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TheStructuralEngineer36 Technical

Number 4

Concrete designApril 2015

Strut-and-tie modelling is a simple method of modelling complex stress patterns in reinforced concrete as triangulated models. It is based on the same truss analogy as the design for shear in Eurocode 21 and can be applied to many elements, but is particularly useful where normal beam theory does not apply, i.e. where plane sections do not remain plane, e.g. in deep beams, corbels and pile caps. EC2 provides information about the use of strut-and-tie modelling and this article is an introduction for engineers who want to take advantage of this useful analysis method.

Regions where normal beam theory cannot be used arise at geometrical discontinuities, supports and concentrated loads and are known as D or disturbed regions. D regions are dimensioned using Bernoullis principle (Figure 1). Regions where normal beam theory can be used are called B or beam or Bernoulli regions. In strut-and-tie modelling, the structure can be split into D and B regions and a strut-and-tie model (STM) prepared for the D regions. The method uses the truss analogy where the actual ow of stresses is represented by a series of struts and ties (Figure 2). The method is an application of the lower bound theory of plasticity. Consequently, there is no unique STM for any given problem. The main requirements are to ensure that equilibrium is satis ed and to proportion the cross-sectional areas of struts and ties such that their resistances are everywhere greater than or equal to the internal forces. Adequate ductility is typically ensured through the provision of code-speci ed minimum reinforcement areas.

Concrete Design GuideNo. 4: An introduction to strut-and-tie modelling

The scope of this article is restricted to strut-and-tie modelling of planar structures. The reader is referred to specialised texts27 for more in-depth treatment of the method.

Development of STMThe rst step is to determine the basic STM geometry within the element. The serviceability limit state (SLS) of cracking is usually satis ed if the ultimate limit state (ULS) STM geometry is based on the elastic stress eld, with members positioned at the centroids of the compressive and tensile elastic stress distributions (Fig. 2a). An elastically based STM can lead to overly conservative designs, since such models do not recognise the redistribution in stress that occurs when the concrete cracks, putting more stress into the reinforcement. Therefore, some deviation from the elastic solution is permissible, as in the design of continuous beams. Consideration of Fig. 2a

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and Fig. 2b shows that the STM geometry changes as the load is increased to failure, due to the internal redistribution of forces that occurs on cracking and subsequent yielding of reinforcement, which in the ultimate case lies within the plastic limit. The STM in Fig. 2b neglects the contribution of the minimal web reinforcement required by EC21 to control cracking, which further increases load resistance.

Load path methodThe load path method2 is a useful tool for developing the STM geometry within elements. The starting point is to determine the stress distribution at the boundaries of the D region after which the STM can be developed using the following rules of Schlaich and Schfer2:

1. Load paths do not cross each other2. Load paths take the shortest streamlined route between the centres of gravity of stress diagrams3. Curvatures concentrate near stress concentrations, reactions and concentrated loads4. In the absence of elastic analysis, position struts from experience and the application of simple rules

N Figure 1Dimensioning of D regions

Dr Robert L Vollum Department of Civil Engineering, Imperial College London

TSE40_36-41 CDG v1.indd 36TSE40_36-41 CDG v1.indd 36 19/03/2015 14:1919/03/2015 14:19

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The procedure is illustrated in Figure 3 for the anchorage zone of a post-tensioned beam. A similar loading arrangement arises when a wall is loaded with a concentrated load at its corner. Note that the forces F1 and F2 are equal and opposite. The geometry of the STM is fully de ned once the dimensions z1 and z2 are de ned. The dimension z2 can be found by assuming that the angle of the diagonal strut is 45o. Alternatively, the strut can be positioned at the centroid of the corresponding elastic compressive stress distribution as done for the deep beam of Fig. 1a.

Figure 4 shows two alternative STMs for a deep beam. Both are lower bound solutions but the STM in Fig. 4b is not optimal, or bad, since the load path is incompatible with the elastic stress eld. The STM in Fig. 4a is more appropriate, or good, because the strut orientation follows the elastic compressive stress trajectories. As noted by Schlaich and Schfer2, it is helpful to realise that loads try to use the path with the least forces and deformations. Consequently, a crude rule of thumb for distinguishing between good and bad models is to choose the model with shortest ties, since ties are much more deformable than struts.

Choice of limiting angle between struts and tiesThe angle between struts and ties should be large enough to avoid strain incompatibilities due to ties extending and struts shortening in almost the same direction. Model Code 19908 gives the following useful rules of thumb for developing STMs which broadly follow the elastic stress eld:

1. Orientation with the elastic stress eld is more important for struts than ties, which can usually be arranged parallel to the edges of the member following practical considerations2. In highly stressed regions, main struts and ties should meet at angles of about 60 and not less than 45 (this is a guideline and not mandatory)3. If the arrangement of the model is made in accordance with the elastic stress eld, the ULS does not require checking

4. The orientation of the model may be allowed to depart from the elastic stress eld at the ULS if the SLS is checked

Figure 5 shows how rule 2 can be used to distinguish between good and bad STMs for a deep beam. The STM in Fig. 5a is bad because insuffi cient tensile reinforcement is provided to control cracking at the SLS. Fig. 5b shows that there is an area of concrete in the top section of the STM which is in tension and therefore this section should be reinforced for the tensile stress. It can be shown from nite-element models that this tensile zone is created. Normally the tensile stress is taken by the nominal reinforcement provided in the deep beam.

Compressive stress eldsCompressive stress elds are the way in which the compressive stress in the idealised struts spreads through the concrete. They are categorised as prismatic, fan- or bottle-shaped (Figure 6). Prismatic stress elds typically arise in B regions, whereas fan- and bottle-shaped stress elds arise in D regions due to the dispersion of the stress paths radiating out from concentrated loads or reactions.

Bottle stress elds are a feature of elastic stress eld analysis. The curvature of the principal compressive stress trajectories of the bottle eld develops signi cant transverse tension. The compressive resistance of bottle stress elds is limited by splitting along the axis of the strut, unless transverse reinforcement is provided to maintain equilibrium after cracking.

The fan-shaped stress eld is an idealisation in which the compressive stress trajectories are assumed to be straight. Consequently, no tension develops at right angles to the stress trajectories of the fan. Fan-shaped stress elds are assumed to develop at supports and concentrated loads in plastic stress eld analysis. Failure is assumed to occur at supports and concentrated loads where stresses are greatest.

Prismatic or parallel-sided stress elds are a special case of the fan- and bottle-shaped stress elds. Figure 7 shows a deep beam loaded with a concentrated load. In reality, the compressive stress eld is bottle-shaped as indicated with the dotted line. However, in practice, struts are often idealised as prismatic (Fig. 7). EC2 requires a minimum amount of horizontal and vertical web reinforcement to be provided in deep beams to control cracking. The code requires an isotropic mesh with a minimum area of 0.001Ac mm2 or 0.001t mm2/mm (where Ac is the cross-sectional area of the member and t is its thickness) to be provided in each face, but the UK National Annex to EC29 increases this area to 0.002Ac.

W Figure 2Development of STM for uniformly loaded deep beama) based on elastic stress eld b) at ULS

S Figure 4Good and bad model based on minimising length of ties

N Figure 3Application of load path method (adapted from EC21)


TheStructuralEngineer38 TechnicalConcrete design

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Compressive strength of strutsEC2 gives design concrete strengths for struts which are based on simple criteria that account for the eff ects of cracking and con nement. In general:

transverse compression is favourable. Especially con nement from transverse reinforcement or surrounding concrete transverse tension is detrimental

Skew cracks are particularly detrimental as equilibrium requires shear forces to be transferred across cracks by aggregate interlock. Struts are checked at their ends where the cross-sectional area is de ned by the node dimensions. EC2 reduces the design concrete strength in prismatic struts with transverse tension to 0.6fcd where = 1fck/250. The strength of 0.6fcd also applies to the ends of bottle-shaped struts like the diagonal struts of Fig. 7 if insuffi cient transverse reinforcement is provided to maintain equilibrium after cracking of concrete.

Classi cation of nodesNodes are de ned as regions were struts change direction or struts and ties intersect (Fig.7). Smeared nodes arise in regions where the orientation of struts is diverted by a band of evenly distributed reinforcement bars and is not checked in design. Concentrated nodes are typically highly stressed and need to be designed to resist the incoming forces without concrete failing in compression. EC2 classi es nodes as C-C-C (three compressive struts), C-C-T (two compressive struts and one tie), and C-T-T (one compressive strut and two ties) (Figure 8). The node where a column meets a pile cap would typically be a C-C-C node, and the node where a two-pile pile cap meets the pile would typically be a C-C-T node.

Forces are transferred from ties into nodes through a combination of bearing and bond stresses within the node. The dimensions of the nodes in Fig. 7 are governed by the dimensions of the loading and support plates as well as the cover to the exural reinforcement. The

node dimensions need to be chosen to ensure that stresses on node boundaries are less than or equal to the design concrete strengths given in Table 1. It is necessary to ensure that the design stresses are less than the design strengths in both nodes and at node-strut interfaces where strut resistance is checked. The design strength of the strut depends on the reinforcement provided within the depth of the element and is often less than the node strength.

Dimensioning of nodesThe dimensions of concentrated nodes are frequently determined by the widths of load and support plates as well as the position of ties (Fig. 8).

Sizing of C-C-C nodesThe width of the loading plate of the deep beam in Figure 9 needs to be chosen to limit the bearing stress to a maximum of 1.0fcd which is the design strength for C-C-C nodes. The depth x of the C-C-C node in Fig. 9 depends upon the assumed exural compressive stress which cannot exceed 1.0fcd. The choice of exural compressive stress in Fig. 9 determines the widths of the incoming diagonal struts. Clause 6.5.4 (8) of EC2 states that C-C-C nodes can normally be sized on the basis that stresses are equal on all node boundaries. In this event, the stress distribution is hydrostatic within the node and the incoming strut centrelines are normal to the node boundaries. In practice, the width of the loaded area is normally governed by other considerations than the design concrete strength. Consequently, the bearing stress can be signi cantly less than fcd. It is suggested that in the absence of shear reinforcement, the exural compressive stress is taken as fcd, to avoid the widths of the incoming struts being overestimated, but this is not an EC2 requirement.

Sizing of C-C-T nodesThe design bearing stress at C-C-T nodes is limited to a maximum of 0.85fcd. The node width normal to the inclined strut centreline in Fig. 8b is given by:

(1)

where Lb is the length of the support plate, u is the height of the back face of the node and is the angle of the strut centreline to the horizontal. EC2 does not require stresses to be checked at the back face of C-C-T nodes.

It is bene cial to provide the tensile reinforcement in several layers in highly stressed concentrated nodes since this increases u and

arrow Figure 5Use of Model Code 19908 rules to distinguish between good and poor STMN Figure 6Compressive stress eldsa) prismaticb) fan-shapedc) bottle-shaped

Table 1: Eurocode 2 node strengths

Code Type of node Strength

EC2 C-C-C 1.0fcdC-C-T 0.85fcdC-T-T 0.75fcd

Note: = 1fck/250



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8. If necessary, increase the design shear resistance by increasing the node dimensions or providing shear reinforcement if C > CRd9. Calculate the area of exural reinforcement required to resist the design force T1. (As = T1/fyd)

Consider the beam in Fig. 9 with L = 5000mm, h = 3000mm, t = 200mm, Lt = 300mm, Lb = 150mm, d = 70mm, P = 500kN and fck = 30MPa.

The design concrete strengths are as follows:

fcd = (1fck/250)(0.85fck/1.5) = 15.0MPa (bearing strength at top node)

0.85fcd = 0.85(1fck/250)(0.85fck/1.5) = 12.7MPa (bearing strength at bottom node)

0.6fcd = 0.6(1fck/250)(0.85fck/1.5) = 9.0MPa (strut strength without calculated transverse reinforcement)

The minimum required widths of the bearing plates are:

Trial and error gives: x = 70mm, = 50, C = 326kN, wb = 205mm, wt = 160mm, and T1 = 209kN.

Therefore, the required area of exural reinforcement is T1/fyd = 480mm2. The inclined strut is overstressed, indicating that shear reinforcement is required. The UK National Annex to EC2 requires nominal horizontal and vertical web reinforcement of area 0.002Ac to be provided in each face. The eff ectiveness of this

hence the strut width w. The reinforcement should be suffi ciently anchored to develop its design tensile force at the node.

Additional recommendations with respect to STMPD6687-1 (2010)10, the background document to the UK National Annex to EC29, gives the design compressive concrete strength as fcd = 0.85fck/c for STM. It also states that for shear, the shear resistance of sections should be veri ed using BS EN 1992-1-1:2004, 6.2.2 at all sections where av > 1.5d, where av is the distance of the section from a concentrated load (or support) and d is the eff ective depth of the section.

Example 1. Design of deep beam with central point loadFig. 9 shows an STM for a deep beam of thickness t loaded with a central point load P. The inclined strut width at the bottom node is given by Equation 1.

At the top node, the inclined strut width is given by:

(2)

where:

(3)

in which:

(4)

where c0 fcd is the exural compressive stress.

cot = (0.5L0.25Lt)/(h0.5xd) (5)The compressive force in the strut is given by:

C = 0.5P/sin (6)

In the absence of minimal reinforcement within the depth of the element, the compressive resistance of the strut is given by:

CRd = 0.6fcdmin(wt,wb)t (7)

EC2 does not de ne how the strut resistance CRd should be calculated. The strut strength is 0.6fcd at the C-C-T node-to-strut interface, but EC2 does not de ne the strut strength at C-C-C nodes. It is suggested that, in the absence of code-prescribed minimal web reinforcement, the strut strength is also taken as 0.6fcd at C-C-C nodes.

Typical design solution procedureEquations 27 can be solved using the following step-by-step procedure:

1. Find the minimum allowable lengths of the top and bottom nodes. (The maximum allowable bearing stress at the top and bottom nodes are fcd and 0.85fcd respectively)2. Choose x (e.g. u)3. Calculate cot with Eq. 54. Calculate T1 with Eq. 45. Calculate co with Eq. 3. If minimal web reinforcement is provided, the chosen value of x is acceptable if P/(Ltt) [bearing stress at top node] co fcd. Otherwise, adjust x by trial and error until co = fcd6. Calculate the strut force C in terms of the applied load P with Eq. 67. Calculate the strut widths wt and wb with Eq. 1 and Eq. 2 and hence the strut capacity CRd with Eq. 7

S Figure 7STM for simple deep beam

arrow Figure 8Classi cation of node typesa) C-T-Tb) C-C-Tc) C-C-C


TheStructuralEngineer40 TechnicalConcrete design

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April 2015

reinforcement can be assessed using the EC2 design equation for full discontinuities, stress eld models or the EC2 design equations for shear enhancement within 2d of supports.

Transverse reinforcement in bottle stress eldsEquations 6.58 and 6.59 of EC2 are used to calculate the area of transverse reinforcement required to equilibrate the transverse tension in bottle-shaped struts with partial and full discontinuity.

Partial discontinuityFor cases of partial discontinuity the design tensile force for the transverse reinforcement (Figure 10) is given by:

T = F(ba)/4b (Exp 6.58)where F = the force in the strut, b = the available strut width and a = the node width.

Full discontinuityFor cases of full discontinuity (b > H/2) (Fig. 10b):

T = F[1.00.7a/H]/4 (Exp 6.59R)in which H is the strut length and a is the width of the loaded area. (Expression 6.59R replaces the term a/h in Expression 6.59 of EC2 with a/H as directed by BSI paper B525/2 11 0034.)

EC2 requires the area of transverse reinforcement resisting T to be provided over the length the compression trajectories are curved, but does not de ne this distance. Schlaich and Schafer2 show the transverse reinforcement being distributed over the central 0.8H of the partial discontinuity shown in Fig. 10b. However, this paper takes the length of the transverse tension zone as 0.6H as suggested by Hendy and Smith11.

Example 2. Inclined strut with full discontinuityThe inclined strut in Fig. 9 is a full discontinuity of length:

Hence, the design tensile force to be resisted by transverse reinforcement over each half of the strut length is:

T = F[1.00.7a/H]/4 = 326[1.00.7160/3776]/4 = 79kNThe tensile force T is provided by a mesh of horizontal and vertical

reinforcement with minimum area of 0.002Ac = 400mm2/m in each face in each direction.

For orthogonal reinforcement, the normal resistance nRd per unit length to the strut centreline is given by:

nRd = (As1sin2/s1 + As2cos2/s2)fyd (8)

where s1/s2 = spacing of horizontal bars of cross-sectional area As1 and vertical bars of cross-sectional area As2 and is the angle of the strut centreline to the horizontal.

For isotropic reinforcement As1 = As2 and nRd = Asfyd/s in all directions.

Hence, the required minimum reinforcement grid of 0.002Ac (i.e. 400mm2/m vertically and horizontally) in each face can resist a transverse force:

Alternatively, the strut capacity corresponding to minimum web reinforcement is given by:

(9)

in which a = wt, H is the strut length and nRd is given by Equation 8.

The maximum possible shear force corresponding to crushing of the concrete at the bottom node of the deep beam of Example 1 is Vmax = 0.85fcdtLb = 381kN. The shear resistance can be increased further by increasing the length of the bearing and loading plates as well as d.

Use of EC2 beam equations for design of shear reinforcementAlternatively, shear reinforcement can be designed using a truss model like that of Schlaich and Schfer12 (Figure 11) or with the EC2 design equations for shear enhancement in beams. In the case of the EC2 shear enhancement equations, no calculated shear reinforcement is required, provided the design shear stress vEd vRdc where vRdc is given by Equation 6.2a in EC2 and = av/2d where av is the clear shear span. The main advantage of using a truss model over the bottle stress eld is to steepen the angle of the inclined strut at the support. For example, the eff ect of introducing the vertical tie in the simpli ed truss model of Fig. 11 is to increase the resultant angle of the inclined strut from:

S Figure 9STM of deep beam S Figure 10Bottle stress eldsa) partial discontinuityb) full discontinuity



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to:

where Model Code 19908 estimates the tie force F1 as:

(10)

where z is the exural lever arm. Clause 6.2.3 (8) of EC2 reduces the design shear force due

to loads applied to the top surface of beams within 0.5d av 2d of supports by = av/2d where av is the clear shear span. Shear reinforcement needs to be provided within the central of av to resist the reduced design shear force. Consideration of Clause 6.2.3 (8) shows that the reduced design shear force is equivalent to the tie force F1 of Fig. 11. Clause 6.2.3 (8) is typically less onerous than Equation 10, with both being less onerous than Expression 6.59 of EC2 for a full discontinuity. It is also possible to design the shear reinforcement using stress eld models7,13.

General applicationThe STM is a very versatile method that can readily be applied to a wide range of structures that would otherwise only be designable with nite-element analysis or empirical design methods. The advantage of an STM over nite-element analysis is that an STM requires load paths to be clearly identi ed. Hence, structures can readily be checked for equilibrium. An STM also has the advantage of clearly identifying the anchorage requirements of reinforcement, unlike nite-element analysis.

SummaryIt is often convenient to base the STM geometry at the ULS on the elastic stress eld, since this typically ensures satisfactory performance at the SLS. However, some deviation from the elastic solution is permissible. If the elastic stress distribution is not available, the STM geometry can be developed using the load path method of Schlaich and Schfer2. Model Code 19908 suggests a 2:1 dispersion rule which is useful for establishing the basic STM geometry. The axial resistance of struts in a planar member equals the minimum of wbtfsb and wttfst (in which wb = strut width at bottom

1) British Standards Institution (2004) EN-1992-1-1:2004. Eurocode 2. Design of concrete structures. Part 1. General rules and rules for buildings, London, UK: BSI

2) Schlaich J. and Schfer K. (1991) Design and detailing of structural concrete using strut-and-tie models, The Structural Engineer, 69 (6), pp. 113125

3) International Federation for Structural Concrete (2011) fib Bulletin No. 61: Design examples for strut-and-tie models, Lausanne, Switzerland: b

4) Thurlimann B., Muttoni A. and Schwartz J. (1989) Design and detailing of reinforced concrete structures using stress elds, Zurich, Switzerland: Swiss Federal Institute of Technology

5) Goodchild C., Morrison J. and Vollum R. L. (2015) Strut-and-tie Models, London, UK: MPA The Concrete Centre

6) Schlaich J., Schfer K. and Jennewein M. (1987) Towards a consistent design of structural concrete, PCI Journal, 32 (3), pp. 74150

7) Sagaseta J. and Vollum R. L. (2010) Shear design of short-span beams, Magazine of Concrete Research, 62 (4), pp. 267282

8) CEB-FIP (1990) Model Code for Concrete Structures, Lausanne, Switzerland: CEB-FIP

9) British Standards Institution (2005) NA to BS EN 1992-1-1:2004 UK National Annex to Eurocode 2. Design of concrete structures. Part 1. General rules and rules for buildings, London, UK: BSI

10) British Standards Institution (2010) PD 6687-1:2010 Background paper to the National Annexes to BS EN 1992-1 and BS EN 1992-3, London, UK: BSI

11) Hendy C. R. and Smith D. A. (2007) Designers Guide to EN 1992 Eurocode 2: Design of concrete structures. Part 2: concrete bridges, London, UK: Thomas Telford

12) Schlaich J. and Schfer K. (2001) Konstruieren im Stahlbetonbau (in German), BetonKalender (Vol 2), Berlin, Germany: Ernst & Sohn, pp. 311492

13) Vollum R. L. and Fang L. (2014) Shear enhancement in RC beams with multiple point loads, Engineering Structures, 80, pp. 389405

Further readingThe Concrete Centre has recently published a guide to Strut-and-tie Models5, which gives more information on how to construct and use STMs. This is available via the Concrete Centre website: www.concretecentre.com

References and further reading

node, wt = strut width at top node, t = member thickness, fsb = strut strength at bottom node and fst = strut strength at top node). In the absence of shear reinforcement, take fst = fsb = 0.6(1fck/250)fcd. The strut strength can be increased if necessary by increasing node dimensions, providing shear reinforcement or a combination of the two. The minimum reinforcement required by the UK National Annex within the depth of deep beams is often suffi cient to increase the strut strength at C-C-T and C-C-C nodes to the node strengths of 0.85fcd and fcd respectively.

S Figure 11Simpli ed truss model according to Schlaich and Schfer13


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