STRUT AND TIE MODELLING Background and New AS 3600 Provisions
Peter Dux The University of Queensland
Background 1: B and D regions •
Strut and tie modelling enables the rational design for the strength limit state of concrete structures in regions of non-linear strain. These are referred to as disturbed or ‘D’ regions. Regions where strains vary linearly are called Bernoulli or ‘B” regions.
•
St Venant’s principle can be used to identify these regions
Sketch of possible strut and tie model of a deep beam - the whole beam is a D region.
• • • • •
The model has struts, a tie and nodes. Struts and nodes have volume. Centrelines of struts and ties bisect the near face of the nodal zone and intersect at a common point or node. Shear is transferred to supports via concrete struts. The orientation of struts is dictated by the shear diagram. The distribution of internal forces and the reactions must satisfy equilibrium. Material capacity must not be exceeded. Internal forces are stress resultants i.e. design is based on stresses in concrete and steel.
Background 2: Plasticity and the Lower-bound Theorem – the Basis of Strut and Tie Design The lower-bound theorem is based on the behaviour of ideal rigid-plastic systems. • The term ‘plasticity’ means strength without stiffness (e.g. a plastic plateau) • The term ‘ductility’ means that the range of plastic deformation is large.
The figure shows a generalised stress space, yield surface and plastic strain increment vector for an ideal plastic material. Normality rule
Stress vector
Incremental plastic strain vector Convex yield surface
This figure shows a generalised rigid-plastic structure with generalised loading, Wi , at the point of collapse. Incremental displacement component in the load direction at a load point during collapse
Consider now a design situation: Suppose a designer has determined some internal stress distribution , σ* , which does not violate the yield surface and which equilibrates a design loading, λWi . The designer has not considered compatibility. Internal stresses, σ, are on or within the material yield surface
This is typical of design for strength.
At collapse: External work = Internal work Σ Wi ui = ∫σ. ε dV where the dot denotes a scalar product. The integral relates only to stresses on the yield surface and the local plastic strain increments as the material is rigid-plastic. However, for small collapse deformations, elastic regions in elastic-plastic materials do not deform (i.e. they are essentially rigid) and only stresses on the yield surface do work. Impose the collapse deformations as a set of virtual deformations on the structure with loads λWi . The work equation becomes Σ λWi ui = ∫σ*. ε dV But, everywhere in the structure, the stress state, σ*, lies on or within the yield surface. Therefore, σ. ε ≥ σ*. ε at all points in the structure. Hence
λ ≤ 1.0
The previous result leads directly to the lower-bound theorem of plasticity:
‘Provided a structure is ductile, each load for which any statically admissible stress state can be given is either the collapse load or a lower bound of the collapse load’ or
‘If a designer can figure out a way for a structure to carry a set of design ultimate loads such that equilibrium is satisfied, yield criteria are not violated and ductile response is ensured, the structure will be able to carry the loads’
Background 3: Concrete structures do not have unlimited ductility •
Ignoring strain-hardening, steel exhibits plasticity and is ductile .
•
Plain concrete exhibits neither plasticity nor ductility.
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Properly designed concrete structures do not have unlimited ductility. Therefore the choice of a ‘statically admissible stress state’ must be done with good judgement.
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The challenge in design of concrete structures for strength is to have yield of steel dominate the failure mode.
•
Concrete stresses must be closely controlled to prevent brittle failure. Otherwise, adoption of the lower-bound theorem ,which assumes a good range of plastic response , is invalidated.
•
The chosen stress state should reflect the way the structure naturally carries loads, for example, as indicated by elastic analysis. The strut and tie approach is a design methodology, not an analysis methodology.
zone of tension
Longitudinal stresses
Transverse stresses
These stress distributions are reasonably consistent with prediction based on application of St Venant’s principle
(a) Centrally located prestressing anchorage or reaction force at column base
zone of compression
Longitudinal stresses
Transverse stresses
Here, the stress distributions are not so intuitively obvious, even though stresses have settled down at the right-hand boundary of the D region.
(b) Eccentrically located prestressing anchorage or reaction force at column base
(a) Centrally located prestressing anchorage or reaction force at column base Likely premature failure here
This design poorly reflects the FE results
(b) Eccentrically located prestressing anchorage or reaction force at column base
Background 4: How strong is concrete in compression?
Uniaxial compression
f cmax
f 'c = ≤ f 'c 0.8 + 170ε 1
Transverse tensile strain reduces compressive strength due to enhanced cracking. Transverse compression increases the compressive strength.
From Mohr’s circle of strain:
ε2
ε 1 = ε s + (ε s − ε 2 ) cot 2 θ s e1 es
As θ s reduces, the principal tensile strain increases and cracks in the transition zone become wider and more skew to the strut.
e2
With ε2 = -0.002 and εs estimated as 0.001 (around half the yield strain of steel) the compressive strength of the strut becomes f 'c f cmax = ≤ f 'c 0.8 + 170ε 1
s
θs
ε1
Strain conditions at a node
45°
0.004
0.68 f ' c
This effect has been confirmed in tests of strut and tie structures.
30°
0.010
0.40 f ' c
25°
0.015
0.30 f ' c
fcmax
AS 3600 provisions 1: Material capacity reduction factors
φconcrete = 0.6
Both are referred to as st in AS 3600
φ
φsteel = 0.8
• The new code uses material capacity reduction factors for strut and tie design. • The concrete factor is lower - concrete has less quality control than steel. Code
φc
φs ( reinforcement)
φs (prestressing)
AS 3600
0.6
0.8
0.8
Eurocode
0.67
0.87
0.87
Canadian
0.6
0.85
0.9
AS 3600 provisions 2: Minimum angle between any strut and any tie at a node
C
For reinforced concrete, θmin = 30o For prestressed concrete, θmin = 20o
•
Cracks in the vicinity of nodes with prestressed ties develop later in loading and are less severe than cracks in the vicinity of nodes with non-prestressed ties.
• The Eurocode specifies a minimum of around 22o for either type of reinforcement. But • For small θ, concrete capacity is severely limited (as has been seen) . •
Forces in flat struts are typically high because the vertical component has to equal the shear being carried i.e. the law of diminishing returns quickly comes into play as θ reduces.
AS 3600 provisions 3: Design strength of struts General: Design strength = φ c βs 0.9f’c Ac i.e. Design stress = 0.54 βs f’c where βs ,the ‘strut efficiency factor’ =
1.0 1.0 + 0.66 cot 2 θ
0.3≤ βs ≤ 1.0
and Ac is the minimum cross-sectional area of the strut Additional requirements where a strut enters a node: Incoming axial stress in the strut (code refers to ‘principal compressive stress on nodal face’, which is the same thing) ≤ 0.54 βn f’c At nodes with no ties (CCC)
βn = 1.0
At nodes with 1 tie (CCT)
βn = 0.8
At nodes with more than 1 tie (CTT)
βn = 0.6
Summary of limiting stress provisions for struts, incorporating nodal stress provisions:
0.5
Design stress (x f'c)
0.4
0.3 AS 3600 .001 0.2
εs
.0015 .002 .0025
0.1
0 20
30
40
50
60
70
Strut angle θs (i.e.θ) Comparison of AS 3600 design stresses for struts with design stresses based on Canadian code formula, for various values of average strain in reinforcement, εs , in nodal zone.
ε 1 = ε s + (ε s − ε 2 ) cot 2 θ s f cmax
f 'c = ≤ f ' c , and design stress = 0.6 fcmax 0.8 + 170ε 1
e1 e2 es
s
Strain conditions at a node
AS 3600 provisions 4: Design strength of ties
Design strength of a tie =
φs (Ast fsy + Apt (σp.ef + Δσp ))
where φ s = 0.8, fsy is typically 500 MPa and the total stress in strand (σp.ef + Δσp ), is ≤ fpy or the 0.1% proof stress obtained from testing.
Ties have to be anchored such that the design strength is achievable at the point where the tie leaves the nodal zone i.e. ties anchor through and beyond the node.
Suppose PE = 1.5 W
Effective prestressing force PE = Apt σp.ef (here, = 1.5 Pw ) applied as a nodal force
Δσp Ap = 0.5W
AS 3600 provisions 5: Bursting reinforcement requirements for bottle-shaped struts
Straight-sided struts are used in design. However, unless a strut is parallel to and immediately adjacent to a free surface (a ‘prismatic’ strut), stresses will diverge as indicated. Near the nodes, the stress field is convex inwards. Transverse compression arises. Over most of the length of the strut the stress field is convex outwards and transverse tension arises. Bursting steel controls the effects of cracking parallel to the strut axis in the bottle zone.
The code has bursting reinforcement requirements for both strength and serviceability
The total bursting force = (force in strut) x Tan(α) For ultimate loading, C*, Tan (α) = 0.2 For service loading, Cs , Tan (α) = 0.5 The bursting tension to cause cracking: Tb.cr = 0.7blb f’ct where f’ct = 0.36 √f’c , lb = lstrut – dc (the ‘convex outwards’ length) and b is the thickness. If 0.2 C* and 0.5Cs ≤ 0.5Tb.cr , no bursting reinforcement is required. Otherwise: • For strength, design reinforcement for the larger of 0.2 C* and Tb.cr • For service, design reinforcement for 0.5Cs • Provide the larger amount and distribute evenly over lb
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Because of the Tan(α) values the service load bursting force will typically be larger than the ultimate load bursting force.
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Except where struts are short and highly stressed (higher strength concrete, high βs and high stress) the default loading case, Tb.cr will dominate the other two load cases.
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An equal orthogonal grid of reinforcement offers the same normal force per unit length for all crack orientations. The code provisions also allow for unequal grids and reinforcement in one direction only.
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To illustrate the amount of reinforcement needed, for an equal grid of horizontal and vertical reinforcement, and assuming a design force for strength of Tb.cr , the area of reinforcement in each direction per unit area of concrete (i.e. ρh = ρv ) is:
Tb.cr = blbφsfsy •
0.252 f ' c 400
For f’c =40 MPa, ρh = ρv = .004. This would be further divided into grids at front and back face, each with ρ = .002 (or 0.2%) in each direction. The grids would extend at least over the length lb.
Design for serviceability: The design force = 0.5Cs The design stress in reinforcement: For a minor degree of crack control
250 MPa
For a moderate degree of crack control
200 MPa
For a strong degree of crack control
150 MPa
Sufficient reinforcement at the chosen stress is required to provide the design force normal to the crack. An equal orthogonal grid provides this force in both directions (e.g. horizontally and vertically). The requirement is then halved for grids at front and back face. The strut length does not feature; hence, the longer the strut the lower the percentage of reinforcement required. Even if the design forces for both ultimate and service load cases are less than 0.5Tb.cr , grids of reinforcement should be provided at front and back faces.
Summary of bursting reinforcement requirements: The bursting force, Tb is defined as follows: For strength, T*b = 0.2C* For serviceability Tbs = 0.5Cs Bursting reinforcement is required if T*b or Tbs > 0.5Tb.cr where Tb.cr = 0.7blb f’ct For orthogonal reinforcement, the provided horizontal and vertical reinforcement must satisfy
Asv Ash Tb 2 2 cos θ + sin θ ≥ sv sh f s lb where θ is the inclination of the strut from the horizontal. For strength, Tb is the larger of T*b and Tb.cr ; fs =φs fsy For service, Tb = Tbs ; fs = 150, 200 or 250 MPa
Conclusions •
The new AS 3600 provisions for strut and ties design are a major change for the better.
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AS 3600 now uses material capacity reduction factors rather than a single capacity reduction factor (0.7). Arguably, the factor for steel (0.8) is still low. The limiting stresses for struts are conservative rather than liberal.
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For code-surfers, the Eurocode provisions are typically less stringent for about the same ultimate loading (1.25G + 1.5Q for most situations). Sources: 1. Standards Australia, Draft standard DR 05252, 2007 2. Chris Calladine, “Engineering Plasticity”, Pergamon,1969 3. Scott Rathie, HonoursThesis, UQ, 2007 4. Nick Stevens, various. 5. Schlaich , J et al, “Towards a consistent design of structural concrete”, PCI ,32,3,1987
