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DESIGN EXAMPLE OF A COLUMN WITH 4 ENCASED STEEL PROFILES
Andr Plumier (Plumiecs & ULg) : [email protected] (Main
Contact) Teodora Bogdan (ULg) : [email protected] Herv Dege
(ULg) : [email protected] Jean-Claude (JC) Gerardy ArcelorMittal
Commercial Sections (Luxembourg) : [email protected]
Abstract Composite mega columns of tall buildings are currently
designed with continuous built-up sections, welded in the
fabrication shop and spliced on the job site without any
prequalified welding procedure. This leads to highly restrained
welds and splices which, under severe dynamic loadings, will likely
crack before exhibiting any ductile behavior. These tall buildings
have not been submitted to severe earthquakes, but it will happen.
The 1994 earthquake in Northridge, California, taught us that
welding procedures, beam-to-column connections and column splices
have to be as simple as possible to properly and reliably work as
anticipated.
Using multiple rolled sections encased into concrete is the
solution for increasing the safety of tall buildings. It
leads to less welding, less fabrication works and reliable
simple splices which have been used for decades in high-rise
projects. AISC allows engineers to design composite sections
built-up from two or more encased steel. But, it doesnt explain how
to perform and check the design. This paper offers a method to do
it. The method is explained by
means of design examples covering combined axial compression,
bending and shear.
Keywords Composite columns, rolled sections, steel shapes, tall
buildings, design method, mega-columns.
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NOTATIONS
Aa = area of 1 steel profile.
Ac = area of concrete.
Ag = gross cross-sectional area of composite section.
As = total area of the steel profiles.
As1 = equivalent steel plate placed along the x-axis. As2 =
equivalent steel plate placed along the y-axis.
Asr = area of the continuous reinforcing bars. Asri =
cross-sectional area of reinforcing bar I. Asrs = area of
continuous reinforcing bars.
Avz = web area of the steel profile. b = width of the steel
profile.
h = height of the steel profile. bs1 = width of As1 plate,
mm.
bs2 = width of As2 plate, mm.
cx = concrete cover, on x direction. cy = concrete cover, on y
direction. db = diameter of the longitudinal reinforcement.
dx = the distance between the two steel profiles HD 400x1299
(W14x16x873), on y - direction. dy = the distance between the two
steel profiles HD 400x1299 (W14x16x873), on x - direction. dsx =
the distance from the local centroid of the steel profile HD
400x1299 (W14x16x873) to the section neutral axis, on x -
direction.
dsy = the distance from the local centroid of the steel profile
HD 400x1299 (W14x16x873) to the section neutral axis, on y
direction. ds2x = the distance from the local centroid of As1 plate
to the section neutral axis, on x - direction. ds1y = the distance
from the local centroid of As2 plate to the section neutral axis,
on y - direction. fc = compressive cylinder strength of concrete.
Fy = specified minimum yield stress of steel shape. Fysr = yield
stress of reinforcing steel. Fu = specified minimum tensile
strength of steel shape.
h1 = height of the concrete section.
h2 = width of the concrete section. hs1 = height of As1 plate,
mm.
hs2 = height of As2 plate, mm. hnx = distance from centroidal
axis (Y-Y) to neutral axis . hny = distance from centroidal axis
(X-X) to neutral axis . Ic = moment of inertia of concrete.
Ir = moment of inertia of reinforcing steel.
Is = moment of inertia of steel shape.
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Isr1x = moment of inertia about x axis of As1 plate, mm.
Isr2x = moment of inertia about x axis of As2 plate, mm. Isr1y =
moment of inertia about y axis of As1 plate, mm. Isr2y = moment of
inertia about y axis of As2 plate, mm.
Isrx = moment of inertia about x axis of equivalent plates, mm.
Isry = moment of inertia about y axis of equivalent plates, mm. n =
number of continuous reinforcing bars in composite section.
nx = number of continuous reinforcing bars on x direction. ny =
number of continuous reinforcing bars on y direction. tf = steel
profile flange thickness.
tw = steel profile web thickness.
Zr1x = full x-axis plastic modulus of As1 plate, mm.
Zr2x = full x-axis plastic modulus of As2 plate, mm. Zr1y = full
y-axis plastic modulus of As1 plate, mm.
Zr2y = full y-axis plastic modulus of As2 plate, mm.
Zsx = full x-axis plastic modulus of steel shape, mm. Zsy = full
y-axis plastic modulus of steel shape, mm. Zcx = full x-axis
plastic modulus of concrete shape, mm. Zcy = full y-axis plastic
modulus of concrete shape, mm.
cxnZ = x-axis plastic modulus of concrete section within the
zone 2hn
r2xnZ = x-axis plastic modulus of As2 plates within the zone
2hn
cynZ = y-axis plastic modulus of concrete section within the
zone 2hn
r1ynZ = y-axis plastic modulus of As1 plates within the zone
2hn.
= steel contribution ratio.
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Preface Mega composite columns of tall buildings in Asia are
typically designed with huge steel continuous caissons built-up
from heavy plates. They are welded together in the steel
fabrication shop and spliced on the job site.
Internationally recognized welding codes such as AWS D1.1
(structural American welding code) and AWS D1.8 (seismic welding
code) or EN 1090-2:2008 (execution of steel structures) and EN
1011-2:2001 (recommendations for welding of metallic materials)
impose the pre-qualification of the welding procedures of such
exotic joints, following strict welding sequences. Required
preheating and interpass temperatures are specified per the
thickness of the steel (>32mm), its composition (CEV/grade), the
type of electrode and the level of restraint in the joint.
Non-destructive tests (ultrasonic test, magnetic particle
examination, radiographic test) performed by certified inspectors
are mandatory to guarantee sound welded connections and a safe
structure.
In practice, even when the welding codes are strictly followed,
it is typical to have to repair up to 10% of the
welds in simple structures. In the case of these huge caissons,
the welding conditions are rather extreme. Heavy thick plates in
typical grade
50 steel (ASTM A572Gr.50 or Q345) must be preheated at 110C in
the steel fabrication shop as well as on the job site prior and
during the welding process. Any lack of preheating when welding
these huge caissons induces sensitive material conditions (hard and
brittle zones) and high levels of restraint (post weld stresses) in
all directions starting in the steel fabrication shop and amplified
on the job site after splicing two caissons together. Applying
adequate preheating during the whole welding process is difficult.
How to preheat such joints at 110C? Correct welding takes days of
work without interruption. Proper controlling and repair of all
welds is so expensive that this solution, when correctly executed,
is not economical at all. There is an economical and safer
alternate to this configuration. AISC design codes allows designers
to use
composite sections built-up from two or more encased steel
shapes provided that the buckling of individual
shapes is prevented before the hardening of the concrete.
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4
The Chinese Institute of Earthquake Engineering is also
recommending the use of multiple jumbo H-shapes rather than large
continuous caissons. The welding procedures and the connection
detailing of single rolled-H-sections are well described in the
above mentioned codes. The use of correct beveling, the so-called
"weld-
access-holes" associated to very precise welding sequences,
including the removal of the backing bars and appropriate grindings
to clean-up the weld surface between passes minimize the amount of
residual stresses after splicing single rolled steel columns.
W14x16 (HD400) rolled sections (jumbos) are today available up to
1299 kg/m (873 lbs/ft) with a flange thickness of 140 mm (5.5 in.)
and W36 (HL920) are available up to 1377 kg/m (925 lbs/ft). These
sizes are not only available in classical grade 345 MPa (ASTM
A992/Grade 345, Q345, S355) which requires to be preheated for
flange thicknesses above 32 mm (1.5 in.) but also in high tensile
modern steel produced by a quenching and self tempering process,
namely ASTM A913 Grade 345 and 450, or per ETA 10-156 (European
Technical Approval) grades Histar 355 and Histar 460. Besides their
higher yields, the main advantages of these high performance steels
are their weldability without preheating (above 0C and with low
hydrogen electrodes) as well as their outstanding toughness. (27J
up to minus 50C). These high performance steels are not only fully
in compliance with American and European standards, they can also
meet
the stringent requirements of the Chinese standards such as the
20% minimum elongation which is mandatory in the Chinese seismic
codes. These QST steels (ASTM A913) have already been successfully
used in the Shanghai World Financial Center. In this paper, a
method for the design of composite sections with multiple encased
steel profiles is presented. It
make use of existing principles and calculations methods, but
the fact is that the method as such does is not
presented up to now in books of structural design.
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Introduction.
The design examples presented hereafter have as main
references:
- ANSI/AISC 360-10 Specification for Structural Steel Buildings,
2010 - AISC DESIGN EXAMPLES Version 14.0,2011
- Building Code Requirements for Structural Concrete ACI 318-08,
2008
Occasionally, reference is made to EN 1994-1-1:2004 Eurocode 4:
Design of composite steel and concrete structures, part 1-1,
general rules and rules for buildings, European Committee for
Standardization (CEN), Brussels, Belgium.
Recall of AISC rules for design of composite members and
introduction to the design examples of composite columns with
several steel profiles encased.
Recall of AISC rule in I4. SHEAR.
For filled and encased composite members, either the shear
strength of the steel section alone, the steel section plus the
reinforcing steel, or the reinforced concrete alone are permitted
to be used in the calculation of available
shear strength.
The explanations and justifications of the design for shear
resistance in the case of a composite column with 4 encased steel
profiles are given within Examples I.X3 and I.X4.
Recall of AISC rule in I5. COMBINED FLEXURE AND AXIAL FORCE.
Design for combined axial force and flexure may be accomplished
using the plastic-distribution method. Several different procedures
for employing the plastic-distribution method are outlined in the
AISC
Commentary to I5.
Each of these procedures is applied for composite steel-concrete
sections concrete with 4 encased steel profiles
in Example I.X1. and Example I.X2.
To help in following these design examples, the interaction
curves which will be used are presented separately
in Fig. I-1e and I-1f. The equations correspond to different
points selected on the interaction curves. Calculations concerning
the slenderness effect are not presented, because they would not be
different of those
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shown in detail in AISC Design Examples V14.0-2011. For design
cases which would be different of the
examples presented (for instance a section with 6 encased steel
profiles, this presentation in Figures I-1e and I-1f shows the way
to develop the appropriate interaction equations.
In the plastic-distribution method, the N-M interaction curves
are convex, because it is assumed that the concrete has no tensile
strength.
Fig.I-1. Axial force (P) bending moment (M) diagram for a
composite cross section.
For a composite cross-section symmetrical about the axis of
bending, Roik and Bergmann (1992) have proposed a simple method to
evaluate its M-N interaction diagram. This method is adopted in
AISC Specifications. As shown in Figure I-1, this method does not
determine a continuous N-M interaction curve, but only a few
key
points. The N-M curve is then constructed by joining these key
points by straight lines. When evaluating these key points,
rigid-plastic material behavior is assumed. Thus, steel is assumed
to have reached yield in either tension or compression. Concrete is
assumed to have reached its peak stress in compression and its
tensile strength is zero. For one equivalent rectangular stress
block the peak stress in
compression is: '
c0.85 0.85 50MPa 42.5 MPaf = =
The key points in Fig.I-1are:
- A - squash load point - B - pure flexural bending point - D -
the maximum bending moment point - C - point with bending moment
equal to the pure bending moment capacity
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PLASTIC CAPACITIES FOR RECTANGULAR, COLUMN WITH 4 ENCASED
PROFILES
BENT ABOUT THE X X AXIS
Section Stress distribution Point Defining Equations
1 1 1= = s x sri s sA n A b h
2 2 2 2= = s y sr s sA n A b h
( )= + sr x y sriA n n A
nx no. of bars on x direction ny no. of bars on y direction bs1=
width of As1 plate hs1= height of As1 plate bs2= width of As2 plate
hs2= height of As2 plate As1 = area of top (bottom) plate
As2 = area of lateral plate Asri = area of one longitudinal
bar
A 0.85A s y s1 ysr s2 ysr c cP A F A F A F A f'= + + +
0AM =
4s aA A=
2C 1 s srA h h A A=
Aa = area of one steel profile As = total area of the steel
shape
C
0.85C c cP A f'=
C BM M=
D 0.852
c c
DA f'
P
=
( ) ( )1 0.852 'Dx sx y r1x r2x ysr cx cM Z F Z Z F Z f= + +
+
Zsx full x- axis plastic modulus of steel shape Zr1x full x-
axis plastic modulus of As1 plates Zr2x full x- axis plastic
modulus of As2 plates
2r1x s1 s1yZ A d=
24
2s2 s2
r2xb h
Z
=
4sx a syZ A d=
r2xZ4
21 2
cx r1x sxh h
Z Z Z
=
Fig. I -1e. Composite member with several encased steel
profiles, X-X -axis anchor points.
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8
B 0BP =
aAb*h
=
where: b* - the width of the equivalent steel rectangle bar Aa
area of one steel profile
For hnx between the two profiles 2nx syhh d
:
( ) ( )1 0.85 2 0.85C
nx
c s2 yrs c
Ph2 h f' b 2 F f'= +
2cxn 1 nx r2xnZ h h Z=
2 2r2xn s2 nxZ b h=
( )1 0.852Bx Dx r2xn yrs cxn cM M Z F Z f'=
For hnx between the two profiles 2 2sy nx syh hd h d < +
:
( )( ) ( ) ( )1 s2
4 0.852
0.85 4 0.85 2 0.85
C sy y c
nx
c y c yrs c
hP d b* 2 F f'h
2 h f' b* 2 F f' b 2 F f'
+
=
+ +
2 2r2xn s2 nxZ b h=
( ) ( ) ( )2
2 2 24
sysxn sx nx sx nx
b* 2 d hZ 2 d +b* h 2 d -b* h
=
2cxn 1 nx r2xn sxnZ h h Z Z=
( )1 0.852Bx Dx r2xn yrs sxn y cxn cM M Z F Z F Z f'=
For hnx above the two profiles the height of vertical
equivalent layer 2 2
s2sy nx
hhd
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PLASTIC CAPACITIES FOR RECTANGULAR, COLUMN WITH 4 ENCASED
PROFILES
BENT ABOUT THE Y Y AXIS
Section Stress distribution Point Defining Equations
A 0.85A s y s1 ysr s2 ysr c cP A F A F A F A f'= + + +
0AM =
4s aA A=
2C 1 s srA h h A A=
Aa = area of one steel profile As = total area of the steel
shape
C
0.85C c cP A f'=
C BM M=
D 0.852
c c
DA f'
P
=
( ) ( )1 0.852 'Dy sy y r1y r2y ysr cy cM Z F Z Z F Z f= + + +
Zsy full y- axis plastic modulus of steel shape Zr1y full y- axis
plastic modulus of As1 plates Zr2y full y- axis plastic modulus of
As2 plates
2s1b2
4s1
r1yd
Z
=
s2x2r2y s2Z A d=
4sy a sxZ A d=
r2yZ4
21 2
cy r1y syh h
Z Z Z
=
Fig. I -1f. Composite member with several encased steel
profiles, Y-Y -axis anchor points.
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B 0BP =
aAh*b
=
where: h* - the height of the equivalent steel rectangle bar Aa
area of one steel profile
For hny between the two profiles 2ny sxbh d
:
( ) ( )2 0.85 2 0.85C
nyc s1 yrs c
Ph2 h f' b 2 F f'= +
22
cyn ny r1ynZ h h Z=
2 2r1yn s1 nyZ b h=
( )1 0.852By Dy r1yn yrs cyn cM M Z F Z f'=
For hny between the two profiles 2 2sx ny sxb bd h d < +
:
( )( ) ( ) ( )2 s1
4 0.852
0.85 4 0.85 2 0.85
C sx y c
nyc y c yrs c
bP d h* 2 F f'h
2 h f' h* 2 F f' b 2 F f'
+
=
+ +
2 2r1yn s1 nyZ b h=
( ) ( ) ( )2
2 2 24
sx
syn sy ny sy ny
h* 2 d bZ 2 d +h* h 2 d -h* h
=
2cyn 1 ny r1yn synZ h h Z Z=
( )1 0.852By Dy r1yn yrs syn y cyn cM M Z F Z F Z f'=
For hny above the two profiles and the height of
vertical equivalent layer 2 2
s1sx ny
hbd h + <
:
( )( ) ( )
2 0.850.85 2 2 0.85
C s y cny
2 c s1 yrs c
P A F f'h
2 h f' b F f'
=
+
2 2r1yn s1 nyZ b h=
syn syZ Z=
2cyn 1 ny r1yn synZ h h Z Z=
( )1 0.852By Dy r1yn yrs syn y cyn cM M Z F Z F Z f'= synZ -
y-axis plastic modulus of equivalent steel
rectangle bar within the zone 2 hny
cynZ - y-axis plastic modulus of concrete section within the
zone 2hny
r1ynZ - y-axis plastic modulus of As1 plates within the zone
2hny
Fig. I -1f. Composite member with several encased steel
profiles, Y-Y -axis anchor points (continued).
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EXAMPLE I.X1 - COMPOSITE COLUMN WITH FOUR ENCASED STEEL PROFILES
IN COMBINED AXIAL COMPRESSION AND FLEXURE ABOUT (X-X) AXIS.
Given: Determine for the encased composite member illustrates in
Fig. I.X1-1 the axial force (P) bending moment (M) diagram.
Fig. I.X1-1. Encased composite member section.
From ArcelorMittal classification, the steel material properties
are:
ASTM A913- 11 Grade 65 Fy = 450MPa; Fu = 550MPa;
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From ArcelorMittal sections catalog, the geometric and material
properties of one steel profile HD 400x1299 (W14x16x873) are:
Aa = 165000 mm2; Avz = 505.2 cm2; b = 476 mm; h = 600 mm; tw =
100 mm; tf = 140 mm;
Zsx = 33250 cm3; Zsy = 16670 cm3; 4 4754600 10 mmHDxI =
4 4254400 10 mmHDyI =
From Fig. I.X1-1, additional geometric properties of the
composite section used for force allocation and load transfer are
calculated as follows: h1 = 3072mm; h2 = 3072mm;
cx = 86mm -14mm = 72 mm; cy = 86mm -14mm = 72 mm; dx = 2500 mm;
dy = 2500mm; dsx = 1012 mm; dsy = 950mm; ds1y = 1400 mm; ds2x =
1450mm;
2(3072mm) (3072mm) 9437184mmg 1 2A h h= = = ; db = 40mm for a
T40 diameter bar;
21256.637mm=sriA ;
n2
i 1321699.09mm
=
= =sr sriA A ;
42 2
i 14 165000mm 660000mms aA A
=
= = =;
2 2 2 6 29437184mm 321699.09mm 660000mm 8.455 10 mmc g sr sA A A
A= = =
( ) 3kg0.043 38007MPa for 2500 m= = =1.5 'c c c cE w f w ; Es =
200000 MPa (AISC I1.3); 2 2
2
321699.09mm 660000mm 0.10494371.84mm
sr s
g
A AA+ +
= =
( )5 2 5 2 6 20.85
6.6 10 mm 450MPa 3.216 10 mm 500MPa 8.455 10 mm 0.85
50MPa817207.653kN
n s y sr ysr c cP A F A F A f'
= + +
= + +
=
2y 660000mm 450MPa 0.363
817207.653kNs
n
A F
P
= = =
where
h1 height of the concrete section, mm. h2 width of the concrete
section, mm. cx concrete cover, on x direction, mm. cy concrete
cover, on y direction, mm. dx the distance between the two steel
profiles HD 400x1299 (W14x16x873), on y - direction, mm. dy the
distance between the two steel profiles HD 400x1299 (W14x16x873),
on x - direction, mm. dsx the distance from the local centroid of
the steel profile HD 400x1299 (W14x16x873) to the
section neutral axis, on x - direction, mm.
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dsy the distance from the local centroid of the steel profile HD
400x1299 (W14x16x873) to the section neutral axis, on y direction,
mm.
ds2x the distance from the local centroid of As1 plate to the
section neutral axis, on x - direction, mm. ds1y the distance from
the local centroid of As2 plate to the section neutral axis, on y -
direction, mm.
Solution:
A simplification of the composite section is made by replacing
the reinforcement by equivalent steel plates, as shown in Fig.
I.X1-1. Horizontal plates include only reinforcement that belongs
to the two main lines. One horizontal plate, As1, replace 52
reinforcement rebars.
52=xn
2 252T40 52 1256.64mm 65364mm= = =s1A ;
For 1003072mm (86mm mm) 2800mm
2= + =s1h
bs1 = 23.338 mm;
2800mm 1400mm2 2s1
s1yhd = = =
Side plates includes besides the two lateral lines, the few
additional rebars. The number of reinforcement which corresponds to
one lateral plate is 76.
30 30 3 2 5 2 76yn = + + + =
2 276T40 76 1256.64mm 95532mms2A = = = ;
For 3072mm 2 86mm 2900mms2h = = bs2 = 32.933 mm;
2900mm 1450mm2 2s2
s2xhd = = =
The moment of inertia of the reinforcing bars about the elastic
neutral axis of the composite section, Isr, is determined for the
two equivalent plates As1 and As2, and is calculated as
follows:
( )3 6 42800mm 23.338mm 2.966 10 mm12 12
3s1 s1
sr1xh b
I
= = =
;
( )3 10 432.933mm 2900mm 6.693 10 mm12 12
3s2 s2
sr2xb h
I
= = =
;
( )26 4 10 4 4 211 4
2 2 2
2 2.966 10 mm 2 6.693 10 mm 2 6.534 10 mm 1400mm3.9 10 mm
2srx sr1x sr2x s1 s1yI I I A d
=
= + + =
+ +
=
;
where
Isr1x moment of inertia about x axis of As1 plate, mm4. Isr2x
moment of inertia about x axis of As2 plate, mm4. Isrx moment of
inertia about x axis of equivalent plates, mm4.
hs1 height of As1 plate, mm.
bs1 width of As1 plate, mm.
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14
hs2 height of As2 plate, mm.
bs2 width of As2 plate, mm. The moment of inertia values of the
entire steel section about X-X is determined as:
( )22 4 4 11 4HDx4 4 I 4 165000mm 950mm 4 754600 10 mm 6.258 10
mm2sx a syI A d= + = + =
where
Isx moment of inertia about x axis of the steel profiles,
mm4.
The moment of inertia values for the concrete about both axes
axis is determined as:
( )3 12 42 3072mm 3072mm 7.421 10 mm12 12
31
gh h
I
= = =
12 4 11 4 11 4 12 47.421 10 mm 3.9 10 mm 6.058 10 mm 6.426 10
mmcx g srx sxI I I I= = =
where
Icx moment of inertia about x axis of the concrete part,
mm4.
Material and Detailing Limitations
Material limits are provided in AISC Specification Sections I1.1
(2) and I1.3 as follows: (1) Concrete strength: 21MPa 70MPacf'
50MPacf' = o. k. (2) Specified yield stress of structural steel:
525MPayF
450MPayF = o. k.
(3) Specified yield stress of structural steel: 525MPaysrF
500MPaysrF = o. k.
Transverse reinforcement limitations are provided in AISC
Specification Section I1.1 (3), I2.1a. (1), I2.1a. (2) and ACI 318
as follows:
(1) Tie size and spacing limitations: The AISC Specifications
requires that either lateral ties or spirals be used for transverse
reinforcement. Where lateral ties are used, a minimum of either 10
mm (No. 3) bar placed at a maximum of 406 mm (12 in.) on center, or
a 13 mm (No. 4) bar or larger spaced at a maximum of 406 mm (16
in.) on center shall be used. 14 mm lateral ties at 75 mm are
provided. o. k. Note that AISC Specification Section I1.1 (1)
specifically excludes the composite column provision of ACI 318
Section 10.13, so it is unnecessary to meet the tie reinforcement
provisions of ACI 318 Section 10.13.8. when designing composite
columns using AISC Specifications Chapter I. If spirals are used,
the requirements of ACI 318 Sections 7.10 and 10.9.3 should be met
according to the User Note at the end of AISC Specification
I2.1a.
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15
(2) Additional tie size limitation: ACI 318 Section 7.10.5.1
requires that all nonprestressed bars shall be enclosed by lateral
ties, at least 10 mm (No. 3) in size for longitudinal bars 32 mm
(No. 10) or smaller, and at least 13 mm (No. 4) in size for 36 mm
(No. 11), 43 mm (No. 14), 57 mm (No. 18), and bundled longitudinal
bars. 14 mm lateral ties are provided for 40 mm longitudinal bars.
o. k.
(3) Maximum tie spacing should not exceed 0.5 times the least
column dimension: 3072mm
0.5 min 1536mm3072mm
1max
2
hs
h=
= = =
75mm 1536mmmaxs s= = o. k.
(4) Concrete cover: ACI 318 Section 7.7 contains concrete cover
requirements. For concrete not exposed to weather or in contact
with ground, the required cover for column ties is 38 mm (1.5 in).
cover 86mm-1T14 86mm-14mm 72mm 38mm= = = >
o. k.
(5) Provide ties as required for lateral support of longitudinal
bars: AISC Design Examples 2011 Part1, page I-96 indicates the
following:AISC Specification Commentary Section I2.1a references
Chapter 7 of ACI 318 for additional transverse tie requirements. In
accordance with ACI 318 Section 7.10.5.3 and Fig. R7.10.5, ties are
required to support longitudinal bars located farther than 6 in.
clear on each side from a laterally supported bar. For corner bars,
support is typically provided by the main perimeter ties. For
intermediate bars, Fig. I.9-1illustrates one method for providing
support through the use of a diamond-shaped tie.
Longitudinal and structural steel reinforcements limits are
provided in AISC Specification Section I1.1 (4), I2.1 and ACI 318
as follows:
(1) Structural steel minimum reinforcement ratio: / 0.01s gA A 5
2
6 26.6 10 mm 0.070mm
9.437 10 mm
=
o. k.
(2) Minimum longitudinal reinforcement ratio: / 0.004sr gA A
5 2
6 23.216 10 mm 0.034mm9.437 10 mm
=
o. k.
(3) Maximum longitudinal reinforcement ratio: / 0.08sr gA A 5
2
6 23.216 10 mm 0.034mm9.437 10 mm
=
o. k.
(4) Minimum number of longitudinal bars: ACI 318 Section 10.9.2
requires a minimum of four longitudinal bars within rectangular or
circular members with ties and six bars for columns utilizing
spiral ties. The intent for rectangular sections is to provide a
minimum of one bar in each corner, so irregular geometries with
multiple corners require
additional longitudinal bars.
256 bars provided o. k. (5) Clear spacing between longitudinal
bars:
-
16
ACI 318 Section 7.6.3 requires a clear distance between bars of
1.5db or 38 mm (1.5in.). 1.5 60mm
max 60mm38mm
bmin
ds
= = =
100mm 40mm 60mins s= = o. k.
(6) Clear spacing between longitudinal bars and the steel core:
AISC Specification Section I2.1e requires a minimum clear spacing
between the steel core and longitudinal reinforcement of 1.5
reinforcing bar diameters, but not less than 38 mm (1.5 in.).
1.5 60mmmax 60mm
38mmb
min
ds
= = =
The distance from the steel core and the longitudinal bars is
determined from Fig. IX1-1, on x direction
as follows:
524mm 60mm 2 40mm 146mm2 minb
s s= = o. k.
The distance from the steel core and the longitudinal bars is
determined from Fig. IX1-1, on y direction
as follows:
586mm 60mm 2 40mm 100mm2 minh
s s= = o. k.
where
h height of HD 400x1299 (W14x16x873) steel profile, mm. b width
of HD 400x1299 (W14x16x873) steel profile, mm.
(7) Concrete cover for longitudinal reinforcement: ACI 318
Section 7.7 provides concrete cover requirements for reinforcement.
The cover requirements for column ties and primary reinforcement
are the same, and the tie cover was previously determined to
be acceptable, thus the longitudinal reinforcement cover is
acceptable by inspection.
Interaction of Axial Force and Flexure
AISC Design Examples 2011 Part1, page I-96 indicates the
following: The interaction between axial forces and flexure in
composite members is governed by AISC Specification Section I5
which, for compact members permits the use of a strain
compatibility method or plastic stress distribution method, with
the option to use the interaction equations of Section H1.1. The
strain compatibility method is a generalized approach that allows
for the construction of an interaction diagram based upon the same
concepts used for reinforced concrete design. Application of
the strain compatibility method is required for
irregular/nonsymmetrical sections.
Plastic stress distribution methods are discussed in AISC
Specification Commentary Section I5 which provides three acceptable
procedures for filled members. Plastic stress distribution methods
are
discussed in AISC Specification Commentary Section I5. The
procedure involves the construction of a piecewise-linear
interaction curve using the plastic strength equations provided in
Fig. I-1-1 located
-
17
within the front matter of the Chapter I Design Examples. The
method is a reduction of the piecewise-
linear interaction curve that allows for the use of less
conservative interaction equations than those
presented in Chapter H. Thereafter are provided approaches
following two methods: a plastic stress distribution method and a
finite
element analysis.
Method1 - Interaction Curves from the Plastic Stress
Distribution Model
Step 1: Construct nominal strength interaction surface A, B, C,
and D without length effects, using the equations provided in Fig.
I-1e for bending about the X-X axis:
Point A (pure axial compression): the available compressive
strength is calculated as illustrated in Design Example I.9.
( )5 2 5 2 6 20.85
6.6 10 mm 450MPa 3.216 10 mm 500MPa 8.455 10 mm 0.85
50MPa817207.653kN
A s y sr ysr c cP A F A F A f'
= + +
= + +
=
0kNmAM =
Point D (maximum nominal moment strength):
( )6 20.85
28.455 10 mm 0.85 50MPa
2179679.054kN
c c
DA f'
P
=
= =
=
The applied moment, illustrated in Fig. I -1e, is resisted by
the flexural strength of the composite section about its X-X axis.
The strength of the section in pure flexure is calculated using the
equations of Fig. I-1e found in
the front matter of the Chapter I Design Examples for Point B.
Note that the calculation of the flexural strength
at Point B first requires calculation of the flexural strength
at Point D as follows: 4 4 8 32 2 6.534 10 mm 1400mm 1.83 10 mm
sr1x s1 s1yZ A d= = = ;
( )2 8 332.933mm 2900mm2 2 1.385 10 mm4 4
2s2 s2
sr2xb h
Z
= = = ;
8 3 8 3 8 31.83 10 mm 1.385 10 mm 3.21 10 mmsrx sr1x sr1xZ Z Z=
+ = + =
5 4 8 34 4 1.65 10 mm 950mm 6.27 10 mmsx a syZ A d= = = ;
( )
22
28 3 8 3 8 3
9 3
43072mm 3072mm
1.83 10 mm 1.385 10 mm 6.27 10 mm4
6.299 10 mm
1cx sr1x sr2x sx
h hZ Z Z Z
=
=
=
;
where
Zsr1x full x-axis plastic modulus of As1 plate, mm3.
-
18
Zsr2x full x-axis plastic modulus of As2 plate, mm3. Zsrx full
x-axis plastic modulus of As1 and As2 plate, mm3. Zsx full x-axis
plastic modulus of steel shape, mm3. Zcx full x-axis plastic
modulus of concrete shape, mm3.
The bending moment of a composite cross-section is taken about
the axis of symmetry. Therefore, the maximum bending moment is
obtained by placing the plastic neutral axis at the axis of
symmetry of the composite cross-section. This conclusion can be
obtained by examining the change in the bending moment of the
composite cross-section by making a small change in the position of
the plastic neutral axis. The coefficient of in front of the
concrete part is a result on the assumption that concrete has no
tensile strength and only the compressive strength contributes to
the bending moment capacity [Nethercot D.A., 2004].
( ) ( )( ) ( )8 3 8 3 8 3 9 3
5
1 0.852
16.27 10 mm 450MPa 1.83 10 mm 1.385 10 mm 500MPa 6.299 10 mm
0.85 50MPa2
5.767 10 kNm
'
Dx sx y sr1x sr2x ysr cx cM Z F Z Z F Z f
=
= + + +
= + + +
Point B (pure flexure): 0kNBP =
The stress distribution point C from Fig. I1e provides the same
moment resistance as B, since the moment from the stress resultants
cancel each other. However, the resulting resistance to axial force
is of the same magnitude
from the pure concrete part 'c0.85 f . This can be seen from
adding up the stress distribution in B and C, with
regard to the equilibrium of forces, by example the resulting
axial force. This follows because the resistance to axial force in
B is zero. Subtracting the stress distributions of B from that of C
it results the value of hnx.
In order to determine the position of the neutral axis on X-X
direction, the HD 400x1299 steel profile has been considered as a
rectangle bar (h x b*) with an equivalent area, as shown in Fig.
I.X1-2a.
165000mm 275mm600mm
aAb*=h
= =
;
where
h the height of the HD 400x1299 steel profile
-
19
Fig. I.X1-2a. Subtracting the components of the stress
distribution combination at point B and C considering normal force
only, when the a.n.is between the profiles.
Assumption 1: the neutral axis is placed between the steel
profiles2nx syhh d
:
( ) ( )10.85 359358.109kN
2 0.85 2 2 0.85C B C c c
nx c nx s2 yrs c
P P P A f' = h h f' h b F f'
= = =
+
( ) ( )( ) ( )
1 0.85 2 2 0.85359358.109kN
3072mm 0.85 50MPa 2 32.933mm 2 500MPa 0.85 50MPa1.108m
Cnx
c s2 yrs c
Ph2 h f' b F f'
=
2
=
+
+
=
Check assumption2nx syhh d
:
1108mm 650mm2nx syhh d= =
assumption not. k.
Assumption 2: the neutral axis is placed within the steel
profiles2 2sy nx syh hd
-
20
Fig. I.X1-2b. Subtracting the components of the stress
distribution combination at point B and C considering normal force
only, when the a.n.is within the profiles.
( ) ( ) ( )1 nx0.85 359358.109kN
0.85 4 2 0.85 2 2 0.852
C B C c c
nx c nx sy y c s2 yrs c
P P P A f'h
= 2 h h f' h d b* F f' h b F f' = = =
+ +
( )( ) ( ) ( )
( )( )
1 s2
4 0.852
0.85 4 0.85 2 0.85
600mm359358.109kN 4 950mm 275mm 2 450MPa 0.852
3072mm 0.85 50MPa 4 275mm 2 450MPa 0.85
C sy y c
nx
c y c yrs c
c
hP d b* 2 F f'h
2 h f' b* 2 F f' b 2 F f'
f' =
2 f
+
=
+ +
+
+ ( ) ( )2 32.933mm 2 500MPa 0.85
767mmc c' f'
+
=
Check assumption2 2sy nx syh hd
-
21
( )( )7 3 7 3 9 3
1 0.8521576734.208kNm 3.878 10 mm 500MPa 9.141 10 mm 450MPa
1.687 10 mm 0.85 50MPa2
480546kNm
Bx Dx r2xn yrs sxn y cxn cM M Z F Z F Z f'
=
=
=
where
cxnZ - x-axis plastic modulus of concrete section within the
zone 2hn, mm3.
sxnZ - x-axis plastic modulus of equivalent rectangle bar within
the zone 2hn, mm3.
sr2xnZ - x-axis plastic modulus of As2 plates within the zone
2hn, mm3.
Point C (intermediate point):
6 2
0.850.85 8.455 10 mm 50MPa359358.109kN
C c cP A f'
=
= =
=
480546kNmCx BxM M= = The available compressive and flexural
strengths are determined as follows:
LFRD ASD Design compressive strength:
0.75C =
whereX" C XP P
X = A,B,C or D=
0.75 817207.65kN 612905.739kNA" C AP P
=
= =
0.75 0kN 0kNB" C BP P
=
= =
0.75 359358.109kN 269518.582kNC" c CP P
=
= =
0.75 179679.054kN 134759.291kND" c DP P
=
= =
Allowable compressive strength:
C 2.00=
where
XX"
C
PP
X = A,B,C or D
=
817207.65kN 408603.826kN2
AA"
c
PP
=
= =
0kN 0kN2
BB
c
PP
=
= =
C
359358.109kN 179679.055kN2
C"c
PP
=
= =
D
179679.054kN 89839.53kN2
D"c
PP
=
= =
-
22
Design flexural strength:
0.90B =
whereX" B XM M
X = A,B,C or D=
0.9 0kNm 0kNmAx" B AxM M
= = =
0.9 480546kNm 432491kNmBx" B BxM M
= = =
0.9 480546kNm 432491kNmCx" B CxM M
= = =
0.90 576734.208kNm 519060.787kNmDx" B DxM M
=
= =
Allowable compressive strength:
b 1.67=
b
where
XX"
MM
X = A,B,C or D
=
b
0kNm 0kNm1.67
AxAx"
MM
=
= =
b
480546kNm 287752kNm1.67
BxBx"
MM
=
= =
b
480546kNm 287752kNm1.67
CxCx"
MM
=
= =
b
576734.208kNm 345349.825kNm1.67
DxDx"
MM
=
= =
The design and allowable strength values are plotted in Fig.
I.X1-3.
Fig. I.X1 -3. Available and nominal interaction surfaces.
-
23
Method2 FEM Results
A numerical model using finite elements is considered with the
purpose of comparison and validation of the simplified method. The
software package is FineLg, developed in collaboration between
University of Lige
and Engineering office Greisch [FineLg Users Manual, V 9.2.
2011]. This numerical tool is continuously being developed since
the 70's and has been validated in a number of PhD theses and
research reports. Specific concrete beam elements have been
developed by Ph. Boeraeve [Boeraeve P., 1991]. 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. I.X1-4) and one relative
longitudinal translational DOF for the node situated at mid-length
of the beam element (node 2 in Fig. I-X1-4). 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 behaviour 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. I.X1 -4. 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. I.X1-6.b). In such fibre elements,
only longitudinal strain and stresses are explicitly modelled. The
shear behaviour 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. I.X1 -5.a). Nodal
values are then extrapolated from this 4-point scheme. At each
longitudinal integration point (LIPi), 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.
-
24
Fig. I.X1 -5. 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 (EN 1992-1-1-Eurocode 2, 2004), as shown
in Fig. I.X1-6, and is analytically defined as follows:
cc 2ccu ccucc c
f
=
2 'cc
fE
=
23
ct cc0.3 4.072MPaf f= =.
where: fcc = 50 MPa compressive strength of unconfined concrete
(AISC I1.2b) fct axial tensile strength of concrete; ccu = 0.003
ultimate compressive strain of unconfined concrete (AISC I1.2b); c
strain at reaching maximum strength;
E = 38007MPa;
Fig. I.X1-6.Parabola-rectangle diagram for concrete in
compression An elastic perfectly plastic law is used to model the
steel material (EN 1994-1-1-Eurocode 4 , 2004), as shown in Fig.
I.X1-7.
Fig. I.X1-7. Bi-linear steel material law.
where:
fy = 450MPA; E = 200000MPa;
(AISC I1.3);
-
25
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 AISC
approach also considering nominal values of the material
properties. This comparison is done in Fig. I-XI-11.
The numerical M-N interaction curve is derived from the
behaviour of a cantilever column with arbitrary length l, as shown
in Fig. I.X1-8. 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. I.X1-8. FineLg - numerical model.
Accounting for the symmetry of the cross-section, only half of
the section is represented, as shown in Fig. I.X1-
9. 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 = 45m. The zone
close to the support is the main zone of interest and needs an
accurate meshing. In total there are 17 nodes, 7 elements with a
length of 6m, and 1 element placed close to the support having 3 m.
This shorter element allows a better localization of the plastic
hinge.
-
26
Fig. I.X1-9. Meshing.
The column is initially loaded by a compressive axial force N.
The compression force is kept constant while a
horizontal load is then increasingly applied until the bending
resistance of the column is overcome (see Fig. I.X1-10). The
corresponding resisting moment in the plastic hinge is calculated
by
maxH l = . The full curve
is then built by considering different values of the compression
force N and by calculating the maximum
bending resistance M corresponding to each value of N.
Fig. I.X1-10. Example of pushover curves obtained with the
numerical model for point C of the interaction
curve.
-
27
The following table summarizes the results obtained with the
Simple method and the Finite Element model.
Nominal LFRD ASD Nominal LFRD ASD FineLg P [kN] 0.75 P [kN] P /2
[kN] Mx [kNm] 0.9 Mx [kNm] Mx/1.67 [kNm] Mx
Point B 0 0 0 480546 432491 287752 516060 Point D 179679 134759
89839 576734 519060 345349 604391 Point C 359358 269518 179679
480546 432491 287752 510177 Point A 817207 612905 408603 0 0 0
0
Fig. I.X1-11. Comparison between the AISC - Plastic Distribution
Method and the FEM method.
Conclusion
Design values of M-N interaction diagram have been obtained on
the basis of a simple general methodology proposed by AISC
Specification and from which explicit expressions have been
developed for the case of composite sections with several encased
steel profiles; these expressions have been presented in Figures
I-1e
and I-1f. The results of a study carried out with a more
accurate FEM model confirm the validity of the results obtained
with the simple method in the case of composite sections with
several encased steel profiles. Results obtained with the simple
AISC method using nominal values of the material properties and FEM
results are compared on Fig. I.XI-11. They are in excellent
agreement for high compression level and the simple method is
reasonably
accurate and safe-sided when bending becomes dominant. The
simple method is thus felt sufficient to evaluate design values of
M-N interaction in the present context.
-
28
EXAMPLE I.X2 - COMPOSITE COLUMN WITH 4 STEEL PROFILES ENCASED IN
COMBINED AXIAL COMPRESSION AND FLEXURE OVER (Y-Y) AXIS.
Given:
Determine for the encased composite member illustrated in Fig.
IX1-1 the axial force (P) bending moment (M) diagram.
Fig. I.X2-1. Encased composite member section.
From ArcelorMittal classification, the steel material properties
are:
ASTM A913- 11 Grade 65 Fy = 450MPa; Fu = 550MPa;
-
29
From ArcelorMittal sections catalog, the geometric and material
properties of one steel profile HD 400x1299 (W14x16x873) are:
Aa = 165000 mm2; Avz = 505.2 cm2 ; b = 476 mm; h = 600 mm; tw =
100 mm; tf = 140 mm;
Zsx = 33250 cm3; Zsy = 16670 cm3; 4 4754600 10 mmHDxI =
4 4254400 10 mmHDyI =
From Fig. I.X1-1, additional geometric properties of the
composite section used for force allocation and load
transfer are calculated as follows: h1 = 3072mm; h2 =
3072mm;
cx = 86mm -14mm = 72 mm; cy = 86mm -16mm = 70 mm; dx = 2500 mm;
dy = 2500mm; dsx = 1012 mm; dsy = 950mm; ds1y = 1400 mm; ds2x =
1450mm;
2(3072mm) (3072mm) 9437184mmg 1 2A h h= = = ; db = 40mm for a
T40 diameter bar;
21256.637mm=sriA ;
n2
i 1321699.09mm
=
= =sr sriA A ;
42 2
i 14 165000mm 660000mms aA A
=
= = =;
2 2 2 6 29437184mm 321699.09mm 660000mm 8.455 10 mmc g sr sA A A
A= = =
( ) 3kg0.043 38007MPa for 2500 m= = =1.5 'c c c cE w f w ; Es =
200000 MPa (AISC I1.3); 2 2
2321699.09mm 660000mm 0.104
94371.84mmsr s
g
A AA+ +
= =
( )5 2 5 2 6 20.85
6.6 10 mm 450MPa 3.216 10 mm 500MPa 8.455 10 mm 0.85
50MPa817207.653kN
n s y sr ysr c cP A F A F A f'
= + +
= + +
=
2y 660000mm 450MPa 0.363
817207.653kNs
n
A F
P
= = =
where h1 height of the concrete section, mm.
h2 width of the concrete section, mm.
cx concrete cover, on x direction, mm. cy concrete cover, on y
direction, mm. dx the distance between the two steel profiles HD
400x1299 (W14x16x873), on y - direction, mm. dy the distance
between the two steel profiles HD 400x1299 (W14x16x873), on x -
direction, mm.
-
30
dsx the distance from the local centroid of the steel profile HD
400x1299 (W14x16x873) to the section neutral axis, on x -
direction, mm. dsy the distance from the local centroid of the
steel profile HD 400x1299 (W14x16x873) to the section neutral axis,
on y direction, mm.
ds2x the distance from the local centroid of As1 plate to the
section neutral axis, on x - direction, mm. ds1y the distance from
the local centroid of As2 plate to the section neutral axis, on y -
direction, mm.
Solution: A simplification of the composite section is made by
replacing the reinforcement by equivalent steel plates, as shown in
Fig. I.X2-1. Horizontal plates include only reinforcement that
belongs to the two main lines. One horizontal plate, As1, replace
52 reinforcement rebars.
52=xn
2 252T40 52 1256.64mm 65364mm= = =s1A ;
For 1003072mm (86mm mm) 2800mm
2= + =s1h
bs1 = 23.338 mm;
Side plates includes besides the two lateral lines, the few
additional rebars. The number of reinforcement which
corresponds to one lateral plate is 76.
30 30 3 2 5 2 76yn = + + + =
2 276T40 76 1256.64mm 95532mms2A = = = ;
For 3072mm 2 86mm 2900mms2h = = bs2 = 32.933 mm;
The moment of inertia of the reinforcing bars about the elastic
neutral axis of the composite section, Isr, is
determined for the two equivalent plates As1 and As2, and is
calculated as follows:
( )3 10 423.338mm 2800mm 4.269 10 mm12 12
3s1 s1
sr1yb h
I
= = =
;
( )3 6 42900mm 32.933mm 8.632 10 mm12 12
3s2 s2
sr2yh b
I
= = =
( )210 4 6 4 4 211 4
2 2 2
2 4.269 10 mm 2 8.632 10 mm 2 9.55 10 mm 1450mm4.87 10 mm
2sry sr1y sr2y s2 s2xI I I A d
= + + +
= + +
=
where Isr1y moment of inertia about y axis of As1 plate,
mm4.
Isr2y moment of inertia about y axis of As2 plate, mm4. Isry
moment of inertia about y axis of equivalent plates, mm4. hs1
height of As1 plate, mm.
bs1 width of As1 plate, mm. hs2 height of As2 plate, mm.
bs2 width of As2 plate, mm.
-
31
The moment of inertia values of the entire steel section about
Y-Y axis is determined as:
( )22 4 2 11 4HDy4 4 I 4 165000mm 1012mm 4 254400 10 mm 6.861 10
mm2sy a sxI A d= + = + =
where
Isy moment of inertia about y axis of the steel profiles, mm.
The moment of inertia values for the concrete about both axes axis
is determined as:
( )3 12 41 3072mm 3072mm 7.421 10 mm12 12
32
gh h
I
= = =
12 4 11 4 11 4 12 4- - 7.421 10 mm 4.87 10 mm 6.861 10 mm 6.428
10 mm
cy g sry syI I I I= = =
where Icy moment of inertia about y axis of the concrete part ,
mm4.
Material and Detailing Limitations
Material limits are provided in AISC Specification Sections I1.1
(2) and I1.3 as follows: (1) Concrete strength: 21MPa 70MPacf'
50MPacf' = o. k. (2) Specified yield stress of structural steel:
525MPayF
450MPayF = o. k.
(3) Specified yield stress of structural steel: 525MPaysrF
500MPaysrF = o. k.
Transverse reinforcement limitations are provided in AISC
Specification Section I1.1 (3), I2.1a. (1), I2.1a. (2) and ACI 318
as follows:
(1) Tie size and spacing limitations: The AISC Specifications
requires that either lateral ties or spirals be used for transverse
reinforcement. Where lateral ties are used, a minimum of either 10
mm (No. 3) bar placed at a maximum of 406 mm (12 in.) on center, or
a 13 mm (No. 4) bar or larger spaced at a maximum of 406 mm (16
in.) on center shall be used. 14 mm lateral ties at 75 mm are
provided. o. k. Note that AISC Specification Section I1.1 (1)
specifically excludes the composite column provision of ACI 318
Section 10.13, so it is unnecessary to meet the tie reinforcement
provisions of ACI 318
Section 10.13.8. when designing composite columns using AISC
Specifications Chapter I. If spirals are used, the requirements of
ACI 318 Sections 7.10 and 10.9.3 should be met according to the
User Note at the end of AISC Specification I2.1a.
(2) Additional tie size limitation: ACI 318 Section 7.10.5.1
requires that all nonprestressed bars shall be enclosed by lateral
ties, at least 10 mm (No. 3) in size for longitudinal bars 32 mm
(No. 10) or smaller, and at least 13 mm (No. 4) in size for 36 mm
(No. 11), 43 mm (No. 14), 57 mm (No. 18), and bundled longitudinal
bars.
-
32
14 mm lateral ties are provided for 40 mm longitudinal bars. o.
k.
(3) Maximum tie spacing should not exceed 0.5 times the least
column dimension: 3072mm
0.5 min 1536mm3072mm
1max
2
hs
h=
= = =
75mm 1536mmmaxs s= = o. k.
(4) Concrete cover: ACI 318 Section 7.7 contains concrete cover
requirements. For concrete not exposed to weather or in contact
with ground, the required cover for column ties is 38 mm (1.5
in).
86mm 1T14 86mm 14mm 72mm 38mmcover = = = > o. k.
(5) Provide ties as required for lateral support of longitudinal
bars: AISC Design Examples 2011 Part1, page I-96 indicates the
following:
AISC Specification Commentary Section I2.1a references Chapter 7
of ACI 318 for additional transverse tie requirements. In
accordance with ACI 318 Section 7.10.5.3 and Fig. R7.10.5, ties are
required to support longitudinal bars located farther than 6 in.
clear on each side from a laterally supported bar. For corner bars,
support is typically provided by the main perimeter
ties. For intermediate bars, Fig. I.9-1illustrates one method
for providing support through the use of a diamond-shaped tie.
Longitudinal and structural steel reinforcements limits are
provided in AISC Specification Section I1.1 (4), I2.1 and ACI 318
as follows:
(1) Structural steel minimum reinforcement ratio: / 0.01s gA A 5
2
6 26.6 10 mm 0.070mm
9.437 10 mm
=
o. k.
(2) Minimum longitudinal reinforcement ratio: / 0.004sr gA A
5 2
6 23.216 10 mm 0.034mm9.437 10 mm
=
o. k.
(3) Maximum longitudinal reinforcement ratio: / 0.08sr gA A 5
2
6 23.216 10 mm 0.034mm9.437 10 mm
=
o. k.
(4) Minimum number of longitudinal bars: ACI 318 Section 10.9.2
requires a minimum of four longitudinal bars within rectangular or
circular members with ties and six bars for columns utilizing
spiral ties. The intent for rectangular sections is to provide a
minimum of one bar in each corner, so irregular geometries with
multiple corners require
additional longitudinal bars.
256 bars provided o. k. (5) Clear spacing between longitudinal
bars: ACI 318 Section 7.6.3 requires a clear distance between bars
of 1.5db or 38 mm (1.5in.).
-
33
1.5 60mmmax 60mm
38mmb
min
ds
= = =
100mm 40mm 60mins s= = o. k.
(6) Clear spacing between longitudinal bars and the steel core:
AISC Specification Section I2.1e requires a minimum clear spacing
between the steel core and longitudinal reinforcement of 1.5
reinforcing bar diameters, but not less than 38 mm (1.5 in.).
1.5 60mmmax 60mm
38mmb
min
ds
= = =
The distance from the steel core and the longitudinal bars is
determined from Fig. I.X1-1, on x
direction as follows:
524mm 60mm 2 40mm 146mm2 minb
s s= = o. k.
The distance from the steel core and the longitudinal bars is
determined from Fig. I.X2-1, on y
direction as follows:
586mm 60mm 2 40mm 100mm2 minh
s s= = o. k.
where
h height of HD 400x1299 (W14x16x873) steel profile, mm. b width
of HD 400x1299 (W14x16x873) steel profile, mm.
(7) Concrete cover for longitudinal reinforcement: ACI 318
Section 7.7 provides concrete cover requirements for reinforcement.
The cover requirements
for column ties and primary reinforcement are the same, and the
tie cover was previously determined to be acceptable, thus the
longitudinal reinforcement cover is acceptable by inspection.
Interaction of Axial Force and Flexure
AISC Design Examples 2011 Part1, page I-104 indicates the
following: The interaction between axial forces and flexure in
composite members is governed by AISC Specification Section I5
which, for compact members permits the use of a strain
compatibility method or plastic stress distribution method, with
the option to use the interaction equations of Section H1.1.
The strain compatibility method is a generalized approach that
allows for the construction of an interaction diagram based upon
the same concepts used for reinforced concrete design. Application
of the strain compatibility method is required for
irregular/nonsymmetrical sections.
Plastic stress distribution methods are discussed in AISC
Specification Commentary Section I5 which provides three acceptable
procedures for filled members. Plastic stress distribution methods
are discussed in AISC Specification Commentary Section I5. The
procedure involves the construction of a piecewise-linear
interaction curve using the plastic strength equations provided in
Fig. I-1-1 located within the front matter of the Chapter I Design
Examples. The method is a reduction of the piecewise-
-
34
linear interaction curve that allows for the use of less
conservative interaction equations than those
presented in Chapter H. Thereafter are provided approaches
following two methods: a plastic stress distribution method and a
finite element analysis.
Method1 - Interaction Curves from the Plastic Stress
Distribution Model
Step 1: Construct nominal strength interaction surface A, B, C,
and D without length effects, using the equations provided in Fig.
I-1f for bending about the Y-Y axis:
Point A (pure axial compression): the available compressive
strength is calculated as illustrated in Design Example I.9.
( )5 2 5 2 6 20.85
6.6 10 mm 450MPa 3.216 10 mm 500MPa 8.455 10 mm 0.85
50MPa817207.653kN
A s y sr ysr c cP A F A F A f'
= + +
= + +
=
0kNmAM =
Point D (maximum nominal moment strength):
( )6 20.85
28.455 10 mm 0.85 50MPa
2179679.054kN
c c
DA f'
P
=
= =
=
The applied moment, illustrated in Fig. I-1f, is resisted by the
flexural strength of the composite section about its Y-Y axis. The
strength of the section in pure flexure is calculated using the
equations of Fig. I-1f found in the
front matter of the Chapter I Design Examples for Point B. Note
that the calculation of the flexural strength at
Point B first requires calculation of the flexural strength at
Point D as follows:
( )2 7 323.338mm 2800mm 9.148 10 mm4 4
2s1 s1
sr1yb h
Z
= = =
4 4 8 32 2 9.55 10 mm 1450mm 2.769 10 mmsr2y s2 s2xZ A d= = = ;
7 3 8 3 8 39.148 10 mm 2.769 10 mm 3.684 10 mmsry sr1y sr1yZ Z Z= +
= + =
5 4 8 34 2 1.65 10 mm 1012mm 6.679 10 mmsy a syZ A d= = = ;
( )2 7 3 8 3 8 39 3
43072mm 3072mm
9.148 10 mm 2.769 10 mm 6.679 10 mm4
6.211 10 mm
21 2
cy r1y r2y syh h
Z Z Z Z
=
=
=
where
Zsr1y full y-axis plastic modulus of As1 plate, mm3.
-
35
Zsr2y full y-axis plastic modulus of As2 plate, mm3.
Zsy full y-axis plastic modulus of steel shape, mm3. Zcy full
y-axis plastic modulus of concrete shape, mm3.
The bending moment of a composite cross-section is taken about
the axis of symmetry. Therefore, the maximum bending moment is
obtained by placing the plastic neutral axis at the axis of
symmetry of the composite cross-section. This conclusion can be
obtained by examining the change in the bending moment of the
composite
cross-section by making a small change in the position of the
plastic neutral axis. The coefficient of in front of the concrete
part is a result on the assumption that concrete has no tensile
strength and only the compressive strength contributes to the
bending moment capacity.
( ) ( )( ) ( )8 3 7 3 8 3 9 3
5
1 0.852
16.679 10 mm 450MPa 9.148 10 mm 2.769 10 mm 500MPa 6.211 10 mm
0.85 50MPa2
6.168 10 kNm
'
Dy sy y sr1y sr2y ysr cy cM Z F Z Z F Z f
=
= + + +
= + + +
Point B (pure flexure): 0kNBP =
The stress distribution type C from Fig. I1f provides the same
moment resistance as B, since the moment from the stress resultants
cancel each other. However, the resulting resistance to axial force
is of the same magnitude
from the pure concrete part 'c0.85 f . This can be seen from
adding up the stress distribution in B and C, with
regard to the equilibrium of forces, by example the resulting
axial force. This follows because the resistance to axial force in
B is zero. Subtracting the stress distributions of B from that of C
it results the value of hny.
In order to determine the position of the neutral axis, on Y-Y
direction, the HD 400x1299 steel profile has been considered as a
rectangle bar (h* x b) with an equivalent area, as shown in Fig.
I.X2-2a.
165000mm 346.639mm476mm
aAh*=b
= =
;
where
b the width of the HD 400x1299 steel profile;
-
36
Fig. I.X2-2a. Subtracting the components of the stress
distribution combination at point B and C considering normal force
only, when the a.n. is between the profiles.
( ) ( )0.85 359358.109kN0.85 2 2 0.85C B C c cny 2 c ny s1 yrs
cP P P A f' = 2 h h f' h b F f'
= = =
+
( ) ( )( ) ( )
2 0.85 2 2 0.85359358.109kN
3072mm 0.85 50MPa 8 23.338mm 2 500MPa 0.85 50MPa1.175m
Cny
c s1 yrs c
Ph 2 h f' b F f'
=
2
=
+
+
=
Check assumption2ny sxbh d
:
1175mm 774mm2ny sxbh d= =
assumption not. k.
Assumption 2: the neutral axis is placed within the steel
profiles2 2sy nx syh hd
-
37
Fig. I.X2-2b. Subtracting the components of the stress
distribution combination at point B and C considering normal force
only, when the a.n. is within the profiles.
( ) ( ) ( )0.85 359358.109kN0.85 2 0.85 2 2 0.85C B C c cny 1 c
s y c ny s1 yrs cP P P A f' = 2 h h f' A F f' h b F f'
= = =
+ +
( )( ) ( ) ( )
( )( )
2 s1
4 0.852
0.85 4 0.85 2 0.85
476mm359358.109kN 4 1012mm 346.639mm 2 450MPa 0.85 50MPa2
3072mm 0.85 50MPa 4 346.639mm 2 45
C sx y c
nyc y c yrs c
bP d h* 2 F f'h
2 h f' h* 2 F f' b 2 F f'
=
2
+
=
+ +
+
+ ( ) ( )0MPa 0.85 50MPa 2 23.338mm 2 500MPa 0.85 50MPa
856mm +
=
Check assumption2 2sy nx syh hd
-
38
( )( )7 3 7 3 9 3
1 0.8521616779.059kNm 3.421 10 mm 500MPa 9.273 10 mm 450MPa
2.124 10 mm 0.85 50MPa2
512804.327kNm
By Dy r1yn yrs syn y cyn cM M Z F Z F Z f'
=
=
=
where
cynZ - y-axis plastic modulus of concrete section within the
zone 2hn,mm3.
synZ - y-axis plastic modulus of equivalent rectangle bar within
the zone 2hn,mm3.
sr1ynZ - y-axis plastic modulus of As1 plates within the zone
2hn,mm3.
Point C (intermediate point):
6 2
0.850.85 8.455 10 mm 50MPa359358.109kN
C c cP A f'
=
= =
=
512804.327kNmCy ByM M= =
The available compressive and flexural strengths are determined
as follows:
LFRD ASD Design compressive strength:
0.75C =
whereX" C XP P
X = A,B,C or D=
0.75 817207.65kN 612905.739kNA" C AP P
=
= =
0.75 0kN 0kNB" C BP P
=
= =
0.75 359358.109kN 269518.582kNC" c CP P
=
= =
0.75 179679.054kN 134759.291kND" c DP P
=
= =
Allowable compressive strength:
C 2.00=
where
XX"
C
PP
X = A,B,C or D
=
817207.65kN 408603.826kN2
AA"
c
PP
=
= =
0kN 0kN2
BB
c
PP
=
= =
C
359358.109kN 179679.055kN2
C"c
PP
=
= =
D
179679.054kN 89839.53kN2
D"c
PP
=
= =
-
39
Design flexural strength:
0.90B =
whereX" B XM M
X = A,B,C or D=
0.9 0kNm 0kNmAy" B AyM M
= = =
0.9 512804kNm 461524kNmBy" B ByM M
= = =
0.9 512804kNm 461524kNmCy" B CyM M
= = =
0.9 616779.059kNm 555101.153kNmDy" B DyM M
= = =
Allowable compressive strength:
b 1.67=
b
where
XX"
MM
X = A,B,C or D
=
b
0kNm 0kNm1.67
AyAy"
MM
=
= =
b
512804kNm 307068kNm1.67
ByBy"
MM
=
= =
b
512804kNm 307068kNm1.67
CyCy"
MM
=
= =
b
616779.059kNm 369328.778kNm1.67
DyDy"
MM
=
= =
The design and allowable strength values are plotted in Fig.
IX2-3.
Fig. I.X2-3. ASD and LFRD interaction surfaces.
-
40
Method2 FEM Results
Table of results obtained with Simple method and Finite Element
model.
Nominal LFRD ASD Nominal LFRD ASD FineLg P [kN] 0.75 P [kN] P /2
[kN] My [kNm] 0.9 My [kNm] My/1.67 [kNm] My
Point B 0 0 0 512804 461524 307068 528385 Point D 179679 134759
89839 616780 555102 369329 622210 Point C 359358 269518 179679
512804 461524 307068 524608 Point A 817207 612905 408603 0 0 0
0
Fig. I.X2-4. Comparison between the Plastic Distribution Method
and the FEM method.
Conclusion
Design values of M-N interaction diagram have been obtained on
the basis of a simple method presented in AISC Specification and
for which explicit expressions have been developed for the case of
composite sections with several encased steel profiles; these
expressions are presented in Figures I-1e and I-1f. The results of
the finite element study made with more refined models confirm the
validity of the results obtained by that the simple method in the
case of composite sections with several encased steel profiles
(compare nominal and FinelG in Table above). The simple method can
be kept to evaluate design values of M-N interaction for that
case.
-
41
EXAMPLE I.X3 COMPOSITE COLUMN WITH 4 ENCASED STEEL PROFILES IN
SHEAR DIRECTION Y.
Given:
Determine if the composite member with 4 encased steel profiles
illustrated in Figure I.X4-1 is adequate for the axial forces,
shears and moments given hereunder, that have been determined in
accordance with the direct analysis method of AISC 2010
Specification Chapter C for the control of ASCE(2010)ASCE/SEI 7-10
load combinations:
Factored bending moment: Mu,X = 450000 kNm Factored axial
(compression) force: Nu = 180000 kN Factored transverse shear Vu,Y
in direction Y: Vu,Y= 20000 kN
The characteristics of the steel profile are:
h = 600 mm b= 476 mm tf = 140 mm tw = 100 mm A = 165000 mm2 Iy =
754600. 104 mm4 Iz = 254400.104 mm4
Solution:
Fig. I.X3-1. Definition of notations.
Available Shear Strength
According to AISC Specification Section I4.1, there are three
acceptable options for determining the available shear strength of
an encased composite member:
-
42
Option 1- Available shear strength of the steel section alone in
accordance with AISC Specification Chapter G. Option 2- Available
shear strength of the reinforced concrete portion alone per ACI
318. Option 3- Available shear strength of the steel section in
addition to the reinforcing steel ignoring the contribution of the
concrete.
Option 1 clearly is a gross underestimation for the section with
4 encased steel profiles, because it would consist in disregarding
the contribution to shear resistance of a net area Ac of concrete
equal to Ac = 8,45 m2. Option 1 is not developed.
Option 2 is envisaged hereunder. Its application however
requires one adaptation for composite sections with several encased
steel profiles, in comparison to, for instance, the procedure
presented in Design Example I.11. The principle of the adaptation
is explained hereunder. It requires separate calculation of shear
strength of sub-sections composing the complete section. This is
presented in detail.
Option 3 will not be used because it would be unsafe for
composite sections with several encased steel profiles. This is
explained below.
Principle of the adaptation of Option 2 to sections with several
encased steel profiles.
The problem to solve in sections with several encased profiles
is that concrete and steel components contributing to shear
resistance are not working in parallel, like in the case of one
central steel profile encased in concrete: they are, for some part,
working in series or chain. This is easier to understand if one
subdivides the column section into 5 smaller sections, each
providing resistance to shear. They are respectively the sections
of width bc3 (2 sections), bs (2 sections) and bc4 (1 section), all
with height hz . They are named section bc3, section bc4 and
section bs in the following. The applied shear force Vu,Y will
distribute itself into Vu,bc3 , Vu,bc4 and Vu,bs between sections
bc3, bc4 and bs, proportionally to the stiffness of those
sections.
Then each section should provide strength greater than the
applied shear force in that section. Sections bc3 and bc4 are
regular reinforced concrete sections and can be treated as
such.
But section bs is a composite steel-concrete section having 2
reinforced concrete flanges, 2 inner steel profiles (the HD
sections) and 1 reinforced concrete web. Section bs is a chain of
components in concrete and steel; its strength should be calculated
on the basis of the weakest link, which is concrete. This is why
the check for transverse shear is made on one section bs
homogenized in concrete. All components of that section are taken
into account. This is valid because the steel profile has more
strength than its equivalent concrete section, as shown by the
following comparison of pure shear resistance:
-
for the steel profile: Vn = 0.6FyAwCv (Spec. Eq G2-1) For webs
of rolled I-shaped members with
-
43
h/tw 2, 24y
EF
=> v = 1.00 and Cv = 1.0 (Spec. Eq G2-2)
The HD section height and web thickness are: h = 600 mm tw=100
mm
h/tw = 6 2, 24y
EF
= 47,2
Aw = 100 x (600- 2 x 140) =32000 mm2 Vn HD = 0.6FyAwCv = 0,60x
450 x 32000 x 1 = 8640 kN vVn = 8640 > Vu,t,S2= 2134 kN where
Vu,t,S2 is the transverse shear in the steel profile as calculated
in the section Resistance to transverse shear of the HD
profile.
- For a concrete section with same height and width of concrete
equivalent to the steel profile, the shear strength in case of pure
shear applied to a section without transverse reinforcement is: Vc
= 2 fc bw d ACI318-08 (11-3) =1,0 for normal weight concrete In
international units (N, mm), ACI318-08 (11-3) expression becomes:
Vc =0,1693 fc bw d =1,1971 bw d for fc=50MPa A similar expression
is defined, which takes into account longitudinal reinforcement
with an upper bound value: Vc = 3,5 fc bw d ACI318-08 (11-5) In
international units (N, mm), it becomes: Vc = 0,2963 fc bw d =
2,095 bw d for fc=50MPa bw = 100 x 200000/38004 =526 mm d = h1 =
600 mm Vc = 2,095 bw d= 661 kN Vc =0,75 x 661,1 = 496 kN
It results: vVn,HD = 0.6FyAwCv = 8640 kN > Vc = 496 kN And it
can be concluded that it is justified to make concrete only checks
for shear resistance: the concrete part of the section bs is weaker
and would fail first. The extra strength of the steel profiles
above concrete strength has no use, as it would only intervene
after crushing the concrete web. [Note: the expression found in the
provisory version of Eurocode 2 or ENV1992 indicated a value for
the shear resistance VRd very similar to Vc = 2,095 bw d :
( )1 1,2 40Rd Rd w lV b dk = + =0,48 x(1,2 +40 x 0,104) bwd =
2,57 bwd with l = 0,104 for the section defined in this
example].
Why Option 3 cannot be applied to sections with several encased
steel profiles.
In option 3, the available shear strength would be found as the
addition of the available shear strength of the steel sections in
addition to the available shear strength of the reinforcing steel,
ignoring the contribution of the concrete. In fact, this way to
present things does not express clearly what is meant. The idea is
that, due to
-
44
cracking, the contribution to shear resistance of concrete
without transverse reinforcement Vc is equal to 0. In such case,
the shear resistance of reinforced concrete in shear can exist, due
to transverse reinforcement and equilibrium between inclined
compression struts of concrete and tension in steel ties, the
stirrups. In such case, the available shear strength indicated in
ACI318-08 is Vs , meaning that the total shear strength is only Vs
instead of being (Vc + Vs). But Vs is limited to an upper bound
value corresponding to crushing of concrete compression struts in
the strut and ties equilibrium recalled above. That limit is: Vs =
8 fc bw d ACI318-08 (11.4.7.9) That expression expressed in
international units, with the data of the section under
consideration, becomes: Vs = 0,676fc bw d = 4,78 bw d for fc =
50MPa [Note: it is remarkable that in option 3, the available shear
strength is said to be the addition of the available shear strength
of the steel section to the available shear strength of the
reinforcing steel, but the available shear strength of the
reinforcing steel is in fact a concrete strength]. However, in a
section with several steel profiles, the applied shear force Vu,Y
will distribute itself into Vu,bc3 , Vu,bc4 and Vu,bs between
sections bc3, bc4 and bs, proportionally to the stiffness of those
sections. Section bs being made of components working in series or
chain, the strength of the chain should be calculated on the basis
of its weakest link, which is concrete. So Option 3 has to be
applied in the same way as Option 2 and adding the shear strength
of the steel profiles to a shear strength of the reinforcement
would lead to an unsafe design.
Distribution of transverse shear in the composite section.
The symbols are defined at Figure I.X3-2. The width bc3, bs and
bc4 are: bc3 = 286mm bs = 476 mm bc4 = 3072 2 x (286+476) = 1548
mm
-
45
Fig. I.X3-2. Definition of sections bc3, bc4, and bs.
Fig. I.X3-3. Position of the reinforcement and the HD
profiles.
The applied shear force Vu,Y is distributed between sections
bc3, bc4 and bs proportionally to their stiffness: Vu,bc3 = Vu,Y x
(EIeff)bc3/EIeff Vu,bc4 = Vu,Y x (EIeff)bc4/EIeff Vu,bs = Vu,Y x
(EIeff)bs/EIeff
The effective bending stiffness EIeff of the column is: EIeff =
Es Is + 0.5Esr Isr + C1Ec Ic (Spec. Eq.I2-6) C1= 0.1+ 2
[(As/(Ac+As)] 0,3 (Spec. Eq.I2-7) In the envisaged section, there
are 4 steel profiles, each with a section A.
-
46
For a HD400x1299: A = 165000 mm2 Total section of 4 profiles: As
= 4 A = 660000 mm2
There are 256 diameter 40mm reinforcing bars. Asr = 256 x 1257 =
321792 mm2
Ac = Ag As Asr
Ac=3072 x 3072 660000 321792 = 8455392 mm2
C1 = 0,1 + 2[(660000/(8455392 + 660000)] = 0,245 0,3
Fig. I.X3-4. Definition of plates equivalent to bars.
The total effective bending stiffness EIeff around the X axis is
the sum of individual EIeff established for sections bc3, bs and
bc4 respectively.
Section bc3: EIeff = 0.5 Esr Isr + C1Ec Ic (Spec. Eq.I2-6) To
calculate Isr, reinforcing bars are replaced in the calculations by
2 equivalent steel side plates. Figure I.X4-4. Each plate has the
same total area As,r,side as the 2 layers of side bars in lines
plus 2x2 bars in the top and bottom lines plus 2x4 inside bars. The
height hp of the plates is the distance between the extreme bars:
hp = 3072 2 x 86 = 2900mm On each side:
- the number of bars is: 30 +30 + 2 x (2+4) = 72 - the area of
those bars and of each equivalent plate is: As,r,side = 72 x 1257=
90504 mm2 - the thickness of the equivalent plate is: tp= As,r,side
/ hp = 90504 / 2900 = 31,20 mm
-
47
- Isr = tp hp 3/12= 31,20 x 29003/12 =6,34.1010 mm4 Icg = bc3 x
hy3/12 = 286 x 30723/12 = 6,9095.1011 mm4 The moment of inertia Isr
corresponding to the fact that there is no concrete where there are
rebars should be deduced from Icg. Ic = Icg - Isr= 6,9095.1011
6,34.1010 = 6,27.1011 mm4 (EIeff)bc3= 0.5 Esr Isr + C1Ec Ic
(EIeff)bc3=0,5 x 200000 x 6,34.1010 + 0,245 x 38007 x
6,27.1011=1,22.1016 mm2
Section bc4: EIeff = 0.5 Esr Isr + C1Ec Ic (Spec. Eq.I2-6) To
calculate Isr, reinforcing bars are replaced in the calculations by
2 equivalent (one top, one bottom) steel plates. Each plate has the
same total area As,r,bc4 as the 2 layers of bars in lines, either
top or bottom side. The area of those bars and of each equivalent
plate is: As,r,bc4 = 32T40 bars =32 x 1257= 40224 mm2
The distance between the center of the top and bottom plates is
: Y= 3072 2 x (86 + 100/2)= 2800 mm Isr= As,r,bc4 x 2 x (Y/2)2 =
40224 x 2 x 14002 = 1,58.1011
Icg = bc4 x hy3/12 = 1548 x 30723/12 = 37,39.1011 The moment of
inertia Isr corresponding to the fact that there is no concrete
where there are rebars should be deduced from Icg. Ic = Icg -
Isr=37,39.1011 1,58.1011 = 35,81. 1011 (EIeff )bc4 = 0.5 Esr Isr +
C1Ec Ic = 0,5 x 200000 x 1,58.1011 + 0,245 x 38007 x 35,81.1011
(EIeff )bc4 = 4,91.1016 Nmm2
Section bs: EIeff = Es Is + 0.5 Esr Isr + C1Ec Ic (Spec.
Eq.I2-6) bs = 476 mm The distance between the centroid of the steel
profiles is dy. dy =2 x 950= 1900 mm dy /2= 950 mm Is=2 x A x
(dy/2)2 + 2 x Iy = 2 x 165000 x 9502 + 2 x754600.104 = 3,1292.1011
mm4 Iy is the moment of inertia of the steel profile, strong
axis.
EsIs=200000 x 3,1292.1011 = 6,258.1016 Nmm2
To calculate Isr, reinforcing bars are replaced in the
calculations by 2 equivalent (one top, one bottom) steel plates.
Each plate has the same total area As,r,bs as the 2 layers of bars
in lines, either top or bottom side above the HD section, meaning 8
bars. As,r,bs = 8x1257= 10056 mm2 The distance between the center
of the top and bottom plates is: Y= 3072 2 x (86 + 100/2)= 2800 mm
Isr= As,r,bs x 2 x (Y/2)2 = 10056 x 2 x 14002 = 3,94.1010 EsrIsr =
200000 x 3,94.1010 = 7,88.1015 N x mm
-
48
Icg = bs x hy3/12 = 476 x 30723/12=1,15.1012
Ic = Icg - Isr -Ics = 1,15.1012 - 3,1292.1011- 3,94.1010=
7.98.1011 Ec Ic = 38007 x 7.98.1011= 3,03.1016 N x mm
(EIeff)bs=6,258.1016 +0,5 x 7,88.1015 + 0,245 x 3,03.1016 =
7,39.1016 N x mm
EIeff =2 (EIeff)bc3 + (EIeff)bc4 +2 (EIeff)bs =2 x 1,22.1016 +
4,91.1016 +2 x 7,39.1016 EIeff =2,21.1017 N x mm
The factored shear Vu,Y = 20000 kN for the complete section is
distributed in the 5 sections (2 bc3, 2 bs, 1 bc4) contributing to
the shear strength of the complete section: Vu,bc3= Vu,Y x
(EIeff)bc3/EIeff = 20000 x 1,22.1016/2,21.1017 = 1104 kN Vu,bs =
Vu,Y x (EIeff)bs/EIeff = 20000 x 7,39.1016/2,21.1017 = 6688 kN
Vu,bc4= Vu,Y x (EIeff)bc4/EIeff = 20000 x 4,91.1016/2,21.1017 =
4430 kN
Section bs is a composite steel-concrete section having 2
reinforced concrete flanges, 2 steel flanges (the HD sections) and
1 reinforced concrete web. To establish longitudinal shear in
section bs, it is convenient to transform the composite section
into a single material section or homogenized section. The single
material can be either steel or concrete. Choosing concrete, the
moment of inertia of the homogenized concrete section Ic* is such
that the stiffness Ec Ic* of the homogenized section is equal to
the stiffness (EIeff)bs : Ic*= (EIeff)bs/Ec = 7,39.1016/38007 =
1,94.1012 mm4
Fig. I.X3 - 5. Homogenized equivalent concrete section bs.
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49
In a homogenized section in concrete (Figure I.X4.5.), the width
of concrete equivalent to the width of steel flanges is bs*: bs*=bs
x Es/Ec =476 x 200000/38007= 2505 mm The width of concrete
equivalent to the width of steel web is tw*: tw*=tw x Es/Ec =100 x
200000/38007= 52,6 mm The homogenized concrete section is presented
at Figure I.X3-5.
Calculation of shear in section bs.
The resultant longitudinal shear force on sections like CC1 and
CC2 at Figure I.X3-5 is: Vu,l = (Vu,bs x S) / Is* S is the section
modulus corresponding to the area limited by a sections CC1 and
CC2. Longitudinal shear is calculated at the interface between
sections C1 and C2 and at the interface between sections C2 and C,
because these steel concrete interfaces are the surfaces where
resistance to longitudinal shear should be checked.
Calculation of longitudinal shear force applied at section
CC1.
S1 is the section modulus for the section C1 ranging from edge
to outer HD flange: For the steel equivalent to concrete, width is
bs*. Height h1 is: h1= 1536 950 600/2 =286 mm The area is:
Area1= bs x h1= 476 x 286 = 136136 mm2 Distance to neutral axis:
950+300+286/2 = 1393 mm Sc1 =136136 x 1393 = 189,6.106 mm3 The
reinforcing bars have been replaced in the calculations by 1
equivalent steel plate (see above) of area As,r,bs : As,r,bs =
8x1257= 10056 mm2 The equivalent area in concrete is: 10056 x
200000/38004 =52920 mm2
The distance between the center of that plate and the axis of
symmetry is: py = Y/2 = 1400 mm Ssr = 52920 x 1400 = 74,1.106
S1= Sc1 + Ssr = 189,6.106 + 74,1.106 = 263,7.106 mm3
The resultant longitudinal shear force on section CC1 is:
Vu,l,CC1 = (Vu,bs x S1) / Ic* = (6688.103x 263,7.106)/ 2,05.1012 =
860 N/mm On 1 m length of column: Vu,l,CC1 = 860 kN/m
Calculation of longitudinal shear force applied at section
CC2.
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50
S2 is the section modulus for the sections C1 plus C2 (the HD
profile). A= 165000 mm2 The equivalent area in concrete for the HD
profile is: 165000 x 200000/38004 =868329 mm2
Distance of HD center to neutral axis: 950 mm SHD= 868329 x 950
= 824,9.106mm3 The concrete between the flanges is not taken into
account as its contribution would require specific stirrups and
connectors welded on or going through the web of the steel profile,
as stated in Eurocode 4 cl.6.3.3(2). S2= Sc1 + Ssr + SHD =
189,6.106 + 74,1.106 + 824,9.106 = 1088,6.106 mm3 Vu,l,CC2 = (Vu,bs
x S2) / Ic* = (6688.103x 1088,6.106)/ 2,05.1012 = 3551 N/mm On 1 m
length of column: Vu,l,CC2 = 3551 kN/m
Resistance to longitudinal shear by headed studs.
The available shear strength of an individual steel headed stud
anchor is determined in accordance with the composite component
provisions of AISC Specification Section I8.3 as directed by
Section I6.3b. Qnv =Fu Asa (Spec. Eq. I8-3) Asa = (19)2/4 = 283 mm2
per steel headed stud anchor diameter 19mm Fu = 450 MPa v = 0,65 v
Qnv = 0,65 x 450 x 283 = 82777 kN = 82,8 kN
In Section CC1, the required number of anchors on the outer
flange of the HD profile is: nanchors= 860/82,8 = 10,3/m Anchors
are placed in pairs on the flange at 120mm longitudinal (vertical)
spacing, meaning: 8,33 x 2 = 16,6 anchors/m. The 120mm longitudinal
(vertical) spacing is justified by spacing of main horizontal
reinforcement. Transverse spacing is 200mm.
In Section CC2, the required number of anchors on the inner
flange of the HD profile is: nanchors= 3551/82,8 = 42,9/m Anchors
are placed in groups of 5 on the flange at 120mm longitudinal
(vertical) spacing, meaning: 8,33 x 5 = 41,7 anchors/m. That value
is considered acceptable due to presence of shear connections on
lateral (=web) side of the section. Transverse spacing is 90mm.
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51
Outer side
Inner side
Fig. I.X3-6. Fig. Headed studs for resistance to longitudinal
shear.
Steel Headed Stud Anchor Detailing Limitations of AISC
Specification Sections I6.4a, I8.1 and I8.3
Steel headed stud anchor detailing limitations are reviewed in
this section with reference to the anchor having a shank diameter,
dsa =19mm. Anchors must be placed on at least two faces of the
steel shape in a generally symmetric configuration.
That limitation applies at sections with one central encased
steel profile. Here the anchors are placed symmetrically with
respect to the axis os symmetry of the complete section. There is
no reason to have
a symmetric configuration for each individual steel profile.
Maximum anchor diameter: dsa 2.5t f 19 < 2.5 x140 = 355 mm.
=> o.k.
Minimum steel headed stud anchor height-to-diameter ratio: h /
dsa 5 The minimum ratio of installed anchor height (base to top of
head), h, to shank diameter, dsa, must meet the provisions of AISC
Specification Section I8.3. For shear in normal weight concrete the
limiting ratio is five.
h / dsa = 100/19=5,26 5
Minimum lateral clear concrete cover = 25,4mm The lateral clear
includes distance bc3=286 mm > 25,4 mm => o.k.
Minimum anchor spacing: smin = 4 dsa=4 x 19 =76mm In accordance
with AISC Specification Section I8.3e, this spacing limit applies
in any direction. Minimum longitudinal spacing is 120mm => o.k.
Minimum lateral spacing is 90mm => o.k.
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52
Maximum anchor spacing:
In accordance with AISC Specification Section I6.4a, the spacing
limits of Section I8.1 apply to steel anchor spacing both within
and outside of the load introduction region. smax= 32 dsa=32 x 19
=608 mm Maximum spacing s =200 mm smax =>o.k.
Clear cover above the top of the steel headed stud anchors:
Minimum clear cover over the top of the steel headed stud anchors
is not explicitly specified for steel anchors in composite
components; however, in keeping with the intent of AISC
Specification Section I1.1 and following the cover requirements of
ACI 318 Section 7.7. for concrete columns, a clear cover of 38 mm
is requested. Clear cover above anchor = 286 -100 =186mm > 38mm
= > o.k.
Concrete Breakout
AISC Specification Section I8.3a states that in order to use
Equation I8-3 for shear strength calculations, concrete breakout
strength in shear must not be an applicable limit state. For the
composite member being designed, no free edge exists in the
direction of shear transfer along the length of the column, and
concrete breakout in this direction is not an applicable limit
state. However, it is still incumbent upon the engineer to review
the possibility of concrete breakout through a side edge parallel
to the line of force. One method for explicitly performing this
check is to analyze transverse reinforcing ties as anchor
reinforcement in accordance with AISC Specification Section
I8.3a(1) : Where anchor reinforcement is developed in accordance
with Chapter 12 of ACI 318 on both sides of the concrete breakout
surface for the steel headed stud anchor, the minimum of the steel
nominal shear strength from Equation I8-3 and the nominal strength
of the anchor reinforcement shall be used for the nominal shear
strength, Qnv, of the steel headed stud anchor.
The reinforcement present in the breakout surface are 2T20
(vertical spacing 120mm) and 6 T14 (vertical spacing 240mm) for a
total area : As = 2x 9,33 x 314 + 6x 9,33 x 154 = 5856 + 8620 =
14481mm2
Nominal strength of the anchor reinforcement:
0,65 x 500 x 14580 mm2 = 4706000N = 4706 kN > 3551 kN =>
ok.
Eurocode 4 provides a rule (clause 6.7.4.2(9)) which reflects
more directly the state of equilibrium in the vicinity of the
connectors: the transverse reinforcement should be designed for the
longitudinal shear that results from the transmission of normal
force from the parts of concrete directly connected by shear
connectors into the parts of the concrete without direct shear
connection. The design and arrangement of transverse reinforcement
should be based on a truss model assuming an angle of 45 between
concrete compression struts and the member axis. Figure I.X3-7. Due
to the 45 angle, the tie design force is equal to one half of the
longitudinal shear Vu,l,CC2.
For a longitudinal shear Vu,l,CC2 = 3551 kN/m, the design force
for
