Strut Tie CHG v1

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Strut & Tie

Webinar 12.30 4 July 2016

Charles Goodchild

BSc CEng MCIOB MIStructE The Concrete Centre

What is Strut and Tie?

A structural element can be divided into:

• B (or beam or Bernoulli) regions ‐ in which

oplane sections remain plane and odesign is based on ‘normal’ beam theory,

and into

• D (or disturbed) regions ‐ in which

oplane sections do not remain plane; o‘normal’ beam theory may be considered inappropriate oStrut & Tie may be used. ousually within hof a discontinuity: 2

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D (or disturbed) regions

3 D regions Pile caps Deep beams Discontinuities Holes Edge beams Pad footings Supports Loads Corbels Wall beams

Why use Strut and Tie?

• Simple Model for complex problems

• Easy to understand

• The engineer stays in control – it is not a

complicated computer analysis

• Can provide more economical solutions

• It is very powerful for the analysis of existing

structures

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What is strut and tie?

Strut‐and‐tie models (STM) are like trusses consisting of struts, ties and nodes. 5 a) Modelling Imagine or draw stress paths which show the elastic flow of forces through the structure b) STM Replace stress paths with polygons of forces to provide equilibrium. A deep beam Conventionally, struts are drawn as dashed lines, ties as full lines and nodes numbered. 5

What is strut and tie?

Strut and tie models consist of: • Struts (concrete) • Ties (reinforcement) • Nodes (intersections of struts and ties). Eurocode 2 gives guidance for each of these. • In the UK this is the first time that S&T has been codified for general use. • It says that in principle, where non‐linear strain distribution exists, i.e. in D‐regions, strut and tie models may be used, e.g. for: • Pile caps • Deep beams • Supports • Concentrated loads • Openings, etc. 6

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Lower Bound method

Strut and tie models are based on the lower bound theorem

of plasticity which states that any distribution of stresses

resisting an applied load is safe providing:

 Equilibriumis maintained and  Stresses do not exceed “yield” 7

Design Process

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Design Process

1. Define and isolate B and D regions

2. Develop an STM

3. Design the members of the STM – struts, ties and

nodes

4. Iterate to optimise the STM as necessary to

minimise strain energy

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1. B & D regions

A structure can be divided into: •B (or beam or Bernoulli) regions and •D (or disturbed) regions 10 D regions 10

As before:

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11 a) Modelling Imagine or draw stress paths which show the elastic flow of forces through the structure b) STM

Replace stress paths with polygons of forces to provide equilibrium.

A deep beam

Conventionally, struts are drawn as dashed lines, ties as full lines and nodes numbered.

2. Develop an STM

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As before:

a) Consider a deep beam with an udl b) Consider elastic stresses (here using FEA) A A Consider elastic stresses in section A‐A

Developing an STM

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Centroid of compressive force

Stress distribution within deep beam with udl

Section A‐A Centroid of tensile force

Developing an STM

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Developing an STM

a) Consider a deep beam with an udl b) Consider elastic stresses (here using FEA) c) Construct STM 14

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Construction of STMs

Elastic analyses of deeper beams with point load

Construction of STMs

Very deep beams

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Construction of STMs

Good and bad model based on minimising length of ties

Minimise tie lengths (minimises energy used) 17 18

a)

Struts

b)

Ties

c)

Nodes

3. Design members

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P1 P2

Question: at failure which is bigger P

₁

or P

₂

?

•Concept by R Whittle/A Beeby, drawn by I Feltham. Used with permission

3a) Struts

Possible answers: A) P1/P2 > 2.0 B) P1/P2 = 2.0 C) P1/P2 = 1.5 D) P1/P2 = 1.0 E) P1/P2 = 0.5 19 P1 P2



2P

₁

3a) Struts

Answer: E) P

₁

/P

₂

= 0.5

20 Possible answers: A) P1/P2 > 2.0 B) P1/P2 = 2.0 C) P1/P2 = 1.5 D) P1/P2 = 1.0 E) P1/P2 = 0.5

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fcd fcd fctd, yy fctd,zz

Bi‐axial strength of Concrete

3a) Struts

0 0 ₂₂ Tension Reduction in compressive strength Where there is no transverse tension _Rd,max= fcd = 0.85fck/1.5 = 0.57 fck Otherwise, where there is transverse tension _Rd,max= 0.6 ’fcd Where: ’ = 1‐fck/250 _Rd,max= 0.6 x (1‐fck/250) x 1.0 x fck/1.5 = 0.4 (1‐fck/250) fck

3a) Struts

EC2 says:

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Discontinuities

Areas of non‐linear strain distribution are referred to as “discontinuities” Partial discontinuity Full discontinuity Curved compression trajectories lead to tensile forces

3a) Struts

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Partial discontinuity

Tension, T, is taken by the reinforcement When b ≤ H/2 T = ¼ [(b – a )/b] F Reinforcement ties to resist the transverse force T may be “discrete” or can be “smeared” over the length of tension zone arising from the compression stress trajectories T T

3a) Struts

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Full discontinuity

Tension, T, is taken by the reinforcement When b > H/2 T = ¼ (1 – 0.7a /h) F Reinforcement ties to resist the transverse force T may be “discrete” or can be “smeared” over the length of tension zone arising from the compression stress trajectories T

3a) Struts

T 26

Note the 2 componentsto the strut dimension

Dimensions of the strut are determined by dimensions of the nodes and assumptions made there.

3a) Struts

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Up to now we have been talking about prism struts.

3a) Fan Struts

In a uniformly loaded deep beam, the flow of internal forces may be visualized either by fan strut‐and‐tie

models or by using more elaborate _{discontinuous stress fields.}

more realistic Usual to assume fan struts OK and check the common governing criterion. . . . . the CCT node Fan struts

Design strength, f

_yd

= f

_yk

/1.15

A

_s

= F

_Ed

/f

_yd

3b) Ties

NB. Reinforcement should be anchored into nodes. Anchorage may start in the extended nodal zone (= strut area) 29

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Nodes are typically classified as: CCC – Three compressive struts CCT – Two compressive struts and one tie CTT – One compressive strut and two ties

3c) Nodes

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CCC nodes

(e.g. column onto two pile cap) The maximum stress at the edge of the node: _Rd,max = k₁’f_cd Where: k1 = 1.0 ’ = 1‐fck/250 _Rd,max = (1‐f_ck/250) x 0.85 x f_ck/1.5 = 0.57 (1‐f_ck/250) f_ck

The stresses _c0& _Rd,2etc are all the same.

3c) Nodes

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CCT nodes

(e.g. beam at end support)

The maximum compressive stress is: _Rd,max = k₂’f_cd Where: k₂ = 0.85 ’ = 1-f_ck/250 _Rd,max = 0.85 (1-f_ck/250) x 0.85 x f_ck/1.5 = 0.48 (1-f_ck/250) f_ck

(based on the more critical of the two struts)

3c) Nodes

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CTT nodes

(e.g. closing corner of a wall) The maximum compressive stress is: _Rd,max = k2’fcd Where: k₂ = 0.75 ’ = 1‐fck/250 σ_Rd,max = 0.75 (1-f_ck/250) x 0.85 x f_ck/1.5 = 0.43 (1-f_ck/250) f_ck

3c) Nodes

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4) Iterate

2‐pile cap – probably not Cantilever wall beam (deep beam) with window – definitely 34

Examples

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Pile‐cap worked example

Using a strut and tie model, what tension reinforcement is required for a pile cap supporting a 500 mm square column carrying 2500 kN (ULS), and itself supported by two‐piles of 600 mm diameter. f_ck= 30 MPa 2 500 kN (ULS) 2700 1400 150 Breadth = 900 mm 36 Angle of strut = tan‐1_(900/1300) = 34.7° Width of strut* = 250/cos 34.7° = 304 mm Force per strut = 1250/cos 34.7° = 1520 kN Force in tie = 1250 tan 34.7° = 866 kN 2 500 kN (ULS) 1800 1400 1 250 kN (ULS) 100 Strut angle 500/2 = 250 1 250 kN (ULS) 34.7o_34.7o 866 kN

STM

*Conventional but simplistic - see later

Pile‐cap worked example

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Check stresses in strut Stress in strut (top) =1520 x 103_{/(304 x 500)} =10.0 MPa Strength of strut (conservatively assuming some transverse tension): _Rd,max = 0.4 (1‐fck/250) fck = 10.6 MPa Tie : Area of steel required: As ≥ 866 x 103/435 ≥ 1991 mm2 Use 5 H25s 866 kN Therefore OK Usually QED But for the sake of thoroughness . . . . .

Pile‐cap worked example

38 • Nodes: top _Rd,2 = 10.0 MPa (as before) _Rd,3 = 10.0 MPa (as before) _Rd,1 = 2500 x 103_/(5002₎ = 10.0 MPa _Rd,max (for CCC node) = 0.57 (1‐fck/250) fck = 15.0 MPa 39 2500 kN 1520 kN 1520 kN 2500 kN 1520 kN 1520 kN (Upside down elevation!) (Elevation)

Pile‐cap worked example

(thorough)

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Nodes: bottom (as a check) Strut above Width of strut* = 600/cos 34.7° = 730 mm Stress in strut (bottom as an ellipse) _Rd,2=1520 x 103_{/(600 x 730 x /4)} = 4.4 MPa _Rd,1= 1250 x 103_{/( x 300}2₎ = 4.4 MPa _Rd,max (for CCT node) = 0.48 (1‐fck/250) fck = 12.7 MPa OK = 1250 kN 1038 kN

*Conventional but simplistic - see later

Pile‐cap worked example

(thorough)

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Comparison:

Compare previously designed pile

cap using bending theory

MEd=2500 x 1.800/4 = 1125 kNm Assume:

25 mm  for tension reinforcement 12 mm link

d = h – c_nom‐ _link‐ 0.5

= 1400 – 75 ‐ 12 – 13 = 1300 mm

Pile‐cap worked example

(thorough)

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Comparison:

A_s= 1125 x 106_{/ (435 x 1270) =}₂₀₃₆_mm2 _{Use 5 H25 (2454 mm}2₎ ' 025 . 0 30 1300 900 10 1125 2 6 ck 2 Ed K f bd M K        208 . 0 ' K







1 1 3.53 0.025



1270mm 2 1300 53 . 3 1 1 2        d K z  K’ 1.00 0.208 0.95 0.195 0.90 0.182 0.85 0.168 0.80 0.153 0.75 0.137 0.70 0.120

c.f. using S&T 1991mm2_{req’d and 5H25 provided}

Pile‐cap worked example

(thorough)

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For the CCT node: wcos  not used in previous calc.

So for the CCC node RE *previous statement that calculated strut dimensions were “Conventional but simplistic ‐ see later” In highly stressed situations in CCCs, fib bulletin 62 allows a local STM to take account of the reinforcement from columns.

Strut dimensions

Pile‐cap worked example

(thorough)

43 Hence struts themselves are rarely critical.

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2, 3 & 4 Pile‐caps

Forces in ties

Pile‐caps etc.

Detailing Detailing of reinforcement anchorage critical – 100% As needed up to extended node (large radius bends may be required) Using S&T, anchorage of 100% As required here cf 25% from here if using bending theory

Pile‐caps etc.

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Detailing

Where flexural design has been used it is common UK practice to provide uniform distribution of reinforcement. However, EN 1992‐1‐1 Clause 9.8.1(3) suggests that “the tensile reinforcement . . . should be concentrated in the stress zones between the tops of the piles”. There is evidence to suggest that bunching orthogonal reinforcement leads to a standard 4‐pile cap being 15% stronger than using the same amount of uniformly distributed reinforcement. The requirement for concentrating reinforcement can be interpreted in different ways but the apparent shortcoming can be alleviated by providing transverse tension and tie‐back reinforcement to distribute forces from bars as indicated in Figure 5.5 (wait for it!). For pile caps supporting structures other than bridges, there would appear to be little reason to deviate from the advice given in BS8110 “ . . only the reinforcement within 1.5 times the pile diameter from the centre of a pile shall be considered to constitute a tension member of a truss”. So in this case, 5 no. H25s distributed across a 900 mm wide pile cap section is considered satisfactory

Tensile force in tie

Pile‐caps etc.

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Problem

Deep beam worked example

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STM

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Deep beam worked example

48 Consider moment about B RA = 2529 x 3.05 / 4.30 = 1794 kN So RB = 2529 ‐ 1794 = 735 kN F12 = 1794 x 1.80 / 1.30 = 2484 kN strut F13 = 2484 x 1.25 / 1.80 = 1725 kN tie F34 = F56 = 735 kN ties Forces Check bearing stresses At node 2, under load F Ed = 2529 x 103 / (450 x 450) = 12.5 MPa CCC node ∴ Rd= 1.0 x (1 – 35/250) x 0.85 x 35 / 1.5 = 17.1 MPa ∴ OK At node 1 at support A (see Figure 5.8) Ed = 1794 x 103 / (475 x 450) = 8.39MPa CCT node ∴ Rd= 0.85 x (1 – 35/250) x 0.85 x 35 / 1.5 = 14.5 MPa ∴ OK At node 7 at support B OK by inspection

Deep beam worked example

49 Ties F13 = 1725 kN: As req’d = 1725 x 103 / (500/1.15) = 3968 mm2 Try 8H25 (3928 mm2 say OK) in two layers i.e. 2 x 4 H25 @ 50 mm cc by inspection provide bobs at end of bars F34 = F56 = 735 kN : As,req’d= 735 x 103 / (500/1.15) = 1690 mm2per tie i.e. per 3.05/3m say 1690 mm2/m. Try H16@225 both sides (1768 mm2_/m) Bursting forces (bottle ties)

Check strut 1‐2: F12 = 2484 KN strut has full discontinuity T = ¼ (1 – 0.7a/H) F

T = ¼ (1 – 0.7 x 0.18) x 2484 = 542.8 kN ∴ As reqd = 542.8 x 103 / (500 / 1.15)= 1248 mm2

To be placed between 0.2H and 0.5H from the loaded surface. i.e. 1248 mm2_{to be placed}

over 0.3 x 1800 = 540 mm ≡ 2311 mm2_{/m over 540 mm at 1.25 in 1.30 slope.}

Try H16@ 175 (1148 mm2/m) both ways both sides (2296 mm2/m both ways (say OK))

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Deep beam worked example

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Summary

Other examples

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• T‐headed bars Vertical section: • Double headed T‐headed bars in shear assemblies Similar plan section: • Single headed T‐headed bars in short lap assemblies or anchorages. 52

Other examples

Other examples (advanced)

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Analysis of two-storey wall beam Analysis and design of a coupling beam (with hole) within a shear wall in a 54-storey block.

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strut inclination method

 cot sw s Rd, z fywd s A V  21.8 <  < 45

Eurocode 2 – Beam shear:

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Other examples

Why use Strut and Tie?

•Simple Model for complex problems •Easy to understand •The engineer stays in control – it is not a complicated computer analysis •Can provide more economical solutions •It is very powerful for the analysis of existing structures 55

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3) International Federation for Structural Concrete (2011) fib Bulletin No. 61: Design examples for strut‐and‐tie models, Lausanne, Switzerland: fi b 4) Thurlimann B., Muttoni A. and Schwartz J. (1989) Design and detailing of reinforced concrete structures using stress fi 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., Schäfer K. and Jennewein M. (1987) ‘Towards a consistent design of structural concrete’, PCI Journal, 32 (3), pp. 74–150 7) Sagaseta J. and Vollum R. L. (2010) ‘Shear design of short‐span beams’, Magazine of Concrete Research, 62 (4), pp. 267–282 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 Schäfer K. (2001) ‘Konstruieren im Stahlbetonbau’ (in German), BetonKalender (Vol 2), Berlin, Germany: Ernst & Sohn, pp. 311–492 13) Vollum R. L. and Fang L. (2014) ‘Shear enhancement in RC beams with multiple point loads’, Engineering Structures, 80, pp. 389–405 AMERICAN CONCRETE INSTITUTE, Building Code Requirements for Structural Concrete and Commentary ACI 318‐08, ACI, Farmington Hills MI 2008. “ACI 318”. 2 SCHLAICH, J., SCHAFER, K.: “Design and detailing of structural concrete using strut and tie models”, The Structural Engineer, Vol. 69, No. 6, March 1991, pp. 113‐125. 3 CEB‐FIP. Model Code for Concrete Structures, CEB‐FIP International Recommendations, 1990, “Model Code 90”. 4 CANADIAN STANDARDS ASSOCIATION (CSA A.23.3‐04). Design of Concrete Structures, 2004. 5 COLLINS M. P. and MITCHELL D. Prestressed Concrete Structures, 1st edn. Prentice Hall, Englewood Cliffs, New Jersey, 1991. 6 BRITISH STANDARDS INSTITUTION. BS EN 1992–1–1, Eurocode 2 – Part 1–1: Design of concrete structures – General rules and rules for buildings. BSI, 2004. 6a National Annex to Eurocode 2 – Part 1–1 incorporating Amendment 1. BSI, 2009. 7 HENDY, C R & SMITH D A. Designer’s Guide to EN 1992‐2, Eurocode 2: Design of concrete structures, Part 2: Concrete Bridges.Thomas Telford, London, 2007. 8 SAHOO K D, SINGH B & BHARGAVA P. Minimum Reinforcement for Preventing Splitting Failure in Bottle shaped Struts, ACI Structural Journal, March April 2011, pp. 206‐216. 9 SIGRIST V , ALVAREZ M & KAUFMANN W. Shear And Flexure In Structural Concrete Beams, ETH Honggerberg, Zurich, Switzerland, (Reprint from CEB Bulletin d’Information No. 223 “Ultimate Limit State Design Models” June 1995). 10 BRITISH STANDARDS INSTITUTION. BS 8004 Code of practice for Foundations, BSI, 1986. 11 BLEVOT, J. L., AND FREMY, R. “Semelles sur Pieux,” Institute Technique du Batiment et des Travaux Publics, V. 20, No. 230, 1967, pp. 223‐295. 12 BRITISH STANDARDS INSTITUTION. BS 8110‐1:1997 Structural use of concrete ‐ Part 1: Code of practice for design and construction, Amd 4, BSI,2007 13 THE INSTITUTION OF STRUCTURAL ENGINEERS. Standard Method of Detailing Structural Concrete. A Manual for best practice. (3rd edition) 2006, ISBN, 978 0 901297 41 9

Strut & Tie

Charles Goodchild