0

DESIGN OF COLUMN WITH SEVERAL ENCASED STEEL PROFILES FOR COMBINED

COMPRESSION AND BENDING

André Plumier (Plumiecs & ULg):

a.plumier@ulg.ac.be

(Main Contact)

Teodora Bogdan (ULg):

teodora.bogdan@ulg.ac.be

Hervé Degée (ULg):

h.degee@ulg.ac.be

Abstract

Concrete sections reinforced by multiple encased rolled sections are a possible solution to realize mega columns

of tall buildings. In comparison to concrete filled caissons, the advantages are less welding, less fabrication

work, the use of simple splices well settled for decades in high-rise projects and possibility of simpler beam to

column connections. All these characteristics, combined to the availability of huge rolled sections in steel which

do not require pre-heating before welding, lead to another advantage: a high potential for reliable ductile

behavior. AISC allows the design of composite sections built-up with two or more encased steel sections, but

the way to perform such design is not explained. This paper defines the principles and an application method

for the design of such columns under combined axial compression and bending. The method is based on well

established theories to which in AISC 2010 Specifications for Structural Steel Buildings refers and it provides

results in agreement with those of recent experimental st udies and of numerical models developed by the

authors in support of the method. A companion paper develops the design approach for the same type of

columns under shear. The paper also states further research steps which would help in refining the method.

Keywords:

Composite columns, rolled sections, steel shapes, tall buildings, design method, mega-columns.

1

1. Possible design of mega composite columns.

Mega composite columns of tall buildings in Asia are typically designed with concrete filled tubes (CFTs) or

concrete filled continuous steel caissons built-up from heavy plates. Both types are spliced on site. All welded

joints have to comply to structural welding codes such as AWS D1.1 and seismic welding codes such as AWS

D1.8, which require the pre-qualification of the welding procedures. Preheating and interpass temperatures are

specified for plate thickness above 1.5 in. depending on steel composition (CE/grade), type of electrode and

level of restraint in the joint. Non-destructive tests (ultrasonic test, magnetic particle examination, radiographic

test) performed by certified inspectors are mandatory. The welding conditions are very requiring for heavy thick

plates or tubes typically made of grade 50 steel (ASTM A572 Gr.50 or Q345) as they must be preheated at

225°F in the fabrication shop and on site, which is difficult. In practice, a significant percentage of welds often

have to be repaired. As proper controlling and repair are expensive, this solution, when correctly executed, may

be not economical. Furthermore, rigid connections of beams to concrete filled tube or to caissons require huge

welded details, in the two categories used for such connections: the strengthening of the caisson or CFT from

outside by means of stiffeners or rings Fig. 1a; the st rengthening of the inside of the caisson or CFT by means

of plates or by making beams continuous through the column Fig. 1b.

1a) strengthening of caisson or CFT by means of outside stiffeners or rings.

1b) strengthening inside CFT or caisson by means of plates or by beams continuous through column.

Fig. 1. Types of CFT or caissons connections.

One possible alternative to CFT or caissons consists in reinforcing concrete sections by several heavy steel

sections Fig.. 2 and 6. It has already been used in proje cts like the Shanghai World Financial Center and the

2

International Finance Center Tower 2 in Hong Kong Fig. 2. AISC 2010 Specifications for Structural Steel

Buildings design codes allows the design of composite sections built-up from two or more encased steel shapes.

Fig. 2. Construction of the International Finance Center Tower 2 in Hong Kong.

On the material side, one advantage is that the welding procedures and connection detailing correspond to those

of single rolled-H-sections, a classical well documented design on all aspects: effective beveling, "weld-access-

holes", welding sequences with removal of backing bars and appropriate grindings between passes. Very heavy

sections are nowadays available to reinforce concrete columns: for instance, as indicated in ASTM A6-12, the

W14x16 rolled sections are available up to 873 lbs/ft with a flange thickness of 5.5 in.; the W36 sections are

available up to 925 lbs/ft. These sizes are not only available in classical grade ASTM A992/Grade 50 which

requires preheating for flange thicknesses above 1.5 in., but also in high tensile steel produced by a quenching

and self tempering process or QST corresponding to ASTM A913 Grade 50 and 65. Those steel are highly

weldable without preheating above 32°F and with low hydrogen electrodes; they possess a high toughness: 20

ft-lbf up to minus 58 °F. These high performance steels comply with American standards and meet requirements

of Chinese standards like the 20% minimum elongation prescribed in the Chinese seismic code. These QST

steels complying with ASTM A913 have a high potential for reliable ductile behavior.

Another advantage of mega column made of composite sections with two or more encased steel shapes is the

simplicity of connections. It is due to the fact that there is no difficulty in having beams framing through the

columns. One plate connecting two steel sections of the mega column offers a direct support to beam shear. Two

plates connecting two steel sections of the mega column provide a couple of forces F

1

and F

2

which equilibrate a

beam moment M Fig. 3.

AISC 2010 Specifications for Structural Steel Buildings does not explain how to design composite sections

built-up with two or more encased steel sections. In the following, the principles and an application method for

the design of such columns under combined axial compression, bending and shear are proposed. The method is

based on well established theories to which AISC 2010 Specifications for Structural Steel Buildings refers. It

provides results in agreement with those of recent experimental studies and of numerical models developed by

the authors in support of the method. The method cannot be considered as an innovative theoretical step, since it

3

makes use of existing principles and calculations methods, but it fills a gap since nothing similar was presented

up to now in books or papers on structural design.

Fig. 3. Beam connection to mega composite column with 4 encased steel sections.

2. Simplifications in the evaluation of contributions of reinforcement by bars to the inertia and to the

plastic moment of a composite mega column section.

2.1 Objective of the simplifications.

The development of a method of calculation of concrete sections with several encased steel sections requires the

calculation of section characteristics like the moment of inertia, the plastic moment, the elastic neutral axis and

the plastic neutral axis of huge mega column sections. Such calculation can be made either by means of

dedicated software in which all the data are given, each reinforcing bar being defined in position and section.

The calculation can also be hand made, in which case the calculation becomes tedious due to the high number

of longitudinal bars in mega columns. In order to facilitate such calculation, some simplifications are proposed

hereafter which replaces lines of rebars by equivalent plates. These simplifications are not fundamental and have

no direct link with the main subject of the paper which is design under compression and bending. They just help

make user friendly calculations in the design examples presented in this paper and the companion one on shear

design of mega columns.

2.2 Flange layers of rebars. Moment of inertia.

In order to make calculations easily, the layers of rebars parallel to one considered neutral axis can be

substituted by an equivalent plate Fig. 4 with the propert ies:

- Plate area A

p :

A

p

= 2n A

b

where A

b

is the cross sectional area of one bar and n the number of bars in one layer.

- Distance of plate geometrical center to neutral axis d

p

: d

p

= (d

1

+ d

2

)/2

4

where d

1

(respectively d

2

) is the distance from the center of rebars of the 1

st

layer (respectively 2

nd

layer) to the neutral axis.

The exact moment of inertia of reinforcing bars, which as usual neglects the bars own inertia, is equal to:

I

b

= n A

b

(d

1

2

+ d

2

2

).The inertia of the proposed equivalent plate, which also neglects the plate own inertia, is

equal to: I

p

= 2n A

b

d

p

2

.

In order to establish to what extent I

p

is equivalent to I

b

, we express: d

1

= d

p

+ ε d

2

= d

p

ε

Then: I

b

= 2n A

b

(d

p

2

+ ε

2

) is compared to: I

p

= 2n A

b

d

p

2.

The error is: ε

2

/

d

p

2

.

It can be considered acceptable if: ε

2

/

d

p

2

< 1% or: ε

/

d

p

< 10%. Expressed with more straightforward data, the

condition is: (d

1

- d

2

) < 0,2 d

p

.

This condition for an error less than 1% on the moment of inertia of reinforcing bars checks for a wide range of

sections, in particular the section considered in the example presented in this paper. It should be mentioned that

the 1% error on the moment of inertia of reinforcing bars induces a still more minor error on other parameters

like the complete section stiffness; the error is 0,1% on EI

eff

in the example.

Fig. 4. Flange layer of rebars and the equivalent plate.

2.3 Flange layers of rebars. Plastic moment.

With the same symbols as above, the exact plastic moment of rebar layers parallel to the neutral axis is:

M

p,b

= F

y,b

n A

b

(d

1

+ d

2

). Again neglecting the contribution of the equivalent plate own plastic moment, the

plastic moment due to one plate is equal to: M

p,b

= F

y,b

2 n A

b

(d

1

+ d

2

)/2 = M

p,b

.

In this case, the simplification is an exact solution.

2.4 Web layers of rebars. Moment of inertia.

Let us consider a layer of (n+1) bars perpendicular to the neutral axis Fig. 5 where s is the step of bars.

The total number of bars in one layer is 2n + 1. The total height h of the layer is: h = 2n s, while A

b

is the cross

sectional area of one bar. The exact moment of inertia I

b

of reinforcing bars is equal to:

I

b

= 2A

b

s

2

(1

2

+ 2

2

+ + n

2

)

I

b

is found equal to: I

b

= 2A

b

s

2

(2n+1) (n+1) n/6 = (1+1/n) (2n+1) A

b

h

2

/12

In order to make calculations easier, the layers of rebars perpendicular to one considered neutral axis can be

substituted by an equivalent plate Fig. 5 with the prope rties:

A

p

= (2n+1)A

b

h

p

= (2n+1)s ([Note: h

p

= h + s ] t

p

= (2n+1)A

b

/[(2n+1)s] = A

b

/s

5

I

p

= t

p

h

p

3

/12 = (2n+1)A

b

h

p

2

/12 = (1+1/2n)(2n+1) A

b

h

2

/12.

Comparing I

b

and I

p

it appears that the error in using I

p

instead of I

b

is equal to: (I

b

I

p

)/I

p

= 1/2n

Fig. 5. Web layer of rebars and the equivalent plate.

With n=12 in the example presented in this paper, the error is equal to: 1/ (2 x 12) =4,2 % on the moment of web

layers of rebars; as those contributes to only 5,7% to the total section stiffness, the error on the total section

stiffness EI

eff

is only: 0,042 x 0,057 = 0,0024 = 0,2%.

This value is acceptable. A simple formulation for the acceptability of the simplification would correspond to a

1% error on EI

eff

. For that, the number of web rebars in a line on one side of the neutral axis should not be less

than: 1/2n x 0,1 ≤ 0,01 meaning n ≥ 5.

This means that the error made on EI

eff

is less than 1% as long as the number of web rebars in a line are not less

than 10 because n is the number of bars for either top or bottom equivalent plate. There are 24 web rebars in a

line in the example presented in this paper.

2.5 Web layers of rebars. Plastic moment.

With the same notations as above, the exact plastic moment due to web rebars is equal to:

M

p,b

= 2 A

b

s (1+2++ n) = n (n+1) A

b

s = (n

2

+ n) A

b

s

The plastic moment of the proposed equivalent web plate is equal to:

M

p,p

= t

p

h

p

2

/4 = A

b

(2n+1)

2

s

2

/ (4s) = (n

2

+ n + 1/4) A

b

s

The error is: 1/[4(n

2

+ n)]

With the minimum s defined in the previous paragraph (n=10/2=5), the error is equal to:

1/ (4x25 + 4x5) = 1/120 = 0,8%

With the number of web rebars in the example presented in this paper, the error is:

1/ (4x12

2

+ 4x12) = 1/120 = 0,1%

2.6 Steel profiles. Moment of inertia.

6

In order to have an easily calculation of the plastic value of the bending moment of the complete section of the

example in which there are 4 encased W 14 x 873 steel shapes (as shown in Fig. 6), equivalent rectangular

plates are replacing the current steel profiles; the rectangular plates have the following dimensions (d

*

x b

*

),

calculated as shown in Table 2:

d

*

= d = 23.62 in.; b

*

= A

s

/d* = 10.859 in.; I

*

= (b*

x d*

3

)/12 = 11925 in.

4

where

d depth of the steel profile;

A

s

one steel profile area;

I* the moment of inertia of equivalent rectangle;

I

sx

the moment of inertia of one steel profile;

The exact moment of inertia due to the 4 encased steel profile is: I

sx

= 4A

s

d²

sy

+4 I

sx

= 1515359 in.

4

The exact moment of inertia due to 4 equivalent EI

eff

rectangular plates is: I*

sx

= 4A

s

d²

sy

+4 I*= 1490513 in

4

The difference between the two values is less than 2%. The error on the effective stiffness EI

eff

of the complete

section is less than 1%.

2.7 Tables

A user friendly presentation of the simplifications defined above is given in Table 1 and 2 for the type of section

which is calculated as example and shown at Fig. 6.

DEFINITION OF PLATES EQUIVALENT TO REBARS FOR BENDING ABOUT x AXIS.

7

CASE OF 2 LAYERS OF REBARS WITH STEP s IN X AND Y DIRECTION

· Equivalent horizontal plate

n rebars on one horizontal layer

1 1 1

= × = ×

s x sri s s

A n A b h

1s

h n s

= ×

1

1

1

s

s

s

A

b

h

=

2

1 2

s1y

d d

d

+

=

2

sr1x s1 s1y

Z A d

= × ×

2

4

sr1x sri s1y

I n A d

= × × ×

· Equivalent vertical plate

2 n+1 rebars on one vertical layer

2 2 2

s y sri s s

A n A b h

= × = ×

(

)

2

sr x y sri

A n n A

= × + ×

2

(2 1)

s

h n s

= + ×

2

2

2

s

s

s

A

b

h

=

2 2

2

4 2

s2 s2 s2 s2

sr2x

b h b h

Z

× ×

= × =

3 3

2

12 6

s2 s2 s2 s2

sr2x

b h b h

I

× ×

= × =

where:

c concrete cover

s

spacing between two vertical rebars

n

x

no. of bars on x direction

n

y

no. of bars on y direction

d

siy

distance from neutral axis of equivalent plate

to neutral axis of entire section

h

s1

depth of A

s1

plate

b

s1

thickness of A

s1

plate

h

s2

depth of A

s2

plate

b

s2

thickness of A

s2

plate

A

s1

area of top (bottom) plate

A

s2

= area of lateral plate

A

sri

= area of one longitudinal bar

A

sr

= total area of longitudinal bars

Z

sr1x

x- axis plastic modulus of horizontal plates

Z

sr2x

x- axis plastic modulus of vertical plates

I

sr1x

moment of inertia of horizontal plates about

x-axis

I

sr2x

moment of inertia of vertical plates about

x-axis

Table 1. Definition of plates equivalent to rebars for bending about x-axis. Definition of symbols.

8

DEFINITION OF PLATES EQUIVALENT TO STEEL PROFILE FOR BENDING ABOUT x AXIS.

CASE OF n ENCASED STEEL PROFILES.

*

d d

=

*

*

a

A

b

d

=

n

i 1

s a

A A

=

=

∑

3

*

12

* *

b d

I

×

=

* 2 *

a

4

sx sy

I n A d I

= × × + ×

where:

d steel profile depth

d

*

equivalent rectangular plate steel depth

b steel profile flange width

b

*

equivalent rectangular plate thickness

d

sy

distance from neutral axis of equivalent

plate to neutral axis of entire section in y

direction

d

sx

distance from neutral axis of equivalent

plate to neutral axis of entire section in x

direction

A

a

area of one steel profile

A

s

total area of steel profiles

I

*

moment of inertia of the equivalent

rectangular plate about x-axis

I

*

sx

moment of inertia of the equivalent

rectangular plate about x-axis

Table 2. Definition of plates equivalent to steel profiles for bending about x-axis. Definition of symbols.

9

3. Composite column with four encased steel profiles in combined axial compression and flexure about

(x-x) axis.

3.1 Introduction

There are two practical ways to check the adequacy of a composite steel-concrete submitted to combined

bending moment M and axial compression P:

- By means of a numerical model in which all components of the section are defined in position,

dimensions and mechanical properties; in that way it is possible to define a complete interaction M-P

curve.

- By means of hand calculated interaction curves established f ollowing the Plastic Distribution

Method. This approach allows in practice only a few particular points of the interaction curve to be

determined; a schematic complete interaction curve is defined by joining the points with straight lines.

In the following, the two approaches are developed and then compared in order to assess to what extent the

hand calculated approach is feasible. This last approa ch is of interest because it is a flexible tool in a pre-

design stage and it can be used safely in the design stage. The comparison between the two types of calculations

is developed on one typical composite concrete steel cross-section with 4 embedded steel profiles, is presented

in Fig. 6. The cross-section is doubly symmetrical.

Fig. 6. Typical concrete section reinforced by steel profiles.

3.2 Definition of M-P interaction curves.

3.2.1 Introduction.

Roik and Bergman [Roik (1992)] have proposed a simple method to determine the interaction between axial

forces P and flexure M in composite members. It is based on the plastic stress distribution method and is similar

to concepts used in reinforced concrete design. The method is defined and recommended in AISC Specification

Section I5. The method does not determine a continuous M-P interaction curve, but only a polygonal shape

10

joining the key points A, B ,C and D shown at figure 7 by straight lines; this gives in a simple way an

approximate, but safe side, interaction curve. A is the squash load point; B is the point of pure flexural bending;

C is a point of bending moment equal to the one in pure bending; D is the point of maximum strength in

bending. Rigid plastic material behavior is assumed in order to evaluate these key points. Steel is assumed to

have reached yield in tension or compression. Concrete is assumed to have reached its peak stress in

compression or have a tensile strength equal to zero; in one equivalent rectangular stress block the peak stress S

in compression is:

S =

'

c

0.85 0.85 7 ksi 5.95 ksi

f

× = × =

A very explicit definition of the calculation steps in the plastic stress distribution method

is given in [Nethercot,

(2004)] where the construction of the interaction curve A, B, C, D is described in detail.

Fig. 7. Points A, B, C, and D defined by the plastic stress distribution method.

In the following, the plastic stress distribution method is extended to composite sections with several encased

steel shapes; all explanations refer to bending of sections about x axis; it is made use of the simplifications

defined in 2.:

- The layers of rebars parallel to the x axis are replaced by equivalent plates A

s1

;

- The layers of rebars perpendicular to the x axis are replaced by equivalent plate A

s2

;

A numerical example is presented in 3. The results obtained are validated by a comparison to results of a non-

linear finite element modeling using sofware FinelG [FinelG, (2011)]. This University of Liege computational

tool has been continuously developed since the 70's, in particular with concrete beam elements [Boeraeve P.

(1991)]. It is shown in 4. that numerical results compare well with experimental ones in the case of walls

reinforced by several encased steel shapes.

The principle of the definition of points A to D of the plastic disstribution method is given hereunder in 3.2.2.

The complete formulation of the equations used is given in the example presented in 4.

3.2.2 Definition of points A, B, C and D

.

Point A.

Point A corresponds to pure axial compression. If local buckling is prevented until concrete reaches its peak

stress, the available compressive strength is the sum of the plastic strength of all components of the section

(Fig.8):

11

(

)

0.85

A s ys s1 ysr s2 ysr c c

P A F A F A F A f'

= × + × + × + × ×

0 kip ft

A

M

= ×

Fig. 8. Point A internal stresses diagrams.

Point D.

Point D is the maximum bending moment point. In the reference method, it is shown that the maximum bending

moment is obtained with a plastic neutral axis or P.N.A at the axis of symmetry of the cross section. This is

demonstrated by testing two different hypotheses on the P.N.A position:

- P.N.A. lower than the axis of symmetry (like in Fig. 10); then a net compressive force is obtained

which acts below the axis of symmetry, causing negative bending moment about the axis of symmetry

(bending moment is positive if the top of the section is in compression) and a reduction in the bending

moment of the composite cross-section in comparison to the one found with the P.N.A. at the axis of

symmetry;

- P.N.A. above the axis of symmetry (like in Fig. 11); the rising of P.N.A. gives a net increase in tension

force that cause a negative bending moment about the axis of symmetry, which reduces the plastic

bending moment of the composite cross section in comparison to the one found with the P.N.A. at the

axis of symmetry.

The conclusion is that the P.N.A. position to obtain the maximum plastic bending moment M

D

is at the axis of

symmetry of the cross-section Figure 9. Then M

D

is given by:

( )

( )

1

0.85

2

'

D sx ys sr1x sr2x ysr cx c

M Z F Z Z F Z f

= × + + × + × × ×

The coefficient ½ in the term expressing the contribution of concrete results from the assumption that concrete

tensile strength is zero and that only the compressive strength contributes to the bending moment. The axial

strength corresponds to the only non-symmetrical contribution in the internal stresses diagram of Figure 9,

which is the one of concrete:

(

)

0.85

2

c c

D

f'A

P

× ×

=

12

Fig. 9. Point D internal stresses diagram.

Compression is negative. The bending moment is positive if the top of the section is in compression.

Point C.

Point C corresponds to a bending moment M axial force P interaction in which the applied bending moment M

is equal to the pure bending moment capacity M

B

. Since it is assumed that the concrete tensile strength is zero,

there is more strength capacity on the compression side than on the tension side and the P.N.A. must be below

the axis of symmetry of the cross section for the resultant axial force P to be equal to zero.

(

)

0.85

C c c

P f'A

= × ×

M

C

= M

B

= M

pl.RD

The complete calculation of M

pl.RD

is presented in 3.3.7.

Fig. 10. Point C - internal stresses diagram.

Point B.

Point B is the pure flexural bending point:

0

B

P

=

M

B

= M

pl.RD

The complete calculation of M

pl.RD

is presented in 3.3.7.

13

a)

b)

Fig. 11. Point B - internal stresses diagram. a) P.N.A. between the rectangular plates;

b) P.N.A. in the rectangular plates.

3.3 Numerical example.

Definition of points of the M-P interaction curve by the Plastic Stress Distribution Method.

3.3.1 Geometrical and material properties of the section

The section analyzed in the example is shown at Figure 6. It complies with all AISC (2010) and ACI 318-08

Specifications, but only the aspects related to the interaction between bending and axial force are presented here.

The symbols are defined at Figure 6 and in Tables 1 and 2.

Steel profile W 14 x 873: F

ys

= 65 ksi;

29000 ksi;

s

E

=

The section characteristics are as follows.

d = 23.62 in.; b

f

= 18.755 in.; t

f

= 5.51 in.; t

w

= 3.935 in.; d

w

= d 2 t

f

= 12.6 in.;

A

a

= 256.5in

2

;

4

2 2

i 1

4 256.5in.1026in.

s a

A A

=

= = × =

∑

where: d, b

f

, t

f

, t

w

, d

w

, depth, width, flange thickness, web thickness, w eb height of the steel profile, in.;

A

a

area of 1 steel profile, in.²;

A

s

total area of steel profiles, in.²;

Reinforcement: F

ysr

= 60 ksi;

29000 ksi;

s

E

=

d

b

= 1.41 in.; n=256 rebars; A

sri

= 1.56 in.

2

;

n

2

i 1

399.66 in.

sr sri

A A

=

= =

∑

where d

b

diameter of longitudinal rebar, in.;

n total number of longitudinal rebars;

A

sri

area of one longitudinal reinforcement, in.²;

14

A

sr

total area of longitudinal reinforcement, in.²;

Concrete : f

c

= 7000psi;

4768962 psi;

c

E

=

2

121 in.121 in.14641 in.

g 1 2

A h h

= × = × =

;

2 2 2 2

14641 in.360.96 in.1026 in.13254.04 in.

c g sr s

A A A A

= = =

where: A

g

gross cross-sectional area of composite section, in.²;

A

c

net concrete area, in.²;

Additional geometric properties presented in the Fig. 9 are calculated as follows:

h

1

= 121 in.; h

2

= 121 in.; c

x

= 3.4 in.; c

y

= 3.4 in.;

d

x

= 98.5 in.; d

y

= 98.5 in.; d

sx

= 40 in.; d

sy

= 37.5 in.; d

s1y

= 55.1in;

where: h

1

, h

2

height and width of the concrete section, in..

c

x

, c

y

concrete cover, on x direction and respectivel y y- direction, in..

d

x

, d

y

distance between two steel profiles W14x873, on y direction and x - direction, in.

d

sx

, d

sy

distance from the centroid of the steel profile W 14x873 to the section neutral axis, on x

direction and respectively y - direction, in..

d

s1y

the distance from the centroid of A

s2

plate to the section neutral axis, on y - direction, in.

Steel percentage in the cross section is:

2 2

2

399.36 in.1026 in.

0.097

14641 in.

sr s

g

A A

A

+ +

= =

Nominal axial strength:

(

)

( )

2 2 2 2

0.85

1026 in.65 ksi 93.6 in.60 ksi 106.08 in.60 ksi 13

215.64 in.0.85 7 ksi 169284 kip

n s ys s1 ysr s2 ysr c c

P A F A F A F A f'

= =

= × + × + × + × ×

× + × + × + × ×

Steel contribution ratio:

2

ys

1026 in.65 ksi

0.394

169284 kip

s

n

A F

δ

P

×

×

= = =

3.3.2 Definition of equivalent top and bottom plates.

For one top plate, there are n

x

= 60#11 rebars or n = 30#11 rebars in one layer, as shown at Fig. 6.

The geometrical characteristics are determined using the formulas of Table 1. The spacing between two

consecutive longitudinal rebars is s = 3.9 in.

2 2

1

60 1.56 in.93.6 in.

s x sri

A n A= × = × =

1

30 3.9 in.=117 in.

s

h n s

= × = ×

2

1

1

1

93.6 in.

0.8 in.

117 in.

s

s

s

A

b

h

= = =

57.1 in.53.2 in.

55.15 in.

2 2

1 2

s1y

d d

d

+ +

= = =

2 3

2 2 93.6 in.55.15 in.10324.08 in.

sr1x s1 s1y

Z A d = =

= × × × ×

( )

2

2 2 3

4 4 30 1.56 in.55.15 in.569373.01 in.

sr1x sri s1y

I n A d = =

= × × × × × ×

3.3.3 Definition of equivalent side plates.

For one vertical plate, there are #11 n

y

= 68 rebars and 2n+1 =26 rebars in one layer, as shown at Fig. 6. There

are in total 16 rebars placed inside the cross section and included. Formulae in Table 1 give the geometrical

characteristics.

15

2 2

2

68 1.56 in.106.08 in.

s y sri

A n A= × = × =

(

)

(

)

2 2

2 2 1.56 in.399.36 in.

sr x y sri x y

A n n A n n= × + × = × + × =

2

(2 1) 26 3.9 in.=101.4 in.

s

h n s

= + × = ×

2

2

2

2

106.08 in.

1.046 in.

101.4 in.

s

s

s

A

b

h

= = =

( )

2

3

1.046 in.101.4 in.

5378 in.

2 2

2

s2 s2

r2x

b h

Z = =

×

×

=

( )

3

3

3

1.046 in.101.4 in.

181785 in.

6 2

s2 s2

sr2x

b h

I = =

×

×

=

3.3.4 Definition of plates equivalent to steel profiles.

The geometrical characteristics of the plates equivalent steel profiles are given in Table 2:

*

23.62 in.

d d= =

2

*

*

256.5 in.

10.859 in.

23.62 in.

a

A

b

d

= = =

4

2 2

i 1

4 256.5 in.1026 in.

s a

A A =

=

= × =

∑

2 3

4 4 256.5 in.37.5 in.38475 in.

sx a sy

Z A d = =

= × × × ×

( )

3

3

* 4

10.859 in.23.62 in.

11925 in.

12 12

* *

b d

I

×

×

= = =

( )

3

* 2 * 2 4 4

a

4 4 4 256.5 in.37.5 in.4 11925.21 in.1490513in.

sx sy

I A d I = == × × + × × × + ×

3.3.5 Additional information for the concrete part.

( )

2

3 3 3 3

121 in.121 in.

10324.08 in.5378.26 in.38475 in.388712 in.

4 4

2

1 2

cx sr1x sr2x sx

h h

Z Z Z Z = =

×

×

=

( )

3

4

121 in.121 in.

17863240 in.

12 4

3

1 2

g

h h

I = =

×

×

=

I*

sx

= 4A

s

d²

sy

+4 I*= 1490513 in.

4

* 4

sx

17863240 569373 181785 1490513 15621568 in.

cx g sr1x sr2x

I I I I I = =

=

where: Z

cx

full x-axis plastic modulus of concrete shape, in.

3

;

I

g

moment of inertia of the gross cross-section, in.

4

;

I

cx

moment of inertia of concrete, in.

4

.

3.3.6 Points of the M-P Interaction curve.

Point A

(

)

( )

0.85

1026 65 93.6 60 106.08 60 13215 0.85 7 169285 kip

A s ys s1 ysr s2 ysr c c

P A F A F A F A f'

= =

= × + × + × + × ×

+ + +

0 kip ft

A

M

= ×

16

Point D

( )

( )

( )

( )

1

0.85

2

1

38475 65 10324 5378 60 388712 0.85 7 383286 kip ft

2

'

D sx ys sr1x sr2x ysr cx c

M Z F Z Z F Z f

= =

= × + + × + × × ×

+ + + ×

(

)

(

)

2

0.85 0.85 7 ksi 13215.64 in.

39317 kip

2 2

c c

D

f'A

P = =

× × × ×

=

Point C

(

)

(

)

2

0.85 0.85 7 ksi 13215.64 in.78633 kip

C c c

P f'A= =

= × × × ×

M

C

= M

pl.RD

Point B

0 kip

B

P

=

M

B

= M

pl.RD

3.3.7 Evaluation of M

pl.Rd

.

In order to compare the value of the plastic bending moment for a cross section with steel shapes and the same

with the proposed simplification of steel shapes into equivalent rectangle, the position of the P.N.A. is

determined taking two hypotheses into account. The value of the plastic bending moment obtained with

equivalent rectangular plates is

*

pl.Rd

M. The value of the plastic bending moment obtained with the exact cross-

section of the steel shapes is

pl.Rd

M.

As shown in Fig. 11, the distance

*

nx

h

can correspond to two different hypotheses:

· assumption a:

*

nx

h

between the rectangular plates:

2

*

nx sy

d

h d

:

Subtracting the stress distribution of B from that of C, the value of

*

nx

h

can be obtained using the distribution

from the Fig. 12.a).

(

)

(

)

1

0.85 78633 kip 2 0.85 2 2 0.85

* *

C B C c c nx c nx s2 ysr c

P P P A f' = h h f'h b F f'

= = × × = × × × × + × × × × ×

( )

( )

( ) ( )

1

78633

46.85 in.

121 0.85 7 2 1.046 2 60 0.85 7

0.85 2 2 0.85

* C

nx

c s2 yrs c

P

h =

2

2 h f'b F f'

= =

+

× × × + × × × ×

The assumption is correct if the following condition is fulfilled:

23.62 in.

46.85 in.37.5 in.25.69 in.

2 2

*

nx sy

d

h d

= = =

If the assumption is not correct, then it should be assumed that the P.N.A. is in the rectangular plates.

· Assumption b: neutral axis within the rectangular p lates:

2 2

*

sy nx sy

d d

d <h d

+

:

( )

( ) ( )

1

0.85 78633 kip

0.85 2 2 0.85 2 2 0.85

2

C B C c c

* * *

C B nx c nx sy ys c nx s2 ysr c

P P P A f'

d

As : P P = 2 h h f'h d b* F f'h b F f'

= = × × =

× × × × + × × × × × + × × × × ×

( )

( ) ( ) ( )

1 s2

2 0.85

2

0.85 2 0.85 2 0.85

C sy ys c

*

nx

c ys c ysr c

d

P d b* 2 F f'

h

2 h f'b* 2 F f'b 2 F f'

+ × × × × ×

=

× × × + × × × × + × × × ×

17

( )

( ) ( ) ( )

23.62

78633 2 37.5 10.859 2 65 0.85 7

2

33.81 in.

121 0.85 7 2 10.859 2 65 0.85 7 2 1.046 2 60 0.85 7

*

nx

h =

2

+

=

+ +

The assumption is verified, as the following condition is fulfilled:

23.62 in.23.62 in.

37.5 in.25.69 in.33.81 in.37.5 in.49.31 in.

2 2 2 2

*

sy nx sy

d d

d <h d

= = = + = + =

a)

b)

Fig. 12. Subtracting the components of the stress distribution combination at point B and C

considering normal force only equivalent rectangu lar plates case

:

a) if P.N.A. is between the rectangular plates; b) if P.N.A. is within the rectangular plates;

Then the plastic bending moment is equal to:

( )

*

1

0.85

2

pl.Rd Dx sr2xn ysr sxn ys cxn c

M M Z F Z F Z f'

= × × × × ×

Where:

cxn

Z

- x-axis plastic modulus of concrete section within the zone 2h

n

, in.

3

sxn

Z

- x-axis plastic modulus of equivalent rectangle bar within the zone 2h

n

, in.

3

sr2xn

Z - x-axis plastic modulus of A

s2

plates within the zone 2h

n

, in.

3

These plastic modulus are calculated as follows:

(

)

( )

2

2

3

2 2 1.046 in.33.81in.2391.91 in.

*

sr2xn s2 nx

Z b h = =

= × ×

( )

(

)

( )

( )

2

2

2

2

3

2

10.859 2 37.5 23.62

2 2 10.859 33.81 10494 in

2 4

sy

* *

sxn nx

b* d d

Z b h =

× ×

= × × =

(

)

( )

2

2

3

121 33.81 2391.91 10494.91 125439 in.

*

cxn 1 nx sr2xn sxn

Z h h Z Z = =

= ×

18

( )

( )

*

* 3 3 3

1

0.85

2

1

383286 kip ft 2391.91 in 60 kip 10494.91 in.65 ki

p 125439.42 in.0.85 7 kip 283381 kip ft

2

pl.Rd Dx sr2xn ysr sxn ys cxn c

pl.Rd

M M Z F Z F Z f'

M = =

= × × × × ×

× × × ×

To determine the exact value of the plastic bending moment M

pl.Rd

, the steel profile is considered divided in

three rectangles (2 flanges and one web). Then, there are three possible P.N.A positions: in the bottom flange, in

the web or in the top flange, as shown in Fig. 13 a), b) and c). The position of

nx

h

is chosen by default and then

verified if the assumption is correct.

a)

b)

c)

Fig. 13. Subtracting the components of the stress distribution combination at point B and C

considering normal force only steel profile case

:

a) P.N.A. in bottom flange; b) P.N.A. in the web; c) P.N.A. in top flange;

· assumption 2a: neutral axis in the bottom flange of the steel profile

2 2

w

sy nx sy

d

d

d <h d :

19

( )

( ) ( )

1 nx

0.85 78633 kip

0.85 2 2 0.85 2 2 0.85

2

C B C c c

nx c f nx sy ys c s2 ysr c

P P P A f'

d

= 2 h h f'b h d F f'h b F f'

= = × × =

× × × × + × × × × × + × × × × ×

( )

( ) ( ) ( )

( )

( ) ( )

1 s2

2 0.85

2

0.85 2 0.85 2 0.85

23.62

78633 2 18.755 37.5 2 65 0.85 7

2

121 0.85 7 2 18.755 2 65 0.85 7 2 1.046 2 60 0.8

C f sy ys c

nx

c f ys c ysr c

d

P b d 2 F f'

h

2 h f'b 2 F f'b 2 F f'

=

2

+ × × × × ×

=

× × × + × × × × + × × × ×

+

+ + ( )

31.299 in.

5 7

=

The assumption is verified if:

23.62 in.12.6 in.

37.5 in.25.69 in.31.299 in.37.5 in.31.2 in.

2 2 2 2

w

sy nx sy

dd

d <h d

= = = = =

The assumption is not correct, it shall be verified if the web the steel shape.

· assumption 2b: neutral axis in the web of the steel profile

2 2

w w

sy nx sy

d d

d <h d +:

( )

( ) ( )

1 nx

0.85 78633 kip

0.85 2 2 2 0.85 2 2 0.85

2

C B C c c

nx c f f w nx sy ys c s2 ysr c

P P P A f'

d

= 2 h h f'b t t h d F f'h b F f'

= = × × =

× × × × + × × + × × × × + × × × × ×

( )

( ) ( ) ( )

( )

( )

1 s2

2 2 0.85

2

0.85 2 0.85 2 0.85

23.62

78633 kip 2 18.755 37.5 2 65 0.85 7

2

121 0.85 7 2 18.755 2 65 0.85

C f f w sy ys c

nx

c f ys c ysr c

d

P b t t d 2 F f'

h

2 h f'b 2 F f'b 2 F f'

=

2

+ × × + × × × × ×

=

× × × + × × × × + × × × ×

+ ×

+ ( ) ( )

31.435 in.

7 2 1.046 2 60 0.85 7

=

+

The assumption is verified if:

12.6 in.12.6 in.

37.5 in.31.2 in.31.435 in.37.5 in.43.8 in.

2 2 2 2

w w

sy nx sy

d d

d <h d

= = = + = + =

Remark: the difference between the values of P.N.A. obtained with the steel shapes or their simplification into

rectangles is less than 10%:

93%

nx

*

nx

h

h

=

The exact plastic value of the bending moment is equal to:

( )

1

0.85

2

pl.Rd Dx sr2xn ysr sxn ys cxn c

M M Z F Z F Z f'

= × × × × ×

Where

cxn

Z

- x-axis plastic modulus of concrete section within the zone 2h

n

, in.

3

sxn

Z

- x-axis plastic modulus of equivalent rectangle bar within the zone 2h

n

, in.

3

sr2xn

Z

- x-axis plastic modulus of A

s2

plates within the zone 2h

n

, in.

3

20

(

)

(

)

(

)

( )

( ) ( )

( )

( )

2 2

2

2 2 2

3

2 2

2

2 2 2

18.755 3.935 2 37.5 12.6 18.755 2 37.5 23.62 3.935 2 31.435

11873.96 in.

2 2 2

f w sy w f sy

w nx

sxn

b t d d b d d

t h

Z

=

× × × ×

× ×

= +

×

= +

( )

2

3 3 3

121 in.31.435 in.2067.56 in.11873.96 in.1056

27.04 in.

2

cxn 1 nx sr2xn sxn

Z h h Z Z = =

= × ×

( )

2

3

2 2 1.046 in.31.435 in.2067.56 in.

2

sr2xn s2 nx

Z b h = =

= × ×

( )

( )

1

0.85

2

1

383286 2067.56 60 11873.96 65 105627 0.85 7 2824

45 kip ft

2

pl.Rd Dx sr2xn ysr sxn ys cxn c

M M Z F Z F Z f'

= =

= × × × × ×

×

The ratio between the values of plastic moment obtained with the simplification of shapes into rectangles or

with the steel shapes is:

*

99.7%

pl.Rd

pl.Rd

M

M

=

.

The difference is less than 1%. It can be concluded that equivalent rectangular plates can be used instead of steel

profiles in order to make the calculation of the section strength easier.

4. Numerical modeling of the composite steel concrete section.

4.1 General presentation.

The chosen finite element is a 2D Bernoulli fiber element with 3 nodes and 7 degrees of freedom (DOF). The

total number of DOF corresponds to one rotational and two translational DOF for the nodes located at beam

element ends (nodes 1 and 3 in Fig. 14) and one relative longitudinal translational DOF for the node situated at

mid-length of the beam element (node 2 in Fig. 14.). The relative translational DOF of the node at beam mid-

length has been proven necessary to take into account the strong variation of the centroid position along the

beam when the behavior of the section is not symmetrical. Such a situation happens for instance in concrete

sections as soon as cracking occurs. The beam elements are able to simulate structures undergoing large

displacements but small deformations. They are developed following a co-rotational total description.

Fig. 14. Strain Plane beam finite element with three nodes.

The model is built using an assembly of concrete (with appropriate reservations at the location of the steel

profiles) and steel fibre elements (see Fig. 15). In such fibre elements, only longitudinal strain and stresses are

21

explicitly modeled. The shear behavior is supposed to remain elastic. Compatibility of longitudinal strains is

assumed at the interface between concrete and steel elements. This translates mathematically a perfectly rigid

longitudinal connection.

For both concrete and steel elements, internal forces in the elements are computed using a longitudinal and

transverse integration scheme. The integration along the beam length is performed using a classical Gauss

scheme with 4 integration points (see Fig. 16). Nodal values are then extrapolated from this 4-point scheme. At

each longitudinal integration point (LIP

i

), a transverse integration is performed using a multilayer scheme. The

section is divided into a number of layers, in which the actual stress state is derived from the strain state and

assuming a uniaxial stress-strain relationship. In this case the cross-section is divided into 29 layers.

Fig. 15. Integration scheme: a) longitudinal integration with 4-point Gauss scheme; b) transversal

integration with multilayer scheme.

A parabola-rectangle constitutive law with tension stiffening is assumed for the concrete, as shown in Fig. 16 a),

and is analytically defined as follows:

2

3

cc ct cc

2

2 0.3

'

ccu ccu c

c c

c c

f

f f f

E

×

= × × = = ×

where: f

cc

= 7 ksi compressive strength of unconfined concre te (AISC I1.2b)

f

ct

axial tensile strength of concrete;

ccu

= 0.003 ultimate compressive strain of unconfine d concrete (AISC I1.2b);

c

strain at reaching maximum strength;

E = 4768 ksi;

Fig. 16. Material laws: a) Parabola-rectangle diagram for concrete in compression.

b) Bi-linear for steel in tension or compression.

An elastic perfectly plastic law is used to model the steel material, as shown in Fig. 16 b), where:

22

f

y

= 60 ksi- for reinforcement and f

y

= 65 ksi- for the steel profile

E = 29000 ksi;

(AISC I1.3);

For both steel and concrete materials, the mechanical properties considered in the numerical simulations are the

nominal values. They should thus compare to the simplified approach defined in 2. and 3.approach, considering

nominal values of the material properties. This comparison is done in Fig 19.

The numerical M-P interaction curve is derived from the behavior of a cantilever column with arbitrary length l,

as shown in Fig. 17. The column is chosen long enough to ensure that shear effects can be neglected but not too

long to avoid stability problems and second-order (i.e. buckling) effects.

Fig. 17. FinelG - numerical model.

Accounting for the symmetry of the cross-section, only half of the section is represented, as shown in Fig. 17.

Results of the FEM analysis are then simply doubled for final post-processing and comparison. The total height

of the composite column is equal to l = 1800 in. The zone close to the support is the main zone of interest and

needs an accurate meshing. In total there are 17 nodes, 8 elements over a 225 in. length.

The column is initially loaded by a compressive axial force P. The compression force is kept constant while a

horizontal load is then increasingly applied until the bending resistance of the column is overcome. The

corresponding resisting moment in the plastic hinge is calculated by

max

H l

= ×

. The full curve is then built by

considering different values of the compression force P and by calculating the maximum bending resistance M

corresponding to each value of P.

4.2 Comparison of experimental and numerical results.

The results of a similar numerical model have been compared to experimental results [Dan D., (2011)]. The

experimental model is shown in Fig. 18. Concrete walls reinforced by several steel profiles were experimentally

studied at Politechnica University of Timi şoara. The structural steel profiles were connected to the concrete

23

web by headed studs. The reinforcement consisted in of #3 vertical bars with spacing 3.9 in.. The height of the

wall was 118 in. and the testing procedure consists in one constant vertical load, N = 22.5 kip, and a cyclically

increasing horizontal lateral load. In this case, due to the presence of boundary horizontal hoops, the value of

ultimate confined strain was taken into account. The measured material properties are listed in Table 3.

Fig. 18. Experimentally tested ductile walls cros s-section definition.

Concrete properties

f

cm

[ksi] E

cm

[ksi]

Steel properties f

y

[ksi] f

u

[ksi] E

s

[ksi]

Specimen CSRCW 3 9.44 5649

Steel rebar #3 79.48 90.21 30600

Specimen CSRCW 4 8.99 5516

I-shaped steel 47.57 74.84 29440

Steel tube 49.60 77.16 30020

Table 3. Material properties in the CRSCW Specimens [Dan D.,(2011)]

A comparison of the force (P) - displacement (∆) curves from the experimental tests and the numerical model is

presented in Fig. 19. The small diference between the experimental and the numerical results is due to the slip

between the concrete and steel, which exists in reality but not taken into account in the numerical model.

Fig. 19.Comparison of numerical and experimental load-displacement curves..

The bending moment axial force interaction curve p resented in Fig. 20 shows that the difference between the

experimental and numerical results is less than 10%.

24

Fig. 20. Ductile walls M-N interaction curve.

4.3 Comparison of strength calculated by the Plastic Distribution Method and the Finite Element Method.

The Plastic distribution Method (P.D.M.) has been used for composite concrete section with several encased

steel profiles in order to calculate the points A, B, C, and D of the M-P interaction curve. Table 4 summarizes

the results obtained.

CSRCW 3

P

P.D.M.

[kip]

M

Rd

P.D.M.

[ kip ft]

M

Rd

FEM [kip ft] M

Rd

- Ratio

Point B 0 283381 310000 91%

Point D 39317 383286 375228 98%

Point C 78633 283381 290612 97%

Point A 169285 0 0 ------

Table 4. Comparison of strength at points A, B, C and D by 2 calculation methods.

Fig. 21. Comparison between the Plastic Distribution Method and the FEM method.

5. Conclusions.

Concrete sections reinforced by multiple encased rolled sections can be an advantageous solution to realize

mega columns of tall buildings, but they are not ye t covered by standard design methods.

Design values of bending moment M-axial force P interaction diagram have been obtained on the basis of the

plastic stress distribution method, a simple meth od presented in the AISC Specification. Explicit expressions

25

have been developed; simplifications of the calculation by replacing reinforcement bars by equivalent plates

have been proposed.

As there is until now no experimental work on mega columns, a direct validation of the proposed method and

simplifications was not possible.

An alternative two steps method was used:

- check of the ability of a finite element method to reproduce correctly the experimental behavior of

walls with several encased steel profiles;

- comparison of the results obtained by the stress d istribution method with those obtained by the fini te

element method in the case of mega columns with several encased steel shapes.

That procedure concludes to the validity of the st ress distribution method and of the simplification s proposed

for their application in the case of mega columns with several encased steel shapes.

A future development should consist in defining the limits of applicability of the plastic stress distribution

method. Indeed, the latter is valid as long as strains implicit to a plastic model do not overcome the deformation

capacity of the materials involved. A work similar to the one presented here made considering a set of different

steel and concrete material properties would clarify this issue.

References

AISC 2011, Design Examples V14 with particular reference to Chapter I: Design of Composite Members,

AISC Chicago, Illinois.

AISC (2010), Specification for Structural Steel Buildings, Chicago, Illinois.

ACI (2008), Building Code Requirements for Structural Concrete and Commentary , ACI 318-08.

Roik K. and Bergman R., (1992), Chapter 4.2.: Composite columns, Constructional Steel Design: An

International Guide, Elsevier Applied Science.

Nethercot D.A. (2004), Composite Construction, Spon Press, ISBN 0-203-45733-1, London.

FinelG Users Manual , V 9.2., (2011) Non linear finite element analysis software,Greisch Info Departement

ArGEnCo Ulg.

BoeraeveP. (1991), ), Contribution à lanalyse statique non linéaire des structures mixtes planes formées de

pouters, avec prise en compte des effets différés et des phases de construction, Doctoral thesis , University of

Liège.

Dan, D., Fabian, A. and Stoian, v. (2011). Nonlinear behaviour of composite shear walls with vertical steel

encased profiles. Engineering Structures 33, 2794-2804.

26

Biography

André Plumier is a Professor at the University of Liege (Belgium), with specialties in steel and composite

steel-concrete structures and seismic design. He led many research projects in these fields. Mr. Plumier is the

inventor of the reduced beam sections concept. He is a consultant in projects in seismic areas and has been full

member of the ECCS TC13 seismic design of steel structures committee since its creation in 1984.

Teodora Bogdan is a research engineer at the University of Liege (Belgium). She obtained a PhD degree in

2011with a thesis in the field of composite steel-concrete structures at Technical University of Cluj-Napoca

(Romania). She is now working at University of Liege (Belgium).

Hervé Degée is research associate at the Belgian Foundation for Research and invited Professor at the

Universities of Liege and Ghent (Belgium). His main research field is structural mechanics and its application in

earthquake engineering for steel, composite steel-concrete and masonry structures. He is active in various

research programs and standardization committees at European level and acts regularly as consultant for

building companies and design offices for stability and seismic questions.

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