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Wednesday, August 9, 2017

Comparative Design of Biaxial R.C. Columns to BS 8110 and EC2


A reinforced concrete column fixed at both ends is subjected to the following loading conditions as given below. We are required to obtain the appropriate longitudinal reinforcement for the column using BS 8110-1:1997 and Eurocode 2.

The column is carrying longitudinal and transverse beams of depth 600mm and width 300mm. It is also supported by beams of the same dimension. The centre to centre height of the column is 3500mm. The plan view of the arrangement of the beams and column is as shown below.



DESIGN ACCORDING TO BS 8110-1:1997
N = 716.88 KN;
TOP: Mx-x = 175.87 KNm; My-y = 35.52 KNm
BOTTOM: Mx-x = -85.832 KNm; My-y = -25.269 KNm


Concrete grade = 30 N/mm2
Yield Strength of reinforcement = 460 N/mm2
Concrete cover = 40mm
Lo = 3.5m, Le = 0.75 × (3500 - 600) = 2175 mm;
Size of column = 400 x 300mm;
Slenderness = 2175/300 = 7.25 < 15. Thus column is short.

Effective depth about X-X axis
h' = (400 - 40 - 12.5 - 10) = 337.5 mm

Effective depth about Y-Y axis
b' = (300 - 40 - 12.5 - 10) = 237.5 mm

Mx/ h' = 521.109 KN; My/ b' = 149.558  KN

Axial-load ratio
Nratio = (N × 1000) / (Fcu × b × h) = (716.88 × 1000) / (30 × 400 × 300) = 0.1991

From Table 3.22 of BS 8110-1:1997, for Nratio = 0.1991
Therefore, β = 1 – 1.1644(0.1991) = 0.7681

As Mx/h' exceeds My/b'
Mx' = Mx + β h'/b'My
Mx' = 175.87 + [0.7681 × (0.3375)/(0.2375 ) × 35.52]

Therefore design moment Mx' = 214.64 KNm

Mimimum Eccentricity in Columns
According to clause 3.8.2.4, at no section in a column should the design moment be taken as less than that produced by considering the design ultimate axial load as acting at a minimum eccentricity, emin equal to 0.05 times the overall dimension of the column in the plane of bending considered but not more than 20 mm.

Where biaxial bending is considered, it is only necessary to ensure that the eccentricity exceeds the minimum about one axis at a time.

In the x-x direction = emin = 0.05 × 400 = 20mm, therefore adopt 20mm, Mx = 716.88 × 0.02 = 14.3376 KNm < 214.64 KNm

In the y-y direction = emin = 0.05 × 300 = 15mm, therefore adopt 15mm, My = 716.88 × 0.015 = 10.7532 KNm < 214.64 KNm

Section design ratios for chart entry
Axial load ratio Nratio = (N × 1000)/(Fcu × b × h) = (716.88 × 1000)/(30 × 400 × 300) = 0.1991

Mratio = M/(Fcu × b × h2) = (214.64 × 106) / (30 × 300 × 4002) = 0.149

With d'/h = 337.5/400 = 0.84375


From chart, (ρ × Fy)/(Fcu) = 0.286 Therefore, ρ = (0.286 × 30) / 460 = 0.01865

ASCreq = 0.01865 × 400 × 300 = 2238.26 mm2

Provide 6Y25mm (Asprov = 2946 mm2)

Maximum area of reinforcement = 0.06bh = 0.06 × 400 × 300 = 7200 mm2

DESIGN ACCORDING TO EUROCODE 2
Clause 5.8.9(2) of EC2 permits us to perform separate design in each principal direction, disregarding biaxial bending,  as a first step. Imperfections need to be taken into account only in the direction where they will have the most unfavourable effect. However in this example, we have carried out imperfection analysis in both directions.

NEd = 716.88 KN

Elastic Moments
Y - direction: M01 = 175.87 KNm; M02 = -85.832 KNm
Z – direction: M01 = 35.52KNm; M02 = -25.269 KNm

Clear column height (L) = 3500 – 600 = 2900 mm

Calculation of the effective height of the column (Lo)
Let us first of all calculate the relative stiffnesses of the members in the planes of bending.

In the y-direction;
Second moment of area of beam 1 (I1) = bh3/12 = 0.3 × 0.63/12 = 0.0054 m4
Stiffness of beam 1 (since E is constant) = 4I1/L = (4 × 0.0054) / 6 = 0.0036

Second moment of area of beam (I2) = bh3/12 = 0.3 × 0.63/12 = 0.0054 m4
Stiffness of beam 2 (since E is constant) = 4I2/L = (4 × 0.0054) / 3.5 = 0.00617

Second moment of area of column (Ic) = bh3/12 = 0.0016 m4
Stiffness of column = 4Ic/L = (4 × 0.0016) / 3.5 = 0.001828

For compression members in regular braced frames, the slenderness criterion should be checked with an effective length l0 determined in the following way:

Lo = 0.5L √[(1 + k1)/(0.45 + k1)) × (1 + k2)/(0.45 + k2))]

Where;
k1, k2 are the relative flexibilities of rotational restraints at ends 1 and 2 respectively
L is the clear height of the column between the end restraints

k = 0 is the theoretical limit for rigid rotational restraint, and k = ∞ represents the limit for no restraint atall. Since fully rigid restraint is rare in practise, a minimum value of 0.1 is recommended for k1 and k2.

In the above equations, k1 and k2 are the relative flexibilities of rotational restraint at nodes I and 2 respectively. If the stiffness of adjacent columns does not vary significantly (say, difference not exceeding 15% of the higher value), the relative flexibility may be taken as the stiffness of the column under consideration divided by the sum of the stiffness of the beams (or, for an end column, the stiffness of the beam) attached to the column in the appropriate plane of bending.

Remember that we will have to reduce the stiffness of the beams by half to account for cracking;

k1 = k2 = 0.001828 / (0.0018 + 0.003085) = 0.3743

Lo = 0.5 × 2900√[((1 + 0.3743)/(0.45 + 0.3743)) × (1 + 0.3743)/(0.45 + 0.3743)] = 2647.77 mm

Compare with BS 8110’s 0.75L = 0.75 × 2900 = 2175 mm

In the z-direction;
Second moment of area of beam 3 (I3) = bh3/12 = 0.3 × 0.63/12 = 0.0054 m4
Stiffness of beam 3 (since E is constant) = 4I1/L = (4 × 0.0054) / 3.5 = 0.00617

Second moment of area of column (Ic) = bh3/12 = 0.0009 m4
Stiffness of column = 4Ic/L = (4 × 0.0009) / 3.5 = 0.00102857

k1 = k2 = 0.00102857/(0.003085) = 0.3334

lo = 0.5 × 2900 √[(1 + 0.3334) / (0.45 + 0.3334) × (1 + 0.3334) / (0.45 + 0.3334)] = 2675.293 mm

Radius of gyration

ix = h/√12 = 400/√12 = 115.47
iz = b/√12 = 300/√12 = 86.602

Slenderness in the x-direction (λx) = 2647.77/115.47 = 22.930
Slenderness in the z-direction (λz) = 2675.293/86.602 = 30.892

Critical Slenderness for the y-direction
λlim = (20.A.B.C)/√n
A = 0.7
B = 1.1
C = 1.7 - M01/M02 = 1.7 - [(-85.832)/175.87] = 2.188
n = NEd / (Ac fcd)
NEd = 716.88 × 103 N
Ac = 400 × 300 = 120000 mm2
fcd = (αcc fck)/1.5 = (0.85 × 30)/1.5 = 17 N/mm2
n = (716.88 × 103) / (120000 × 17) = 0.3514
λlim = (20 × 0.7 × 1.1 × 2.188 )/√0.3514 = 56.842

22.930 < 56.842, second order effects need not to be considered in the y-direction

Critical Slenderness for the z-direction
A = 0.7
B = 1.1
C = 1.7 - M01/M02 = 1.7 - [(-25.269)/35.52] = 2.411
n = NEd / (Ac fcd)
NEd = 716.88 × 103 N
Ac = 400 × 300 = 120000 mm2
fcd = (αcc fck)/1.5 = (0.85 × 30)/1.5 = 17 N/mm2

n = (716.88 × 103) / (120000 × 17) = 0.3514

λlim = (20 × 0.7 × 1.1 × 2.411 )/√0.3514 = 62.634

30.892 < 62.634, second order effects need not to be considered in the z - direction

Design Moments (y-direction)

Y - direction: M01 = 175.87 KNm; M02 = -85.832 KNm

e1 is the geometric imperfection = (θi l0/2) = (1/200) × (2647.77/2) = 6.619 mm

Minimum eccentricity e0 = h/30 = 400/30 = 13.333 mm. Since this is less than 20mm, take minimum eccentricity = 20mm (clause 6.1(4) EC2).

Minimum design moment = e0NEd = 20 × 10-3 × 716.88 = 14.3376 KNm

First order end moment M02 = MTop + eiNEd

eiNEd = 6.619 × 10-3 × 716.88 = 4.745 KNm

M02 = MTop + eiNEd = 175.87 + 4.754 = 180.624 KNm

Longitudinal Steel Area
d2 = Cnom + ϕ/2 + ϕlinks = 40 + 12.5 + 10 = 62.5 mm

d2/h = 62.5/400 = 0.156

So we will read from chart; d2/h = 0.15;

MEd/(fck bh2) = (180.624 × 106)/(30 × 300 × 4002 ) = 0.125

NEd/(fck bh) = (716.88 × 103) / (30 × 300 × 400) = 0.199


From the chart, (As Fyk)/(bhfck) = 0.23

Area of longitudinal steel required (As) = (0.23 × 30 × 400 × 300)/460 = 1800 mm2

As,min = 0.10 NEd/fyd = (0.1 × 716.88)/400 = 0.179 mm2 < 0.002 × 400 × 300 = 240 mm2
As,max = 0.04bh = 4800 mm2

Provide 4Y25mm (Asprov = 1964 mm2)



Design Moments (z-direction)
Z - direction: M01 = 35.52 KNm; M02 = -25.269 KNm

e1 is the geometric imperfection = (θi l0/2) = (1/200) × (2675.293/2) = 6.688 mm

Minimum eccentricity e0 = h/30 = 400/30 = 13.333 mm. Since this is less than 20mm, take minimum eccentricity = 20mm (clause 6.1(4) EC2)

Minimum design moment = e0NEd = 20 × 10-3 × 716.88 = 14.3376 KNm

First order end moment M02 = MTop + eiNEd

eiNEd = 6.688 × 10-3 × 716.88 = 4.794 KNm

M02 = MTop + eiNEd = 35.52 + 4.794 = 40.314 KNm

Longitudinal Steel Area
d2 = Cnom + ϕ/2 + ϕlinks = 40 + 12.5 + 10 = 62.5 mm
d2/h = 62.5/400 = 0.156

Reading from chart No 1; d2/h = 0.156;
MEd/(fck bh2) = (40.314 × 106)/(30 × 300 × 4002) = 0.0279
NEd/(fck bh) = (716.88 × 103) / (30 × 300 × 400) = 0.199

From the chart, (As Fyk)/(bhfck) = 0.00 (Nominal reinforcement required)

Biaxial Effects


Check if λy / λz ≤ 2.0 and λzy ≤ 2.0

17.203/18.159 = 0.9473 < 2.0, and 18.159/17.203 = 1.0556 < 2.0

Furthermore, let us also check;
(ey/heq) / (ez/beq) ≤ 0.2 or (ez / beq) / (ey / heq) ≤ 0.2

The definition of eccentricity is given in Figure 5.8 of EC2


ey = MEd,y/NEd = (180.624 × 106) / (716.88 × 103) = 251.958 mm
ez = MEd,z/NEd = (40.314 × 106) / (716.88 × 103) = 56.235 mm

heq = iz.√12 = 300 mm
beq = iy.√12 = 400 mm

(ey/heq) ÷ (ez/beq) = 251.958/300 ÷ 56.235/400 = 5.9739 > 0.2

Therefore we have to check for biaxial bending interaction;

(ME,dz/MR,dz )a + (MEd,y/MRd,y )a ≤ 1.0

(As Fyk)/(bhfck) = (1964 × 460) / (30 × 300 × 400) = 0.251

Therefore from the chart; MRd/(fckbh2) = 0.13

MRd = (0.13 × 30 × 300 × 4002) × 10-6 = 187.2 KNm

NRd = Acfcd + Asfyd
NRd = [(300 × 400 × 17) + (1964 × 400)] × 10-3 = 2825.6KN
NEd/NRd = 716.88 / 2825.6 = 0.2537

To evaluate the value of a, let us look at the table below as given in Clause 5.8.9(4) of EC2


By linear interpolation, a = 1.0 + [(0.2537 – 0.1 )/(0.7 – 0.1)] × (1.5 - 1.0) = 1.128

 (ME,dz/MR,dz )a + (MEd,y/MRd,y )a ≤ 1.0

(40.314/187.2)1.128 + (180.624/187.2)1.128 = 0.1769 + 0.9606 = 1.1375 > 1.0. This is not ok, and this shows that 4Y25 is inadequate for the biaxial action on the column.

Let us increase the area of steel to 6Y25mm (Asprov = 2946 mm2)

Let us check again;
(As Fyk)/(bhfck) = (2964 × 460) / (30 × 300 × 400) = 0.3787
Therefore from the chart; MRd/(fckbh2) = 0.175
MRd = (0.175 × 30 × 300 × 4002) × 10-6 = 252 KNm

NRd = Acfcd + Asfyd
NRd = [(300 × 400 × 17) + (2964 × 400)] × 10-3 = 3225.6 KN

NEd/NRd = 716.88 / 3225.6 = 0.222

By linear interpolation, a = 1.0 + [(0.222 – 0.1 )/(0.7 – 0.1)] × (1.5 - 1.0) = 1.102

(ME,dz/MR,dz )a + (MEd,y/MRd,y )a ≤ 1.0

(40.314/252)1.102 + (180.624/252)1.102 = 0.132 + 0.6928 = 0.825 > 1.0.

This shows that 6Y25mm is adequate for the column, and shows some agreement with result from BS 8110-1:1997.



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3 comments:

  1. Dear Ubani,

    I was doing a research about this topic and I came across your blog.

    I'm trying to understand the right combination of design moments for biaxial check.

    As you mentioned EC2 states that the imperfections need to be considered only in the direction where they will have the most unfavourable effect, Cl 5.8.9(2)

    Now, since I'm dealing with high axial force and relative small bending moment(flat slab) the minimum design moment due to eccentricity Cl 6.1(4) dictates the design.

    At the biaxial bending moment check which combination would you consider;

    _Minimum moment due to eccentricity in both axis.
    _Minimum moment due to eccentricity in the unfavourable axis and moment from analysis in the other one.
    _Minimum moment due to eccentricity in the unfavourable axis and moment from imperfection in the other one.

    I noticed that there is confusion about this aspect and because I've just started working with a new software which consider the third option I want to find out the right procedure.

    Best Regards

    Alessio

    ReplyDelete
  2. • It is appropriate time to make a few plans for the future and it is time to be happy. I have learn this publish and if I may just I desire to recommend you some attention-grabbing things or tips. Perhaps you can write next articles relating to this article. I want to read more issues about it!.

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  3. Dear Ubani,
    is there a typographic error in writing the formula for effective length (lo)?
    I think, it should be 0.5*L*sqrt((1+K1/(0.45+K1))*(1+K2/(0.45+K2))) than of (1+K1)/(0.45+K1) or (1+K2)/(0.45+K2)

    ReplyDelete