design example of a column with 4 -...
TRANSCRIPT
DESIGN EXAMPLE OF A COLUMN WITH 4 ENCASED STEEL PROFILES
André Plumier (Plumiecs & ULg) : [email protected] (Main Contact)
Teodora Bogdan (ULg) : [email protected]
Hervé Degée (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 doesn’t
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.
1
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.
f’ c = 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.
2
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.
δ = steel contribution ratio.
3
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 110°C 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
110°C? 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.
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 0°C and
with low hydrogen electrodes) as well as their outstanding toughness. (27J up to minus 50°C). 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.
5
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 AXIA L 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
6
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
7
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 0.85
2c c
D
A f'P
⋅ ⋅=
( ) ( )10.85
2'
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
r2x
b hZ
⋅=
4sx a syZ A d= ⋅ ⋅
r2xZ4
21 2
cx r1x sx
h hZ Z Z
⋅= − − −
Fig. I -1e. Composite member with several encased steel profiles, X-X -axis anchor points.
8
B 0BP = aA
b*h
=
where: b* - the width of the equivalent steel rectangle bar Aa – area of one steel profile
For hnx between the two profiles 2nx sy
hh d ≤ −
:
( ) ( )1 0.85 2 0.85C
nx
c s2 yrs c
Ph
2 h f' b 2 F f'=
⋅ ⋅ ⋅ + ⋅ ⋅ ⋅ − ⋅ 2
cxn 1 nx r2xnZ h h Z= ⋅ − 2 2
r2xn 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 sy
h 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'
h2 h f' b* 2 F f' b 2 F f'
+ ⋅ − ⋅ ⋅ ⋅ − ⋅ =
⋅ ⋅ ⋅ + ⋅ ⋅ ⋅ − ⋅ + ⋅ ⋅ ⋅ − ⋅
2 2r2xn s2 nxZ b h= ⋅ ⋅
( ) ( ) ( )2
2 2 24
sy
sxn 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 <h + ≤
:
( )( ) ( )
2 0.85
0.85 2 2 0.85
C s y c ynx
1 c s2 yrs c
P A F f'h
2 h f' b F f'
− ⋅ ⋅ − ⋅=
⋅ ⋅ ⋅ + ⋅ ⋅ ⋅ − ⋅
2 2r2xn s2 nxZ b h= ⋅ ⋅ sxn sxZ Z=
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'= − ⋅ − ⋅ − ⋅ ⋅ ⋅
sxnZ - x-axis plastic modulus of equivalent steel
rectangle bar within the zone 2hnx
cxnZ - x-axis plastic modulus of concrete section
within the zone 2hnx
r2xnZ - x-axis plastic modulus of As2 plate within the
zone 2hnx
Fig. I -1e. Composite member with several encased steel profiles, X-X -axis anchor points. (continued)
9
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 0.85
2c c
D
A f'P
⋅ ⋅=
( ) ( )10.85
2'
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
2s1b
24
s1r1y
dZ
⋅= ⋅
s2x2r2y s2Z A d= ⋅ ⋅
4sy a sxZ A d= ⋅ ⋅
r2yZ4
21 2
cy r1y sy
h hZ Z Z
⋅= − − −
Fig. I -1f. Composite member with several encased steel profiles, Y-Y -axis anchor points.
10
B 0BP = aA
h*b
=
where: h* - the height of the equivalent steel rectangle bar Aa – area of one steel profile
For hny between the two profiles 2ny sx
bh d ≤ −
:
( ) ( )2 0.85 2 0.85C
ny
c s1 yrs c
Ph
2 h f' b 2 F f'=
⋅ ⋅ ⋅ + ⋅ ⋅ ⋅ − ⋅
22
cyn ny r1ynZ h h Z= ⋅ − 2 2
r1yn 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 sx
b bd h d − < ≤ +
:
( )( ) ( ) ( )2 s1
4 0.852
0.85 4 0.85 2 0.85
C sx y c
ny
c y c yrs c
bP d h* 2 F f'
h2 h f' h* 2 F f' b 2 F f'
+ ⋅ − ⋅ ⋅ ⋅ − ⋅ =
⋅ ⋅ ⋅ + ⋅ ⋅ ⋅ − ⋅ + ⋅ ⋅ ⋅ − ⋅
2 2r1yn s1 nyZ b h= ⋅ ⋅
( ) ( ) ( )2
2 2 24
sxsyn 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 0.85
0.85 2 2 0.85
C s y c
ny
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).
11
EXAMPLE I.X1 - COMPOSITE COLUMN WITH FOUR ENCASED ST EEL 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;
12
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 1
321699.09mm=
= =∑sr sriA A ;
42 2
i 1
4 165000mm 660000mms aA A=
= = ⋅ =∑;
2 2 2 6 29437184mm 321699.09mm 660000mm 8.455 10 mmc g sr sA A A A= − − = − − = ⋅
( ) 3
kg0.043 38007MPa for 2500
m= ⋅ = =1.5 '
c c c cE w f w;
Es = 200000 MPa (AISC I1.3);
2 2
2
321699.09mm 660000mm0.104
94371.84mmsr s
g
A A
A
+ += =
( )5 2 5 2 6 2
0.85
6.6 10 mm 450MPa 3.216 10 mm 500MPa 8.455 10 mm 0.85 50MPa
817207.653kN
n s y sr ysr c cP A F A F A f'
= ⋅ + ⋅ + ⋅ ⋅
= ⋅ ⋅ + ⋅ ⋅ + ⋅ ⋅ ⋅=
2y 660000mm 450MPa
0.363817207.653kN
s
n
A Fδ
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.
13
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 100
3072mm (86mm mm) 2800mm2
= − + =s1h bs1 = 23.338 mm;
2800mm1400mm
2 2s1
s1y
hd = = =
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;
2900mm1450mm
2 2s2
s2x
hd = = =
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 23.338mm2.966 10 mm
12 12
3s1 s1
sr1x
h bI
⋅⋅= = = ⋅
;
( )3
10 432.933mm 2900mm6.693 10 mm
12 12
3s2 s2
sr2x
b hI
⋅⋅= = = ⋅
;
( )26 4 10 4 4 2
11 4
2 2 2
2 2.966 10 mm 2 6.693 10 mm 2 6.534 10 mm 1400mm
3.9 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.
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 mm2
sx 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 423072mm 3072mm
7.421 10 mm12 12
31
g
h hI
⋅⋅= = = ⋅
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.
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:
3072mm0.5 min 1536mm
3072mm1
max2
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 reinforcement’s 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 2
6.6 10 mm0.070mm
9.437 10 mm
⋅ =⋅
o. k.
(2) Minimum longitudinal reinforcement ratio: / 0.004sr gA A ≥
5 2
6 2
3.216 10 mm0.034mm
9.437 10 mm
⋅ =⋅
o. k.
(3) Maximum longitudinal reinforcement ratio: / 0.08sr gA A ≤
5 2
6 2
3.216 10 mm0.034mm
9.437 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 60mmmax 60mm
38mmb
min
ds
⋅ = = =
100mm 40mm 60 mins 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 min
bs 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 min
hs 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 2
0.85
6.6 10 mm 450MPa 3.216 10 mm 500MPa 8.455 10 mm 0.85 50MPa
817207.653kN
A s y sr ysr c cP A F A F A f'
= ⋅ + ⋅ + ⋅ ⋅
= ⋅ ⋅ + ⋅ ⋅ + ⋅ ⋅ ⋅=
0kNmAM =
Point D (maximum nominal moment strength):
( )6 2
0.85
28.455 10 mm 0.85 50MPa
2179679.054kN
c cD
A 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 mmsr1x s1 s1yZ A d= ⋅ ⋅ = ⋅ ⋅ ⋅ = ⋅ ;
( )2
8 332.933mm 2900mm2 2 1.385 10 mm
4 4
2s2 s2
sr2x
b hZ
⋅⋅= = = ⋅ ;
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
2
8 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 ½ 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
10.85
21
6.27 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. I–1e 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.
165000mm275mm
600mmaA
b*=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 sy
hh d ≤ −
:
( ) ( )1
0.85 359358.109kN2 0.85 2 2 0.85
C 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 0.85
359358.109kN
3072mm 0.85 50MPa 2 32.933mm 2 500MPa 0.85 50MPa
1.108m
Cnx
c s2 yrs c
Ph
2 h f' b F f'
= 2
=⋅ ⋅ ⋅ + ⋅ ⋅ ⋅ − ⋅
⋅ ⋅ ⋅ + ⋅ ⋅ ⋅ − ⋅=
Check assumption2nx sy
hh d ≤ −
:
1108mm 650mm2nx sy
hh d= ≤ − = assumption not. k.
Assumption 2: the neutral axis is placed within the steel profiles2 2sy nx sy
h hd <h d − ≤ +
:
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 nx
0.85 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 0.85
23072mm 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'
h2 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 sy
h hd <h d − ≤ +
:
650mm 767mm 1250mm2 2sy nx sy
h hd h d− = < = < + = assumption o. k.
( )2
7 3
232.933mm 767mm
3.878 10 mm
2sr2xn s2 nxZ b h
=2 =
= ⋅ ⋅⋅ ⋅
⋅
( ) ( ) ( )
( ) ( ) ( ) ( ) ( )
2
2 2
22 2
7 3
24
275mm 2 950mm 600mm2 1012mm 275mm 767mm 2 767mm - 275mm 1108mm 2
49.141 10 mm
sy
sxn sx nx sx nx
b* 2 d hZ 2 d +b* h 2 d -b* h
=
⋅ ⋅ −= ⋅ ⋅ − ⋅ ⋅ − ⋅
⋅ ⋅ −= ⋅ + ⋅ − ⋅ ⋅ − ⋅
⋅
( )2 7 3 7 3
9 3
3072mm 767mm 3.878 10 mm 9.141 10 mm
1.687 10 mm
2cxn 1 nx sr2xn sxnZ h h Z Z
=
=
= ⋅ − −
⋅ − ⋅ − ⋅
⋅
21
( )
( )7 3 7 3 9 3
1 0.8521
576734.208kNm 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
0.85
0.85 8.455 10 mm 50MPa
359358.109kN
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 =φ
whereX" C XP P
X = A,B,C or D
= ⋅φ
0.75 817207.65kN 612905.739kNA" C AP P
= ⋅= ⋅ =
φ
0.75 0kN 0kNB" C BP P
= ⋅= ⋅ =
φ
0.75 359358.109kN 269518.582kNC" c CP P
= ⋅= ⋅ =
φ
0.75 179679.054kN 134759.291kND" c DP P
= ⋅= ⋅ =
φ
Allowable compressive strength:
C 2.00=Ω
where
XX"
C
PP
X = A,B,C or D
=Ω
817207.65kN408603.826kN
2
AA"
c
PP
=
= =
Ω
0kN0kN
2
BB
c
PP
=
= =
Ω
C
359358.109kN179679.055kN
2
C"c
PP
=
= =
Ω
D
179679.054kN89839.53kN
2
D"c
PP
=
= =
Ω
22
Design flexural strength:
0.90B =φ
whereX" B XM M
X = A,B,C or D
= ⋅φ
0.9 0kNm 0kNmAx" B AxM M
φ= ⋅= ⋅ =
0.9 480546kNm 432491kNmBx" B BxM M
φ= ⋅= ⋅ =
0.9 480546kNm 432491kNmCx" B CxM M
φ= ⋅= ⋅ =
0.90 576734.208kNm 519060.787kNmDx" B DxM M
= ⋅= ⋅ =
φ
Allowable compressive strength:
b 1.67=Ω
b
where
XX"
MM
X = A,B,C or D
=Ω
b
0kNm0kNm
1.67
AxAx"
MM
=
= =
Ω
b
480546kNm287752kNm
1.67
BxBx"
MM
Ω=
= =
b
480546kNm287752kNm
1.67
CxCx"
MM
Ω=
= =
b
576734.208kNm345349.825kNm
1.67
DxDx"
MM
=
= =
Ω
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 Liège
and Engineering office Greisch [FineLg User’s 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 ccuc
c c
fε εσε ε
= ⋅ ⋅ −
2 'c
c
f
Eε ⋅
=
2
3ct cc0.3 4.072MPaf f= ⋅ = .
where:
fcc = 50 MPa –compressive strength of unconfined concrete (AISC I1.2b)
fct – axial tensile strength of concrete;
εccu = 0.003 – ultimate compressive strain of unconfined concrete (AISC I1.2b);
εc – 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 lΜ = ⋅ . 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 PROFIL ES 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 1
321699.09mm=
= =∑sr sriA A ;
42 2
i 1
4 165000mm 660000mms aA A=
= = ⋅ =∑;
2 2 2 6 29437184mm 321699.09mm 660000mm 8.455 10 mmc g sr sA A A A= − − = − − = ⋅
( ) 3
kg0.043 38007MPa for 2500
m= ⋅ = =1.5 '
c c c cE w f w;
Es = 200000 MPa (AISC I1.3);
2 2
2
321699.09mm 660000mm0.104
94371.84mmsr s
g
A A
A
+ += =
( )5 2 5 2 6 2
0.85
6.6 10 mm 450MPa 3.216 10 mm 500MPa 8.455 10 mm 0.85 50MPa
817207.653kN
n s y sr ysr c cP A F A F A f'
= ⋅ + ⋅ + ⋅ ⋅
= ⋅ ⋅ + ⋅ ⋅ + ⋅ ⋅ ⋅=
2y 660000mm 450MPa
0.363817207.653kN
s
n
A Fδ
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 100
3072mm (86mm mm) 2800mm2
= − + =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 2800mm4.269 10 mm
12 12
3s1 s1
sr1y
b hI
⋅⋅= = = ⋅
;
( )3
6 42900mm 32.933mm8.632 10 mm
12 12
3s2 s2
sr2y
h bI
⋅⋅= = = ⋅
( )210 4 6 4 4 2
11 4
2 2 2
2 4.269 10 mm 2 8.632 10 mm 2 9.55 10 mm 1450mm
4.87 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 mm2
sy 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 413072mm 3072mm
7.421 10 mm12 12
32
g
h hI
⋅⋅= = = ⋅
12 4 11 4 11 4 12 4- - 7.421 10 mm 4.87 10 mm 6.861 10 mm 6.428 10 mmcy 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:
3072mm0.5 min 1536mm
3072mm1
max2
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 2
6.6 10 mm0.070mm
9.437 10 mm
⋅ =⋅
o. k.
(2) Minimum longitudinal reinforcement ratio: / 0.004sr gA A ≥
5 2
6 2
3.216 10 mm0.034mm
9.437 10 mm
⋅ =⋅
o. k.
(3) Maximum longitudinal reinforcement ratio: / 0.08sr gA A ≤
5 2
6 2
3.216 10 mm0.034mm
9.437 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 60 mins 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 min
bs 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 min
hs 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 2
0.85
6.6 10 mm 450MPa 3.216 10 mm 500MPa 8.455 10 mm 0.85 50MPa
817207.653kN
A s y sr ysr c cP A F A F A f'
= ⋅ + ⋅ + ⋅ ⋅
= ⋅ ⋅ + ⋅ ⋅ + ⋅ ⋅ ⋅=
0kNmAM =
Point D (maximum nominal moment strength):
( )6 2
0.85
28.455 10 mm 0.85 50MPa
2179679.054kN
c cD
A 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 2800mm9.148 10 mm
4 4
2s1 s1
sr1y
b hZ
⋅⋅= = = ⋅
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 3
9 3
43072mm 3072mm
9.148 10 mm 2.769 10 mm 6.679 10 mm4
6.211 10 mm
21 2
cy r1y r2y sy
h hZ 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 ½ 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
10.85
21
6.679 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. I–1f 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.
165000mm346.639mm
476mmaA
h*=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 359358.109kN
0.85 2 2 0.85C B C c c
ny 2 c ny s1 yrs c
P P P A f' = 2 h h f' h b F f'
− = = ⋅ ⋅ =⋅ ⋅ ⋅ ⋅ + ⋅ ⋅ ⋅ ⋅ − ⋅
( ) ( )
( ) ( )
2 0.85 2 2 0.85
359358.109kN
3072mm 0.85 50MPa 8 23.338mm 2 500MPa 0.85 50MPa
1.175m
Cny
c s1 yrs c
Ph
2 h f' b F f'
= 2
=⋅ ⋅ ⋅ + ⋅ ⋅ ⋅ − ⋅
⋅ ⋅ ⋅ + ⋅ ⋅ ⋅ − ⋅=
Check assumption2ny sx
bh d ≤ −
:
1175mm 774mm2ny sx
bh d= ≤ − = assumption not. k.
Assumption 2: the neutral axis is placed within the steel profiles2 2sy nx sy
h hd <h d − ≤ +
:
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 359358.109kN
0.85 2 0.85 2 2 0.85C B C c c
ny 1 c s y c ny s1 yrs c
P 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 50MPa
23072mm 0.85 50MPa 4 346.639mm 2 45
C sx y c
ny
c y c yrs c
bP d h* 2 F f'
h2 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 sy
h hd <h d − ≤ +
:
774mm 856mm 1250mm2 2sx ny sx
b bd h d+ = < = ≤ + = assumption o. k.
( )2
7 3
2
23.338mm 856mm3.421 10 mm
2sr1yn s1 nyZ b h
=2 =
= ⋅ ⋅⋅ ⋅
⋅
( ) ( ) ( )
( ) ( ) ( ) ( ) ( )
2
2 2
22 2
7 3
24
346.639mm 2 950mm - 476mm2 950mm 346.639mm 856mm 2 950mm 346.639mm 856mm - 2
49.273 10 mm
sxsyn sy ny sy ny
h* 2 d bZ 2 d +h* h 2 d -h* h
⋅ ⋅ −= ⋅ ⋅ − ⋅ ⋅ − ⋅
⋅ ⋅= ⋅ + ⋅ − ⋅ − ⋅ ⋅
= ⋅
( )2 7 3 7 3
9 3
3072mm 1009mm 3.421 10 mm 9.273 10 mm
2.124 10 mm
2cyn 1 ny r1yn synZ h h Z Z
=
=
= ⋅ − −
⋅ − ⋅ − ⋅
⋅
38
( )
( )7 3 7 3 9 3
1 0.8521
616779.059kNm 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
0.85
0.85 8.455 10 mm 50MPa
359358.109kN
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 =φ
whereX" C XP P
X = A,B,C or D
= ⋅φ
0.75 817207.65kN 612905.739kNA" C AP P
= ⋅= ⋅ =
φ
0.75 0kN 0kNB" C BP P
= ⋅= ⋅ =
φ
0.75 359358.109kN 269518.582kNC" c CP P
= ⋅= ⋅ =
φ
0.75 179679.054kN 134759.291kND" c DP P
= ⋅= ⋅ =
φ
Allowable compressive strength:
C 2.00=Ω
where
XX"
C
PP
X = A,B,C or D
=Ω
817207.65kN408603.826kN
2
AA"
c
PP
=
= =
Ω
0kN0kN
2
BB
c
PP
=
= =
Ω
C
359358.109kN179679.055kN
2
C"c
PP
=
= =
Ω
D
179679.054kN89839.53kN
2
D"c
PP
=
= =
Ω
39
Design flexural strength:
0.90B =φ
whereX" B XM M
X = A,B,C or D
= ⋅φ
0.9 0kNm 0kNmAy" B AyM M
φ= ⋅
= ⋅ =
0.9 512804kNm 461524kNmBy" B ByM M
φ= ⋅
= ⋅ =
0.9 512804kNm 461524kNmCy" B CyM M
φ= ⋅
= ⋅ =
0.9 616779.059kNm 555101.153kNmDy" B DyM M
φ= ⋅
= ⋅ =
Allowable compressive strength:
b 1.67=Ω
b
where
XX"
MM
X = A,B,C or D
=Ω
b
0kNm0kNm
1.67
AyAy"
MM
=
= =
Ω
b
512804kNm307068kNm
1.67
ByBy"
MM
Ω=
= =
b
512804kNm307068kNm
1.67
CyCy"
MM
Ω=
= =
b
616779.059kNm369328.778kNm
1.67
DyDy"
MM
=
= =
Ω
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
E
F => Φv = 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
E
F= 47,2
Aw = 100 x (600- 2 x 140) =32000 mm2
Vn HD = 0.6FyAwCv = 0,60x 450 x 32000 x 1 = 8640 kN
ΦvVn = 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 λ √f’ c bw d ACI318-08 (11-3)
λ=1,0 for normal weight concrete
In international units (N, mm), ACI318-08 (11-3) expression becomes:
Vc =0,1693√ f’c bw d =1,1971 bw d for f’c=50MPa
A similar expression is defined, which takes into account longitudinal reinforcement with an upper
bound value:
Vc = 3,5 λ √f’ c bw d ACI318-08 (11-5)
In international units (N, mm), it becomes:
Vc = 0,2963√ f’c bw d = 2,095 bw d for f’c=50MPa
bw = 100 x 200000/38004 =526 mm
d = h1 = 600 mm
Vc = 2,095 bw d= 661 kN
Φ Vc =0,75 x 661,1 = 496 kN
It results: ΦvVn,HD = 0.6FyAwCv = 8640 kN > Φ 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 dkτ ρ= + =0,48 x(1,2 +40 x 0,104) bwd = 2,57 bwd with ρl = 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 λ √f’ c bw d ACI318-08 (11.4.7.9)
That expression expressed in international units, with the data of the section under consideration, becomes:
Vs = 0,676√f’ c bw d = 4,78 bw d for f’ c = 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.
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*=t w 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.
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 = π (19)2/4 = 283 mm2 per steel headed stud anchor diameter 19mm
Fu = 450 MPa
Φv = 0,65
Φv 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.
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.
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 ties in the strut and tie model is thus:
As,tie fy ≥ Vu,l,CC2 /2= 3551/2 = 1775 kN
53
As,tie fy ≥ 1775000/500 = 3551 mm2
The reinforcement present in the immediate vicinity of the connectors are 1T20 (vertical spacing 120mm) and
2T14 (vertical spacing 240mm) providing:
9,33 x (314 + 154)= 4366 mm2 > 3551 mm2 => ok.
Fig. I.X3-8. Struts and ties model for the design of transverse reinforcement.
Checks of resistance to transverse shear of section bs following Option 2—Available shear strength of the
reinforced concrete portion alone per ACI 318.
Design of reinforced concrete cross sections subject to shear is based on:
ΦVn ≥ Vu ACI318-08 (11-1)
Vu is the factored shear force at the section considered.
Vn is nominal shear strength computed by: Vn = Vc + Vs ACI318-08 (11-2)
Vc is the nominal shear strength provided by concrete and Vs the nominal shear strength provided by shear
reinforcement.
In ACI318-08 , Vc should be computed by different expressions related to the type of factored actions applied to
the section. The more refined expressions have upper bound limitations:
- shear Vu and flexure Mu : Eq.(11-3) or Eq.(11-5) with limitation;
- shear Vu and axial compression Nu : Eq.(11-4);
- shear Vu and axial tension Nu : Vc = 0 or Eq.(11-8) with limitation;
- shear Vu and flexure Mu and compression Nu: Eq.(11-5) completed by Eq(11-6) with limitation
expressed by Eq.(11.7).
A comparison of those various shear stress equations for members submitted to a combination of axial loads,
bending moment and shear, which is typically the case for the columns with HD profiles (see R11.2.2.2), is
presented in ACI318-08 Comments. It indicates that Eq.(11-4) is a correct safe side estimate of Vc. Figure I.X4-
54
8. Also, it obviates the determination of individual values of bending moment Mu and compression Nu attributed
to each zone bc3, bc4 and bs.
Fig. I.X3-8. Figure R.11.2.2.2 from ACI318-08.
2 1 '2000
uc c w
g
NV f b d
Aλ
= +
ACI318-08 (11-4)
λ=1,0 for normal weight concrete
f’ c = 50 MPa (or N/mm2)
Nu =180000 kN
Ag = 3072mm x 3072mm = 9437.103 mm2
In international units (N, mm), ACI318-08 (11-4) expression becomes for the considered gross section Ag and
the applied compression force Nu=180000kN:
Vc = 0,3974√ f’c bw d = 2,81 bw d for f’ c=50MPa
Resistance to transverse shear in sections bc3.
Sections bc3 are reinforced concrete sections. They are checked using ACI318-08 Eq.(11-4) for Vc.
bw = bc3= 286 mm
d = 0,8 hz = 0,8 x 3072mm = 2457 mm
Vc = 2,81 bw d = 1974 kN ACI318-08 Eq.(11-4)
Φ Vc =0,75 x 1974 = 1480 kN > Vu,bc3= 1104 kN
No transverse reinforcement, but the imposed minimum, is required in sections bc3.
Minimum Reinforcing Limits
Check that the minimum shear reinforcement is provided as required by ACI 318, Section 11.4.6.3.
',min
500,75 w w
v cyr yr
b s b sA f
f f
= ≥
ACI 318-08 Eq. (11-13)
55
In international units (N, mm), ACI318-08 (11-13) expression becomes:
'0,0623 0,007v w wc
yr yr
A b bf
s f f
= ≥
286 2860,0623 50 0,25 0,007 0,004
500 500vA
s = = ≥ =
This is realized with 1T14 bar close to column edge (external hoops) at step s =240mm
Av = 78,5mm2 Av /s=78,5/240= 0,32 > 0,25
Resistance to transverse shear in section bc4.
Section bc4 is a reinforced concrete section. It is checked using ACI318-08 Eq.(11-4) for Vc.
bw = bc4= 1548 mm
d = 0,8 hz = 0,8 x 3072 = 2457 mm
Vc = 2,81 bw d= 10687 kN ACI318-08 Eq.(11-4)
Φ Vc =0,75 x 10867 = 8150 kN > Vu,bc4= 4430 kN
No transverse reinforcement, but the imposed minimum, is required in section bc4.
1548 15480,0623 50 1,36 0,007 0,0217
500 500vA
s = = ≥ =
This can be realized for instance with 4T8 stirrups at step s =240mm
Av = 8 x 50,3 = 402 mm2 Av /s=402/240= 1,67 > 1,36
In the final design, T14 are placed to fulfill other criteria.
Resistance to transverse shear in section bs.
Vu,bs = 6688 kN
The area considered for shear resistance of the composite section of section bs is only the web of the complete
section bs presented at Figure I.X3-2. .
Φ(Vc +Vs) ≥ Vu,bs
Vs ≥ (Vu,bs- ΦVc)/ Φ=(6688 – 2464)/0,75= 5632 kN
Vs = Av fyt d/s
fyt = 500 MPa
For s = 120 mm: Av ≥ s x Vs/ fyt d =(120 x 5632000)/(500 x 2457) = 550 mm2
567 mm2 is realized by a T20 stirrup: 2 x 314 = 628 mm2 > 550 mm2 .
Resistance to transverse shear of the HD profile.
56
The shear force Vu,l per unit length in CC2 correspond to the maximum longitudinal shear stress in the HD
profile. The HD profile adequacy can be checked in shear under a shear force Vu,t,S2 calculated on the basis ofthat
safe side value Vu,l ,CC2 and on the fact that transverse and longitudinal shear stresses are equal at one point:
Vu,t,S2= Vu,l ,CC2 x h2= 3551 x 600= 2130600 N = 2131 kN
Vn = 0.6FyAwCv
Φv = 1.00 and Cv = 1.0 (see above)
h = 600 mm tw=100 mm
Aw = 100 x (600- 2 x 140) =32000 mm2
Vn = 0.6FyAwCv = 0,60x 450 x 32000 x 1 = 8640 kN
ΦvVn = 8640 kN > Vu,t,S2= 2131 kN
[Note:@100 for instance means that the vertical spacing is 100mm].
Fig. I.X3-8. Definition of transverse reinforcement resulting from all requirements in Examples I.X3 and I.X4 .
57
EXAMPLE I.X4 COMPOSITE COLUMN WITH 4 ENCASED STEEL PROFILES IN SHEAR
DIRECTION X.
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 controlling ASCE(2010) ASCE/SEI 7-10 load
combinations:
Factored bending moment: Mu,Y = 450000 kNm
Factored axial (compression) force: Nu = 180000 kN
Factored transverse shear Vu,X in direction X: Vu,X= 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 Ix = 254400.104 mm4
Solution:
Fig. I.X4-1. Definition of notations.
Example I.X4 studies the shear resistance of the same section as the one presented in Example I.X3. The line of
the developments presented here is similar to the one of Example I.X3. The differences are the direction of
shear, the orientation of the encased steel profiles and the way to provide resistance to longitudinal shear. The
introduction presented in Example I.X3 on available shear strength, principle of the adaptation of Option 2 to
58
sections with several encased steel profiles and why Option 3 cannot be applied to sections with several encased
steel profiles are not repeated here.
Distribution of transverse shear in the composite section.
The symbols and dimensions are defined at Figures I.X4-1, I.X4-2 and I.X4-3 .
The width bc1, hs and bc2 are:
bc1 = 286mm
hs = 600 mm
bc2 = 3072-2 x (286+600)=1300 mm
Fig. I.X4-2. Definition of sections bc1, bc2, and hs.
59
Fig. I.X4-3. Position of the reinforcement and the HD profiles.
The applied shear force Vu,Z is distributed between sections bc1, bc2 and hs proportionally to their stiffness:
Vu,bc1 = Vu,X x (EIeff)bc1/EIeff
Vu,bc2 = Vu,X x (EIeff)bc2/EIeff
Vu,hs = Vu,ZX x (EIeff)hs/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.
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
The total effective bending stiffness EIeff around the X axis is the sum of individual EIeff established for sections
bc1, hs and bc2 respectively.
Section bc1: 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 = 60
- the area of those bars and of each equivalent plate is: As,r,side = 60 x 1257= 75420 mm2
- the thickness of the equivalent plate is: tp= As,r,side / hp = 75420 / 2900 = 26 mm
- Isr = tp hp 3/12= 26 x 29003/12 =5,28.1010 mm4
Icg = bc1 x hx3/12 = 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 – 5,28.1010 = 6,38.1011 mm4
(EIeff)bc1= 0.5 Esr Isr + C1Ec Ic
60
(EIeff)bc1=0,5 x 200000 x 5,28.1010 + 0,245 x 38007 x 6,38.1011=1,12.1016 mm2
Fig. I.X4-4. Definition of plates equivalent to bars.
Section bc2: 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,bc2 = 32T40 bars =32 x 1257= 40224 mm2
The distance between the center of the top and bottom plates is :
X= 3072 – 2 x (86 + 100/2)= 2800 mm
Isr= As,r,bc2 x 2 x (X/2)2 = 40224 x 2 x 14002 = 1,58.1011 mm4
Icg = bc2 x hx3/12 = 1548 x 30723/12 = 37,39.1011 mm4
The moment of inertia Icsr 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 mm4
(EIeff )bc2 = 0.5 Esr Isr + C1Ec Ic = 0,5 x 200000 x 1,58.1011 + 0,245 x 38007 x 35,81.1011
(EIeff )bc2 = 4,91.1016 mm2
Section hs: EIeff = Es Is + 0.5 Esr Isr + C1Ec Ic (Spec. Eq.I2-6)
hs = 600 mm
The distance between the centroids of the steel profiles is dx.
dx =2 x 1012= 2024 mm
dx/2 = 1012 mm
Is=2 x A x (dx/2)2 + 2 x Ix = 2 x 165000 x 10122 + 2 x 254400.104 = 3,43.1011 N mm2
61
Ix is the moment of inertia of the steel profile, weak axis.
EsIs=200000 x 3,43.1011 = 6,86.1016 N x mm
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 2 x 6 + 3 =15 bars.
As,r,bs = 15x1257= 18855 mm2
The distance between the center of the top and bottom plates is:
X= 3072 – 2 x (86 + 100/2)= 2800 mm
Isr= As,r,bs x 2 x (X/2)2 = 18855 x 2 x 14002 = 7,39.1010
EsrIsr = 200000 x 7,39.1010 = 14,8.1015 N x mm
Icg = bs x hz3/12 = 476 x 30723/12=1,15.1012
Ic = Icg - Isr –Is = 1,15.1012 - 7,39.1010 – 3,43.1011 = 7,33.1011
Ec Ic = 38007 x 7,33.1011 = 2,79.1016 N x mm
(EIeff)hs=6,86.1016 +0,5 x 1,48.1016 + 0,245 x 2,79.1016 = 8,28.1016 N x mm
EIeff =2 (EIeff)bc1 + (EIeff)bc2+2 (EIeff)hs =2 x 1,12.1016 + 4,91.1016 +2 x 8,28.1016
EIeff =2,37.1017 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,bc1= Vu,X x (EIeff)bc1/EIeff = 20000 x 1,12.1016/2,37.1017 = 945 kN
Vu,hs = Vu,X x (EIeff)hs/EIeff = 20000 x 8,28.1016/2,37.1017 = 6987 kN
Vu,bc2= Vu,X x (EIeff)bc2/EIeff = 20000 x 4,91.1016/2,37.1017 = 4143 kN
Section hs 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)hs :
Ic*= (EIeff)hs/Ec = 8,28.1016/38007 = 2,18.1012 mm4
In a homogenized section in concrete (Figure I.X4-5), the width of concrete equivalent to the thickness of the 2
steel flanges is bs*:
bs*=2 x tf x Es/Ec =2 x 140 x 200000/38007= 1473 mm
The width of concrete equivalent to the height of the steel web is hw*:
bw*=h w x Es/Ec =320 x 200000/38007= 1684 mm
The homogenized concrete section is presented at Figure I.X4-5.
62
Fig. I.X4-5. Homogenized equivalent concrete section hs.
Calculation of shear in section bs.
The resultant longitudinal shear force on sections CC1 and CC2 at Figure I.X4-5 is:
Vu,l = (Vu,bs x S) / Ic*
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= bc3 =286 mm
The area is:
Area1= hs x h1= 600 x 286 = 171600 mm2
Distance to neutral axis: 950+476/2+286/2 = 1331 mm
Sc1 =171600 x 1331 = 228,4.106 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:
63
px = X/2 = 1400 mm
Ssr = 52920 x 1400 = 74,1.106
S1= Sc1 + Ssr = 228,4.106 + 74,1.106 = 302,4.106 mm3
The resultant longitudinal shear force on section CC1 is:
Vu,l,CC1 = (Vu,hs x S1) / Ic* = (6987.103x 302,4.106)/ 2,18.1012 = 969 N/mm
On 1 m length of column: Vu,l,CC1 = 969 kN/m
Calculation of longitudinal shear force applied at section CC2.
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: 1012 mm
SHD= 868329 x 1012 = 878,7.106mm3
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. Eurocode 4 cl.6.3.3(3)
S2= Sc1 + Ssr + SHD = 228,4.106 + 74,1.106 + 878,7.106 = 1181,2.106 mm3
Vu,l,CC2 = (Vu,hs x S2) / Ic* = (6987.103x 1181,2.106)/ 2,18.1012 = 3786 N/mm
On 1 m length of column: Vu,l,CC2 = 3786 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 = π (19)2/4 = 283 mm2 per steel headed stud anchor diameter 19mm
Fu = 450 MPa
Φv = 0,65
Φv Qnv = 0,65 x 450 x 283 = 82777 kN = 82,8 kN
In Section CC1, the required number of anchors on the web of the HD profile is:
nanchors= 969/82,8 = 11,7 = 12/m
Anchors are placed in pairs on the web with a longitudinal spacing = 150mm, 13,3 anchors/m.
Lateral spacing is 120 mm
In Section CC2, the required number of anchors on the web of the HD profile is:
nanchors= 3786/82,8 = 45,7/m
There are difficulties in placing the connectors in a single plane.
The connectors could be placed in 2 lines on the web and one line on each interior side of flanges, with a
longitudinal spacing = 90mm , for a total 44 anchors/m. Lateral spacing of web connectors would be 120 mm.
64
However, such a layout is not supported by experience; it creates a preferential failure plane AA-Figure IX4-6
and requires additional reinforcing hoops within the depth of the web.
Placing the connectors on the tip of the flanges, so that transverse tying bars can be effective raise another
problem: it is not possible to reach the required number of connectors/m, at least on the inner side, so that
additional hoops remain necessary.
Inner side Outer side
A
A
Fig. I.X4-6. Headed studs for resistance to longitudinal shear. Potential shear failure surface AA (top).
Required additional hoops (bottom).
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.
65
• 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 90mm => o.k.
Minimum lateral spacing is 100mm => o.k.
• 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 =150 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. 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.
Design checks relevant for concrete breakout are similar to those in Example I.X3, due to the symmetry of the
section.
Direct Bearing
One method of utilizing direct bearing as a load transfer mechanism is through the use of internal bearing plates
welded between the flanges of the encased W-shape as indicated in Figure I.X4-7. Where multiple sets of
bearing plates are used, it is recommended that the minimum spacing between plates be 2 times the width of the
66
plates so that concrete compression struts inclined at 45° can develop. That spacing also enhances
constructability and concrete consolidation.
Elevation view Plan view
Fig. I.X4-7. The strut and tie equilibrium justifying direct bearing on stiffeners.
Width of plate a: a = ( bf – tw)/2= (476 -100)/2= 188 mm
Length of plate b: b =h – 2tf = 600 – 2 x 140 = 320 mm
Width of clipped corners c : c= 15,4mm
Plate bearing area: Al = ab-c2 = 59922 mm2
The available strength for the direct bearing force transfer mechanism is:
Rn = 1,7 fc’ Al (Spec. Eq. I6-3)
That expression expresses that a force is a pressure times an area and is the same in international units (N, mm),
Rn = 1,7 x 50 x 59922 = 5093370 N = 5093 kN
The required resistance to longitudinal shear can be achieved by few more loaded and thicker stiffeners and
welds or more numerous less loaded thinner stiffeners and welds.
The expressions governing the design are:
ΦB Rn ≥ Vr’
Pressure wu on plate: wu =Vr’ /Al
Required bearing plate thickness tp (assuming b ≥ 2a and for tp < tf):
( )( )
22 3 2
3 6u
py
a w b at
F a b
−=
+φ (AISC Design Examples V14.0, Example I8)
For the longitudinal shear on the inner side of the HD sections:
Vr’= Vu,l,CC2 = 3786 kN/m
With stiffeners placed every 400mm (the minimum spacing being 2 x 188 = 376mm +tstiffener ≈ 400mm), the
required bearing force per stiffener is:
Vr’=3786 x 400/1000=1514 kN
ΦB Rn = 0,65 x 5093 = 3310 kN> 1514kN= Vr’
Pressure wu on plate: wu =Vr’ /Al = 1514000/59922=25,2 MPa
Required bearing plate thickness tp:
67
( )( )
22 3 2
3 6u
py
a w b at
F a b
−≥
+φ=
( )( )
22 188 24, 2 3 320 2 18827,1
3 0,90 345 6 188 320
× × × − ×=
× × × +mm
The bearing plates should be connected to the encased steel member using welds designed in accordance with
AISC Specification Chapter J to develop the full strength of the plate. For fillet welds, a weld size of 5/8tp will
serve to develop the strength of S355 plate (see AISC Manual Part 10).
For the longitudinal shear on the outer side of the HD sections:
Vr’= Vu,l,CC1 = 969 kN/m
With stiffeners placed every 800mm (the minimum spacing being 2 x 188 = 376mm +tstiffener ≈ 400mm), the
bearing force per stiffener is:
Vr’=969 x 800/1000= 775kN
ΦB Rn = 0,65 x 5093 = 3310 kN> 775kN= Vr’
Pressure wu on plate: wu =Vr’ /Al = 775000/59922=12,9 MPa
Required bearing plate thickness : tp = 19,2 mm.
Direct Bearing. Additional check in the strut and tie equilibrium justifying direct bearing by stiffeners.
The direct bearing which is provided by stiffeners welded between the flanges of a steel section requires an
equilibrating strength brought in by horizontal ties. See FigureI.X4-7. The tie design force for 1 stiffener is
equal to the longitudinal shear force Vr’ supported by that stiffener. It should be provided by the horizontal ties
in form of stirrups passing around both of the encased steel profiles in the section hs. As this tie force is just the
expression of the tie force in a global “strut and tie” mechanism of the complete hs section, this force should not
be added to the general transverse shear force which is taken into account to define the transverse reinforcement
in section hs. But it should be checked that the strength of the transverse reinforcement in section hs is great
enough:
Φ Av Fy ≥ Vr’
For the longitudinal shear on the inner side of the HD sections:
Vr’=1514 kN /stiffener
Vr’=1514 kN/0,4=3785 kN/m as stiffeners are placed every 0,4 m
As transverse reinforcement defined by the check for transverse shear on the complete section requires T20 with
spacing s=150mm (see further down),
Av holds for 1000/150 = 6,6 stirrups/m, meaning 13,3 T20 on 1 m, meaning :
Av = 13,3 x 314 = 4176 mm2
With S500 stirrups and Φ=0,75 ACI318-08 (9.3.2.6)
Φ Av Fy = 0,75 x 4176 x 500 = 1566000 N = 1566 kN > Vr’= 1514 kN
The stirrups defined in the check for transverse shear on the complete section are OK as ties for the bearing
force on stiffeners placed on the outer side of the HD profiles.
For the longitudinal shear on the outer side of the HD sections:
68
Vr’=775 kN /stiffener
Vr’=775 kN/0,8 = 968 kN
as stiffeners are placed every 800mm
As transverse reinforcement defined by the check for transverse shear on the complete section requires T20 with
spacing s=120mm (see further down), Av holds for 8 stirrups, meaning 12 T20 on 1 m, meaning :
Av = 12 x 314 = 3770 mm2
With S500 stirrups and Φ=0,75 ACI318-08 (9.3.2.6)
Φ Av Fy = 0,75 x 3770 x 500 = 1413000 N = 1413 kN > Vr’= 969 kN
The stirrups defined in the check for transverse shear on the complete section are OK as ties for the bearing
force on stiffeners placed on the outer side of the HD profiles.
Checks of resistance to transverse shear following Option 2—Available shear strength per ACI 318.
Design of reinforced concrete cross sections subject to shear is based on:
ΦVn ≥ Vu ACI318-08 (11-1)
Vu is the factored shear force at the section considered.
Vn is nominal shear strength computed by: Vn = Vc + Vs ACI318-08 (11-2)
Vc is the nominal shear strength provided by concrete and Vs the nominal shear strength provided by shear
reinforcement.
In ACI318-08 , Vc should be computed by different expressions related to the type of factored actions applied to
the section. The more refined expressions have upper bound limitations:
- shear Vu and flexure Mu : Eq.(11-3) or Eq.(11-5) with limitation;
- shear Vu and axial compression Nu : Eq.(11-4);
- shear Vu and axial tension Nu : Vc = 0 or Eq.(11-8) with limitation;
- shear Vu and flexure Mu and compression Nu: Eq.(11-5) completed by Eq(11-6) with limitation
expressed by Eq.(11.7).
A comparison of those various shear stress equations for members submitted to a combination of axial loads,
bending moment and shear, which is typically the case for the columns with HD profiles (see R11.2.2.2), is
presented in ACI318-08 Comments. It indicates that Eq.(11-4) is a correct safe side estimate of Vc. Figure I.X4-
8. Also, it obviates the determination of individual values of bending moment Mu and compression Nu attributed
to each zone bc3, bc4 and bs.
69
Fig. I.X4-8. Figure R.11.2.2.2 from ACI318-08.
2 1 '2000
uc c w
g
NV f b d
A
= +
λ ACI318-08 (11-4)
λ=1,0 for normal weight concrete
f’ c = 50 MPa (or N/mm2)
Nu =180000 kN
Ag = 3072 x 3072 = 9437.103 mm2
In international units (N, mm), ACI318-08 (11-4) expression becomes for the considered gross section Ag and
the applied compression force Nu=180000kN:
Vc = 0,3974 √ f’c bw d = 2,81 bw d for f’c = 50 MPa
Resistance to transverse shear in sections bc1.
Sections bc1 are reinforced concrete sections. They are checked using ACI318-08 Eq.(11-4) for Vc.
bw = bc1= 286 mm
d = 0,8 hz = 0,8 x 3072 = 2457 mm
Vc = 2,81 bw d = 1974 kN ACI318-08 Eq.(11-4)
Φ Vc =0,75 x 1974 = 1480 kN > Vu,bc1= 945 kN
No transverse reinforcement, but the imposed minimum, is required in sections bc3.
Minimum Reinforcing Limits
Check that the minimum shear reinforcement is provided as required by ACI 318, Section 11.4.6.3.
',min
500,75 w w
v cyr yr
b s b sA f
f f
= ≥
ACI 318-08 Eq. (11-13)
In international units (N, mm), ACI318-08 (11-13) expression becomes:
'0,0623 0,007v w wc
yr yr
A b bf
s f f
= ≥
286 2860,0623 50 0,25 0,007 0,004
500 500vA
s = = ≥ =
70
This is realized for instance with 1T10 bar close to column edge (external hoops) at step s =300mm
Av = 78,5mm2 Av /s=78,5/300= 0,26 > 0,25
Resistance to transverse shear in section bc2.
Section bc2 is a reinforced concrete section. It is checked using ACI318-08 Eq.(11-4) for Vc.
bw = bc2= 1300 mm
d = 0,8 hz = 0,8 x 3072 = 2457 mm
Vc = 2,81 bw d= 8975 kN ACI318-08 Eq.(11-4)
Φ Vc =0,75 x 8975 = 6731 kN > Vu,bc2= 4143 kN
No transverse reinforcement, but the imposed minimum, is required in section bc4.
1300 13000,0623 50 1,14 0,007 0,0182
500 500vA
s = = ≥ =
This is realized for instance with 4T8 stirrups at step s =300mm
Av = 8 x 50,3 = 402 mm2 Av /s=402/300= 1,34 > 1,14
Resistance to transverse shear in section hs.
Vu,hs = 6987 kN
The area considered for shear resistance of the composite section of section bs is only the web of the complete
section presented at Figure.
bw =600 mm
d = 0,8 x 3072 = 2457 mm
Vc = 2,81 bw d= 4142 kN
Φ Vc =0,75 x 4142 = 3107 kN < Vu,t,S1= 6987 kN
Rebars are needed to provide Vs such that:
Φ(Vc +Vs) ≥ Vu,hs
Vs ≥ (Vu,hs- ΦVc)/ Φ=(6987 – 3107)/0,75= 5173 kN
Vs = Av fyt d/s
fyt = 500 MPa
For s = 140 mm: Av ≥ s x Vs/ fyt d =(150 x 5173000)/(500 x 2457) = 631 mm2
631 mm2 is realized by a T20 stirrup: 2 x 314 = 628 mm2 ≈ 631 mm2 .
Note: as similar T20 bars spacing in Y direction is 120mm, this unique spacing is kept in both directions in the
drawing at Figure I.X4-9. Definition of reinforcement .
Resistance to transverse shear of the HD profile.
The shear force Vu,l per unit length in CC2 correspond to the maximum shear stress in the HD profile. The HD
profile adequacy can be checked in shear under a shear force Vu,t,S2 based on that safe side value:
71
Vu,t,S2= Vu,l ,CC2 x h2= 3786 x 476= 1802000 N = 1802 kN
Vn = 0.6FyAwCv
Φv = 1.00 and Cv = 1.0 (see above)
h = 476 mm t= 2 x tf =2 x 140 = 280 mm
Aw = 476 x 280 =133280 mm2
Vn = 0.6FyAwCv = 0,60x 450 x 133280 x 1 = 35985 kN
ΦvVn = 35985 kN > Vu,t,S2= 1802 kN
[Note:@100 means that a vertical spacing is 100mm].
Fig. I.X4-9. Definition of transverse reinforcement resulting from all requirements in Examples I.X3 and I.X4 .
72
References
EN 1994-1-1 – Eurocode 4 (2004), “Design of composite steel and concrete structures, Part 1.1 – General
Rules for buildings”, European Committee for Standardizations, Brussels.
ENV 1992-1-1 – Eurocode 2 (1994), “Design of concrete structures, Part 1.1 – General Rules for buildings”,
Provisory Version, European Committee for Standardizations, Brussels.
EN 1992-1-1 – Eurocode 2 (2004), “Design of concrete structures, Part 1.1 – General Rules for buildings”,
European Committee for Standardizations, Brussels.
Nethercot D. A. (2004), “Composite Construction, Spon Press”, ISBN 0-203-45733-1, London.
FineLg User’s Manual, V 9.2. ,(2011) Greish Info – Departament ArGEnCo – ULg.
FINELG User’s Manual (1999), “Non linear finite element analysis software”. Version 8.2.
Boeraeve P. (1991), “Contribution à l’analyse 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.
A 913/A 913M – 11, (2011) “Standard Specification for High-Strength Low-Alloy Steel Shapes of Structural
Quality, Produced by Quenching and Self-Tempering Process (QST)”, ASTM;
ETA 10/0156, (2010) Long products made of HISTAR 355/355L and HISTAR 460/460L, DIBT;
AISC (2010), “Specification for Structural Steel Buildings”, Chicago, Illinois.
AISC (2011), “Design Examples V14”, with particular reference to Chapter I: Design of composite
members; AISC, Chicago, Illinois.
ACI (2008), “Building Code Requirements for Structural Concrete and Commentary”, ACI 318-08.
ArcelorMittal Long Carbon Europe (2008), “Sections and Merchant Bars – Sales Progrmame”, available
in PDF format or in paper format upon request at www.arcelormittal.com/sections.
Plumier A., Doneux C., Castiglioni C., Brescianini J., Crespi A., Dell’Anna S., Lazzarotto L., Calado L.,
Ferreira J., Feligioni S., Bursi O., Ferrario F., Sommavilla M., Vayas I., Thanopoulos P., Demarco T.
(2006), “Two innovations for earthquake-resistant design: the INERD project” – Science Research
Development – EUR 22044 EN.
73
A. Plumier, T. Bogdan, H. Degée, (2012), “Design of composite mega-column with several encased Jumbo
profiles”, Internal report for ArcelorMittal LCE, Plumiecs & Ulg.
A.Plumier, T. Bogdan, H. Degée, (2012), “Design example of a rectangular composite mega-column with 4
encased Jumbo profiles”, Internal report for ArcelorMittal LCE, Plumiecs & ULg. (downloadable on
http://www.arcelormittal.com/sections).
ASCE (2010), Minimum Design Loads for Buildings and Other Structures, ASCE/SEI 7-10, American
Society of Civil Engineers, Reston, VA.
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 has led many research projects in these fields. Mr. Plumier was
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.
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.
Teodora Bogdan is a research engineer at the University of Liege (Belgium). She prepared a PhD thesis in the
field of composite steel-concrete structures at Technical University of Cluj-Napoca (Romania) and obtained her
PhD degree in 2011. She is now working at University of Liege (Belgium).
Jean-Claude (JC) Gerardy is senior project sales manager of ArcelorMittal Commercial Sections
(Luxembourg). He is in charge of promoting and selling steel sections and HISTAR steels (high strength steels)
in large projects worldwide. He has been based many years in the USA, Asia, Near East and Africa for
ArcelorMittal.