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03/06/2014 Calculation sheet Page 1of 4Reference file:
BiaxialBending(2).xls
Biaxialbending(rectangularcrosssection)
Doublyreinforcedsection
Geometrycrosssection
Rectangular shape:Width b = 400 mm (width in respect of minor axis x-x)Height h = 650 mm (depth in respect of major axis y-y)
Analysis (biaxialbendingwitha 0)
2ndorder effects in each direction (x-x and y-y) taken into account: bending moment about x-x axis (ULS): MEdx= 370 kNm ( > 0) bending moment about y-y axis (ULS): MEdy= 70 kNm ( > 0) axial load on column (ULS): NEd= 700 kN (compressive).
Materialscharacteristic/designstrengths
Concrete
Strength class for concrete:fck= 30 N/mm
2= 30 MPafcd= accfck/gc= 1,00fck/1,50 = 20,00 N/mm
2
Steel
Strength of reinforcement (Class C): uk= 7,50%ud= 0,9uk= 6,75%
fyk= 500 N/mm2= 500 MPa
fyd= fyk/s= fyk/1,15 = 0,87fyk= 435 N/mm2= constant
Es= 200 GPa; yd= fyd/Es= 0,002174 [-].
Stressstraindiagrams
Grade of concrete C30/37 with cu3= 3,5/103, c3= 1,75/10
3
= 1,00, = 0,80.
Ref. EN 1992-1-1Sec. 9.5Sec. 5.8.9 Exp. (5.39)Sec. 6.1 Figure 6.1
Cl. 9.5.1(1): b/h 0,25
See a in Exp. (5.39)
Table 3.1
Table C.1 Annex C
Cl. 3.2.7(2) b)
Figure 3.5 modified
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03/06/2014 Calculation sheet Page 2of 4Reference file:
BiaxialBending(2).xls
Durabilityandcovertoreinforcement
Nominal cover to enclosing link:
cnom= cmin+ cdev= 30 mm
Geometry
Maximum size of aggregate: dg= 25 mm
Diameter of enclosing link: link= 10 mm (shear requirements)
Mainlongitudinalreinforcement(provided)
Top steel: 5H18vertical bars on the right side: 3H18vertical bars on the left side: 3H18Bottom steel: 5H18
AreasofreinforcementBRESLERMethod
For bending moment about x-x axis :Top: As2x= 5H18 = 1272 mm (d2= 49 mm)Bottom: Asx= 5H18 = 1272 mm(d1= 49 mmd = h d1= 601 mm).
For bending moment about y-y axis:Top: As2y= 3H18 + (1H18 + 1H18) = 1272 mm (d2= 49 mm)Bottom: Asy= 3H18 + (1H18 + 1H18) = 1272 mm(d1= 49 mmd = h d1= 351 mm).
Exp. (4.1)
Cl. 8.2(2) Note
[@section beingconsidered]
Sec. 5.8.9(4)
Effective depth x-x
Effective depth y-y
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03/06/2014 Calculation sheet Page 3of 4Reference file:
BiaxialBending(2).xls
700; 370
Mmin
A
BCD
E
F
G
H
I J
K
L
M
N
O
MRdx
-200
-100
0
100
200
300
400
500
600
700
800
-2000 -1000 0 1000 2000 3000 4000 5000 6000 7000
Moment,M
Edx
[kNm]
Axial compression, NEd[kN]
Note:(biaxial bending with a 0).
700; 70
Mmin
A
BC
D
E
F
G
H
I J
K
L
M
N
O
MRdy
-100
-50
0
50
100
150
200
250
300
350
400
450
-2000 -1000 0 1000 2000 3000 4000 5000 6000 7000
Moment,
MEdy
[kNm]
Axial compression, NEd[kN]
MRd,x: NRdinteractionchart for b = 400 mm xh = 650 mm
LEGENDA
s2
s s2 ud
s ud c
s ud c c3
s ud c cu3
s c c3
s c cu3
s c cu3
s c cu3
s
A) 0,01; 0 @ x h
B)
C) ; 0
D) ;
E) ;
F) 0,01;
G) 0,01;
H) 0,005;
I) 0,0025;
J)
yd s yd c cu3
s yd c cu3
s c cu3
s c cu3
c cu3
s s2 c c3
E / f ;
K) ( 0,0001);
L) 0,0012;
M) 0,0002;
N) ; x h
O)
MRd,y: NRdinteractionchart for b = 650 mm xh = 400 mm
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03/06/2014 Calculation sheet Page 4of 4Reference file:
BiaxialBending(2).xls
Reinforcementdetails
As,tot= 5H18 + 5H18 + 3H18 + 3H18 = 4072 mm.As,min= MAX[0,10NEd/fyd; 0,002bh] = 575 mm
2As,tot(satisfactory)As,max= 0,04bh = 10400 mm
2As,tot(satisfactory).
BRESLERMethod
NEd= 700 kN (compressive).NRd= bhfcd+ As,totfyd== [(400)(650)(20,00) + (4072)(434,8)]/1000 = 6970 kN.
Biaxial bending for 0,1 < NEd/NRd< 1,0 (rectangular sections).In this case:NEd/NRd= 0,10a = 1,00 (biaxial bending with a 0).
With NEd= 700 kN (compressive):
MRd,x= 493 kNm (ULS)MRd,y= 285 kNm (ULS)
MEdx= 370 kNm (ULS)MEdy= 70 kNm (ULS)
aa
EdyEdx
Rd,x Rd,y
MM
M M
(370/493)1,00+ (70/285)1,00=
= 0,750 + 0,245 = 0,995 1 (satisfactory)(biaxial bending with a 0).
Note
Ed Edd
c cd cd
N N
A f bhf 0,135.
Cl. 9.5.2(2) NoteCl. 9.5.2(3) Note
Cl. 5.8.9(4)
Cl. 5.8.9 Exp. (5.39)
Exp. (5.39)
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