013 - strut and tie method (2014 11 09) (1)

13. Design for RC structures with

strut-and-tie models

Prof Tan Kang Hai

Email: [email protected]

Director of Protective Technology Research Centre (PTRC)

School of Civil & Environmental Engineering

Outline

1. Introduction

2. Design requirements

3. Worked examples

2

3

D-Regions (D: Discontinuity / Disturbed): Regions in

structures that are adjacent to discontinuities, concentrated

loads, holes, or changes in cross-section. In these regions,

strain distribution is not linear, normal flexure theory cannot

be used;

Definition of adjacent: In general these regions extend from the discontinuities up to a distance that equals to one

section depth of the member (EC2 Clause 5.6.4(1))

B-Regions (B: Beam / Bernoulli): Regions in which there is

linear distribution of strains and normal flexure theory can be

applied.

Introduction

Introduction

Design

requirements

Worked

Examples

4

Introduction

Design

requirements

Worked

Examples

F

hc hc

hb

hc

hc

hc

hc

hb hb hb hb

D-Regions B-Regions

Introduction

Introduction

Design

requirements

Worked

Examples

F

Concrete strut

Steel tie

Nodal

zone

Concrete strut

Nodal zone

Nodal zone

In D-Regions, elastic stress field is disrupted by cracking, then internal forces

are reoriented so that their major portions are transferred directly to the

supports following strut-and-tie model, which is a hypothetical truss consisting

of:

- Concrete struts in compression (in bottle shapes but idealised as prismatic);

- Reinforcing steel ties in tension; and

- Nodal zones joining the struts and ties together.

Introduction

- Crushing of the struts;

- Failure of nodal zones;

- Anchorage failure of the ties

More ductile failure desirable

6

Introduction

Design

requirements

Worked

Examples

Strut-and-tie models are assumed to fail due to:

- Yielding of the ties;

F

Introduction

7

Introduction

Design

requirements

Worked

Examples

Introduction

1. Introduction


3. Worked examples

8

Outline

9

Design requirements

Introduction

Design

requirements

Worked

Examples

Without transverse tension

Rd,max

Rd,max = fcd

(there is transverse

compressive stress or no

transverse stress)

Rd,max

With transverse tension

Rd,max = 0.6 fcd

= 1 fck/250

(concrete struts in cracked

compression zones)

Rd,max Rd,max

Design strength of concrete struts

10

Introduction

Design

requirements

Worked

Examples

Design strength of transverse ties and reinforcement:

fyk/1.15 (Clause 6.5.3(1))

Reinforcement should be adequately anchored in the nodes

(Clause 6.5.3(2))

Design requirements

11

Design requirements

Introduction

Design

requirements

Worked

Examples

12

Design requirements

Introduction

Design

requirements

Worked

Examples

Design value of compressive stresses within nodes

1. Compression nodes where no ties are anchored at the node

Rd,max = k1 fcd

= 1 fck/250

k1= 1

Rd,max = max (Rd,1, Rd,2,Rd,3)

13

Design requirements

Introduction

Design

requirements

Worked

Examples


2. Compression-tension nodes with anchored ties provided in

one direction

Rd,max = k2 fcd

= 1 fck/250

k2= 0.85

Rd,max = max (Rd,1, Rd,2)

14

Design requirements

Introduction

Design

requirements

Worked

Examples



more than one direction

Rd,max = k3 fcd

= 1 fck/250

k2= 0.75

Rd,max = max (Rd,i)

15

Design requirements

Introduction

Design

requirements

Worked

Examples

16

Design requirements

Introduction

Design

requirements

Worked

Examples

17

Design requirements

Introduction

Design

requirements

Worked

Examples

Detailing rules for frame corners

18

Design requirements

Introduction

Design

requirements

Worked

Examples


19

Design requirements

Introduction

Design

requirements

Worked

Examples


20

Design requirements

Introduction

Design

requirements

Worked

Examples


Outline

1. Introduction


3. Worked examples

21

Worked Example #1

Introduction

Design

requirements

Worked

Examples

Design a wall beam shown below with strut-and-tie model:

Material: Concrete C25/30 fck=25 MPa; fcd=0.85x25/1.5=14.17 MPa

Reinforcing steel B450C fyk=450 MPa; fyd=450/1.15=391.3 MPa

22

400 4004600

5400

250

P = 405 kN CL P = 405 kN

1500 1200 15001200

1200

Column 250x400Column 250x400

P = 405 kN P = 405 kN

A

A

Section A-A

2460

2460

Worked Example #1

Introduction

Design

requirements

Worked

Examples

(Source: [1])

23

Reactions & Applied loads Strut-and-tie model

P

2

3 4

C2

T2C3

T1

1

200

180

P = 405 kN CL P = 405 kN

1500 1200 15001200

P

R R

12001300 13001200 200

57

2460

2000

280

400 4004600

5400

P = 405 kN CL P = 405 kN

1500 1200 15001200

1200

P = 405 kN P = 405 kN

1300 13001200

R = 2P = 810 kN R = 2P = 810 kN

2460

Worked Example #1

Introduction

Design

requirements

Worked

Examples

(Source: [1])

24

Force equilibrium at nodes Strut-and-tie model

P

2

3 4

C2

T2C3

T1

1

200180

P = 405 kN CL P = 405 kN

1500 1200 15001200

P

R R

12001300 13001200 200

57

2460

2000

280

Worked Example #1

Introduction

Design

requirements

Worked

Examples

(Source: [1])

25

Strut-and-tie model

Required area of reinforcing

bars at T1:

As1=T1/fyd=526x103/391.3=1344

mm2

Provide 6H18 As1=1524 mm2

Required area of reinforcing

bars at T2:

As2=T2/fyd=405x103/391.3=1035

mm2


T1=526 kN

T2=405 kN

P

2

3 4

C2

T2C3

T1

1

200

180

P = 405 kN CL P = 405 kN

1500 1200 15001200

P

R R

12001300 13001200 200

57

2460

2000

280

Worked Example #1

Introduction

Design

requirements

Worked

Examples

(Source: [1])

26

Arrangement of reinforcement Determination of nodal zones

400

CL

1500 1200 15001200

12001100 1200

2460

4001100

4H20 4H20

6H18

400

CL

1500 1200 15001200

12001100 1200

2460

4001100

265.8

265.8

280

125 125 125 125

280

Worked Example #1

Introduction

Design

requirements

Worked

Examples

(Source: [1])

27

Node 2: Compression-tension

nodes with anchored ties

provided in one direction

Rd,max = k2 fcd

= 1 fck/250

k2= 0.85

Compressive strength:

2Rd,max = 0.85x(1-25/250)x14.17

Verification of Node 2

c1 = 405x103/(250x250)=6.48 < 10.83 MPa

c2 = 526x103/(560x250)=3.74 < 10.83 MPa

2Rd,max = 10.83 MPa

c3 = 966x103/(531.6x250)=7.27 < 10.83 MPa

280

280

265.8

265.8

Worked Example #1

Introduction

Design

requirements

Worked

Examples

(Source: [1])

28

Node 3: Compression-tension

nodes with anchored ties

provided in one direction

Verification of Node 3

cR = 810x103/(400x250)=8.1 < 10.83 MPa

c3 = 966x103/(531.6x250)=7.27 < 10.83 MPa

Rd,max = k2 fcd

= 1 fck/250

k2= 0.85

Compressive strength:

2Rd,max = 0.85x(1-25/250)x14.17

2Rd,max = 10.83 MPa

Worked Example #1

Introduction

Design

requirements

Worked

Examples

29

Arrangement of reinforcement

Worked Example #1

Introduction

Design

requirements

Worked

Examples

30

Worked Example #2

Introduction

Design

requirements

Worked

Examples

(Ref: [1])

31

Design a corbel shown below with strut-and-tie

model:

Material: Concrete C35/45 fck=35 MPa; fcd=0.85x35/1.5=19.83 MPa

Reinforcing steel B500 fyk=500 MPa; fyd=500/1.15=434.8 MPa

Worked Example #2

Introduction

Design

requirements

Worked

Examples

(Ref: [1])

32

Determine compression strength of nodes:

1. Compression nodes where no ties are anchored at the

node:

1Rd,max = k1 fcd = 1.0x(1-35/250)x19.83 = 17.05 MPa


one direction

2Rd,max = k2 fcd = 0.85x(1-35/250)x19.83 = 14.50 MPa


more than one direction

3Rd,max = k3 fcd = 0.75x(1-35/250)x19.83 = 12.79 MPa

Worked Example #2

Introduction

Design

requirements

Worked

Examples

(Ref: [1])

33

Equivalent applied loads Strut-and-tie model

Determination of x at node 2 under compression: C2=FEd=600 kN;

2x > C2/(b1Rd,max)=600x103/(400x17.05)=88 mm.

Choose x=50 2x=100 mm

Worked Example #2

Introduction

Design

requirements

Worked

Examples

(Ref: [1])

34

Force equilibrium at nodes Strut-and-tie model

Required area of reinforcing bars at T1:

As1=T1/fyd=324.07x103/434.8=745 mm2


Worked Example #2

Introduction

Design

requirements

Worked

Examples

(Ref: [1])

35

cF = 600x103/(300x150)=13.3 < 2Rd,max = 14.5 MPa

c3 = 681.93x103/(400x128)=13.3 < 2Rd,max =14.5 MPa

Node 1:

Compression-

tension nodes with

anchored ties

provided in one

direction

Worked Example #2

Introduction

Design

requirements

Worked

Examples

(Ref: [1])

36

Node 2:

Compression node

without anchored ties

provided in one

direction

c1 = 324.7x103/(400x72)=11.2 < 1Rd,max = 17.05 MPa

c2 = 600x103/(400x100)=15.0 < 1Rd,max = 17.05 MPa

c3 = 681.93x103/(400x128)=13.3 < 1Rd,max =17.05 MPa

Worked Example #2

Introduction

Design

requirements

Worked

Examples

(Ref: [1])

37

Determination of secondary layer of reinforcement:

Required area of reinforcing bars at T2:

As2=T2/fyd=305.3x103/434.8=702 mm2 Provide 5x2H10 As2=785 mm

2

Worked Example #2

Introduction

Design

requirements

Worked

Examples

38 Arrangement of reinforcement

(Ref: [1])

Worked Example #3

Introduction

Design

requirements

Worked

Examples

39

Geometry: 325mmx300 mm cantilever beam (width b=400 mm),

150mmx220mm load plate, 400mmx400mm column

325mmx300mm cantilever beam

and the strut-tie model

Materials:

concrete

C40/50 fck=40MPa,

steel fyk=500MPa

Design a deep cantilever beam with strut-and-tie

model:

22

0

Worked Example #3

Introduction

Design

requirements

Worked

Examples

40

The model proposed in EC2 is indeterminate with one more boundary

condition needed to evaluate the stress values in the model.

Stress in vertical reinforcement Fwd is evaluated assuming a linear relation

between Fwd and the value in the range Fwd=0 when a=z/2 and Fwd=FEd

when a=2z.

When a z/2 (a corbel or a deep beam), truss 1 is the only model and when

a 2z (cantilever short beam) truss 2 is the only model.

Worked Example #3

Introduction

Design

requirements

Worked

Examples

41

The expression for Fwd is:

When the two conditions Fwd (a=z/2 )=0 and Fwd(a=2z) =FEd are imposed,

some trivial algebra leads to:

In conclusion, the expression for Fwd as a function of a is the following:

Material strength:

fcd=0.85fck/1.5=0.85x40/1.5=22.67MPa

fyd=fyk/1.15=500/1.15=434.8MPa

1

2

3

1

3

1

3

2

z

aFF

z

aFF

EdEdEdwd

EdwEdwFF

z

aFF

3

1;

3

2

21

21 wwwdFaFF

Worked Example #3

Introduction

Design

requirements

Worked

Examples

42

Nodal compression resistance

Compression nodes:

1Rd, max=k1(1fck/250)fcd=1(140/250)22.67=19.04MPa

Tied-compressed nodes with tension rods in one direction:

2Rd, max=k2(1fck/250)fcd=0.85(140/250)22.67=16.19MPa

Tied-compressed nodes with tension rods in different directions:

3Rd, max=k3(1fck/250)fcd=0.75(140/250)22.67=14.28MPa

Worked Example #3

Introduction

Design

requirements

Worked

Examples

43

Actions

FEd=500 kN

Load eccentricity with respect to the column outer

side: e=200 mm

The column vertical strut width is evaluated setting the

compressive stress equal to 1Rd, max:

x1=Fed/(1Rd, max b)=500000/(19.04x400)=63mm

Node 1 is located x1/2=31.5mm from the outer edge of the

column

The upper reinforcement is located 40mm from the beam top.

Hence d = 300 -40 =260 mm and the internal lever arm z is

set at at 0.8d=208mm. For the bottom nodal zone 1,

y1=0.2d=52mm (from beam soffit to centre of Fc)

Fed

Ft

Fc

Fwd

e

F

z

d

x1

a

Node 2

Node 1

Worked Example #3

Introduction

Design

requirements

Worked

Examples

44

Rotational equilibrium:

FEd(e1+x1/2)=Fcz

Fc=Ft=500000x(200+31.5)/208=556.5kN

Node 1 verification

=Fc/(2by1)=556.5x1000/(2x400x52)

=13.37MPa

Worked Example #3

Introduction

Design

requirements

Worked

Examples

45

Awd=Fwd/fyd=204.3x1000/434.8=470mm2

EC2 (J.3(3)) suggests a minimum secondary reinforcement of

Awd=k2FEd/fyd=0.5x500x1000/434.8=575mm2

we use 3 stirrups 12 (As=678mm2)

Node 2 verification, below the load plate

Node 2 is a compressed-stressed node, in which the main

reinforcement is anchored, the compressive stress below the load

plate is

=FEd/(150X220)=500x1000/(150x220)

=15.15MPa 2Rd, max=16.19MPa

Worked Example #4

Introduction

Design

requirements

Worked

Examples

46

Design an inverted T beam with a strut-and-tie model:

Material: Concrete C35/45, fck=35MPa

The inverted T beam supports 12 m long, 3.6 m wide double T beam. The

width of the double T is 120mm and the bearing plate is 150mm long. The

dead load of the double T is 3kPa, including self-weight, and the beam

carries a live load of 2kPa. A horizontal force is equal to 20 percent of the

vertical reaction.

50

03

00

450 25

840

100 100145100 145

Reinforcing steel fyk=450MPa

Worked Example #4

Introduction

Design

requirements

Worked

Examples

47

Material strength:

fcd=0.85fck/1.5=0.85x35/1.5=19.83MPa

fyd=fyk/1.15=450/1.15=391.3MPa

Factored loads on the beam stem for 1.8 m wide single T beam and 12

m span are

qu=1.35x3 + 1.5 x2=7.05 kPa

Ru=7.05x1.8x12/2=76.14 kN

and Tu=0.2x76.14=15.23 kN

.

Ru=76.14 kN

Tu=15.23 kN

Ru=76.14 kN

Tu=15.23 kN

Worked Example #4

Introduction

Design

requirements

Worked

Examples

48

The strut-tie mode is shown below:

Tb

d=

94

.03

kN

b

d

20

0

145

Ru=76.14 kN

Tab=70.43 kN Tu=15.23 kN

Tdf=76.14 kN

Fcd=55.2 kN

9797

13

3

50

03

00

450 25

840

100 100145100 145

a b

c d

e f

Worked Example #4

Introduction

Design

requirements

Worked

Examples

49

At node b:

The bearing area under the double T leg is 120mm by 150mm =

18000mm2, giving a nodal bearing stress of

n=79.74x1000/18000=4.43 MPa

Strut bd:

bd=Fbd/Abd=94.03x1000/(97x150)=6.46MPa

The nominal capacity of the node is:

1Rd, max=k1x(1fck/250)xfcd

=0.75x(135/250)x19.83=12.79MPa

> n= 4.43 MPa,

>bd=6.46MPa

Worked Example #4

Introduction

Design

requirements

Worked

Examples

50

At node d:

Strut bd:

db=Fdb/Adb=94.03x1000/(97x150)=6.46MPa

Strut dc:

dc=Fdc/Adc=55.2x1000/(133x150)=2.77MPa

The nominal capacity of the node is:

1Rd, max=k1x(1fck/250)xfcd

=0.85x(135/250)x19.83=14.50MPa

> db= 6.46MPa,

>dc=2.77MPa

Worked Example #4

Introduction

Design

requirements

Worked

Examples

51

The required area for horizontal tie ab is

Ats=Tab/fy=70.43x1000/391.3=180mm2

Provide two No. 13 bars welded to each bearing plate.

For vertical tie df,

Ats=Tdf / fy=76.14x1000 / 391.3= 195mm2

Provide two No. 13 closed stirrups at 100 mm centre at

each load point.

Worked Example #5

Introduction

Design

requirements

Worked

Examples

(Ref: [4])

52

Strut-and-tie model for single deep beam

Worked Example #6

Introduction

Design

requirements

Worked

Examples

(Ref: [5])

53

Strut-and-tie model for single deep beam

subjected to unequaled loads

Worked Example #7

Introduction

Design

requirements

Worked

Examples

(Ref: [6])

54

Strut-and-tie model for continuous deep beam

Worked Example #8

Introduction

Design

requirements

Worked

Examples

(Ref: [1])

55

Strut-and-tie model for variable height beam

Worked Example #9

Introduction

Design

requirements

Worked

Examples

(Ref: [1])

56

Strut-and-tie model for wall with opening

57

References

1. Eurocode 2 Worked Examples European Concrete Platform, May 2008.

2. British Standard Institution. BS EN 1992-1-1:2004: Eurocode 2: Design of

concrete structures Part 1-1: General rules and rules for buildings. London, BSI, 2004.

3. Reinforced concrete Mechanics and Design J.G MacGregor et al. (2005)

4. Examples for the design of structural concrete with Strut-and-Tie models K.H. Reineck (2002)

5. Single-span deep beams subjected to unsymmetrical loads N. Zhang; KH Tan (ASCE 2009)

6. Direct strut-and-tie model for single span and continuous deep beams - N.

Zhang; KH Tan (Engineering structures 2007)

Thank You !

013 - strut and tie method (2014 11 09) (1)

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