strut and tie - konstruktionsteknik | konstruktionsteknik university / structural engineering strut...

Strut and tieBetongbyggnad VBKN05

Lund University / Structural Engineering

Strut and Tie

• Design the deep beam (hög balk) in the figure.

qd = 200 kN/m

3 m

3 m


Strut and Tie


Strut and Tie

• Describe the stresses in a concrete structure with a strut and tie

model

• Are normally used in the ULS

• Based on plasticity theory


B- and D-regions

The structure is divided into B- and D-regions


B- regions (continuity region)

• Plane sections remain plane under loading.

• Plane strain distribution is assumed to be valid (Bernoulli

hypothesis)


D- region (discontinuity region)

• Non-linear strain distribution due to geometry and external load.

Wall on concentrated supports, deep beam (hög balk)


Extension of D-region - St Venant’s princip

Study a body that is subjected to a system of forces in self

equilibrium. On the global level the body is not subjected to a

force resultant and will not move. Locally, stresses will occur in

the body. The extension d of the stressed region is equal to the

maximum distance h between the forces.


Example of forces/stresses in a body

=

=


Example of forces/stresses in a body


B- and D-regions in simply supported beam

Distributed load

Concentrated load


Example on D-regions


Principles for Strut and tie method

Compressive

struts in concrete

Reinforcement that

acts as ties,

subjected to tension

Struts and ties are

”connected” in nodes

Describes the stress field in a structure with strut and tie modell

(fackverksanalogi)


Deep beam without cracks

The stresses can be determined with linear elastic analysis under the

assumption that the material structure is homogeneous, concrete.

Stress field

Concrete arch

Simplifed stress condition Strut and tie modell


Deep beam with cracks

When cracks have occurred the stresses will be redistributed since the

cracked regions will have a decreased stiffness.


Deep beam– ultimate state

Ultimate state = non-linear material behaviour. Advanced analysing methods

are needed.

These models show examples of stress fields that can be

realistic in the plastic state.

Large cracks in tensile zone


Deep beam – ultimate state

Zones that are critical for the capacity of the deep beam


Plasticity theory

Assumption

– The stress field is supposed to be in equilibrium with the external loads.

– The plastic capacity of the structure is large enough so that a redistribution of

stresses are possible.

– Indefinite plastic deformation capacity


Plastic analisys

Reality

• Concrete doesn´t have indenfinite deformation capacity

• Ultimate strain in concrete is low, that is there is only a limited capacity for

redistribution of stresses

Use a strut and tie model that is close to the linear elastic stress field.


Design

1. Determine the external loads and the reactions at supports.

2. Divide the structure into B- and D-regions

3. Design the B-regions

4. Determine the contact forces between the B- and D-regions.

5. A stress field in the discontinuity region is assumed by sketching and further

analyzed by the load path method. Alternatively it can be determined by linear

FE analysis.

6. Choose a suitable strut and tie model based on stress field

7. Determine the forces in the struts and ties

8. Design the reinforcement and control the capacity of the struts and the nodes.


Forces between B- and D-regions

Prestressed beam

Q

Column with two concentrated loads


Choice for strut and tie model

Two different methods

1. Load path method

2. Linear FE-analysis


Load path method

Streamlined load paths are inserted to simulate the stress field in a simplified way.

• Two load paths cannot cross

• Forces on one side of the structure must be in equilibrium with the forces on

the other side of the structure


Load paths – load dividers

Load dividers = section where the shear force is zero


Load paths in prestressed beam

Prestressed beam

Q


Load paths in prestressed beam

U-sväng

The curvature of the paths leads to transverse forces


Strut and tie model (prestressed beam)

Node

Ties (solid line) and struts (dotted line)


Column with two concentrated loads D-region

Load paths Strut and tie model


Strut and tie-models


Strut and tie models based on FEM

Result from FE-analysis presented as main stresses.


Strut and tie model based on FEM

Result of FE-analysis, stress field.


FEM - problems

Required anchorage capacity can be underestimated if the design is

based only on FE-analysis

FEM shows tensile stresses (forces) distributed over a large area. In

reality the reinforcement will take the tensile forces and the concrete

will not contribute to any capacity. The forces must then be anchored

in the end of the reinforcement.


Collapse of the oil platform Sleipner A 1991

The oil platform collapsed when it was towed to the oil field.

Failure took place in one of the walls.

Collins et.al. Concrete International, v 19, n 8, p 28-35, Aug 1997


Investigations showed that the designer had used a FE-analysis and

that there were flaws in the model that lead to failure. A new platform

was designed, basically with hand calculations.

Failure was initiated near the corner


Collins et.al. Concrete International, v 19, n 8, p 28-35, Aug 1997



New design

Initial design


Design – Strut and tie

Design

1. The angle between the tie and the strut should be larger than 45˚, often 60˚

is appropriate

2. When two ties and a strut meets the angles should be larger than 30˚, 45˚ is

recommended

3. The stresses from concentrated loads are to distributed in the structure,

recommended angles are 30˚ , not larger than 45˚

1 2 3


Design

- The forces in the ties and the struts are determined based on equilibrium

- Some of the angles between ties and struts are chosen according to the

requirements. Other angles and lengths of struts and ties are given by

geometry.

- The reinforcement area is given from the forces in the ties.

- The stresses in the nodes and the struts are controlled.


Nodes

Concentrated node:

Nodes where concentrated forces meet, normally near the boundary of the D-region

Due to the limited area the stresses has to be checked

Distributed node:

Node where distributed stress fields meet. No control is needed

since there is a capacity for stress redistribution within a larger area.


Concentrated nodes

Compression node (CCC) - node where three compression struts meet

Stress should be limited: (EC2)

where: cdmax,Rd fk 1250

1 ckf

, k1 = 1,0


Concentrated nodes

Compression–tension node (CCT) – two struts and one tie meet


cdRdfk

2max,

2501 ck

f , k2 = 0,85where:


Concentrated nodes

Compression-tension node(CTT) – one strut and two (or more) ties meet


cdRdfk

3max,

2501 ck

f , k3 = 0,75where:


Ties

Tie = normal or prestressed reinforcement

Distributed ties:

Distribute the reinforcement evenly over this height


Ties

Concentrated ties:

Place the reinforcement close to each other


Ties

Anchorage of reinforcement. Important to control.

bdla bd

la


Anchorage of ties

Different ways to increase the anchorage capacity


Struts

The stresses in the struts should fulfill the following requirements (EC 2):

cdRdf 0.1

max,

cdRdf 6.0

max,

– Uni-axial conditions:

– For a concrete strut in a cracked

compression zone:


Optimization of models

Try to find the model that gives the least amount of reinforcement.

Preferred model

Optimization criterion: min miii lF

Fi = force in part i

li = length on part i i

εmi = average strain in part i

Less appropriate model


Optimization of reinforcement

Use straight bars

Place reinforcement parallell or transverse to the boundaries of the structure


Serviceability limit state

Normally, the same stress field can be used in both serviceability and ultimate

limit state.

The structure that is controlled in serviceability limit state (tension and crack

widths), with the same strut and tie model used for the ultimate limit.

Different strut and tie models can be used if the linear elastic stress field gives

uneconomical or impractical reinforcement.


Strut and tie models, B-regions

Shear reinforcement


Example

Design the deep beam with the help of a strut and tie-model

qd = 200 kN/m

3 m

3 m

Concrete C25

Reinf. B500B ᴓ12

Thickness of beam = 300 mm

Width of support = 120 mm

Design the deep beam with the help of a strut and tie-model

Concrete C25

Reinf. B500B ᴓ12

Thickness of beam = 300 mm

Width of support = 120 mm

fcd = 16,7 MPa

fyd = 435 MPa

Use = 60

Node N1

T = 300/tan30 = 173,2 kN

C1 = 300/sin60 = 346,4 kN

Reinforcement: A = 173,2 /435*103 = 398 mm2 i.e. 4 bars

Controll Node CCT-node:

Upper node contact pressure 𝜎𝑅𝑑,𝑚𝑎𝑥 = 𝑘2𝑓𝑐𝑑

k2 = 0,85

= 1 - fck /250 = 1 - 25/250 = 0,9

𝜎𝑅𝑑,𝑚𝑎𝑥 = 0,85 ∙ 0,9 ∙ 16,7 = 12,8 𝑀𝑃𝑎

𝜎𝑐 =300 ∙ 10−3

0,3 ∙ 0,12= 8,3 𝑀𝑃𝑎 < 12,8 𝑀𝑃𝑎 𝑂𝐾!

Control strut C1

s = 10 + 12 + 12/2 = 28 mm (with respect to cover)

u = 2*28 = 56 mm

a

C1

T

120

s u

T

C1

300 kN

x1 = 56cot60 = 32 mm

x2 = 120 +32 = 152 mm

a = 152 cos 30 =132 mm

Requirement

cdRd f 6.0max, = 0,6*0,9*16,7 = 9 MPa

𝜎𝑐1 =346,4 ∙ 10−3

0,3 ∙ 0,132= 8,7 𝑀𝑃𝑎 < 9 𝑀𝑃𝑎 𝑂𝐾!

Height to next node: h

h0 = (750-120/2)*tan60 = 1195 mm

h = h0 + s = 1195 + 28 = 1223 mm

Node N2

C2 = 346,4*cos60 = 173,2 kN i.e. C2 = T

C2 = T

The stresses are not concentrated in this node, no control is needed.

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