1. NAME – ABHIJIT SAHA
ROLL NO – 26901321038
REG NO – 212690101320001 OF 2021-22
DEPT – CIVIL ENGINEERING
SUBJECT – PRESTRESSED CONCRETE , CE(PE)702A
SEM -7TH , CA – 1ST
TOPIC – Load Balancing Method
2. Content
•Load Balancing Method
•Concept of Load Balancing Method
•Life History Of Prestressed Member
•Mechanism & Explanation Of Balanced load
Concept
•ADVANTAGES OF LOAD BALANCING METHOD
•SIMPLE & CANTILEVER BEAM WITH LOAD
BALANCING METHOD
•EXAMPLE PROBLEM OF CANTILEVER BEAM
•Conclusion
•Acknowledgement
3. Load Balancing Method
3
⚫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
underdirect stress and thus greatlysimplifies both the
design and analysisof structure.
⚫ The application of this method requires taking the
concrete as a free bodyand replacing the tendonswith
forces acting on theconcrete along the span.
4. Concept of Load Balancing Method
4
⚫There are three basic concepts in prestressed concrete
design
1.Stressconcept : Treating prestressed concrete as an
elastic material
2.Strength concept: Considering prestressed concrete as
reinforced concrete dealing with ultimatestrength.
3.Balanced load concept: Balancing a portion of the load
on the structure.
Load Balancing method follows the third one.
5. Life History Of Prestressed Member
5
⚫ Analysing the life history of the prestressed member
under flexure leads to understanding the balanced load
concept relative toother twoconcept.
⚫So lets find out the load deflection relationshipof a
memberas a simple beam..
6. ⚫K1 = Factorof
safetyapplied to
working load to
obtain minimum
yield point.
⚫K2= Factorof
safetyapplied to
ultimatestrength
design toobtain
minimum ultimate
load.
6
7. ⚫The load deflection relationship of the above figure
leads to several critical points. Such as..
⚫ 1. Pointof nodeflection which indicatesrectangular
stress block.
⚫ 2. Pointof no tensionwhich indicates triangularstress
block with zero stress at the bottom fiber.
⚫
7
8. 3. Pointof cracking which occurswhen the extreme fiber
is stressed to the modulusof rupture.
4.Pointof yielding at which steel is stressed beyond its
yield point so thatcompleterecovery is not possible.
5.Point of ultimate load which represents the
maximum load carried by the memberat failure.
8
9. Mechanism & Explanation Of Balanced
load Concept
9
According to figure there are three stages of beam
Behavior :
Applied Loadings Stagesof beam behavior
⚫ DL+k3LL ⚫ Nodeflection
⚫ DL+LL ⚫ Notension
⚫ K2(DL+LL) ⚫ Ultimate
10. ⚫ Where,
⚫DL+LL is the stress concept with some allowable
tensionon 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.
10
11. ⚫In balanced load conceptdesign isdone by pointof no
deflection.
⚫So prestressing is done in such awayso thateffective
prestress balances the sustained loading & beam
remain perfectly level withoutdeflecting.
11
12. ADVANTAGES OF LOAD BALANCING METHOD
12
⚫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.
13. SIMPLE & CANTILEVER BEAM WITH LOAD
BALANCING METHOD
13
Figure illustrates how to balance a concentrated load
bysharply bending the c.g.s. at midspan , thuscreating
an upward component
V=2Fsinθ
14. ⚫If V exactly balancesaconcentrated load P
Applied at midspan the beam is not subjected toany
transverse load.
The stresses in the beam atany section are simplygiven
by
Any loading addition to P will cause bending and
additional stressescomputed by
14
15. ⚫Figure illustratesthe balancing of a uniformly
distributed load by means of a paraboliccable
whose upward component is given by
15
16. ⚫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
16
17. ⚫Figure represents a cantilever beam. Any vertical
component at the cantilever end C will upset the
balance, unless there is an externallyapplied load at
that tip.
17
18. ⚫To balancea uniform load w, the tangent to the c.g.s. at
C will have to be horizontal. Then the parabola for the
cantileverportion can best be located bycomputing
And the parabola for the anchorarm by
18
19. EXAMPLE PROBLEM OF CANTILEVER BEAM
19
⚫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 amountof prestress, assuming that the c.g.s
must haveaconcrete protection of at least 76.2mm. If a
concentrated load of 62KN is added at the mid span
compute the maximum top and bottom fiberstresses
20. SOLUTION:
20
To use the least amountof prestress, theecentricity over the
support should bea maximum that is , h=300mmoro.3m.
The prestress required is F = wL2 / 2h
=(23.3x62)/(2x0.300)=1395KN
Thesag for the parabola must be h1 = w(L1)2/8F
=(23.3x14.82)/(8*1395)
=0.46m
21. Uniform compressivestress
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 fiberstresses are
f=Mc/I=6M/bd2 = (6x229x106)/(300x7602)
= 7.93MPa
Stressat mid span are
ftop = -6.12-7.93=-14.05MPacompression
fbottom = -6.12+7.93 =+1.81 MPa tension
21
22. EXAMPLE PROBLEM OF
CONTINUOUS BEAMS:
22
⚫For thesymmetrical continuous beam prestressed
with
F= 1420KN along a paraboliccableas
shown, compute theextreme fiberstresses overthe
centersupport DL+LL=23.0KN/m
23. SOLUTION:
23
The upward transversecomponentof prestress is
Wb= 8Fh/L2 = (8x1420x0.3)/(15x15)=15.1KN/m
The beam is balanced under uniform stress
f= (1420x103)/(300x760)= -6.2MPa
Forapplied load w=23.0 KN/m, the unbalanced
downward load = (23.0-15.1) = 7.9 KN/m
This load produces a negative momentover 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
fbottom = -6.2-7.68 =+13.88 MPa compression
24. conclusion
Load balancing is a design method for prestressed
structural members, primarily beams and slabs,
that provides efficient use of materials. When using
load balancing, the prestressing tendon is designed
for a prestressing force and eccentricity in order to
'balance' all or a portion of the dead load.
25. Acknowledgement
I would like to express my sincere thanks to Mr. Sudip
Gupta sir for her valuable guidance and support in
completing my project .
