Prestressed Concrete 3

Prestressed Concrete Design
(SAB 4323)
Preliminary Design for Flexure
Assoc. Prof. Baderul Hisham Ahmad
1
http://www.tekniksipil.org
Analysis or Design?
Analysis
Check if the specified design criteria at every
section along the member are satisfied
Beams description and characteristics given
(loading, span, cross sectional dimensions,
material properties etc)
Design :
Reverse process of analysis
Involves finding of member size required and
details of prestressing force and tendon profile
2
2
Basic Inequalities
Inequalities At Transfer
Consider at mid span of a simply supported beam
aPi/A - aPie/z1 + Mi/z1
y1
kt
kb
> = ftt
cgc
hp
y2 e
Aps
hc
cgs
aPi/A + aPie/z2 - Mi/z2
< = fct
4
4
Inequalities At Service
Consider at mid span of a simply supported beam
y1
kt
A
kb
bPi/A - bPie/z1 + Ms/z1
< = fcs
bPi/A + bPie/z2 - Ms/z2
> = fts
cgc
hp
y2 e
Aps
hc cgs
Inequalities At Service
Writing down all the inequalities:
Pi/A - Pie/z1 + Mi/z1 > = ftt (1)
Pi/A + Pie/z2 - Mi/z2 < = fct ...(2)
Pi/A - Pie/z1 + Ms/z1 < = fcs (3)
Pi/A + Pie/z2 - Ms/z2 > = fts (4)
By combining inequalities (1) & (3) and (2) & (4)
z1 > = (Ms Mi) / ( fcs ftt )..(5)
z2 > = (Ms Mi) / ( fct fts )..(6)
Beware of +ve and ve values!
Derive (5) & (6)!
Section Selection
From (5) & (6), a suitable section can be selected
Both z1 and z2 depend on Mi and Ms
Mi and Ms can be determined if the member self
weight is known
However, the self weight can only be determined
if the section size (hence z1 and z2) is known
In general, the solution can be obtained using
trial and error method or using standard section
7
Section Adequacy Flowchart

Span & Loading
Choose
a section
Calculate Ws,Mi & Ms

Calculate Z1 & Z2
from 5 & 6
No
Z req <=
Zprov
Yes
Section Adequate
8
Examples of Standard Section
9
9
10
10
11
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Example 3-1
A 20m span simply supported beam for a bridge
construction is to be designed using class 1 post-tensioned
prestressed concrete. The beam is subjected to a
characteristic live load of 20kN/m in addition to its own
self weight. The initial prestressing force is 2000kN with
an eccentricity of 500mm. The short and long term losses
of prestress are estimated to be 10% and 20% respectively.
With fci = 30 N/mm2 and fcu = 50 N/mm2 select a suitable
section for the beam using,
1. Rectangular section
2. Standard M beams
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12
Solution
Given:
Span = 20m; fci = 30 N/mm2 ; fcu = 50 N/mm2 and class 1
category
Pi = 2000kN and e = 500 mm
= 0.9 , = 0.8
Stress Limits:
At transfer
fct = 0.5fci = 15 N/mm2 and ftt = 1.0 N/mm2
At service
fcs = 0.33fcu = 16.5 N/mm2 and fts = 0 N/mm2
13
13
Solution
1)
Rectangular Section
try: b = 300mm and h = 1300 mm
A = 390000 mm2 ; z1 = z2 = bh2/6 = 84.5 x 106 mm3
Self wt, Wsw = 24 x 0.39 = 9.36 kN/m
Mi = 9.36 x 202/8 = 468 kNm
Total service load, Ws = 20 + 9.36 = 29.36 kN/m
Ms = 29.36 x 202/8 = 1468 kNm
Design as RC
Size: 200 x 2500
2 layers of 3T25
l/d actual = 8.2
l/d all = 11.9
Required Section Modulus

from (5): z1 > = (0.9x14680.8x468)x106/(0.9x16.50.8(-1))
> = 60.50 x 106 mm3
z1 provided = 84.5 x 106 mm3 Ok
from (6): z2 > = (0.9x14680.8x468)x106/(0.8x15.00.9(0))
> = 78.90 x 106 mm3
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14
Solution
2)
Standard Section M beams

try: M6 beams
A = 387050 mm2 ; z1 = 75.39 x 106 mm3 ; z2 = 116.23 x 106 mm3
Self wt, Wsw = 9.42 kN/m
Mi = 9.42 x 202/8 = 471 kNm
Total service load, Ws = 20 + 9.43 = 29.42 kN/m
Ms = 29.42 x 202/8 = 1471 kNm
Required Section Modulus
from (5): z1 > = (0.9x14710.8x471)x106/(0.9x16.50.8(-1))
> = 60.52 x 106 mm3
from (6): z2 > = (0.9x14710.8x471)x106/(0.8x15.00.9(0))
> = 78.93 x 106 mm3
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15
Design of Prestress Force
Rearranging inequalities (1) to (4) will yield inequalities for the

required prestress force, for a given value of eccentricity
Thus the new inequalities are:
Pi > = (z1 ftt Mi) / (z1/A e).(7)
Pi < = (z2 fct + Mi) / (z2/A + e).(8)
Pi < = (z1 fcs Ms) / (z1/A e).(9)
Pi > = (z2 fts + Ms) / (z2/A + e).(10)
The inequalities sign in (7) & (9) will be reversed if the denominator
16
becomes -ve
16
Example 3-2
A post-tensioned prestressed concrete bridge deck is in
the form of a solid slab with a depth of 525 mm and is
simply supported over 20 m. It carries a service load of
10.3 kN/m2. If the maximum eccentricity of the
tendons at midspan is 75 mm above the soffit, find the
minimum value of the prestress force required. Use the
following information:
fct = 20.0 N/mm2 and ftt = 1.0 N/mm2
fcs = 16.7 N/mm2 and fts = 0 N/mm2
= 0.9 , = 0.8
17
17
Solution
z1 = z2 = 5252 x 103 / 6 = 45.94 x 106 mm3/m
A = 525 x 1000 = 5.25 x 105 mm2/m; e = 525/2 75 = 188 mm
Mi = 24 x 0.525 x 202 / 8 = 630 kNm/m
Ms = 630 + (10.32 x 202 / 8) = 1145 kNm/m
Pi < = 7473.4 kN ....(7)
Inequalities sign reversed
Pi < = 6426.31 kN.(8)
Pi > = 4699.25 kN.(9)
Pi > = 5195.01 kN...(10)
The minimum value of Pi which lies within the limits is 5195.01kN
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18
Solution
19
19
Magnel Diagram
First explored by Magnel, a Belgian engineer

Plot of e versus Pi produced a hyperbolic curve
Plot of e versus 1/Pi produced a straight line
Therefore, we will use e versus 1/Pi
Sign convention:
X-axis represents 1/Pi
Y-axis represents e
+ve Y-axis (e values) pointing downwards (if possible)
+ve X-axis (1/Pi values) pointing to the right
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20
Magnel Diagram
Rearranging inequalities (7) to (10):

Pi + z1/A.(11)m = (Mi-z1ftt)/ , c = z1/A
e <= (Mi z1ftt)/
e <= (Mi + z2fct)/
Pi - z2/A.(12)m = (Mi-z2fct)/ , c = -z2/A
e >= (Ms z1fcs)/
Pi + z1/A(13)m = (Ms-z1fcs)/ , c = z1/A
e >= (Ms + z2fts)/
Pi - z2/A.(14)m = (Ms-z2fts)/ , c = -z2/A
Note that z1/A = kb and z2/A = kt i.e lower and upper
limits of the central kern respectively
The above inequalities can be written as:
e <= mx + c or e >= mx + c
where m is the gradient and c is the vertical axis intercept
21
21
Magnel Diagram
The maximum permissible eccentricity,

emp = y2 (hc)min............................(15)
Where (hc)min is the minimum concrete cover to
c.g.s. which must conform to durability and fire
protection requirements
Therefore, e < = emp
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22
Cover & Eccentricity

hc = Aps * y / Aps
e = y2 - hc
CGC
y2
emp
hc
(hc)min
23
23
Magnel Diagram
This is the maximum
permissible
prestresing force
kt
All four stress

y1
inequalities
are 13
Ineq.
satisfied within this
region. Thus all
1/Pi
combinations of e and
1/Pi within this region
are safe
H
Safe
zone
kb
This is the minimum

permissible
prestresing force
y2
kt
A
cgc
kb
hp
e
Aps
Ineq. 11
Ineq. 12
cgs
hc
Ineq. 14
24
24
Magnel Diagram
Top face
Toward minimum Pi
y1
y2
kt
kb
*
*
emp
pt14
x
x
pt12
x
pt13
x
pt11
1/Pi
Practical feasible domain
Minimum practical Pi
(hc)min
Bottom face
Minimum Pi
e
25
25
Example 3-3
It is required to construct a building floor using standard
precast, pre-tensioned units of double T-section (Class 2) as
shown on next slide. Given the following information:
fcu = 50 N/mm2; fci = 36 N/mm2
Span = 10m (simply supported)
Dead load due to floor finish = 1.5 kN/m2
Live load = 3.0 kN/m2
(a) Choose a suitable double T-section
(b) Construct a Magnel diagram to determine the
minimum prestressing force for the tendon.
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26
Example 6
27
27
Example 3-3
x 103
x 109
Try this section
28
28
Solution
Try Section 250 x 2400
Wsw = 0.202 x 24 = 4.85 kN/m; Mi = 4.85 x 10 x 10 / 8 = 60.6 kNm

Ws = (1.5+3.0) x 2.4 + 4.85 = 15.65 kN/m
Ms = 15.65 x 100/8 = 195.6 kNm
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29
SolutionSolution
Try Section 300 x 2400
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30
Solution
31
31
Solution Manual Plotting
32
32
Solution Using Graph v4.3
33
33
Solution Using Inequalities
34
34
Example 3-4
A post-tensioned precast concrete beam (shown in next
slide), simply supported over a span of 29.4m carries a
total uniformly distributed service load of 35.8 kN/m in
addition to its own self weight. The following information
is given:
Class 1 category; fci = 45 N/mm2; fcu = 50 N/mm2
A = 723700 mm2; y2= 876 mm
I = 255.34 x 109mm4 ; cover to tendon = 152 mm
Take unit weight of concrete, g as 25 kN/m3
Construct a Magnel diagram and find the minimum
prestress force. Compare your results with those obtained
35
using the inequalities.
35
Example 3-4
36
36
Solution
37
37
Solution
Magnel Diagram
38
38
Solution Using MS Excel

Magnel Diagram
-450
Ineq 11
Ineq 12
Ineq 13
Ineq 14
-350
-250
emax
e (mm)
-150
-0.1
-506E-16
50
0.1
0.2
0.3
0.4
0.5
0.6
1/P x 10-6 (N-1)
150
250
350
450
550
650
750
39
39
Solution Using Inequalities
40
40
Solution Using Graph V4.3
41
41

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Prestressed Concrete Design

aPi/A + aPie/z2 - Mi/z2

bPi/A - bPie/z1 + Ms/z1

bPi/A + bPie/z2 - Ms/z2

Section Adequacy Flowchart

Calculate Ws,Mi & Ms

Examples of Standard Section

Examples of Standard Section

Examples of Standard Section

Required Section Modulus

Standard Section M beams

Design of Prestress Force

Rearranging inequalities (1) to (4) will yield inequalities for the

First explored by Magnel, a Belgian engineer

Rearranging inequalities (7) to (10):

The maximum permissible eccentricity,

Cover & Eccentricity

All four stress

This is the minimum

Try this section

Wsw = 0.202 x 24 = 4.85 kN/m; Mi = 4.85 x 10 x 10 / 8 = 60.6 kNm

Solution Manual Plotting

Solution Using Graph v4.3

Solution Using Inequalities

Solution Using MS Excel

1/P x 10-6 (N-1)

Solution Using Inequalities

Solution Using Graph V4.3

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