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
2. Design requirements
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
Design value of compressive stresses within nodes
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
Design value of compressive stresses within nodes
3. Compression-tension nodes with anchored ties provided in
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
Detailing rules for frame corners
19
Design requirements
Introduction
Design
requirements
Worked
Examples
Detailing rules for frame corners
20
Design requirements
Introduction
Design
requirements
Worked
Examples
Detailing rules for frame corners
Outline
1. Introduction
2. Design requirements
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
Provide 4H20 As2=1257 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
2. Compression-tension nodes with anchored ties provided in
one direction
2Rd,max = k2 fcd = 0.85x(1-35/250)x19.83 = 14.50 MPa
3. Compression-tension nodes with anchored ties provided in
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
Provide 3H20 As1=942 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 !