CIVL6003 – Advanced Reinforced

Concrete Structure Design

THE UNIVERSITY OF HONG KONG

Ir Dr. Ray Su
Department of Civil Engineering

(office: HW 6-6, Tel. no.: 2859 2648

Strut & Tie Modelling - 03 Email: klsu@hku.hk)
Nodes
• Nodes are the connections of struts and ties in truss
models.
• There are two main types of nodes, they are concentrated
nodes and smeared nodes. The concentrated nodes are
idealized as points whereas the smeared nodes is spread
and may extend throughout a significant part of the D-
region.
Type of nodes

CCC CCT CTT TTT

Strut
Tie
2
Concentrated Nodes and Smeared Nodes
Idealized model Actual behaviour

Uniform load on a
Smeared node
deep beam
Bottle-shaped str
Conc. node

Uniform load on a deep

beam with single central
Smeared node
support Bottle-shaped str
Conc. node

Without steel bars

Opposing concentrated Conc. node

Bottle-shaped str
loads on a deep beam Smeared node
Bottle-shaped str
Conc. node
3
The behaviour of node depends on the arrangement and position of ties.
Concentrated Nodes and Smeared Nodes

• The smeared nodes are rarely of concern in

design (don’t need to design normally)
• Concentrated nodes are often the most highly
stressed and hence most critical parts of the D-
region. Bearing plate
• In determining the size and shape of a
concentrated node the locations of individual
reinforced bars and the size of any adjacent
bearing plate have to be considered carefully.
• When the node contains steel bars which have
to be anchored, the anchorage length should
also be considered in relation to the size of the
node.
• The shape and size of the node are likely to Check anchorage length
govern the size of the adjacent struts.

4
Node in Hydrostatic & Force Equilibrium
Node in Hydrostatic Equilibrium Node in Force Equilibrium
a2 is known a1 & a2 are known

a3 C1 a3 C1

σ a1 σ1 a1
σ σ3
C3 C3 α

σ σ2 C2
C2
a2 a2

force C 3 C 2 C1
  =σ×b C3=(C12+C22)1/2
size of node a 3 a 2 a1
a3=a1cosα+a2sinα
a2
a1  C1 σ3=C3/(a3×b) 5
C2
Strength of Nodes
The strength of concrete in the nodal zones depends on a number of
factors such as
Tensile strain at node
• the confinement of the zones by the reactions, would weaken its strength
compression struts, anchorage plates for
prestressing, and hoop reinforcement;
• the effects of strain discontinuities within the
nodal zone when ties strained in tension are
anchored in, or cross, a compressed nodal zone;
and
• the splitting stresses and hoop-bearing stresses Increase the anchorage
resulting from the anchorage of the reinforcing length
bars of a tension tie in or immediately behind a
nodal zone.

Splitting
stresses
Hoop-bearing stress
6
Design Strength of Node
The Design Strength of Node can be expressed as,

f node   n f cu (2)

where n is the node efficiency factor given in Table 1.

Table 1. Node efficiency factor

Condition Node efficiency factor (n)
of Node
Triaxial CCC 0.55
Uniaxial CCC 0.45
CCT 0.40
CTT 0.36
Minimum 0.28
Partial safety factor of 1/0.67 is allowed

Type of nodes

7
CCC CCT CTT TTT
Design Strength of Bearing
When the size of support is smaller than that of the node, higher stresses would be
developed at the bearing face. A plinth is introduced to spread the load.

Bearings and plinths

Node
A2
2 A1 < A2
Plinth 1 > 2
Bearing 1
1 2
A1
8
Stress developed on the bearing
Lateral Confinement

Lateral confinement can increase the compressive strength and hence the
bearing stress of concrete.

Loaded area
concrete face

Confinement by the Confinement by the

surrounding concrete reinforcement

9
 Design Strength of Plain Concrete Node
with Bearing Plate
Sectional
According to CoP for Precast Concrete Construction 2016
area A
(cl.2.7.9.4), the design ultimate bearing stress is based on the weaker
of the two bearing surfaces and is calculated as follows:

•0.4 fcu for dry bearing on the concrete;

•0.6 fcu for bedded bearing on concrete; or
•0.8 fcu for contact face of a steel bearing plate cast into a member or
support, with each dimension not exceeding 40% of the
Loaded area Ab corresponding concrete dimension.
X-X

Bedding
Halving Joint mortar
Pile cap
direct in X X
contact of
concrete faces Steel H pile

Dry bearing Bedded bearing Cast-in-plate

10
Design Strength of Ties
• The strength of a steel tie may be taken as 0.87fy, where 0.87 is
the partial safety factor of steel reinforcement as according to
British Standard BS8110 or Hong Kong Code.
• For control cracks at compressive struts, a minimum
transverse reinforcement of 0.4% should be evenly distributed
at each face of the section considered.
• For beam sections the amount of minimum transverse
reinforcement could be reduced to 0.2% when the concrete
grade is less than 45 MPa, as suggested in Table 9.1 of Hong
Kong Code (Table 3.27 of BS8110).
• For brackets, corbels or nib design, to improve crack control, a
minimum amount of horizontal steel equal to 50% of the main
tensile reinforcement should be distributed over 2/3 (close to
the tension side) of the effective depth of the section
considered (cl. 9.8.3 Hong Kong Code).

11
Comments on Ties

RC detailed according to the

simplified S&T model.

main rebar As

2/3d
d

50% As

Tensile stress will generate Horizontal links are added

Simplified
in bottle-shaped strut and to control cracking of
S&T model
wide cracks will form. compressive strut during
working condition and to
reduce the stresses of the
critical reinforcement.
12
Unreinforced Concrete
The strength of the ties solely provided by the tensile strength of the
concrete for tie without reinforcement is not recommended. Progressive
type of failure may occur.

Example 1. A beam section subject to bending

compression
The induced tensile stress
keeps increasing, crack
would propagate
continuously
Crack occur at tension
tension zone
Tensile stresses migrating from a failure zone into adjoining area
may result in (zipper-like) progressive failure

13
 r
Anchorage Radius of bent bar

Safe anchorage of ties in the node has to be assured ; to achieve this,

• The use of minimum radii of bent bars and anchorage Effective area
lengths of bars should follow the code recommendations. =a×w
• The tension tie reinforcement must be uniformly Not advised
distributed over an effective area of concrete which is at
least equal to the tie force divided by the concrete stress a
limit for the node.
• The anchorage begins where the transverse compression La
stress trajectories meet the bars and are deviated.
Smaller
• The anchorage must be located within and ‘behind’ the diameter
nodes. rebars

• If the required anchorage length is insufficient, the bar

may be extended beyond the node region. La

Anchor
block
La
14
Anchorage

Case 1: Anchorage of reinforcement within the node

Rebar dia. 
Tie force sn
Depth
Node of node
sn
Conc. load
sn ≤ 3 
Depth of node

15
Anchorage

Case 2: Anchorage of reinforcement within the node using a plinth

Rebar dia. 
Tie force

Node

≥ 30o A plinth
Conc. load

Depth of node

If the bursting stress is high, additional longitudinal steel should

be provided in the nodal region.

16
Anchorage

Case 3: Anchorage of reinforcement behind a node

Back
face of
Tie force Reinforcing bars node

Node

Anchorage length lb

The tensile forces introduced behind the node can resist the
remaining forces developed within the nodal regions.

17
Anchorage

Case 4: The node is wider than the bearing

Extended
nodal zones

Transverse Transverse reinforcement Transverse tension

tension

bearing
Tie bars Node Node

Provided anchorage length ≥ lb+be be Transverse

PLAN reinforcement
lb - ultimate anchorage bond length
ELEVATION SECTION

In such case, transverse reinforcement (i.e. the horizontal legs

perpendicular to the tie bars) should be added to resist the
bursting stress and transverse tension developed. 18
Design Example 1- Design of deep
beam subjected to unequal
concentrated loads
In this example, a 400 mm wide and 4000 mm deep rectangular deep beam subjected to
two ultimate concentrated loads, 3000 kN and 6000 kN respectively as shown in Figure
1 is considered. The characteristic cube strength of concrete fcu = 45 MPa and yield
strength of ribbed steel bar fy = 500 MPa are assumed.

Figure 1. Deep beam subjected to two

unequal concentrated loads.
19 self-
(Note: For simplicity, the uniform
weight of beam is neglected)
Following the procedures of STM described, the design of the deep beam can be
accomplished in the following steps.

Step 1. Isolate non-flexural component

As the distributed regions due to the concentrated forces and reactions, assumed
equal to the lateral dimension of 4m, are almost overlapping with each other, the
whole deep beam is considered as a D-region.
Step 2. Compute the internal forces on the boundaries of D-region

By considering the global equilibrium of the applied and reaction forces, the support
reactions at A and B are found to be 3941 kN and 5059 kN, respectively.

Maximum bearing stress is therefore equal to 5059×103/(400×1400) = 9.0MPa. From

cl.2.7.9.4, CoP Precast Concrete Construction 2016 (or cl.5.2.3.4, BS8110), the
allowable bearing stress is 0.4×45 = 18 MPa (assuming dry bearing) which is greater
than the bearing stress of 9.0 MPa.

The maximum bearing stress at the loading point is 6000×103/(400×850) = 17.6 MPa.

Referring to Table 1, the node efficiency factor with uniaxial CCC condition is 0.45.
The allowable bearing stress is then 0.45×45 = 20.25 MPa which is greater than the
bearing stress of 17.6 MPa.
20
 Step 3. Idealize the deep beam by STM

The idealized STM for the deep beam is shown in Figure 2.

3200 3800 3200

3000 kN 6000 kN17.6 MPa< 0.45×45=20.25 MPa

D a
γ
C
4000

β α
a1.6a
A B
Legend 700
700 Strut
3941 kN Tie 5059 kN 9 MPa
< 18 MPa
Figure 2. Idealized STM of deep beam.

21
 Table 1. Node efficiency factor
Condition Node efficiency factor (n)
of Node
Triaxial CCC 0.55
Uniaxial CCC 0.45
CCT 0.40
CTT 0.36
Minimum 0.28
Partial safety factor of 1/0.67 is allowed

22
Step 4. Dimension and check of struts, ties and nodes

Due to the presence of a tension tie, the nodal zone stress at the beam support is
0.4fcu according to Table 1, under CCT condition.
The strength of strut, according to Equation (1) is 0.4 fcu which is the same as
the design strength of nodes.
The horizontal compressive force at node D is equal to the tensile force at node B.
When the depth of node D is equal to ‘a’, the depth over which the tie force
distributed is also a.

From the geometry, tan α = [4000-(a+a)/2]/(3200-850/4) 3200

The tie force AB = 5059/tan α = 5059×2988 / (4000-a).
6000 kN

D a
C = 0.4 fcu × a × b
As the allowable stress at node B is limited to 0.4×45 = 18 MPa,
therefore, the tension stress in the tie AB should be
850/4
AB≦ 18×103×(a×400)×10-6 α
a1.6a
5059×2988/(4000-a) ≦ 18×10-3×(a×400) B
T = 0.4 fcu × a × b 700

Solving the above equation, the minimum a is 621 mm and α = 48.5o. 5059 kN

23
Step 4. Dimension and check of struts, ties and nodes

Having determined α = 48.5o, by equilibrium of the nodes,

the tie force AB=5059/ tan α = 4476 kN

and the strut force DB=5059/ sin α = 6754 kN.

By the Pythagoras theorem, the strut force AC = √(AB2 + 39412) = 5963 kN.

By considering the equilibrium of node C,

the strut force CD = √[(3941-3000)2 + AB2] = 4574 kN.

3200 3800 3200

3000 kN 6000 kN

D a
C γ
AB C
4000 3941-3000 kN

β α 5059 kN
α
1.6a
a
A B
A Legend B
700 Strut
700 24
3941 kN Tie 5059 kN
Choosing reinforcement for tension tie

The tensile force at the main tie has been found to be 4476 kN, hence the required area of
steel is equal to As = 4476×103/(0.87×500) = 10,289 mm2.

Providing 15T32 bars (Aspro’= 12,060 mm2) will be sufficient.

(a)
As the minimum depth of the nodal zones A and B is 621 mm (say 620 mm),
the vertical spacing between the main reinforcement should be equal to 120 mm as shown
in Figure 3.

Note that as the required development length (according to Table 8.4, HK Code) for T32
bars is 33D=33×32 = 1056 mm, the tension tie can be transferred to the nodal zone within
the bearing length of 1400 mm.

70

Effective depth of nodal

4@120 zone = 620 mm
Figure 3. Arrangement of tension ties.
25
54 mm ≈ cover + links 70
26
Checking stresses at nodes and struts

As node C is an internal node (smeared node), the dimensions of compressive struts

joined at this node can be chosen such that the node and the strut stresses are all less
than the corresponding allowable values. Further check for this node is not required.

At node D, the nodal zone is in equilibrium under a hydrostatic stress condition.

The length of the faces of the nodal zone must be proportional to the loads applied to
these faces and the faces must be perpendicular to the loads.
a a /cosα
Thus the stress in all struts at Node Zone D is equal to
α
f1 = 6754×103/(400×621/cosα) = 18.0 MPa

Crushing strength of struts = 0.4×45 =18 MPa which is the

6754 kN
same as the stress of strut (18.0 MPa).

27
 6754kN
Checking stresses at nodes and struts
α=48.5o
At the bottom of strut DB, the stress in the strut is a= 620
=980
B
f2 = (6754×103)/ [400×(620×cos48.5o+1400×sin48.5o)] = 11.6 MPa 1400

which is less than the ultimate compressive strength of strut (i.e. 5059 kN

18 MPa).

Node in Force Equilibrium

a1 & a2 are known

a3 C1

C3=(C12+C22)1/2
σ1 a1
a3=a1cosα+a2sinα σ3
σ3=C3/(a3×b)
C3 α

σ2 C2
28
a2
The required steel area for distribution bars is 400×1000×0.2%= 800mm2/metre.
Using T10-200 each face and both ways is sufficient. The R.C. detail for the deep
beam is summarized in Figure 4.

A
CLof support CL of s upport

4T16 T10-20 0 E. F.
T10-200E.F.

5T10-U Bars A-A

3T32(5 Layers) 5T10-U Bars
A

Figure 4a. RC details for deep beam by STM approach.

29
To be continued