Prestressed Concrete 3

Preliminary Design for Flexure

Prestressed Concrete Design

(SAB 4323)

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Preliminary Design for Flexure

Assoc. Prof. Baderul Hisham Ahmad

http://www.tekniksipil.org

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


Inequalities At Transfer

• Consider at mid span of a simply supported beam

y1 k

aPi/A - aPie/z1 + Mi/z1 > = ftt

4

4

cgs

cgcH

y1

y2

A

Aps

kt

kb hp

hc

e

+

aPi/A + aPie/z2 - Mi/z2 < = fct


• 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


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)!


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


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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10

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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

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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


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l/d all = 11.9


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


from (6): z2

> = (0.9x1471–0.8x471)x106/(0.8x15.0–0.9(0))

> = 78.93 x 106 mm3

z2


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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


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


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


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


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


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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Prestressed Concrete 3

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