3 strut and tie

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Strut-and-tie models forReinforced Concrete Structures


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Typical example of B-regionsand corresponding truss models

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2D and 3D Structural Elements A.Y. 2009/2010


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Typical example of element withoutclearly-defined B-regions



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Strut-and-tie models

In contrast to the B-regions (or main regions) introduced before, alocal region is a portion of the structure, where there is a strongvariation of stresses and strains

These regions are also referred to as D-regions, where D stands fordiscontinuity, disturbance or detail

Typical D-regions are: connections between a beam and a column

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end of a beam or a column corbels regions adjacent to a concentrated load recesses and holes


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Typical geometrical discontinuities 6

holerecess


frame corner

footing


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Typical statical discontinuities 7

beam end

concentrated


load

deep beam


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Example of concentrated loadsGround anchors

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Example of concentrated loadsGround anchors

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Identification of D-regionsExample of a column (FE analysis)

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plots of the minimum


(a) (b) (c)

principal stresses


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Identification of D-region

a) Replace the real structure by a fictitious structure, which is loaded insuch a way that (a) it complies with Bernoullis hypothesis; and (b) itsatisfies equilibrium with sectional forces: this structure consistsentirely of B-regions, but violates boundary conditions.

b) Introduce a self-equilibrating state of stress which, if superimposedon the fictitious structure, allows to satisfy the boundary conditions.

c) Apply the principle of De Saint Venant: the influence of the self-

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equ ra ng sys em ecomes neg g e a a cer a n s ance romthe equilibrating forces; this distance is roughly equal to the distancebetween the equilibrating forces.

d) The D-regions (and B-regions) of the real structure are thusidentified.


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Identification of D-regionsExample of a beam with direct supports

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Identification of D-regionsT-beam

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Identification of D-regionsExample of different types of beams

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In this case, the beam would be considered slender (L 4h) on the basisof the usual limits for beams; the point load, however, increases theextension of the D-regions.


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17General procedure for modelling

Generally, it is very time-consuming to model a whole structure bymeans of a truss; the first important thing to do is to understandwhether there are more B-regions or D-regions.

Most structures contain a substantial part of B-regions: beams and continuous frames slabs and shells

For the above-mentioned cases a linear elastic analysis is sufficient,


and allows to evaluate the generalized forces (N, M, V and T) at eachsection of the structure. If there are D-regions (e.g. the corners in a frame), the sectional

forces can be used as boundary conditions for a detailed study of thedisturbed portion of the structure.

If the structure consists of a single D-region (e.g. a deep beam) thensolving for the sectional forces is not necessary, and only theprocedures outlined in the following should be applied.


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18Modelling of individual D-regions

If a D-region is in the uncracked state, standard methods of analysiscan be used (typically linear elastic finite elements analysis).

If the tensile stresses in individual D-regions exceed the tensilestrength of concrete, the inner forces can be determined as follows:1. A strut-and-tie model is developed, by condensing the continuous

stress fields (in compression and tension) into resultant straightlines.


2. The strut and tie forces are calculated, on the basis of the typicalprocedures used for trussworks.

3. The struts and ties are dimensioned, together with the nodes, inaccordance to the materials properties, and with dueconsideration of crack width limitations.

4. Generally, the struts are represented by stress flows in theconcrete , and the ties by the reinforcing bars .


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20Crack pattern at ultimate


Note that the previous stress flow is a good representation of thebehaviour of the deep beams at ultimate; therefore, this type ofanalysis is most suited for the ultimate conditions; its application tothe serviceability limit states is more questionable...


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21The Load Path Method

First, the outer equilibrium of the considered D-region has to besatisfied, by determining all the loads and reactions (support forces)acting on it.

The stress diagram is then subdivided in such a way, that the loadson one side of the structure find their counterpart on the other.

An important thing to consider is that, generally, the load paths tend tobe of minimum length; therefore, they do not cross each other.



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23Model optimization

Being based on the lower bound theorem of plasticity, the Load PathMethod allows multiple solutions for one single problem.

Once a model is chosen, an open question is whether the givenproblem has been solved in the most proper way

It is useful to remember that the loads flow along the pathcharacterized by the least forces and deformations; since thedeformability of ties (usually represented by rebars) is by far largerthan that of concrete struts, the model with the least and shortest


es s e es . s cr ter on can e expresse as o ows:

iF iLi mi = minimum

where F i = force in tie i, L i = length of member i, and mi = mean strainof member i. This condition implies also that the stiffness of the chosen truss is

the maximum possible .


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Choice between different models 24


good choice bad choice


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Choice between different models (2) 25


Model with inclined reinforcement


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Choice between different models (3) 26


Model with straight reinforcement

In this case, the model with straight reinforcement should bepreferred, because of the higher simplicity of placing rebars withoutchanges of direction.

f d d


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27Dimensioning of struts, ties and nodes

The load transfer has to be ensured not only through a properdimensioning of the struts and ties, but also by checking the loadtransfer between the different elements.

A close relationship can be established between the detailing of thenodes, and the flow of forces through the members of the model (bethey struts or ties).

Therefore, it is mandatory to check whether the model assumed is still


va a er e a ng or nee s correc on.

S d i


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Struts and ties

The T s forces (ties) are essentially linear; on the contrary, C c (and T c,if they are taken into account) are 2-D or 3-D stress fields, spreadingout across finite zones of concrete, from one node to the other.

The spreading of the struts implies transverse tensile andcompressive stresses, that must properly be accounted for.

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Oth l f t t b l i


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Other examples of struts bulging

The struts in the model are resultants of the stress fields. Generally, the curvatures and deviations of the forces are

concentrated at the nodes, connecting straight elements. It may be argued, however, that this idealization of the reality is too

crude; the nodes can then be smeared over a larger portion of thestructural element.

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T i l fig ti 30


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Typical configurationsof compression fields

To cover all cases of compression fields, three typical configurationsare sufficient:

a) the fan;b) the bottle;c) the prism.

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Eurocode 2 provisions for the bottle 31


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Eurocode 2 provisions for the bottle(section 6.5 of EN 1992-1-1)

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T = (b a)/b F T = (1 0.7a/h) F

32Failure Criteria for Concrete


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32Failure Criteria for Concrete

It is well known that concrete strength is affected by the presence ofmultiaxial states of stress. The most authoritative source on this issueis the research carried out by Kupfer et al. (1969).


33Main findings on concrete strength


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33Main findings on concrete strength

Transverse compression is favourable, especially if it acts in bothtransverse directions, as it is the case in confined regions. A properconfinement degree can be attained by:

having some bulk concrete surrounding the stressed region; providing transverse reinforcement (stirrups, spirals).


34Basic Behavior of Reinforced Concrete


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34Basic Behavior of Reinforced Concrete

At the beginning of the 80s, Vecchio and Collins (at the University ofToronto) developed a unique testing device (Panel Element Tester),to study the behavior of reinforced concrete under various loadingconditions.


35Main findings on compression softening


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35Main findings on compression softening

Transverse tensile stresses and the ensuing cracks are detrimental. The compressive strength is greatly reduced, if the transverse tension

causes cracks parallel to the principal compression stresses ( thecompression prisms between the cracks are narrow and ragged).

Cracks which are not parallel to the principal compressive stressesare particularly detrimental.



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Eurocode 2 provisions 37


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p(section 6.5 of EN 1992-1-1)

The design strength of a concrete strut in a portion where transversecompressive stresses (or any tensile stresses) are expected, can becalculated through the following expressions.

Rd,max = fcd




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p(continued)

It is recomended that the design strength of concrete struts bereduced in the cracked compressed zones, and, when no accurateinvestigations are carried out, can be determined by means of thefollowing equation.

Rd,max = fcd ( = 1 f ck /250)


Local verification of the nodes 39


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The nodal zones connecting several struts are generally subjected tobiaxial states of stress.

A simple procedure, based on Mohrs circle, allowing to quantify thestate of stress in the nodal zones was proposed by Marti (1985).


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The same construction can be applied to the points corresponding tothe struts B and C.

The final result is the dashed circle, that represents the state of stressin the nodal zone.



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(section 6.5 of EN 1992-1-1)


Rd,max = k1 fcd

(k1 = 1.0, = 1 f ck /250)



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(section 6.5 of EN 1992-1-1)


Rd,max = k2 fcd

(k2 = 0.85; = 1 f ck /250)



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(section 6.5 of EN 1992-1-1)


Rd,max = k3 fcd

(k2 = 0.75; = 1 f ck /250)

Worked-out example 1(Ch 1982 Ch 7)

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(Chen, 1982; Chapter 7)


Main assumptions:

NO beam theory

point load is distributed (NO concrete crushing)

steel reinforcement is unbonded (end plates are provided)

As is such, that the neutral axis depth is half the beam depth

Worked-out example 1S i h

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Static approach


The maximum resistant bending moment at midspan is

Mmax = fcbd 2 /4 = f yAsd/2

Since concrete stresses are everywhere below (or equal to) thecompressive strength, and steel is at its yield point, this stressdistribution gives a lower bound of the ultimate load.

Worked-out example 1St ti h

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Static approach


Now, if we enforce free body equilibrium conditions, we obtain:

1/2 P u L/2 = f yAs d/2

P u = 2fyAs d/L

(or P u = fcbd 2 /L)


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Worked-out example 1Kinematic approach

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Kinematic approach


The outward displacement at the ends of the beam, as well as theinward displacement at midspan, can be thought of as crushing ofthe compressed concrete (caused by the loading and end plates).

The equation of virtual works becomes:

P u = (2 bd/2 d/L + bd/2 2 d/L ) fc

P u = fcbd 2 /2

Worked-out example 1Kinematic approach

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Kinematic approach


The outward displacement at the ends of the beam is nowassumed to cause stretching (and yielding) of the reinforcing tie.The internal energy dissipation is now split into two parts.

The equation of virtual works becomes:

P u = (2 As d/L ) fy + (bd/2 2 d/L ) fc

P u = fcbd 2 /2 (same as before)

Worked-out example 1Final considerations

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Final considerations


The static and the kinematic approach give the same value ofthe ultimate load; therefore, this is the exact value of thecollapse load.

The static approach, in this simple case is surely more handy

and quick.

Worked-out example 2Marti (1985)

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Marti (1985)


Materials and loads:

reinforcing steel (FeB44k): f yd = 430/1.15 = 374 MPa

concrete (Rck 35): f cd = 0.83 35 0.85/1.6 = 15 MPa

concentrated loads: P = 250 kN

distributed load: q = 40 kN/m

Worked-out example 2 53


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Reaction forces and maximum bending moment:

R A = R B = qL/2 + P = 40 5 + 250 = 450 kN

Mmax = R A L/2 q L2 /8 P (L/2 a) = 1000 kNm

The compression zone required to resist the maximum bendingmoment is approximately 250 mm; the bottom reinforcement isassumed to be 75 mm above the bottom of the beam.



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Design the stirrups in the first portion of the beam:

As,tot = Nmax /fyd = 410000/374 = 1096 mm 2

Choose 10 stirrups (two legs) spacing = 115 mm



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Design the stirrups in the center portion of the beam:

As,tot = Nmax /fyd = 96000/374 = 257 mm 2

Choose 8 stirrups (two legs) spacing = 310 mm

Worked-out example 2Final considerations

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e gure s ows a poss e arrangement o t e re n orcement. ote

the following: the presence of longitudinal bars in the web (cracking control);

the reinforcement detailing in the support area (D-region).


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60Outline of solution procedure


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1. The reaction forces are determined, by means of simple equilibriumequations.

2. A FE analysis of the structure is carried out, assuming a linear elasticbehaviour, in order to highlight the stress distribution.

3. The structure is subdivided into two parts, whose dimensions aredetermined on the basis of the values of the reaction forces; on thesetwo parts, equilibrium in the vertical direction is enforced.


4. Across the subdivision line (characterized by V = 0) the interaction

between the two parts is enforced, by enforcing the horizontalequilibrium.

5. The structure is then represented by a series of struts and ties; theaxial forces of these elements is determined using simple equilibrium

equations.6. Once the axial force in each element of the model is calculated, the

ties and struts can be dimensioned/checked.

61External equilibrium


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The reaction forces are determined by

two simple rotation equilibriumequations:


R

B= 3000 450 / 700 = 1929 kN

62Finite element analysis


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undeformed shape deformed shape(M.F. = 200)

63Minimum principal stresses(from FE analysis)


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64Maximum principal stresses(from FE analysis)


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66Right part of the structure


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The position and dimensions of the truss elements aredetermined, if the depth of the compression zone at the top andthe portion occupied by the reinforcing bar on the bottom are

known. Distance of T from the bottom = 500/2 = 250 mm

Distance of C from the top = 350/2 = 175 mm

68Left part of the structure


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The position and dimensions of the truss elements aredetermined, by taking into consideration the previousassumptions concerning T and C.

Distance of N 2 from the top right hole corner = 100 mm (for bond)

The other elements are simply determined, by considering theintersections of the elements with the load paths from support Ato the applied concentrated load.



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71Superposition of the two trusses


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Value of T at the bottom = 1100 kN required A s : 12 16

Check of stresses in the top strut: = C/A = 7.65 MPa (< f cd )

73Checks at node B


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Check of strut N 1 (= -2206 kN): = N1/A1 = 8.11 MPa (< f cd )

Load on the bearing plate: = R B/AB = 9.65 MPa

It is also very important to check the anchorage length, in order to allowa proper transfer of the bond forces from the bars to the concrete.

74Possible reinforcement arrangement


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3 strut and tie

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