DESIGN FOR SHEAR OF COLUMNS WITH SEVERAL ENCASED STEEL 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:
fbd =
π db2 Fy 4 x 47 xπ db2
=
60 4 x 47
= 0,32 ksi
2
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
Concrete: f’c = 7000 psi
Ec=57000√7000 = 4768ksi 3
or with wc’=148lbs/ft 2.3.2
Fy=60ksi
Ec=148 √7 = 4768ksi 1.5
Es=29000ksi (ACI318-08. Clause 8.5.1) (AISC Spec. I2-1b)
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.
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bc4 = 61.25in.
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*=tw 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 subsections. 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.
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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
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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
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-
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.
13
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:
4.
Vu,bs = 1744 kip
Vu,c = 1511kip
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.
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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.
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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:
Or:
vu,l,a,ext = 8.7 kip/in.
vu,l,a,int = 31.0 kip/in.
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.
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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 2
Asa = π (0.75) /4 = 0.44 in. Fu = 63.3 ksi
2
per steel headed stud anchor diameter 0.75in.
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 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
