This document provides information on cable layout and load balancing methods for prestressed concrete beams. It discusses layouts for simple, continuous, and cantilever beams. For simple beams, it describes layouts for pretensioned and post-tensioned beams, including straight, curved, and bent cable configurations. It also compares the load carrying capacities of simple and continuous beams. The document concludes by explaining the load balancing method for design, using examples of how to balance loads in simple, cantilever, and continuous beam configurations.

1. CABLE LAYOUT
CONTINUOUS BEAM
LOAD BALANCING METHOD
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2. Group 5B
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Arafat Hossain
Sadik Hasan
Sadia Mahajabin
Raihan Mannan
Sadia Mannan
Tanvir Alam
Ikhtiar Khan
Ifat Hasan Shawon
Md. Asif Rahman
Sarani Reza
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4. What is Cable Layout
 Cable : A strong thick rope, usually twisted hemp or wire
 Layout : The arrangement or plan of something
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5. Simple Beam Layout
 Simple Beam: a structural beam that rests on a support
at each end
 Simple Beam can be two types :
1. Pretensioned
2. Posttensioned
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6.  Pre-tensioning : When the steel is tensioned before concrete
placement, the process is called pre-tensioning.
 Post-tensioning : When the steel is tensioned after concrete
placement, the process is called post-tensioning.
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7. Layouts for pre-tensioned beam
1. Straight cables are preferred since they can be more
easily tensioned between two abutments.
2. Such a section can not often economically designed
because of conflicting requirements of the midspan and
end section.
3. At the maximum moment section generally occuring at
mid span, it is best to place the cable as near to the
bottom as possible.
4. Since there is no external moment at the end it is best to
arrange the tendons so that c.g.s will coincide with c.g.c
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8. Bent Soffit
Curved Soffit
1. For both layouts c.g.s at mid span can be depressed as low as desired.
2. The end can be kept near c.g.c
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9. Bent Extrados
Curved Extrados
1. When it is possible to vary the extrados of concrete it can be advantageously
used.
2. These will give a favorable height at mid span where it is most needed and yet
yield a concentric or nearly concentric prestress at end section.
3. (d) is simpler in formwork than (e).
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10. 1. Most pretensionning plans in USA have buried anchores along the stressing
bed so that tendons for a pretensioned beam can be bent.
2. It may be economical to do so if the beam has to be of straight and uniform
section.
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11. Layouts For Post tensioned Beam
1. Most of the layouts for pre tensioned beams can be used for post
tensioned once as well.
2. For a beam of straight and uniform section the tendons are very often
curved.
3. Curving the tendons will permit favorable position of c.g.s to be
obtained at both the end and mid span sections and other points as
well.
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12. combination of curved tendons with curved
soffits
1. A combination of curved tendons with curved soffits is frequently used
when straight soffits are not required.
2. This will permit a smaller curvature in tendons thus reducing the friction.
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13. 1. Curved or bent cables are also combine with beams of variable depth.
1. Combination of straight and curved tendons are sometimes found convenient.
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14. 1. Some cables are bent upwards and anchored at the top flanges.
1. Some cables are stopped part way in the bottom flange.
2. This arrangements will save some steel but may not be justified unless
the saving is considerable as for very long span carrying heavy loads.
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15. Cantilever Beam
 A cantilever is a beam anchored at only one end. The beam carries the
load to the support where it is forced against by a moment and shear
stress.
 Two general layouts are possible for cantilevers :
1.Single Cantilevers
2. Double Cantilevers
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16. LAYOUTS FOR SINGLE CANTILEVER
1. For a short span with a short cantilever a straight and uniform section may be
the most economical.
2. It is only necessary to vary the c.g.s profile so that it will confirm with the
requirements of the moment diagrams.
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17. 1. When the cantilever span become longer it is advisable to taper the beam.
2. If the anchore span is short compared to the cantilever it may be entirely
subjected to the negative moments and the c.g.s may have to be located
above the c.g.c
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18. 1. For longer span it may be desirable to haunched them.
2. The c.g.s profile can be properly curved or may remain practically
straight.
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19. Layouts for double cantilever
1. For short double cantilevers a straight and uniform section can be adopted.
1. When the cantilevers are long they may be tapered.
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20. 1. If the anchor span is long it may be haunched.
1. If the anchor span is short compared with the cantilevers the c.g.s
line may lie near the top of the beam.
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21. CONTINIOUS BEAM
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22. CONTINUOUS BEAM
A Continuous beam is one, which is
supported on more than two supports.
For usual loading on the beam
hogging ( -ve ) moments causing
convexity upwards at the supports
and sagging ( +ve ) moments causing
concavity upwards occur at mid span.
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23. ASSUMPTIONS OF CONTINUOUS
BEAM
1. The ecentricities of the
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prestressing cable are small
compared to length of the
members.
2. Frictional loss of prestress is
negligible.
3. The same tendons run through
the entire length of the member.
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24. Advantages of Continuous Beam
Over Simply Supported Beam
 The maximum bending moment in case of
continuous beam is much less than in case of
simply supported beam of same span carrying same
loads.
 In case of continuous beam, the average bending
moment is lesser and hence lighter materials of
construction can be used to resist the bending
moment.
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25. COMPARING OF LOAD
CARRYING CAPACITY BETWEEN
SIMPLY SUPPORTED BEAM AND
CONTINIUOUS BEAM
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26. LOAD CARRYING CAPACITY OF
SIMPLE BEAM
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27. LOAD CARRYING CAPACITY OF
SIMPLE BEAM
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28. LOAD CARRYING CAPACITY OF
CONTINUOUS BEAM
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29. LOAD CARRYING CAPACITY OF
CONTINUOUS BEAM
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30. DISADVANTAGES OF CONTINUOUS BEAM
1. Frictional loss is significant
2. Shortening of long continuous
beam under prestress.
3. Concurrence of maximum moment
and shear over support.
4. Difficulties in achiving continuity
for precast elements.
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31. CABLE LAYOUT OF
CONTINUOUS BEAM
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32. Curbed Tendon In Straight Beam
•This lay out is often used for slabs or short span
beams
•The main objections here are the heavy frictional
loss
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33. Straight Tendon In Curbed
Beam
•Used for longer span and heavier loads.
•Often difficult to get the optimum eccentricities
along the beam if the tendons are to remain entirely
straight
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34. Curbed Tendon In Haunched Or
Curbed Beams
•This would permit optimum depth of beam as well as
ideal position of steel at all points,aboidin excessive
frictional loss.
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35. Overlapping Tendons
•Offer a possibility of varying prestressing force along
the beam
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36. Determination Of Resisting
Moment Of Continuous Beam
1. Plot the primary moment diagram for the entire
continuous beam as produced only by pre-stress
eccentricity. As if there were no support to the beam.
2. Plot shear diagram.
3. Plot loading diagram.
4. Plot moment diagram corresponding to loading diagram
considering all support.
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38. 1/24/2014
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39. Load Balancing Method
 Load in the concrete is balanced by stressing the steel.
In the overall design of prestressed concrete
structure, the effect of prestressing is viewed as the
balancing of gravity load . This enables the
transformation of a flexural member into a member
under direct stress and thus greatly simplifies both the
design and analysis of structure.
 The application of this method requires taking the
concrete as a free body and replacing the tendons with
forces acting on the concrete along the span.
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40. Concept of Load Balancing Method
 There are three basic concepts in prestressed concrete
design
1. Stress concept : Treating prestressed concrete as an
elastic material
2. Strength concept: Considering prestressed concrete as
reinforced concrete dealing with ultimate strength.
3. Balanced load concept: Balancing a portion of the load
on the structure.
Load Balancing method follows the third one.
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41. Life History Of Prestressed
Member
 Analysing the life history of the prestressed member
under flexure leads to understanding the balanced load
concept relative to other two concept.
 So lets find out the load deflection relationship of a
member as a simple beam..
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42.  K1 = Factor of
safety applied to
working load to
obtain minimum
yield point.
 K2= Factor of
safety applied to
ultimate strength
design to obtain
minimum ultimate
load.
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43.  The load deflection relationship of the above figure
leads to several critical points. Such as..
 1. Point of no deflection which indicates rectangular
stress block.
 2. Point of no tension which indicates triangular stress
block with zero stress at the bottom fiber.

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44. 3. Point of cracking which occurs when the extreme fiber
is stressed to the modulus of rupture.
4. Point of yielding at which steel is stressed beyond its
yield point so that complete recovery is not possible.
5. Point of ultimate load which represents the
maximum load carried by the member at failure.
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45. Mechanism & Explanation Of
Balanced load Concept
According to figure there are three stages of beam
Behavior :
Applied Loadings
Stages of beam behavior
 DL+k3LL
 No deflection
 DL+LL
 No tension
 K2(DL+LL)
 Ultimate
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46.  Where,
 DL+LL is the stress concept with some allowable
tension on beam or no tension.
 K2(DL+LL) is the strength concept consists with the
ultimate strength of the beam.
 DL+K3LL is the balanced load concept with the point
of no deflection where k3 is zero or some value much
less than one.
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47.  In balanced load concept design is done by point of no
deflection.
 So prestressing is done in such a way so that effective
prestress balances the sustained loading & beam
remain perfectly level without deflecting.
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48. ADVANTAGES OF LOAD BALANCING
METHOD
 1.simplest approach to prestressed design and
analysis for statically indeterminate structures.
 2.It has advantages both in calculating and in
visualizing.
 3.Convenience in the computation of deflections.
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49. SIMPLE & CANTILEVER BEAM
WITH LOAD BALANCING
METHOD
Figure illustrates how to balance a concentrated load
by sharply bending the c.g.s. at midspan , thus creating
an upward component
V=2Fsinθ
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50.  If V exactly balances a concentrated load P
Applied at midspan the beam is not subjected to any
transverse load.
The stresses in the beam at any section are simply given
by
Any loading addition to P will cause bending and
additional stresses computed by
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51.  Figure illustrates the balancing of a uniformly
distributed load by means of a parabolic cable
whose upward component is given by
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52.  If the externally applied load w is exactly balanced by
the component Wb there is no bending in the beam.
The beam is again under a uniform compression with
stress
External load produced moment M and corresponding
stresses
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53.  Figure represents a cantilever beam. Any vertical
component at the cantilever end C will upset the
balance, unless there is an externally applied load at
that tip.
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54.  To balance a uniform load w, the tangent to the c.g.s. at
C will have to be horizontal. Then the parabola for the
cantilever portion can best be located by computing
And the parabola for the anchor arm by
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55. EXAMPLE PROBLEM OF
CANTILEVER BEAM
 A double cantilever beam is to be designed so that its
prestress will exactly balance the total uniform load of 23.3
KN/m normally carried on the beam. Design the beam
using the least amount of prestress, assuming that the c.g.s
must have a concrete protection of at least 76.2mm. If a
concentrated load of 62KN is added at the mid span
compute the maximum top and bottom fiber stresses
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56. SOLUTION:
To use the least amount of prestress, the ecentricity over the
support should be a maximum that is , h=300mm or o.3m.
The prestress required is F = wL2 / 2h
=(23.3x62)/(2x0.300)=1395KN
The sag for the parabola must be h1
= w(L1)2/8F
=(23.3x14.82)/(8*1395)
=0.46m
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57. Uniform compressive stress
f= F/Ac = 1395/(2.28x105) = 6.12 Mpa
Moment M at the mid span due to P=62KN
M=PL/4=(62x14.8)/4=229KN-m
Extreme fiber stresses are
f=Mc/I=6M/bd2 = (6x229x106)/(300x7602)
= 7.93MPa
Stress at mid span are
ftop = -6.12-7.93=-14.05MPa compression
f bottom = -6.12+7.93 =+1.81 MPa tension
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58. EXAMPLE PROBLEM OF
CONTINUOUS BEAMS:
 For the symmetrical continuous beam prestressed
with
F= 1420KN along a parabolic cable as
shown, compute the extreme fiber stresses over the
center support DL+LL=23.0KN/m
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59. SOLUTION:
The upward transverse component of prestress is
Wb= 8Fh/L2 = (8x1420x0.3)/(15x15)=15.1KN/m
The beam is balanced under uniform stress
f= (1420x103)/(300x760)= -6.2MPa
For applied load w=23.0 KN/m, the unbalanced
downward load = (23.0-15.1) = 7.9 KN/m
This load produces a negative moment over the center
support, M= wl2/8 =(7.9x152)/8 = 222KN-m
And fiber stresses, f = Mc/I = (6x222x106)/(300x7602)
=
7.68 Mpa
ftop = -6.2+7.68=+1.48MPa tension
f bottom = -6.2-7.68 =+13.88 MPa compression
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