Journal of Stability v2

PHAN SOPHEA CIVIL ENGINEERING JOURNAL CIVIL ENGINEERING September 2015 [email protected]

1 | P a g e

Journal of Stability: Comparison study between ACI and BS EN1992-1-1 (Eurocode 2)

Sway and Non-sway effects | Global and local second order effect

Slenderness effect | local second order effect

Stability is always among the important aspect of analyses of a building. But it is also the main concern for structural designer. This paper aims to illustrate how to check stability or taking into account its effect in building analysis by trying to compare two main international codes (ACI and Eurocode).

The first word that you will hear from stability problem is "sway" or "non-sway" structure. ACI uses "sway" or "non-sway" to describe the stability while Eurocode on the other hand uses "Global second order effect" referred as "sway" effect. If a structure is classified as sway, the global second order effect cannot be ignored. Otherwise, if it is classified as non-sway, the global second order effect can be ignored. So what is sway or global second order effect?

Global second order effect is the effect that taking into account the deformation of a structure in analysis. Usually, global second order effect is well-known called P-delta effect (P*∆). When a structure is loading by gravity load dead, live named P with or without lateral load, the structure will be deformed vertically and horizontally. The horizontal deformation is ∆. The P multiple by ∆ which is P-Big delta effect (P*∆) becomes the additional moment to the first order moment in elastic analysis for design. This is P*∆ is called global second order since it is deformed of the whole structures. On the other hand, each compression members also deform locally (δ). This will create another second order effect which is called local second order effect (P-small delta: P*δ).

According to ACI, the horizontal deformations (∆) are calculated as story to story drift. Hence, there are some floors are sway sensitive and some are not depending on each story's stability index. On the other hand, according to Eurocode, there is no concept of story drift, ∆ is considered as the overall deformation of the structure. The whole building is classified only once whether the global second order effect cannot be ignored or has to be considered in analysis.

Some definitions and important key terms:

Braced members or systems: structural members or subsystems, which in analysis and design are assumed not to contribute to the overall horizontal stability of a structure.

Bracing members or systems: structural members or subsystems, which in analysis and design are assumed to contribute to the overall horizontal stability of a structure.

Buckling: failure due to instability of a member or structure under perfectly axial compression and without transverse load

Note. “Pure buckling” as defined above is not a relevant limit state in real structures, due to imperfections and transverse loads, but a nominal buckling load can be used as a parameter in some methods for second order analysis.

Buckling load: the load at which buckling occurs; for isolated elastic members it is synonymous with the Euler load.

Effective length: a length used to account for the shape of the deflection curve; it can also be defined as buckling length, i.e. the length of a pin-ended column with constant normal force, having the same cross section and buckling load as the actual member.


First order effects: action effects calculated without consideration of the effect of structural deformations, but including geometric imperfections

Sway: a building is sway sensitive when the lateral deformation cannot be ignored in analysis; otherwise it is non-sway.

How to classify as sway or non-sway? ACI

How to check whether the global second order effect cannot be ignored? Eurocode

Base on ACI 10.10.5.2, a story within a structure is assumed as non-sway if the stability index Q is small or equal to 0.05.

Load combination:

Base on BS EN1992-1-1 art 5.8.2 (6), the global second order effect can be ignored if it is less than 10% of the corresponding first order effects. This principle can be used for all types of structural both braced or unbraced.

Here, EUROCODE does not have a stability index formula like ACI; however, the art 5.8.2 (6) does represent the global second order effects can be ignored if it is less than 10%. Hence, it is stability index Q of overall building which is smaller or equal to 10% (Q ≤0.10) if global second order can be ignored.

On the other hand, if the structure is braced, Eurocode has an alternative method which is specified in BS EN1992-1-1 art 5.8.3.3 (1).


An alternative method can be done to check whether the building is sway or non-sway sensitive is to do the buckling analysis. The result of bucking analysis is amplification factor (λ) where λ = 1/Q. Hence, for ACI, the structure is considered as non-sway if λ≥20. Otherwise it is sway. For EUROCODE, the global second order effect can be ignored if λ≥10. Otherwise it has to take in to account in analysis (read below how to take it into account in analysis).

What is braced and unbraced structure/members?

In reality, there is no structure fully braced or unbraced. However both codes emphasize a structure as braced when there are bracing elements which have sufficient lateral rigidity such as shear wall, core walls, stair case, and masonry wall. In this kind of structure, columns do not participate in bracing the structure, hence are considered as braced ("braced-column"). Those braced columns can be designed to received gravity loads only while the lateral load such as wind load will be transferred to bracing elements though the rigid slab diaphragm. On the other hand, if a structure does not have bracing elements with sufficient lateral rigidity as mentioned above, columns themselves are considered bracing elements. Those columns take part in stabilize the structure, hence called "un-braced columns" or "bracing columns". In this un-braced structure, the columns are designed to receive both gravity and lateral loads.

Braced members or systems: structural members or subsystems, which in analysis and design are assumed not to contribute to the overall horizontal stability of a structure.

Bracing members or systems: structural members or subsystems, which in analysis and design are assumed to contribute to the overall horizontal stability of a structure.

How to take into account sway effect if it has too? ACI

How to take into account global second order effect if it has too? Eurocode

Base on ACI 10.10.7, the sway effect has to be taken into account by amplifying the sway moment (Ms). The sway moments (M1s or M2s) are moment resulting from sway tendency load such as lateral loads. The non-sway moment (M1ns or M2ns) are moment resulting from non-sway tendency load such as gravity loads. The design moment (M1 or M2) are the sum of non-sway moment and sway moment at end 1 and 2 of column.


Why no local amplification should be applied to the non-sway moments in ACI?

According to Nilson, Darwin, et Dolan (2010):

According to ACI 10.10.7.3, the sway amplification factor is calculated as:

If δs exceeds 1.5, the alternative formula in 10.10.7.4 shall be used:


Base on BS EN1992-1-1 Annex H.2, equation (H.7) represent the amplification of horizontal force (FH,Ed) to take into account the global second order effect and is written as:

The ratio of total vertical load on bracing and braced members to nominal global buckling load is equal to the stability index Q as described about.

FH,Ed = FH,0Ed / (1 - Q) ≥ FH,0Ed

After the horizontal load is amplified, the design of unbraced columns are the same to braced column with respective buckling lengths for unbraced and braced condition. (read below to design a braced column).

There are also second possibility to take into account the second order effect by doing a P-delta analysis which is a non-linear analysis using sophisticated commercial programs like advance design, robot, sap, etabs, or staad pro,... etc. However, this usually a time consuming analysis.

On the other hand, there are also third option to take into account or not global second order effect. It is add bracing system to get the structure non-sway sensitive, so that global second order effect can be ignored.


How effective length of column is calculated in a braced and unbraced structure?

Effective length is a length used to account for the shape of the deflection curve; it can also be defined as buckling length, i.e. the length of a pin-ended column with constant normal force, having the same cross section and buckling load as the actual member

According ACI fig. R10.10.1.1, the effective length (k.Lu) is calculated from effective length factors (k) multiply by unsupported length (Lu) of the column. The effective length factors (k) is written from ACI Fig R10.10.1.1:


According to BS EN1992-1-1 art 5.8.3.2 (3), the effective length (L0) can be calculated as:

θ/M = 1 / [4*(EI/L)b] for a beam with fix end (Fig. a) => k=(EI/L)c / [4*(EI/L)b]

θ/M = 1 / [3*(EI/L)b] for a beam with pinned end (Fig. b) => k=(EI/L)c / [3*(EI/L)b]

(EI/L)c : rigidity of columns ; (EI/L)b: rigidity of beams

It is worth noting that in ACI, effective length is smaller than unsupported length in a no sway structure and bigger than unsupported length in a sway structure. However, in Eurocode, the effective length is smaller than column length if it is braced otherwise it is bigger than column length. Eurocode calculate effective length of a column according to its condition as braced or unbraced members.

What is slenderness effects? ACI

What is local second order effect? Eurocode

Slenderness effects and local second order effects are P*δ (P*small delta). P*δ effect is local effect in it sense that column is buckled to load locally.

According to ACI 10.10.1, slenderness effects [column is short] can be ignored if it smaller than the limit as described below:


In non-sway structure, if the column has a slenderness effect that can be ignored (klu/r ≤22) , the column is designed using the end moment M2. Otherwise, if slenderness effect cannot be ignored, the design moment is Mc=δns M2.

In non-sway structure, if the column has a slenderness effect that can be ignored (klu/r ≤34-12*(M1/M2)) , the column is designed using the end moment M2=M2ns + M2s. Otherwise, if slenderness effect cannot be ignored, the design moment is M2=M2ns + δn M2s.

According to BS EN1992-1-1 art 5.8.3.1 (1), the slenderness effects may be ignored if the slenderness λ is below a certain value λlim.

Local second order effect (P*δ) can be ignored if the slenderness of the column is smaller than the limit λlim according to its condition as braced or unbraced column. If local second order effect can be ignored, the columns is design with the compression bending Ned and moment M0,Ed including geometry imperfection. On the other hand, if slenderness of the column is bigger than the limit λlim , local second order effect cannot


be ignored and final total design moment can be calculated from three methodes specify in BS EN1992-1-1 art 5.8.5 and 5.8.6:

1) General method based on non-lineaire analysis: the calculation is complex it need a sofisticated softwares or computer run. Sometimes, it is called P*delta analysis in some commercial softwares. This text will not cover this method. However, the next two methods will be disussed in details.

2) Method based on nominal stiffness, design moment M is: EI → NB → M=M0Ed [1+ β /(NEd / NB -1)]

Remark: this method is used to design a braced members in a braced structure or design an bracing wall (please note that L0 of wall is smalle than 1). This second method is applicable to desig members of a structure braced by shear wall where global bending effect gavornsd (see figure below on the left). On the other hand if global shear deformation is important such as an unbraced frame or a system of shear walls with grand opening (see figure below on the right), the second method is no longe applicable. It comes application of the third method based on noinbal curvature where all type of structure regardless global bending and shear deformation in nature.

3) Method based on nominal curvature, design moment M is: L0 → e2 → M=M0,Ed + NEd* e2

As discuss in remark above; the third method based on nomial curvature is used to design columns or walls in both braced and unbraced conditions. On the other hand, the second method based on nominal stiffness can be used to desin only a braced column, but not design an unbraced column. However, if it is an unbraced column but with b≤4*h, Hence a wall (bracing wall), second method is again applicable. It is because the global bending govens rather than shear deformation when the column is a wall. That is why we disginguish column from wall regardless their similarity in nature and design.

How to design a column for non-sway frames? ACI moment magnifier method

How to design a column for braced and non-braced frames? Eurocode 2: (method based on nominal stiffness and method based on nominal curvature)

According to ACI 10.10.6, the design moment in a non-sway frame is Mc = δ*M2 and summarized as below:


According to BS EN1992-1-1 5.8.7.3, the total design moment, including [local] second order moment, may be expressed as a magnification of the bending moments resulting from linear analysis, namely:



According to BS EN1992-1-1 5.8.8, the total design moment, including [local] second order moment, may also be expressed as a sum of the bending moments resulting from linear analysis and second order moment M2, namely:

It is another worth noting that ACI amplifies the maximum end moment M2 by the amplified factor δ to get the total design moment Mc. Different to ACI, Eurocode do not use the maximum end moment for amplification, but the moment equivalent M0e. According to BS EN1992-1-1 at 5.8.8.2 (2), M0e is written:

However, the final total design moment is the maximum of (MEd , M02 , M01 + 0.5*M02) where MEd depend on method used:

Or


Also, as mentioned above, Eurocode do not distinguish different method to design a braced and unbraced column. The design step of the two kind of column is the same by using either method based on nominal stiffness or method based on nominal curvature, but with different effective length condition and different slenderness limit (λlim). On the other hand, ACI clearly separates the design procedure according to condition of sway and non-sway frames.


Numerical application: .(Nilson, Darwin, et Dolan, 2010)




Use the same input in example 9.1, but design column C3 according to EC2.

As described in the example 9.1, column is braced against sideway. L0 is calculated by:




Since the column C3 is in the middle of frame, beams are fix at their end, then k=(EI/L)c / [4*(EI/L)b]

k1=(EI/L)c / [4*(EI/L)b] ; k2=(EI/L)c / [4*(EI/L)b]

Calculate k1: Beams: Rotational stiffness of the two beams left and right:

L=7.3m, section 1.20x0.30 m2

Inertia: I = (1.2x0.33)/12 = 2.7x10-3 m4

2x4x(I/L)b = 2x4x2.7x10-3 / 7.3 = 2.96x10-3

m3

column: Rational stiffness of the top column and design column:

L=3.95m, section 0.45x0.45m2

Inertia: I = (0.45*0.453 /12) = 3.417x10-3 m4

2x(I/L)c = 2 x 3.417x10-3 / 3.95 = 1.73x10-3 m3

k1 = (EI/L)c / [4*(EI/L)b] = 1.73x10-3 / (2.96x10-3) = 0.592

calculate k2: since the section of beams and columns at each floor are the same. k2=k1=0.585

k11

k2

C3


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Hence, the effective length is calculated as:

L0 = 0.5 x L x sqrt { [1+k1/(0.45+k1)] x[1+k2/(0.45+k2)] } = 0.5*Lx sqrt(2x1.568) = 0.784xL

L0 = 3.09m

It is reminded ACI found k=0.783 while EC2 found 0.88. It is a good match!

Primary moments and axial load on column C3 according to EC2: at ULS: 1.35*D+1.5*L Geometric imperfection:

Effective length L0 = 3.49m, actual length L=3.95m

αh = 2 / sqrt(L) = 2 / sqrt (3.95) = 1.006, but 2/3≤αh ≤1, then take αh = 1

θL = θ0 αh αm = (1/200) x 1 x 1 = 5x10-3

ei = θLx L0 / 2 = 5x10-3 x 3.49 /2 = 8.725x10-3 = 8.72 mm

also EC2 art 6.1(4), ei = max(20mm, h/30, ei , L0/400) = max(20mm, 450/30, 8.72mm , 3.09/400)

= max(20mm, 15mm, 8.72mm, 7.65mm) = 20mm

Moment resulting from imperfection: Mei = NEd ei = 2454 x 20x10-3 = 49. kN.m

At ULS, design moment including geometric imperfection:

NEd= 1.35x990 + 1.5x745 = 2454 kN.

M02 = 1.35x3 + 1.5x146 + NEd ei= 1.35x3 + 1.5x146 + 49 = 273 kN.m

M01 = 1.35x(-3) + 1.5x135 + NEd ei = 1.35x(-3) + 1.5x135 + 49 = 247 kN.m

According BS EN1992-1-1 art 5.8.8.2 (2), the difference first order end moment M01 and M02 may be replaced by an equivalent first order end moment M0e.

M0Ed = M0e = 0.6 M02 + 0.4 M01 = 0.6x273 + 0.4x247 = 262 kN.m > 0.4 M02 = 0.4x273 = 109 kN.m OK!

Verification if the local second order [column is short] effect can be ignored if λ ≤ λlim : λ = L0 / i where L0 = 0.88L = 3.49m and raduis of gyration of the unbracked section: i = sqrt ( I / A )

i = sqrt[ b*h3/12 / (b*h) ] = h / sqrt(12)

=> λ = L0 *sqrt(12) / h = 3.09 x sqrt(12) / 0.45 = 23.80 compared to ACI (29.3)

Effective creep coefficient calculation:

Age of loading, t0 = 28days (assumed)

Assume all four sides are exposed o atmosphere

u, perimeter of the column = 2x(450 + 450) = 1800mm

Ac, area of concrete in the cross section = 450x450 = 2025x102 mm2\

h0 = 2 Ac / u = 2x2025x102 / 1800 = 225mm

Assuming class S cement, fck = 25MPa, relative humidity RH=50% from

Nqp0 = 1x990 + 0.3x745 = 1213.5 kN

Mqp0= 1x0.3 + 0.3x146 = 44.1 kN.m

e1 = e0+ei = Mqp0 / Nqp0 + ei = 44.1 / 1213.5 + 0.02 = 0.056m

Nqp1 = Nqp0 = 1213.5 kN

Mqp1 = Nqp1*e1 = 1213.5 x 0.066 = 71.07 kN.m

And M0Ed = M0e = 2 kN.m


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Φef = Φ(∞,t0 ) x (M0Eqp / M0Ed) = Φ(∞,t0 ) x (71.07 / 262)

Φ(∞,t0 ) =φRH * β(fcm) * β(t0)

φRH=1+(1-RH/100) / [0.1*h0 ^(1/3) ] = 1+(1-50/100) / [0.1*225 (̂1/3)] = 1.82

β(fcm)= 16.8 / sqrt(fck+8) = 16.8 / sqrt(28+8) = 2.80

β(t0) = 1 / (0.1 + t00.2) = 1 / (0.1+280.2) = 0.49

Considering concrete loading at t0 = 28 days

Φ(∞,t0 ) =φRH * β(fcm) * β(t0) = 1.82 x 2.80 x 0.49 = 2.492

Φef = Φ(∞,t0 ) x (M0Eqp / M0Ed) = 2.482 x 71.07 / 262 = 0.68

λlim = 20*A*B*C / sqrt(n) where n = NEd / (Ac fcd) = 2454x10-3 MN / (0.45x0.45x28/1.5) = 0.649

Calculate A: A = 1 / (1+0.2 Φef) where Φef = 0.68

A = 1 / (1+0.2x0.68) = 0.88

Calculate B B = sqrt (1+2ω) = sqrt [ 1+2*As*fyd / (Ac*fcd) ]

As = 4HA32 + 4HA29 = 58.56 cm2; fyd = fyk / 1.15 = 420 / 1.15 (Mpa) = 365 Mpa Ac = 45x45cm2 = 2025 cm2; fcd = fck /1.5 = 28/1.5 Mpa = 18.67 Mpa

B = sqrt (1+2x58.56x365 / (2025x18.67)] = 1.46

Calculate C C = 1.7 - rm = 1.7 - 0.90 = 0.80

rm = M01 / M02 = 247 / 273 = 0.90

Calculate n n = Ned / (Ac*fcd) = 2454x10-3 / (0.45x0.45x28/1.5) = 0.65

λlim = 20*A*B*C / sqrt(n) = 20 x 0.88 x 1.46 x 0.80 / sqrt(0.65) = 25.28 > λ = 23.80 m. Thus, the local second

order effect can be ignored. This λlim compares to ACI (23.3) < λ = 25.5, slenderness effect cannot be ignored.

Hence, since slenderness limit in Eurocode 2 is bigger than column's slenderness, local second order effect

can be ignored and column shall be design using (NEd =2454 kN, MEd = M02 = 272 kNm),

Compression bending design (Jean-Pierre, 2006): inputs: fck=28Mpa, fyk=420Mpa, b=0.45m, h=0.45m, c=0.05, NEd =2454 kN, MEd = M02 = 272 kNm

results: d=0.40m, d' = 0.05m, Nbmax = 0.85 b h fcd = 0.85 x 0.45 x 0.45 x 28 / 1.5 = 3.213 MN

ψ1 = NEd / Nbmax = 2.454 / 3.213 = 0.764 thus smaller than 0.81

and ψ1 = 0.764 > 2 / 3=0.667 thus

ζ = (3*ψ1 - 1)*(1-ψ1) / (4*ψ1) = (3 x 0.764 - 1) x (1 - 0.764) / (4 x 0.764) = 0.0998

eNC = ζ x h = 0.0998 x 0.45 = 0.045 m < e = MEd / NEd = 272 / 2454 =0.111m

Thus, the compression bending has the section as tension-compression.

Calculation of tension-compression section (Jean Roux, 2009): Pure bending design

MuA = MEd + NEd (d - h/2) = 272 - 2454 x (0.40 - 0.45 / 2) = 702 kN.m

μlim = 0.3717 considering column is class XC1

μcu = MuA / (b d2 fcd) = 702 / (0.45 x 0.402 x 28 / 1.5) = 0.522 > μlimi

Thus, compresison rebar is need (A's).

Find A's

Mlu = μlim b d2 fcd = 0.3171 x 0.45 x 0.402 x 28 / 1.5 = 499.6 kN.m


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α1 = 1.25 (1 - sqrt(1 - 2 μlim ) = 1.25 (1 - sqrt( 1 - 2 x 0.3171)) = 0.62

εsc = 3.5 /1000 (α1 - d'/d) / α1 = 3.5 / 1000 x (0.62 - 0.05 / 0.4) / 0.62 = 2.79x10-3

σ'sc = min(fyd / 1.15 , εsc Es) = (420 / 1.15, 2.79x10-3 x 2x105 = min (365 , 558)

σ'sc = 365 Mpa

σ'sc2 = min [ fyk / 1.15 , 9γ fck - f1 x d'/d x (0.5 / 15 + 13) fck + (6517 / 15 + 1) ]

f1 = 0.9 for fyk <=400 otherwise, f1 =1

σ'sc2 = min [ fyk / 1.15 , 9γ fck - 1 x d'/d x (0.5 / 15 + 13) fck + (6517 / 15 + 1) ]

σ'sc2 = min (365 , 253) Mpa = 253 MPa

εyd = fyd / Es = 420 / 1.15 / (2x105) = 1.83x10-3 Mpa < εsc = 2.79x10-3

If εsc > εyd , then σ's2 = σ'sc otherwise σ's2 = σ'sc2

=> in our case σ's2 = σ'sc = 365 Mpa

A's = (MuA - Mlu) / [ (d - d') σ's2 ] = (702 - 499.6)x10-3 / [(0.40 - 0.05) x 365 ]

A's = 15.80cm2 Find As for pure bending design

μcu = min (μcu , μlim ) = min (0.522 , 0.3717) = 0.3717

αcu = 1.25 ( 1 - sqrt(1 - 2 μcu) = 1.25 (1 - sqrt( 1 - 2 x 0.3717)) = 0.616

zc = d (1 - 0.4 αcu ) = 0.40 x (1 - 0.4 x 0.616 ) = 0.3013

zc ≈ 0.75d when A's <> 0

As = Mlu / (zc fyd ) + A's σ's2 / fyd = 499.6x10-3 / (0.3013 x 365) + 15.80x10-4 x 365 / 365

As = 45.40 + 15.80 = 61.2 cm2

Find As final for compression bending:

As = As bending - NEd / fyd = 61.2 - 2454x10-3x104 / 365 = -5.99 cm2 < 0

As = As mini = max (0.26 fctm / fyk b h , 0.0013 b d) = 3.47 cm2

Thus, the final rebar to put in column is:

As = 2 x max(A's , As) = 2 x max (15.8 , 3.47) = 31.6 cm2

Since As = 31.6 cm2 < As initial, = 58.56 cm2, there are two possibilities:

1) if we want to refine the As for more economic design purpose, we may need to restart

the calculation with a new As initial = 31.6 cm2 and follow all the steps above to find a

new As and compare again to As initial = 31.6cm2. If the former is smaller than the

former with small difference, we can stop calcualtion and take As = As initial = 31.6cm2.

However, if As found is bigger than As initial, thus, the third step is needed by again

taking As initial = As. This iteration is end when As found equal to As initial. This need help

from computer. See annex A for excel calculation, and the final As for column

accoridng to Eurocode 2 is As = 33.4cm2 compared to ACI rebar = 58.56.

2) if we just want to verify the stability of the column with As initial = 58.56cm2, it is

sufficient to obtain: As < As,initial . Then the stability of columns with As = 54.68cm2 is

satisfied.


It is to keep in mind that if As initial is changed and lower than 58.56 in this example, this will

result in lower slenderness limit λlim then, somehow λ might be bigger than λlim. Thus, the

local second order effect cannot be ignored. If it is the case, before we design a

compression bending, we need to amplify the design moment by nominal stiffness or

nominal curvature method.

In conclusion, the results are summarized as below: EC2 ACI

As 33.4cm2 54.68cm2 nominal stiffness EI: 20.71 MN.m2 22.66 MN.m2 Critical buckling load NB: 21.39 MN 18.9 MN Amplified factor δ: 1.16 1.15

In a braced frame, both codes give the same amplified factor δ to get total design moment. Thus the

method based on nominal stiffness in EC2 is comparable to the amplification method in ACI 10.10.6.

EC2 ACI




Use the same input in example 9.2, but design column C3 according to EC2.

As described in the example 9.1, column is braced against sideway. L0 is calculated by:




Since the column C3 is in the middle of frame, beams are fix at their end, then k=(EI/L)c / [4*(EI/L)b]

k1=(EI/L)c / [4*(EI/L)b] ; k2=(EI/L)c / [4*(EI/L)b]


Calculate k1: Beams: Rotational stiffness of the two beams left and right:

L=7.3m, section 1.20x0.30 m2

Inertia: I = (1.2x0.33)/12 = 2.7x10-3 m4

2x4x(I/L)b = 2x4x2.7x10-3 / 7.3 = 2.96x10-3

m3

column: Rational stiffness of the top column and design column:

L=3.95m, section 0.45x0.45m2

Inertia: I = (0.45*0.453 /12) = 3.417x10-3 m4

2x(I/L)c = 2 x 3.417x10-3 / 3.95 = 1.73x10-3 m3

k1 = (EI/L)c / [4*(EI/L)b] = 1.73x10-3 / (2.96x10-3) = 0.584

calculate k2: since the section of beams and columns at each floor are the same. k2=k1=0.584

Hence, the effective length is calculated as:

L0 = Lx max{ sqrt { [1+10 k1 k2 / (k1+k2)] ; ( (1+k1/(1+k1) ) ( 1+k2/(1+k2) ) }

L0 = 3.95 max (1.98, 1.88) = 1.98 x 3.95 = 7.821m

It is reminded ACI found k=1.64 while EC2 found 1.98. It is a good match!

Primary moments and axial load on column C3 according to EC2: at ULS: 1.35*D+1.5*L Geometric imperfection:

Effective length L0 = 7.821m, actual length L=3.95m

αh = 2 / sqrt(L) = 2 / sqrt (3.95) = 1.006, but 2/3≤αh ≤1, then take αh = 1 θL = θ0 αh αm = (1/200) x 1 x 1 = 5x10-3

ei = θLx L / 2 = 5x10-3 x 7.821/2 = 19.55x10-3 = 19.55 mm

ei > min(20mm, h/30) = min(20mm, 450/300) = min(20mm, 15mm) = 20mm

Take ei = 20mm

Moment resulting from imperfection: Mei = NEd ei = 2454 x 20 x10-3 = 49 kN.m

Moments including geometric imperfection:

NEd= 1.35x990 + 1.5x745 + 0.9x2.6 = 2477 kN. M02 = 1.35x3 + 1.5x146 + 0.9x102 + 49 = 364 kN.m

M01 = 1.35x(-3) + 1.05x135 + 1.5x(-92) + 49 = 165 kN.m

k11

k2

C3


26 | P a g e

M0e = 0.6 M02 + 0.4 M01 = 0.6x364+ 0.4x165= 285 kN.m > 0.4 M02 = 0.4x364=145.6 kN.m OK!

M0Ed = 285 kN.m

Verification if the local second order [column is short] effect can be ignored if λ ≤ λlim : λ = L0 / i where L0 = 0.88L = 3.49m and raduis of gyration of the unbracked section: i = sqrt ( I / A )

i = sqrt[ b*h3/12 / (b*h) ] = h / sqrt(12)

=> λ = L0 *sqrt(12) / h = 7.821 x sqrt(12) / 0.45 = 60.23 compared to ACI (48)

Effective creep coefficient calculation:

Age of loading, t0 = 28days (assumed)

Assume all four sides are exposed o atmosphere

u, perimeter of the column = 2x(450 + 450) = 1800mm

Ac, area of concrete in the cross section = 450x450 = 2025x102 mm2\

h0 = 2 Ac / u = 2x2025x102 / 1800 = 225mm

Assuming class S cement, fck = 25MPa, relative humidity RH=50% from

Nqp0 = 1x990 + 0.3x745 = 1213.5 kN

Mqp0= 1x0.3 + 0.3x146 = 44.1 kN.m

e1 = e0+ei = Mqp0 / Nqp0 + ei = 44.1 / 1213.5 + 0.02 = 0.056m

Nqp1 = Nqp0 = 1213.5 kN

Mqp1 = Nqp1*e1 = 1213.5 x 0.066 = 71.07 kN.m

And M0Ed = M0e = 285 kN.m

Φef = Φ(∞,t0 ) x (M0Eqp / M0Ed) = Φ(∞,t0 ) x (71.07 / 285) according to BN EN1992-1-1 annex B

Φ(∞,t0 ) =φRH * β(fcm) * β(t0)

φRH=1+(1-RH/100) / [0.1*h0 ^(1/3) ] = 1+(1-50/100) / [0.1*225 (̂1/3)] = 1.82

β(fcm)= 16.8 / sqrt(fck+8) = 16.8 / sqrt(28+8) = 2.80

β(t0) = 1 / (0.1 + t00.2) = 1 / (0.1+280.2) = 0.49

Considering concrete loading at t0 = 28 days

Φ(∞,t0 ) =φRH * β(fcm) * β(t0) = 1.82 x 2.80 x 0.49 = 2.492

Φef = Φ(∞,t0 ) x (M0Eqp / M0Ed) = 2.482 x 71.07 / 285= 0.62

λlim = 20*A*B*C / sqrt(n) where n = NEd / (Ac fcd) = 2454x10-3 MN / (0.45x0.45x28/1.5) = 0.649

Calculate A: A = 1 / (1+0.2 Φef) where Φef = 0.62

A = 1 / (1+0.2x0.51) = 0.89

Calculate B B = sqrt (1+2ω) = sqrt [ 1+2*As*fyd / (Ac*fcd) ]

As = 4HA32 + 4HA29 = 58.56 cm2; fyd = fyk / 1.15 = 420 / 1.15 (Mpa) = 365 Mpa Ac = 45x45cm2 = 2025 cm2; fcd = fck /1.5 = 28/1.5 Mpa = 18.67 Mpa

B = sqrt (1+2x58.56x365 / (2025x18.67)] = 1.46 Calculate C C = 1.7 - rm = 0.7 for unbraced column

Calculate n n = Ned / (Ac*fcd) = 2493x10-3 / (0.45x0.45x28/1.5) = 0.66


27 | P a g e

λlim = 20*A*B*C / sqrt(n) = 20 x 0.89x 1.46 x 0.7 / sqrt(0.66) = 22.45 < λ = 60.23 m. Thus, the local second

order effect (slenderness effect) cannot be ignored. This λlim compares to ACI (23.3).

Since the column in this second example is unbraced, amplification method is not applicable. Curvature method is used next:

Estimation of curvature

1/r= 0.00836 m-1 =kr*kΦ*1/r0

d= 0.40 = h-c

1/r0= 0.01014 =εyd / (0.45 d)=fyd / Es / (0.45d)

Kr= 0.78093 =(nu - n) / (nu -nbal ) <=1

n= 0.6554 =NEd / (Ac fcd )

ω= 0.5658 =(As fyd ) / (Ac fcd ) = 58.56x10-4 x 365 / (0.45 x 0.45 x 28 / 1.5) = 0.5658

nu= 1.5658 =1+ω = 1 + 0.5658 = 1.5658

nbal= 0.4 (for rectangular section)

KΦ= 1.05502 = 1+β*Φef = 1 + 0.08845 x 0.62 >= 1 OK!

β= 0.08845 = 0.35 + fck /200 - λ / 150

Φef= 0.62 (see calculation above)

Curvature e2:

e1 = 0.115 m = MEd / NEd

e2 = 0.052 = 1/r*L02 / c = 0.00836 x 7.8212 / 9.87

e total et = 0.167 m = e1 + e2

c= 9.87 =^2

M2 = 128.4 kN.m = NEd et

Total moment: MEd = M0Ed + M2 = 128.4 + 28.47 = 413.2 kN.m

Since local second order effect can be ignored and column shall be design using:

(NEd =2477 kN, MEd = 413.2 kNm),

Compression bending design (Jean-Pierre, 2006): inputs: fck=28Mpa, fyk=420Mpa, b=0.45m, h=0.45m, c=0.05, NEd =2477 kN, MEd = 413.2 kNm

results: d=0.40m, d' = 0.05m, Nbmax = 0.85 b h fcd = 0.85 x 0.45 x 0.45 x 28 / 1.5 = 3.213 MN

ψ1 = NEd / Nbmax = 2.447 / 3.213 = 0.771 thus smaller than 0.81

and ψ1 = 0.771 > 2 / 3=0.667 thus

ζ = (3*ψ1 - 1)*(1-ψ1) / (4*ψ1) = (3 x 0.771 - 1) x (1 - 0.771) / (4 x 0.771) = 0.0975


28 | P a g e

eNC = ζ x h = 0.0975 x 0.45 = 0.044 m < e = MEd / NEd = 413.2 / 2477 =0.167m

Thus, the compression bending has the section as tension-compression.

Calculation of tension-compression section (Jean Roux, 2009): Pure bending design

MuA = MEd + NEd (d - h/2) = 413.2 - 2477 x (0.40 - 0.45 / 2) = 847 kN.m

μlim = 0.3717 considering column is class XC1

μcu = MuA / (b d2 fcd) = 847 / (0.45 x 0.402 x 28 / 1.5) = 0.630 > μlimi

Thus, compression rebar is need (A's).

Find A's

Mlu = μlim b d2 fcd = 0.3171 x 0.45 x 0.402 x 28 / 1.5 = 499.6 kN.m

α1 = 1.25 (1 - sqrt(1 - 2 μlim ) = 1.25 (1 - sqrt( 1 - 2 x 0.3171)) = 0.62

εsc = 3.5 /1000 (α1 - d'/d) / α1 = 3.5 / 1000 x (0.62 - 0.05 / 0.4) / 0.62 = 2.79x10-3

σ'sc = min(fyd / 1.15 , εsc Es) = (420 / 1.15, 2.79x10-3 x 2x105 = min (365 , 558)

σ'sc = 365 Mpa

σ'sc2 = min [ fyk / 1.15 , 9γ fck - f1 x d'/d x (0.5 / 15 + 13) fck + (6517 / 15 + 1) ]

f1 = 0.9 for fyk <=400 otherwise, f1 =1

σ'sc2 = min [ fyk / 1.15 , 9γ fck - 1 x d'/d x (0.5 / 15 + 13) fck + (6517 / 15 + 1) ]

σ'sc2 = min (365 , 253) Mpa = 253 MPa

εyd = fyd / Es = 420 / 1.15 / (2x105) = 1.83x10-3 Mpa < εsc = 2.79x10-3

If εsc > εyd , then σ's2 = σ'sc otherwise σ's2 = σ'sc2

=> in our case σ's2 = σ'sc = 365 Mpa

A's = (MuA - Mlu) / [ (d - d') σ's2 ] = (847 - 499.6) x 10-3 / [ (0.40 - 0.05) x 365 ] > 0

A's = 27.16 cm2

Find As for pure bending design

μcu = min (μcu , μlim ) = min (0.630 , 0.3717) = 0.3717

αcu = 1.25 ( 1 - sqrt(1 - 2 μcu) = 1.25 (1 - sqrt( 1 - 2 x 0.3717)) = 0.616

zc = d (1 - 0.4 αcu ) = 0.40 x (1 - 0.4 x 0.616 ) = 0.3013 zc ≈ 0.75d when A's <> 0

As = Mlu / (zc fyd ) + A's σ's2 / fyd = 499.6x10-3 / (0.3013 x 365) + 27.16x10-4 x 365 / 365

As = 45.40 + 27.16 = 72.6 cm2

Find As final for compression bending:

As = As bending - NEd / fyd = 72.6 - 2477x10-3x104 / 365 = 4.72 cm2 > As mini OK!

As mini = max (0.26 fctm / fyk b h , 0.0013 b d) = 3.47 cm2

Thus, the final rebar to put in column is:

As = 2 x max(A's , As) = 2 x max (27.16 , 4.72) = 54.3 cm2


29 | P a g e

Since As = 54.3 cm2 < As initial, = 58.56 cm2, Thus, the As = As initial = 58.56cm2.

Remark: for curvature method, As initial can be set to equal zero at the beginning if unknown, and the result As is the

final answer. In the example above, As initial is set to 58.56cm2, thus we only want to verify buckling if it is okay with the

section As initial = 58.56cm2. Then we verify if As calculation < As initial.

If the As initial = 0 cm2 from the beginning, the As = 52.8cm2. More economically. See annex B for excel calculation.

In conclusion, the results are summarized as below:

EC2 ACI

1) As 58.56 or 52.8 cm2 58.56cm2

2) NEd 2477 kN 1975 kN (including 1.6 Pwind )

3) moment M0Ed 28.5 kN.m M2ns =150 kN.m¸ M2s =163 kN.m

4) additional moment (2nd order) M2 = 128.4 kN.m M2 = 0.23xM2s = 37 kN.m

5) total design moment 413.2 kN.m M=M2ns+M2s+37.49 = 350 MN.m2

Conclusion

Reference BS EN1992-1-1:2004, Eurocode 2: Design of concrete structures, Part 1-1: General rules and

rules for building, Jean-Pierre, M. (2006), Béton armé BAEL91 modifiée 99 et DTU associés, Eyrolles Jean Roux, (2009), Pratique de l'Eurocode 2 guide d'application, Afnor Edition, Eyrolles


Annex A:


Annex B:

Documents

Journal of Stability v2