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DESIGN FOR SHEAR OF COLUMNS WITH SEVERAL ENCASED STE EL PROFILES . A PROPOSAL.
André Plumier (Plumiecs & ULg) : [email protected] (Main Contact)
Teodora Bogdan (ULg) : [email protected]
Hervé Degée (ULg) : [email protected]
Abstract
Concrete sections reinforced by multiple encased rolled sections are a possible solution to realize mega columns
of tall buildings. In comparison to concrete filled caissons, the advantages are less welding, less fabrication
work, the use of simple splices well settled for decades in high-rise projects and possibility of simpler beam to
column connections. All these characteristics, combined to the availability of huge rolled sections in steel which
do not require pre-heating before welding, lead to another advantage: a high potential for reliable ductile
behavior. AISC allows the design of composite sections built-up with two or more encased steel sections, but
the way to perform such design is not explained. This paper develops two methods for the design of such
columns under combined shear and bending, one based on the theory of beams and the other one on the strut and
tie method. The paper also states further research steps which would help in refining the method. A companion
paper develops the design approach for combined axial compression and bending.
Keywords
Composite columns, shear, rolled sections, steel shapes, tall buildings, design method, mega-columns.
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1. Introduction.
1.1 Type of section envisaged.
The type of section envisaged is shown at Figure 1. They can be walls or columns with several encased steel
sections. The advantages of such design are presented in a companion paper (Plumier, 2013). Section heights
should be rather large to justify such type of reinforcement, with an order of magnitude of 4 times the encased
steel section depth. In the example presented in this paper, we have: hx = hy = 121 in., but the proposal is
general for rectangular sections.
Fig. 1. Type of section for which the method is proposed.
1.2 Specific problems and principle of methods to evaluate shear in columns with several encased
steel profiles
The sections envisaged are reinforced concrete sections with a peculiarity: some reinforcement are steel
sections. This should not basically change the design procedure, as long as the steel sections possess the
characteristics required from classical rebars: adequate yield stress, elongation capacity and bond. Yield stress
and elongation of steel are similar for sections and rebars, but bond requires specific thinking because there are
no indentations in steel sections. This is translated in very different available bond and this might be a problem,
as bond is needed to prevent relative displacement between steel and concrete at the steel concrete interface or,
in other words, to resist longitudinal shear at the steel concrete interface.
The order of magnitude of design bond stress for rebars can be deduced from ACI318-08 expression (12-1)
which for fc’= 4000 psi and Grade 60 rebars reduces to: ld= 47db. Expressing that the design bond force should
at least be equal to the bar yield strength provides a value of a design bond stress fbd:
260
0,322 4 474 47
d Fybfbd xx x db
π
π= = = ksi
3
AISC 2010 Specification does not indicate what the bond stress is for an encased steel section; it is defined for
concrete filled sections (LRFD):
ΦFin=0.45 x 0.06 =0.027 ksi (AISC Spec. I6.3c)
This bond stress is around 10 times smaller than the one of rebars and it would even be smaller for an encased
steel section. For this reason, resistance to longitudinal shear at a steel sections-concrete interface of an encased
steel shape is generally realized by means of shear connectors like headed studs or welded bars or plates.
If enough shear resistance is provided at the steel section – concrete interface, a global section of the type
envisaged at Figure 1 behaves like a standard reinforced concrete section. Two problems need to be solved in
the design for shear of concrete sections with several encased steel profiles.
The first one is the distribution of the applied transverse shear in a section which has a very irregular density of
reinforcement: the steel content is much higher where te steel shapes are.
The second problem is the evaluation of the applied shear at steel section-concrete interface. It is not a problem
with classical rebars, because it has been assessed once for all that bond is high enough to skip the checks of
resistance to longitudinal shear.
In order to solve the problem, two methods will be used.
The first method is an application of the classical beam theory, which allows the calculation of shear forces in
continuous elements without cracks. Reinforced concrete sections have cracks. In particular, in the envisaged
types of sections, the steel content is high and steel prevents concrete shrinkage more than in classical RC
sections, which favor shrinkage cracks in addition to regular cracks due to bending. However, the sections
studied here are meant for columns and the compression stresses tend to close the cracks so that there is some
logic in considering the hypothesis of a homogeneous section without cracks.
The second method considers sections with cracks. It is an application of the “strut and tie” method or STM.
The flow of internal forces in concrete is idealized as a truss carrying the imposed loading through the region to
its supports. Like a real truss, a strut-and-tie model consists of struts and ties interconnected at nodes. This truss
is a statically admissible stress field in lower-bound (static) solutions and applies to components with cracks. As
the STM backs the classical reinforced concrete design for shear, it is natural to try to apply it to the type of
sections envisaged.
In order to better figure the design problems, each step of the developments is illustrated by calculations on an
example which is the section of Figure 1.
2. Method based on the theory of elastic beams.
2.1 Evaluation of applied longitudinal shear. Principle.
The first approach in the evaluation of shear in columns with several encased steel sections makes use of the
theory of elastic beams, following which the longitudinal shear force Vl in one horizontal plane cut of a section,
for instance plane aa at Figure 2, is equal to: Vl = VS/I (1)
where V is the applied transverse shear, S the first moment of area of the area between the exterior of the section
and I the inertia (or second moment of area) of the section.
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Fig.2. Determination of longitudinal shear on plane aa.
In order to calculate the longitudinal shear force Vl around a steel shape used as reinforcement, it is necessary to
divide the complete section of Figure 1 into sub-sections amongst which some sub-sections have a width limited
to the width of the steel shape. Then expression (1) allows to establish the longitudinal shear at the faces of the
top and bottom flanges of each of the encased steel shapes, provided that the part Vi of the total transverse shear
V applied to each sub-section is known. This an elementary problem: as all sub-sections have the same
deformed shape over the column length, Vi applied to a sub-section is proportional to that sub-section relative
flexural stiffness EIi,/ ΣEIi, neglecting, the shear deformations being neglected as it is usual in the application of
the theory of beams.
It should be noted that the applied longitudinal shear is independent of the applied axial force in the element,
because the applied transverse shear is independent of that axial force.
Based on that and considering as example the case of a design shear Vu,Y applied in the Y direction to the section
presented at Figure 1, the steps of calculations are:
a. Division of the column section into reinforced concrete sub-sections and composite sub-sections; the
height of those sub-sections is parallel to the direction of shear. Looking at Figure 1, the sub-sections
for a Y oriented design shear Vu,Y are 3 reinforced concrete sub-sections of width bc3 and bc4 and 2
composite sub-section of width bs ; all sub-sections have the same height hy.
b. Calculation of the effective bending stiffness (EI)eff of each sub-section using in each sub-section the
expressions given in AISC 2010 Specification:
EIeff = Es Is + 0.5Esr Isr + C1Ec Ic (AISC Spec. Eq.I2-6)
C1= 0.1+ 2 [(As/(Ac+As)] ≤ 0,3 (AISC Spec. Eq.I2-7)
In each sub-section, EIeff is computed with the same expression and one single value of the coefficient
C1, which the one of the global section: C1 = 0,244
To facilitate the calculation of Isr, reinforcing bars can be replaced by equivalent steel plates as shown
at Figure 3 and explained in the companion paper (Plumier, 2013). There, a numerical example with
the same data is presented.
c. Distribution of the total design shear Vu into design shear Vu,Y,i applied to sub-sections.
Vu,Y,i = Vu,Y x(EIeff,i,/ ΣEIeff,i)
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The total effective bending stiffness EIeff around the X axis is the sum of individual EIeff established for
sub-sections bc3, bs and bc4 respectively.
d. Evaluation by using expression (1) of longitudinal shear Vu,Y,l,i,j applied at the different interfaces j
between steel profiles and concrete in the sub-section i.
2.2 Design checks.
Once the applied transverse shear in the different sub-sections is calculated, checks of resistance can be made:
- Check in each reinforced concrete subsection i that the shear resistance ΦVn,Y,i is not smaller than the
design shear Vu,Y,i: ΦVn,Y,i ≥ Vu,Y,i
The check is a standard shear check according to ACI318-08. It results in a certain required amount of
horizontal reinforcement which should be present in each of those reinforced concrete subsections. The
shear resistance ΦVn,Y,i is not independent of the applied axial force.
- Check composite subsections:
• standard shear check under transverse shear for what concern reinforced concrete strength and the
definition of horizontal reinforcement which should be present in each of those composite subsections;
this horizontal reinforcement is placed around the steel sections – Figure 4;
• calculation of the required connectors such that, on interfaces j between steel profiles and concrete in
sub-section i , the longitudinal shear resistance ΦVn,Y,i is not smaller than the applied shear Vu,Y,l,i,j:
ΦVn,Y,i ≥ Vu,Y,l,i,j
• calculation of shear action effect in the steel profiles and check that the shear strength of the profile
is not smaller than the shear action effect.
2.3 Design example.
2.3.1 Data.
The section presented at Figure 1 is submitted to a factored bending moment Mu,X , a factored
axial compression force Nu and a factored transverse shear Vu,Y in direction Y:
Mu,X = 331907 kip-feet Nu = 40500 kip Vu,Y= 5000 kip
The encased steel sections are W14x873 (W360X1299) as defined in (ASTM A6/A6M -12):
Grade 65 Fy=65ksi Es=29000ksi Area A = 256,5 in.2
Depth d = 23,62 in. Flange width bf= 18.755 in.
Flange thickness tf =5.51in. Web thickness tw =3.935in.
Inertia or second moment of area Iy strong axis = 753837 in.4
Steel reinforcing bars: Grade 60 Fy=60ksi Es=29000ksi
Concrete: f’ c = 7000 psi Ec=57000√7000 = 4768ksi (ACI318-08. Clause 8.5.1)
or with wc’=148lbs/ft3 Ec=1481.5√7 = 4768ksi (AISC Spec. I2-1b)
2.3.2 Distribution of transverse shear in the section.
The width bc3, bs and bc4 of the subsections defined at Figure 3 are:
bc3 = 11.123 in. bs = 18.755in. bc4 = 61.25in.
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Fig. 3. Definition of sections bc3, bc4, and bs.
Fig. 4. Position of the reinforcements and the W14x873shapes( top).
Calculated distribution of transverse shear in the section (bottom).
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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)
Total area of 4 steel sections: As = 4 Aa = 1026in.2
Total area of 256#11 reinforcing bars: Asr = 256 x 1.56 = 399in.2
Net area of concrete: Ac = Ag – As – Asr =1212 – 1026 – 399 = 13216in.2
C1 = 0,1 + 2[(1026 /(13216 + 1026)] = 0,244 ≤ 0,3
Fig. 5. Definition of plates equivalent to reinforcing bars.
The numerical values established in a companion paper for equivalent steel plates (Figure 5) are given
in the example hereunder without repeating the explanations.
Bending stiffness EIeff of sub-section bc3
The part of the horizontal plates inertia corresponding to sub-section bc3 width is:
Isr1xp =(bc3/hx)xIsr1x=(11.123/121)x569373=52340in.4
One vertical plate inertia is equal to: Isr2x=90892 in.4
The inertia of the gross concrete section is: Icg = bc3 x hy3/12 = 11.123 x 1213/12 = 1642082 in.4
To obtain Ic , the moment of inertia Isr corresponding to the fact that there is no concrete where there
are rebars is deduced from Icg:
Ic = Icg - Isr1x,p - Isr2x = 1642082 - 52340 - 90892 = 1498850in.4
(EIeff)bc3= 0.5 Esr Isr + C1Ec Ic
(EIeff)bc3=0,5 x 29000 x (52340+90892) + 0,244 x 4768 x 1498850 =3,820,614,099kip- in.2
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Bending stiffness EIeff of sub-section bc4
The part of the horizontal plates inertia corresponding to bc3 sub-section width is:
Isr1xp =(bc4/hx)xIsr1x=(61.25/121)x569373=288216in.4
The inertia of the gross concrete section is: Icg = bc4 x hy3/12 = 61.25 x 1213/12 = 9042342 in.4
To obtain Ic , the moment of inertia Isr corresponding to the fact that there is no concrete where there
are rebars is deduced from Icg:
Ic = Icg - Isr1x,p = 9042342-288216=8754126in.4
(EIeff )bc4= 0.5EsrIsr + C1Ec Ic =0,5x29000 x 288216 +0,244x4768x8754126 =1.436 x 1010 kip-in.2
Bending stiffness EIeff of sub-section bs.
The part of the horizontal plates inertia corresponding to one bs sub-section width is:
Isr1xp =(bs/hx)xIsr1x=(18.755/121)x569373=88252in.4
Steel sections: bs = 18.755in. Isx=1515359/2 = 757679in.4
The inertia of the gross concrete section is: Icg = bs x hy3/12 = 18.755 x 1213/12 = 2768802 in.4
To obtain Ic , the moment of inertia Isr corresponding to the fact that there is no concrete where there
are rebars and steel sections is deduced from Icg:
Ic = Icg - Isr1x,p - Isx= 2768802-757679-88252=1922871in.4
(EIeff)bs= EsIs + 0.5 Esr Isr + C1Ec Ic = 29000 x757679 +0,5 x29000 x 88252 + 0,244 x 4768x1922871=
2.549 x 1010 kip-in.2
Bending stiffness EIeff of the complete section.
EIeff =2 (EIeff)bc3 + (EIeff)bc4 +2 (EIeff)bs =2x3.82x109+1.436x1010+ 2 x2.549 x1010=7.298 x 1010 kip-in.2
Distribution of transverse shear in sub-sections bc3, bs and bc4.
The factored shear Vu,Y = 5000 kip applied to the complete section is distributed in the 5 sub-sections
(2bc3, 2bs, 1bc4) proportionally to their bending stiffness.
Vu,bc3= Vu,Y x (EIeff)bc3/EIeff = 5000 x 3.82x109 /7.298 x 1010 = 261.7 kip
Vu,bc4 = Vu,Y x (EIeff)bc4/EIeff = 5000 x 1.436 x1010 /7.298 x 1010 = 983.8 kip
Vu,bs= Vu,Y x (EIeff)bs/EIeff = 5000 x 2.538 x1010 /7.298 x 1010 = 1738.8 kip
The distribution of transverse shear is very uneven in the section. This is better set forward expressing
the transverse shear v per unit width of sub-sections:
vu,bc3= 261.7/11.123 = 23.5 kip/in.
vu,bc4 = 983.8/61.25= 16.1 kip/in.
vu,bs= 1738.8/18.755= 92.7 kip/in.
This distribution of transverse shear shown at Figure 4 sets forward that:
o the input to flexural stiffness due to the encased steel sections is high;
o design checks in shear have to be made separately in the different subsections because these
subsections are submitted to very different levels of shear in the concrete and because one
limit in concrete shear strength is on the concrete side;
o subsection bs is the most critical one for concrete;
o to be effective, transverse reinforcement should follow the distribution of transverse shear in
the section.
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2.3.3 Longitudinal shear at steel-concrete interfaces in subsection bs.
Section bs is a composite steel-concrete section having 2 reinforced concrete flanges (section C1 at
Figure 6), 2 steel “flanges” (the HD profiles, sections C2 at Figure 6) 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 bs section is equal to the stiffness (EIeff)bs :
Ic*= (EIeff)bs/Ec = 2.538 x 1010 /4768 = 5.322 x 106 in.4
Fig. 6. Homogenized equivalent concrete section bs.
In a homogenized section in concrete (Figure 6), the width of concrete equivalent to the width of steel flanges is
bs*: bs*=bs x Es/Ec =18.755 x 29000/4768= 114 in.
The width of concrete equivalent to the width of steel web is tw*:
tw*=t w x Es/Ec =3.935 x 29000/4768= 23.93 in.
The homogenized concrete section is showed at Figure 6.
The resultant longitudinal shear force on sections like CC1 and CC2 at Figure 6 is:
Vu,l = (Vu,bs x S) / Is*
S is the first moment of the areas between sections CC1 or CC2 and the upper face of the section, that is sections
C1 and C2 at figure 6.
Longitudinal shear is calculated at the steel-concrete interfaces between sections C1 and C2 and between
sections C2 and C, because resistance to longitudinal shear should be checked on those interfaces.
Calculation of longitudinal shear force applied at section CC1.
S1 is the section modulus for the section C1:
The width of concrete equivalent to steel is bs*. Height h1 is: h1= 11.19 in.
The area is: Area1= bs x h1= 18.755 x 11.19 = 209.86in.2
Distance from CC1 to neutral axis: 121/2 -11.19/2=54.90 in.
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Moment of area: Sc1 =209.86 x 54.90 = 11523 in.3
The top and bottom reinforcing bars are replaced in the calculations by 1 equivalent steel plate of area As,1.
The part As,1 in bs subsection is: As1=(18.755/121) x93.6=14.5 in.2
The equivalent area in concrete is: 14.5x 29000/4768 =88.2 in.2
The distance between the center of that plate and the axis of symmetry is dsy1=55.15 in.
Moment of area: Ssr =88.2 x 55.15 = 4866 in.3
S1= Sc1 + Ssr = 11523 + 4866 = 16389 in.3
The resultant longitudinal shear force on section CC1 is:
Vu,l,CC1 = (Vu,bs x S1) / Ic* = (1738.8 x 16389)/ 5.322 x 106 = 5.35kip/in.
Calculation of longitudinal shear force applied at section CC2.
S2 is the section modulus for the sections C1 plus C2 (the HD profile). Steel section: A= 256.5 in.2
The equivalent area in concrete for the HD profile is: 256.5 x 29000/4768= 1560 in.2
Distance of W14x873 center to neutral axis: dsy= 37.5 in.
SW = 1560 x 37.5 = 58503 in.3
Concrete between the flanges: area = (18.755 – 3.935) x(23.62-5.51)/2 =134 in.2
Sc=134 x 9.05=1213 in.3
S2= S1 + SW + Sc= 16389 + 58503+ 1213= 76105 in.3
Vu,l,CC2 = (Vu,bs x S2) / Ic* = (1738.8 x76105)/ 5.322 x 106 = 24.86 kip/in.
Distribution of longitudinal shear.
The longitudinal shear at the inner flange-concrete interface is 4.6 times greater than at the outer flange.
The shear at mid-height of the steel shapes can be estimated as the average between the values of shear in
sections CC1 and CC2 : (5.35+24.86)/2=15.1 kip/in
2.3.4 Comment on the design procedure based on the theory of beams.
The procedure considers that there is no longitudinal shear at the interfaces between the different sub-sections.
This is exact only if the expression given in the AISC Specification to calculate the effective flexural stiffness
EIeff of each subsection is correct in each subsection. Otherwise, the calculated shear in each sub-section is
approximate, in particular due to the different steel contents in the reinforced concrete and composite sub-
sections. It is unlikely that the expression of EIeff be exact in all cases. However, the total shear resistance is
correctly calculated and there is a huge potential of resistance to longitudinal shear at the sub-sections interfaces,
meaning that the redistribution of shear between subsections takes place without problem and the design is safe.
3. Strut and tie method.
3.1 Evaluation of longitudinal shear force at the steel shape-concrete interface. Principle.
In the strut and tie method, the structural element is modeled as a concentric bracing with depth z equal to the
lever arm of internal forces in the reinforced concrete section. The standard inclination of diagonals is 45°. For
the column under consideration, it has been shown that the lever arm of internal forces in the reinforced concrete
section is approximately equal to the distance between the axis of two encased steel sections, which results in
panels of the bracing having a length equal to z if diagonals are inclined at 45°. The internal forces are
established by simple equilibrium equations. The equilibrium under axial compression does not involve shear.
11
In a panel submitted to a bending moment Mu and shear force Vu , the force in the diagonal Fdiag is established by
expressing the horizontal equilibrium in the section aa – Figure 7:
Fdiag x cos45°=Vu
Fig.7. Applied Mu and Vu and the equivalent forces F and Fdiag. in section aa (left).
Body in equilibrium used to relate ∆F and Vu (right).
Expressing the evolution of bending moment along the column by the variation of forces in chords of the
bracing and considering the equilibrium of moment around point B in a portion of bracing – Figure 7 -
indicates: ∆F x z = Fdiag x z sin45° = Vu x z sin45°/cos45° = Vu x z
It means that the change ∆F in chord force along a panel of length z is equal to Vu.
In the reality of the reinforced concrete column, the diagonal force Fdiag is not concentrated at a node A, but
applied as a distributed force along the length of chords, which can be expressed in the variation ∆F being
induced in a continuous way at the concrete struts-chord interface: ∆F is the longitudinal shear force Vl at the
concrete struts-chord interface over a panel length z.
It results: Vu,l = ∆F = Vu
This is in fact the expression at a panel scale of the general principle of equality of tangential stresses on
orthogonal faces.
The tension chord in a strut and tie model represents all the longitudinal steel sections which are in tension:
rebars and steel sections. Making for simplification the hypothesis that all steel rebars and sections are at the
level of the tension chord of the model, the elongation under Vu,l in those tension elements is equal to:
ε = Vu,l/At = Vu /At
where At represents the sum of the rebars and profiles area on the tension side. Given that there are n steel
profiles of cross section Aa on one side of the section:
As = n Aa At = As + Asr= nAa +Asr
12
The part of Vu,l applied to one steel profiles on the tension side is: Vu,l,s = Vu,l x Aa/( nAa + Asr)
The part of Vu,l applied to one steel profile on the tension side finally is, over a length z:
Vu,l,a = Vu,l x Aa /( nAa + Asr)
3.2 Evaluation of longitudinal shear force at the steel shape-concrete interface. Example.
Cross section area Aa of one W14x873: Aa =256.5 in.2
For 2 w14x873: As= 2 x 256.5 = 513 in.2
Cross section area of rebars on the tension side:
- One full horizontal plate of area: Asr = As1= 60 x 1,56 = 93.6 in.2
- One half of the side plates on each side: 2 As2/2= As2=106.1in.2
Estimation of z, lever arm of internal forces: z = 0.75hy = 0.75 x 121 = 90.75in.
Vu,l,a applied to one steel profile on the tension side over length z:
Vu,l,a= Vu,Y x Aa /( nAa + Asr) = 5000 x 256.5/(513 + 93.6+106.1) = 1799 kip
Longitudinal shear force per unit length of steel section : Vu,l = 1799/90.75 = 19.8 kip/in.
This value is the average for the whole steel shape. In spite of the difference in approach, this value is
reasonably close to the average longitudinal shear force per unit length of steel section obtained by the beam
method (in 2.3.3: Vu,l = 15.1 kip/in.); the difference is 30% on the safe side. This deserves further comments, see
3.5.
3.3 Distribution of longitudinal shear around the steel profiles in the strut and tie method.
The strut and tie approach defined in 3.1 and 3.2 does not give the distribution of longitudinal shear around the
profiles: it provides an average resultant shear per unit of surface around the shape.
The elastic distribution of longitudinal shear obtained in 2.3.3 in application of the theory of beams is partial as
it consists only in two tangential shear forces at the flange faces of the steel section, but it sets forward an image
of the real distribution and one important fact: longitudinal shear stresses are significantly higher at the flange
face at the interior of the global section than at the exterior, the ratio being 4.6 in the example.
As the strut and tie method is more direct than the theory of beams, it is interesting to establish a method of
evaluation of the distribution of longitudinal shear around a section in the strut and tie approach. The way used
in the beam theory, with calculation of moment of area, can be exported to the strut and tie approach, with two
additional remark allowing simplifications:
- In a global reinforced concrete section in which cracks are present, the first moments of areas S in
the cracked zone are limited to the steel components of the section. This is a first simplification.
- In some cases, like the one of the example, the distances from the center of area of rebars and steel
shapes in tension to the neutral axis are little different. Considering a single value for that distance
only influences the distribution of longitudinal shear around the steel shapes, not the resultant
shear; furthermore, the influence is minor.
The calculation steps are:
- Calculation of S of the rebars included in As1 to obtain Sext at the face of the flange on the exterior
side of the complete section: Sext= (As1 d1+ As2d2)
- Calculation of S of the rebars included in As1 plus S of the n steel sections present on the tension
side to obtain S at the face of the flange on the interior side of the complete section:
Sint== (As1 d1+ As2d2) + nAada
13
- Definition of an average S: Sav= (Sint+ Sext)/2
- Distribution the longitudinal shear Vl,a around the steel shape in relation with Sint and Sext.
Vu,l,a,ext = Vl,a x Sext/ Sav Vu,l,a,int = Vl,a x Sint/ Sav
3.4 Example.
Due to the dimensions of the section, we can admit: d1 =d2= da=1.0
Sext= (As1 + As2) = 93.6 +106.1 = 199.7 in.3
Sint= (As1 + As2) + nAa = 199.7+ 2 x 256.5 = 712.7 in.3
Sav= (Sint+ Sext)/2= 456.2
Vl,a applied to one steel profile over a length z:
Vu,l,a,ext = Vl,a x Sext/ Sav =1799 x 199.7/456.2= 787.5kip
Vu,l,a,int = Vl,a x Sint/ Sav =1799 x 712.7/456.2= 2810.5kip
The results expressed in longitudinal shear vu,l,a applied per unit length of steel profile are:
Vu,l,a,ext = 787.5/90.75 = 8.7 kip/in.
vlu,l,a,int = 2810.5/90.75=31.0 kip/in.
The longitudinal shear at the inner flange-concrete interface is 3.5 times greater than at the outer flange; it is 4.6
with the beam theory. The results obtained by the strut and tie method are on the safe side in comparison to
those obtained by the theory of beams:
- 8.7 kip/in. instead of 5.4kip/in. for Vu,l,a,ext
- 31.0 instead of 24.9 kip/in. for Vu,l,a,int
This tends to show that the proposal based on the strut and tie method gives safe results for the longitudinal
shear force at the steel shapes, which is the parameter used in the design of shear connectors.
3.5 Longitudinal shear at shapes closer to the axis of symmetry.
With the strut and tie method, there is only a single value of the longitudinal shear in the steel shapes because
steel shapes like concrete are submitted to a single shear force which corresponds to a single Fdiag , see Figure 8
or 12. However, one remark should be made. In 3.1 to 3.4, the explanation has been focused on the definition of
longitudinal shear at the steel-concrete interface of encased steel shapes which are close to the periphery of the
section and can act effectively as tension reinforcement. As shown at Figure 8, those peripheral steel shapes are
submitted to Fdiag on one side, while more central shapes are submitted to Fdiag on two sides. This remark is
important when coming to the design of connectors: there are twice more numerous for those central shapes,
because the shear force Vu,l= Fdiag cos45° has to transit from the concrete to the steel shape and from the steel
shape to the concrete, see Figure 8.
14
Fig.8. Transmission of compression struts forces Fdiag by longitudinal shear to the steel shapes.
3.6 Distribution of transverse shear. Principle.
It is necessary to know the distribution of transverse shear in the global column section, because one limit in
concrete shear strength is on the concrete side and the stress level is much higher in the vicinity of the steel
shapes, as explained in 3. The distribution of transverse shear is directly known from the developments in 3.1:
- Vu =Vu,l over a length z;
- the part of Vu,l applied to one steel profile on the tension side is: Vu,l,a = Vu x Aa /( nAa + Asr)
- the part of Vu applied to a sub-section including one steel profile is thus:
Vu,bs = Vu,l,a = Vu x Aa /( nAa + Asr)
- in the strut and tie approach, there is a single concrete sub-section of width equal to the total
section width minus the width of the n subsections bs: bc= hx-n x bs
- the applied shear in the concrete sub-section is: Vu,c = Vu - nVu,a
3.7 Distribution of transverse shear. Example.
The applied shear Vu,a to a sub-section bs including one steel profile on the tension side is equal to:
Vu,bs = Vu,l,a = 1799 kip
The applied shear Vu,c to the complete concrete sub-section is: Vu,c = Vu - nVu,bs= 5000 – 2 x 1799 = 1402 kip
These values are close to the results obtained by the beam theory: Vu,bs = 1744 kip Vu,c = 1511kip
4. Practical implementation of shear connectors on the steel shapes.
4.1 Principle of the method.
In reality, for steel profiles as well as for rebars, longitudinal shear is present all around the steel sections and
the evaluations of the distribution of shear presented in 2.3.2 and in 3.3 are approximations since they define
shear forces only at the inner and outer faces of the steel shape. The direct application of such results may be
adequate in the case of walls with encased steel sections, because there may be too little space to place side
connectors in walls like the one at Figure 8.
15
But in the case of a column like the one of the example shown at Figure 1, the implementation of shear
connectors around the steel section can be made recognizing that longitudinal shear is also present on the sides
and can be resisted in part on those sides. This solution has the advantage to provide connectors which are able
to resist longitudinal shear resulting from a global transverse shear in direction X as well as in direction Y,
which concludes in an economy on the total number of shear connectors if the transverse shear in direction X
and Y are not applied simultaneously.
The proposed guidance for the implementation of shear connectors is the following one.
Case of profiles oriented with web parallel to the applied shear
- the connectors can all be placed on the flanges with different densities on the inner and outer
flange so that the calculated shear on each flange face be properly resisted;
- the connectors can be placed both on the flanges and on the web; in such case, the design should be
such that the connectors on the inner half of the section can support the shear calculated at the
inner flange, with not less that 50% of resistance realised by flange connectors; the design should
be such that the connectors on the outer half of the section can support the shear calculated at the
outer flange, with not less that 50% of resistance realised by flange connectors;
- the design of web connectors should be such that failure on a weak plane through concrete be
avoided.
Case of profiles oriented with web perpendicular to the applied shear
- the connectors can all be placed on the web with different densities on the inner and outer side so
that the calculated shear on each side be properly resisted;
- the design of web connectors should be such that failure on a weak plane through concrete be
avoided.
- the connectors can be placed both on the flanges and on the web; in such case, the design should be
such that the connectors on the inner half of the section can support the shear calculated at the
inner side, with not less that 50% of resistance realised by web connectors; the connectors on the
outer half of the section can support the shear calculated at the outer side, with not less that 50% of
resistance realised by web connectors.
4.2 Design of connectors. Some peculiarities.
The design of connectors is classical and should be made following AISC Specification. Only some aspects
which are peculiar to the type of sections or connectors are mentioned hereunder.
AISC Specification indicates that steel headed stud anchors should be placed on the steel shape in a generally
symmetric configuration. That limitation applies at sections with one central encased steel profile. For the type
of sections envisaged, there is no reason for a symmetric configuration in each individual steel profile. So the
rule should be expressed in a different way: the anchors are placed symmetrically with reference to the axis of
symmetry of the complete section.
16
Elevation view Plan view
Fig. 9. Internal bearing plates acting as connectors and the strut and tie equilibrium.
Internal bearing plates welded between the flanges of encased W shapes as shown at Figure 9 can be connectors.
They use the direct bearing of concrete compression struts to transfer loads from concrete to steel. Where
multiple sets of bearing plates are used, it is recommended that the minimum spacing between plates be twice
the width of the plates plus the plate thicknes (spacing ≥ 2a + tstiffener –Figure 9), so that concrete compression
struts inclined at 45° can develop. That spacing also enhances constructability and concrete consolidation. There
is an optimum spacing to define. It can be greater than the minimum, if the bearing pressure remains acceptable;
in that way useless multiplication of plates is avoided. But, like with headed studs, spacing should be to limited
to an upper bound value; in order to avoid too much inclined compression struts, the proposal is a maximum
spacing = 6a. The definition of spacing should also be made to avoid too high bearing pressure requiring too
high plate thickness.
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 Figure 9. The tie design force for one stiffener is equal
to the longitudinal shear force Vr’ supported by that stiffener. Strength should be provided by horizontal stirrups
passing around the two or more encased steel profiles serving as supports to the concrete struts. The bearing
plates should be connected to the encased steel shape to develop the full strength of the plate.
4.3 Design of connectors. Example.
Figure 10a) indicates the applied longitudinal shear calculated in 3.4 that for a 5000kip applied transverse shear
in direction Y in the bottom left encased steel shape of Figure 4:
vu,l,a,ext = 8.7 kip/in. vu,l,a,int = 31.0 kip/in.
Or: vu,l,a,ext = 104.4 kip/ft vu,l,a,int = 372 kip/ft
Figure 10b) indicates a possible distribution of those longitudinal shear forces which follows the guidance
explained in 4.1. Figure 10c) indicates the design forces for plate connectors welded to the web; for that type of
connectors, the value of design shear has finally to be equal in each half of the welded plate because it is one
single plate.
17
Fig.10. Distribution of longitudinal shear in kip/in.: a ) calculated; b) distributed; c) used in design.
If combining direct bearing on plates and headed studs is acceptable, connectors for longitudinal shear forces of
Figure 10c) due to Vuy only can be designed as follows:
Headed studs connectors on flanges.
Qnv =Fu Asa AISCI2010.Spec. Eq. I8-3
Asa = π (0.75)2/4 = 0.44 in.2 per steel headed stud anchor diameter 0.75in.
Fu = 63.3 ksi Φv = 0,65
Φv Qnv = 0.65 x 63.3 x 0 .44 = 18 kip
The required number of anchors on the inner side flange of the HD shape is: nanchors= 200/18 = 11.1/ft
Anchors are placed in pairs on the flange at 2in. longitudinal (vertical) spacing. The transverse spacing is 8in.
The required number of anchors on the outer side flange of the HD shape is: nanchors= 54/18 = 3/ft
Anchors could be placed in one line with a maximum 5.5in. spacing, but AISC Specification requirement is an
absolute maximum of 6 diameters, meaning 4.5in.
Plate connectors welded to the web of the steel shape.
Width of plate a: a = ( bf – tw)/2= (18.755 -3.935)/2= 7.41in.
Length of plate b: b =h – 2tf = 23.62 – 2 x 5.51 =12.6in. Width of clipped corners c: c= 0.6in.
Plate bearing area: Al = ab-c2 = 93in.2
For the longitudinal shear on both sides the web of the HD sections, as from Figure 10c): Vr’= 172 kip/ft
The available strength for the direct bearing force transfer mechanism is:
Rn = 1,7 fc’ Al=1.7 x 7 x 93=1106kip/plate stiffener (AISC Spec. Eq. I6-3)
The minimum spacing of stiffeners is: 2a + tstiffener ≈ 2 x 7.41 = 14.82in.=1.235ft
With stiffeners placed every 24in., the required bearing force per stiffener is: Vr’=100x24/12= 200kip <1106kip
Pressure wu on plate: wu =Vr’ /Al= 200/93=2.15ksi
It should be checked that: ΦB Rn ≥ Vr’ ΦB Rn = 0,65 x 1106 = 719 kip > 200kip
The required bearing plate thickness tp is calculated on the basis of the expression used in AISC2011- Example
I8, modified to take into account that b=1.7a<2a which means yield lines shorter than 2a√2 – Figure 11. The
correction refers to the actual yield lines length which is: b√2+a-b/2 = a+0.9b .
18
The required bearing plate thickness tp is:
( )( )
22 3 22.8
0.9 3 6u
py
a w b aat
a b F a bφ−
≥+ +
=( )( )
22 7.41 2.15 3 12.6 2 7.412.8 7.410.86
7.41 0.9 12.6 3 0.90 60 6 7.41 12.6
× × × − ×× =+ × × × × +
in.
Fig.11.Plate connectors welded to the web of the steel shape (left).
Yield lines used in calculation of plate strength (right).
In case a transverse shear Vux in direction X also of 5000kip is applied, the longitudinal shear at the steel shape-
concrete interface would be, due to the symmetry of the section, very close to Vl,a,ext and Vl,a,int calculated above;
similar design steps of the connectors would result in the design forces shown at Figure 10d), e) and f).
5. Design for transverse shear.
5.1 Design option.
AISC 2010 Specification for Structural Steel Buildings Section I4.1 indicates three possibilities for determining
the available shear strength of a member with encased composite sections:
1) The available shear strength of the steel section alone as defined in AISC Specification Chapter G.
2) The available shear strength of the reinforced concrete portion alone per ACI 318.
3) The available shear strength of the steel section in addition to the reinforcing steel, ignoring the
contribution of the concrete.
Consideration of the only shear resistance of the steel profiles as proposed the first possible approach certainly
does not correspond to the real behavior in shear of column sections which are essentially reinforced concrete.
Disregarding the contribution to shear resistance of the net area Ac of concrete would clearly be a gross
underestimation for a large section with several encased steel profiles.
The second design possibility is best adapted to mega column which are reinforced concrete sections, but it
should be adapted in two ways. First, the section considered for shear resistance should be the gross concrete
section, like in reinforced concrete. Steel profiles are just another type of rebars and no one would think of
checking RC sections taking out the rebars area. Second, as was shown in 2. and 3., the distribution of shear in
the section is not regular. A global section check like in a standard RC section would ignore this fact. So care
19
should be taken to identify and check the most stressed zones. In 2.1 or 3.6. they have been identified into
“composite sub-sections”. Checking those zones individually provides the required safety.
The third design possibility is meaningless for composite sub-sections with several encased steel profiles: the
stiffness EI of a single steel shape is 50 times smaller than the stiffness of the sub-section bs in which it is
encased: EI = 18129 x 29000=0.52 x 109 kip-in.2<< (EIeff)bs= 25.38 x 109 kip-in.2
As the stiffer system always catches the forces, the section bs (and the complete section) will work compositely,
not as a sum of individual components. Furthermore, the third design approach would be unsafe for composite
sub-sections with several encased steel profiles, because these ones are chains of concrete and steel components
and their strength is the one of the weakest links, the concrete links. Figuring the composite sub-section as a
concrete web in a classical Morsch concentric bracing representation of a reinforced concrete beam help
understand this reality: replacing locally concrete by steel blocks in the diagonal concrete struts in compression
does not improve the diagonal strength –Figure 12. The extra strength of the steel profiles would only intervene
after crushing the concrete links and this would result in the qualitative shear V -shear deformation γ diagram at
Figure 12.
Fig.12. Sub-section bs chain of components represented by a diagonal in Morsch truss bracing model of RC
elements (left). Qualitative shear V -shear deformation γ diagram representing the behavior of sub-section bs
under progressively increased displacements(right).
In conclusion, the 2nd design possibility mentioned in the AISC Specification, which consists in checking the
section as a reinforced concrete one, with some modifications in order to take into account the peculiarities of
sections with several encased steel profiles, is the best option for sections with several encased steel shapes.
An uncertainty remains on the definition of the width of the composite sub-sections. The proposal is the width
(or depth, depending on the axis) of the steel shape, which is safe side, but the reality may be a wider effective
width of web, due to a transverse distribution of stresses. This could in particular be envisaged for the design of
walls with encased steel shapes which could be more simply checked as a single section.
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5.2 Design checks for transverse shear. Expression used.
The design of reinforce sections is based on (ACI, 2008). The design expression is:
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. It should be computed by different expressions related to the type of factored actions applied to
the section. 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 ACI318-08 Eq.(11-4) is a correct safe side
estimate of Vc: 2 1 ' 398.76 2000
uc c w w
g
NV f b d b d
Aλ
= + =
(Nu /Ag in psi).
λ=1,0 for normal weight concrete; f’ c = 700psi ; Nu =40500kip; Ag = 121 x 121 = 14641 in.2
Standard shear checks in RC sub-sections are classical and do not need to be presented in detail. The
reinforcements at Figure 1 comply with all requirements. Only the shear checks in a composite sub-section bs
are presented hereunder. They consist in:
- the standard shear check of subsection bs considered as being fully reinforced concrete;
- a check that the steel shape substitute with safety the concrete shear strength of the area which is in
fact occupied by the steel shape; this check is normally superfluous with heavy steel shapes, but
could be a problem with slender shapes or unfilled hollow sections.
5.3 Design checks for transverse shear. Example in sub-section bs.
From 3.7, we have: Vu,bs = 1799 kip
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 6:
bw= bf =18.755in. d =0.8 x 121=96.8 in.
Vc=398.76bwd=723,959 lbs =723.9 kip
Vs ≥ (Vu,bs - ΦVc)/ Φ=(1799 –0.75 x723.9)/0,75= 1675 kip
For a spacing s of stirrups: s= 3in.: Av ≥ s x Vs/ fyt d =(3 x 1675)/(60 x 96.8) = 0.86in.2
0.86in.2 is realized by a #6 stirrup: 2 x 0.44= 0.88in.2 > 0.86in.2
Shear strength of concrete area occupied by the steel shape:
Vc=398.76bwd =398.76 x 18.755 x 23.62= 176647lbs=177 kip
Steel shape shear strength:
Vn = 0.6FyAwCv Φv = 1.00 Cv = 1.0
Aw = 3.935 x (23.62- 2 x 5.51) =49.58in.2
ΦvVn = 0.6FyAwCv = 0,6x 65 x 49.58 x 1 = 1933kip
Check: ΦvVn = 1933kip >> 177 kip
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6. Conclusions.
6.1 General.
Concrete sections reinforced by multiple encased rolled sections can be an advantageous solution to realize
mega columns of tall buildings, but they are not yet covered by standard design methods.
The design for shear of composite steel-concrete sections has to deal with two types of shear: transverse shear in
the section and longitudinal shear at the section components interfaces.
Two reference theories have been used to come out with design methods for shear of the proposed type of
columns: the classical theory of elastic beams and the strut and tie method or STM. The first theory applies to
continuous elements without cracks, the second to elements with cracks. It is shown that both can provide the
results necessary to design for transverse and longitudinal shear. Each theoretical development made is followed
by an example of application on a typical mega column, which helps setting forward peculiarities of the design
of such sections and the advantages of each method developed.
The application of the theory of elastic beams is made sub-dividing the section into sub-sections, either
reinforced concrete or composite. As the deformed shape in bending is single, the applied shear is distributed
proportionally to the bending stiffness of the sub-sections. It appears that the sub-sections with the encased steel
shapes attract a high proportion of the applied shear, which justifies a design for shear made individually into
each sub-section. It is a way to take into account the real flow of stresses into the section. The longitudinal shear
at the interface between concrete and steel reinforcing bars or steel shapes is deduced. It is needed to design the
shear connectors which are always necessary on the steel shapes as the bond stress is not reliable.
The strut and tie method provide results which are close to those of the elastic beam approach, but in a more
direct way. There are two reasons for this efficiency: the theory behind the STM method expresses reinforced
concrete structural elements as macro elements rather than a continuum; simplifications and short cuts are used
in the evaluation of mechanical characteristics of the section.
The similarity of the results obtained by the two proposed methods, in spite of the different hypothesis behind
the methods and the different simplifications proposed, tends to demonstrate that the proposals are adequate to
evaluate both the distribution of transverse shear into the section and the values of longitudinal shear at the steel
shapes-concrete interface.
6.2 Discussion on the calculated longitudinal shear forces at the steel shape-concrete interface.
The results obtained by the strut and tie method, both for the global longitudinal shear force at the steel shape-
concrete interface and for the distribution of that force around the sections, are reasonably close to those
obtained by the theory of beams, but always some 30% on the safe side.
This could be explained by the differences between the two methods. It should first be noted that, in the
example, the steel shapes are situated close to the perimeter of the section. The first approach in the evaluation
of shear in columns with several encased steel sections makes use of the theory of elastic beams and the
longitudinal shear force Vl in one horizontal plane cut of a section is proportional to S: Vl = VS/I
In a section with encased steel shapes closer to the axis of symmetry like the one of Figure 8, the values of S of
those more central profiles would be greater than the S of peripheral shapes, resulting in greater longitudinal
shear force Vl at the central shapes.
With the strut and tie method, there is only one value of the longitudinal shear force at the steel shapes-concrete
interfaces, see Figure 12, because steel shapes like concrete are submitted to a single shear force which
22
corresponds to a single Fdiag. The theory of beams gives a distribution of longitudinal shear force Vl, while the
strut and tie method thus gives an average Vu,l = Fdiag.cos45°; this average can be greater than the value of Vl
established by the theory of beams for peripheral steel shapes, but smaller for central shapes. The question best
method is the one which correspond to the hypothesis of the method: cracked or uncracked section, which may
depend on the loading case.
6.3 Future developments.
Until now, experimental tests have been focused on the behavior of sections with several encased steel shapes in
bending, demonstrating the adequacy of the concept and the calculation methods for bending combined to
compression; this explained in the companion paper (Plumier, 2013). But testing has been only on walls, not on
mega columns and it would be interesting to confirm the adequacy of calculation methods in that case; this
requires large scale experimental tests.
The existing tests provide little information on the behavior and the most adequate calculation methods in the
case of shear. Experimental campaigns should be made in order to calibrate the two calculation proposals
presented here. In particular, the following aspects should be looked at:
- The best way to estimate the internal lever arm of forces z in the application f the strut and tie
method; this is of importance, because the parameter z directly influences the value found for
longitudinal shear;
- The distribution of longitudinal shear in section, both in the direction of the applied shear and in
the perpendicular direction; the discussion in 6.2 sets forward some uncertainties and suggest a
logic for clarification which should be checked.
A research project led by the University of Liege on different problems related to the local use of steel shapes in
reinforced concrete sections started recently (Smartcoco, 2012) will contribute to a better general understanding
of the field, but more research on mega columns with several encased steel profiles, in particular submitted to
high shear, is needed.
23
References
AISC (2010), “Specification for Structural Steel Buildings”, Chicago, Illinois.
AISC (2011), “Design Examples V14”, Chicago, Illinois.
ACI (2008), “Building Code Requirements for Structural Concrete and Commentary”, ACI 318-08.ASTM
ASTM A6/A6M - 12. Standard Specification for General Requirements for Rolled Structural Steel Bars,
Plates, Shapes, and Sheet Piling
SMARTCOCO : Smart Composite Components – concrete structures reinforced by steel profiles”, RFSR-CT-
2012-00031 project of the European Fund for Steel Research, 2012.
Biography
André Plumier is a Professor at the University of Liege (Belgium), with specialties in steel and composite
steel-concrete structures and seismic design. He led many research projects in these fields. Mr. Plumier is the
inventor of the reduced beam sections concept. He is a consultant in projects in seismic areas and has been full
member of the ECCS TC13 seismic design of steel structures committee since its creation in 1984.
Teodora Bogdan is a research engineer at the University of Liege (Belgium). She obtained a PhD degree in
2011with a thesis in the field of composite steel-concrete structures at Technical University of Cluj-Napoca
(Romania). She is now working at University of Liege (Belgium).
Hervé Degée is research associate at the Belgian Foundation for Research and invited Professor at the
Universities of Liege and Ghent (Belgium). His main research field is structural mechanics and its application in
earthquake engineering for steel, composite steel-concrete and masonry structures. He is active in various
research programs and standardization committees at European level and acts regularly as consultant for
building companies and design offices for stability and seismic questions.
