hoang - on the conceptual basis of the crack sliding theory

Årgang LXXIX, Nr. 1-2, marts 2008

BYGNINGSSTATISKE MEDDELELSER

udgivet af

DANSK SELSKAB FÓR BYGNINGSSTATIK

Proceedings of the Danish Society for Structural Science and Engineering

Y. ZHAO, L. C. HOANG, M. P. NIELSEN On the Conceptual Basis of the Crack Sliding Theory…………………1-48

KØBENHAVN 2008

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Copyright © 2008 ”Dansk Selskab for Bygningsstatik”, København

ISSN 0106-3715 (trykt udgave)

ISSN 1601-6548 (online)

Årgang LXXIX, Nr. 1, marts 2008

BYGNINGSSTATISKE MEDDELELSER

udgivet af

DANSK SELSKAB FOR BYGNINGSSTATIK

Proceedings of the Danish Society for Structural Science and Engineering

Y. ZHAO, L. C. HOANG, M. P. NIELSEN On the Conceptual Basis of the Crack Sliding Theory…………………1-48

KØBENHAVN 2008

Redaktionsudvalg

Lars German Hagsten (Redaktør) Rasmus Ingomar Petersen

Finn Bach Morten Bo Christiansen

Jørgen Nielsen Mogens Peter Nielsen Sven Eilif Svensson

Artikler offentliggjort i Bygningsstatiske Meddelelser har gennemgået review.

Papers published in the Proceedings of the Danish Society for Structural Science and Engineering have been reviewed.

On the Conceptual Basis of the Crack Sliding Theory

1 Introduction 1

2 Review of the classical plasticity approach 1

3 The Crack Sliding Model 3

4 Model with arbitrary curved yield lines 8

4.1 Dissipation formula for curved yield lines 8

4.2 Crack Sliding Model for curved diagonal cracks 10

5 Calculation based on observed crack patterns 15

6 Conclusions 31

References 32

Notations 33

Appendix A. Calculations for short beams 36

Appendix B. Experimental parameters for test series in [1]. 44

BYGNINGSSTATISKE MEDDELELSER Proceedings of the Danish Society for Structural Science and Engineering Edited and published by the Danish Society for Structural Science and Engineering Volume 79, No. 1-2, 2008, pp. 1-48

On the Conceptual Basis of the Crack Sliding Theory

Y. Zhao1

L. C. Hoang2

M. P. Nielsen3

1. Introduction

In the original formulation of the crack sliding theory developed by Zhang [5] and Hoang [9], very simple yield line patterns consisting of straight yield lines were used in order to simplify calculations. These calculations rendered very good re-sults for not too small shear spans. However, for short shear spans, the results were rather conservative.

In this paper, more accurate curved yield lines, in fact those observed in tests, are calculated using the same values of the effectiveness factors as those formed originally. By taking into account whether sliding failure takes place in an already existing crack or in uncracked concrete, remarkably accurate results are found in the whole relevant shear span interval. Both concentrated loading and uniform loading are treated.

2. Review of the classical plasticity approach

In the classical plasticity approach to beam shear problems, concrete is identified as a homogeneous and rigid plastic Coulomb material without tensile strength. The lack of perfectly plastic compressive behaviour is accounted for by introduc-

1 PhD student, Department of Bridge Engineering, Tongji University, Shanghai, China 2 Professor, Institute of Industrial and Civil Engineering, University of Southern Denmark 3 Professor Emeritus, Technical University of Denmark

2 Y. Zhao et al.: On the conceptual basis of the crack sliding theory

ing a so-called effectiveness factor ν into the theoretical solutions. Any possible strength reduction due to cracking is also covered by the effectiveness factor. Therefore, models based on the classical approach are neither capable of quantify-ing the strength reduction due to cracking nor account for the influence of crack-ing on the failure mechanism. An example on this drawback can be seen in the model developed for shear strength of beams without stirrups. For the standard case of a simply supported beam symmetrically loaded by two concentrated forces, an exact plastic solution for the maximum shear capacity is obtained by considering the stress field and the failure mechanism shown in figure 2.1, [2]. The solution appears in formula (1).

Figure 2.1 Failure mechanism and stress field rendering the exact plastic solution.

−

+==h

a

h

a

bhf

P

f c

u

c

u

2

12

1

νντ

(1)

In order to determine the effectiveness factor ν, solution (1) was compared with a large number of test results, [2]. It turned out that, unlike the case of shear rein-forced beams, good agreement with test results could only be obtained when the effectiveness factor ν had a dependency on the shear span to depth ratio a/h. This result was rather unsatisfactory for two reasons: 1) a physical explanation of the a/h-dependence was lacking. 2) The dependency on a/h is not practical for design purposes.

The a/h dependency indicated that the classical plasticity approach was unable of fully capturing the shear failure mechanism of beams without stirrups. Figure 2.2 schematically shows the process at the onset of a typical shear failure for beams with a/h larger than approximately 2. At a load level near the failure load, the crack pattern is as shown in figure 2.2 (left). Of particular interest prior to failure is the last formed diagonal crack, which typically initiates perpendicular to the bottom face and propagates towards (and end very near) the loading point. This crack will in the following be termed the critical diagonal crack and characterised by its horizontal projection x (or xo). For beams with a/h larger than 2 we have x < a. When the beam collapses, figure 2.2 (right), a sliding failure takes place in a

Y. Zhao et al.: On the conceptual basis of the crack sliding theory

3

curved yield line. This yield line runs along a large portion of the last formed di-agonal crack. Near the bottom face of the beam, the yield line deviates from the path of the last formed diagonal crack and runs to the bottom face in almost alignment with the longitudinal reinforcement. In contrary to the upper part, this lower part of the yield line does not run along an existing crack. Rather, this part is formed in uncracked concrete at the very onset of failure.

A mechanically transparent explanation of the disagreement between the actual failure mode and the one predicated by the classical plasticity approach was given by Zhang [5; 7], who studied the strength reduction due to diagonal cracking and subsequently developed a modified upper bound plasticity approach – the so called Crack Sliding Model. Description of this model is given in the next chapter. A model based on lower bound considerations and taking into account strength reduction due to cracking has been developed by Muttoni & Schwartz [4] and Muttoni & Fernández [10].

Figure 2.2 Typical crack pattern prior to failure (left). Failure mechanism involving sliding in part of the last formed diagonal crack (right).

3. The Crack Sliding Model

The basic principles of the Crack Sliding Model will be described in the follow-ing. For more details, see Zhang [5; 7], Nielsen [8] and Hoang [9].

As already mentioned a typical shear failure in concrete beams without stirrups is characterized by the formation of a critical diagonal crack and subsequently slid-ing failure in parts of this crack. The phenomenon of crack sliding has been ob-served by Muttoni [3] who measured the relative displacements along the critical diagonal crack at different loading levels, see figure 3.1. When the crack is formed, the relative displacement is mainly perpendicular to the crack. At the load level corresponding to failure there is a displacement component parallel to the crack. In other words, the critical diagonal crack is transformed into a yield line. Such a yield line has less sliding resistance than the yield line formed in un-cracked concrete.


Figure 3.1 Relative displacements along a diagonal crack at different load levels, Muttoni [3].

The distinction between yield lines formed in uncracked concrete and yield lines formed in cracked concrete may be illustrated by figure 3.2 which schematically shows two different ways of analyzing the shear failure of an overreinforced beam without stirrups. Case a) shows a straight yield line formed in uncracked concrete as assumed in the classical plastic approach. This is characterized by the fact that the relative displacement between the two parts on both sides of the yield line is described by only one displacement parameter u. In this case there is no displace-ment discontinuity prior to failure. Case b) illustrates a straight diagonal crack transformed into a yield line. Here the displacement is described by two paths, namely w followed by u, where w represents the opening (formation) of the crack prior to failure and u represents the shear failure along the crack, i.e. transforma-tion of the crack into a yield line. It is evident that the assumptions in case b) is in closer agreement with experimental observations, figure 3.1, than that of case a). Note that the mechanism of figure 3.2 b) is still a simple approximation. As indi-cated in figures 2.2 and 3.1, diagonal cracks are not straight but curved. In addi-tion, only a part of the diagonal crack is transformed into a sliding yield line. These details will be taken into account in chapter 4 and 5.

Figure 3.2 Yield line formed in uncracked concrete a). Crack transformed into yield line b).


5

To account for crack sliding, two Modified Coulomb failure criteria are used; see figure 3.3. For uncracked concrete, the failure criterion corresponds to the dotted lines with cohesion c and angle of friction ϕ. In planes where cracks are devel-oped, the failure criterion is assumed to shrink such that the tensile strength disap-pears and the cohesion reduces to c ́ while the angle of friction ϕ remains un-changed.

For uncracked concrete the angle of friction and the cohesion are taken to be ϕ = 37o and c = 0.25ν0fc. Here fc is the uniaxial compressive strength and ν0 is an ef-fectiveness factor taking into account softening and micro-cracking, see formula (3). For cracked concrete the cohesion is assumed to reduce to c’ = νsc where νs is called the crack sliding reduction factor, which may be taken to be νs = 0.5. This value has been confirmed by Zhang [6] using a micro mechanical model.

Figure 3.3 Failure criteria for uncracked concrete and for cracked concrete.

Having defined the failure criterion along planes of cracks, it is now possible to calculate the dissipated energy (internal work) in a crack suffering sliding failure. The dissipation per unit length of the crack reads; see for instance [8].

( )01

1 sin ; 2 s cW f buν ν α ϕ α π ϕ= − ≤ ≤ −

l (2)

Here α is the angle between the yield line and the displacement u and b is the width of the beam. The constraints on α are due to the fact that crack sliding must be treated as a plane strain problem, [5]. The effectiveness factor ν0 may be taken from (3) in which the compressive strength fc is inserted in MPa and the depth h in


meters. The parameter ρ = As/bh is the longitudinal reinforcement ratio. The factor λ depends on the loading configuration. For beams subjected to concentrated loading λ = 1.0 whereas λ = ¾ for beams subjected to uniform loading [5].

( )0 00.88 1

1 1 26 ; 1cf h

ν λ ρ ν = + + ≤

(3)

For a straight diagonal crack transformed into a yield line, the total dissipated en-ergy WI can be calculated by use of (2). Using the notations from figure 3.2(b) and assuming the displacement u to be vertically directed, we find:

2

0

11

2I s c

x xW f bh u

h hν ν

= + −

(4)

Inserting (4) into the work equation, the external work being uP u and νs = 0.5, we

find the following expression for the shear capacity as a function of x.

2

0

11

4u c

x xP f bh

h hν

= + −

(5)

The remaining problem is to determine the actual horizontal projection x of the yield line/critical diagonal crack. According to the Crack Sliding Model, this problem is solved by requiring that the load needed to develop the diagonal crack must be equal to the load needed to cause sliding failure in the crack.

Using a plasticity approach, a crack may develop when the effective tensile strength of concrete, ftef, is reached along the crack path, figure 3.4. Hence, the load required to form a crack may either be determined by considering a “crack-ing mechanism” as shown in figure 3.4 or simply by considering moment equilib-rium around the crack tip. A more refined model would require a fracture mechan-ics approach. After formation of the crack, the longitudinal reinforcement will be activated and prevent the cracking mechanism to further develop into a flexural failure. If the crack is critical, a shear failure will follow immediately. If not, the load may be further increased and another crack with smaller inclination (larger x) is to be considered. Using the notations from figure 3.4, the cracking load Pcr may be expressed as follows:

( )2 21

2cr tef

b x hP f

a

+= (6)

The effective tensile strength includes a size effect factor and may be taken to be, [5]:


7

0.302/30.156

0.1tef ch

f f−

=

( cf in MPa and h in meters) (7)

Figure 3.4 Equivalent plastic stress distribution along a developing crack.

By means of (5) and (6) we may now qualitatively illustrate the variation of the forces Pcr and Pu (corresponding to the cracking load and the crack sliding capac-ity, respectively) as a function of x/h, see figure 3.5.

Figure 3.5 Cracking load and crack sliding capacity versus x/h.

Using figure 3.5 the shear failure mechanism in a beam without stirrups may now be explained as follows: When the applied load P increases, cracks with increas-ing horizontal projection x are developed. However, as long as the load is less than PA, sliding failure in the cracks developed will not occur since sliding re-quires a higher load level. Only at the load level PA corresponding to the intersec-tion point between the two curves, sliding failure may take place in the crack just developed. Thus the shear capacity of the beam has been reached.


Therefore, by solving the equation Pu = Pcr, the horizontal projection x of the critical diagonal crack is determined and consequently the load carrying capacity may be calculated as well. If x is found to be larger than a, then x = a must be in-serted into (5) when calculating the shear capacity. Further, crack sliding is only possible for x/h ≥ tanϕ = 0.75. This is due to the constraint, imposed by the nor-mality condition of the theory of plasticity, on the angle between the relative dis-placement and the yield line when plane strain problems are treated.

Using the procedure outlined, Zhang has compared the calculated shear strength with a large number of test results, [5]. The agreement was found to be good. On the average, the model underestimates the shear strength by approximately 10%. The results obtained in chapter 4 and 5 by considering curved yield lines suggest that parts of the deviation found by Zhang are most probably due to the assump-tion of straight diagonal cracks.

4. Model with arbitrary curved yield lines

As mentioned in chapter 3 only straight diagonal cracks are considered in the Crack Sliding Model. This is an approximation which adds to the simplicity of the model and makes it suitable for practical use. However, as explained previously the shear failure mechanism is more complex. The critical crack is curved and the actual failure does not take place entirely as crack sliding, but rather as sliding in a yield line consisting partly of a portion of the last formed diagonal crack and partly of a portion formed at the onset of failure. The last portion may be inter-preted as a yield line formed in uncracked concrete.

In view of these facts, it appears interesting to investigate whether the concept of crack sliding may still be applicable if some of the complexity of the shear failure mode is included in the analysis of the ultimate load. More generally, it is also interesting to outline how the Crack Sliding Model can be extended to include arbitrary curved yield lines. For both purposes, there is a need of developing a dissipation formula for arbitrary curved yield lines.

4.1 Dissipation formula for curved yield lines

Consider a shear failure mechanism as shown in figure 4.1 where part I moves away from part II by the downwards displacement u. A curved yield line separates the two rigid parts. For the time being, the yield line may be formed in uncracked concrete or it may be the result of a diagonal crack suffering sliding failure. For both cases, similar dissipation formulas are obtained. The only difference will be the effective compressive strength which takes the form *

0sc cf fν ν= in case of

crack sliding and *0c cf fν= in case of sliding failure through uncracked concrete.


9

Figure 4.1 Shear failure in a curved yield line.

Assume that the curved yield line is described by the function y = f(x) with f’(x) > 0. The dissipation in a unit length ds of the yield line may be expressed as, see also formula (2):

*1(1 sin )

2I cdW f bu dsα= − (8)

In case we are dealing with crack sliding, the constraints on α as given in (2) also apply to (8). If the yield line is formed in uncracked concrete, the dissipation may be calculated by assuming plane stress condition and α can therefore take any possible value. From geometrical considerations, the following relations are ob-tained:

' 2

1sin

1 ( ( ))f xα =

+ (9)

' 21 ( ( ))ds f x dx= + (10)

By inserting (9) and (10) into (8) and integrating over the whole length of the yield line, we obtain the following simple expression for the dissipation in a curved yield line:

0* ' 2

0

11 ( ( ))

2

x

I c oW f bu f x dx x = + − ∫ (11)

As can be seen from formula (11), the first part in the bracket is the total length of the yield line while the second part in the bracket simply is the horizontal projec-tion xo of the curve. Formula (11) can therefore be rewritten as:


*1( )

2I c oW f bu l x= − (12)

Here l is the length of the yield line:

0 ' 2

01 ( ( ))

xl f x dx= +∫ (13)

According to (12) we find that for any fixed value of horizontal projection ox a

straight yield line always renders minimum dissipation. Thus, a straight yield line is, as expected, also the correct result when dealing with sliding failure through uncracked concrete. For crack sliding problems, however, the form of the crack can not be determined by minimizing (12). Calculation of the cracking path must be based on fracture mechanical crack growth analysis. This is, however, not the scope of this paper. In the following, we will carry out analyses based on prede-fined crack shapes and based on crack patterns observed in tests.

4.2 Crack Sliding Model for curved diagonal cracks

Having established (12) it is now possible to extend the crack sliding model to deal with arbitrary curved diagonal cracks.

By inserting (12) into the work equation, the external work being Puu, we find the following expression for the shear capacity as a function of xo:

*1( )

2u c oP f b l x= − (14)

The cracking load and the shape of the curved diagonal crack should, as men-tioned above, ideally be determined using fracture mechanical crack growth analyses. However, in a much more simplified approach the shape of the diagonal cracks may be assumed (predefined) and the cracking load can be calculated as outlined in chapter 3, i.e. by assuming the normal stresses along the crack path to be constant and equal to ftef. This approach implies, as indicated in figure 4.2, that the cracking load only depends on the horizontal projection of the crack but not on the shape of the crack. Thus, the cracking load may be calculated using formula (6).


11

Figure 4.2 Models rendering identical cracking load.

To illustrate the concept, we will in the following carry out calculations based on diagonal cracks having parabolic shape, i.e. 2( )y f x Ax Bx C= = + + (see figure 4.3). To determine the constants A, B and C, we have the following geometrical conditions.

[ ]o

o

xxforxf

hxf

f

;03/4)('0

)(

0)0(

∈≤≤=

= (15)

The first two conditions are obvious. The third condition ensures that 0 ( )f x h≤ ≤ and α ϕ≥ = 37o. The situation for which f´(0) = 4/3 and at the same time f´(xo) = 0 is found for xo = 3/2h. When xo < 3/2h, we always have f´(0) = 4/3 whereas f´(xo) = 0 is a sufficient replacement of the third condition in (15) when xo > 3/2h. For xo = 3/4h, A = C = 0 resulting in a straight diagonal crack.

Having determined f(x) as described, formulas (13) and (14) are then used to cal-culate the crack sliding capacity as a function of xo. The results appear as shown in (16) and (17) with *

0sc cf fν ν= . As mentioned, the cracking load only depends

on the horizontal projection and can therefore be calculated using (6).

Figure 4.3 Beam with parabolic diagonal crack.


2 2

0 0

0

*

2

0 0

0

1 2 4 1 2 4 51 1

2 3 3 334

1

2

2 4 2 41

3 31 1ln 1

4 343

u c o

h h

x xh

x

P f bx

h h

x x

h

x

+ − + + − − − =

− + + − + − −

;2

3

4

3 ≤≤h

xo (16)

2

0

*

0

2

0 0

1 11 1 4

2 2

1

21 1

ln 14

2 1 4

u c o

h

x

P f bx

x

h h h

x x

− − + = − − + +

; h

xo≤23

(17)

Formulas (6), (16) and (17) have been used to calculate the shear strength of two beams tested by Leonhardt and Walther [1], namely beam No. 6 and No. D4/1. Data for these test specimens can be found in appendix B. Based on the test data, the cracking load curve and the crack sliding capacity curve have been calculated, see figure 4.4. In the figure, the sliding capacity curve for straight diagonal cracks is also shown.

The intersection points provide the results shown in table 4.1.

Straight yield line Parabolic yield line Test results Beam No.

xo/h Pu [kN] xo/h Pu [kN] Ptest [kN] 6 2.3 60 2. 6 70 62

D4/1 2.03 73.5 2. 2 85.5 75.5

Tabel 4.1 Results based on different yield lines for beam No.6 and beam No. D4/1.


13

As expected, calculations based on parabolic cracks give higher shear capacity. For these two particular tests, calculations based on straight diagonal cracks can be seen to be in closer agreement with experiments. However, as mentioned in the preceding, approximately 10% underestimation of the capacity was found for a large number of tests with a/h > 2 when straight diagonal cracks are considered, [5]. It is therefore expected that on the average closer agreement with tests can be obtained when using curved diagonal cracks.

Figure 4.5 shows the calculated positions of the cracks. It can be seen that the assumed parabolic shape to some extent captures the actual cracking path better than the assumption of straight crack.

0

50

100

150

200

0 250 500 750 1000 1250 1500

x (mm)

P (kN)Pu(straight yield line)

Pu(parabolic yield line)

Pcr

a) Beam No. 6, [1].

0

50

100

150

200

0 250 500 750 1000 1250 1500

x (mm)

P(kN)

Pu(straight yield line)

Pu(parabolic yield line)

Pcr

b) Beam No.D4/1, [1].

Figure 4.4 Calculation of shear capacity assuming different shape of yield lines


a) Beam No. 6, [1].

b) Beam No. D4/1, [1].

Figure 4.5 Actual yield lines and calculated yield lines with predefined shape.


15

5. Calculation based on observed crack patterns

The formulas developed for curved yield lines will in the following be used to investigate the applicability of the crack sliding concept when observed shear failure modes are used in the analysis of the ultimate load.

Data used in the following stems from the well-known test series by Leonhardt and Walther [1]. As explained in the previous chapters, we consider the shear failure as the result of sliding in a yield line partly formed along a crack and partly formed in uncracked concrete. This is indicated in figure 5.1, where the part formed in uncracked concrete is shown with the length l1 and horizontal projec-tion xo1. The part of the diagonal crack suffering sliding failure is shown with the length l2 and horizontal projection xo2. The point where the yield line deviates from the path of the diagonal crack marks the transition between crack sliding failure and failure through uncracked concrete. This transition point is usually easy to localize from photographs of test specimens. In cases where the transition point does not appear clearly, the point at which the diagonal crack forms the an-gle α = ϕ with the vertical displacement vector u will be used as the transition point. The reason for this is, that α = ϕ marks the point where crack sliding ac-cording to (2) is possible.

Figure 5.1 Failure mode with combination of crack sliding and yield line through uncracked concrete.

To reflect the failure mode described above and shown in figure 5.1, formula (14) must be rewritten as follows:

[ ]0 1 1 2 2

1( ) ( )

2u c o s oP f b l x l xν ν= − + − (18)

For beams subjected to uniform loading, a formula similar to (18) can be obtained by use of the work equation. In this case, the internal work is taken as the right


hand side of (18) and the external work is q(a-ao). Using the notation from figure 5.2, the result reads:

[ ]0 1 1 2 2

1( ) ( )

2u c o s oo

aP qa f b l x l x

a aν ν= = − + −

− (19)

Figure 5.2 Beam subjected to uniformed loading

The tests by Leonhardt and Walther [1] include a number of beams without stir-rups. These tests are in the following grouped in three series, 1, 2 and 3. The main test data and calculation data are summarized in appendix B.

The first series consists of 10 beams subjected to four point bending. The shear span to depth ratio a/h was the main varying parameter. In this series, 8 beams failed in shear. Figure 5.3 shows photographs of the test specimens. Based on the photographs, the geometries of the yield lines have been extracted and shown in a drawing of each beam4. From the drawings, l1, l2, xo1 and xo2 are measured (see appendix B) and inserted into formula (18) in order to calculate the shear capacity. Figure 5.4 shows the comparison of calculated values and test results. The agree-ment is remarkably good. It should be mentioned that for beams No. 1 and No. 2, the calculations have not been based on formula (18). The reason is that this for-mula and the associated failure mechanism only applies to beams with a/h larger than approximately 2. For beam No. 1, the geometry of the beam and of the load-ing plates make it impossible for crack sliding to take place (α < ϕ). In this case, the calculation must be based on yield line through uncracked concrete. The fail-

4 Note that the loading and support plates shown in the drawings have the correct dimensions. According to [1] the width of the loading plates was 130 mm.


17

ure mechanism of beam No. 2 is completely different from that considered in fig-ure 5.1.

If for beam No. 2 a failure mechanism consisting of a straight yield line in un-cracked concrete from support to load is calculated, almost exact agreement is found. However, this failure mechanism has not developed, as seen in figure 5.3. If the observed mechanism is calculated the lower bound solution deviates rather much from the upper bound solution. However, the average gives a very accurate value. It may happen that the observed mechanism is the result of some compli-cated fracture dynamics at the onset of a sudden failure, a phenomenon which can not yet be treated theoretically in an accurate way. Detailed calculations for beams No. 1 and 2 are given in appendix A.

The second series consists of 8 beams with varying length and subjected to uni-form loading. Seven beams failed in shear. Two of the seven beams have been omitted because of difficulties in determining the yield lines (photographs show large areas with damages making it difficult to identify the yield lines). Yield lines observed and used in calculations are shown in figure 5.5. Comparison of calcula-tions with tests appears in figure 5.6. The agreement is seen to be good, especially for a/h larger than 4.

The third series from [1] consists of beams, all having shear span to effective depth ratio equals to 3. The depth varies from 80 mm to 670 mm. Photographs of 8 specimens are provided in [1]. These and drawings of yield lines used in calcu-lations are shown in figure 5.7. Comparison of calculations with tests can be seen in figure 5.8. The agreement is good.

Summary of the calculations for all three series are shown in figure 5.9. The agreement with tests is seen to be very good.

The results obtained strongly indicate that the crack sliding concept is capable of capturing the essential phenomena in the shear failure mechanism of reinforced concrete beams without stirrups.


Figure 5.3 Series 1, test specimens and drawings with yield lines. Tests by Leonhardt & Walther [1].


19

Figure 5.3 (continued).


21

Figure 5.3 (continued)


0

100

200

300

400

500

0 1 2 3 4 5 6

a/h

Pu(kN)

cal test

0

100

200

300

400

1 2 3 4 5 6 7-1 8-1

Specimen No.

Pu (kN)

cal test

Figure 5.4 Results for beams subjected to concentrated load, series 1.


23

Figure 5.5 Series 2, test specimens and drawings with yield lines. Tests by Leonhardt & Walther [1]


25


0

100

200

300

400

11-1 12-1 13-1 14-1 15-1Specimen No.

Pu(kN)cal test

Figure 5.6 Results for beams subjected to uniform loading, serie 2.


Figure 5.7 Series 3, test specimens and drawings with yield lines. Tests by Leonhardt & Walther [1].


27



29



0

20

40

60

80

100

D1-1 D2-2 D3-2 D4-1

Pu(kN)

cal test

0

60

120

180

C1 C2 C3 C4

Pu(kN)

cal test

Figure 5.8 Results for beams subjected to concentrated load, series 3.

0

50

100

150

200

250

300

350

400

450

0 50 100 150 200 250 300 350 400 450

Vcal.(kN)

Vtest (kN)

series 1

series 2

series 3

Figure 5.9 Comparison of tests and calculations for all test series.


31

6. Conclusions

In the classical plasticity approach, failure mechanisms are determined solely by energy minimization. The Crack Sliding Model on the other hand determines the critical yield line by evaluating both the sliding strength and the cracking load of the considered crack pattern. Since cracking is a fracture mechanical phenome-non, it is not possible within the framework of plastic theory accurately to calcu-late the form of the crack suffering sliding failure. In the original formulation of the Crack Sliding Model, diagonal cracks are predefined to be straight.

The objective of this paper has been to investigate the applicability of the crack sliding model when arbitrary curved diagonal cracks are taken into account. For this purpose, a general dissipation formula for arbitrary yield lines has been de-veloped.

By means of this formula, a simple extension of the Crack Sliding Model involv-ing cracks of parabolic shape has been carried out. Results of two simple calcula-tions suggest that good agreement with tests can be obtained.

Further, it has been shown that the developed formula in a very simple way may be used to take into account the actual shear failure mechanism when analyzing the ultimate strength. Remarkably good agreement with test results has been found.

The results obtained first of all demonstrate that there is a sound physical basis behind the simple straight yield line calculations used in the crack sliding theory.

The results obtained further may be used as a basis for future research as well as immediate implementation in practice. In future research, it is obvious that the concept of crack sliding in arbitrary curved cracks should be combined with a fracture mechanical approach to evaluate the crack formation. With regards to practical implementation, the results found in chapter 5 evidently may be used to evaluate the vulnerability towards sliding failure in observed cracks in existing concrete structures. This is a task often met when dealing with strength assess-ment of existing concrete bridges.


References

[1] Leonhardt, F. & Walther, R.: Schubversuche an einfelddrigen

Stahlbetonbalken mit und ohne Schubbewehrung. Deutscher Ausschuss für Stahlbeton, Heft 151, 1962.

[2] Nielsen, M. P., Braestrup, M. W., Jensen, B. C. & Bach, F. : Concrete

plasticity---Beam shear-Shear in joints-Punching shear. Special Publica-tion. Danish Society for Structural Science and Engineering, Structural Research Laboratory, Technical University of Denmark, Lyngby, 1978.

[3] Muttoni, A. : Die Anwendbarkeit der Plastizitätstheorie in der

Bemessung von Stahlbeton. Institut für Baustatik und Konstruktion, ETH Zurich, Bericht Nr.176, Juni 1990.

[4] Muttoni, A. & Schwartz, J.: Behaviour of Beams and Punching in Slabs

without Shear Reinforcement. IABSE Colloquium, Vol. 62, pp. 703-708, Stuttgart, Germany, 1991.

[5] Zhang, J. P.: Strength of cracked concrete. Part 1---Shear strength of

conventional concrete beams, deep beams, corbels, and prestressed rein-forced concrete beams without shear reinforcement. Technical Univer-sity of Denmark, Department of Structural Engineering, Report R 311, Lyngby, 1994.

[6] Zhang, J. P.: Strength of cracked concrete. Part 2---Micromechanical

modeling of shear failure in cement paste and in concrete. Technical University of Denmark, Department of Structural Engineering, Report R 17, Lyngby, 1997.

[7] Zhang, J. P.: Diagonal cracking and shear strength of reinforced con-

crete beams. Magazine of Concrete Research, Vol. 49, No.178, 1997, pp. 55-65.

[8] Nielsen, M. P. : Limit analysis and concrete plasticity. Second edition,

CRC Press, 1998. [9] Hoang, L. C.: Extension and application of the crack sliding model. Pro-

ceedings of the Danish Society for Structural Science and Engineering, Vol. 72, No.4, 2001, pp. 71-118.


33

[10] Muttoni, A. & Fernández, R. M.: Shear strength of members without transverse reinforcement as function of critical shear crack width, ACI Structural Journal, Vol. 105, No. 2, pp. 163-172, 2008.

Notations

a Shear span/half of span length ao

Distance between the end of the yield line and the support in uniformly loaded beams

As Cross section area of longitudinal reinforcement

b Web width of the beam

c Cohesion in uncracked concrete

c' Cohesion in cracked concrete

d Effective depth of cross section

fc Uniaxial compressive strength of concrete

fc* Effevtive compressive strength of concrete, defined as fc

*=νfc

ftef Effetive plastic tensile strength of concrete

fy Yield strength of longitudinal reinforcement

h Depth of beam

l Total curve length of yield line/crack

l1 Curve length of yield line in uncracked concrete

l2 Curve length of yield line/critical diagonal crack

Lo Length of support/loading plate

P External load

Pcr Cracking load


Pu Ultimate load/load carrying capacity

Pucal Calculated ultimate shear force

Putest Experimental ultimate shear force

q Load per unit length

u Relative displacement in yield line

V Reaction at support

w Displacement in diagonal crack at onset of cracking

WE External work at failure

WI Internal work at failure

x Horizontal projection of yield line

xo Horizontal projection of yield line

xo1 Horizontal projection of yield line formed in uncracked concrete

xo2 Horizontal projection of yield line/critical diagonal crack

α Angle between yield line and displacement direction

β Inclination of yield line

θ Rotation angle

λ Loading configuration factor

ν Effectiveness factor for compressive strength of concrete

ν0 Effectiveness factor due to softening effects and microcracking

νs Sliding reduction factor

ρ Reinforcement ratio(=As/bh)


35

σ Normal stress

σc Concrete stress

τ Shear stress

ϕ Angle of friction for concrete

ϕ´ Angle of friction in a crack


Appendix A. Calculations for short beams

For the short beams No. 1 and No. 2 in the Leonhardt-Walther test series [1], the failure mechanism assumed in figure 5.1 of the paper does not apply. This appen-dix contains calculations specifically for these two beams.

A.1. Specimen No. 1

For beam No. 1 the shear span to depth ratio is: 0.27/0.32 = 0.84. When the widths of the loading plate (130 mm) and the support plate (100 mm) are taken into account and when the plates are assumed to be rigid, the largest possible horizontal projection of the yield line becomes 155 mm. This means that the angle α = Arctan (155/270) = 30o. This is smaller than 37o meaning that the assumption of crack sliding can not be used. Instead, the calculation must be based on the classical plasticity approach. For this purpose, formula (18) can still be used if l2 and xo2 are put to zero. From figure A-1 the following length are measured: l1 = 365 mm and xo1 = 162 mm. Note that xo1 is a bit larger than the theoretical values (155mm). The reason is that the loading and support plates are not perfectly rigid. The effectiveness factor for this specimen is ν0 = 0.65 (see also appendix B). In-serting into (18), we find:

[ ]

[ ]

0 1 1 2 2

1( ) ( )

2

10.65 29.47 190 (365 162

2

369

u c o s oP f b l x l x

MPa mm mm mm

kN

ν ν= − + −

= ⋅ ⋅ −

=

(A.1)

Figure A-1 Failure mode of beam No. 1, photo and drawing of yield line.


37

A.2 Specimen No. 2

Upper bound Analysis:

From the photograph of the failure mechanism of beam No. 2, it appears that it is completely different from that considered in figure 5.1. For the crack patterns shown on the drawing in figure 5.3, a failure mechanism involving a simple verti-cal displacement is not possible. The beam must therefore be analyzed using an-other failure mechanism.

What we notice from the photograph of the test specimen is that the compression zone between the applied loads is crushed. A mechanism allowing for crushing of this zone and shear failure in the diagonal cracks can be seen in figure A-2.

This mechanism consists of a rotation θ of part I about the point of intersection between the diagonal crack and the longitudinal reinforcement. The rotation causes the compression zone between the loads to crush. Dissipation in the crushed zone may be calculated by considering a vertical yield line undergoing a linear displacement field with constant α = -90o. Since the reinforcement is not yielding (due to the position of the point of rotation) the mechanism also involves a downward “punching” of part II. The resulting displacement field in the diago-nal crack and in the horizontal yield line is composed of the vertical displacement of part II and the rotation of part I. This means that the displacement vector varies along the yield lines. This is illustrated in figure A-3.

The diagonal crack (or parts of it) is assumed to have been developed prior to failure whereas the rest of the yield lines are assumed to develop at the onset of failure. This means that for the horizontal yield line, failure through uncracked concrete is assumed. Dissipations for the three different yield lines are derived in the following.

Figure A-2 Assumed failure mechanism for beam No. 2 based on photograph of specimen.


Figure A-3 Relative displacements along yield lines.

A.2.1 Dissipation in diagonal crack

Figure A-4 Sliding failure along the diagonal crack, (left) the part above rotation point, (right) the part below rotation point.

As shown in figure A-4 (left), there are two components of displacement for any point along the diagonal crack; namely v1 stemming from the rotation of part I and u1 from the vertical displacement of part II. The relative displacement u between the two parts along the crack is the vector sum of v1 and u1. For a point at the distance x from the point of rotation, we have:

2 2 2 21 1 1 1 1 2 1 22 cos 2 ( ) cos ( )u u v u v x y y x y yβ β θ= + − ⋅ ⋅ ⋅ = − ⋅ + ⋅ ⋅ + + ⋅ (A.2)


39

From geometrical conditions, we further find:

1 1sin cosu u vα β= − ; 1 1 2( )u y y θ= + ; 1v xθ= (A.3)

The dissipation for a length dx of the yield line is:

*1(1 sin )

2 cdW f bu dxα= − (A.4)

Inserting (A.2) and (A.3) into (A.4) and integrating along the yield line from the point of rotation to the end point A, we find:

*1 0

1(1 sin )

2

R

cW b f u dxα= −∫

[ ] [ ]

[ ]

2 2*1 2 1 20

*1 20

( ) cos ( )sin1

2 ( )cos

R

c

R

c

f x y y y y dxb

f y y x dx

β βθ

β

− + + + = − + −

∫

∫ (A.5)

Note that in (A.5) the effective compressive strength may vary along the yield line depending on whether α is smaller or larger than ϕ. Using figures A-2 and A-4 (left) and values in table A-1, it can be shown that 37oα ϕ= = at a distance x1

from the point of rotation. This distance is:1 1 2 2 1

3( )( ) /

4x y y y z R= + − . Thus, for

10 x x≤ ≤ we have 37oα ≥ while 37oα < for 1x x R< ≤ . Therefore, in (A.5) *

0c s cf fν ν= must be inserted for 10 x x≤ ≤ and *0c cf fν= applies when1x x R< ≤ .

For the part of the yield line below the rotation point, see figure A-4 (right), we have:

2 21 2 1 22( ) cos ( )u x y y x y yβ θ= + + + +

1 2sin cosu u vα β⋅ = + (A.6)

2v x θ= ⋅

Similar to (A.5) we find the dissipation for the part of the crack below the point of rotation to be:


[ ] [ ]2 2

1 2 1 20*

22

1 2

( )cos ( )sin1

2 ( )cos2

r

c

x y y y y dx

W b f ry y r

β βθ

β

+ + + + = − + +

∫ (A.7)

In (A.7) *0c s cf fν ν= may be used as 37oα ≥ for 0 x r≤ ≤ .

A.2.2 Sliding failure in horizontal part of yield line

Figure A-5 Displacement in horizontal yield line.

For the horizontal part of the yield line AB in figure A-5, we find the following geometrical conditions:

2 2 21 2 1 2 12 ( ) ( )u x y y x y y z θ= − ⋅ + ⋅ + + + ⋅ ; 1 3sin cosu u vα β⋅ = − ⋅ ;

3 cos

xv θ

β= ⋅ (A.8)

The dissipation along the horizontal yield line is:

1 2

2

*3

1(1 sin )

2

y y

c yW f b udxα

+= −∫

[ ]{ }1 2 1 2

2 2

2* 21 2 1 1 2

1( ) ( )

2

y y y y

c y yf b x y y z dx y y x dxθ

+ += − + + − + −∫ ∫ (A.9)

As this part of the yield line is assumed to develop at the onset of failure, *

0c cf fν= may be used.


41

A.2.3 Concrete crushing

Figure A-6 Displacement in vertical yield line.

For the vertical yield line representing concrete crushing, see figure A-6, we have

constant α = -90 degrees. The dissipation is:

1 3

14

11 sin( 90 )

2

z z ob c z

W v f b x dxθ+ = − − ∫

2 2 23 1 1 3 3 1( ) 2

2 2b c b c

z z z z z zf b f bν θ ν θ

+ − += =

(A.10)

For concrete crushing in zones with constant moment, the effectiveness factor may be taken as the one valid for bending, [8]:

0.975000 300

y cb

f fν = − − (A.11)

With fc = 29.5 MPa and fy = 355 MPa we find νb = 0.80.

The total dissipation (for half of the beam) is the sum of (A.5), (A.7), (A.9) and (A.10):


[ ] [ ] [ ]( )[ ] [ ] [ ]( )

[ ] [ ]

1 1

1 1

1 2 3 4

2 2

0 1 2 1 2 1 20 0

2 2

0 1 2 1 2 1 2

22 2

0 1 2 1 2 1 20

( )cos ( )sin ( )cos

( )cos ( )sin ( )cos1

2 ( )cos ( )sin ( ) cos2

I

x x

s

R R

x x

c r

s

W W W W W

x y y y y dx y y x dx

x y y y y dx y y x dxf b

rx y y y y dx y y r

ν ν β β β

ν β β βθ

ν ν β β β

= + + + =

⋅ ⋅ − + + + − + −

+ − + + + − + −

+ ⋅ ⋅ + + + + − + ⋅ +

∫ ∫

∫ ∫

∫

[ ]( )1 2 1 2

2 2

22 2 3 3 1

0 1 2 1 1 2

2( ) ( )

2

y y y y

by y

z z zx y y z dx y y x dxν ν

+ +

++ − + + − + − + ∫ ∫

(A.12)

The external work for half of the beam is:

1 2( )EW P a a θ= ⋅ + ⋅ (A.13)

By setting up the work equation E IW W= , an upper bound is found for the shear

capacity of the beam. The geometrical parameters to be used in (A.12) and (A.13) are measured from photograph and drawing of the test specimen. These are listed in table A-1. In the table, strength parameters are also shown (see also appendix B).

By inserting the parameters into (A.12) and (A.13) and by performing numerical integration, we obtain the following upper bound for the shear capacity:

369uP KN= (A.14)

a1 a2 y2 y3 R r y1 z3 z1 z2

cm cm cm cm cm cm cm cm cm cm

25.32 14.87 26.81 7.64 32.89 9.37 16.21 7.49 19.05 5.43

x1 b h fc ν0 νs νb cos sin cm cm cm MPa - - - - -

16.38 19 32 29.47 0.649 0.5 0.8 0.82 0.58

Table A.1 Parameters for beam No. 2, [1]. .


43

Lower bound analysis:

Beam No. 2 has also been analyzed using a lower bound approach. Figure A-7 shows an admissible stress field with diagonal struts avoiding the observed crack pattern. The geometrical parameters for this stress field are shown in the figure, where the width of the bearing stress block is Lo = 41 mm. The triangular parts are in plane hydrostatic stress conditions.

Since the diagonal struts are not crossing any cracks, a compressive stress corre-sponding to σc = ν0fc may be transferred throughout the system. A lower bound for Pu can therefore be calculated as below:

0

0.65 29.5 41 190

149

u c oP f L b

MPa mm mm

kN

ν=

= ⋅ ⋅ ⋅

=

(A.15)

Figure A-7. Stress field rendering lower bound solution

Averaging the results:

From the upper and lower bound analyses, an estimate of the shear capacity of the beam can be obtained by taking the average value of the two results. Thus:

( )1149 369 259

2uP kN kN= + = (A.16)

Notice that the capacity obtained in test is 265 kN.


Appendix B. Experimental parameters for test series in [1].


45


Resumé

I forbindelse med anvendelse af plasticitetsteoriens øvreværdiprincip til beregning af forskydningsstyrken af betonbjælker er der behov for at bestemme den plasti-ske dissipation i brudlinierne. De hidtil udviklede beregningsformler er hovedsa-geligt baseret på antagelsen om rette brudlinjer. Det fremgår dog af utallige for-søg, at bruddet sker i krumme forskydningsrevner. Denne artikel præsenterer en procedure til beregning af dissipationen i vilkårlige krumme brudlinier. Resultatet viser sig at være en simpel formel velegnet til praktisk anvendelse. I artiklen an-vendes den udviklede formel til forskydningsberegning iht. revneglidingsteorien. Først foretages beregninger, hvor det antages at den krumme forskydningsrevne har parabolske form. Dernæst foretages beregninger baseret på brudlinieforløb observeret ved forsøg. Der opnås gode overensstemmelser mellem beregnede og eksperimentelle styrker.

Summary

When the upper bound plasticity approach is used to calculate the shear strength of concrete beams, there is a need of determining the energy dissipated in the yield lines. Plastic design formulas developed until now are mostly based on the assumption of straight yield lines. However, many experimental results have shown that shear cracks in beams are curved. This paper presents a procedure to calculate the dissipation in arbitrary curved yield lines. The result turns out to be a simple formula suitable for practical use. The derived formula is used to carry out shear strength analyses within the framework of the plasticity based Crack Sliding Model. First, an analysis based on the assumption of shear failure in cracks with predefined parabolic shape is carried out. There next, the formula is used to carry out calculations based on crack patterns observed in tests. Very good agreements with test results are obtained.


47

Artikel modtaget december 2008

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