Failure criterion for reinforced concrete beams and plates subjected to membrane force, bending and shear

European Journal of Mechanics A/Solids 27 (2008) 1161–1183

Failure criterion for reinforced concrete beams and plates subjected to membrane force, bending and shear Pierre Koechlin a,∗ , Stéphane Andrieux a , Alain Millard b , Sergueï Potapov c a LaMSID UMR CNRS-EDF 2832, 1, av. du Général de Gaulle BP 408, 92141 Clamart cedex, France b CEA-LM2S, CEA Saclay, 91191 Gif-Sur-Yvette cedex, France c Dept. Analyses Mécaniques et Acoustique EDF R&D, 1, av. du Général de Gaulle BP 408, 92141 Clamart cedex, France

Received 3 April 2007; accepted 19 December 2007 Available online 4 March 2008

Abstract In this paper, a failure criterion for reinforced concrete plates is derived through the kinematic method in the framework of the limit analysis theory. This criterion is expressed in terms of the stress resultant variables: membrane force, shear force and bending moment at once. The aim of the authors is to be able to predict the failure of reinforced concrete plate structures in statics or in slow dynamics using directly the internal forces (membrane and shear forces and moment) resulting from a finite-element computation. In a first step, a beam criterion is derived. The closed form expression of the criterion shows that it is made up of two parts, one independent of the moment (i.e. depending only on the normal force and the shear force) and one depending on the normal force, the shear force and the bending moment. This structure of the criterion allows to determine two failure modes: shear failure and bending failure. Then in a second step, the beam criterion is extended to the case of reinforced concrete plates. The obtained criterion is partly numerical and partly a close form expression. It gives an upper bound of the load, and when this limit load is reached, the criterion is able to supply, on one hand, the failure mode (as seen in the beam case) and, on the other hand, the angles of the failure plane in the reinforced concrete plate section. Thirdly, the criterion is implemented in the finite element software Europlexus and validated with respect to punching experimental tests. We show that the criterion must be used with an effectiveness factor applied on the concrete compressive strength. © 2008 Elsevier Masson SAS. All rights reserved. Keywords: Reinforced concrete; Concrete slabs; Bending; Shear; Yield surface; Failure criterion

1. Introduction The determination of limit loads is a basic task of structural design. Before the development of numerical computation and the finite element method, the limit analysis theory supplied relatively simple tools for yield design to * Corresponding author.

E-mail addresses: [email protected] (P. Koechlin), [email protected] (S. Andrieux), [email protected] (A. Millard), [email protected] (S. Potapov). 0997-7538/$ – see front matter © 2008 Elsevier Masson SAS. All rights reserved. doi:10.1016/j.euromechsol.2007.12.009

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determine the ultimate loads a structure is able to withstand. Indeed the limit analysis is a simple, but very efficient tool, because it starts from assumptions on local criteria and gives a global result on the failure of the whole structure, without the need to use any discretization or to know the evolution of the material state. Since concrete is a complex material and its behavior is difficult to model in 3D, limit analysis has been widely employed to study the load bearing capacity of reinforced concrete structures. And it is still used in the civil engineering field: – to build basic design rules. Its influence can be found for example in the Eurocode (Eurocode2, 1991); – in connection with a finite element discretization: the limit analysis theory is associated with linear and non-linear programming to compute limit loads; this approach is called finite element limit analysis, numerical limit analysis or automated yield line analysis (Anderheggen and Knöpfel, 1972; Liu et al., 1995); – to build global constitutive models for beams and plates; for example, the elastic domain can be deduced from the yield criterion obtained through limit analysis (Koechlin and Potapov, 2007). Thus it can be still useful to get analytical solutions of limit analysis problems and to derive analytical yield conditions, in terms of stress resultant variables (bending moment, membrane forces and shear forces). Johansen (1962) started with a criterion for concrete slabs under bending, with isotropic reinforcement. Nielsen (1999) completed the research with the derivation of the criterion for a different reinforcement along x and y. A criterion for plates under membrane and bending load has been proposed in Cookson (1979), Koechlin and Potapov (2007). Regarding the transverse shear, up to now, it has been considered separately (see, Nielsen, 1999, for maximum shear capacity, or punching shear strength of beams and plates). But this paper deals with a combination of bending moments, membrane forces and transverse shear forces. An analytical criterion is derived first for a reinforced concrete beam, and then we extend it to reinforced concrete plates. Since shear is considered, it is necessary to use a local 3D criterion that accounts for the frictional behavior of concrete, which means a pressure dependent criterion. The simplest failure criteria of this kind are the Coulomb criterion and the Drucker–Prager criterion; only the latter is considered in this paper, but our global criterion could be similarly derived using the Coulomb criterion. It is a common practice to associate the theory of limit analysis with perfect plasticity. However in this paper, we follow the more general theoretical framework of Salençon (2001) for limit analysis. In this way it is not necessary to assume any rigid-plastic behavior for the concrete or the steel. The criterion is derived through the kinematical method (approach from outside): it gives an upper bound of the loading. Firstly, the main assumptions are detailed: failure mechanism, local strength of the materials. Secondly, the principle of the kinematic method is used, leading to a maximization problem for the plate. To solve it, a two steps procedure is used: a beam criterion is first obtained, and then is extended to plates. Finally, this paper shows the validation of the plate criterion by comparison with experimental punching tests. 2. Notations and assumptions The kinematical method in limit analysis is base, on one side, on a geometry and a kinematical mechanism and, on the other side, on local stress criteria of the concrete and steel materials. 2.1. Geometry and failure mode Let us consider a reinforced concrete plate P with the following characteristics: – – – – –

h: thickness of the plate Ωx sup and Ωx inf : steel area of the bending reinforcement layers in the x direction (m2 /m) Ωy sup and Ωy inf : steel area of the bending reinforcement layers in the y direction (m2 /m) ΩT : shear reinforcement density (m2 /m2 ) zsup = ρsup h/2 and zinf = ρinf h/2: location of the two bending reinforcement layers (−1  ρinf < ρsup  1)

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Fig. 1. Failure mechanism in the plate.

The plate is supposed to yield in a plane Σ inclined with an α angle with respect to the x-axis, and with a θ angle with respect to the xy-plane (see Fig. 1). It means, there is a jump in the velocity field on Σ . We are looking for a failure criterion depending on the force tensors N, T , M, denoting respectively the membrane forces, the transverse shears and the bending moments. In the x, y, z axes their components are Nx , Ny , Nxy , Tx , Ty , Mx , My , Mxy , and in the X, Y, Z axes, which are associated with the α angle vertical crosssection, they are denoted NX , NY , NXY , TX , TY , MX , MY , MXY . Definition of the membrane forces: h/2 Nx =

h/2 σxx dz,

−h/2

Ny =

h/2 σyy dz,

Nxy =

−h/2

σxy dz

(1)

−h/2

Definition of the bending moments: h/2 Mx =

h/2 zσxx dz,

My =

−h/2

h/2 zσyy dz,

Mxy =

−h/2

zσxy dz

(2)

−h/2

Definition of the transverse forces: h/2 Tx = −h/2

h/2 σxz dz,

Ty =

σyz dz

(3)

−h/2

The components in the different axes are linked by: ⎧ 2 2 ⎪ ⎨ NX = Nx cos α + Ny sin α + 2Nxy cos α sin α TX = Tx cos α + Ty sin α ⎪ ⎩ MX = Mx cos2 α + My sin2 α + 2Mxy cos α sin α

(4)

The force tensors N, T , M will be denoted NO , TO , MO when their values correspond to the vertical section at point O and NQ , TQ , MQ when they are taken at point Q (see their components NQX , TQX , MQX in Fig. 1). If ΩX inf and ΩX sup are defined as the effective steel area of the lower and upper reinforcement layers in the X direction, they can be expressed from Ωx inf , Ωy inf , Ωx sup , Ωy sup (Save et al., 1997) as:

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ΩX sup = Ωx sup cos2 α + Ωy sup sin2 α

(5)

ΩX inf = Ωx inf cos2 α + Ωy inf sin2 α 2.2. Local strength criterion

The yield strength σY for steel is identical in tension and compression. There is no shear failure of the steel reinforcement. Thus the local one-dimensional criterion for steel can be expressed as: fS (σu ) = |σu | − σY  0

(6)

For the concrete, the Drucker–Prager criterion is used. Usually it is expressed as: fDP (σ ) = σeq + a tr σ − σ0  0

(7)

with: a > 0 and σ0 > 0 

1 1 2 tr(s ) σeq = s is the deviatoric stress: s = σ − tr σ I 2 3 Following Salençon (2001), we rewrite (7) using parameters similar to those of Coulomb (where C is the cohesion and φ the friction angle): fDP (σ ) = σeq + 

sin φ 3(3 + sin φ) 2

tr σ − 

3 cos φ

C

(8)

3(3 + sin φ) 2

3. The kinematical method With a virtual velocity field admissible for the failure mechanism described in Section 2.1, the principle of virtual work for the plate P of boundary ∂P can be expressed as:    (σ · n) · u˙ dS = σ : ε˙ dP (9) ∂P

P

The left part of this equation can be expressed as a function of the load tensor NO , TO , MO and of the velocity jump components. If we find an upper bound of the right part of the equation, depending only on the virtual velocity jump, it will possible to derive an upper-bound of the load. That is the principle of the so-called kinematical method (or upper-bound theorem) of the limit analysis. First we are going to detail the virtual velocity used, and then to find an upper-bound of σ : ε˙ . 3.1. Virtual velocity and virtual strain rate field Let us consider a virtual velocity field, which is zero in the left part of the solid and corresponds to a rigid body motion in the right part. Consequently the velocity jump at a point A of Σ can be described with the following equation: −→ ˙ O ∧ − ˙ ˙ O + ω OA u(A) =u

(10)

In the X, Y, Z axes, associated with an α angle oriented and vertical cross-section (Fig. 1), and assuming that there ˙ has is no sliding (slipping) along Y axis (because we do not want a criterion for in-plane shear), the velocity jump u two translation components u˙ OX and u˙ OZ along the X and Z axes and a rotation component ω˙ OY around the Y axis. Then we can write the velocity jump as: −−→ ˙ u(A) = u˙ OX eX + u˙ OZ eZ + ω˙ OY eY ∧ OA

or as follow:

(11)

P. Koechlin et al. / European Journal of Mechanics A/Solids 27 (2008) 1161–1183

˙ n + u˙ t (A)t u(A) = u˙ n (A)

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(12)

or also as: ˙ u(A) = u˙ X (A)eX + u˙ Z (A)eZ

(13)

All these components are linked through: u˙ n = u˙ OX sin θ + u˙ OZ cos θ + s ω˙ OY u˙ t = −u˙ OX cos θ + u˙ OZ sin θ and by:

u˙ X = u˙ OX + s ω˙ OY sin θ

(14)

(15)

u˙ Z = u˙ OZ + s ω˙ OY cos θ with a condition on the abscissa s along t on Σ : h h s 2 sin θ 2 sin θ According to Eq. (12), the strain rate field in the plane Σ has the following expression: −

1  ˙ = u˙ n n ⊗ n + u˙ t [ ˙ ⊗ n + n ⊗ u) n ⊗ t + t ⊗ n] ε˙ = (u 2 2 But it should be noted that ε˙ = 0 anywhere else in the plate.

(16)

(17)

3.2. Upper-bound of σ : ε˙ Let us define the support function Π of the f criterion as:  

Π(˙ε ) = sup σ : ε˙ f (σ )  0

(18)

σ

In Salençon (2001) it is shown that for a virtual strain rate field described by Eq. (17), and for an appropriate choice of parameters (as written in Eq. (8)), the support function ΠC for the concrete ruled by the Drucker–Prager can be written as follows: C u˙ n if u˙ n  |u| ˙ sin φ (19) ΠC (˙ε ) = tan φ ΠC (˙ε ) = +∞ if u˙ n < |u| ˙ sin φ20 (20) For the steel, using the criterion (6), the support function ΠS can be written for the longitudinal reinforcement as: ΠS long (˙ε ) = σY |u˙ X |

(21)

and for the transversal reinforcement as: ΠS trans (˙ε ) = σY |u˙ Z |

(22)

3.3. Upper-bound of the work of the external forces It is now possible to write the principle of virtual work (9) for the strain rate field (17) admissible for the kinematics:  NQX u˙ QX + TQX u˙ QZ + MQX ω˙ QY = σ : ε˙ dP (23) P

If we try to express the work of the external forces as a function of variables at point O on the failure plane Σ , Eq. (23) becomes:   ˙ dΣ = σ : ε˙ dP (σ · n) · u (24) Σ

P

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because ε˙ = 0 in the right part of the plate. Using the definition (18) of the support function Π , it gives:   ˙ dΣ  Π(˙ε ) dΣ for ε˙ such as Π(˙ε ) < +∞ (σ · n) · u Σ

(25)

Σ

The left part of the inequality (25), which corresponds to the work of the external forces is equal to: 

˙ dΣ = (σ · n) · u

h/(2  sin θ)

h/(2  sin θ)

(u˙ OX σXX sin θ ) ds + −h/(2 sin θ)

Σ

(u˙ OZ σZX sin θ ) ds

−h/(2 sin θ)

h/(2  sin θ)

h/(2  sin θ)

(ω˙ OY σXX s sin θ ) ds +

+

2

−h/(2 sin θ)

(u˙ OX σXZ cos θ ) ds

(26)

−h/(2 sin θ)

If the stress is assumed to be independent of the abscissa X, the work of the external forces can be expressed from the stress resultant variables:  ˙ dΣ = NOX u˙ OX + TOX u˙ OZ + MOX ω˙ OY + cot θ TOX u˙ OX (σ · n) · u (27) Σ

This last expression (27) can be used for derivation of a beam criterion. In case of derivation of a plate criterion, the work of the external force (26) and (27) can be written as:  ˙ dΣ = (NOx cos2 α + NOy sin2 α + 2NOxy cos α sin α)u˙ OX + (TOx cos α + TOy sin α)u˙ OZ (σ · n) · u Σ

+ (MOx cos2 α + MOy sin2 α + 2MOxy cos α sin α)ω˙ OY + cot θ (TOx cos α + TOy sin α)u˙ OX The right-hand side member of the inequality (25) can be split into three parts:     Π(˙ε ) dΣ = ΠC (˙ε ) dΣ + ΠS long (˙ε ) dΣ + ΠS trans (˙ε ) dΣ Σ

Σ

Σ

(28)

(29)

Σ

If ΩT is the density of the transverse reinforcement (in m2 /m2 ), the amount of yielding steel over a length h/| tan θ | is ΩT h/| tan θ | (in m2 /m). Thus, the density of yielding steel over the discontinuity surface Σ can be written as follow (see Fig. 2): ΩT

h sin θ = ΩT | cos θ | | tan θ | h

(30)

Consequently, Eq. (29) becomes: 

h/(2  sin θ)

Π(˙ε ) dΣ = Σ

−h/(2 sin θ)

 



    ρsup h  C  u˙ X ρinf h  + σ Ω u˙ n (s) ds + σY ΩX sup u˙ X Y X inf   tan φ 2 sin θ 2 sin θ 

h/(2  sin θ)

+

  σY ΩT | cos θ |u˙ Z (s) ds

−h/(2 sin θ)

The use of Eqs. (14) and (15) gives:

(31)

P. Koechlin et al. / European Journal of Mechanics A/Solids 27 (2008) 1161–1183

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Fig. 2. Density of transversal reinforcement.



C Π(˙ε ) dΣ = tan φ

Σ

h/(2  sin θ) −h/(2 sin θ)

   ρsup h   (u˙ OX sin θ + u˙ OZ cos θ + s ω˙ OY ) ds + σY ΩX sup u˙ OX + ω˙ OY 2 

   ρinf h   + σY ΩX inf u˙ OX + ω˙ OY + σY ΩT | cos θ | 2 

h/(2  sin θ)

|u˙ OZ + s ω˙ OY cos θ | ds

(32)

−h/(2 sin θ)

Since: x2 |ax + b| dx = |ax2 + b|

ax2 + b ax1 + b − |ax1 + b| 2a 2a

(33)

x1

we have finally:        ρsup h  ρinf h  hC   + σY ΩX inf u˙ OX + ω˙ OY [u˙ OX + u˙ OZ cot θ ] + σY ΩX sup u˙ OX + ω˙ OY Π(˙ε ) dΣ = tan φ 2  2  Σ  

   σY ΩT | cos θ |  hω˙ OY hω˙ OY  u˙ OZ + + cot θ  u˙ OZ + cot θ − u˙ OZ  2ω˙ OY cos θ 2 2 

  hω˙ OY h ω ˙ OY cot θ  u˙ OZ − cot θ − 2 2

(34)

The condition Π(˙ε ) < +∞ implies (cf. (19)) u˙ n (s)  |u(s)| ˙ sin φ for −h/(2 sin θ )  s  h/(2 sin θ ). But u˙ n (s)  |u(s)| ˙ sin φ is equivalent to u˙ n (s)  |u˙ t (s)| tan φ, which can be expressed as: u˙ OX sin θ + u˙ OZ cos θ + s ω˙ OY  |u˙ OX cos θ − u˙ OZ sin θ | tan φ

(35)

Finally the condition that has to be satisfied is: u˙ OX sin θ + u˙ OZ cos θ ±

h ω˙ OY  |u˙ OX cos θ − u˙ OZ sin θ | tan φ 2 sin θ

(36)

3.4. The plate problem Now Eq. (25) can be written in an expanded form using Eqs. (28) and (34). So, for a reinforced concrete plate, the problem consists in finding the domain in the (NOx , NOy , NOxy , TOx , TOy , MOx , MOy , MOxy ) space such that the following inequality holds:   NOx cos2 α + NOy sin2 α + 2NOxy cos α sin α + cot θ (TOx cos α + TOy sin α) u˙ OX + (TOx cos α + TOy sin α)u˙ OZ + (MOx cos2 α + MOy sin2 α + 2MOxy cos α sin α)ω˙ OY    ρsup h  hC 2 2  [u˙ OX + u˙ OZ cot θ ] + σY (Ωx sup cos α + Ωy sup sin α)u˙ OX + ω˙ OY  tan φ 2 

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   ρinf h   + σY (Ωx inf cos α + Ωy inf sin α)u˙ OX + ω˙ OY 2   

 σY ΩT  hω˙ OY hω˙ OY  + sgn(cos θ ) u ˙ u ˙ cot θ cot θ + + OZ  OZ 2ω˙ OY  2 2  

  hω˙ OY hω˙ OY σ Y ΩT   u ˙ u ˙ cot θ cot θ − − − sgn(cos θ ) OZ  OZ 2ω˙ OY  2 2 2

2

(37)

for all u˙ OX , u˙ OZ , ω˙ OY , θ, α which match the conditions:  h u˙ OX sin θ + u˙ OZ cos θ ± ω˙ OY  |u˙ OX cos θ − u˙ OZ sin θ | tan φ (38) 2 sin θ 0<θ <π This problem is in fact the searching of a maximum for a function depending on the five variables: u˙ OX , u˙ OZ , ω˙ OY , θ and α. Since:   (39) max F (a, b) = max max F (a, b) a,b

a

b

it is possible to solve the problem step by step, which means to set one or two variables as constants in order to find the maximum for the others. Actually the first step consists in finding a beam criterion. 3.5. The beam problem If α is set as a constant, the formulation of the problem leads to a beam criterion. The domain of the (NOX , TOX , MOX ) space that has to be found is such that: (NOX + cot θ TOX )u˙ OX + TOX u˙ OZ + MOX ω˙ OY       ρsup h  ρinf h  hC  + σ u ˙  Ω + ω ˙ [u˙ OX + u˙ OZ cot θ ] + σY ΩX sup u˙ OX + ω˙ OY Y X inf  OX OY tan φ 2  2   

 σY ΩT  hω˙ OY hω˙ OY u˙ OZ + cot θ  u˙ OZ + cot θ + sgn(cos θ )  2ω˙ OY 2 2  

 σY ΩT  hω˙ OY  u˙ OZ − hω˙ OY cot θ u ˙ cot θ − sgn(cos θ ) − OZ  2ω˙ OY  2 2 for all u˙ OX , u˙ OZ , ω˙ OY , θ which match the conditions:  h u˙ OX sin θ + u˙ OZ cos θ ± ω˙ OY  |u˙ OX cos θ − u˙ OZ sin θ | tan φ 2 sin θ 0<θ <π

(40)

(41)

4. Derivation of the beam criterion In the last section, using the kinematical method, we have expressed two maximization problems that need to be solved in order to get a beam criterion and then a plate criterion. In the present section, the beam problem (40) is solved: finding the maximum on the variables u˙ OX , u˙ OZ , ω˙ OY , θ will give strength conditions only expressed in terms of NOX , TOX , MOX . 4.1. Maximization on the velocity field components The variable θ is considered constant between 0 and π . We start with the case: u˙ OX cos θ − u˙ OZ sin θ  0 Let define the function f as:

(42)

P. Koechlin et al. / European Journal of Mechanics A/Solids 27 (2008) 1161–1183

f (x, y, z) = a1 x + b1 y + c1 z − |a2 x + b2 y + c2 z| − |a3 x + b3 y + c3 z| k k − |a4 x + b4 y + c4 z|(a4 x + b4 y + c4 z) − |a5 x + b5 y + c5 z|(a5 x + b5 y + c5 z) z z

1169

(43)

The problem described in Section 3.5 by Eqs. (40) and (41) can be written as the search of conditions on parameters ai , bi , ci (i = 1 to 5), k, d, e, f, g, h such that: ⎧ ⎨ dx + ey + f z  0 f (x, y, z; ai , bi , ci , k)  0 ∀x, y, z with dx + ey − f z  0 ⎩ gx + hy  0

(44)

Appendix A gives six necessary conditions on parameters ai , bi , ci , k, d, e, f, g, h for (44) to be satisfied (Eqs. (114)– (118) and (119)). Going back to the original problem expressed with the NOX , TOX , MOX variables, the six conditions are equivalent to the three following conditions:

⎧   2MOX ⎪ ⎪ N − σ Ω + (1 + ρ ) + Ω (1 + ρ ) + hσ Ω OX Y X sup sup X inf inf Y T sin θ ⎪ ⎪ h ⎪ ⎪

⎪ ⎪ 1 hC ⎪ ⎪ ⎪ + 2T + hσ 0 (45) cos θ − Ω OX Y T ⎪ ⎪ tan φ sin θ ⎪ ⎪

⎪ ⎪   ⎪ 2MOX ⎪ ⎪ N Ω − (1 − ρ ) + Ω (1 − ρ ) + hσ Ω sin θ − σ OX Y X sup sup X inf inf Y T ⎪ ⎪ h ⎨

hC 1 ⎪ + 2TOX cos θ − + hσY ΩT 0 (46) ⎪ ⎪ tan φ sin θ ⎪ ⎪ ⎪



⎪ ⎪ hC hC ⎪ ⎪ NOX + cot θ TOX − (cos θ + tan φ sin θ ) − TOX − cot θ (sin θ − tan φ cos θ ) ⎪ ⎪ ⎪ tan φ tan φ ⎪ ⎪ ⎪ ⎪ − σY (ΩX sup + ΩX inf )| cos θ + tan φ sin θ | ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ − h σY ΩT | sin θ − tan φ cos θ || cos θ | − (sin θ − tan φ cos θ ) cos θ   0 (47) sin θ 2 Let us look at the second case: instead of (42), we suppose now: u˙ OX cos θ − u˙ OZ sin θ  0 Again, this case gives three necessary conditions:

⎧   2MOX ⎪ ⎪ N − σ Ω + (1 + ρ ) + Ω (1 + ρ ) + hσ Ω OX Y X sup sup X inf inf Y T sin θ ⎪ ⎪ h ⎪ ⎪

⎪ ⎪ 1 hC ⎪ ⎪ ⎪ + 2T + hσ 0 cos θ − Ω OX Y T ⎪ ⎪ tan φ sin θ ⎪ ⎪

⎪ ⎪   ⎪ 2MOX ⎪ ⎪ N sin θ − σ Ω − (1 − ρ ) + Ω (1 − ρ ) + hσ Ω OX Y X sup sup X inf inf Y T ⎪ ⎪ h ⎨

1 hC ⎪ + 2TOX cos θ − + hσY ΩT 0 ⎪ ⎪ tan φ sin θ ⎪ ⎪



⎪ ⎪ ⎪ hC hC ⎪ ⎪ (cos θ − tan φ sin θ ) + TOX − cot θ (sin θ + tan φ cos θ ) − NOX + cot θ TOX − ⎪ ⎪ ⎪ tan φ tan φ ⎪ ⎪ ⎪ ⎪ − σY (ΩX sup + ΩX inf )| cos θ − tan φ sin θ | ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ − h σY ΩT | sin θ + tan φ cos θ || cos θ | + (sin θ + tan φ cos θ ) cos θ   0 sin θ 2

(48)

(49)

(50)

(51)

Since the first two conditions (Eqs. (49) and (50)) have already been obtained in the preceding case (Eqs. (45) and (46)), we finally get the four conditions for 0 < θ < π :

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⎧

  2MOX ⎪ ⎪ N sin θ Ω + (1 + ρ ) + Ω (1 + ρ ) + hσ Ω − σ ⎪ OX Y X sup sup X inf inf Y T ⎪ ⎪ h ⎪ ⎪

⎪ ⎪ 1 hC ⎪ ⎪ ⎪ ⎪ + 2TOX cos θ − tan φ + hσY ΩT sin θ  0 ⎪ ⎪ ⎪

⎪ ⎪   2MOX ⎪ ⎪ ⎪ NOX − − σY ΩX sup (1 − ρsup ) + ΩX inf (1 − ρinf ) + hσY ΩT sin θ ⎪ ⎪ h ⎪ ⎪

⎪ ⎪ 1 hC ⎪ ⎪ ⎪ + 2T + hσ 0 cos θ − Ω OX Y T ⎪ ⎪ tan φ sin θ ⎪ ⎪



⎨ hC hC N (cos θ + tan φ sin θ ) − T + cot θ T − − cot θ (sin θ − tan φ cos θ ) OX OX X ⎪ tan φ tan φ ⎪ ⎪ ⎪ ⎪ ⎪ − σY (ΩX sup + ΩX inf )| cos θ + tan φ sin θ | ⎪ ⎪ ⎪ ⎪  h σ Y ΩT  ⎪ ⎪ ⎪ | sin θ − tan φ cos θ || cos θ | − (sin θ − tan φ cos θ ) cos θ  0 − ⎪ ⎪ ⎪ ⎪ sin θ 2



⎪ ⎪ hC hC ⎪ ⎪ − N (cos θ − tan φ sin θ ) + T cot θ (sin θ + tan φ cos θ ) + cot θ T − − ⎪ OX OX X ⎪ ⎪ tan φ tan φ ⎪ ⎪ ⎪ ⎪ − σY (ΩX sup + ΩX inf )| cos θ − tan φ sin θ | ⎪ ⎪ ⎪ ⎪  ⎪ h σ Y ΩT  ⎪ ⎩ − | sin θ + tan φ cos θ || cos θ | + (sin θ + tan φ cos θ ) cos θ  0 sin θ 2

(52)

(53)

(54)

(55)

4.2. Maximization on the θ angle The first two conditions (Eqs. (52) and (53)) are NOX , TOX , MOX dependent, whereas the last two (Eqs. (54) and (55)) are independent of the moment MOX . 4.2.1. Conditions on NOX , TOX , MOX The first two conditions (Eqs. (52) and (53)) have the shape: f (θ )  0 ∀θ ∈ ]0, π]

(56)

with a function f defined as: f (θ ) = a cos θ + b sin θ +

c sin θ

with c < 0

(57)

The function f can also be expressed as: f (θ ) =

a sin 2θ − b cos 2θ + b + 2c 2 sin θ

(58)

From (58), we deduce:    max 2 sin θf (θ ) = a 2 + b2 + b + 2c

θ∈]0,π]

(59)

And the maximum is reached for: θopt =

1 b π + arctan 4 2 a

(60)

Since c < 0, the inequality (56) is satisfied for all 0 < θ < π if and only if: a 2  4c2 + 4bc Thus the two conditions (52) and (53) are equivalent to:

(61)

P. Koechlin et al. / European Journal of Mechanics A/Solids 27 (2008) 1161–1183

⎧

2 hC ⎪ 2 ⎪ ⎪ TOX  + hσY ΩT ⎪ ⎪ tan φ ⎪ ⎪



⎪ ⎪   hC 2MOX ⎪ ⎪ ⎪ − σ + hσ − N + (1 + ρ ) + Ω (1 + ρ ) + hσ Ω Ω Ω OX Y X sup sup X inf inf Y T Y T ⎨ h tan φ

2 ⎪ hC ⎪ 2 ⎪ ⎪ TOX + hσY ΩT  ⎪ ⎪ tan φ ⎪ ⎪



⎪ ⎪   hC 2MOX ⎪ ⎪ ⎩ − NOX − − σY ΩX sup (1 − ρsup ) + ΩX inf (1 − ρinf ) + hσY ΩT + hσY ΩT h tan φ And the angle of the optimal kinematics is given respectively for each condition by: ⎧ NOX + 2MOX / h − σY [ΩX sup (1 + ρsup ) + ΩX inf (1 + ρinf )] + hσY ΩT π 1 ⎪ ⎪ ⎨ θopt = + arctan 4 2 2TOX N − 2M / h − σ [Ω (1 ⎪ 1 π OX Y X sup − ρsup ) + ΩX inf (1 − ρinf )] + hσY ΩT ⎪ ⎩ θopt = + arctan OX 4 2 2TOX 4.2.2. Conditions on NOX and TOX Concerning the last two conditions (Eqs. (54) and (55)), they can be expressed as: ⎧ NOX cos(θ − φ) sin θ − TOX sin(θ − φ) sin θ + TOX cos(θ − φ) cos θ ⎪ ⎪ ⎪ ⎪ ⎪  hC cos φ + σY (ΩX sup + ΩX inf )| cos(θ − φ)| sin θ ⎪ ⎪ ⎪ ⎪  hσY ΩT  ⎪ ⎪ | sin(θ − φ) cos θ | − sin(θ − φ) cos θ + ⎨ 2 ⎪ − N cos(θ + φ) sin θ + TOX sin(θ + φ) sin θ − TOX cos(θ + φ) cos θ OX ⎪ ⎪ ⎪ ⎪ ⎪  hC cos φ + σY (ΩX sup + ΩX inf )| cos(θ + φ)| sin θ ⎪ ⎪ ⎪ ⎪  hσY ΩT  ⎪ ⎩ | sin(θ + φ) cos θ | + sin(θ + φ) cos θ + 2 Let us define the function F as:   F (θ ) = hC cos φ + σY (ΩX sup + ΩX inf )cos(θ − φ) sin θ

(62)

(63)

(64) (65)

(66)

(67)

  hσY ΩT  sin(θ − φ) cos θ  − sin(θ − φ) cos θ 2 The problem is now to find, in the space (NOX , TOX ), the boundary of the convex that satisfies: +

NOX cos(θ − φ) sin θ − TOX sin(θ − φ) sin θ + TOX cos(θ − φ) cos θ  F (θ )

1171

∀θ ∈ ]0, π]

(68)

(69)

This problem is solved in Appendix B. Actually if we notice that the derivative of F is continuous for all θ ∈ [0, π], except when θ = φ, θ = π/2 and θ = π/2 + φ, it is possible to prove that the boundaries of the convex domain defined by the inequality (66) are enclosed by one of the three following curves of the (NX , TX ) space: −TOX  hC + hσY ΩT tan φ

(70)

NOX sin φ − TOX cos φ  hC cos φ + σY (ΩX sup + ΩX inf ) sin φ ⎧   hC ⎪ ⎪ ⎪ NOX (θ ) = tan φ + σY (ΩX sup + ΩX inf ) sgn cos(θ − φ) ⎪ ⎪ ⎪

  2 ⎪ ⎨   hC hσY ΩT  cos θ + sin2 (θ − φ) + 1 − sgn sin(θ − φ) cos θ − 2 ⎪ sin2 θ + cos2 (θ − φ) tan φ ⎪ ⎪

  ⎪ ⎪   hC hσY ΩT  cos θ sin θ − sin(θ − φ) cos(θ − φ) ⎪ ⎪ ⎩ TOX (θ ) = + 1 − sgn sin(θ − φ) cos θ tan φ 2 sin2 θ + cos2 (θ − φ)

(71)

(72)

This last curve (72) is a parametric surface defined for θ ∈ ]0, π[ with θ = φ, θ = π/2 and θ = π/2 + φ. The angle of the mechanism is given by the parameter θ . The second inequality (67) gives the three following curves:

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TOX  hC + hσY ΩT tan φ

(73)

NOX sin φ + TOX cos φ  hC cos φ + σY (ΩX sup + ΩX inf ) sin φ ⎧ hC ⎪ ⎪ NOX (θ ) = − σY (ΩX sup + ΩX inf ) sgn(cos(θ + φ)) ⎪ ⎪ tan φ ⎪ ⎪

  ⎪ ⎨   hC hσY ΩT  cos2 θ + sin2 (θ + φ) + 1 + sgn sin(θ + φ) cos θ − 2 ⎪ sin2 θ + cos2 (θ + φ) tan φ ⎪ ⎪

  ⎪ ⎪   hC hσY ΩT  cos θ sin θ − sin(θ + φ) cos(θ + φ) ⎪ ⎪ ⎩ TOX (θ ) = + 1 + sgn sin(θ + φ) cos θ tan φ 2 sin2 θ + cos2 (θ + φ)

(74)

(75)

for θ ∈ ]0, π[ with θ = π − φ, θ = π/2 et θ = π/2 − φ. 4.3. Results Finally it is possible to summarize the results (conditions (62), (63), (70)–(74) and (75)) for the beam configuration and to present them in a non-dimensional way. For this purpose, we set: n=

NOX , hfc

t=

TOX , hfc

m=

MOX h2 fc

and fc =

2C cos φ 1 − sin φ

The beam criterion is divided into three parts: 1. Two parabolic cylinders (Eqs. (62), (63)) (ξ = ±1) generated by the line n + 2ξ m = 0:

1 − sin φ σY ΩT 2 + t2  2 sin φ fc



 σ Y ΩT 1 − sin φ σY ΩT σY  + ΩX sup (1 + ξρsup ) + ΩX inf (1 + ξρinf ) + − n + 2ξ m − hfc fc 2 sin φ fc

(76)

(77)

The angle of the mechanism is given by: n + 2ξ m − (σY / hfc )[ΩX sup (1 + ξρsup ) + ΩX inf (1 + ξρinf )] + σY ΩT /fc 1 π + arctan 4 2 2t 2. Four planes (independent of the moment m). The first two planes are (with ξ = ±1): θopt =

ξt 

1 − sin φ σY ΩT + tan φ 2 cos φ fc

with an angle: π θopt = − ξ φ 2 The last two planes are (with ξ = ±1): n sin φ + ξ t cos φ 

σY 1 − sin φ + (ΩX sup + ΩX inf ) sin φ 2 hfc

In that case, the angle of the mechanism is: π θopt = 2 3. Two parametric symmetric curves (with ξ = ±1): ⎧   1 − sin φ σY ⎪ ⎪ n(θ ) = (ΩX sup + ΩX inf ) sgn cos(θ + ξ φ) −ξ ⎪ ⎪ 2 sin φ hfc ⎪ ⎪

 2  ⎪ ⎨   cos θ + sin2 (θ + ξ φ) 1 − sin φ σY ΩT  + − 1 + ξ sgn sin(θ + ξ φ) cos θ 2 sin φ 2fc ⎪ sin2 θ + cos2 (θ + ξ φ) ⎪ ⎪

  ⎪ ⎪   cos θ sin θ − sin(θ + ξ φ) cos(θ + ξ φ) 1 − sin φ σY ΩT  ⎪ ⎪ ⎩ t (θ ) = + 1 + ξ sgn sin(θ + ξ φ) cos θ 2 sin φ 2fc sin2 θ + cos2 (θ + ξ φ)

(78)

(79)

(80)

(81)

(82)

(83)

P. Koechlin et al. / European Journal of Mechanics A/Solids 27 (2008) 1161–1183

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Fig. 3. Beam criterion and failure modes.

The two surfaces are partly superimposed. Consequently, the parameter θ can take values only in the range [ π4 − φ π π ξ φ2 ; 3π 4 − ξ 2 ] with θ = 2 and θ = 2 + ξ φ. The angle of the optimal mechanism is given by the value of the parameter: θopt = θ . This beam criterion is quite complex, but it is expressed in a closed form. It can be seen in Fig. 3 in the space (n, t, m). It is made up of several conditions that can be classified in two categories: – the conditions which are functions of the moment, the normal force and the shear force (parabolic cylinders (77)); – the conditions independent of the moment and expressed only in terms of normal force and shear force (planes (79) and (81) and parametric surfaces (83)). It is possible to interpret these two parts of the criterion as characterizing respectively a bending failure mode and a shear failure mode. Note that, because of the appropriate choice of parameters for the Drucker–Prager concrete criterion (as written in Eq. (8)), the expression of the global criterion would have been identical if the Coulomb criterion had been chosen instead of Drucker–Prager. 5. Criterion extended to plates The proposed criterion can be extended to reinforced concrete plates. The derivation consists in an optimization of the beam conditions (77), (79), (81) and (83) on the α parameter. Contrary to the preceding part, it is not possible to find a complete closed form solution; the beam conditions (81) and (83) will necessitate a numerical solving. 5.1. Closed form expressions In order to simplify the notations, let us set: Tmax = hC + hσY ΩT tan φ and (from Eqs. (62) and (63)):

(84)

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fξbeam (NOi , TOi , MOi )



  Tmax 2 Tmax 2MOi 2 = TOi + NOi + ξ − σY Ωi sup (1 + ξρsup ) + Ωi inf (1 + ξρinf ) + hσY ΩT − tan φ h tan φ

(85)

where i stands for x, y or X. The parabolic cylinders (77) (equivalent to Eqs. (62) and (63)) can be written more easily: fξbeam (NX , TX , MX )  0

(86)

The (X, Y, Z) axes are oriented by the α angle. We are going to write the condition (86) in the (x, y, z) axes as a condition depending on the α parameter. Using the formulas (4) and (5), the plate criterion becomes the domain of (NOx , NOy , NOxy , TOx , TOy , MOx , MOy , MOxy ) for which ∀α ∈ [0, 2π]: fξbeam (NOx , TOx , MOx ) cos2 α + fξbeam (NOy , TOy , MOy ) sin2 α

  2MOxy Tmax NOxy + ξ cos α sin α  0 + 2 TOx TOy + tan φ h

(87)

Since:

 (a − b)2 + c2 + a + b max [a cos α + b sin α + c cos α sin α] = α∈[0,2π] 2 2

2

(88)

the condition (87), which must be verified for all α ∈ [0, 2π], is equivalent to: 

   beam 2 2MOxy 2 Tmax beam NOxy + ξ fξ (NOx , TOx , MOx ) − fξ (NOy , TOy , MOy ) + 4 TOx TOy + tan φ h + fξbeam (NOx , TOx , MOx ) + fξbeam (NOy , TOy , MOy )  0

(89)

The angle of the mechanism in the plate is given by: αopt =

fξbeam (NOx , TOx , MOx ) − fξbeam (NOy , TOy , MOy ) 1 π − arctan 4 2 2[TOx TOy + (Tmax / tan φ)(NOxy + ξ(2MOxy / h))]

(90)

This condition (89) is quite complicated, but his main advantage is to be expressed in a closed form. If we consider now the beam condition (79) written in the (x, y, z) axes, the following inequality has to be verified for all α ∈ [0, 2π]: |TOx cos α + TOy sin α|  Tmax Eliminating α, it gives another plate criterion:  2 +T2 T TOx max Oy

(91)

(92)

And the angle of the optimal mechanism is: αopt = arctan

TOy TOx

(93)

5.2. Computational criterion For the other two parts of the beam criterion described by the planes (81) and the parametric surfaces (83), it is difficult to get a closed form expression of the plate criterion. But it is possible to use numerical methods to predict failure. This can be done in two different ways: 1. For a specific reinforced concrete plate, we try to find the failure conditions (which means a surface) directly in the (Nx , Ny , Nxy , Tx , Ty , Mx , My , Mxy ) space. It implies to discretize the 8-dimensions space with the n points (Nxi , Nyi , Nxyi , Txi , Tyi , Mxi , Myi , Mxyi )1in and to find λ such that the point (λNxi , λNyi , λNxyi , λTxi , λTyi , λMxi , λMyi , λMxyi ) satisfies the conditions (37) and (38) defining the plate failure problem. This method does not take advantage of the results obtained in the beam configuration.

P. Koechlin et al. / European Journal of Mechanics A/Solids 27 (2008) 1161–1183

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Fig. 4. Discretization of the potential mechanisms for the computational criterion.

2. For a specific reinforced concrete plate and for a specific set of forces (Nx , Ny , Nxy , Tx , Ty , Mx , My , Mxy ), we are going to check if the beam criterion is reached in an α angle (in the plate plane: see Fig. 1). It means a finite number of potential beam failure directions. This method takes advantage of the closed form expression of the beam failure criterion: that is why it is preferred to the previous one. Let us describe the second method more in detail. A finite number of αi angles are chosen. In each of these αi directions, the beam conditions are checked for the following forces NOXi TOXi MOXi : ⎧ 2 2 ⎪ ⎨ NOXi = NOx cos αi + NOy sin αi + 2NOxy cos αi sin αi TOXi = TOx cos αi + TOy sin αi (94) ⎪ ⎩ 2 2 MOXi = MOx cos αi + MOy sin αi + 2MOxy cos αi sin αi with a reinforcement:  ΩX sup i = Ωx sup cos2 αi + Ωy sup sin2 αi ΩX inf i = Ωx inf cos2 αi + Ωy inf sin2 αi

(95)

Actually, only the beam conditions (81) and (83) should be checked numerically using this method with discretized failure mechanism defined by the αi directions, because the other beam conditions (77) and (79) have already given closed form expressions for the plate (see Section 5.1). Now the only difficulty comes from the fact that the parametric curve (83) needs to be also discretized in a finite number of θj (see Fig. 4). The curve is then described by the points (NOαi (θj ), MOαi (θj )). Let define the function ξα (NOX , MOX ) that is equal to −1 when the point (NOX , MOX ) is inside the curve and +1 when it is located outside. Let (NOα (θj ), MOα (θj ));(NOX , MOX ) be the distance function between the points (NOα (θj ), MOα (θj )) and (NOX , MOX ). With these notations, the numerical criterion can be written as:       (96) max ξαi (NOXi , MOXi ) min NOαi (θj ), MOαi (θj ) ;(NOXi , MOXi )  0 i

j

6. Validation and numerical application 6.1. Reinforced concrete beam The beam failure criterion defined in Section 4.3 is the first one in the literature which deals with bending moment, shear and normal forces at once. The comparison is then possible only with partial existing criteria. In the following is shown the comparison with the Nielsen’s criterion, widely used for reinforced concrete beams subjected to shear (Nielsen, 1999). It marks the boundary of the non-dimensional shear force t as follow: 

σ Y ΩT σ Y ΩT t 1− (97) fc fc Note that it is a very simple criterion, which is usually an advantage. But the drawback is that it is independent of the concrete friction angle and especially of the beam longitudinal reinforcement. The only condition on the longitudinal

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Fig. 5. Comparison between Nielsen and the new criterion depending on the reinforcement. Table 1 Characteristics of the slabs tested by Yamada et al. (1992) Test

1 2 3 4 5 6 7

Concrete

Longitudinal reinforcement

Compressive strength

Steel area per meter of slab

Transversal reinforcement Yield stress

Steel area per square meter of slab

Yield stress σY T

fc

Ωx inf = Ωy inf = Ωx sup = Ωy sup

σY L

ΩT

MPa

m2 /m

MPa

m2 /m2

MPa

26,0 27,17 25,90 27,37 26,00 26,39 27,76

0,00306 0,00306 0,00306 0,00306 0,00306 0,00306 0,00306

568 568 568 568 568 568 568

0,00 0,25 0,50 0,55 1,11 0,99 1,98

347 347 317 317 330 330

reinforcement is that it should be large enough to prevent any bending failure. The comparison with our new criterion is carried out for a friction angle φ which matches tan φ = 0, 75 and for several longitudinal reinforcement ratio ψ . The longitudinal reinforcement ratio is defined as ψ = σY (Ωsup + Ωinf )/(hfc ). Fig. 5 shows both new criterion and Nielsen’s criterion. When the transverse reinforcement is low, the gap between the two criteria comes from the prediction of the Nielsen’s criterion, which is not relevant in that case (see Nielsen, 1999). Nevertheless, when the transverse reinforcement ratio increases, the new criterion fits the Nielsen’s curve well. 6.2. Reinforced concrete plate In the scientific literature, systematic experimental tests which give directly the critical combined forces in membrane, bending and shear for a simple reinforced concrete plate are very rare (a few test data are available in Jau et al. (1982), Adebar and He (1994)). Thus, we validate our criterion on a more complex structure. The validation is based on a comparison with an experimental program carried out by Yamada et al. (1992). The tested structure is a column which is embedded in a reinforced concrete slab. A symmetrically distributed load is applied on the slab sides in order to produce a punching shear at the connection with the column (Fig. 6). In this way a combination of bending moments and shear forces in both directions is obtained. The tests characteristics are given in Table 1.

P. Koechlin et al. / European Journal of Mechanics A/Solids 27 (2008) 1161–1183

1177

Fig. 6. Geometrical configuration of the punching tests performed by Yamada et al. (1992) (dimensions in cm).

Fig. 7. Concrete behavior.

Actually, on both sides of the Atlantic ocean, when a limit analysis criterion is applied to predict failure (specially in the case of a shear criterion), it is recommended to use an effectiveness factor on the compressive strength of concrete (Eurocode2, 1991; Wang, 1995; Nielsen, 1999). This effectiveness factor is defined by (with fc in MPa):

fc νfc = max 0, 8 − ;0, 5 (98) 200 Three main reasons can justify the use of such a corrective factor: – It corrects the fact that the derived criterion is an upper bound. – Since the criterion is global, it is unrealistic to suppose that the concrete reaches locally the maximum stress fc everywhere. – Theoretically the criterion is valid for any material behavior. Nevertheless, yield design are well adapted to ductile or perfectly plastic materials. Yet, the concrete cannot be considered as ductile, contrary to the steel. The effectiveness factor enables to correct this error, defining an equivalent perfectly plastic behavior for the concrete (Fig. 7). Linear computations with the new criterion are performed using Europlexus, a fast dynamics software (Europlexus, 2006). The results are shown in Fig. 8 and fit well the experimental curve. As a general remark, since the Drucker– Prager local concrete criterion is comparatively worse than the steel criterion, the more the reinforcement in the plate, the better the global criterion is. Fig. 8 shows that the criterion can predict not only the failure load, but also the failure mode.

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Fig. 8. Failure load and failure mode. Comparison between the new criterion and the experiment.

Fig. 7 shows also the curve that would be obtained without using the effectiveness factor. In that case, the computed failure load and the measured one evolve in a similar way, but there is little gap between the two curves. It proves that the use of the effectiveness factor is necessary. 7. Conclusion In this paper, we have derived using limit analysis: – a criterion for reinforced concrete beams in an analytical closed form expression; – a criterion for reinforced concrete plates, partly in an analytical closed form, partly discretized. These two criteria define an upper-bound for the load expressed in terms of membrane (or normal) forces N , shear (transverse) forces T , and bending moments M. When the limit is reached, they give the angles of the failure plane in the reinforced concrete section. The shape of the criterion (one part depending on N, T , M and another independent of the moment M) allows to distinguish two failure modes: a bending failure and a shear failure. The comparison with experimental tests of a punching slab has shown that the criterion can predict very well the failure of a reinforced concrete slab, provided that an effectiveness factor is used on the compressive strength of the concrete. These results open the way to use this criterion in static or in dynamic computations (e.g., Koechlin et al., 2006; Koechlin, 2007). Acknowledgements The authors would like to thank Prof. Patrick de Buhan for his helpful remarks on the content of this paper. Appendix A Let define a function f as: f (x, y, z) = a1 x + b1 y + c1 z − |a2 x + b2 y + c2 z| − |a3 x + b3 y + c3 z| k k − |a4 x + b4 y + c4 z|(a4 x + b4 y + c4 z) − |a5 x + b5 y + c5 z|(a5 x + b5 y + c5 z) z z

(99)

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1179

We would like to find conditions on parameters ai , bi , ci (i = 1, . . . , 5), k, d, e, f, g, h such as: f (x, y, z; ai , bi , ci , k)  0 ∀x, y, z ∈ R which satisfy: ⎧ ⎨ dx + ey + f z  0 dx + ey − f z  0 ⎩ gx + hy  0 The resolution is made as follows: If x, y, z satisfy the conditions (101), then it means ∃t, u, v  0 such as: ⎧ ⎨ dx + ey + f z = t dx + ey − f z = u ⎩ gx + hy = v Thus it is possible to express x, y, z as functions of t, u, v, provided that (dh − eg)f = 0: ⎧ ht + hu − 2ev ⎪ x= ⎪ ⎪ ⎪ 2(dh − eg) ⎪ ⎪ ⎨ −gt − gu + 2dv y= ⎪ 2(dh − eg) ⎪ ⎪ ⎪ ⎪ t − u ⎪ ⎩z = 2f

(100)

(101)

(102)

(103)

Let define f˜ as:   f˜(t, u, v) = f x(t, u, v), y(t, u, v), z(t, u, v)

(104)

The function f˜ can also be written as: f˜(t, u, v) = A1 t + B1 u + C1 v − |A2 t + B2 u + C2 v| − |A3 t + B3 u + C3 v| K − |A4 t + B4 u + C4 v|(A4 t + B4 u + C4 v) t −u K − |A5 t + B5 u + C5 v|(A5 t + B5 u + C5 v) t −u where Ai , Bi , Ci , K, are parameters depending on ai , bi , ci , k, d, e, f, g, h (i = 1, . . . , 5): ⎧ ai h − bi g ci ⎪ ⎪ Ai = + ⎪ ⎪ ⎪ 2(dh − eg) 2f ⎪ ⎪ ⎪ ⎪ ⎨ B = ai h − bi g − ci i 2(dh − eg) 2f ⎪ ⎪ ⎪ −ai e + bi d ⎪ ⎪ ⎪ Ci = ⎪ (dh − eg) ⎪ ⎪ ⎩ K = 2f k

(105)

(106)

Our problem comes down to find conditions on parameters Ai , Bi , Ci , K, such as: f˜(t, u, v)  0 ∀t, u, v  0 If K = 0, the conditions come easily: ⎧ ⎨ A1 − |A2 | − |A3 |  0 B1 − |B2 | − |B3 |  0 ⎩ C1 − |C2 | − |C3 |  0

(107)

(108)

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But if K = 0 (which means f k = 0), the problem is more difficult because it is almost impossible to find some necessary and sufficient conditions. But we are going to find necessary conditions that are relevant. If K = 0, the condition f˜(t, u, v)  0 must be satisfied in the following specific cases: ⎧ u = v = 0 and t > 0 ⎪ ⎪ ⎪ ⎪ t = v = 0 and u > 0 ⎪ ⎪ ⎪ ⎨ t − u > 0 and v → +∞ ⎪ t − u < 0 and v → +∞ ⎪ ⎪ ⎪ ⎪ + ⎪ ⎪ ⎩ v = 0 and t − u → 0 v = 0 and t − u → 0−

(109)

All these different cases give the respectively the conditions:   ⎧ A1 − |A2 | − |A3 | − K |A4 |A4 + |A5 |A5  0 ⎪ ⎪   ⎪ ⎪ ⎪ B1 − |B2 | − |B3 | + K |B4 |B4 + |B5 |B5  0 ⎪ ⎪ ⎪ ⎨ −|C |C K − |C |C K  0 4 4 5 5 ⎪ ⎪ ⎪ |C4 |C4 K + |C5 |C5 K  0 ⎪ ⎪ ⎪ −|A4 + B4 |(A4 + B4 )K − |A5 + B5 |(A5 + B5 )K  0 ⎪ ⎪ ⎩ |A4 + B4 |(A4 + B4 )K + |A5 + B5 |(A5 + B5 )K  0

(110)

Since K = 0, this is equivalent to:   ⎧ A1 − |A2 | − |A3 | − K |A4 |A4 + |A5 |A5  0 ⎪ ⎪ ⎪ ⎨ B − |B | − |B | + K |B |B + |B |B   0 1 2 3 4 4 5 5 ⎪ C4 + C 5 = 0 ⎪ ⎪ ⎩ A4 + B4 + A5 + B5 = 0

(111)

But, knowing these equations, if we look again at the cases:

t − u > 0, t − u < 0,

v → +∞ v → +∞

(112)

it gives two other conditions: 

  C1 − |C2 | − |C3 | − K |C4 |A4 + |C5 |A5 + C4 |A4 | + C5 |A5 |  0   C1 − |C2 | − |C3 | + K |C4 |B4 + |C5 |B5 + C4 |B4 | + C5 |B5 |  0

(113)

In summary, if K = 0, the asymptotic behavior supplies some necessary conditions ((111) and (113)):   ⎧ A1 − |A2 | − |A3 | − K |A4 |A4 + |A5 |A5  0 ⎪ ⎪ ⎪ ⎪ B − |B | − |B | + K |B |B + |B |B   0 ⎪ 1 2 3 4 4 5 5 ⎪ ⎪ ⎪ ⎨ C4 + C 5 = 0 ⎪ A4 + B4 + A5 + B5 = 0 ⎪ ⎪   ⎪ ⎪ ⎪ C1 − |C2 | − |C3 | − K |C4 |A4 + |C5 |A5 + C4 |A4 | + C5 |A5 |  0 ⎪ ⎪   ⎩ C1 − |C2 | − |C3 | + K |C4 |B4 + |C5 |B5 + C4 |B4 | + C5 |B5 |  0

(114)

Transferring the conditions (114) on parameters ai , bi , ci , k, d, e, f, g, h, if f k = 0, the necessary conditions can be written as:

P. Koechlin et al. / European Journal of Mechanics A/Solids 27 (2008) 1161–1183

    ⎧ a1 h − b1 g c1  a2 h − b2 g c2   a3 h − b3 g c3  ⎪ ⎪ ⎪ dh − eg + f −  dh − eg + f  −  dh − eg + f  ⎪ ⎪ ⎪   

 ⎪ ⎪  a4 h − b 4 g c4  a 4 h − b4 g c4  a5 h − b5 g c5  a5 h − b5 g ⎪ ⎪    ⎪ −fk  +  + + +  + ⎪ ⎪ dh − eg f dh − eg f dh − eg f dh − eg ⎪ ⎪     ⎪ ⎪ ⎪ a1 h − b1 g c1  a2 h − b2 g c2   a3 h − b3 g c3  ⎪ ⎪ − − − − −  ⎪ ⎪ f dh − eg f dh − eg f ⎪ dh − eg ⎪    

⎪ ⎪  a4 h − b 4 g c4  a 4 h − b4 g c4  a5 h − b5 g c5  a5 h − b5 g ⎪ ⎪     ⎪ + f k − − + − − ⎪  dh − eg  dh − eg ⎪ f  dh − eg f f  dh − eg ⎪ ⎪ ⎪ ⎪ ⎪ −a4 e + b4 d − a5 e + b5 d = 0 ⎪ ⎪ ⎪ ⎪ ⎪ a 4 h − b4 g + a5 h − b5 g = 0 ⎪ ⎪     ⎨ −a1 e + b1 d  −a2 e + b2 d   −a3 e + b3 d  − − ⎪ dh − eg dh − eg   dh − eg  ⎪ ⎪   



⎪ ⎪  −a4 e + b4 d  a4 h − b4 g c4  −a5 e + b5 d  a5 h − b5 g c5 ⎪ ⎪     ⎪ − f k + + + ⎪  dh − eg  dh − eg  dh − eg  dh − eg ⎪ f f ⎪ ⎪  



 ⎪ ⎪  a4 h − b4 g c4   a5 h − b5 g c5  −a −a e + b d e + b d ⎪ 4 4 5 5 ⎪     ⎪ + ⎪  dh − eg + f  +  dh − eg + f   0 ⎪ dh − eg dh − eg ⎪ ⎪     ⎪ ⎪ ⎪ −a1 e + b1 d  −a2 e + b2 d   −a3 e + b3 d  ⎪ ⎪ − − ⎪ ⎪ dh − eg dh − eg   dh − eg  ⎪ ⎪  



 ⎪ ⎪  −a5 e + b5 d  a5 h − b5 g c5  −a4 e + b4 d  a4 h − b4 g c4 ⎪ ⎪     ⎪ + +fk  − − ⎪ ⎪ dh − eg  dh − eg f dh − eg  dh − eg f ⎪ ⎪  



 ⎪ ⎪     ⎪ −a5 e + b5 d  a5 h − b5 g c5  −a4 e + b4 d  a4 h − b4 g c4  ⎪ ⎩ +  dh − eg − f  +  dh − eg − f   0 dh − eg dh − eg

1181

c5 f

c5 f



0

(115)

0

(116)



(117) (118)

(119)

(120)

Appendix B Let be a function F defined by: F (x, y, θ ) = x cos(θ − φ) sin θ − y sin(θ − φ) sin θ + y cos(θ − φ) cos θ − f (θ )

(121)

The following assumptions are made: – f is defined for θ ∈ [0, π], is continuous for θ ∈ [0, π] and its derivative is also continuous except in a finite number of points θi , i = 1, . . . , n – f (θ ) > 0 for θ ∈ [0, π] – 0 < φ < π2 Let us find the boundary of the domain D defined by the points (x, y) which satisfy: F (x, y, θ )  0 ∀θ ∈ ]0, π]

(122)

Obviously, D is identical to the domain defined by the points (x, y) which satisfy: F (x, y, θ )  0 ∀θ ∈ [0, π]

(123)

D, as the intersection of several convex domains, is a convex domain. Moreover if the point (x, y) belongs to the boundary of the convex D, then it is possible to prove that: ∃θ0 ∈ [0, π]

such as F (x, y, θ0 ) = 0

(124)

And if θ0 = θi and θ ∈ ]0, π], we have: ∂F (x, y, θ0 ) = 0 ∂θ

(125)

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P. Koechlin et al. / European Journal of Mechanics A/Solids 27 (2008) 1161–1183

Consequently (x, y) is the solution of the two equations:    x cos(θ0 − φ) sin θ0 + y cos(θ0 − φ) cos θ0 − sin(θ0 − φ) sin θ0 = f (θ0 )     x cos(θ0 − φ) cos θ0 − sin(θ0 − φ) sin θ0 − 2y cos(θ0 − φ) sin θ0 + sin(θ0 − φ) cos θ0 = f (θ0 )

(126)

Since 0 < φ < π2 , this system of equation is equivalent to: ⎧ 2[cos(θ0 − φ) sin θ0 + sin(θ0 − φ) cos θ0 ]f (θ0 ) + [cos(θ0 − φ) cos θ0 − sin(θ0 − φ) sin θ0 ]f (θ0 ) ⎪ ⎪ x = ⎪ ⎨ cos2 (θ0 − φ) + sin2 θ0 (127) ⎪ [cos(θ0 − φ) cos θ0 − sin(θ0 − φ) sin θ0 ]f (θ0 ) − cos(θ0 − φ) sin θ0 f (θ0 ) ⎪ ⎪ ⎩y = cos2 (θ0 − φ) + sin2 θ0 Thus the boundaries of D are included: 1. either in the parametric curve defined by: ⎧ 2[cos(θ − φ) sin θ + sin(θ − φ) cos θ ]f (θ ) + [cos(θ − φ) cos θ − sin(θ − φ) sin θ ]f (θ ) ⎪ ⎪ x(θ ) = ⎪ ⎨ cos2 (θ0 − φ) + sin2 θ0 ⎪ [cos(θ − φ) cos θ − sin(θ − φ) sin θ ]f (θ ) − cos(θ − φ) sin θf (θ ) ⎪ ⎪ ⎩ y(θ ) = cos2 (θ − φ) + sin2 θ

(128)

for θ ∈ ]0, π], θ = θi 2. or in the n straight lines, boundaries of the half space: x cos(θi − φ) sin θi − ysin(θi − φ) sin θi + y cos(θi − φ) cos θi − f (θi )  0 for i = 1, . . . , n 3. or in the two straight lines, boundaries of the following half space: y cos φ − f (0)  0 y cos φ − f (π)  0

(129)

(130)

References Adebar, P., He, W., 1994. Influence of membrane forces on transverse-shear reinforcement design. ASCE Journal of Structural Engineering 120 (4), 1347–1366. Anderheggen, E., Knöpfel, H., 1972. Finite element limit analysis using linear programming. International Journal of Solids and Structures 8 (12), 1413–1431. Cookson, P.J., 1979. A general yield criterion for orthogonally reinforced concrete slab elements. In: Proc. of IABSE Colloquium, Session I, Plasticity in Reinforced Concrete, Final Report, Vol. 29, pp. 43–50. Eurocode 2, 1991. Design of Concrete Structures – Part 1, General rules for buildings, European Prestandard ENV 1992-1-1:1991, European Committee for Standardisation TC 250, Brussels. (Europlexus, 2006). A Computer Program for the Finite Element Simulation of Fluid–Structure Systems under Transient Dynamic Loading – User’s manual, CEA-JRC. Available on: http://europlexus.jrc.it/public/manual_html/index.html. Jau, W.C., White, R.N., Gergely, P., 1982. Behavior of Reinforced Concrete Slabs Subjected to Combined Punching Shear and Biaxial Tension, Rapport NUREG/CR-2920, Department of Structural Engineering, Cornell University, Prepared for US Nuclear Regulatory Commission. Johansen, K.W., 1962. Yield-Line Theory, Cement and Concrete Association, London. Koechlin, P., 2007. Modèle de comportement membrane-flexion et critère de perforation pour l’analyse de structures minces en béton armé sous choc mou, PhD Thesis, Université Pierre et Marie Curie Paris VI. Koechlin, P., Andrieux, S., Millard, A., Potapov, S., 2006. Perforation analysis of reinforced concrete plates under soft impact. In: Cirne, J., Dormeval, R., et al. (Eds.), Proc. of 8th International Conference on Mechanical and Physical Behaviour of Materials under Dynamic Loading. EURODYMAT 2006, Dijon, France, J. Phys. IV France 134, 1295–1300. Koechlin, P., Potapov, S., 2007. Global constitutive model for reinforced concrete plates. ASCE Journal of Engineering Mechanics 133 (3), 257– 266. Liu, Y.H., Cen, Z.Z., Xu, B.Y., 1995. A numerical method for plastic limit analysis of 3-D structures. International Journal of Solids and Structures 32 (12), 1645–1658. Nielsen, M.P., 1999. Limit Analysis of Concrete Plasticity, second ed.. CRC Press, Boca Raton, FL.

P. Koechlin et al. / European Journal of Mechanics A/Solids 27 (2008) 1161–1183

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Salençon, J., 2001. Elastoplasticité et calcul à la rupture. Presses de l’Ecole Polytechnique, Palaiseau. Save, M.A., Massonet, C.E., de Saxcé, G., 1997. Plastic limit analysis of plates, shells and disks. In: Achenbach, J.D., Budiansky, B., Lauwerier, H.A., Saffman, P.G., Van Wijgaarden, L., Willis, J.R. (Eds.), Applied Mathematics and Mechanics, vol. 43. Elsevier Science, Amsterdam. Wang, W., 1995. Shear strength and fatigue crack propagation in concrete by energy method, PhD Thesis, New Jersey Institute of Technology. Available on: http://www.library.njit.edu/etd/1990s/1995/njit-etd1995-012/njit-etd1995-012.html. Yamada, T., Nanni, A., Endo, K., 1992. Punching shear resistance of flat slabs: Influence of reinforcement type and ratio. ACI Structural Journal 89 (5), 555–563.