling modes
Lateral torsional buckling check using the simplified assessment methods for
beams with restraints in buildings:
8*1.5m
Lateral restraints
(purlins) IPE 500
● 4.19m ● Lateral restraints
(bracing system) 3*4.18m
Bracing system
In buildings , members with discrete lateral restraint to the compression flange are not susceptible to
lateral-torsional buckling if the length Lc between restraints or the resulting equivalent compression
flange slenderness f
satisfies:
,
,0
,, 1
c Rdccf c
y Edf z
Mk Li M
[6.3.2.4]
Where
My,Ed
is the maximum design value of the bending moment within the restraint spacing
kc
is a slenderness correction factor for moment distribution between restraints, see EN 1993-1-1
Table 6.6;
if,z
is the radius of gyration of the compression flange including 1/3 of the compressed part of the web
area, about the minor axis of the section;
,0c is the slenderness parameter of the above compression element:
,0 ,0 0.10c LT
,0 0.4LT then ,0 0.4 0.10 0.5c
1 93.9y
E
f and
2
235
yNf
mm
[6.3.2.3]
3
,
2* *3 12
2
wz
f z
tdI
I
then
3
4
,
42.6 1.022142 2* *
3 121069.74
2f zI cm
,
12* *
2 3f z w
dA A t
then
2
,
1 42.6116 2* *1.02 43.52
2 3f zI cm
,
,
,
1069.744.96
43.52
f z
f z
f z
Ii cm
A
3
, 2194y pl yW W cm
1 93.9 93.9y
E
f
3
,
1
2194*235*10515.59
1.0
y y
c Rd
M
W fM KNm
Combination 1.35 1.5G Q
MB=MEd=366.87 KNm
We consider that the coefficient is the same if the rafter is unrestraint then
239.330.65235
366.87
C
B
M
M
Then 1 1
0.6471.33 0.33 1.33 0.33*0.65235
CK
table 6.6
But between restraints in the centre of the rafter where the moment are maximum,
the moment distribution may be considered as constant :KC=1.0 table 6.6
, 1
1.0*1500.322
4.96*93.9
C Cf
f z
K L
i
The maximum bending moment is at the origin B of the rafter then the lateral torsional buckling may
be also in the origin
,366.87
y EdKNmM
,
,0
,
515.590.5* 0.703
366.59
c Rd
c
y Ed
MM
, 1
1.0*1500.322
4.96*93.9
C Cf
f z
K L
i
0.322 0.703
Combination 31.35 1.8G W
MB=,
388.37y Ed
KNmM
,
,0
,
515.590.5* 0.6637
388.37
c Rd
c
y Ed
MM
, 1
1.0*4180.8975
4.96*93.9
C Cf
f z
K L
i
Not verified
It’s necessary to add other bracing systems each 3m spacing then
LC=3m
,388.37
y EdKNmM
,
,0
,
515.590.5* 0.6637
388.37
c Rd
c
y Ed
MM
, 1
1.0*3000.6441
4.96*93.9
C Cf
f z
K L
i
0.644 0.663
Then the lateral torsional buckling is satisfactory
A detailed procedure to do verification for the rafter is shown below as for column
When the above procedure is not satisfactory.
NOTA
The real comportement of the rafter is shown in the figure
1 tension flange
2 elastic section
3 plastic stable length
4 plastic stable length
5 elastic section
6 plastic hinge
7 restraints
8 bending moment diagram
9 Compression flange
10 plastic stable length
11 plastic stable length
12 elastic section
Annex A
y
4.5/ the haunch verification
C +
O - D
F S
x
the equation of the bending moment curve is a parabolic form 2Y aX
the point F is considered the limit of elastic moment
3
0
1928*10 *235453.8
1.0
el y
el
M
W fM KNm
X 0 m S=12.55
Y MC=242.15 MD+MC=388.37+242.15=630.52
Then 2
630.524
157.5
Ya
X
The bending moment curve equation will be 24Y X
For 12.55X F then 3
0
1928*10 *235453.8
1.0
el y
el
M
W fY M KNm
Then 2
454 4 12.55 F then 2 100.4 176 0F F
Conclusion 1.78F m
Length of the rafter F=2m
The same verification for buckling 1/about yy
2/about zz
3/lateral torsional buckling
as for column in section 5 may be used
5/COLUMN
The verification of the column is carried out for the combination 1011.35 1.5G Q
106.97EdN KN (assumed to be constant along the column)
57.23EdV KN
(assumed to be constant along the column)
366.87EdM KNm (at the top of the column)
5.1/Classification of the section
Web: the web slenderness is 437.6
42.910.2w
c
t §5.5 (tab5.2)
10697044.63
10.2*235
EdN
w y
Nd
t f
426 44.630.552 0.50
2 2*426
w N
w
d d
d
Then the limit for the class is
396 396*164.119
13 1 13*0.552 1
Until 42.9 ≤ 64.119 the web is class 1
Flange: the flange slenderness is
273.92 4.61816
w
f f
b t rc
t t
§5.5 (tab5.2)
The limit of the class is
9 9*1.0 9
Until 4.618 ≤ 9.0 the flange is class 1
So the section is Class 1. The verification of the member will be based on the plastic
resistance of the cross-section. 5.2 /Resistance
Verification for shear force
Shear area
max 2 2 ;V f w f w wA A bt t r t h t §.6.2.6
max 11600 2*200*16 10.2 2*21 *16;1.0*426*10.2VA
max 6035.2;4345.2VA
26035.2VA mm
3
,
0
2356035.23 3
*10 818.841.0
yV
pl Rd
M
fA
V KN
,
57.230.07 0.5
818.84
Ed
pl Rd
V
V § 6.2.8(2)
The effect of the shear force on the moment resistance may be neglected
Verification to axial force
3
,
0
11600*235*10 2726
1.0
y
Pl Rd
M
AfN KN
§6.2.4
106.97EdN KN
,0.25 0.25*2726 681.5Pl RdN KN 6.2.9.1(4)equ 6.33
0
0.5 0.5*426*10.2*235510.56
1.0*1000
w w y
M
h t fKN
6.2.9.1(4)equ 6.34
Since 106.97 ≤ 681.5 and 106.97 ≤ 510.56
The effect of the axial force on the moment resistance may be neglected
Verification to bending moment
,
,
0
2194*235515.59
1.0*1000
Pl Rd y
pl Rd
M
W fM KNm
,
366.870.711 1.0
515.59
Ed
pl Rd
M
M Ok! §6.2.5
5.3 serviability limit state
Horizontal Deflection
Horizontal Deflection at the top of the column must be verified for two combinations
and
Combination 201: G+Q combination 202:G+W1
Combination 201 G+Q
The moment at a point x in the column is .x AM H x
G+Q
M
x
HA
By introducing a virtual force P at the summit of the column AB
This effort generate the following forces
R
C
I h hk
s I s
f
h
23 3 3k
3 211
2
3 21
2 2A
PR
BM Ph
3 211
2
E AR P R
CM Ph
1 3 21
2 2
A E
PhV V
l DM Ph
For an IPE 500 column we obtain :
0.17004 0.511k 4.10786 0.56913 0.43087 0.04458
Then we have the results
0.534AR P 0.466ER P
The moment in the point M is
0.534XM Px
The resultant moment under the two actions is
0.534X AM H x Px
the internal potential energy of the column is:
2
0
10.534
2
h
AW H x Px dxEI
2 2
0
10.534
2
h
AW H P x dxEI
23
0
1 10.534
2 3
h
AW x H PEI
3
20.534
6A
hW H P
EI
31.070
6A
C
dWP h H
dP EI
3
6
1.07*6.41 *3980.41.847
6*2.1*10 *48200cm
6412.137
300 300
lcm
Since 1.847 ≤ 2.137 OK!
Combination 202 G+W
16.494 39.9 23.406AH KN
By application a similar resolution as the above
2
0.5342
X A
xM H x q Px
the internal potential energy of the column is:
22
0
10.534
2 2
h
A
xW H x q Px dx
EI
Using a similar calculation we have
22
0
10.534
2 2
h
A
xW H x q Px dx
EI
23 2 21 1 1
0.134 0.534 0.052 4 3
A AW x qx H P H P q xEI
23 2 21 1 1
0.134 0.534 0.052 4 3
A AW h qh H P H P q hEI
0dW
PdP
4 310.067 0.178 Aqh H h
EI
4 2 3
6
0.067*234*641 *10 0.178*2340.6*6411.345
2.1*10 *48200cm
P B
M
qx2/2 x
q
HA A
RA
5.4 Buckling Resistance
The buckling resistance of the column is sufficient if the following conditions are fulfilled
(no bending about the weak axis, MZ,Ed=0): §6.3.3
,
1 1
1Ed Edyy
y RK LT y RK
M M
N Mk
N M
equation 6.61
,
1 1
1Ed Edzy
y RK LT y RK
M M
N Mk
N M
equation 6.62
Buckling about yy
, 6.41CR yL m
5002.5 1.2
200
h
b 16 40ft mm mm buckling curve :a(αy=0.21) table 6.2
42 2
, 2 2 3
,
210000*48200*1024313.64
6410 *10
y
cr y
cr y
EIN KN
L
3
,
11600*2350.335
24313.64*10
y
y
cr y
Af
N §6.3.1.2
2
20.5 1 0.2 0.5 1 0.21 0.335 0.2 0.335 0.5703yy y
2 22 2
1 10.9691
0.5703 0.5703 0.335y
y y y
Buckling about zz
Buckling curve :b (αz=0.34)
2 2 4
, 2 2
,
*210000*2142*101080.5
6410
zcr z
cr z
EIN KN
L
3
,
11600*2351.5883
1080.5*10
yz
cr z
Af
N
2
0.5 1 0.2z zz z
20.5 1 0.34 1.5883 0.2 1.5883 1.9973z
2 22 2
1 10.3117
1.9973 1.9973 1.5883z
z z z
1 tension flange
2 plastic stable length
3 elastic section
4 plastic hinge
5 restraints
6 bending moment diagram
7 compression flange
8 plastic with tension flange restraint,
9 elastic with tension flange
Column with restraints by cladding rail
along long span
Annex A
Lateral torsional Buckling Annex A
5002.5 2
200
h
b then buckling curve c(αLT=0.49)
Moment diagram with linear variation : 0 then 1 1.77C
The simplification of critical moment may be used:
22,
1 2 2
,
cr LT twzcr
cr LT z z
L GIIEIM C
L I EI
2 4 6 2 4
2 6 4 2 4
210000*2142*10 1249000*10 6410 *80770*89.29 *101.77
6410 *10 2142*10 210000*2142*10crM
2 5 2
2 2
2.1*21420 1249*10 641 *8077*89.291.77 676.32
641 2142 21*21420crM KNm
3,
6
2194*10 *2350.8731
676.32*10
pl y yLT
cr
W f
M
2
,00.5 1 yLT LT LT LT
20.5 1 0.49 0.8731 0.4 0.75*0.8731 0.9663LT
With a values of ,0 0.4LT and 0.75
2 22 2
1 10.6377
0.9663 0.9663 0.75*0.8731LT
LT LT LT
For 0 then
10.7519
1.33 0.33cK
6.3.2.3 table 6.6
Bending moment diagram and the
coefficient
2
1 0.5 1 1 2 0.8LTcf K
2
1 0.5 1 0.7519 1 2 0.8731 0.8 0.8773 1f
,mod
0.63770.7269 1
0.8773
LTLT
f
Calculation of the factor Kyy
,
1
1
y
yy my mLTEd yy
cr y
K C CN C
N
annex A
,0 ,01
1
y LT
my my my
y LT
aC C C
a
annex A
2
,
, ,
1
1 1
LTm LT my
Ed Ed
cr z cr T
aC C
N N
N N
annex A
,
,
1
1
Ed
cr y
yEd
y
cr y
N
N
N
N
annex A
2 ,2 2
max max
,
1.6 1.61 1 2
el y
yy y my my pl LT
y y pl y
WC w C C n b
w w W
Calculation of y
106.971
24313.64 0.9998106.97
1 0.969124313.64
y
,
,
21941.138 1.5
1928
pl y
y
el y
Ww
W
Critical axial force in the torsional buckling mode
2
, 2
0 ,
wcr T t
cr T
EIAN GI
I L
For a doubly symmetrical section
2 2 4
0 0 0 48200 2142 50342y zI I I y z cm
2 5 4 64
, 4 3 2
11600 *2.1*10 *124.9*10 *1080770*89.29*10
50342*10 *10 6410cr TN
, 3113.56cr TN KN
22,
,0 1 2 2
,
cr LT twzcr
cr LT z z
L GIIEIM C
L I EI
,0crM is the critical moment foe the calculation of 0 for uniform bending moment as specified in
annex A . Then we have C1=1
2 4 6 2 4
,0 2 6 4 2 4
210000*2142*10 1249000*10 6410 *80770*89.29*101* 382.1
6410 *10 2142*10 210000*2142*10crM KNm
3,
06
,0
2194*10 *2351.162
382.1*10
pl y y
cr
W f
M
40,lim 1
, ,
0.2 1 1Ed Ed
cr z cr TF
N NC
N N
For doubly symmetrical section , ,cr TF cr TN N
40,lim
106.97 106.970.2 1.77 1 1 0.2569
1080.5 3113.56
Then 0 0,lim
Calculation of myC
,0 ,011
y LT
my my my
y LT
aC C C
a
3,
3
,
366.87*10 1160020.635
106.97 1928*10
y Ed
y
Ed el y
M A
N W
89.291 1 0.928
1249
tLT
w
Ia
I
Calculation of Cmy,0
,0
,
0.79 0.21 0.36 0.33 Edmy y y
cr y
NC
N table A2
With a value 0y then
,0
106.970.79 01188 0.7895
24313.64myC
20.635 *0.928
0.7895 1 0.7895 0.95961 20.635 *0.928
myC
2
,
, ,
1
1 1
LTm LT my
Ed Ed
cr z cr T
aC C
N N
N N
2
,
0.9280.9596 0.9457 1
106.97 106.971 1
1080.5 3113.56
m LTC
Then , 1m LTC
Calculation of Cyy
2 ,2 2
max max
,
1.6 1.61 1 2
el y
yy y my my pl LT
y y pl y
WC w C C n b
w w W
max max ;y z z
, 0 0z Ed LTM b
1
1069700.03924
11600*235
1.0
Edpl
Rk
M
Nn
N
2 2 21.6 1.61 1.138 1 2 *0.9596 *1.5883 *0.9596 *1.5883 *0.03924 0.978
1.138 1.138yyC
,
,
19280.8787
2194
el y
pl y
W
W
Since 0.978 ≥ 0.8787 Ok!
Calculation of Kyy
,
1
1
y
yy my mLTEd yy
cr y
K C CN C
N
0.9998 10.9596*1* 0.9853
106.97 0.9781
24313.64
yyK
Verification with interaction formula
,
1 1
1Ed Edyy
y RK LT y RK
M M
N Mk
N M
6
3
106970 366.87*100.9853* 0.766 1
0.9691*11600*235 2194*10 *2350.9663*
1.0 1.0
OK!
The buckling resistance of the section is satisfactory
This figure illustrate different categories of buckling modes
A similar method of calculation of the factor yzK in the equation 6.62 mentioned above
May be used for the verification of the second formula(not treated for this sheet)
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