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2nd

Part

Ductility and fracture of reinforcedconcrete structural members

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2.1 Ductility

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2. Ductility and fracture of reinforced concrete structural members

2.1 Ductility

The basic reason most buildings do not collapse under seismic action is a material

property called ductility. It is the property of a material to deform permanently without

loosing its strength, i.e. without decreasing its ability to resist during deformation. A

piece of wire, e.g. an office clip bends but not brake. Due to ductility a system

resists mobilizing all its reserves (Fig. 2.1).

2

P/Py

1

0 1

3/2

/y

E D B

A

Figure 2.1Ductility of a steel beam under bending

Ductility of metals occurs due to the relocation of zones of molecules while they are

still bonded with tensile forces. This behaviour occurs under tensile and compressive

stress. Concrete exhibits satisfactory ductility in compression caused only by the

slide mechanism (friction) when it is simultaneously laterally under compression.When concrete is not laterally compressed it exhibits reduced ductility since the

fracture mechanism is functioning [1] that quickly exhausts the limits of its further

deformation (Fig. 2.2). The behaviour of concrete in tension and in shear is not

ductile but brittle. Deformation mechanisms of concrete will be discussed in a

following chapter.

[]0 -10 -20

400

200

F[kN]

-40-30

600

-50

2

1

1 2 slide mechanism

fracture mechanism

Figure 2.2Conventional concrete under compression without (1) or with confinement (2) [14]

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90

Other materials e.g. fiber-reinforced mortars, high-strength concretes (Fig. 2.3)

exhibit ductility under compression and under tension due to the slip (friction) of fibers

within the mortars matrix.

40 2

/f

c

1.00

-[]

6 8

0.75

0.50

0.25

(4)

(3)

(2)

(1)

without fibers (1)with synthetic fibers (2)with steel fibers (3)with a mixture of fibers (4)

Ultra high strength Concrete

Figure 2.3Ductility of ultra-high-strength concrete with fibers

Compressive reinforcement used in vertical members (columns, walls) and in beams

as well, increases ductility since it stabilizes the compression zone of concrete.

The soil is ductile [40] as a granular material through friction (Fig. 2.4) effect that

facilitates the smoothening of support reactions in the foundation.

F

F

Figure 2.4Foundation soil ductility

Ductility is also assured through friction in the case of brick masonry units (Fig. 2.5)

and the pullout of metal elements from the body of concrete (Fig. 2.6).

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2.1 Ductility

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Figure 2.5 & 2.6Brick wall inelastic deformation and reinforcement pullout from concrete

Generally ductility is connected with energy absorption / dissipation and itstransformation to heat. When gaps occur that have to close during motion reversal

the absorbed energy is reduced (pinching). Metals do not exhibit this effect hence

there is higher energy dissipation without noticeable damages (disruption of the

continuity of structure). In high-strength concretes without confinement ductility is

limited. For this reason fiber-reinforced or even confined concrete is frequently used

(Fig. 2.7).

0 0.005 0.010 0.015 0.020

50

100

150

200

250

300

350

fc[N/mm

]

-[m/m]

UHPC without steel fibers - not confined

UHPC with steel fibers - not confined

UHPC without steel fibers - confined

UHPC with steel fibers - confined

C25/30

Figure 2.7Ductility of ultra-high-performance concrete with fibres and confinement and cylinder failure

mode

The systems used in earthquake resistant design should have ductility since there is

always the risk of stronger action that will lead them to the inelastic region no matter

how strong we are going to make them. Then the increased energy dissipation ability

offered by ductility will protect the system from damages and collapse. There is no

reason to use even ultra-high strength components since they exhibit brittle

behaviour.

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Reinforced concrete exhibits ductile behaviour when:

- its longitudinal reinforcement is low (under-reinforced).

- it is over-reinforced transversely (stirrups).

- it sustains limited compressive stress.

- it contains everywhere the minimum constructional reinforcement to be protected

against tensile / shear brittle failure.

- it is confined and has compressive reinforcement in highly compressed zones and

in plastic hinges regions.

- it is appropriately designed as a system, i.e. does not receive major concentrated

deformations in particular positions (ground floor pilotis, short columns).

- reinforcements are sufficiently anchored to prevent loss of concrete cover or

spalling at these positions.

Finally it should be emphasized that current understanding of earthquake resistant

design of structures is the assurance of the ductile behaviour of the system through

the definition of appropriate positions where inelastic deformations will occur. This

method is called capacity design and comprises the basis of many modern

earthquake resistant design Codes (e.g. EC 8, EAK 2003).

1600

-1040

400

0 20

200

1000

1200

1400

600

800

S500N

1800

2000

Prestress strands

S500H

120

S420

1008060

Fe360

Fe510

160 180140

S220

Bolts 8.8

Prestress bars

Prestress wires

260

220

10200

C16/20 (With confinement).2.9

Figure 2.8Steel under tension

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2.1 Ductility

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Figure 2.8 illustrates simplified bilinear curves of structural steel, reinforcement steel

and prestressed steel of various quality classes, where relations of elasticity, strength

and ductility under tension are shown. For comparison reasons the concrete under

compression curves of Fig. 2.9 are plotted in the same scale.

C16/20 ( With confinement)

C16/20

40 2

20

MB(90)MB(45)

50

106 8-10

Ductal

C100/115

120

C30/37

60

80

100

40

-(MPa)

180

200

140

160

Figure2.9Concrete and brick wall (MB) under compression

Figure 2.9 illustrates simplified brick wall (MB) bilinear curves under various stress

directions and various quality classes of concrete curves where relations of elasticity,

strength and ductility under compression are shown.

Comparing diagrams 2.8 and 2.9 results that steel ductility under tension andcompression is a multiple of concrete ductility under compression. Concrete in

tension has practically no ductility. Particularly reinforcement steel exhibits a plastic

deformation between 25 to 100 while unconfined concrete does not practically

exceed 4 and this only under compression. If confinement is applied the internal

deformation mechanism of concrete transforms from spalling mechanism to a friction

one and the plastic deformation under compression may reach 50, i.e. increase

more than ten times. In the event of cyclic loading concrete does not return to the

initial condition since it doesnt deform plastically under tension. If it contains

reinforcement it cracks during the reversal of loading and energy dissipation is

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94

reduced with the progress of cycles due to the continuously declining contribution of

concrete.

Table 2.1Indicative property values of materials used in Fig. 2.8, 2.9 and 2.10.

Material (GPa)fy or fc

(MPa)u ()

1 (45

) brick wall 4.5 2.40 2

2 (90

) brick wall 4.5 8 2

3 C16/20 28 16 3.5

4 C16/20 (With confinement) 32 23 50

5 C30/37 32 30 3.5

6 C100/115 45 100 3

7 Ductal 58 200 5

8 S220 200 220 180

9 Fe360 200 235 260

10 Fe510 200 355 220

11 S420 200 420 100

12 S500H 200 500 50

13 S500N 200 500 25

14 bolts 8.8 200 640 120

15 Prestress bars 200 940 50

16 Prestress wires 200 1390 70

17 Prestress strands 200 1600 60

18 GFRP 50 1500 30

19 AFRP 96 2200 24

20 CFRP 175 2800 16

Figure 2.10 illustrates simplified linear curves of fibre-reinforced polymers with

carbon-fibres (CFRP), aramid (AFRP) and glass (GFRP) where the relation between

elasticity and tensile strength is shown. In the same diagram bilinear curves of

structural steel and reinforcement steel are plotted (in the same scale) for

comparison reasons.

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2.1 Ductility

95

50

800

16

200

600

400

12840

Fe360

S500H

24 28 3220 4836 4440

AFRP

2000

1400

1200

1800

1600

1000

3000

(MPa)

2400

2600

2800

2200

CFRP

GFRP

260

52

Figure 2.10Fibre-reinforced polymers

2.2 Application of plasticity theory to reinforced concrete

The design of a structure aims the resistance to the specified in Codes service loads

with sufficient safety against potential failure. The global safety factor is expressed as

a product of load and materials factors. Considering that service loads assumptionsof Codes are generally conservative the conclusion is that structures have adequate

strength reserves under the actually imposed loads. Other factors producing strength

reserves are members design that frequently is based on stiffness criteria

(deformation, deflection) or architectural ones (binding dimensions). Strength

reserves also have as origin the fact that members design is based on envelopes of

elastic stresses from different loadings. Thus maximum stress in various cross-

sections does not necessarily result from the same loading and therefore do not

occur simultaneously. Statically overdetermined structures have ample strength

reserves in adjoining members that are activated when overstressed and contributeto the increase of bearing capacity. What is frequently said that concrete is

generous is therefore directly related also to ductility that allows stress redistribution

among members..

The function of ductility in dynamic loadings is to dissipate energy preventing failure

of the structure that is actually avoiding collapse compensated with some residual

deformations. Efforts to interpret experimental results and understand the function of

ductility leaded early enough (the interwar years) to the development of the theory of

plasticity. Theory of plasticity is today fully documented theoretically and

experimentally and is the unique and basic means for the estimation of the maximum

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96

(ultimate) strength of a structure and of a structural system generally. Its basic

differences from the theory of elasticity are illustrated in Figure 2.11.

M Elastic stiffness M

Plastic strength

My

My: Yield moment

K : Stiffness

Theory of elasticity is based on stiffness

relations

Theory of plasticity is based on strength

relations

Figure

2.11Elastic stiffness and plastic strength

Therefore is obvious that the most basic factor of the material performance is the

available ductility. It is also essential for the interpretation of the strength of structures

under static loads and understanding the performance of structures under dynamic

stress. Historically academic studies commence with elasticity, the Hookes law and

the explicit elastic analysis of overdetermined structures. Maybe in the future studies

will start from plasticity and the essence of static and dynamic loads with emphasis

on non-linear and dynamic effects in order to understand the real performance ofstructures. Calculations may be simpler in theory of elasticity but our aim should be

the deeper understanding of natural phenomena. It is the only way to get to synthesis

and through this to the solution of the practical problems.

Computers today are the most modern analysis tool. Decisions concerning geometry

and materials that will be used for the design of a project are products of a synthetic

procedure demanding knowledge, experience and imagination. The basis for

understanding the performance of a structure under increasing load is theory of

plasticity or better elasto-plasticity combining both theories. In the following a fixed on

both ends beam under uniform load will be analyzed elastically and elasto-plastically

(Fig. 2.12). Under load q fixed-end moment reaches yield limit. The load may be

further increased by 33% to reach plastic strength of the beam. At their ends occurs a

residual plastic rotation.

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97

DZ=-1.04

-0.65%

-1.58

5.17

-10.42

q 1.33q

'Rotations'

'Moments'

after stress

redistribation

mmmmDZ=-2.78

%

kNm

kNm

kNm-10.42

10.35 kNm

plastic

rotation

hinges

residual

plasticrotation

Figure 2.12Elastic and elastoplastic analysis of a fixed on both ends beam

Theory of plasticity is strength and not stiffness based like theory of elasticity. Many

reasonable and useful practical conclusions resulted from this theory and these are

the following:

1. upper bound theorem: if a plastic deformation mechanism exists satisfying the

compatibility of deformations and yield conditions then corresponding load is an

upper bound for the strength of the structure (kinematic theorem)

2. lower bound theorem: if an internal forces distribution exists satisfying

everywhere the equilibrium conditions not exceeding anywhere the structuralmembers strength then the corresponding load is a lower bound for the strength

of the structure (static theorem).

3. Eigen stresses(temperature, prestressing) do not affect the overall strength of

the structure since sufficient ductility is available.

4. addition of new members or strengthening of the existing ones never

decreases the strength of the structure. The resistance of the strengthened

structure is not affected by the fact that strengthening took place later on pre-

loaded members. Essential condition though is the sufficient ductility and the

insignificant increase of masses.

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98

5. imposed displacements of supports (settlements etc.) do not decrease the

strength of the structure but simply result to residual deformations, assuming

these inelastic deformations vary within acceptable limits that do not result toannihilation of ductility and lead to failure and collapse.

6. under a dynamic action (earthquake) on the structure, certain inelastic

deformations occur after the first cycles. After its readjustment and if the new

elastic limits allow it the structure will elastically vibrate within these limits without

any further plastification (shakedown).

The overall strength of a system under static load, such as gravity loads, should have

sufficient safety margins to prevent yield in several positions and form a mechanism.In a static loading the load is a permanent force of constant value and direction, time-

independent unlike seismic action that is a cyclic imposed displacement.

If a system carrying gravity loads is subjected at the same time to a dynamic stress

(earthquake) its design should assure that members carrying significant gravity loads

will not plastify and if they plastify to have sufficient ductility margins without suffering

strength loss.

The last half of the century the estimation of the ultimate strength of reinforced

concrete structures is based on adequate truss models representing simply internalforces equilibrium models. Concrete undertakes the function of struts and

reinforcement the function of ties. The scope of design is the selection of

reinforcement cross-sections that will fully cover the tensile forces. Usually during this

procedure neither stresses nor deformations of concrete are checked. The selection

of the exact position of the resultant force of the strut is usually based on empiric

rules (see Figure 2.13).

Q

Q/2 Q/2 fc

fc

fc fc

fc

d z

Figure 2.13 Struts and ties models (wire frame models or models with thickness for struts)

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Nowadays we know that this method is based on lower bound theorem of the theory

of plasticity (static theorem). The validity though of this theorem depends exclusively

on the sufficient ductility of concrete.

Thuerlimann and his associates in Zurich [9, 10, 11, 12, 1, 29, 32] proved during the

period between 65 and 85 after systematic experimental research that in most cases

of practice, up to a rather high (longitudinal) reinforcement ratio, the experimentally

measured strength is really higher than the theoretical lower bound of theory of

plasticity. These cases were described as under-reinforced unlike the over-

reinforced where concrete fails in a brittle mode at a level lower than the limit

strength according to the theory of plasticity. The question is what is the maximum

reinforcement that distinct under-reinforced and over-reinforced members in

practice? From what factors it depends? What is the available ductility of under-

reinforced members and how could it be increased? What proportion of the materials

always leads to the optimal performance?

A method widely used for bending with axial force was based on the Bernoulli

hypothesis (plane sections remain plane after bending). The stress-strain curve of

concrete is non-linear (parabolic-linear) and the one of steel elastic ideally plastic.

Failure criterion for concrete is assumed the maximum contraction of the edge fiber

of compression zone. Thus two distinct cases result:

- concrete fracture after yielding of reinforcement and since the steel experienced

large deformation (ductile behaviour).

- early concrete fracture before yielding of reinforcement (brittle behaviour).

In the reality the Bernoulli hypothesis is not verified due to the evident flexural cracks

and the diagonal cracks in the event of flexural shear. Besides this model considers

only the longitudinal and not the transversal deformation of concrete that reaches

significant values in inelastic region and is critical for the strength since it affects the

deformation mechanism of concrete.

Anyway this method introduced the concept of the limitation of inelastic deformations

and exhibits satisfactory correspondence with experimental results. It offers though

no physical explanation for the concrete failure. The physical explanation may result

through the consideration of descending branches and the concentration / increase

of inelastic deformation on a thin zone where the ductility is exhausted (material

instability). Another form of stable inelastic behaviour of concrete under compression

is the one occurring under confinement or generally under lateral compression. Due

to lateral compression the deformation mechanism is transformed from a spalling

mechanism to a slip (friction) mechanism as we will see in the following sections,

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which is stable and demonstrates significant remaining strength even for

contractions of approximately 4%, i.e. ten times more than that without confinement.

2.3 Inelastic deformation and fracture mechanism of reinforced concrete

under compression

The behaviour of concrete under tension is known that is brittle unless it is fibre-

reinforced. Therefore in the event of tension the failure criterion is based on the limit

stress. For the case of inelastic deformation under compression extensive

experimental research has taken place in the past where frequently also transversal

deformations have been measured even for high values of deformation.

Characteristic are the experiments of Stckl [14] on cylindrical specimens with 150

mm diameter and 600 mm height under axial compression (see Figure 2.14) with or

without spiral confinement reinforcement. The spiral reinforcement was rather dense

with a pitch of 25 mm and sections diameter 5 mm. In the case of confinement the

transversal deformations of concrete were measured at the position of reinforcement

(qs) and at the region between consecutive reinforcements (qc) as well.

150

l

qsqc l qc

qs

l qc qs

300

62.5

62.5

100

100

300

300

300

150

300

300

100

100

l q

ql

q

l

Figure 2.14Specimens of Stckl tests with and without confinement

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2.3 Inelastic deformation and fracture mechanism of reinforced concrete under compression

101

0 10 20 30 40

10

20

-310

q

10

SpallingSlip

qcqs

Figure 2.15Transverse deformation of concrete with confinement vs. longitudinal contraction

q

10

0

10

-310642

Spalling

Slip

8

20

Figure 2.16Transverse deformation of concrete without confinement vs. longitudinal contraction

Observing experimental record lead to the conclusion that when transverse

expansion is constrained (see Figure 2.15) concrete demonstrates a completely

different inelastic deformation mechanism than that of when its laterally free (see

Figure 2.16). The ratio of transverse deformation to longitudinal contraction is in the

first case much lower.

For a better understanding of the effect two different deformation possibilities of the

internal structure of concrete are initially examined based on a model (Fig. 2.17 and

2.18). This consists of 4 grains of aggregates lying on the vertices of a rhomb and

the matrix of the bonding mortar (cement paste) filling the intermediate space which

is in charge of the equilibrium of the internal forces of the system. The inclination of

the sides of the rhomb is assumed to be everywhere 2:1.

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102

Forces

Slip lines

Figure 2.17Mechanical model of bond-slip mechanism

Forces

Spalling

Cracks

Figure 2.18Mechanical model of spalling mechanism

In the presence of lateral restraint (confinement) a slip mechanism occurs (Fig.

2.17). In the slip interfaces occurs only slide and no dilatancy therefore the volume

remains constant.

0V

dV= (2.1)

0ddd321

=++

1dd

d

21

3 =+

(2.2)

Without lateral restraint a spalling mechanism occurs (Fig. 2.18). The relationships

between deformations are derived as follows:

4

1

d

d

2

1

2

1

AB/A'A

CB/'CC

1

3 =

= (d2= 0)

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2.3 Inelastic deformation and fracture mechanism of reinforced concrete under compression

103

Superimposing the components of deformation for the other transverse direction

results:

4

1

dd

d

21

3 ==+

(d20) (2.3)

3

1

CC''

C'

B

A''

A' A

Figure 2.19Mechanical model of deformation of a spalling mechanism

The dilatancy results as:

0d3dddV

dV3321 >=++= (2.4)

Therefore spalling mechanism exhibits a significant expansion in the volume of the

material.

In the experimental results diagrams of Figures 2.15 and 2.16 the lines resulting from

the sliding and spalling models are plotted as well. The agreement in the case of

sliding (Fig. 2.15) is good enough for values of deformation up to q= 20 and l=

40. The longitudinal elastic deformation of concrete of approximately 3el= 1 is

neglected.

In the second diagram (Fig. 2.16) without confinement the longitudinal elastic

deformation (compressive strain) of 3el= 1 was considered too. We may observe

that spalling mechanism based on the model describes satisfactorily the effect up to

a value of the transverse strain of q= 4. In the following chapters will be proved

that around this value the effect of concentration of deformations in a thin zone starts

therefore further measurements are not reliable due to inhomogenity of the field ofdeformations.

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2.4 Triaxial stress and residual strength

In the following concrete strength is examined. The concrete strength of cylindrical

specimens under uniaxial compression is expressed as fc. Under triaxial compression

(Fig. 2.20) we observe that the presence of the transverse compressive stress2=1

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2.4 Triaxial stress and residual strength

105

dF1

dF3

2

1a3

a1

3

1

Figure 2.22Mechanical model for the triaxial strength of concrete

4d

d

dFa

dFa2

dF

dF,2

a

a

1

3

11

33

1

3

1

3 ==

== , (2.5)

Above expression (2.5) shows that longitudinal resistance increases by the quadruple

(in absolute values) of the lower external transverse compression.

However, except external transverse forces there are internal forces as well bonding

the aggregate grains, i.e. intermolecular attractive forces, bonding forces responsiblealso for uniaxial strength (Fig. 2.23).

Transverse

compression

Resistance

to compression

Cohesion

3

1

Figure 2.23Mechanical model for the influence of cohesion and transverse compression to the

compressive strength of concrete

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106

With the increase of transverse deformation concrete microstructure experiences

successive tension failures. This leads to a decrease of the total cohesion stress and

correspondingly to the residual stress fc* (Fig. 2.24). In this occasion we talk about

concrete strength loss (descending branch). Important is that the strength component

3 due to transverse compression is not decreased by the increase of the

transversal deformation.

The residual strength under uniaxial compression fc* depends on the maximum

transverse deformation maxq=max(1,2). Thus triaxial compressive strength will be

derived by an expression of the following form:

3q

*

c3 )(maxf += (2.6)

where 0)(maxf q*

c >

To model the residual strength of concrete following simplifications are assumed:

The strength loss is assumed to occur abruptly when the maximum transverse

deformation reaches the value qR=4 (Fig. 2.24). Until then the residual strength of

concrete under compression is assumed to be equal to the strength of the cylindrical

specimen fc. After the loss of strength the residual strength is assumed to be constant

with a value of fc/2 up to the transverse deformation of qo=20. These values are

documented in the next chapters through experimental results.

0 4=qR

0.5

fc*/f

c

1.0

20=qo

-q10

Figure 2.24Assumption for the residual strength of concrete in relation to the transverse deformation

It remains the modeling of the elastic part of deformations.

It is assumed that concrete behaves linear-elastically up to a compressive strain of

cel=1 when inelastic deformation begins. Thus the elasticity modulus c is

approximated in relation to the compressive strength of cylinder as:

ccelcc f1000/fE == (2.7)

The transverse elastic deformation in elastic region is assumed to be equal to zero.

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2.5 Concentration of deformations and fracture

107


Various researchers tried occasionally to measure the inelastic deformations of

concrete on compressed specimens. The difficulty is that these vary from region to

region after the occurrence of the loss of strength (descending branch).

Roy and Sozen in their tests separated the middle part of the prismatic concrete

specimen in two regions for measurements. Through imposed deformation they

managed to trace even the descending branch of the F-curve (Fig. 2.25).

Since the strength loss begins regions 2 and 3 do not any longer demonstrate the

same deformation behaviour (Fig. 2.25). While region 3 exhibits further increase of

the compressive strain, in region 2 takes place a decrease of the compressive strain

(elastic unloading). From the side of mechanical behaviour we may say that inelastic

deformation of the specimen was concentrated in a region with limited dimensions(region 3). In such cases we discuss about localization of deformations. The energy

quantity consumed in this region for the further disruption of molecular bonds comes

partly from the external force and partly from the elastic energy stored in the rest part

of the specimen as we may conclude from the observed elastic discharge in region 2.

0 10 20 30

-10

40

F

50

2 1

3

F

123

Figure 2.25Roy and Sozen tests where the differentiation of deformations in regions 2 and 3 were

measured [13]

Based on a simple bar model (Fig. 2.26) we may show that strength loss constitutes

a necessary condition for the concentration of deformations.

equilibrium: 32 dd =

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108

material law: 3t32e2 dEd,dd ==

substituting: 3t2e dEdE =

compatibility condition: ( ) +== tdtld0dl 32

tl

t

d

d

3

2

=

Substituting :

tl

t

E

e

t

= (2.8)

in order to get t>0, it should be Et

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109

DC

EDtan = (2)

(1), (2) 1

3

1

32

tan

tan ==

and since2

1

4

1tan

4

1

d

d

1

3 === (2.9)

/2

C

D

3

1

F

-3

D

CA

B

2=0

1

d3d1

=41-3 1

2

3

Figure 2.27 Deformations discontinuity plane of the spalling mechanism

Therefore Region 3 where the increase of inelastic deformations occur may belimited by two close parallel deformation discontinuity planes with an inclination of

(equation 2.9) to the direction of compressive stress.

/2

C

2=0

3

2

2

-3

3

3

1

t 0

Figure 2.28Concentration of inelastic deformations zone of the spalling mechanism

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110

O 4

0.5

-3

/fc

1.0

20

A B

C E

FD

2 : O-A-B-C-D

3 : O-A-B-C-E-F

110

Figure 2.29Strength loss diagram

Before the initial strength loss the whole body is governed by homogeneous strain 2

(Fig. 2.28). With the beginning of the strength loss in point B (Fig. 2.29) the inelastic

deformations increase abruptly in a thin zone (zone 3) while in the rest regions

remain constant. Thus, within this zone of concentration of deformations the total

ductility of the material is exhausted.

The slope of the strain discontinuity line is dictated by the proportion between the

inelastic deformation components )d/d( 13 .

Experimental results prove that the concentration of deformations occurs only in the

spalling mechanism. Such effect does not take place in the slide mechanism.

2.6 Confinement through transverse reinforcement

The ability of concrete to sustain inelastic deformations is of utmost practical

importance. It is a decisive factor for the ultimate strength of the structure and the

further performance of a reinforced concrete structure under seismic loads as well.

Structures with the ability of plastic deformation may undertake seismic or impactloads without a failure risk. Even non-uniform settlements under above mentioned

conditions do not lead to failure.

In the preceding chapter it was stated that when concretes deformation occurs

through a slide mechanism it possesses the ability for large inelastic deformations.

The deformations of the spalling mechanism are concentrated in a thin zone and lead

after a relatively small inelastic compressive strain to failure.

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111

To increase the inelastic deformation ability in practice an adequate reinforcement is

used consisting of closed stirrups. Through the restraint of transversal strain concrete

is deformed inelastically in the event of a slide (friction) mechanism. Simultaneously

the transverse compression contributes to the increase of the concretes resistance

to compression in the longitudinal direction.

The contribution of the transverse reinforcement (stirrups or circular spirals) to the

resistance of concrete is not easy to estimate. The reason is that the stress field due

to the imposed concentrated forces is complex enough and not uniform (Fig. 2.30).

The material performance in inelastic region is also complex since a descending

branch occurs (loss of strength).

Stirrup

Figure 2.30Trus model diagram of the flow of compressive forces in a confined with ties (stirrups)

square-sectioned prism

In the following, the way the confinement through tie reinforcement (stirrups) acts will

be investigated using simple stress fields in the inner of the body.

In the angles of stirrups concentrated forces are imposed to concrete directed to the

inner of its body. These forces cause a deviation of the longitudinal compressive

forces flow field towards the inner of the body. A transverse compression is resulting

there that deviates for a second time the flow of the field towards the outer surface to

meet the next stirrup. Thus the region between two consecutive stirrups is partially

under tri-axial compression (Fig. 2.31).

The truss of Fig. 2.30 represents the axes of the resultant compressive forces.

Transforming the truss to a stress field we may estimate the concrete stresses (Fig.

2.31). Concrete is imposed horizontally to the tie (stirrups) forces and vertically to theload resultants.

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112

Tie

forces

3

1

2

Load

s

d

Figure 2.31Stress field for the assessment of confinement on a square-sectioned prism with stirrups

It is important to observe (Fig. 2.31) that the tri-axial stress in the internal of a body is

caused through the deviation of the uni- or bi-axial stress field.

2.6.1 Prismatic members with confinement

From the stress field of Fig. 2.31 we may verify that the distance between

consecutive stirrups (spacing) does not affect the extent of the region under tri-axial

stress. Critical factor for the extent is the application of the horizontal forces of the

stirrups. In cylindrical test specimens where horizontal force is applied uniformly the

dimensions of the tri-axially stressed region are increased when the spacing of ties is

decreased.

Concrete strength in the bi-axially stressed region longitudinally is limited under large

inelastic deformations to the value of the residual strength (fc/2). In the transverse

direction the strength loss is expected to be less but even there the strength may be

conservatively assumed for practical applications equal to the residual strength fc/2.

For the interpretation of test results the value of the residual strength transversally

may reach the value fc. The influence of these two stress limitations to the

development of the maximum possible transversal compression may be evaluated as

follows:

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113

D

P

A

K1

3

M

'

1

3d

s

'

Figure 2.32Stress field on the external surface of the square-sectioned prism between two

consecutive confinement ties

The ratio of the principal stresses in the bi-axially stressed region (Fig. 2.32) is

expressed by the following equation:

( )

2

2

2

2

3

1

s2

dtan

cos'

sin'

==

= (2.10)

Critical for the verification of stresses is point K where stresses are double than these

of centre M.

( )

2c

23M

qc

3s

d

16

f

s2

d

2

2

f

=

== (2.11)

4

f

2

2

f c

1M

qc

1 === (2.12)

The two limitations are intersecting for:

2

1

d

s= (2.13)

Previous expression indicates that the maximum possible transversal compression

may not further increase for tie spacing s

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114

d

Ac=0.20Ac=0.20d

Figure 2.33Tri-axially stressed region in a square-sectioned prism

For the assessment of stresses similar fields may be used as in previous example.

The limitations formulae derived are also here valid if d is the distance between two

consecutive points where tie forces are applied.

Square with internal stirrup: 56.0A/ cc = (2.15)

Hexagon: 40.0A/ cc = (2.16)

Ac/Ac=0.56 Ac/Ac=0.40

d

d

d

d

Figure 2.34Tri-axially stressed region for various stirrup arrangements

The mean resistance 3 of a prismatic member is expressed by the following

equation:

c

slsl

c

cq

c3

A

Af

A

4

2

f

++= (2.17)

where the last term corresponds to the contribution of longitudinal reinforcement.

For the square-sectioned prism with square stirrups from the equilibrium of

transverse forces results:

ds

fA sqsqq

= (2.18)

Through replacement results the dimensionless form:

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115

lq

c

3 8.05.0f

++= (2.19)

wherec

sqsq

q

fds

fA

= (2.20)

and

cc

slsll

fA

fA

= (2.21)

while previously derived stress limitations are also valid:

if 50.0d

s> :

2

qs

d

16

1

< (2.22)

if 50.0d

s< :

25.0q< (2.23)

Above relations may be represented graphically (Fig. 2.35).

0 .1 .4 .6

s/d

q

.2 .3 .5 .7

.05

.10

.15

.20

.25

Figure 2.35Maximum mechanical confinement ratio relation to the stirrups spacing in a square cross-

section

Practical conclusions are :

- confinement through polygonal stirrups is more effective when lateral force

application points are closer. This results to the increase of the tri-axially stressed

region in the internal of the body. For this reason frequently multiple ties are used

(external stirrups with internal) or even polygonal ties (with 6-8 vertices) in

corresponding column cross-sections.

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116

- lateral pressure may not exceed the limit of q=0.25 (equation 2.23) since bi-axially

stressed regions of the external surface of the prism fail. Therefore the maximum

contribution (increase) of the tri-axial stress due to the compressive resistance for a

square-sectioned prism is 3/f

c= 0.800.25=0.20.

- if the distance between two consecutive ties (spacing) is greater than d/2 then

maximum limit for qis (d/s)2/16 (equation 2.22), i.e. for s/d=0.7 the maximum value

of qis 0.128 and the maximum possible contribution of the tri-axial stress effect is

3/fc=0.800.128=0.10. Therefore the distance between consecutive stirrups

should be less than d/2.

-The value of maxqis practically already very high. E.g. if fs/fc=20 then:

%25.1

20

25.0

20

f

f

q

s

cqs ====

if sd=815=120 cm2then:

As=s120=1.50 cm2,

i.e. 14/8 stirrups with 15 cm leg distance in plan view.

The contribution of tri-axial stress effect will be 3/fc=0.800.25=0.20. Therefore the

main part of the resistance (0.50.fc) is owed to the residual strength when slide

mechanism is developed.

Finally the longitudinal reinforcement contribution for a ratio of 1% is:

20.0f

f%1

f

c

s

c

3 ==

The additional contribution of longitudinal reinforcement lies though on its stabilizing

action due to the lateral restraint of concrete in angles.

2.6.2 Cylindrical members with confinement

In the following stress fields in cylindrical test specimens confined with circular

stirrups are discussed. Unlike square-sectioned prisms, critical for the extent of the

tri-axially stressed region (Fig. 2.36) of cylindrical members is the distance between

(spacing) consecutive stirrups (pitch). The area of the region is increased with the

decrease of spacing.

The transverse stress qresults from the equilibrium:

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117

4

ssd

fA

2

sqsq

q

= (2.24)

3

d

sP

4q

q

Figure 2.36 Stress field for the assessment of confinement in a cylindrical test specimen with stirrups

The ratio of the area of the tri-axially stressed region to the total area is:

( )[ ] 22c

c )d/s1(dsdA

== (2.25)

From the previous equation results the dimensionless mean resistance (strength)

l

2

q

c

3 )d4(s1

)ds1(45.0

f

+

+= (2.26)

where q, lare the same expressions used in equations 2.20 and 2.21.

Critical for the verification of stresses is point P (Fig. 2.36) where stress field is uni-

axial. This is the reason why transverse compression may not exceed the residual

stress. The maximum transverse reinforcement ratio for square-sectioned prisms

may be consequently calculated as follows:

( )d4s1f

2

f

cqcq

== (2.27)

therefore:

( )d8s5.0q < (2.28)

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118

The total mean resistance (strength) of prismatic and cylindrical test specimens is

graphically represented in relation to the transverse reinforcement for various values

of stirrups spacing ratio s/d (1/2, 1/4 and 1/8). In the diagram of Figure 2.38 stress

limitations are also considered while the contribution of longitudinal reinforcement is

not taken into account.

0 .1 .4 .6

s/d

.2 .3 .5 .7

.2

.1

.5

.4

.3

q

Figure 2.37Maximum mechanical confinement ratio stirrups spacing relation in cylindrical test

specimens

Figure 2.38 illustrates also that the resistance (strength) increase due to tri-axial

stress in cylindrical specimens with the same transverse reinforcement ratio q

depends from the s/d ratio as well.

.5

1.0

Residual Strength f c / 2

fc

1/2

1/2

0 .1 .2 .3

q

.4

3

/fc

.5

.4

.8

1.2

1.6

2.0

2.4

1/2

1/2

1/4

1/8

Figure 2.38Strength of prismatic and cylindrical test specimens confined with stirrups

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119

General conclusions are: concrete compressive resistance (strength) in large

deformations results partially from its residual strength and partially from the

contribution of tri-axial stress. Through the proposed models quantitative indications

for these two effects may result. For the experimental confirmation square-sectioned

specimen tests were evaluated and the contribution of the residual strength varied

between 80 to 90% of the total strength, while in cylindrical specimens tested this

ratio was approximately 45%. In both cases agreement with experimental results was

very good. A further modeling confirmation is the verification of stress limitations

resulting to the stirrup spacing (s/d) and quantity limitation. When above restrictive

conditions where not followed concrete ductility proved to be limited. Experimentally

has been proved that the minimum transverse confinement reinforcement for the

occurrence of a slide (friction) mechanism is approximately q=0.05.

2.7 Deformation and fracture of the compression zone

In the next chapters will be discussed problems where occurrence of strain

concentration leads to brittle fracture of concrete (without residual strength). The

interest will be concentrated on the detection of cases where a member fails before

the development of its full flexural strength (longitudinal reinforcement yield). In these

cases there is no ductility available and they should be treated with special care

because:

- they lead to brittle failure without warning that is dangerous for the safety of the

structure.

- Plasticity theory for the calculation of ultimate strength is not applicable.

- Elastoplastic analysis used for the verification of the structural performance under

earthquake (pushover) is not applicable.

- Measures should be taken during the design or re-design (strengthening) of the

structure to avoid such dangerous for the safety of the structure cases.

At this point it should be emphasized that brittle failure modes should not only beavoided but sufficient ductility should be assured beyond flexural ultimate strength.

Conditions to assure sufficient ductility are described in aseismic design codes (EAK

2003) and in re-design interventions recommendations ( 2005), based

mainly on empiric relationships derived from the evaluation of experimental results. In

cases of verification of existing structures the detection of probable occurrences of

brittle failure is crucial in order to take all necessary structural strengthening

measures. The typical case of strain concentration in concrete struts is discussed in a

previous chapter, where the assumption was formulated that this concentration

occurs under a critical value of the transverse strain. Furthermore it was pointed out

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120

that sufficient condition for its occurrence is that inelastic deformation mechanism of

concrete should be a spalling mechanism and not a slide (friction) mechanism, i.e.

without the presence of transverse compression through confinement.

Strut and tie models to study the mechanical behaviour of reinforced concrete iscommonly known that are used more than half a century. The adequacy of these

models is based on the following facts:

- concrete tensile strength is 10 to 20 times less than its compressive strength. After

the occurrence of tensile failures (cracked state) the reinstalment of a tensile forces

field in concrete is impossible. Therefore compression is not distributed and simple

struts with practically linear stress trajectories are formed (Fig. 2.39).

c

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121

fc-NR -NRa

b

d/2

d/2x/2x/2

Figure 2.40Prisms with rectangular cross-section under eccentric compression according Rsch [19]

d

x

d

a21

fdb

Nnorfb2a

2

dN

c

RRcR ==

=

=

This assumption may be verified from the tests (Fig. 2.41). Strength value fc varied

between 10 and 57 N/mm2.

.5

0

1.0

.1 .2 .3

1-x/d

.4 .5

-nR

1.5

Figure 2.41Experimental verification of strut strength assuming constant fcstrength

Displacements of a reinforced concrete structure may be estimated with sufficient

accuracy using truss models. The distribution of inelastic deformations and therefore

the failure mode of a concrete strut depend directly from the form of cracking of the

member. The analysis of this problem may be based on the examination of two

different effects:

1) converging cracks or theory of the failure angle.

2) dense parallel cracks or theory of curvature.

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122

2.7.1 Theory of the bend angle

The first case refers to components where either only one crack is formed (un-

reinforced components, components without bonded reinforcement, Fig. 2.42) or

more converging cracks (beams, slabs in the vicinity of direct supports, Fig. 2.65).

Concretes inelastic deformation occurs then locally in the front of cracks where a

concentrated change of angle or bend takes place.

A simplified assumption is the formation of a homogeneous deformed region

separated from the neighboring undeformed regions through deformation

discontinuity planes (Fig. 2.42).

2=0

1

-3

x

a

fc

3

1 2

Figure 2.42Theory of the strut bend angle

Inclination of discontinuity planes is related to the ratio of the strain components

(equation 2.9):

( ) 2/113 /tan =

The bend angle may be calculated as follows:

A'

AA''

B'

B

/2

C

31

a

x

Figure 2.43Homogeneous deformation region in bend angle theory

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123

From the similar triangles (Fig. 2.43) results:

1x''A'A,2/a

''A'A

AC

'AA

2

'AA

''A'A

AC2

a==== ,

1ax4 = and with

2tan

ax =

results:

2/1

1

311

2tan2

== (2.29)

When the maximum inelastic transverse tensile strain 1reaches the value of 4 a

strain concentration in a thin zone II (Fig. 2.44) occurs leading to brittle failure without

residual strength (Fig. 2.45). Through the previous equations the corresponding

ultimate bend angle of the strut is calculated:

42

142

4

12

2/1

qRR ==

= (2.30)

3

2

2=0

c

R

I II III

1

x

t 0

Figure 2.44Inelastic deformations concentration zone according to the bend angle theory

0 4

0.5

c

/fc

1.0

20110

A C

D F

GE

I : A-C-D-EII : A-C-D-F-G

III : A-H-B

H

B

Figure 2.45Strength loss diagram

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124

It should be emphasized that in this case inelastic deformation reaches at once all

over the depth of the compression zone x the critical value of qR=4. This leads to a

sudden formation of the whole deformations concentration zone II and therefore to

the complete failure of the strut. This exceptionally brittle failure mode may be also

verified from the corresponding tests.

2.7.2 2.7.2 Curvature theory

The second case refers to components exhibiting dense parallel cracks practically

perpendicular to the compression zone axis (Fig. 2.46). Typical cases are reinforced

concrete members with good bonded longitudinal reinforcement under pure bending

with or without compressive stress.

fc

3

12

l

Figure 2.46Strut curvature theory

Inelastic deformations of concrete occur in this case not locally but practically

uniformly distributed along the compression zone. Compression zone is subjected to

a uniformly distributed angle change or a constant curvature /l (Fig. 2.47).

Distribution of inelastic deformations along the depth of the compression zone may

be described accepting the Bernoulli hypothesis (plane sections remain plane after

bending) due to the relatively dense cracking.

Strain components are linearly distributed along the compression zone depth:

11

3 xl

r

x == (2.31)

where the curvature is:

l

r

1= (2.32)

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125

1

3 2

1

x

b

x1 x1

b b

r/2

r

l

Figure 2.47 Compression zone deformations in accordance with the strut curvature theory

When the spalling mechanism is activated (equation 2.3):

( ) 4 213 += and 1=2

and replacing 3in equation 2.31 results:

121 xl

2 == (2.33)

When the maximum transversal strain 1=2of the compression zone occurring in the

upper boundary of the cross-section reaches the critical value qR=4 then splitting

of the consecutive layers starts through spalling mechanism. Fracture energy spent

comes partially from the elastic unloading of the upper boundary fibres of the

compression zone (Fig. 2.48).

II

1

3

c

1>qR

1

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126

Such splitting effects in compression zone may be observed in corresponding tests.

These effects result to a strength loss of the compression zone represented in the

descending branch of the moment-curvature curve. It should be emphasized that in

this case failure is not sudden and complete as in the first case of converging cracks

but is expressed as a gradual strength loss. This case is more interesting to assure

ductility in the design of reinforced concrete members. Curvature at the beginning of

failure is expressed using conditions 1= qR= 4 and x1= x in equation 2.33:

x

2

x2

4

x2

l

qRR === (2.34)

The ultimate plastic rotation due to curvature for a length d (Fig. 2.49) is:

d/x

2d

l

R = (2.35)

0 .4 .6

(Rd

/l)1

0

.2

20

40

x/d

.8

30

10

Figure 2.49Maximum inelastic rotation for length d according to the curvature theory

When the section has compressive reinforcement and is sufficiently confined to

assure the slide (friction) mechanism the plastic rotation capacity increases

significantly.

2.8 Bending with axial force

In this Chapter analytical relations will be developed for the events when concrete

fails before the steel yields and therefore concrete exhibits a non-ductile behaviour.

Objective is the formulation of simple equations based on a reinforcement

exploitation factor . When this factor has values

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127

estimation of the ultimate strength while for

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128

Additional longitudinal reinforcement strain when concrete fails s is calculated as

follows:

( ) ( ) []1xdx

2xd

l

ces =+=

or []3)x/d2( s = (2.36)

Steel force Fswhen concrete fails consists of two parts:

ssps FFF +=

where:

Fsp: is the prestressing force for zero strain of the neighboring concrete and

Fs: the force increase due to the additional steel strain s

The reinforcement force increase is:

( )[ ]1000

AE3xd2AEF ssssss

== for s < sy

or sss fAF = s > sy (2.37)

where for prestressing reinforcement value fs corresponds to the additional stress

until reinforcement yields (Fig. 2.51). In dimensionless form:

( )[ ] ( )[ ] [],1/3x/d2f1000

E3xd2

fA

Fsysy

s

s

ss

s

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129

fsp

fs

sp

sp sy

Figure 2.51Stress-strain relation of prestressed reinforcement

Formulating equilibrium conditions:

Moments:

=

2

xdfbxM c

or dimensionless

==

d2x1

d

xmfdb

M

c

2 (2.41)

axial forces: cssp fbxFFN +=

or dimensionless

+

== c

c

c

s

c

sp

c fdb

fbx

fdb

F

fdb

Fn

fdb

N

nd

x

d

xn +=+= (2.42)

Failure mode verification:

Quantities n, , are given, assume =1 and calculate the x/d ratio:

nd

x+=

Verification of the validity of relation 2.38:

1

13

x

d2

sy

, where sy []

If

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130

For over-reinforced sections where

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131

F's N'

M'

+ =N

MF'sd'

(F's)'s

s

dd'

Figure 2.52Contribution of compression reinforcement to bending with axial force

The verification of the failure type of the section may be based on n and m as it was

presented up to now for sections without compression reinforcement. If on the

contrary the ultimate moment of the section should be verified then initially it is

assumed n=n-and the ultimate moment m results as the sum of the contribution of

the compression reinforcement to the value of m (without compression

reinforcement).

d

dmm

+=

The influence of early concrete rupture to the ultimate moment of a member

subjected to bending with axial force may be graphically represented as a relation of

the amount of tensile reinforcement (Fig. 2.53). Values on vertical axis represent the

ratio of the ultimate moment (with limited concrete compressive strain) to the plastic

moment in accordance with the theory of ideally plastic materials (with no limitation of

concrete strain) and on the horizontal axis the mechanical ratio of tensile

reinforcement. Axial force at failure state is used as parameter. For steel is assumed:

s= 2105/mm2, fs= 460 /mm

2

From Figure 2.53 we may verify that for uni-axial bending and reinforcement ratios up to approximately 0.4 (fs / fc 20, < 2%) critical is steel yield therefore some

ductility is available. In the presence of axial force nR 0.4 always critical isconcrete therefore the available ductility is limited. It should be emphasized that all

these are applicable when concrete compressive strain ductility is low (without

confinement).

The next diagram (Fig. 2.54) represents the favorable influence of prestress to the

resistant moment. This influence takes place only when concrete fails early before

the steel yields (over-reinforced members).

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132

0

.5

.5

1.0

mR

/mp

l

1.0

1.5

0.-.2-.4-.6-.8-1.

.2

.4.6

-.2 0. .2 .4 .6 .8 1.nR

Figure 2.53 Failuremoment to plastic moment relation for bending with axial force

For the untensioned steel the same assumptions are made, as in the previous

diagram (Fig. 2.53):

0 .5

+1.0

mR/mR(=0)

2.0

1.5

1.0

=.4

.2

0.

.6

.81.0

.4

Figure 2.54Influence of prestress on the failure moment ratio

We may verify from Figure 2.54 that for usual in practice values, i.e. for =0.1 to 0.4

the favourable influence of prestress to the resistant moment does not exceed 10%.

For under-reinforced members theoretical results are compared with 364

experimental ones (Fig. 2.55). Resistant moment is calculated theoretically from

equations m=0.9(approximate practical formula) resulting from equations 2.41 and

2.42 for n==0 and =1 and equation 2.46 as well.

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133

=

2

1m (2.46)

It is obvious that both formulae approximate satisfactorily the experimentally

measured failure moments. For a better approximation maybe strain hardening ofsteel should be taken into account.

0

.1

.1 .2 .3

.4

mR

equation 2.46

mR=.9

.2

.3

Figure 2.55Comparison of experimental results with theoretical curves for under-reinforced members

subjected to bending

Finally, a series of experimental results for over-reinforced elements is evaluated

where concrete is the critical factor (Fig. 2.56). These results are graphically

represented in relation to tensile reinforcement mechanical ratio.

Test specimens of concrete with lower strength (fc< 18 N/mm2) tend to exhibit higher

experimental to theoretical strength ratios (Fig. 2.56). This happens because inelastic

deformation capacity of concrete increases with its strength decrease. For

comparison reasons flexural strength with fc< 18 N/mm2was calculated for a second

time on the assumption that concrete transversal compressive strain is not 4 but

8 (qR=8). Approximation proved to be much better (Fig. 2.56).

Increase of inelastic deformation capacity with the decrease of strength is

qualitatively confirmed also from uni-axial compression tests as shows Wischers

comparison illustrated in Figure 2.57. When high or ultra-high strength concretes are

used in aseismic design the required ductility should be assured through the

arrangement of the adequate confinement reinforcement.

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134

0 .5

1.0

mR

ex

/mRth

1.5

1.0

1.1

.9

fc18 N/mm2

Figure 2.56Comparison of experimental results with theoretical relation for over-reinforced members

under bending

0 2

10

-104 6 8

B55

B35

B15

Figure 2.57Ductility of concretes with various strengths [25]

Ductility increase in ultra-high strength concretes is also achieved with the addition of

fibres (metallic or composite).

Finally we may conclude that for the estimation of the ultimate moment of reinforced

and prestressed members, the assumption of a constant stress fcacross the width of

compression strut leads to very good results compatible with the experimental ones.

Failure mode depends mainly from the members deformations affected from bond,

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135

reinforcements arrangement and the type of stress. For this reason two models are

examined:

The first case of converging cracks leading to local bend of the compressive strut is

applicable on bad bond conditions or near direct supports of beams and slabs.Inelastic deformations occur locally and lead to a brittle failure of the component.

Applications will be examined in the next Chapter.

The second case of parallel cracks leading to the curvature of the compressive strut

is appropriate for members with good bond conditions subjected to bending with or

without axial force. Inelastic deformation appears to be distributed along the strut and

failure progresses gradually in form of peelings in the most compressed fibre of

concrete. Evaluation of tests confirmed that in under-reinforced members the ultimate

moment in accordance with the theory of plasticity is always achieved. Reinforced

and prestressed concrete members in practice are usually executed as under-

reinforced in order to have increased ductility. In over-reinforced members where

resistant moment according to the theory of plasticity is not achieved or in

compressed members with symmetrically arranged reinforcement when compressive

stress ratio n>0.40 the theoretical approach presented leads to satisfactory

coincidence with experimental results. Obviously rotational ductility in these cases is

very limited.

2.9 Flexural shearin members without shear reinforcement

To study this problem a simply supported one-span beam is selected with a

concentrated load in its middle. Available is only longitudinal reinforcement in the

bottom tensile flange (Fig. 2.58):

2Q

Q QAs

b

d

Figure 2.58Simply supported one-span beam without shear reinforcement

With the increase of load tensile zone of concrete cracks first in the middle region

and longitudinal reinforcement bears the tensile force. A simplified representation of

the flow of forces may be a truss consisting of struts and ties (Fig. 2.59):

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136

Q Q

2Q

Reinforcement

Concrete under tension Concrete under compression

Figure 2.59Flow of forces in a one-span beam without shear reinforcement

In the region between compression and tension flange (region of shear) tensile stress

of concrete is still much lower than its tensile strength. Thus, tensile force of

reinforcement is decreasing through shear coming closer to the support. This forces

transfer reminds the beam bending theory and for this reason we may describe it as

flexural behaviour. Under a specific load concrete experiences tensile failure in the

region of shear. As failure criterion a value for the mean principal tensile stress may

be considered. Such a strength criterion is used also in practice as lower limit of

shear strength (Fig. 2.60).

t

=

R

fct

Figure 2.60Lower limit of shear strength and concrete principal stresses

It is experimentally confirmed that the mean shear failure stress depends mainly on

the quality of concrete and the depth of the cross-section. In the following anexperimental relation for the failure shear stress R (equation 2.47) is presented

taking into account the depth of the members cross-section (size effect):

R= (0.6 + 0.03fc)(1.2 d) [N/mm2] (2.47)

fcin [N/mm2] and d in [m] (d < 0.6, otherwise d = 0.6)

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0 10

fc(N/mm)20

R

(N/m

m)

2

30

1

d=10cm

20

40

>60

40

Figure 2.61 Lower shear strength limit of concrete in accordance with equation 2.47

In the event of earthquake (cyclic load) and yield of tensile reinforcement (plastic

hinge formation) Priestley proposes a decrease of above strength proportional to the

required rotational ductility of the cross-section in accordance with relations 2.48.

20,f29.0 ccR

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138

experimental results. American FEMA recommendations for the verification of

columns practically coincide with above equations 2.47 and 2.48.

After the shear failure of concrete, when stress reaches the value =Ra new force

transfer system is formed. It is made up by two inclined compression struts and a tie(the longitudinal reinforcement, Fig. 2.63).

2Q

Q QAs

l l

d

Figure 2.63Flow of forces in a one-span directly supported beam without shear reinforcement

Assuming constant cross-section of the longitudinal reinforcement and good

anchorage at its ends the ultimate moment at the middle-span section should be

always achieved. Experimental research proved that the ultimate load of flexural

strength is not reached in many cases. This leads to the conclusion that above flow

of forces through direct support is subjected to certain restrictions.

Before the clarification of this problem is useful to see which are the cases of shear

failure of concrete prior to the steel yield in the middle-span section. For this reason

the yield stress of the longitudinal reinforcement is used and mean shear stress is

calculated from the equilibrium of moments (equation 2.49):

l

d

db

fA

zbl

Afz

zb

Q ssss

pl

pl

=

=

=

or ,a

l

dv

f

or

l

d

f

s

pl

c

pl

s

pl ==== (2.49)

where asis the shear ratio.

In Figure 2.64 is plotted for comparison reasons the value R/fs=2.5. This value is

approximately corresponding to a concrete strength fc= 20 N/mm2and a structural

depth d=250 mm. We may observe in Figure 2.64 that the shear strength of concrete

is sufficient until steel yield for shear ratios as>8 when =2% or for as>4 when =1%

or for as>2 when =0.5% etc.

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0 1/2

d / l1/1

(pl

/fs)10

5

10

R/fs []

2

=1%

.5

.25

1/41/8

Shear Strength

of concrete

Figure 2.64Concrete shear stress relation to and d/l ratio when flexural strength is reached

Therefore the slenderer the structure and the lower its longitudinal reinforcement ratio

is the easier is the relatively decreased shear force transfer through concretes shear

strength and stirrups could be avoided (e.g. in case of slabs).

On the other hand the alternative shear force mechanism through the direct support

is not always possible as already mentioned. Condition is the avoidance of an early

failure of concrete strut, i.e. prior to the longitudinal reinforcement yield. Critical for

the struts failure is inelastic deformation occurring near node A (Fig. 2.65).

Q

Q

Fs

I

II

A

l

d

Figure 2.65Concentration of inelastic deformation in the strut in case of direct support

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140

Inelastic deformation due to struts rotation occurs in region I (Fig. 2.65) due to lack of

confinement in accordance with the spalling mechanism. The maximum possible

concentrated rotation (bend) of the strut is R= 4 (equation 2.30)

The angles change may be calculated from the truss, on the assumption of smalldeformations, constituting of two parts. The first part is made from the

longitudinal reinforcements elongation s (Fig. 2.66)

sd

l = (2.50)

A

C

B'

B''B

'

l

l

d

Figure 2.66Struts rotation due to elongation of longitudinal reinforcement at the direct support

The second part appears due to the elastic contraction of the strut celeading to thechange of its length by BB (Fig. 2.67).

ced

l = (2.51)

A

C

B'

B

''

B''

l

d

Figure 2.67Struts rotation due to concrete compressive strain at the direct support

Total rotation angle is calculated by the following equation:

( )ces dl

+=+= (2.52)

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The mean elastic compressive strain of concrete may reach at the very most

approximately the value of ce = -1. Through the distribution of compressive

stresses in the intermediate region it may even be noticeably lower, i.e.

approximately ce

= -0.5. In the following, since all values refer to rupture, index R

is omitted. The additional elongation of longitudinal reinforcements steel s at

concretes rupture is equal to:

(2.30, 2.52) []1l

d4

l

d ces

== (2.53)

From equation 2.53 we obtain the increase of the force in reinforcement:

1000

A1

l

d4AF ssssss

== when s < sy

or Fs= As.

fs when s > sy (2.54)

where fsis the increase of steel stress until yielding of reinforcement.

In dimensionless form :

[],1

11

l

d4

f1000

E1

l

d4

fA

Fsy

sys

s

ss

s

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We may observe (Fig. 2.68) that for usual yield limits of steel of approximately 2 to

2.5 the limit of l/d ratio for a viable direct support until yielding of steel is

approximately 1.3 to 1.1. Therefore direct support is practically achievable only for l/d

ratios of approximately 1.0. For higher ratio values shear force transfer should be

effectuated through transverse reinforcement (stirrups). The ultimate load of the

structure may be calculated as follows (Fig. 2.69):

b

d

x

Q

Q

Fs

fc

fc

l

Figure 2.69Strength calculation for a direct support

Using equilibrium conditions we obtain:

ss F2

x

dFzlQ

== (2.56)

and cs fbxF =

In case of prestressing force Fsis made of two parts (Fig. 2.51). Substituting in above

equations we obtain:

( )

++=

c

ssp

sspfdb2

FF1FF

l

dQ

and in dimensionless form:

( )

++=

=2

1

l

d

fdb

Qv

c

R (2.57)

Kani [26] performed numerous tests on beams without transverse reinforcement. For

the evaluation of these tests following assumptions are made :

l)21(d

l)21(d

Q

Q

m

m

pl

R

pl

R

== (2.58)

s= 2.10

5N/mm

2,fs= 400 N/mm

2

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On the evaluation diagrams (Fig. 2.70 and 2.71) failure lines are also plotted referring

to the shear force transfer through concretes strength.

s

R

plss

RR

pl

R

pl

R

f

d

l

z)lfA(

zb

Q

Q

m

m

== (2.59)

assuming: plR zz

asas0 2 4 6 8 0 2 64 8

mR

/mp

l

.4

.8

.2

.6

1.0

=0.5%

0.

1.2

=0.8%

Figure 2.70Failure moment to plastic moment relation in beams without shear reinforcement

(Kani experiments evaluation) for various longitudinal reinforcement ratios =0.5, 0.8%

=1.9%

0

.6

.2

.4

1.2

1.0

.8

2 4 6 8

-ce10=0.5

=2.8%

0

1.5

1.0

642as

8

Direct support Shear strength of concreteChange of concrete's

elastic strain

as

Figure 2.71Failure moment to plastic moment relation in beams without shear reinforcement

(Kani experiments evaluation) for various longitudinal reinforcement ratios =1.9, 2.8%

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In the last diagram (Fig. 2.71) for =2.8% two failure curves are presented for

comparison reasons, where the mean elastic compressive strain of concrete ce

varies 50%. It seems that for higher values of the shear ratio as compression

distribution is wider resulting to the decrease of the mean elastic compressive strain

of concrete.

Application of prestressing certainly affects favourably failure load since for the same

strain a greater force corresponds to the reinforcement than the force when it is

untensioned. To evaluate this effect following diagrams 2.72 to 2.74 were prepared

for various values of the shear ratio as. The assumed material parameters are

Es= 2.105N/mm

2, fs= 460 N/mm

2and fc= 20 N/mm

2.

0 .1+

.2

1.0

.3

.5mR

/m

pl

as= 1.5

=.2

.1

0.

Shear strength of concrete

Direct support

Figure 2.72Failure moment to plastic moment for beams with shear ratio as=1.5 without transverse

reinforcement, with prestressed longitudinal reinforcement

0 .1+

.2

1.0

.3

.5m

R

/m

pl

as= 2

=.2

.1

0.

Direct support

Shear strength

of concrete



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=.2

.1

.10 .2 .3+

m

R

/m

pl

.5

1.0

as= 3

Shear strengthof concrete



For low slenderness (as3, Fig. 2.74) due to the

flexural behaviour (shear resistance of concrete). For intermediate values and

especially for high prestressing degrees the result is a significant increase of the

strength of the structure.

2.10 Shear transfer mechanisms shear reinforcement

Shear transfer mechanisms in reinforced concrete members (slabs, beams, columns,

walls) are more than one and usually develop simultaneously. Most important is

always to examine the capacity shear force, i.e. the one corresponding to the flexural

strength of the members and not any value resulting from an elastic analysis. Brittle

failure modes as shear are verified in terms of forces and not in terms of

deformations.

2.10.1 Shear strength of concrete

Shear forces may be transferred through the shear strength of concrete as discussed

in the previous Chapter (Fig. 2.75). In the present Chapter a constant value for the

R/fcratio will be used for comparison reasons, corresponding to a concrete strength

fc=20 N/mm2and a structural depth d=40 cm.

%505.0vf

R

c

R === (2.60)

In this case when a plastic hinge is formed a decrease of shear strength results in

accordance with equation 2.48. It is a brittle failure mode. Flexural strength is

reached only when shear strength is sufficient. In practice, shear strength of concrete

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146

is utilized in slender beams, slabs, shells and generally in thin structures subjected

mainly to bending, in most cases plane structures.

Fs

''Reinforcement's force''

VV

V

Figure 2.75Reinforcement force during shear force transfer through the shear strength of concrete

Using equilibrium equations for the example of Figure 2.76:

lVM =

dl

dVMas ==

, where as: shear ratio (2.61)

s

llpl

c

ss

a

l

dv

f

l

d

db

fA

db

V ===

=

=

That means when flexural strength will be reached the shear force ratio will be:

s

llpl

a

l

dv == (2.62)

From equation 2.62 results that if the shear strength ratio is vR=0.05 and ratio d/l=1/2

then the maximum longitudinal reinforcement to avoid early shear failure will be:

10.005.01

2v

d

lmax Rl === ,

corresponding to a geometric longitudinal reinforcement ratio =5.

In the presence of axial force (Fig. 2.77) the shear force ratio when flexural strength

will be reached will be (n positive in compression):

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M

V

VV

d

V

V

Asfs

l

Figure 2.76Shear force transfer through the shear strength of concrete

N

V

M

V

V

V

MN

fc

fc

d

l

Asfs

Figure 2.77Shear and compression force transfer through the shear strength of concrete

( ) ( )[ ]n1nl

d

l

dn1n

l

dv llpl +=+= (2.63)

The shear force ratio at failure will be:

( ) ldn1n

fv

c

RR += (2.64)

For symmetrically arranged reinforcement on both ends of the member shear force

transferred due to bending is higher (Fig. 2.78):

d2

l

dV

MaIVM2 s =

== , where as: shear ratio (2.65)

s

llpl

c

ss

a

l

d2v

f

l

d

db

fA2

db

V ===

=

= (2.66)

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Asfs

l

Asfs

l

Asfs

Asfs

VV

V

V

VV

V

M

N V

V

V N

M

d

d

Figure 2.78Shear force transfer through concretes shear strength and additionally through an

inclined compression strut in a member restrained on both ends

d2

l

dV

MaIVM2 s =

== , where as: shear ratio (2.65)

s

llpl

c

ss

a

l

d2v

f

l

d

db

fA2

db

V ===

=

= (2.66)

Consequently equation 2.62 is applicable again with the only exception that shear

ratio asis half than before.

To reach the flexural strength the shear force ratio at failure will be vR= 0.05 and the

ratio d/l=1/2:

05.005.012

2v

d2

lmax Rl =

=

=

corresponding to a geometric ratio 2.5.

In the presence of axial force (Fig. 2.78) the shear force ratio when flexural strength

will be reached will be:

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( ) ( )[ ]ss

lllpl

a2

)n1(n

a

n1n2

l

d

l

dn1n

l

d2v

+=+=+= (2.67)

The shear force ratio at failure will be (n positive by compression):

( )sc

R

c

RR

a2

)n1(n

f

l

dn1n

f

v

+=+= (2.68)

Therefore relatively low longitudinal reinforcement ratios will result to early shear

failure since shear strength of concrete will be exceeded. In slabs with shear ratios as

of approximately 10:

maxl= asvR= 100.05= 0.5,

corresponding to geometric ratio =2.5%, i.e. there is no problem for early shearfailure.

Equations for members without transverse reinforcement (stirrups) of EC2/

2000 [6] and of the new DIN 1045-1 [16] for the verification of failure shear force at

cross-sections level and not members level differentiate in a certain degree.

Specifically EC2 / 2000 provide equation:

) db15.0)4020.1(kV wcplRd1Rd ++= (2.69)

1d60.1k = (d in m),

cp> 0 for compression, due to load or prestressing,

that could be rewritten to be comparable with the previous approach (equation 2.68):

( ) ++=

=c

cp

l

c

Rd

cw

1Rd1Rd

f

15.04020.1k

f

fdb

Vv

( ) n15.04020.1kf

v l

c

Rd1Rd ++= (2.70)

Equation 2.70 includes the contribution of shear strength of concrete, the influence oflongitudinal reinforcement contributing to direct support formation (see Chapter

2.10.2) and the contribution of a potential compressive stress or prestressing at

cross-sections level. It does not though depend on asratio that refers to a members

property and therefore does not sufficiently correspond to the mechanical behaviour

as in the case of equation 2.68.

New DIN 1045-1 (07/01) provides for concretes shear stress at designs level

following equation:

Rd,ct= 0.10 k (100 lfck)

1/3

+ 0.12 cd, fck [N/mm

2

] (2.71)

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2d

2001k += , d in mm

l=Asl/(bwd) 0.02 and cd= NEd/Ac, NEd > 0 for compression

d= axial force due to load or pres

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