Solution for bond distribution in asymmetric R.C. structural members

Engineering Structures 31 (2009) 633–641

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Solution for bond distribution in asymmetric R.C. structural members G. Russo, M. Pauletta ∗ , D. Mitri Department of Civil Engineering and Architecture, University of Udine, Viale delle Scienze 208 – 33100 Udine, Italy

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Article history: Received 29 November 2007 Received in revised form 6 August 2008 Accepted 3 November 2008 Available online 16 December 2008

Keywords: Bond problem RC member Asymmetric condition Anchorage Shear stress Maximum slip

a b s t r a c t The bond problem regarding bottom reinforcing longitudinal continuous bars or anchorages or lapped splices of RC beams is analyzed. This is an asymmetric problem both for the geometry and the mechanics. The geometrical asymmetry is due to the difference in the extension of the concrete areas surrounding the bars: the concrete area below the horizontal plane crossing the bars’ axes, and the concrete area over this plane up to the plane connecting the tips of the cracks. The mechanical asymmetry is due to the loading asymmetry with respect both to the above mentioned horizontal plane and to the vertical plane crossing the beam at the midspan of two consecutive cracks. Whenever a moment gradient exists, the equilibrium of the beam bottom portion bounded by two consecutive transversal cracks needs that the horizontal components of shear stress, acting on the plane connecting the tips of the cracks, balance the forces gradient acting along the bar. However, these components of shear stress have never been considered until now. The analytical solution to bond problems under asymmetrical conditions is worked out. Long and short anchorages are considered. Comparisons between 132 experimental results and values calculated by means of the proposed asymmetrical solution and an available symmetrical one are performed. The proposed solution provides more accurate and consistent predictions than the symmetrical unreliable one, and it is useful for studying serviceability behavior of reinforced concrete members. © 2008 Elsevier Ltd. All rights reserved.

1. Introduction The classical solution to the bond problem involves finding steel-to-concrete slip distribution along the embedded bar, through equilibrium and compatibility equations and constitutive laws. From this slip distribution it is possible to obtain bond stress, and steel and concrete strains, whose distributions completely describe the anchorage behavior. A local bond stress–slip law must be known in order to determine the bar response (slip distribution). As this law is nonlinear [1,2], so the second order differential equation governing the bond problem results as non-linear [3,4]. This equation is valid under axial symmetry conditions, but the symmetry is usually solely assumed for convenience [1,3–5], often disregarding the fact that great asymmetry exists in bond problems. The most representative among the asymmetric problems are those concerning the bottom reinforcing longitudinal continuous bars or anchorages or lapped splices of RC beams. The correct solution to these cases requires additional considerations and equations that are proposed, for the first time, in this paper. Reference is made to the general case of RC beams. As it is known, by increasing the loading on a beam results in cracks

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Corresponding author. Tel.: +39 0432 558065; fax.: +39 0432 558052. E-mail address: [email protected] (M. Pauletta).

0141-0296/$ – see front matter © 2008 Elsevier Ltd. All rights reserved. doi:10.1016/j.engstruct.2008.11.003

appearing in the zone under tension. For a cracked beam (Fig. 1(a)) the steel forces at two consecutive cracked sections are different, due to the variation of the external moment (moment gradient). This difference of forces (1F in Fig. 1(b)) can be equilibrated only by the horizontal component of the force due to the concrete shear stresses (τc in Fig. 1(b)) acting on the plane connecting the crack tips (called ‘‘tip crack plane’’ here). Consequently the bond–slip behavior of the reinforcing bars depends on the equilibrium of the beam bottom portion bounded by two consecutive flexural cracks at sides and by the ‘‘tip crack plane’’ at top, shown in Fig. 1(b). This portion is asymmetric both from the geometrical and loading points of view. As regards the beam under consideration, first cracks appear when concrete reaches the tensile concrete strength [6] at the beam bottom surface. In the case of continuous bottom reinforcement, the crack distribution depends on the loads acting on the beam: if loads are symmetrical, cracks are symmetrical too, and, first appear at the sections with the highest tensile stresses on concrete. In the cases of anchorages or lapped splices placed in beam zones under tension, first transversal cracks often appear at the sections in line with the free ends of the bars [5,7]. Indeed in these sections, due to the steel area variation, there is a sudden increase in the steel tensile stresses and, consequently, in the concrete stresses.

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Fig. 2. Double reinforced concrete rectangular beam section under (a) service and (b) ultimate stress distributions.

Fig. 1. (a) Cracked beam under uniformly distributed load, and (b) bottom portion of the beam bounded by two consecutive flexural cracks at sides and by the ‘‘tip crack plane’’ at top.

2. Analytical solution for the asymmetrical bond problem The asymmetrical bond problem is analyzed with reference to the general case of a double reinforced beam with not constant moment. The steps that are followed to attain the solution are described in this section: - the neutral axis depth is derived by means of compatibility and equilibrium equations (Section 2.1); - an expression for the shear stress distribution on the plane connecting the tips of the cracks is provided (Section 2.2); and - the equations governing the asymmetrical bond problem of the beam longitudinal bars and the corresponding solution are worked out (Section 2.3). 2.1. Neutral axis depth

1

fcmax b · kd − mAs fs + m0 A0s fs0 = 0 (1) 2 where fcmax = maximum concrete compressive stress (= M · kd/In , with M = acting moment and In = moment of inertia of the reacting section with respect to the neutral axis); b = width of the section; m0 and m = number of compression and tension bars, respectively, A0s and As = areas of one compression and one tension reinforcing bar, respectively; fs0 = n · fcmax (d − dc 0 )/kd and fs = n · fcmax (d − kd)/kd, with n = ratio between the elastic moduli of steel and concrete, d and d0c = distance from the extreme compression fiber of concrete to the centroid of tension steel area and to the centroid of compression steel area, respectively. As regards ultimate loading conditions (Fig. 2(b)), the neutral axis depth ku d is obtained from the equilibrium of the moments with respect to the centroid of tension steel area

 d−

for strengths above 28 MPa β1 shall be reduced continuously at a rate of 0.05 for each 7 MPa, but β1 shall not be taken less than 0.65. The value of fs can be determined by means of the equilibrium of the horizontal forces at the ultimate limit state fs =

The neural axis depth, kd, is determined on the basis of the assumption that plane sections before bending remain plane after bending. For service loading conditions, linear constitutive laws for steel (fs = Es · εs ) and concrete (fc = Ec · εc ) are considered [8], where fc , fs = concrete and steel stresses, Ec , Es = concrete and steel elastic moduli, εc , εs = concrete and steel strains, respectively. For ultimate loading conditions, a linear behavior, until yielding, for steel and an equivalent rectangular stress distributions (stress block) for concrete are considered [6]. The tensile strength of concrete is ignored. As regards service loading conditions (Fig. 2(a)), the neutral axis depth kd is obtained from the equilibrium of the horizontal forces

0.85fc0 b · β1 ku d ·

Fig. 3. Steel forces and mean horizontal tangential stresses on a beam bottom portion bounded by two consecutive cracks at sides and by the tip crack plane at top; (a) asymmetrical and (b) symmetrical model.

β1 ku d 2

 = Mu

(2)

where Mu = ultimate moment and β1 = factor [6] that shall be taken as 0.85 for concrete strengths fc0 up to and including 28 MPa,

0.85 · fc0 b · β1 ku d mAs

.

(3)

2.2. Shear stress distribution at the tip crack plane When the moment is not constant along the beam (Fig. 1) ultimate conditions are reached only at the section subjected to the maximum moment Mu . But, since in the most stressed zone the cracks are very close one to the other, it can be assumed that the moment M at the crack nearest the crack under the maximum moment Mu is equal to Mu . As a consequence of this, it can also be assumed that the tips of adjacent cracks lie on the same level and that those tips are connected by an ideal horizontal plane surface (Fig. 3(a)), ‘‘tip crack plane’’, not by an inclined one (Fig. 1(b)). For the equilibrium of the beam bottom portion bounded by two consecutive cracks at sides and by the tip crack plane at top (Fig. 3(a)), the force due to shear stresses τchm acting on the tip crack plane must equalize the variation of force in the steel. The shear stresses are taken as evenly distributed on the tip crack plane and they can be calculated from the following translational equilibrium equation

τchm = mAs

fs0 − fsL b·L

(4)

with fs0 , fsL = steel stresses at x = 0 and x = L respectively, where the origin of the abscissa x is taken as in Fig. 3. From Eq. (4) and Fig. 3(a), it is apparent that, when a gradient of steel stress exists (i.e. fs0 6= fsL ), the tangential stresses on the tip crack plane, that are different from zero, guarantee the equilibrium of the horizontal forces acting in the steel (τchm b · L − fs0 mAs + fsL mAs = 0). Moreover it can be observed that, for the beam under consideration, a symmetric model (Fig. 3(b)), given by a single bar embedded in a concrete cylinder [3,9] with thickness equal to the minimum cover to the bar, cmin , is not reliable. In fact, when

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where εc = concrete strain at the bar to concrete interface; strains caused by tension are positive. Taking advantage of Eqs. (6) and (7), the derivation of Eq. (9) gives s00 − χb · τ [s (x)] = 0

(10)

where s = d s/dx and 00

2

χ b = Σs ·

Fig. 4. Stresses on a RC beam bottom portion of length dx.

a moment gradient exists in the beam, a consequent steel stress gradient develops (fs0 6= fsL ) and, the equilibrium of the cylindrical element in Fig. 3(b) is not possible, being fs0 mAs − fsL mAs 6= 0 and the perimetric surface unloaded. 2.3. Governing equations For a symmetrical R.C. element, as the one represented in Fig. 3(b), the equilibria of horizontal forces and of moments coincide, hence only a single equilibrium equation can be written [1,3,4]. In contrast, given the problem under consideration (Fig. 3(a)), the equilibrium of horizontal forces differs from the equilibrium of moments, hence two equilibrium equations hold. To write these equations, reference is made to the beam bottom portion bounded by two consecutive flexural cracks or by the support and the crack nearest the support. The conventional ‘‘horizontal tip crack plane’’ is set at the depth, dt in Fig. 4 (= h − kd with h = section height in Fig. 2), between two consecutive cracks, or at the crack tip for the end beam span between a crack and the support. The equilibrium equations are written for an infinitesimal beam length dx with width b as shown in Fig. 4. The concrete stresses perpendicular to the cross section are considered evenly distributed. The equilibria of horizontal forces and of moments respectively give



2

1 As · Es

+



m · ds

ψ · Acb · Ec · dG

.

(11)

For the case of an axial-symmetric RC element the corresponding χ expression is given by [9]

χ = Σs ·



1

+

As · Es



1

ψ · Ac · Ec

.

(12)

The difference between Eqs. (11) and (12), due to the different schematizations (asymmetrical and symmetrical) used to solve the bond problem, is in the second term of these expressions, which takes the shape of the concrete area surrounding the reinforcing bars into account. By using Eq. (8) for the τ (s) expression, the first integral of Eq. (10) is

p

s0 = ∓ 2 · (γb · s1+α + C ).

(13)

where the minus sign implies tensile steel condition; C = arbitrary constant of integration and

γb = χb ·

τ1 . (1 + α) · sα1

(14)

Eq. (13) represents two concave functions, one decreasing (minus sign) and the other increasing (plus sign). The integration of Eq. (13) leads to equations similar to the solving equations of the symmetrical bond problem [4,11], but different in the expression of χb (Eq. (11)), holding only for the asymmetrical bond problem 1 s 1 α+2 γb · Hyp2F 1 x=B∓ √ , , , − · s1+α C (α + 1) 2 (α + 1) 2C



Acb · ψ

dfc

+ mAs ·

dx dfc

Acb · ψ ·

dx

dfs dx

= −τchm · b

· dG + mAs ·

dfs dx

· ds = 0

(5) (6)

where ψ = ratio of the average tensile stress fcmn (Fig. 4) to the maximum tensile stress fc (dc , x) that occurs at the steel–concrete interface [10]; Acb = b · dt − mAs ; dG = [bd2t /2 − mAs (dt − dc )]/Acb distance between the resultant of mean concrete stresses and the conventional horizontal tip crack plane; ds = dt − dc distance between the bar axis and the conventional horizontal tip crack plane, with dc = distance between the bar axis and the beam bottom surface. The reinforcing bar equilibrium is given by As ·

dfs dx

= −Σs · τ (s)

(7)

where Σs = reinforcing bar circumference (= π db , with db = bar diameter) and τ (s) = bond stress at the steel–concrete interface as a function of the relative slip s. The bond stress–slip relationship of Ciampi et al. [1] is used

τ = τ1 ·



s

α (8)

s1

where τ1 = maximum bond stress, s1 = the slip corresponding to τ1 and α = exponent ranging from 0.2 to 0.5. The strain–displacement relation is given by

εs − εc = −

ds dx

= −sc 0

(9)

1 1 C 1+α C 1+α for − < s ≤ γb γb x=B∓ √

2·s

(15)

1−α 2

2γb · (1 − α)

· Hyp2F 1

α−1 3α + 1 C , , − · s−(1+α) 2 2 (α + 1) 2 (α + 1) γb 1 1 C 1+α C 1+α for ≤ s ∪ s < − γb γb 

×

1



,



(16)

where B is a constant of integration and Hyp2F 1 represents 2F 1 hypergeometric function which, used in codes of calculus, makes it possible to express the series of functions with infinite terms of the sum in a compact and convergent form. Expression (15) represent two convex functions, expression (16) two concave ones. It should be noted that if the constant of integration C is zero, the solution of Eq. (13) can be written in a closed form [4]

 s= ∓

1−α 2

·

2  1−α

2γb · (x − B)

p

.

(17)

This case occurs when the s(x) function exhibits one or more points with s = 0 and s0 = 0, namely when there is a portion of the bar not affected by bond stress and slip, hence when no rigid motion of the bar occurs.

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The integration of Eq. (5), by means of Eq. (4), provides

εc = −ξ · εs − ζ · x + D (18) where D = integration constant; ξ = ρ n/ψ ; ρ = mAs /Ac ; and ζ =

τchm b . ψ Ac Ec

(19)

The term ζ takes account of the shear (tangential) stresses, which balance the difference of steel forces acting at the cracked sections (Fig. 3). Being fs0 6= fsL , the beam portion in Fig. 3(b), symmetrical with respect to the reinforcing layer, is not in equilibrium. The tangential stresses, although necessary for the equilibrium, have always been neglected in the previous literature, to simplify the approach to the bond problem. Eq. (9), by means of Eqs. (13) and (18), provides the steel strain

εs =

D−ζ ·x±

p

2 · (γb · s(1+α) + C )

(1 + ξ )

.

(20)

Eq. (18), by means of Eq. (20), provides the concrete strain

εc =

D−ζ ·x∓ξ

p

2 · (γb · s(1+α) + C )

(1 + ξ )

.

(21)

All bond problems can be solved using the previous equations and determining the constants B, C and D on the basis of the boundary conditions. The sign of Eqs. (15)–(17), (20) and (21) depends on the physical behavior of the case under consideration. 3. Bond behavior of beam reinforcing bars under tension A reinforced concrete beam that has reached a stabilized cracked regime is considered. The beam bottom portion bounded by two consecutive cracks at sides and by the tip crack plane at top (Fig. 3(a)) is analyzed in the following. The direction of the x-axis is assumed to enter the considered beam portion and the origin is set at the extreme left section; tensile stresses are positive. At the cracked sections boundary conditions for steel are:

εs (0) = εs0 =

fs0

εs (L) = εsL =

fsL

Es

Es and for concrete

εc (0) = εc (L) = 0.

(22) (23)

(24)

By means of Eqs. (22) and (24), Eq. (18) gives the first constant of integration D = ξ · εs0

(25)

while, by means of conditions (18), (23) and (24) gives D = ζ · L.

(26)

By taking Eq. (4) and the expressions of ξ , ρ and ζ into account, it can be proved that Eqs. (25) and (26) give the same D value. Bars embedded in beam regions under tension can be continuous bars, anchorages or lapped splices. Boundary conditions (22)–(24) are useful for the general case of a continuous bar, while, if an anchorage is considered or a lapped splice bar, one end of the bar is unloaded (for example at x = L), hence Eqs. (22) and (24) still hold, while Eq. (23), becomes εs (L) = 0. 3.1. Solution for beam anchorages under tension Let’s consider a beam ending anchorage or a lapped splice bar, in both cases the asymmetrical bond problem can be represented as in Fig. 5(a). As regards the ending anchorage, the schematization in Fig. 5(a) represents the beam bottom portion between the end section of the reinforcing bars (on the right at x = L in Fig. 5) and the closest crack (on the left at x = 0 in Fig. 5). This beam portion

Fig. 5. (a) Asymmetrical and (b) symmetrical model for beam ending anchorages or lapped bars in a splice.

is bounded at the top by the ideal horizontal plane tangent to the tip of the crack occurring at x = 0. As regards the bond problem of the lapped splice bar, the schematization in Fig. 5(a) represents the beam bottom portion bounded by the two cracks that usually grow up just outside the lapped splice length. In this case the beam portion is bounded at the top by the ideal horizontal plane set at the depth of the two cracks. In Fig. 5(b) the same bond problem is schematized in a radial symmetrical way, with respect to the bar axis. It must be observed that, in this schematization, the absence of any tangential stress and the presence of the end force, F /m, do not allow any possible equilibrated solution of the problem. To identify the boundary conditions necessary to find the constants of integration (B, C and D) of the proposed general solution of the bond problem (Eqs. (15)–(17), (20) and (21)), the physical behavior of the bar in Fig. 5(a) has to be analyzed. Moving into the schematized element from x = 0 to x = L, the relative slip between steel bars and concrete must decrease due to bond stresses that develop at steel–concrete interface. The steel strain, which has its maximum value at the loaded end (x = 0), must decrease with x up to zero at the unloaded end (x = L). Consequently, dfs /dx ≤ 0 and, by means of Eq. (7), τ (s) ≥ 0. Moreover, because of Eq. (10), s00 = d2 s/dx2 = χb · τ [s(x)] ≥ 0. It follows that s(x) is a concave function, hence Eq. (16) or (17) must be used. If the anchorage is long enough or the load is not high, a bar portion, close to the unloaded end, does not undergo to slip (s = s0 = 0). Hence, by means of Eq. (13), C = 0 and Eq. (17) can be used. In contrast, if the anchorage is short or the load is high, the whole bar slip, and Eq. (16) must be employed. The acceptation ‘‘long anchorages’’ and ‘‘short anchorages’’ are used here for the former and the latter cases, respectively. Because of the slip decrease, the minus sign holds in Eq. (16) or (17). The steel strain distribution, also decreasing, is given by Eq. (20), with the plus sign, while concrete strain is given by Eq. (21), with the minus sign. The cases of long and short anchorages and the discriminating condition between them are explained in the following. 3.1.1. Long anchorages In this case a bar portion (RZ in Fig. 6(a)) does not slip (s = 0). By means of Eq. (8), it follows that in this bar portion also bond

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Table 1 Specimen geometrical and mechanical characteristics [12]. Characteristics

Notation

Value

Anchorage length Longitudinal bottom bar diameter Diameter of stirrups within the splice Section height Section effective depth Section width Bottom concrete cover (from the bar centroid) Specific concrete strength Steel elastic modulus Steel stress at the splice extremities

L (mm) db (mm) dst (mm) h (mm) d (mm) b (mm) dc (mm) fc0 (MPa) Es (MPa) fs0 (MPa)

305 20 10 305 260 240 35 55.4 200000 203

The integration constant B is given by Eq. (16) with the minus sign calculated for x = 0 Fig. 6. Typical slip and steel strain distributions in a bar pulled-out with concrete under tension when rigid motion of the bar (a) does not occur or (b) occurs.

stresses are equal to zero (τ = 0), and, by means of Eq. (9), it follows that εs = εc = 0 too. Since sR = s0R = 0 at point R, Eq. (13) gives C = 0, for x = xR . It follows that the solution for the portion OR is given by Eq. (17), with the minus sign. At the cracked section (x = 0), concrete strain is equal to zero (εc (0) = 0), hence, for x = 0, Eqs. (9) and (13) give 1  1+α 2 εs0 s0 = (27) 2 · γb where s0 = s(0) and εs0 = εs (0) are the maximum slip and the



maximum strain in the bar portion OR. Eq. (17), for s = 0, yields B = Lrm (Fig. 6(a)), and Eq. (17) taking account of Eq. (27), for x = 0, gives B = Lrm =

2 1−α

 ·

1 2γb

1  1+α

1−α 1+α

· εs0 .

(28)

The boundary conditions of the problem (Eqs. (22) and (24)) give the constant D. 3.1.2. Discriminating condition between long and short anchorages On the basis of the applied load it is possible to determine a priori whether the case in Fig. 6(a) occurs, or the whole bar slip (Fig. 6(b)). The discriminating condition between the two cases is L − Lrm ≥ 0, where L = anchorage length. If this condition is verified, no bond stresses (i.e. slip) occur in a portion of the steel reinforcement, no rigid motion of the bar occurs and the closed form solution (Eq. (17)) can be used. In the other case, bond stresses develop along the whole steel reinforcement, rigid motion of the bar occurs and the solution by series of functions (Eq. (16)) must be used. If L = Lrm , the discriminating condition is equal to zero, hence Lrm is called here ‘‘rigid motion length’’, i.e. the length that prevent any rigid motion of the bar and allows one to distinguish between long and short anchorages. 3.1.3. Short anchorages In this case (Fig. 6(b)), the slip differ from zero along the whole anchorage and s0 = 0 only at x = L, because of Eq. (9), where εs (L) = εc (L) = 0. For x = L, Eq. (13) gives the integration constant C = −γb sL1+α

(29)

where sL = s(L) slip at the free end of the bar (minimum slip). By means of Eqs. (20), (22), (24), (25) and (29), it is possible to express the slip at the loaded bar end as a function of sL



s0 = sL1+α +

2 εs0 2γb

1  1+α

.

(30)

1−α

2 · s0 2

B= √ · Hyp2F 1 2γb · (1 − α)

" ×

 1+α # α−1 3α + 1 sL , , , . 2 2 (α + 1) 2 (α + 1) s0 1

(31)

Since Eq. (16) with the minus sign, for s = sL , must give the value L, by substituting Eqs. (29)–(31) into Eq. (16) calculated for x = L, Eq. (16), iteratively solved, gives sL . By means of this value and Eq. (30), it is possible to calculate the slip at the loaded end of the anchorage s0 and, hence, the integration constants C (Eq. (29)) and B (Eq. (31)). For s varying between s0 and sL , Eq. (16) with the minus sign provides the abscissa at which the value of s occurs. The steel strain is given by Eq. (20), with the plus sign, while the concrete strain is given by Eq. (21), with the minus sign. The distribution of bond stress along the bar, obtained from the slip by means of Eq. (8), is similar to the slip distribution (Fig. 6(b)): bond stress is maximum at the loaded bar end (for x = 0), then decreases. 4. Application example of the calculation process The analytical solution for the asymmetrical bond problem developed in the previous sections allows to obtain the distributions of slip (Eqs. (15), (16) or (17)), bond stress (Eq. (8)) and steel and concrete strains (Eqs. (20) and (21), respectively) along the reinforcing bar. An application example to one of the beams used in Section 5 for the experimental comparisons is developed, showing how to follow the calculation process to obtain the slip, by means of the proposed asymmetrical solution or the symmetrical one. The geometry, the loading arrangement and the reinforcing layout of the beam [12] considered for the example are represented in Fig. 7. The geometrical and mechanical characteristics of the beam necessary for the calculations are reported in Table 1. Three steel bars, overlapped in the midspan and surrounded by stirrups, reinforce the beam, which is made of high strength concrete. The authors of the test [12] measured the opening of the cracks that appeared in the constant moment region on the tension side of the beam just outside the splice length, at various loading steps. The authors provided the steel stress versus the average crack width for every step of loading. The step at which steel stress fs0 = 203 MPa occurred is considered here. Since the crack opening is nearly equal to twice the slip between the bar and the adjacent concrete [13], twice the slip calculated at the crack should be equal to the measured crack opening. The steps that have to be followed to calculate this slip are explained in the following. 4.1. STEP 1 - Identification of the bond stress–slip parameters First of all, proper parameters (τ1 , s1 and α ) of the local bond stress–slip law (Eq. (8)) have to be chosen. The same parameters

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G. Russo et al. / Engineering Structures 31 (2009) 633–641

Fig. 7. Geometry, reinforcement layout and loading arrangement of the beam [12] used for the example.

hold both for the asymmetrical and the symmetrical bond solutions. Since the beam is made of high strength concrete, the FIB parameters [14] are used. It has to be reminded that these parameters are different in relation to different bond conditions: ‘‘good’’ conditions, or ‘‘other’’ conditions. Since the specimen, even under service loads, showed splitting cracks in the concrete cover in the lapped splice region, the bond conditions should be evaluated not good. But the specimen had also stirrups in the lapped splice length, that, in any case, improve bond conditions. Hence actual bond condition of the considered beam stay in the middle course between ‘‘other’’ and ‘‘good’’, and, as a consequence, bond parameters (τ1 , s1 and α ) have to be interpolated between these two conditions. No interpolation is needed if no stirrups are present in the lapped splice region. The values of s1 and α are respectively equal to 0.5 mm and 0.3, both for ‘‘good’’ and ‘‘other’’ bond conditions [14], while τ1 is linearly interpolated [2] between τ1 = 0.225fcm , holding for ‘‘other’’ conditions, and τ1 = 0.45fcm , holding for ‘‘good’’ conditions, where fcm = fc0 + 8 MPa [14] is the mean value of concrete compressive strength. The interpolation is performed on the basis of the area of stirrups (two legs) over a length equal to the anchorage length, Ast , in the cases where Ast ,min < Ast < mAs , with Ast ,min = 0.25 · m · As [2]. For the considered beam the anchorage length is the lapped splice length (L in Fig. 7 and Table 1) and the number of stirrups over the anchorage length is 4, hence Ast = 2 · (2π d2st /4) · 4 = 628 mm2 , with dst = stirrup diameter (Table 1). The longitudinal bar area is As = π d2b /4 = 314 mm2 and the number of longitudinal bars in the anchorage length is m = 6, hence Ast ,min = 0.25 · 6 · π 202 /4 = 471 mm2 and mAs = 6 · 314 = 1884 mm2 ; fcm = 55.4 + 8 MPa = 63.4 MPa, hence τ1 is equal to 0.225fcm = 14.265 MPa for ‘‘other’’ bond conditions, and to 0.45fcm = 28.53 MPa for ‘‘good’’ bond conditions. The linear interpolation of τ1 between 14.265 and 28.53 MPa on the basis of Ast gives 15.864 MPa.

4.3. STEP 3 - Calculation of the parameters χ and γ

4.2. STEP 2 - Calculation of the neutral axis depth

Eq. (27) gives values of the slip at the crack equal to 0.0417 mm or 0.0284 for the asymmetrical or the symmetrical solutions, respectively. The opening of the crack measured by the authors of the test [12] was 0.09041 mm. Since twice the slip should be equal to the crack opening, the asymmetrical solution predicts the best value of the slip in comparison with the symmetrical one, indeed 2 × 0.0417 = 0.0834 mm, obtained by means of the former, is much closer to 0.09041 than 2 × 0.0284 = 0.0568 mm, obtained by means of the latter.

The neutral axis depth is calculated only for the asymmetrical bond solution. Because the considered beam is subjected to service loads, the equation to be used to calculate the neutral axis depth kd is Eq. (1), where, for the particular case under consideration m0 = A0s = 0 (no reinforcement at the beam top between the applied loads P, Fig. 7), fs = fs0 = 203 MPa is reported in a graph by the authors [12], fcmax = fs · kd/[n · (d − kd)] from Fig. 2(a), with n = Es /Ec = 200 000/(4700 · fc00.5 ) = 5.72 [6]. Moreover in Eq. (1) m = 3, because the bar slip is calculated just outside the lapped splice where only three bars are present, and b = 240 mm (Table 1). The obtained value for the neutral axis depth is kd = 46.1 mm.

Parameter χ is calculated by means of Eq. (11) (χ = χb ) or (12) for the asymmetrical or the symmetrical bond solutions, respectively. Parameter γ is calculated by means of Eq. (14) for both solutions by using the proper value of χ : χb from Eq. (11) for asymmetrical solution, and χ from Eq. (12) for the symmetrical one. The terms common to Eq. (11) and Eq. (12) are: Σs = π db = 62.83 mm, Es , Ec = 4700 · fc00.5 = 34 983 MPa, As = 314 mm2 , and ψ = 0.75 (a suitable value for tension concrete [10]). Moreover, as regards (11), m = 3 and the terms ds , Acb and dG can be easily calculated knowing the neutral axis depth obtained in step 2: ds = dt − dc , where dt = h − kd, Acb = bdt − mAs , and dG = [bd2t /2 − mAs (dt − dc )]/Acb with h in Fig. 2 and ds , dt , and dc in Fig. 4. Hence dt = 228.9 mm, ds = 198.9 mm, Acb = 61 184 mm2 , and dG = 144.8 mm. On the basis of these values Eq. (11) gives χb = 1.186 × 10−6 and Eq. (14) gives γb = 3.203 × 10−5 . As regards Eq. (12), Ac = π d2c − As = 2512 mm2 (Fig. 5(b)). On the basis of the above calculated values Eq. (12) gives χ = 1.953 × 10−6 and Eq. (14) gives γ = 5.276 × 10−5 . 4.4. STEP 4 - Calculation of the rigid motion length Lrm The value of the rigid motion length Lrm allows to discriminate between long and short anchorages and to select which equation has to be used to calculate the slip s0 at the cracks, Eq. (27) for the former case or Eq. (30) for the latter. The value of Lrm is given by Eq. (28), where εs0 is the steel strain at the section just outside the splice length equal to fs0 /Es = 1.015 × 10−3 . By using the other parameters, already calculated, Eq. (28) gives values of the rigid motion length equal to 117 mm or to 80 mm, for the asymmetrical or the symmetrical (γb = γ ) solutions, respectively. In both cases Lrm is less than the anchorage length L = 305 mm (Table 1), hence Eq. (27) can be used to calculate the slip. 4.5. STEP 5 - Calculation of the slip

5. Experimental comparisons The analytical results obtained by means of the proposed asymmetrical solution and by means of the symmetrical one are compared with experimental results obtained from tests available

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Table 2 Specimens considered for the experimental comparisons. Authors NSC (26 beams) Thompson et al. (1979) [15]

Treece and Jirsa (1989) [16] HSC (8 beams) Hamad et al. (2001) [7] Hamad and Najjar (2002) [12]

Name of specimens

No. of results

6-12-4/2/2-6/6; 8-18-4/3/2-6/6; 8-18-4/3/2,5-4/6; 8-36-4/1/2-6/6; 8-36-4/1/2,5-4/6; 8-36-4/1/4-6/6; 8-24-4/2/2-6/6; 8-24-4/2/2,5-4/6; 8-24-4/2/4-6/6; 8-24-4/2/2-6/6-TC; 11-45-4/1/2-6/6; 11-30-4/2/2-6/6; 11-30-4/2/4-6/6; 11-30-4/2/2,7-4/6; 11-25-6/2/3-5/5; 11-30-4/2/2-6/6-TC; 14-60-4/2/2-5/5; 14-60-4/2/4-5/5; 8-15-4/2/2-6/6-S5;11-20-4/2/2-6/6-S5; 11-20-4/2/2-6/6-S2.9; 11-30-4/2/2-6/6-S5; 11-20-4/2/2-6/6-SP; 14-40-4/2/2-5/5-S3; 14-40-4/2/2-5/5-S5.7 0-6-4

43

B20F0; B25F0; B32F0 B20S0; B20S1; B20S2; B20S3; B20S4

21 46 Total 132

Fig. 8. Geometry and loading arrangement of the R.C. beam experimental specimens.

in the literature. The comparisons are performed to evaluate the accuracy and consistency of the two solutions and to assess which of them is the best. The geometry and the loading arrangement of the 34 beam specimens taken under consideration for the comparisons are traceable back to the two schematizations in Fig. 8. The beams have reinforcing bars overlapped in the midspan, where the acting moment is constant and the shear equal to zero. In some cases there are stirrups in the splice length. For all the considered beams the experimenters measured the opening of the cracks that appeared in the constant moment region, on the tension side of the beam, just outside the splice length, at various loading steps. The authors, the names of the considered specimens and the number of experimental results for each group of specimens are reported in Table 2. Some specimens (26 beams [15,16]) were made by Normal Strength Concrete, NCS, and others (8 beams [7,12]), by High Strength Concrete, HSC. The total number of considered experimental results is 132 (65 for NSC and 67 for HSC). Crack openings less than 0.05 mm are not taken under consideration, because they are too small and not significant from the design point of view. The values of measured crack openings are compared to twice the slip at the crack, calculated by using both the asymmetrical and the symmetrical solutions, following the calculation process shown in Section 4. It must be observed that 18 of the considered results [15] reported in Table 2 refer to collapse conditions, while the other 114 refer to service loading conditions. Hence, to calculate the neutral axis depth, Eq. (2) has been used in the former cases, while Eq. (1)

22

in the latter ones. Only cases of collapse due to bond failure are taken into account. As regards the calculation of the bar slip, the bond parameters (τ1 , s1 and α ) provided by the CEB [2] are used for NSC specimens, while those provided by the FIB [14] for HSC ones. As regards thepCEB [2] parameters, those for unconfined concrete (τ1 = 2 fc0 , s1 = 0.6 mm, α = 0.4) are employed for specimens not having stirrups in the splice length, while the parameters are linearly interpolated between unconfined p condition to confined one (τ1 = 2.5 fc0 , s1 = 1 mm, α = 0.4), for specimens having stirrups. The linear interpolation is performed on the basis of the area of stirrups, Ast , analogously to what done for the example in Section 4. As regards the FIB parameters [14], those for ‘‘other’’ bond conditions (τ1 = 0.225fcm , s1 = 0.5 mm, α = 0.3) are employed for specimens not having stirrups in the splice length, while the parameters are linearly interpolated between ‘‘other’’ conditions to ‘‘good’’ ones (τ1 = 0.45fcm , s1 = 0.5 mm, α = 0.3), for specimens having stirrups, as done in Section 4. To calculate the slip s0 for each case of loading, the calculation process explained in Section 4 is followed, i.e.: the neutral axis depth is calculated by means of Eq. (1) or (2) depending on the loading stage, constant ψ is always taken equal to 0.75 [10], parameter χb is calculated by means of Eq. (11) for the asymmetrical solution and by means of Eq. (12) for the symmetrical one, parameter γb by means of Eq. (14), the length Lrm by means of Eq. (28), and the slip s0 by means of Eq. (27), if the splice length L is greater than Lrm , while by means of Eq. (30), if L < Lrm . The comparisons between the measured values of crack opening, wfmeas , and the corresponding calculated ones (twice the slip, 2s0 ), are performed in Figs. 9–11. The calculated values of crack opening (2s0 ) are reported on the abscissa and the measured ones (wfmeas ) on the ordinate. The average and the coefficient of variation of calculated to measured values are reported on the diagrams. Data relevant to NSC beams are shown in Fig. 9, those relevant to HSC beams are reported in Fig. 10. In Fig. 9(a) and 10(a) the measured values are plotted versus the values calculated by means of the asymmetrical solution, while in Fig. 9(b) and 10(b), versus those calculated by means of the symmetrical solution. Fig. 11 shows the comparisons for all the 132 experimental results by using the asymmetrical solution (Fig. 11(a)) and the symmetrical one (Fig. 11(b)). The percentage of error in the average accuracy (AVG) of the slip prediction is 7.5% by means of the asymmetrical solution, while is 16.4% by means of the symmetrical one, hence the asymmetrical solution gives a |7.5 − 16.4| · 100/16.4 = 54% improvement in the accuracy. As regards the

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Fig. 9. Measured crack widths versus values calculated by means of (a) asymmetrical and (b) symmetrical solutions, for 65 loading conditions obtained from 26 NSC tested beams [15,16].

Fig. 10. Measured crack widths versus values calculated by means of (a) asymmetrical and (b) symmetrical solutions, for 67 loading conditions obtained from 8 HSC tested beams [7,12].

Fig. 11. Measured crack widths versus values calculated by means of (a) asymmetrical and (b) symmetrical solutions, for all the 132 loading conditions obtained from the 34 tested beams [7,12,15,16].

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uniformity of the prediction (COV), the asymmetrical solution gives a |0.217 − 0.369| · 100/0.369 = 41% improvement. 6. Conclusions The analytical solution for the bond problem occurring in longitudinal bars of R.C. beams is provided. This solution makes good the lack of a bond theory for geometrical and mechanical asymmetric problems, which have been analyzed under physically unrealistic symmetry until now. The proposed solution can be applied, for example, to reinforcing bars, or anchorages, or lapped splices of RC beams. It takes account of the horizontal shear stresses occurring at the neutral axis depth, which are necessary for the equilibrium of the asymmetric RC element. The solution for beam lapped splices under tension is fully developed. An application example to explain the calculation process, both for the asymmetrical and the symmetrical solutions, is reported. The comparisons between 132 experimental crack openings available in the literature and the twice the bar slip predictions show that the new asymmetrical solution is more accurate and consistent than the symmetrical one. The asymmetrical solution yields a 54% improvement in accuracy and a 41% in uniformity. Acknowledgements The research has been funded by the Italian Department of Civil Protection in the frame of the national project RELUIS.

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References [1] Ciampi V, Eligenhausen R, Bertero VV, Popov EV. Analytical model for concrete anchorages of reinforcing bars under generalized excitations. Rep. no. UCB/EERC. Berkley: Univ. of California; 1982, p. 82–3. [2] CEB-FIP Model Code 1990. Bullettin d’information no. 195; First Draft, Chapters 1–5, 3–1—3–3. [3] Rehm G. Über die Grundlagen des Verbundes zwischen Stahl und Beton. Deutscher Ausschuss für Stahlbeton, Berlin (Germany); 1961. (138), p. 1–59 (in German). [4] Russo G, Zingone G, Romano F. Analytical solution for bond–slip of reinforcing bars in R.C. joints. J Struct Eng, ASCE 1990;116(2):336–55. [5] Tepfers R. A theory of bond applied to overlapped tensile reinforcement splices for deformed bars. Pub. 73:2 Division of Concrete Structures. Goteborg (Sweden): Chalmers University of Technology; 1973. p. 328. [6] ACI 318M-05, Building Code Requirements for Structural Concrete and Commentary. American Concrete Institute, Farmington Hills, MI; 2005. [7] Hamad BS, Harajili MH, Jumaa G. Effect of fiber reinforcement on bond strength of tension lap splices in high-strength concrete. ACI Struct J 2001;98(5): 638–47. [8] Park R, Paulay T. Reinforced concrete structures. NY: John Wiley and Sons; 1975. p. 426–95 [Chapter 10]. [9] Russo G, Romano F. Cracking response of R.C. members subjected to uniaxial tension. J Struct Eng, ASCE 1992;118(5):1172–90. [10] Edwards A, Picard A. Theory of cracking in concrete members. J Struct Div, ASCE 1972;98(12):2687–700. [11] Russo G, Pauletta M. A simple method for evaluating the maximum slip of anchorages. Mater Struct 2006;39:533–46. [12] Hamad BS, Najjar S. Evaluation of the role of transverse reinforcement in confining tension lap splices in high strength concrete. Mater Struct 2002;35: 219–28. [13] Gambarova G, Rosati GP, Zasso B. Steel-to-concrete bond after concrete splitting: Test results. Mater Struct 1989;22:35–47. [14] FIB BULLETIN 1, MC90 approach to modelling of bond. Structural Concrete – Textbook on Behaviour. Design and Performance Updated Knowledge of the CEB/FIP Model Code 1990, 1999; vol. 1, p. 30–5. [15] Thompson MA, Jirsa JO, Breen JE, Meinheit DF. Behavior of multiple lap splices in wide sections. ACI Journal; 1979. Symposium Paper. Title no. 76–12. [16] Treece RA, Jirsa JO. Bond strength of Epoxy-coated reinforcing bars. ACI Mater J 1991;86(2):167–74.