Biaxial folding for Eurocode 2 guide
Support design for non-thin supportsBiaxial bending
Dimensioning of a support between the ground floor and the upper floor. Suppose the foundation is at ground level. Therefore, the moments at ground level are considered as zero restraint conditions of the diversion pin.
- 400mm square column
- Axial load N Ed. = 3000kN
- Moments at the top M j = 110kNm and M for example = 130 kNm
- Moments on the ground are zero
- Fck = 30N/mm2
- F Yes = 500N/mm2
- Nominal cap 25mm
- Floor to floor height 4000 mm
- Depth of the beam supported on the pillar: 500 mm
The following column design is calculated for two-axis bending. This method is done using the trial and error method because it needs to be done until (M Edz / M Rdz ) A + (M Swirl / M Rdy ) A < 1. However, the Eurocode 2 Handbook for the Design of Concrete Structures and The book “Eurocode 2 Reinforced Concrete Structures” by Bill Mosley, John Bungey and Ray Hulse recommend the old method used in BS 8110 for the design of a column for biaxial flexible supports. The old method is simpler than the method considered in the following calculations because, once the biaxial bending moments have been determined, we can easily determine the area of the reinforcement.
M for example = 130 kN·m
M j = 110 kN m
N Ed. = 3,000 kN
Free height = 4000-500 = 3500 mm
Effective length = l Ó = factor * l
Factor = 0.85 (abbreviated Eurocode 2, Table 5.1. This may be more conservative).
l Ó = 0.85 * 3500 = 2975 mm
Slenderness λ = l Ó /L
i = radius of gyration = h/√12
λ = l Ó /( h/√12 ) = 3.46*l Ó /h = 3.46*2975/400 = 25.73
Limit thinness λlim
λ lemon = 20ABC/√n
A = 0.7 if the effective creep factor is unknown
B = 1.1 if the degree of mechanical reinforcement is unknown
C = 1.7 – rm = 1.7 points o1 /M o2
M o1 = 0 kN·m
M o2 = 130kNm where lMo2l ≥ lMo1l
C = 1.7 – 0/130) = 1.7
n = N Ed. / (A C *F CD )
F CD =f ck / 1.5 = (30/1.5)*0.85 = 17
n = 3,000*1,000 / (400*400*17) = 0.94
λ lemon = 20*0.7*1.1*1.7/√0.94 = 27
λ lemon > λ therefore the column is not slender .
Calculation of design moments
The column has moments in both directions. First, determine the critical moment.
Look at mine,
Assume that no moment occurs at the ground floor level because the foundations are at this level.
M Swirl = Max{M o2 M oed +M 2 M o1 + 0.5 million 2 }
M o2y = Max {M Optimal M below } + a*N Ed. = 110 + (2.975/400)*3000 ≥ Max.(400/30, 20)*3000 = 110+22.3 ≥ 60 = 132, 3kNm > 60kNm
M o1y = 0
M oEdy = 0.6*M o2 + 0.4*M o1 ≥ 0.4*M o2 = 0.6*132.3 + 0.4* (0) ≥ 0.4*132.3 = 79, 4≥ 52.9
M 2 years = 0, the column is not narrow
M Swirl = Max{M o2 M oed +M 2 M o1 + 0.5 million 2 } = Max{132.3, 79.4+0, 0 + 0.5*0} = 132.3 kNm
Consider Mz,
Assume that no moment occurs at the ground floor level because the foundations are at this level.
M Edz = Máx{M o2 M oed +M 2 M o1 + 0.5 million 2 }M o2z = Máx {Moben, Munten} + ei*NEd = 130 + (2.975/400)*3000 ≥ Max(400/30 .20)*3000 = 130+22.3 ≥ 60 = 152.3 kNm > 60 kNm
Mo1z = 0
M oEdz = 0.6*M o2 + 0.4*M o1 ≥ 0.4*M o2 = 0.6*152.3 + 0.4*0) ≥ 0.4*152.3 = 91.4 ≥ 60.9
M 2z = 0 , the column is not narrow
M Edz = Max{M o2 M oed +M 2 M o1 + 0.5 million 2 } = Max{152.3, 91.4+0, 0 + 0.5*0} = 152.3 kNm
Imperfections only need to be taken into account in one direction, i.e. the one that has the most unfavorable effect.
Therefore,
M Edz = 132.3kNm and M Swirl = 130 kNm
Consider the critical moment
M Edz = 132.3 kN m
M Ed. / (bra 2 )*F ck ) = (132.3*10^6) / (400*(400^2)*30) = 0.07 = (3000*10^6) / (400* 400*30 = 0.63
Adopt 25mm diameter bars as main reinforcement and 10mm diameter bars as shear connections.
D 2 = 25+10+25/2 = 47.5 mm
D 2 /h = 47.5 / 400 = 0.12
Note: The d2/h = 0.15 diagram is used to determine the area of the reinforcement, but it is more conservative. Interpolation between the graphs d2/h = 0.10 and d2/h = 0.15 can be used to determine more accurate answers.
A S *F Yes / bitch ck = 0.3A S = 0.3*400*400*30/500 = 2880 mm2
Includes six 25mm posts (provides 2,940mm2). Six 25mm bars are provided for bending, but the spine needs to be reinforced symmetrically. Therefore, the total number of bases allocated to the column is eight.
Check for biaxial bending
Additionally, an examination is not necessary if:
0.5 ≤ (λ j /λ for example ) ≤ 2.0 For rectangular column
AND
0.2 ≥ (e j /H equation )/(e for example /B equation ) ≥ 5.0
Here λ j and λ for example are slenderness proportions
λ j is equal to λ for example since the beam height is similar in each direction.
Therefore j /λ for example = 1
Therefore j /λ for example < 2 und > 0.5 OK
t j /H equation = (M Edz / N Ed. ) / H equation
t for example /B equation = (M Swirl / N Ed. ) / b equation
(e j /H equation )/(e eg /B equation ) =M Edz / M Swirl
Here h=b=h equation =b equation The column is square
M Edz = 130 kN m
M Swirl = 132.3 kN m
(e j /H equation )/(e for example /B equation ) = 130/132.3 = 0.98 > 0.2 e < 5
Therefore, a two-axis test is necessary.
(M Edz / M Rdz ) A + (M Swirl / M Rdy )A ≤ 1
M Edz = 130 kNm
M Swirl =132.3kNm
M Rdz and M Rdy are the moment-bearing capacity in the respective direction, corresponding to an axial load N Ed. .
For symmetrical reinforcement cross section
M Rdz =M Rdy
As stated = 2940 mm2
As*fyk / b*h*fck = 2940*500/(400*400*30) = 0.31
N Ed. / (bitch ck ) = 0.63
From the table d 2 /h =0.15
M Ed. / (bra 2 )* fuck ) = 0.075
M Ed. = 0.098*400*400*400*30 = 144kNm
a = an exponent
a = 1.0 for N Ed. /N Road = 0.1
a = 1.5 for N Ed. /N Road = 0.7
N Ed. = 3000kN
N Road = Ac*f CD + Como*f yard
N Road = 400*400*(0.85*30/1.5) + 3920* (500/1.15) = 4424kN
N Ed. /N Road = 3000/4424 = 0.68
By interpolation you get a = 1.48
(M Edz / M Rdz ) A + (M Swirl / M Rdy ) A = (130/144) 1.48 + (132/144) 1.48 = 1.74 > 1
Therefore, biaxial bending test is not acceptable
Therefore, increase the number of bars to 12. (In this design, only the reinforcement along both sides was considered, without taking into account the effect of the other reinforcement bars in the opposite direction.) Therefore, the number of reinforcements and the capacity of effective load for the moment is eight (3,920 mm²) and twelve (5,880 mm²) for the axial load capacity.
A S *F Yes / bitch ck = 3920*500/(400*400*30) = 0.41
N Ed. / (bitch ck ) = 0.63
From the table d 2 /h =0.15
M Ed. / (b*(h^2)*f ck ) = 0.105
M Ed. = 0.105*400*400*400*30 = 202kNm
a = an exponent
a = 1.0 for N Ed. /N Road = 0.1
a = 1.5 for N Ed. /N Road = 0.7
N Ed. = 3000kN
N Road = Ac*f CD + As*f yard N Road = 400*400*(0.85*30/1.5) + 3920* (500/1.15) = 4850kN
N Ed. /N Road = 3000/4850 = 0.62
Through interpolation
a = 1.43
(M Edz / M Rdz ) A + (M Swirl / M Rdy ) A = (130/202) 1.43 + (132/202) 1.43 = 1 (corresponds to almost one)
So as long as there are reinforcements, it's fine.
For more information on biaxial bending for Eurocode 2, see the Structural Guide.
