1 University of Technology, Sydney Faculty of Engineering & IT 49150-Prestressed Concrete Design Autumn 2015 Coordinator and Lecturer: Dr. Shami Nejadi CB11.11.220, [email protected]The main notes are taken from ‘Concrete Structures’ by RF Warner, BV Rangan, AS Hall, KA Faulkes; and ‘Design of Prestressed Concrete’ by R.I. Gilbert and N.C. Mickleborough. Some slides and figures were developed by Ken Faulkes and Zora Vrcelj. Introduction History of Prestressing LECTURE 1 - OUTLINE Methods of Prestressing Material Properties Transverse Forces Caused by Draped Tendons Equivalent Loads Calculation of Elastic Stresses: 2 - Combined Load Approach - Load Balancing Approach Stress Distribution - Special Cases Reinforced concrete is one of the most widely used structural materials in construction. Due to the low tensile strength of concrete, t lb it d dt lli t lt il f INTRODUCTION steel bars are introduced to carry all internal tensile forces. w Most reinforced concrete beams are cracked under service loads. Cracked cross-sections resist the applied moment by a force in the concrete, C (compressive) & a force in the steel, T (tensile). linear stresses C S C T w BEAM UNDER SERVICE LOADING Section Reinforcing bars cracked section INTRUDUCTION • Professor Gustav Magnel from the • The books on the bottom are like pre- compressed concrete: using a compressive University of Ghent in Belgium. 4 force, they support their own weight… plus significant superimposed loads, represented by the books on top.
Transcript
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University of Technology, Sydney Faculty of Engineering & IT
49150-Prestressed Concrete Design Autumn 2015
Coordinator and Lecturer: Dr. Shami Nejadi CB11.11.220, [email protected]
The main notes are taken from ‘Concrete Structures’ by RF Warner, BV Rangan, AS Hall, KA Faulkes; and ‘Design of Prestressed Concrete’ by R.I. Gilbert and N.C. Mickleborough. Some slides and figures were
developed by Ken Faulkes and Zora Vrcelj.
Introduction History of Prestressing
LECTURE 1 - OUTLINE
Methods of Prestressing Material Properties Transverse Forces Caused by Draped Tendons Equivalent Loads Calculation of Elastic Stresses:
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- Combined Load Approach- Load Balancing Approach
Stress Distribution- Special Cases
Reinforced concrete is one of the most widely used structural materials in construction.
Due to the low tensile strength of concrete, t l b i t d d t ll i t l t il f
INTRODUCTION
steel bars are introduced to carry all internal tensile forces.
w
Most reinforced concrete beams are cracked under service loads.Cracked cross-sections resist the applied moment by a force in the
concrete, C (compressive) & a force in the steel, T (tensile).
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linear stresses
C
S
C
T
w
BEAM UNDER SERVICE LOADING
SectionReinforcing bars
cracked section
INTRUDUCTION• Professor Gustav Magnelfrom the
• The books on the bottom are like pre-compressed concrete: using a compressive
University of Ghent in Belgium.
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force, they support their own weight… plus significant superimposed loads, represented by the books on top.
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PRESTRESSED CONCRETE is a particular form of reinforced concrete, which involves the application of an
initial compressive load on a structure to reduce or
WHAT IS PRESTRESSED CONCRETE
initial compressive load on a structure to reduce or eliminate the internal tensile forces and thereby control or
eliminate cracking.
The initial compressive force is imposed and sustained by highly tensioned steel reinforcement reacting on the concrete.
A prestressed concrete section is considerably
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stiffer than the equivalent cracked reinforced section.
Prestressing may also impose internal forces which counterbalance external loads and may reduce or
eliminate deflection.
• Early attempts foiled by losses due to the creep and shrinkage strains in the concrete. F i t f F (1930’ )
HISTORY OF PRESTRESSING
• Freyssinet of France (1930’s).• Advent of high-strength and high-ductility steel.• Later, Leonhardt of Germany, Mikhailov of Russia,
and T. Y. Lin of USA, also contributed a great deal to the art and science of prestressed concrete design.
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g• Australia - Warragamba ice tower was constructed
in 1950’s to serve as an ice tower for use in the production of concrete at the Warragamba Dam site.
EUGENE FREYSINNET who proposed methods
HISTORY OF PRESTRESSING
to overcome prestress losses through the use of high-strength and high-ductility steels.
YVES GUYON, a student of Freysinnet once summarized the importance of the method saying:
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“ There is probably no structural problem to which prestress cannot provide a solution, and
often a revolutionary one”
Why prestress ?• To improve service loadservice load performance
-- control deflection and cracking.control deflection and cracking.• To achieve economyTo achieve economy
- Allows for thinner more economical floor slabsystems.
• Smaller sections• Slender members
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• Longer spans
•• Especially in structures where Especially in structures where selfself--weight weight is a high is a high proportion of the total load.proportion of the total load.
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When Should You Prestress?• Each job needs to be considered on its own merits, but
as a guide, this graph gives a good comparison of costs versus span :versus span :
For spans up to about 7.5 m it is generally more economical to adapt
i f d l ti
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a reinforced solution. Over 7.5 m, post tensioning will become cheaper.
METHODS OF PRESTRESSINGPrestressing is applied to a concrete member by highly tensioned steel reinforcement (wire, strand, or bar)
ti th treacting on the concrete.
The high strength steel is most often tensioned using hydraulic jacks.
The tensioning operation may occur before or after the concrete is cast and results in two classifications:
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concrete is cast and results in two classifications:
(i) PRE - TENSIONED
(ii) POST - TENSIONED
T d t d
The three stages required for pret-tensionedconcrete member are as follow:
1 Tendons stressed between supports
Concrete cast and cured2
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and cured2
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Tendons released and prestress transferred.
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hollow duct
The three stages required for post-tensionedconcrete member are as follow:
1. Concrete cast and cured
uplift forcesTENSILE FORCE
COMPRESSIVE FORCE2. Tendons stressed &
t t f d
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prestress transferred
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dead endlive end3.Tendons anchored
& duct grouted
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Material properties
• Concrete
For prestressed construction, strengths typically 35 - 50 MPa
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=0.002
Typical uniaxial stress-strain response of concrete in compression
Reinforcing steel• There are some assumptions that are made initially:
- The strain distribution in the section is linear.- Steel behaves in an elastic-plastic mannerSteel behaves in an elastic plastic manner.
500 MPa
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0.0025
Es = 200,000 MPa
•• Prestressing steelPrestressing steel
Often idealised to elastic-plastic:
Approx. values
For strandstrand:fpy ≈ 1600 MPa
Ep ≈ 195,000 MPa
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For barbar:fpy ≈ 950 MPa
Ep ≈ 170,000 MPa
Transverse forces caused by draped tendons
In addition to the longitudinal force, P exerted on a prestressed member at the anchorages transverseprestressed member at the anchorages, transverse
forces are also exerted on the member wherever curvature exists in the tendons.
Assumptions:• Angles are assumed SMALL and are
d i RADIANS th
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measured in RADIANS, thenCos = 1 & Sin = Tan =
• Can also use ARC = RADIUS x or (radians ) = ARC / RADIUS
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FORCES AT ANCHORAGESRemember: P cos = P, P sin = P
P
PP
P
P
M = Pee
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in radians) = tan = h / Lh
L
Transverse forces caused by draped tendons (cable with a kink)
Bent up tendon RP
Small θ (θ)0
e
L/2 L/2
PP Cosθ = 1
sin tan e /(L/2)
R = 2Psin
P P
FBD
P P cosP P cos
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R = 2Psin Pθ P sinP θP sin
Bending Moment Diagram
- PeLP 2
sin
From statics
Transverse forces caused by draped tendons (cable with a kink)
2/sin
Le
LPe
LePPR 4
2/2sin2
At each anchorage the cable has a horizontal component of P cos θP and a vertical component of P sin θ = 2Pe/L.
Th b i id t b lf t d
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The beam is said to be self stressed.
No external reactions are induced at the supports (if ws.w.=0).
The beam suffers curvature and deflects upward owing to the internal bending moment caused by prestress.
FORCE DUE TO A PARABOLICCABLE
L/2 L/2
PP
w
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L/2 L/2
PARABOLIC TENDON PROFILE
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The parabola shown below has length L and at its mid-length has an offset (sag), relative to the line joining its ends (chord), of h.
A D C
GENERAL PARABOLAGENERAL PARABOLA
B
C
Ih
L/2 L/2
CCAA h X
Y
21AI and CI are the tangents at A and C respectively
AA
CC
A A = = CC = 2h/(L/2) = 4h/L radians= 2h/(L/2) = 4h/L radians
CABLE CABLE SLOPESLOPE +_
Parabolic Cable Profile(Symmetric)
2xx
4
Lx
Lxhy = “ cable profile”
Lx
Lh
dxdy 214 = “ slope”
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2 8L
hdx
ydp
= “ curvature”
The curvature or rate of change of the cable’s slope is constant for a parabola
Transverse forces caused by draped tendons
hdy 4Lh
dxdy 4
At anchorages x = 0 and x = L
Vertical force at anchorage: PhP 4
Horizontal force at anchorage = P PPh
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Small θ = sin tan θ
Vertical force at anchorage:4Ph/L = Pθ
PLhPv
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w o = vertical
Equivalent Loadwp = Equivalent Load per unite length due to prestress
PP
R distributed forces
p - curvature
unit length(very small length) ∑w = w = R
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(very small length)
p
P
P R = P sin p= P p
cable forces
The curvature is the angular change in direction of the
cable per unite length
∑wo wp R
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Equivalent LoadAn upward force exists in the cable where R = wp = P p.
This upward force is an equivalent distributed load along the member such that:
2
8LPhPw pp (Eq. 1.1)
w is the force per unit length exerted by the concrete on
2
8L
hp
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wp is the force per unit length exerted by the concrete on the cable, which is of course equal and opposite to the
cable exerts on the concrete.
It acts radially, but for small angles can be considered to act vertically with negligible error.
Equivalent LoadThe negative sign indicates that ‘wp’ acts upwards when the sag ‘h’ is downwards. The equivalent load for any parabolic
cable segment can be obtained from Eq 1.1, provided that the
wp
P P
2
8LPhwp
g q , pmid-span sag ‘h’ is measured from the straight line joining the
two ends of the segment (chord).
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L/2 L/2
hP
4Pe/L4Pe/L
cable(e = h)
Calculation of elastic stresses
The components of stress on a prestressed concreteThe components of stress on a prestressed concrete section caused by the prestress, the self-weight, and the external loads are usually calculated using simple beam
theory and assuming linear elastic behaviour.
The section properties of the gross-section are used, provided it is uncracked.
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provided it is uncracked.
If an elastic calculation indicates that cracking may occur at service loads, a cracked section analysis is then
necessary. (this will be considered later in the course)
Calculation of elastic stresses
There exists two methods to calculate stresses:
(i) Combined load approach
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(ii) Load balancing approach
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i) COMBINED LOAD APPROACH
Cross section Elevation Elevation
Stresses due to prestress only (at any arbitrary section with e):
e P
P
Mp=Pey
Stresses due to prestress P/A -Pey/I P/A-Pey/I
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y+ =e
P/A+Pey/IDue to P
Moment due to Mp= Pe Resultant
i) COMBINED LOAD APPROACH
Due to Prestress only: PeyP
Total stress due to prestressDue to Prestress only:
Iy
A
Due to External Moment only: I
My M
Total stress due
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to loading
Combined Stresses:I
MyI
PeyAP
Total stress
Within the service load range, prestressed members are usually designed so that they behave essentially
linear elastically
ii) LOAD BALANCING APPROACH
Therefore, superposition procedures are valid, and analysis at service loads is conveniently related to
conditions at the balanced load.
Complete load balancing would be achieved if the d f d t bl t tl
linear elastically.
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upward forces due to cable curvature exactly counteracted a selected service load.
At balanced condition bending moments, shear forcesand deflections would then all be zero.
ii) LOAD BALANCING APPROACHTherefore, if w = wp, the bending moment and shear force
caused by the gravity load on every cross-section are balanced by the equal and opposite values caused by wp.
0 pub www
wpP
wP
h
2
8LPhwp (Eq. 1.1)
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L/2 L/2wpL/2
wL/2 wL/2
wpL/2A Bor 4Ph/L or 4Ph/L
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ii) LOAD BALANCING APPROACH
If w = wp LOAD BALANCING OCCURS
Thus stress due to Mub = M – Mp is ‘0’
calculate corresponding stresses at balanced
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2wLM 8
2LwM p
p =
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Beam does not deflectA
P
p gcondition:
If w wp
ii) LOAD BALANCING APPROACH
Unbalanced Load, wub = (w - wp)
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22 LwwLwM pubub
Unbalanced Moment, Mub caused by unbalanced load must be calculated:
Add corresponding stresses to the balanced
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yM ubAP
Add corresponding stresses to the balanced condition:
Calculate the elastic stress distribution at mid-span of the simpl s ppo ted beam sho n belo
EXAMPLE 1.1 (Gilbert & Mickleborough)
the simply supported beam shown below:
30 kN/m (includes self weight)
e = 250 mmparabolic curve
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curve
6000 mm 6000 mm
23 mm10220A kN1760P 46 mm1020000IP is assumed to be uniform along the beam.
Stresses due to prestress plus applied loads (including self-weight)
+ =C
T
CC
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Total stresses due to eccentric
prestress
Stresses due to applied bending
moment
Combined stresses
T
Stress distributions with increasing bending moment, prior to cracking (Special Cases)
Figure 1.1
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• Zero curvature moment (Figure 1.1-d)
• Decompression moment (Figure 1.1-f)
• Cracking moment (Figure 1.1-g)
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Zero Curvature Moment
PA
= P/A = Mp .y/I = MzeroCurv .y/I
MzeroCurv = Mp
+ =
T
C
C
A
+
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Prestressalone
Zero curvatureZero curvaturemomentmoment
Stress at all levels σ = P/A
T
Decompression moment
T
+ =C
CC
T
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Prestressalone
DecompressionDecompressionmomentmoment
Zero bottom fibre stress
T
Cracking moment
T
+ =C
CC
T
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Prestressalone
Cracking Cracking momentmoment
Bottom fibre tensile stress σb = fct
TT
Example 1.2(W.R.H.F. Book, Page 120)
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resultant/
compressive---------
49 50
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SuccessSuccess is not final,is not final, FailureFailure is not fatal.is not fatal.It is theIt is the Courage to ContinueCourage to Continue that counts.that counts.SuccessSuccess is not final,is not final, FailureFailure is not fatal.is not fatal.
It is theIt is the Courage to ContinueCourage to Continue that counts.that counts.
