Transcript
Preliminary Design for Flexure
Prestressed Concrete Design
(SAB 4323)
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Preliminary Design for Flexure
Assoc. Prof. Baderul Hisham Ahmad
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Analysis or Design?
Analysis
• Check if the specified design criteria at every
section along the member are satisfied
• Beam’s description and characteristics given
(loading, span, cross sectional dimensions,
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(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
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Basic Inequalities
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Basic Inequalities
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Inequalities At Transfer
• Consider at mid span of a simply supported beam
y1 k
aPi/A - aPie/z1 + Mi/z1 > = ftt
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cgs
cgcH
y1
y2
A
Aps
kt
kb hp
hc
e
+
aPi/A + aPie/z2 - Mi/z2 < = fct
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• Consider at mid span of a simply supported beam
< = fcs
y1
bPi/A - bPie/z1 + Ms/z1
Inequalities At Service
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> = ftsbPi/A + bPie/z2 - Ms/z2
cgs
cgcH
y1
y2
A
Aps
kt
kb hp
hc
e
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Writing down all the inequalities:
αPi/A - αP
ie/z
1+ M
i/z
1> = f
tt………………(1)
αPi/A + αP
ie/z
2- M
i/z
2< = f
ct……………...(2)
βPi/A - βP
ie/z
1+ M
s/z
1< = f
cs………………(3)
Inequalities At Service
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βPi/A - βP
ie/z
1+ M
s/z
1< = f
cs………………(3)
βPi/A + βP
ie/z
2- M
s/z
2> = f
ts………………(4)
By combining inequalities (1) & (3) and (2) & (4)
z1
> = (αMs
– βMi) / (α f
cs– β f
tt)………………..(5)
z2
> = (αMs
– βMi) / (β f
ct– α f
ts)………………..(6)
Beware of +ve and –ve values!Derive (5) & (6)!
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Section Selection
• From (5) & (6), a suitable section can be selected
• Both z1
and z2
depend on Miand M
s
• Miand M
scan be determined if the member self
weight is known
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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
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Span & LoadingSpan & Loading
Choose a sectionChoose
a section
Section Adequacy Flowchart
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Calculate Ws,Mi & MsCalculate Z1 & Z2
from 5 & 6
Calculate Ws,Mi & MsCalculate Z1 & Z2
from 5 & 6
Z req <= Zprov
Z req <= Zprov
No
Section AdequateSection Adequate
Yes
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Examples of Standard Section
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Examples of Standard Section
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Examples of Standard Section
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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
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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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Solution
Given:
Span = 20m; fci = 30 N/mm2 ; fcu = 50 N/mm2 and class 1
category
Pi = 2000kN and e = 500 mm
α = 0.9 , β = 0.8
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α = 0.9 , β = 0.8
Stress Limits:
At transfer
fct
= 0.5fci = 15 N/mm2 and f
tt= 1.0 N/mm2
At service
fcs
= 0.33fcu = 16.5 N/mm2 and f
ts= 0 N/mm2
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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
M = 29.36 x 202/8 = 1468 kNm
Design as RCSize: 200 x 25002 layers of 3T25l/d actual = 8.2l/d all = 11.9
Solution
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Ms
= 29.36 x 202/8 = 1468 kNm
Required Section Modulus
from (5): z1
> = (0.9x1468–0.8x468)x106/(0.9x16.5–0.8(-1))
> = 60.50 x 106 mm3
z1
provided = 84.5 x 106 mm3 �Ok
from (6): z2
> = (0.9x1468–0.8x468)x106/(0.8x15.0–0.9(0))
> = 78.90 x 106 mm3
z2
provided = 84.5 x 106 mm3 �Ok
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l/d all = 11.9
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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
M = 29.42 x 202/8 = 1471 kNm
Solution
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Ms
= 29.42 x 202/8 = 1471 kNm
Required Section Modulus
from (5): z1
> = (0.9x1471–0.8x471)x106/(0.9x16.5–0.8(-1))
> = 60.52 x 106 mm3
z1
provided = 75.39 x 106 mm3 �Ok
from (6): z2
> = (0.9x1471–0.8x471)x106/(0.8x15.0–0.9(0))
> = 78.93 x 106 mm3
z2
provided = 116.23 x 106 mm3 �Ok
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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> = (z
1 f
tt– M
i) / α(z
1/A – e)………………….(7)
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Pi< = (z
2 f
ct+ M
i) / α(z
2/A + e)………………….(8)
Pi< = (z
1 f
cs– M
s) / β(z
1/A – e)………………….(9)
Pi> = (z
2 f
ts+ M
s) / β(z
2/A + e)………………….(10)
• The inequalities sign in (7) & (9) will be reversed if the denominator becomes -ve
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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
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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
α = 0.9 , β = 0.8
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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
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s
Pi< = 7473.4 kN ..………………..(7)
Pi< = 6426.31 kN………………….(8)
Pi> = 4699.25 kN………………….(9)
Pi> = 5195.01 kN..……………….(10)
The minimum value of Piwhich lies within the limits is 5195.01kN
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Inequalities sign reversed
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Solution
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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
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• 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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Magnel Diagram
• Rearranging inequalities (7) to (10):
• e <= (Mi– z
1ftt)/ααααP
i+ z
1/A………….(11)
• e <= (Mi+ z
2fct
)/ααααPi- z
2/A………….(12)
• e >= (Ms
– z1fcs
)/ββββPi+ z
1/A…………(13)
m = (Mi-z1ftt)/α α α α , c = z1/A
m = (Mi-z2fct)/α α α α , c = -z2/A
m = (Ms-z1fcs)/α α α α , c = z1/A
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s 1 cs i 1
• e >= (Ms
+ z2fts
)/ββββPi- z
2/A………….(14)
• Note that z1/A = kb and z2/A = kt i.e lower and upper
limits of the central kern respectively
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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
m = (Ms-z2fts)/α α α α , c = -z2/A
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Magnel Diagram
• The maximum permissible eccentricity,
emp = y2 – (hc)min............................(15)
• Where (hc)min is the minimum concrete cover to
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• 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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Cover & Eccentricity
⅀ Aps * y / ⅀ Aps
e = y2
- hc
hc = ⅀ Aps * y / ⅀ Aps
e = y2
- hc
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CGC
y2
(hc)min
emp
e
hc
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H
cgc
y1
A
kt
k h
1/Pi
ktIneq. 13
All four stress inequalities are satisfied within this region. Thus all combinations of e and 1/Pi within this region are safe
Magnel Diagram
This is the maximumpermissible
prestresing force
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H
cgs
y2
A
Aps
kb hp
hc
e
e
kb
Ineq. 14Ineq. 12
Ineq. 11
Safezone This is the minimum
permissible prestresing force
are safe
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1/Pi
*kt
xpt14
x
Top face
y1
Toward minimum Pi
Magnel Diagram
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1/Pi
e
*kb
*emp
xpt13
xpt12
xpt11
Bottom face
y2
(hc)min
Practical feasible domain
Minimum practical Pi
Minimum Pi
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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
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cu ci
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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Example 6
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x 103 x 109
Example 3-3
Try this section
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Solution
Try Section 250 x 2400
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Wsw = 0.202 x 24 = 4.85 kN/m; Mi = 4.85 x 10 x 10 / 8 = 60.6 kNmWs = (1.5+3.0) x 2.4 + 4.85 = 15.65 kN/mMs = 15.65 x 100/8 = 195.6 kNm
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Solution
Try Section 300 x 2400
Solution
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Solution
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Solution – Manual Plotting
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Solution – Using Graph v4.3
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Solution – Using Inequalities
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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
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– 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
using the inequalities.35
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Example 3-4
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Solution
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Solution
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Magnel Diagram
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-450
-350
-250
-150
-50-0.1 6E-16 0.1 0.2 0.3 0.4 0.5 0.6
e (m
m)
Magnel Diagram
Ineq 11 Ineq 12
Ineq 13 Ineq 14
emax
Solution – Using MS Excel
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50
150
250
350
450
550
650
750
e (m
m)
1/P x 10-6 (N-1)
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Solution – Using Inequalities
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Solution Using Graph V4.3
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