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
Roof loads such as workmen, snow or hail
Wind 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
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 l oose structural
strength
2) Wind subjects the portal frame to uplift forces(the roof tends to fly-off)like an
overturning forces on the sides and ends of the building,
plane wing,to
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 do wnwards”. 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
E=210000N/mm 2 G=80770N/mm 2 Steel:S235
Height
h= 7.5 m
Roof slope
α= 5°
Purlin
spacing
sp=1.5 m
Cladding
rail
spacing
Sp’=2.0m
Articulated purlin
purlin
cladding rail
column
I nternal portal frame
door 4*5m
α= 5°
7.5m
6.41m
25.0m
Int ernal portal frame
2.2 Loads
2.2.1 Permanent loads
Self-weight of the beam
Roofing with purlins
For an internal frame
G= 0.35 KN/m 2
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/m 2
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
-3.74 (J)
-6.55 (G)
-3.74 (H)
-3.74 (I)
-8.88(F)
92.0 (F1 etF2 )
w1
-2.34 (E)
+2.34(D)
1.5
11
1.5
11
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.
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
k
IR h
will be
S IC
k
h
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
VA
VE
Rigidity coefficient at B and D
K 
I h
stiffness.of .rafter
 R
stiffness.of .comumn
S IC
In application of Castigliano théorem and with the structure symetry

M dM
ds  0
ABCDE 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
dM
 y and
dH
h
1 

0
Hy
1
. y.dy 
EI R
EI R
h
2
 Hy dy 
0
Displasment  2 in BC rafter
The moment expression at abscissa ,x,is:
x 2 cos2 
M  H  h  x sin    q
 Vx cos 
2
dM
 h  x sin 
dH
and


x 2 cos2 
2    H  h  x sin    q
 Vx cos    h  x sin   dx
2

0
s
HE
1 Hh3
EI R 3
x
cos 
We have
1
EI R
2 
l
and
2s
f
then
s
sin  



f 2s
1
 5 2

.  H .  h 2 .s 
 h. f .s   q 
l . f .s 
hl 2 s  
3
12
 96




Then the equation   1  2  0 give the result
ql 2 s
32 I R
H 
3
h
IC
5 f  8h
h s 
f 2
f 

3

3


2
IR 
h
h 
2
and may be reduced if we use the rigidity
k
h
s
k
IR h
s IC
in place of the real expression
We obtain the simplified expression
H
ql 2
5 f  8h
2
32 h  k  3  f  3h  f

Conclusion
M B  M D  Hh
ql 2
MC 
 H h  f 
8
H A  HE  H 
VA  VE 
ql 2
5 f  8h
2
32 h  k  3  f  3h  f 
ql
2
2/Case 2 Vertical actions( wind up to fly)
M B  M D   Hh
MC  
ql 2
 H h  f 
8
H A  HE  H  
ql 2
5 f  8h
2
32 h  k  3  f  3h  f 
VA  VE  
ql
2
Y
q
C
-
-
+ B
HA
+ D
A
E
VA
VE
3/Case 3 Horizontal actions( wind 1 pressure)
qh2
MB 
 H E .h
2
M D  H E .h
HE
5kh  6  2h  f 
qh2
HE 
16 h2  k  3  f  3h  f 
H A  q.h  H E
x
qh2
MC 
 HE h  f 
4
qh2
VE  VA 
2l
Y
C
-
+
+ B
+ D
q
-
HA
A
E
VA
HE
x
VE
4/ Case 4 Horizontal actions( wind 1 succions)
MD  
qh2
 H A .h
2
HA 
M B  H A .h
5kh  6  2h  f 
qh2
2
16 h  k  3  f  3h  f 
H E  q.h  H A
qh2
MC  
 H A h  f 
4
qh2
VE  VA 
2l
Y
C
+
+ B
+
D
q
-
HA
A
VA
E
HE
VE
x
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
wc ,1  2.34 KN / m see fig above
wc ,2  2.34KN / m
wc,3  3.74KN / m
W2 wind in long span (internal depressure)
Wc,3
Wc,1
Take
c p ,i  0.3
Wc,2
EN 1991-1-4(2005) (7.2.9 (6)note 2)
wc ,1   c p ,e  c p ,i  q p s   0.7   0.3  0.668*7  4.68 KN / m
wc,2  0KN / m
We have choose the max value of G zone for wind calculation but not the better
wc ,3   c p ,e  c p ,i  q p s   1.2   0.3  0.668*7  4.21KN / m
And For the zones H;I,and J this value is
wc ,3   c p ,e  c p ,i  q p s   0.6   0.3  0.668*7  1.4 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
wc,1  wc,2  0.668*7  4.676  4.68KN / m
wc,3  1.0*7  7 KN / m
Calculus actions
It is to determinate:
--the support reactions
HA ; H E ;VA and VE
-- the max bending moments M B ;M C and M D
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
2.45*252
5*1.09  8*6.41
H A  HE 
 16.494KN
32 6.412  0.511  3  1.09 3*6.41  1.09 
VA  VE 
2.45* 25
 30.625KN
2
2.45*252
MC 
 16.494  6.41  1.09   67.7 KNm
8
M B  M D  16.494*6.41  105.73KNm
actions
case
q(KN/m)
H A(KN)
HE(KN)
VA(KN)
VE(KN)
M B(KNm) M C(KNm) M D(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
W 1 ;W c,1
3
2.34
11.39
3.61
1.92
1.92
24.933
3.04
23.14
W 1; W c,2
4
2.34
3.61
11.39
1.92
1.92
23.14
3.04
24.933
W 1 W c,3
2
3.74
24.9
24.9
46.75
46.75
159.61
105.44
159.61
39.90
17.12
50.59
42.91
207.68
105.44
111.54
Total
W 2 ;W c,1
3
4.68
22.69
7.21
3.85
3.85
49.93
6.0
46.22
W 2 ;W c,2
4
0
0
0
0
0
0
0
0
W 2 ;W c,3
2
4.21
28.03
28.03
52.63
52.63 179.67
118.68
179.67
50.72
35.24
56.48
48.78
229.6
124.68
133.45
Total
W 3 ;W c,1
4
4.68
22.79
7.21
3.85
3.85
49.93
6.0
46.22
W 3 ;W c,2
4
4.68
7.21
22.79
3.85
3.85
46.22
6.0
49.93
W 3 ;W c,3
2
7
46.61
46.61
87.5
87.5
298.77
197.3
298.77
31.03
31.03
87.5
87.5
295.06
185.3
295.06
Total
3 Load combinations
Partial factor
 G max  1.35
permanent loads
 G min  1.0
permanent loads
 Q  1.50
variable loads
When there is more then one variable action acting,requiring the actions to be combined, the
expression is
ULS
:

j
g, j
GK , j  0.9  Q ,i QK ,i
i 1
SLS
G
K, j
j
 0.9 QK ,i
i 1
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)
HA
101 1.35G  1.5Q
HE
57.23
102 1.35G  1.5W1
VA
Bending moments (KNm)
VE
MB
MC
MD
57.23
106.97
106.97
366.87
239.33
366.87
useless
103 1.35G  1.5W2
53.81
30.59
43.38
31.83
201.66
95.63
57.44
104 1.35G  1.5W3
24.28
24.28
89.91
89.91
299.85
186.56
299.85
105 G  1.5W1
43.36
9.19
45.26
33.74
205.79
90.46
61.58
106 G  1.5W2
59.59
36.37
54.09
42.55
238.67
119.32
94.45
107 G  1.5W3
30.05
30.05
100.63
100.63
336.86
210.25
336.86
108 1.35G  1.8W1
49.55
8.55
49.72
35.89
231.09
98.4
109 1.35G  1.8W2
69.03
41.17
60.32
46.46
270.54
133.03
97.47
110 1.35G  1.8W3
33.59
33.59
116.17
116.17
388.37
242.15
388.37
The maximum values are collected in the table
Reactions (KN)
HA
Bending moments (KNm)
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 : M B= M D=-366.87 KNm
- Eave connection : M C=+239.33 KNm
The expression
With
M Rd 
M  M Rd
M
W pl . f y
M0
must be verified for bending
.We have
W pl . f y
M0
then
W pl 
M . M 0
fy
58.04
For
366.87
apex connection Wpl  235000
239.33
eave connection Wpl  235000
- In apex connection
Wpl  1561.1cm3
- In eave connection
Wpl  1018.4cm3
IPE 360+(1/2) IPE 360
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 Zanp = 270 mm
Yane = 85 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
q 
 ql
M x  M B   x  x2 
2 
2
By integration of the equation
d2y
M


dx 2
EI
We have
l
2
dy
M
1

dx  
dx 0 EI
EI
For
x
1
y
EI
l
2
l
2
ql
q 2

M

x

0  B 2 2 x dx
dy
 0
we have dx
then
l
2

ql 2 q 3
l ql 3
0  M B .x  4 x  6 x  M B . 2  24

 dx

For x=0 we have y=0 then
ymax 
1
384 EI
 5ql
4
 48M B .l 2 
E=210000 MPa=210000N/mm 2=2.1x108 KN/m 2
I=16270 cm 4
q= G  Q =2.45+3.5=5.95KN/m
L=25.1m
M B=105.73+149.42=255.15 KNm
For IPE 360 the vertical deflection is
 5*5.95* 25.1

4
ymax
 48* 255.15* 25.12 
384 * 2.1*108 *16270 *10 8
 0.3119m  31.2cm
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
5*5.95* 25.1


4
ymax
f adm 
 48* 255.15* 25.12 
384* 2.1*108 * 48200*10 8
 0.1052m  10.52cm
l
2510

 12.55cm
200 200
CARACTERISTIQUES GEOMETRIQUES
h = 500 mm
b = 200 mm
tw = 10,2 mm
tf = 16 mm
r = 21 mm
d = 426 mm
IPE 500
CARACTERISTIQUES MECANIQUES
g = 90,70 kg/m
A = 116,00 cm 2
Iy = 48 200,00 cm 4
Wel.y = 1 928,00 cm 3
Wpl.y = 2 194,00 cm 3
iy = 20,43 cm
Avz = 59,87 cm 2
Iz = 2 142,00 cm 4
Wel.z = 214,20 cm 3
Wpl.z = 335,90 cm 3
iz = 4,31 cm
It = 89,29 cm 4
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 L c between restraints or the resulting equivalent compression
flange slenderness
f 
k L
i 
c
c
f ,z
1
f
satisfies:
 c ,0
M
M
c , Rd
[6.3.2.4]
y , Ed
Where
M
y,Ed
is the maximum design value of the bending moment within the restraint spacing
k is a slenderness correction factor for moment distribution between restraints, see EN 1993-1-1
c
Table 6.6;
i , is the radius of gyration of the compression flange including 1/3 of the compressed part of the web
fz
area, about the minor axis of the section;
c ,0 is the slenderness parameter of the above compression element:
c ,0  LT ,0  0.10
LT ,0  0.4
1  
c ,0  0.4  0.10  0.5
E
 93.9 and
fy
 
fy
I f ,z
then
235
N



mm 2 

3

 d  tw  
  I z   2*  *  
 3  12  


2
[6.3.2.3]
then
I f ,z
3

 42.6  1.02  
  2142   2*

*
3  12  




 1069.74cm4
2
1
 d  
Af , z    A   2*  * tw  
2
 3    then
1

 42.6 
2
I f , z   116   2*
 *1.02   43.52cm
2
3 


I f ,z
i f ,z 
Af , z

1069.74
 4.96cm
43.52
Wy  Wpl , y  2194cm3
E
 93.9  93.9
fy
1  
M c , Rd 
Wy f y
 M1
2194* 235*103

 515.59 KNm
1.0
Combination 1.35G  1.5Q
MB=MEd=366.87 KNm
We consider that the coefficient  is the same if the rafter is unrestraint then

M C 239.33

 0.65235
M B 366.87
Then KC 
1
1

 0.647
1.33  0.33 1.33  0.33*0.65235
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 :K C=1.0
f 
table 6.6
KC LC
1.0*150

 0.322
i f , z 1 4.96*93.9
The maximum bending moment is at the origin B of the rafter then the lateral torsional buckling may
be also in the origin
M
y , Ed
 366.87 KNm
M
M
c ,0
c , Rd
 0.5*
y , Ed
515.59
 0.703
366.59
KC LC
1.0*150

 0.322
i f , z 1 4.96*93.9
f 
0.322  0.703
Combination
MB= M
c ,0
M
M
f 
y , Ed
c , Rd
1.35G  1.8W3
 388.37 KNm
 0.5*
y , Ed
515.59
 0.6637
388.37
KC LC
1.0* 418

 0.8975 Not verified
i f , z 1 4.96*93.9
It’s necessary to add other bracing systems each 3m spacing then
LC=3m
M
y , Ed
c ,0
M
M
f 
 388.37 KNm
c , Rd
y , Ed
 0.5*
515.59
 0.6637
388.37
KC LC
1.0*300

 0.6441
i f , z 1 4.96*93.9
0.644  0.663
Then the lateral torsional buckling is satisfactory
A detailed procedure to do verification for the rafter is sho wn 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
7 restraints
2 elastic section
8 bending moment diagram
3 plastic stable length
9 Compression flange
4 plastic stable length
10 plastic stable length
5 elastic section
11 plastic stable length
6 plastic hinge
12 elastic section
Annex A
y
4.5/ the haunch verification
C
+
- D
O
F
S
x
the equation of the bending moment curve is a parabolic form
Y  aX 2
the point F is considered the limit of elastic moment M el 
Wel f y
M0
1928*103 * 235

 453.8KNm
1.0
X
0m
S=12.55
Y
M C=242.15
M D+MC=388.37+242.15=630.52
a
Then
Y
630.52

4
2
X
157.5
The bending moment curve equation will be
For X  12.55  F then Y  M el 
Wel f y
Then 454  4 12.55  F  then
2
Conclusion
M0

Y  4X 2
1928*103 * 235
 453.8 KNm
1.0
F 2  100.4 F  176  0
F  1.78m
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 101 1.35G  1.5Q
N
Ed
 106.97KN
V
Ed
 57.23KN
(assumed to be constant along the column)
(assumed to be constant along the column)
M Ed  366.87 KNm (at the top of the column)
5.1/Classification of the section
Web: the web slenderness is
dN 

c
437.6

 42.9
tw
10.2
N Ed
106970

 44.63
tw f y 10.2* 235
d w  d N 426  44.63

 0.552  0.50
2d w
2* 426
Then the limit for the class is
§5.5 (tab5.2)
396
396*1

 64.119
13  1 13*0.552  1
42.9 ≤ 64.119
Until
the web is class 1
Flange: the flange slenderness is
c

tf
 b  t w  2r 
tf
2  73.9  4.618
16
§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
AV  max  A  2bt f   t w  2r  t f ; hwt w 
§.6.2.6
AV  max 11600  2*200*16  10.2  2*21 *16;1.0*426*10.2 
AV  max  6035.2; 4345.2 
AV  6035.2mm2
V pl , Rd
f

AV  y
6035.2  235 

3
3




*103  818.84 KN
M0
1.0
VEd
57.23

 0.07  0.5
V pl , Rd 818.84
§ 6.2.8(2)
The effect of the shear force on the moment resistance may be neglected
Verification to axial force
N Pl , Rd 
N
Ed
Af y
M0

11600* 235
*103  2726 KN
1.0
§6.2.4
 106.97KN
0.25N Pl , Rd  0.25*2726  681.5KN
6.2.9.1(4)equ 6.33
0.5hwtw f y
M0
Since

0.5* 426 *10.2 * 235
 510.56 KN
1.0 *1000
106.97 ≤ 681.5 and
6.2.9.1(4)equ 6.34
106.97 ≤ 510.56
The effect of the axial force on the moment resistance may be neglected
Verification to bending moment
M pl , Rd 
WPl , Rd f y
M0

2194* 235
 515.59 KNm
1.0*1000
M Ed
366.87

 0.711  1.0 Ok!
M pl , Rd
515.59
§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
G+Q
M
x
HA
M x  H A.x
By introducing a virtual force P at the summit of the column AB
This effort generate the following forces
k 
IR h
h

s IC
s

  k  3  3  3 2
f
h

1    3  2  
1 

2


RA 

1    3  2  
1 

2


RE  P  RA


1 
2
1    3  2  
2
P    3  2  
1 

2
2

VA  VE  


M B   Ph
M C   Ph
M D   Ph
Ph
l
For an IPE 500 column we obtain :
k  0.511
  0.17004
  4.10786
Then we have the results
RA  0.534P
RE  0.466P
The moment in the point M is
M X  0.534Px
The resultant moment under the two actions is
M X  H A x  0.534Px
the internal potential energy of the column is:
1
W
2EI
h
  H A x  0.534Px  dx
2
0
  0.56913
  0.43087
  0.04458
h
1
W
2EI
  H A  0.534P 
1
W
2 EI
2
1 3
x
H

0.534
P


A
 3

0
2
x 2 dx
0
h
h3
2
W
 H A  0.534P 
6 EI
dW
1.07 3
h HA
 P  0   
dP
6 EI C
1.07*6.413 *3980.4

 1.847cm
6
6*2.1*10 *48200
l
641

 2.137cm
300
300
Since
1.847 ≤ 2.137 OK!
Combination 202 G+W
H A  16.494  39.9  23.406KN
By application a similar resolution as the above
x2
M X   H A x  q  0.534 Px
2
the internal potential energy of the column is:
2
h

1 
x2
W

H
x

q

0.534
Px

 dx
A

2 EI 0 
2

Using a similar calculation we have
1
W 
2 EI
2


x2
0   H A x  q 2  0.534Px  dx
h
W
1 3  1

2
 1
x  qx   H A  0.134 P    H A  0.534P   0.05q 2 x 2 
2 EI   4
 3

W
1 3  1

2
 1
h  qh   H A  0.134 P    H A  0.534 P   0.05q 2 h 2 
2 EI   4
 3



1
4
3
dW
 P  0      0.067qh  0.178H A h 
EI
dP
 0.067*234*6414 *102  0.178*2340.6*6413 
  1.345cm
 
6
2.1*10 *48200
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, M Z,Ed=0):
N Ed
M Ed
 k yy
1
 y N RK
 LT M y , RK
 M1
§6.3.3
equation 6.61
 M1
N Ed
M Ed
 k zy
1
 y N RK
 LT M y , RK
 M1
equation 6.62
 M1
Buckling about yy
L
CR , y
 6.41m
h 500

 2.5  1.2
b 200
Ncr , y  
y 
2
EI y
L2cr , y
Af y
N cr , y
t f  16mm  40mm buckling curve :a(αy=0.21)
table 6.2
210000*48200*104

 24313.64 KN
64102 *103
2

11600 * 235
 0.335
24313.64 *103
§6.3.1.2


y  0.5 1   y  0.2   y   0.5 1  0.21 0.335  0.2   0.3352   0.5703

y 
1
y    
2
y

2
y
2

1
0.5703  0.5703  0.335
2
2
 0.9691
Buckling about zz
Buckling curve :b (αz=0.34)
Ncr , z 
z 
 2 EI z
2
cr , z
L
Af y
N cr , z


 2 *210000*2142*104
6410
2
 1080.5KN
11600* 235
 1.5883
1080.5*103
z  0.5 1   z   z  0.2    z 
2


z  0.5 1  0.34 1.5883  0.2   1.58832   1.9973
z 
1
z    
2
z

2
z
1
1.9973  1.9973  1.5883
2
2
 0.3117
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
h 500

 2.5  2
b 200
Annex A
then buckling curve c(αLT=0.49)
Moment diagram with linear variation :
 0
then C1  1.77
The simplification of critical moment may be used:
M cr  C1
 2 EI z
L2cr , LT
2
I w Lcr , LT GI t

Iz
 2 EI z
 2 210000*2142*10 4 1249000*10 6 6410 2*80770*89.29 *104
M cr  1.77

64102 *106
2142*104
 2 210000* 2142*104
M cr  1.77
 2 2.1* 21420 1249*105
6412
2142

6412 *8077 *89.29
 676.32 KNm
 2 21* 21420
 LT 
Wpl , y f y
M cr

2194*10 3 * 235
 0.8731
676.32*10 6


2
LT  0.5 1   LT LT  LT ,0   y 


LT  0.5 1  0.49 0.8731 0.4  0.75*0.8731 2  0.9663
With a values of
 LT 
LT ,0  0.4
1

LT     
For   0
2
LT
then
2
LT
Kc 
  0.75
and
1
0.9663  0.96632  0.75*0.87312
 0.6377
1
 0.7519
1.33  0.33
Bending moment diagram and the

6.3.2.3 table 6.6
coefficient


2
f  1  0.5 1  K c  1  2  LT  0.8 


2
f  1  0.5 1  0.7519  1  2  0.8731  0.8    0.8773  1



0.6377
 LT ,mod  LT 
 0.7269  1
f
0.8773
Calculation of the factor K yy
K yy  Cmy CmLT
y
1
N Ed
N cr , y
Cmy  Cmy ,0  1  Cmy ,0 
2
Cm , LT  Cmy
1
C yy
annex A
 y aLT
1   y aLT
aLT

N Ed
1 
N cr , z


N Ed
  1 
N cr ,T




annex A
1
annex A
N Ed
N cr , y

N Ed
1  y
N cr , y
1
y
annex A

 Wel , y
1.6 2
1.6 2 2 
C yy  1   wy  1  2 
Cmy  max 
Cmy  max  n pl  bLT  


w
w

 Wpl , y
y
y

Calculation of
y
106.97
24313.64
y 
 0.9998
106.97
1  0.9691
24313.64
1
wy 
Wpl , y
Wel , y

2194
 1.138  1.5
1928
Critical axial force in the torsional buckling mode
N cr ,T
 2 EI w
A

 GI t  2
I 0 
Lcr ,T



For a doubly symmetrical section
I 0  I y  I z   y02  z02   48200  2142  50342cm 4
Ncr ,T

11600
 2 *2.1*105 *124.9*104 *106 
4

 80770*89.29*10 

50342*104 *103 
64102

Ncr ,T  3113.56KN
M cr ,0  C1
 2 EI z
L2cr , LT
L2cr , LT GI t
Iw

Iz
 2 EI z
M cr ,0 is the critical moment foe the calculation of  0
annex A . Then we have
M cr ,0  1*
C1=1
 2 210000*2142*104 1249000*106
64102 *106
for uniform bending moment as specified in
2142*104
64102 *80770*89.29 *104

 382.1KNm
 2 210000*2142*104
0 
W pl , y f y
M cr ,0
2194*103 * 235

 1.162
382.1*106

 0,lim  0.2 C1 4 1 

N Ed
N cr , z
For doubly symmetrical section


106.97 

N Ed
 1 
N cr ,TF




Ncr ,TF  Ncr ,T
106.97 

 0,lim  0.2 1.77 4 1 
1 
  0.2569
1080.5
3113.56
Then

 0   0,lim
Calculation of
Cmy
Cmy  Cmy ,0  1  Cmy ,0 
y 
M y , Ed
N Ed
aLT  1 
 y aLT
1
 y aLT
A
366.87*103 11600

 20.635
Wel , y
106.97 1928*103
It
89.29
 1
 0.928
Iw
1249
Calculation of C my,0
Cmy ,0  0.79  0.21 y  0.36  y  0.33
With a value
y 0
table A2
then
Cmy ,0  0.79  01188
106.97
 0.7895
24313.64
Cmy  0.7895  1  0.7895 
2
Cm , LT  Cmy
N Ed
N cr , y
20.635 *0.928
 0.9596
1  20.635 *0.928
aLT

N Ed
1



N cr , z


N Ed
1




N cr ,T





1
Cm, LT  0.95962
Then
0.928
106.97 
 106.97  
1 
 1 

 1080.5   3113.56 
 0.9457  1
 Cm, LT  1
Calculation of C yy
 1.6 2
 Wel , y
1.6 2 2 
C yy  1   wy  1  2 
Cmy  max 
Cmy  max  n pl  bLT  

wy
wy

 Wpl , y



 max  max  y ;  z   z
M z , Ed  0  bLT  0
n pl 
N Ed
106970

 0.03924
N Rk 11600* 235
1.0
 M1

1.6
1.6


C yy  1  1.138  1  2 
*0.95962 *1.5883 
*0.95962 *1.58832  *0.03924  0.978
1.138

 1.138

Wel , y
Wpl , y
Since

1928
 0.8787
2194
0.978 ≥ 0.8787 Ok!
Calculation of K yy
K yy  Cmy CmLT
K yy  0.9596*1*
y
1
N Ed
N cr , y
1
C yy
0.9998
1
 0.9853
106.97 0.978
1
24313.64
Verification with interaction formula
N Ed
M Ed
 k yy
1
 y N RK
 LT M y , RK
 M1
 M1
106970
366.87 *106
 0.9853*
 0.766  1
0.9691*11600* 235
2194*103 * 235
0.9663*
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
K yz
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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