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Module 4D8Prestressed Concrete
Lent Term 2010
Lecture 11 – Continuous Beams
Consider a continuous prestressed beam
The cable will induce moment –PeThis will cause a curvature –Pe/EIThe beam will tend to move relative to the supports
(Shown here as upwards at central support but can be either way)
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But the beam does not lift off the supportInstead, the support reactions will alter to bring the beam back to the support
These reactions will cause bending moments
These bending moments are usually called Secondary Moments, but they are not small
They are sometimes called Parasitic Moments, but they are not bad
The best name is Reactant Moments, since they are caused by a redistribution of the dead load support reactions
They can only occur in indeterminate structures
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Reactant Moments• Reactant moments are not necessarily small –
they can easily be about 50% of the dead load moments but are distributed differently
• Reactant moments are not bad• Good designers use them to assist with the design
• Most designers try to minimise them to avoid complications
• Bad designers ignore them, which is very dangerous – the moments are real and can cause very large forces
ExampleDouble-T beams are very good at resisting sagging bending but very poor for hogging
They are widely used for precast simply-supported beams
They are easy to form for in-situ construction since there is no internal formwork, but in a continuous viaduct there will be hogging moments over the piers
The solution is to generate large sagging secondary moments to eliminate most of the hogging
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Effect of prestressWe need to define two cable profiles• es is the actual cable profileThe primary moment caused by the prestress
itself is – PesThere are also secondary moments M2So the total effect of the prestress is
– Pes + M2 = – Pepep is the place where the cable appears to act• ep is known as the line of thrust of the cable
ep or es ?• It is the total effect of the prestress (i.e. the line
of thrust ep) which must satisfy the stress inequalities, and thus which must be plotted on the Magnel diagram
• But ep can only be found by analysing the whole cable profile
• So paradox – you need to know es before you can find ep to enable you to design the prestress at a cross-section
• We must find a sensible way to design these structures
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Constraints• The line of thrust ep must satisfy the stress
limits• but it need not lie wholly within the section
since it is only a notional line• The actual cable profile must lie within the
section and be subject to limits on the cover
• This gives great freedom to the designer
Calculate ep for a given es
There are two methods for analysing a cable profile
1. Equivalent load method – calculate the forces the cable causes on the beam and analyse
2. Virtual WorkWe can then devise methods for
designing the beams
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θeP
Anchorage
Equivalent load methodThe cable will cause forces on the concrete at anchorages and when the cable changes direction
P.sinθ
P.cosθ
P.sinθ
P.cosθPe.cosθ
θ1P
θ2
P P(sinθ1+sinθ2)
P(cosθ1-cosθ2)
Where cable changes direction sharply
Cable inclination normally small so cosθ ~1
P(sinθ1+sinθ2)
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2
8LhPp =
When cable is parabolic – uniformly distributed load, as with load balancing
p
(x1,y1,z1)(x2,y2,z2)
(x3,y3,z3)
θd
v1
v2
For general 3-D profile • divide into segments • calculate change in angle at corner• calculate friction loss due to change in angle
and wobble• find resultant force at change in angle• apply force to concreteCan apply 3-D forces in finite element analysis
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Worked exampleBeam with three unequal spans – uniform EIConstant Prestress = 1000 kN (all dims. in m, forces in kN)
Assume cable profile made from 5 parabolae with slope continuity and anchorages on the centroidal axis
Forces from tendon to concrete will be 5 uniformly distributed loads plus reactions at anchorage (don’t forget these!)
The resulting bending moment will be –PepThe reactions will be self-equilibrating
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– Pes + M2 = – Pep so ep – es = – M2/P
ep - es -.428-.535
14.45 20.51
17.13 17.83
Reactions
both methods give Moment M2
428 535
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Linear Transformation• M2 is caused by a redistribution of the
support reactions• It must be zero at end simple supports
(and in any cantilever overhangs)• It must vary linearly between supports• If P is constant, then ep – es must also vary
linearly between supports• This is known as a linear transformation
of the cable profile
Use of Linear Transformation• If we have a cable profile (es) and/or its
corresponding line of thrust (ep), then:-• We can vary es by any linear transformation, and
the line of thrust will not alter• We will have changed the indeterminate support
reactions, but not the total effect of the cable• This gives great freedom to the designer; once a
satisfactory ep or es has been found, linear transformations can be applied at will, for example to fit the tendon into the section
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Virtual Work Method
∫ ∑ Δ=L
WdxM0
**κ
We can find the secondary moments directly
Real compatibility system due to primary and secondary effects of prestress
Fictitious equilibrium system
Unknowns will be the secondary moment Qj at the internal supports j = 1,n
L1 L3L2
Q1
β1Q1
Q2
β2Q2
Q3
β3Q3
Total ∑=j
jjQM β2
Real Curvatures due to the Prestress
EI
PeQ sj
jj ⎟⎟⎠
⎞⎜⎜⎝
⎛−
=∑β
κ
Add primary effect of prestress and find curvature
Secondary Moments only
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Q*1
β1Q*1
R1j*
Q*2
β2Q*2
R2j*
Fictitious equilibrium systems –moments and reactions
L1 L3L2
System 1
L1 L3L2
System 2
M*i =βiQ*i
One such system for each internal support
∫ ∑ Δ=L
WdxM0
**κAll values of Δ are zero since they are the real deflections at supports, so the values of W* = R*ij are not needed
0.* =⎟⎟⎠
⎞⎜⎜⎝
⎛−∫ ∑ dxEIPeQQ s
jjjii ββ i = 1,2 … n
For each equation Q*i is a constant so can be cancelled
∫∫∑ = dxEIPedxEIQ siijj
j .. βββ i = 1,2 … n
This is a set of linear simultaneous equations with unknowns Qj
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⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡=
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
∫∫M
M
M
LLL
LL
LLL
dxEIPeQ
QdxEI si
n
ji ..1
βββ
Solve for Qi
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡=
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
∫∫M
M
M
LLL
LL
LLL
dxPeQ
Qdx si
n
ji ..1
βββ
Iff EI is constant, then
Many of the terms in the l.h.matrix will be zero since either βior βj will be zero
βiβj
siLi
For j = i -1
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.1.
i
iiji
L
dsLs
Lsdx
=
⎟⎟⎠
⎞⎜⎜⎝
⎛−=∫ ∫ββ
6. 1+∫ = i
jiLdxββSimilarly, if j = i + 1
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For j = i βi=βj
siLi
si+1
Li+133. 1++=∫ ii
jiLLdxββ
( )( )
( ) ∫=⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
++
+
dxPeQ
etcLLLLL
LLLLLLL
sij ..
.0020
02002
61
4
4433
3322
221
β
So, finally, we get a simple result
The integrals on the r.h.s. can be found numerically
The integrals can be performed using Simpson’s Rule on a spreadsheet
In this case choose an interval of 2.5 m and use Simpson’s Rule coefficients (s) 1,4,2,4,…4,2,4,1
Chainage Eccent. β(1) β(2) s β(1).e.s β(2).e.s0.0 0.0000 0 0 1 0 02.5 0.3281 0.1 0 4 0.1313 05.0 0.5625 0.2 0 2 0.2250 07.5 0.7031 0.3 0 4 0.8438 0
10.0 0.7500 0.4 0 2 0.6000 012.5 0.7031 0.5 0 4 1.4063 015.0 0.5625 0.6 0 2 0.6750 017.5 0.3281 0.7 0 4 0.9188 020.0 0.0000 0.8 0 2 0.0000 022.5 -0.2813 0.9 0 4 -1.0125 025.0 -0.3750 1 0 2 -0.7500 027.5 -0.2813 0.9375 0.0625 4 -1.0547 -0.070330.0 0.0000 0.8750 0.1250 2 0.0000 0.000032.5 0.3438 0.8125 0.1875 4 1.1172 0.257835.0 0.6250 0.7500 0.2500 2 0.9375 0.312537.5 0.8438 0.6875 0.3125 4 2.3203 1.054740.0 1.0000 0.6250 0.3750 2 1.2500 0.750042.5 1.0938 0.5625 0.4375 4 2.4609 1.914145.0 1.1250 0.5000 0.5000 2 1.1250 1.125047.5 1.0938 0.4375 0.5625 4 1.9141 2.460950.0 1.0000 0.3750 0.6250 2 0.7500 1.250052.5 0.8438 0.3125 0.6875 4 1.0547 2.320355.0 0.6250 0.2500 0.7500 2 0.3125 0.937557.5 0.3438 0.1875 0.8125 4 0.2578 1.117260.0 0.0000 0.1250 0.8750 2 0.0000 0.000062.5 -0.2813 0.0625 0.9375 4 -0.0703 -1.054765.0 -0.3750 0 1.0000 2 0 -0.750067.5 -0.2813 0 0.9167 4 0 -1.031370.0 0.0000 0 0.8333 2 0 0.000072.5 0.3375 0 0.7500 4 0 1.012575.0 0.6000 0 0.6667 2 0 0.800077.5 0.7875 0 0.5833 4 0 1.837580.0 0.9000 0 0.5000 2 0 0.900082.5 0.9375 0 0.4167 4 0 1.562585.0 0.9000 0 0.3333 2 0 0.600087.5 0.7875 0 0.2500 4 0 0.787590.0 0.6000 0 0.1667 2 0 0.200092.5 0.3375 0 0.0833 4 0 0.112595.0 0.0000 0 0 1 0 0
15.4125 18.4063
2for 153394063.1835.21000
1for 128444125.1535.21000
===
===
i
i
sePhdxPenodes
sisi ∑∫ = ββ3
.
Totals
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⎥⎦
⎤⎢⎣
⎡=⎥
⎦
⎤⎢⎣
⎡
⎥⎦
⎤⎢⎣
⎡=⎥
⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡+
+
04.53516.428
1533912844
)3040(24040)4025(2
61
2
1
2
1
gives Moment M2
428 535
Substitute into equation for Q
(as before)
Equivalent Load Method
• Intuitive• Need a beam
analysis program• Can be used to
obtain local forces as well as global moments
Virtual Work Method• Less intuitive• Need to solve
simultaneous equations
• Only gives global moments
• Easy to build into an interactive cable analysis system
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Concordant Profile• A cable profile that causes no secondary
moments is called a concordant profile
• The line of thrust (ep) of any cable profile (es) is itself a concordant profile because the curvatures due to –Pep are compatible with zero displacement at supports
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡=
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
∫∫M
M
M
LLL
LL
LLL
dxEIPeQ
QdxEI si
n
ji ..1
βββ
From the Virtual Work analysis
For a concordant profile all the Qi values are zero, so:-
0. =∫ dxEIPepiβ
The cable must have zones with –ve e as well as zones with +ve e
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M2 sagging or hogging?
Road bridge • large top flange• high centroid• easy to provide +ve es• secondary moments +ve
(sagging)• ep will be above es and
possibly above top of section
Rail bridge• large bottom flange• low centroid• difficult to provide +ve es• secondary moments will
be –ve (hogging)• ep will be below es and
possibly below bottom of section
Line of Thrust DesignFor the design of continuous beams, we obtain the bounds on the line of thrust from our normal stress inequalities
• These define upper and lower bounds on ep
The required cable profile must have a line of thrust (which will be concordant) that satisfies these limits
Can we design the line of thrust directly?
PM
PZf
AZe
PM
PZf
AZ t
pc ++−≥≥++− 2222e.g.
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Notional loadConsider any load on a continuous beam
This will give rise to a bending moment M
The curvature M/EI must be compatible with zero displacement at the supports
Now consider a beam with a prestressing tendon with a profile e = M/P
e
This will give rise to curvatures Pe/EI
These curvatures are the same as those caused by the original load.
• So the cable will not induce any secondary moments• So the cable profile will be concordant
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Thus, the bending moment from any load, suitably scaled by P, will give a concordant profile
Loads do not have to be forces, nor do they have to be gravity loads
End couple to give eccentricity at one end
Distributed load at support to eliminate kink
Superposition
• Superposition applies to notional loads, so required load can be built-up from a series of basic moment diagrams
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Finding line of thrust• So the selection of a line of thrust that
satisfies the limits on ep is equivalent to finding a notional load whose bending moment M, when scaled by the prestressing force P, satisfies those limits
• Note that the notional loading has nothing to do with the real load on the structure
Automated Design of ep
• It is relatively easy to design a suitable profile, since most engineers have a feel for the loads needed to cause a bending moment at a given point
• The process can be automated by using an iterative procedure
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• Guess a notional load• Find ep• See where it most
departs from required bounds
• Find moment influence line at that position
• Apply additional load in proportion to the I.L.
• Repeat until ep within bounds
Ref. Burgoyne C J, Automated determination of concordant profilesProc. ICE, 1988, 85, 333-352
Alternative design process• Many designers say they “never
bother” to design a concordant profile
• Instead, they choose the amount of secondary moment they want (or expect to get), and treat this M2 as a load
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Cable Profile DesignObtain the bounds on the cable profile from our normal stress inequalities, rewritten to include M2
• These define upper and lower bounds on es
The required cable profile must satisfy these limits and generate the required secondary moment
Designers choose a cable profile, calculate M2 and then adjust as necessary
( ) ( )P
MMPZf
AZe
PMM
PZf
AZ t
sc 222222 +
++−≥≥+
++−
e.g.
Equivalent methodsThe two methods are exactly equivalent
( ) ( )P
MMPZf
AZe
PMM
PZf
AZ t
sc 222222 +
++−≥≥+
++−
A cable profile that satisfies these limits:-
and also generates M2
PM
PZf
AZe
PM
PZf
AZ t
pc ++−≥≥++− 2222
will have a concordant line of thrust that satisfies these limits:-
even if the designer never calculates it!
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Logical design sequence1. Choose M2 required and add to effect of applied
loads2. Design section and choose prestress force3. Find limits on es
4. Use Linear transformation to find limits on ep
5. Find notional loads by automated method and hence ep
6. Transform back to get actual cable profile
Can we always find a solution?Assume that a concordant profile exists between the upper and lower bounds on ep
If a tendon is placed at the lower limit it will cause the beam to hog, inducing sagging M2
If placed at the upper limit it will cause hogging M2
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For a solution to exist • a cable placed at the upper limit must give
hogging M2, and• a cable placed at the lower limit must give a
sagging M2• For a particular value of P, the two limits can
be checked for this condition• Possible to produce expressions for the
limiting value of P
0.min =∫ dxEIPeiβ
For example, in a road bridge it is easy to provide +ve eccentricity but difficult to provide –ve eccentricity.
So limiting condition is usually when emin gives zero secondary moments – i.e. it is concordant.
0.min =∫ dxeiβIf P and EI are constant then:-
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Imagine a typical internal span in a long multi-span bridge
e
The bending moment M will be symmetrical, so emin will be symmetrical
1
β
But β will be skew symmetric, with average value ½ , so:- 0.5.0. minmin == ∫∫ dxedxeiβSo the average value of emin in a span must be zero
t
b
t
tw
RPM
RPZf
AZe ++−= 22
min
Commonly, emin is governed by tensile stresses in bottom fibre at the working load
0.122 =+⎟⎟⎠
⎞⎜⎜⎝
⎛−− ∫ dxM
RPRPfZ
AZL b
tt
tw
If the average value of emin is to be zero, then
bMIf the average value of Mb is
AZfZMRP twb
t /2
2−=
If no tension allowed2ZAMRP b
t =
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Easy to show that this is a minimum value for required P
Low called this value P3
(His P1 corresponds to our Pmin from the Magnel diagram and his P2is a value chosen to ensure that the cable fits within the section at both the mid-span region and over the piers)
If P < P3 it will be impossible to obtain a concordant line of thrust forthe cable no matter what adjustments are made to the profile.
N.B. Many assumptions made in this example but easy to rework if other conditions apply.
Ref. Low A M, The preliminary design of prestressed concrete viaducts, Procs ICE, 1982, 73, 351-364and Burgoyne C J, Cable design for continuous prestressed concrete bridges, Procs ICE, 1988, 85, 161-184
Beam with tendon force just larger than minimum value
Can be difficult to provide smooth
profile over support
Tendon forced close to top of allowable zone
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Does M2 exist at the ultimate load?• When a structure collapses it must be a mechanism• When the penultimate hinge forms it must be statically
determinate• So M2 should not exist at the ultimate load• But in practice ultimate load checks are carried out on
factored elastic solutions, which do include M2• Tests in the US (by measuring reactions in indeterminate
beams) have shown that M2 only disappears when the beam is actually collapsing
• So the consensus is that beams should be checked for the applied moments + M2
Trapped Moments
• Similar to secondary prestressing moments in that they cause a redistribution of dead load reactions
Caused by• Sequential construction• Temporary supports• Temporary cables
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Monolithic momentThe dead load moment that would be caused if the beam were cast in one pieceConsider beam with 3 No. 40 m spans, dead load 100 kN/m
-16000
12800
4000
Monolithic moment
Span-by span construction
• Assume built in three stages, with a 10 m overhang in the first two stages
-5000
17578
Stage 1
-6836 -5000
11728Stage 2
2157
-8625
13768Stage 3
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Total moment from staged construction differs from monolithic moment
Difference is the Trapped Moment
Must be straight between supports and zero at ends (like M2)
spans 40 & 60 m, 100kN/m
Temporary props Placing props in structure and removing them when the support conditions have altered will also cause trapped moment
Stage 1, with prop
Stage 2, prop removed
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Temporary prestress• Common to place temporary cables
during construction, which are later removed when permanent cables are fixed
• Wrong to assume that they leave no effect
Stage 2 – complete structure (3 No. 40 m spans) and remove temporary cables
(Other prestress will be applied to the completed structure but that can be considered in the normal way)
Stage 1 – build side spans (40m) and instal temporary cables
Assume equivalent load 100 kN/m upwards in each span
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Stage 1 moments due to prestress only
Stage 2 moments caused by removal of temporary cables
Will leave residual trapped moment
-20000
-8000
16200
What is effect of creep?• If creep causes a uniform change in Elastic
Modulus, then deflections will change but distribution of moments will not alter
• So secondary and trapped moments are not affected by creep if concrete is all of a similar age when loads are applied
• Effects of staged construction can be altered by creep since concrete will be at different ages when loads are applied