CE0061

Professional Course 4 – (Specialized 2) StE Track:

Prestressed Concrete Design
Analysis of Prestressed Concrete Beams using Load
Balancing Method
 Module 5
OBJECTIVES

■At the end of the chapter, the learner should be able to:
- Apply the basic assumptions in load balancing method to post –
tensioned.
− Analyze prestressed concrete beams subjected to different loading
and tendon profiles.
− Design post – tensioned joist and girder using load balancing method.
ANALYSIS OF PRESTRESSED CONCRETE BEAMS USING LOAD
BALANCING METHOD

INTRODUCTION TO
ANALYSIS OF PRESTRESSED CONCRETE BEAMS
USING LOAD BALANCING METHOD
Concept of Load Balancing
It is possible to select suitable cable profiles in a prestressed
concrete member such that the transverse component of the cable force
balances the given type of external loads. This can be readily illustrated
by considering the free – body of concrete, with the tendon replaced by
forces acting on the concrete beam as shown in the figure below.
Concept of Load Balancing
Concept of Load Balancing
The various types of reactions of a cable upon a concrete member depend upon
the shape of the cable profile. Straight portions of the cable do not induce any reactions
except at the ends, while curved cables result in uniformly distributed loads. Sharp
angles in a cable induce concentrated loads. The concept of loading – balancing is useful
in selecting the tendon profile, which can supply the most desirable system of forces in
concrete.
Concept of Load Balancing
In general, this requirement will be satisfied if the cable profile in a prestressed
member corresponds to the shape of the bending moment diagram resulting from the
external loads. Thus, if the beam supports two concentrated loads, the cable should
follow a trapezoidal profile. If the beam supports uniformly distributed loads, the
corresponding tendon should follow a parabolic profile. The principle of load –
balancing is further amplified with the following examples.
Concept of Load Balancing
Load Balancing Method
In figure 1, the forces exerted on a prestressed beam by a parabolic cable were
shown and the uniformly distributed load “wp” may be calculated by Equation 1.

8Pe
wp = 2
L
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”. With the
transversed loads balanced, the beam is subjected only to the longitudinal prestress, “P”
applied to the anchorage. If it is located at the centroid of the section, a uniform stress
distribution of intensity P/A occurs on each section and the beam does not deflect.
Load Balancing Method
If w≠ wp, the bending moment, Mb caused by the unbalanced load (w-wp) must be
calculated and the resultant stress distribution must be added to the stress caused by the
axial prestress (P/A).

Force caused by a parabolic

tendon on a concrete beam
Load Balancing of Prestressed Beams
I. THEORY OF LOAD BALANCING
1. If the upward component prestressed is balanced by the downward external applied
loads, the beam will have no bending and deflection at any location of the span. The
entire section is only subjected to uniform axial compressive stress.
P
Concrete Stress: fc = -
A
Load Balancing of Prestressed Beams
I. THEORY OF LOAD BALANCING
2. If the external downward load is not balanced by the upward prestress component,
bending and deflection will now occur due to the difference in the above vertical
loads.
P Mc
Concrete Stress: fc = - ±
A I
Load Balancing of Prestressed Beams
I. THEORY OF LOAD BALANCING
3. If balanced:
A. Any load in addition will now cause bending and deflection.

Bending Moment due to Prestress = Bending Moment due to External Loads

Load Balancing of Prestressed Beams
II. COMPUTATION OF THE UPWARD PRESTRESS COMPONENTS
1. Simple beams under uniform loads.
Most efficient cable profile is a parabolic curve.

8Pe
wp =
L2
✓ If w = wp, the beam is balanced.

wub = w - wp
wub L2
M ub =
8
Load Balancing of Prestressed Beams
II. COMPUTATION OF THE UPWARD PRESTRESS COMPONENTS
2. Simple beams under concentrated loads.
Load Balancing of Prestressed Beams
II. LOAD BALANCING OF SHORT CANTILEVER
To balance the external loads (so that the beam will have no bending and deflection anywhere in the
span), the tendon shall be placed in the position defined by the profile shown in the figure.
Load Balancing of Prestressed Beams
II. LOAD BALANCING OF SHORT CANTILEVER
1. At the beam ends:
✓ Prestress must be horizontal
✓ Tendon must be placed at the concrete centroid

2. At the cantilever support

✓ Tendon shall be above the concrete centroid at a distance of
wL2
h=
2P
Note:
For the maximum/minimum prestress, place the tendon in such a manner that it must have a
minimum concrete covering at the cantilever support.
Load Balancing of Prestressed Beams
II. LOAD BALANCING OF SHORT CANTILEVER
3. At the midspan of the anchor arm
✓ tendon must be at a distance of

wL12
h1 =
8P
ANALYSIS OF PRESTRESSED CONCRETE BEAMS USING LOAD
BALANCING METHOD

SAMPLE PROBLEMS TO
ANALYSIS OF PRESTRESSED CONCRETE BEAMS
USING LOAD BALANCING METHOD
Example 1
A prestressed concrete 400mm by 800mm has a span of 24m and carries a
uniform load of 22 kN/m that produces an upward load of 17.5 kN/m. The eccentricity
of the section is 375 mm.
1. Determine the force in the tendon.
2. Determine the net moment at midspan.
3. Determine the final stresses.
Example 1 - Solution
Given: Wp = 17.5 kN/m
A. Force in the Tendon
8Pe
wp =
L2

æ 375 ö
8P ç
è 1000 ÷ø
17.5 =
24 2

P = 3360kN
Example 1 - Solution
Given: Wp = 17.5 kN/m
B. Net Moment at the Midspan

M net =
( w )( L ) ( w - w )( L )
ub
=
2
p
2

8 8

M net =
( 22 -17.5 ) ( 24 )
2

8
M net = 324kN.m
Example 1 - Solution
Given: Wp = 17.5 kN/m
C. Final Stresses
i. Top Fiber
3360x10 3 324x10 6 ( 400 )
ft = - -
400 ( 800 ) 400 ( 800 )
3

12
ft = -18.09MPa
ii. Bottom Fiber
3360x10 3 324x10 6 ( 400 )
fb = - +
400 ( 800 ) 400 ( 800 )
3

12
fb = -2.91MPa
Example 2
An 18-m post tensioned beam is loaded by a 135 kN and 66 kN
loads at third points of the span. The tendon defines a bent up profile with
zero eccentricity at ends and 0.38m eccentricity under bigger load.
Neglecting the weight of the beam, determine the required cable force and
complete the tendon profile if no deflection is to occur upon the
application of the given loads.
Example 2 - Solution
A. Solve for the Prestressing Force
M EX = M PS

M EX = M A = 112 ( 6 )

M EX = M A = 672kN.m

P = 1768.42kN
Example 2 - Solution
B. Solve for “y”
M EX = M PS

M EX = M B = 112 (12 ) -135 ( 6 )

M EX = M B = 534kN.m

æ y ö
534 = 1768.42 ç
è 1000 ÷ø

y = 301.96mm
Example # 3
A rectangular concrete beam 300mm wide and 800mm deep
supports two concentrated loads of 20kN each at the third point of a span
of 9m.
a) If the eccentricity of the cable profile is 100mm for the middle third
portion of the beam, calculate the prestressing force required to
balance the bending effect of the concentrated loads (neglect the self –
weight of the beam)
b) For the same cable profile, find the effective force in the cable if the
resultant stress due to self – weight, imposed loads and prestressing
force is zero at the bottom fiber of the mid – span section. Unit wt. of
conc = 24 kN/cu.m
Example 3 - Solution
1. Calculate the prestressing force required to balance the bending effect of the
concentrated loads (neglect the self – weight of the beam)
M EX = M PS
20 ( 9 )
= P ( 0.100 )
3
P = 600kN
Example 3 - Solution
2. Find the effective force in the cable if the resultant stress due to self – weight,
imposed loads and prestressing force is zero at the bottom fiber of the mid – span
section.
kN
wDL = 24 ( 0.3) ( 0.8 ) wDL = 5.76
m
5.76 ( 9 )
2

M DL = M DL = 58.32kN.m
8
fb = fbDL + fbLL + fbPS = 0

58.32x10 6 ( 400 )
fbDL = +
300 ( 800 )
3

12
fbDL = +1.823MPa
Example 3 - Solution
2. Find the effective force in the cable if the resultant stress due to self – weight,
imposed loads and prestressing force is zero at the bottom fiber of the mid – span
section.
20 ( 9 )
M LL = = 60kN.m
3
60x10 6 ( 400 )
fbLL = + fbLL = +1.875MPa
300 ( 800 )
3

12
P M PS c
fbPS =- -
A I

fbPS =-
Px10 3
-
( Px0.1) x10 6 ( 400 )
( 300 ) ( 800 ) 300 ( 800 )
3

12
Example 3 - Solution
2. Find the effective force in the cable if the resultant stress due to self – weight,
imposed loads and prestressing force is zero at the bottom fiber of the mid – span
section.
fb = fbDL + fbLL + fbPS = 0

1.823 +1.875 -
Px10 3
-
( Px0.1) x10 6 ( 400 )
=0
( 300 ) ( 800 ) 300 ( 800 )
3

12

P = 507.154kN
Example 4
A concrete beam with a single overhang is simply supported at A
and B over a span of 8m and the overhang BC is 2m. The beam is of
rectangular section 300mm wide by 900mm deep and supports a
uniformly distributed live load of 3.52 kN/m over the entire length in
addition to its self – weight. Determine the profile of the prestressing
cable with an effective force of 500kN which can balance the dead and
live loads on the beam.
Example 4
The single overhang beam ABC
supporting uniformly distributed load is
shown in the figure below.
Example 4
Given:
Pe = 500 kN, wDL = 6.48kN/m,
wLL=3.52kN/m, RA=37.5 kN, RB=62.5kN

A. Eccentricity at the location of Maximum Positive

Moment
æ e1 ö
70.3 = 500 ç e1 = 140.6mm
è 1000 ÷ø
Example 4
Given:
Pe = 500 kN, wDL = 6.48kN/m,
wLL=3.52kN/m, RA=37.5 kN, RB=62.5kN

B. Eccentricity at the location of Maximum Negative

Moment
æ e2 ö
20 = 500 ç e2 = 40mm
è 1000 ÷ø
 ASK ANY QUESTION
RELATED TO OUR TOPIC
FOR TODAY.
Do the Exercises about
the Analysis of
Prestressed Concrete
Beams using Load
Balancing Method
Reference:
Raju, N. K. (2006). Prestressed concrete. Tata McGraw-Hill Education.