UNIT II - DESIGN CONCEPTS - WordPress.com · UNIT 1 INTRODUCTION – THEORY AND BEHAVIOUR Basic...

Sl.No Contents Page No.

PRESTRESSED CONCRETE STRUCTURES

UNIT I - THEORY AND BEHAVIOUR

1.1 Pre-stressed concrete 2

1.2 Types of pre-stressing 2

1.3 Losses 3

UNIT II - DESIGN CONCEPTS

2.1 Analysis of beam section - concept 18

2.2 Elastic Design for flexure 38

2.3 Permissible stresses for flexure member 41

2.4 End block 54

UNIT III CIRCULAR PRESTRESSING

3.1 Design Procedure for circular tanks Computations 61

3.2 Circular pre-stressing 66

3.3 Design of pipes 70

3.4 Design of circular water tanks 74

UNIT IV COMPOSITE CONSTRUCTION

4.1 Introduction 78

4.2 serviceability limit state 78

4.3 Ultimate strength 81

4.4 Horizontal shear 81

UNIT V PRESTRESSED CONCRETE BRIDGES

5.1 Prestressed concrete bridges 85

CE PRESTRESSED CONCRETE STRUCTURES L T P C 3 0 0 3

OBJECTIVE

A t the end of this course the student shall have knowledge of methods of prestressing advantages ofprestressing concrete, the losses involved and the design methods for prestressed concrete elementsunder codal provisions.

UNIT 1 INTRODUCTION – THEORY AND BEHAVIOUR

Basic concepts – Advantages – Materials required – Systems and methods of prestressing – Analysis of

sections – Stress concepts – Strength concepts – Load balancing concept – Effect of loading on the

tensile stresses in tendons – Effect of tendon profile on deflections – Factors influencing deflections –Calculation of deflections – Short term and long term deflections – Losses of prestress – Estimation ofcrack width .

UNIT II DESIGN CONCEPTS

Flexural strength – Simplified procedures as per codes – strain compatibility method – Basic concepts inselection of cross section for bending – stress distribution in end block, Design of anchorage zonereinforcement – Limit state design criteria – Partial prestressing – Applications.

UNIT III CIRCULAR PRESTRESSING

Design of prestressed concrete tanks – Pipes.

UNIT IV COMPOSITE CONSTRUCTION

Analysis for stresses – Estimate for deflections – Flexural and shear strength of composite members.

UNIT V PRESTRESSED CONCRETE BRIDGES

General aspects – pretensioned prestressed bridge decks – Post tensioned prestressed bridge decks –Principle of design only.

TOTAL: 45 PERIODS

TEXT BOOKS

1. Krishna Raju N., Prestressed concrete, Tata Mcgraw Hill Company, New Delhi, 19982. Mallic.S.K. and Gupta A.P., Prestressed concrete , Oxbord and IBH publishing Co.Pvt.Ltd 1997.

3. Rajagopalan, N” Prestressed Concrete”, Alpha Science, 2002.REFERENCES1. Ramaswamy G.S.Modern prestressed concrete design, Arnold Heinimen, Newdelhi, 19902. LinT.Y., Design of prestressed concrete structures, Asia Publishing House, Bombay, 1995

3. David A.Sheppard, William R and Philphs, Plant Cast precast and prestressed concrete – Adesign guide, McGraw Hill,Newdelhi 1992

CE2404 Prestressed Concrete Structures

SCE Dept of Civil

Chapter -1INTRODUCTION – THEORY AND BEHAVIOUR

Basic concepts – Advantages – Materials required – Systems and methods of prestressing – Analysis of

sections – Stress concepts – Strength concepts – Load balancing concept – Effect of loading on the

tensile stresses in tendons – Effect of tendon profile on deflections – Factors influencing deflections –Calculation of deflections – Short term and long term deflections – Losses of prestress – Estimation ofcrack width .

1.1 Pre-stressed concreteDefinition: Concrete in which there have been introduced internal stresses of such magnitude anddistribution that the stresses resulting from given external loadings are countered to a desired degree -ACI

1.2 Types of pre-stressing

1.2.1 Pre-tensioning & Post-tensioningIn pre-tensioning the tendons are tensioned before the concrete is placed. The tendons are temporarilyanchored to abutments or stressing beds. Then the concrete member is cast between and over thewires. After the concrete has attained the required strength, the wires are cut from the bulkhead andpre-stress is transferred to the concrete member.

In post-tensioning the concrete member is cast with ducts for the wires. After concrete has attainedsufficient strength, wires are threaded into the ducts, tensioned from both or one end by means ofjack/jacks and at the precise level of pre-stress the wires are anchored by means of wedges to theanchorage plates at the ends.

1.2.2 Bonded & Un-bonded tendonIn post-tensioned members, the wires are either left free to slide in the ducts or the duct is filled withgrout. In the former, the tendon is un-bonded and in the latter it is bonded.

Stages of loading

Initial stageThe member is under pre-stress but is not subjected to any superimposed external loads. Furthersubdivision of this stage is possible.

1. Before pre-stressing: Concrete is weak in carrying loads. Yielding of supports must be prevented.2. During pre-stress:

a. Steel: This stage is critical for the strength of tendons. Often the maximum stress towhich the wires will be subjected throughout their life may occur at this stage.

b. Concrete: As concrete has not aged at this stage, crushing of concrete at anchorages ispossible, if its quality is inferior or the concrete is honeycombed. Order of pre-stressingis important to avoid overstress in the concrete.


SCE Dept of Civil

3. At transfer of pre-stress: For pre-tensioned members, where transfer is within a short period,and for post-tensioned members where transfer may be gradual, there are no external loads onthe member except its own weight.

4. De-shuttering: The removal of form-work must be done after due considerationThus the initial pre-stress with little loss imposes a serious condition n the concrete and often controlsthe design of the member.

Final stageThis is the stage when actual working loads come on the structure. The designer must consider variouscombinations of live loads on different parts of the structure with lateral loads such as wind andearthquake forces and strain loads produced by settlement of supports and temperature. The majorloads in this stage are:

1. Sustained load: It is often desirable to limit the deflection under sustained loads sue to its ownweight and dead loads.

2. Working load: The member must be designed for the working load. Check for excessive stressand deflection must be made. But this design may not guarantee sufficient strength to carryoverloads.

3. Cracking load: Cracking in a pre-stress member signifies a sudden change in bond and shearingstresses. This stage is also important

4. Ultimate load: This strength denotes the maximum load the member can carry before collapse.

1.3 Losses

Elastic Shortening (ES) – Cl 18.5.2.4Shortening in steel that occurs as soon as Fi is transferred to the concrete member and the member as

a whole shortens.

Fi = Pre-stress just before transfer

F = Final stress after losses

Fo = Immediately after transfer – very difficult to estimate

Note: The value of Fo may not be known, but it is not necessary, as the losses from Fi to Fo is only a

small percentage of Fi . Total accuracy is relative anyway, as Ec – the young’s modulus of concrete –cannot be determined accurately.

Therefore


SCE Dept of Civil

S

ES = E s

f c

EC

F O

AC EC

where is the shortening in steel that occurs as soon as Fi is transferred to the

concrete member and the member as a whole shortens. Thus is the shortening in the

member due to Fi at the level of steel.

FES O E

AC EC

Since

S

f c is the stress in concrete at level of steel and isFO

AC

FES O E

ESTaking nEC

AC EC

F n O

AC

As Fo cannot be estimated, Fi can be used to calculate ES.

at level of steel Fi

AC EC AS ES

ES Es

Es

Fi

AC EC AS ES

nFi

AC nAS

Taking At AC EC AS ES


SCE Dept of Civil

Ac

nFES i

AT

whichever way the ES is calculated

ES = n (concrete stress at level of steel)

If external loads are acting on the member, then concrete, then concrete stress due toall loads at level of steel must be determined.

F F e 2 M ef c

O

AG

O

I G

I

Note: AG , the gross-area, instead of the transformed sectional area, leads to simpler calculations and

fairly accurate results.

Fo 0.9Fi for pre-tensioned member

f FO

G

ES nf c

Creep (CR) Cl 18.5.2.1Among the many factors affecting creep are volume to surface ratio, age of concrete at time of pre-stress, relative humidity, type of concrete (lightweight / normal). Creep is assumed to occur in themember after permanent loads are imposed after pre-stress. Creep occurs over a long period of timeunder sustained load. Part of initial compressive strain induced in concrete immediately after transfer isreduced by the tensile strain produced by superimposed permanent loads.

Therefore for bonded members, loss due to creep

CR n f cir f cds f c

En S

EC

= Creep coefficient – Cl 4.5.3 & Cl 5.2.5.1

f cir = concrete stress at level of steel immediately after transfer.


SCE Dept of Civil

fcds = stress in concrete at steel level due to superimposed dead loads applied to the

member after transfer of pre-stress

Shrinkage of concrete (SH) Cl 18.5.2.2Factors like volume to surface ratio, relative humidity, time from end of moist curing to application ofpre-stress, affect shrinkage in concrete. Shrinkage is time-dependant and about 80% of the final loss dueto shrinkage occurs in the first year and 100% after several years.

Shrinkage strain

sh 0.0003 for pretensioned member

0.0002

log10 t 2 for posttensioned member and

may be increased by 50% in dry condition

but not more than 0.0003

Cl 5.2.4.1

Relaxation of steel (RE) Cl 18.5.2.3When elongation is sustained over pre-stressing cable for a long time, the pre-stress will decrease

gradually. The RE – loss due to relaxation depends on type of steel, time, as well as the ratio off i wheref p

f i is the initial pre-stress and f p is the characteristic strength of steel.

RELAXATION LOSSES FOR PRESTRESSING STEEL AT 1 000 H AT 27°C

INITIAL STRESS RELAXATION

INITIAL STRESS RELAXATION LOSS

N/mm2

0.5 fp

0.6 fp

0.7 fp

0.8 fp

0

35

70

90


SCE Dept of Civil

Anchorage slip (ANC) Cl 18.5.2.5In post-tensioning, when the jack is released, the full pre-stress is transferred to the anchorage and theytend to deform, allowing the tendon to slacken. Friction wedges will slip a little before they grip the wirefirmly. So, in post-tensioning the wedges are positively engaged before the jack is released. In pre-tensioning also, the anchorage slip is compensated for during stressing operation.

The loss is caused by a fixed shortening of the anchorages, so the percentage loss ishigher in shorter wires than in long ones.

If a tendon is stressed to 1035 MPa, with Es 2105 MPa and the anchorage slips by 2.5 mm,

Total 1035

0.0051752105

In a cable of 3m length, elongation l 0.005175 3000 15.53 mm , ie % l 2.5

15.53100 16%

But in a cable of 30 m length, elongation l 0.005175 30000 155.30 mm , ie

% l 2.5

100 1.6% only155.30

Frictional loss Cl 18.5.2.6Frictional loss comprise of two parts: (1) The length effect and (2) The curvature effect.

The length effect or the wobble effect of the duct is the friction that will exist between straight tendonand the surrounding material. This loss is dependant on the length and stress in the tendon, thecoefficient of friction between the contact materials, the workmanship and the method used in aligningand obtaining the duct.

The curvature effect is the loss due to intended curvature of the tendon. This again depends on thecoefficient of friction between the materials and the pressure exerted by the tendon on the curvature.

For un-bonded tendon, lubrication, in the form of grease and plastic tube wrapping can be used toadvantage.

For bonded tendon lubricant in the form of water soluble oils are used during stressing operation andflushed off with after before grouting.

Jacking from both ends of the beam will also reduce loss due to friction.


SCE Dept of Civil

5 5

For straight or moderately curved structures, with curved or straight cables, the value of pre-stressingforce Px at a distance x meters from tensioning end and acting in the direction of the tangent to thecurve of the cable, shall be calculated as below:

Px = Poe– ( μα + kx ).

Where Po = pre-stressing force in the pre-stressed steel at the tensioning end acting in the direction of

the tangent to the curve of the cable, α = cumulative angle in radians through which the tangent to thecable profile has turned between any two points under consideration, μ = coefficient of friction in curve;

unless otherwise proved by tests, μ may be taken as: 0.55 for steel moving on smooth concrete, 0.30 forsteel moving on steel fixed to duct, and 0.25 for steel moving on lead, k = coefficient for wobble or waveeffect varying from 15 × 10–4 to 50 × 10–4 per meter. The expansion of the equation for Px for small

values of (μα + kx) may be Px = Po (1 – μα – kx).

Examples

To calculate ES in Pre-tensioned beam - eccentric tendonA pre-tensioned beam of 100 mm x 300 mm is pre-stressed by straight wires with Fi = 150 kN at an e =

50 mm. ES = 2.1x10 MPa, EC = 0.35x10 MPa and AP = 188 mm2. Estimate ES.

AG = 100 x 300 = 30000 mm2

1003003

I =12

= 225x106 mm4

2.10n =

0.35= 6.0


SCE Dept of Civil

c

F F e2

f i i

AG I

f c = 150103

30000

150103 5050

225106= -6.67 MPa

ES nf c

ES = 66.67 = 40.02 MPa

Loss =40.02188

150103= 5.02%

ES in Pre-tensioned beam - concentric tendonA straight pre-tensioned beam 12 m long of 380 mm x 380 mm is concentrically pre-stressed with 780

mm2 wires anchored to bulkheads with a f = 1035 MPa. E = 2x105 MPa, E = 0.33x105 MPa.

Estimate ES at transfer.i S C

Fi = 1035 x 780 = 807.30 kN

AG

AC

= 380 x 380

= AG - AS

= 144400 – 780

= 144400 mm2

= 143620 mm2

AT = AC + nAS


SCE Dept of Civil

Ac

Ac

5 5

= 143620 + 6x780 = 148300 mm2

2.00n =

0.33= 6.0

Ff i

T

807.30103

f c = = -5.44 MPa148300

ES nf c

ES = 65.44 = 32.66 MPa

If Fo 0.9Fi

FO = 0.90x807.30 = 726.57 kN

Ff O

G

726.57103

f c = = 5.03 MPa144400

ES = 5.03x6 = 30.18 MPa

ES in Pre-tensioned beam - Eccentric tendons at top & botA pre-tensioned beam of 200 mm x 300 mm is pre-stressed with 15#5mm wires located at 65 mm

from the bottom of the beam and 3#5mm wires located at 25 mm from the top of the beam. f i =

840 MPa. ES = 2.1x10 MPa, EC = 0.315x10 MPa. Estimate ES at transfer.


SCE Dept of Civil

Aw = Area of one wire

52

=4

= 19.63 mm2

Fi = 18x19.63x840 = 296.81 kN

200 3003

I =12

= 450x106 mm4

n =

eeq =

2.10

0.315

1519.6384085 319.63840125

1819.63840

= 6.67

= 50 mm

fCTop Fi AG

Fi e yI t

fCTop

296.81103

= 60000

296.81103 50

450106125 = -0.824 MPa

fCBot Fi AG

Fi e yI t


SCE Dept of Civil

i

5 5

fCBot

296.81103

= 60000

296.81103 50 85450 106

= -7.75 MPa

ES nf c

ESTop = 6.670.824

ESBot = 6.677.75

= 5.50 MPa

= 51.69 MPa

To calculate ES in Post-tensioned beamA straight post-tensioned beam 12 m long of 380 mm x 380 mm is concentrically pre-stressed with 780

mm2 wires made up of 4 tendons with 195 mm2 and the tendons are pre-stressed sequentially with a f

= 1035 MPa. ES = 2x10 MPa, EC = 0.33x10 MPa. Estimate ES at transfer.

The loss in the 1st tendon is due to the shortening of concrete by the pre-stressing of the previous 3

Fitendons. We can assume that Fi in each of these tendons are constant and f c n .AG

for the 1st tendon

31951035ES1 = 6 = 25.16 MPa

380380

for the 2nd tendon

21951035ES2 = 6 = 16.77 MPa

380380


SCE Dept of Civil

st

nd

rd

th

for the 3rd tendon

11951035ES3 = 6 = 8.39 MPa

380380

There is no loss in the 4th tendon

The average loss

ESav =25.16 16.77 8.39

4= 12.58 MPa

When there are many cables, it is quite enough to assume that ESav1

of the loss in the 1st cable.2

Thus ESav =125.16

2= 12.58 MPa

If it is desired that there should be no loss at all, then the cables can be overstressed before anchorage.

So,

f i in 1 cable = 1035+25.16 = 1060.16 MPa

f i in 2 cable = 1035+16.77 = 1051.77 MPa

f i in 3 cable = 1035+8.39 = 1043.39 MPa

f i in 4 cable = 1035.00 MPa

But this stressing pattern is highly theoretical.

To calculate CR, SH and RE in post-tensioned beamA straight post-tensioned beam of size 100 mm x 300 mm is pre-stressed with 5 wires of 7 mm . The

5average pre-stress after short-term losses is f pe = 1200 MPa. The gae at loading is 28 days. ES = 2x105MPa, EC = 0.35x10 MPa. Estimate CR, SH and RE assuming fp = 1715 MPa.


SCE Dept of Civil

= 1.6 T.2c.1, Cl 5.2.5.1

2.00n =

0.35= 5.71


7 2

=4

= 38.45 mm2

Fpe = 5x38.45x1200 = 230.7 kN

AG = 300 x 100 = 30000 mm2

1003003

I =12

= 225x106 mm4

F F ef cir

pe

AG

pe yI

= 230700

23070050

50 = -10.25 MPa30000 225106

CR = 1.6 5.7110.25 = 93.64 MPa


SCE Dept of Civil

t = 28 days

sh 0.0002

log10 t 2Cl 5.2.4.1

0.0002 -4=log10 28 2 = 1.35x10

SH = sh ES

= 1.35x10-4 x (2x105) = 27 MPa

f pe = 1200 MPa

f pe

f p

1200=

1715= 0.699

70 %

f pe = 0.70 f p

RE = 70 MPa T.4, Cl 18.5.2.3

To calculate frictional losses – tensioned from one endA post-tensioned beam 100 mm × 300 mm of le = 10 m is stressed by successive tensioning andanchoring of 3 cables A, B, and C respectively as shown in figure. Each cable has cross section area of

200 mm2 and has initial stress of 1200 MPa. If the cables are tensioned from one end, estimate the

percentage loss in each cable due to friction at the anchored end. Assume μ = 0.35, K = 0.0015 / m.


SCE Dept of Civil

8y, the cable being considered a parabola of segment length = x and y = central sag.

x

L = 10000 mm for all cables

= 0.35

K = 0.0015 / m

Let F1 be the pre-stress at beginning of the 1st segment

Cable Lmm

KL y

mm

rad

KL eKL Stress @end of seg

A 10000 0.015 100 0.08 0.028 0.043 0.958 0.958F1

B 10000 0.015 50 0.04 0.014 0.029 0.971 0.931F1

C 10000 0.015 0 0 0 0.015 0.985 0.917F1

Loss = 1 – 0.917 = 0.08

= 8%

To calculate frictional losses – tensioned from both endsA pre-stressed concrete beam is continuous over two spans and its curved tendon is to be tensionedfrom both ends. Compute the percentage of loss of pre-stress due to friction from one end to the centerof the beam (A-E). The coefficient of friction between the cable and the duct is 0.40 and the averagewobble or length effect is represented by k = 0.0026/m. The cable is straight between A-B and C-D. Thechange in angle between BC is 0.167 radians and that between DE is 0.100 radians.


SCE Dept of Civil

Segment L

m

KL

rad

KL eKL Stress @end of seg

AB 5.334 0.014 0 0 0.014 0.986 0.986F1

BC 7.620 0.020 0.167 0.067 0.087 0.917 0.904F1

CD 5.334 0.014 0 0 0.014 0.986 0.892F1

DE 3.048 0.008 0.100 0.040 0.048 0.953 0.850F1

Loss = 1 – 0.850 = 0.15

= 15%


SCE Dept of Civil

Chapter 2

DESIGN CONCEPTS

Flexural strength – Simplified procedures as per codes – strain compatibility method – Basic concepts inselection of cross section for bending – stress distribution in end block, Design of anchorage zonereinforcement – Limit state design criteria – Partial prestressing – Applications.

2.1 Analysis of beam section - concept

Sign convention1. Tension is (+)2. Compression is (-)

Different conceptsDifferent concepts can be applied to the analysis if PSC concrete beams, namely

1. Pre-stressing transforms concrete into an elastic material.2. Pre-stressing is a combination of high-strength steel and concrete.3. Pre-stress balances loads.

Elastic materialThis concept treats concrete as an elastic material and is the most common among engineers. Hereconcrete is visualized as being subjected to:

1. Internal pre-stress2. External loads.

So long as there are no cracks in the section, the stresses, strains and deflections of the concrete due tothe two systems of forces can be considered separately and superimposed if needed.

Due to a tensile pre-stressing force F, thee resulting stress at a section is given below.


SCE Dept of Civil

f F

Fey

My

A I I

The concrete stress at a section due to pre-stress f is dependant only on the magnitude and location of

pre-stress at that section, ie., F and e, regardless of how the tendon profile varies elsewhere along the

beam.

[Note: Stresses are calculated with force and eccentricity in steel.]

Ex 1A pre-stress concrete rectangular beam of size 500 mm x 750 mm has a simple span of 7.3 m and isloaded with a udl of 45 kN/m including its self-weight. An effective pre-stress of 1620 kN is produced.Compute the fiber stresses in concrete at mid-span section.


SCE Dept of Civil

F = 1620 kN

A

e

= 500 x 750 = 375000 mm2

= 145 mm

5007503

I =12

= 1.758 x1010 mm4

750y =

2= 375 mm

45 7.32

M =8

= 299.76 kN-m

f F

Fey

My

A I I

f = 1620000

375000

1620000145375

1.7581010

299.76106 375

1.7581010

= 4.32 5.01 6.39


SCE Dept of Civil

f top = 4.32 5.01 6.39 = -5.70 MPa

f bot = 4.32 5.01 6.39 = -2.94 MPa

High strength steel and concreteThis considers the pre-stressed concrete as combination of steel and concrete similar to RCC. Tensionexists in steel and compression in concrete. These two form a internal resisting couple against external

moment produced by loads.

f C

Cey

A I

[Note: Stresses are calculated with force and eccentricity in concrete. e is the eccentricity of C, thecompressive force in concrete.]

Ex 2Solve Ex.1 using this concept.


SCE Dept of Civil

45 7.32

M =8

= 299.76 kN-m

C = T = 1620 kN

MLever arm a =

C

299.76106

=1620103

= 185 mm

C acts at = 185 + 230 = 415 mm from top

750e for C = 415 = 40 mm

2

f C

Cey

A I

f = 1620000

162000040375

375000 1.7581010

= 4.32 1.38

f top = 4.32 1.38 = -5.70 MPa

f bot = 4.32 1.38 = -2.94 MPa

Load balancingThe effect of pre-stressing is considered as the balancing of gravity loads so that the member underbending will not be subjected to flexural stresses under a given loading condition.

wup 8Pe

L2


SCE Dept of Civil

wup 4Pe

L

wup Pe

aL

Due to a parabolic tendon of length l, sag h and stressed to F, wb 8Fh

l 2

f F

Mywhere M is the moment due to net loads.

A I

Ex 3Solve Ex.1 using this concept.


SCE Dept of Civil

w 8Fh

bl 2

wb =816200.145

7.32= -35.30 kN/m (up)

Net udl = 45.0 – 35.30 = 9.70 kN/m (down)

9.7 7.32

M =8

= 64.60 kN-m

f F

Myc A I

f c =1620000

375000

64.60106 375

1.7581010

= 4.32 1.38

f top = 4.32 1.38 = -5.70 MPa

f bot = 4.32 1.38 = -2.94 MPa


SCE Dept of Civil

G

Analysis of beam section for flexure

Stress in concrete and steel due to pre-stress only

If

F = pre-stress (whether initial or final) and applied through the centroid.

Ff c =

Awhere A is the area of concrete

Using the transformed method, stress in concrete is uniform even at the level of steel

f c = Fi

AC nAS

F F= i or i

AT AG

Stress in steel

f s = n f c

nF= i

AC nAS

nF= i Which represents the immediate reduction in pre-stress in steel at transfer.

AT

But is approximated to,

nF= i where A

AG

is the gross area, the error being about 2% to 3 %

Example - Pre-tensioned member – concentric tendonA pre-tensioned beam of size 200 mm x 300 mm is concentrically pre-stressed with 520 mm2 wires

anchored to bulkheads with a f i = 1035 MPa. Assuming n = 6, compute the stresses in concrete and

steel immediately after transfer due to pre-stress only.


SCE Dept of Civil

AG = 200 x 300 = 60000 mm2

Fi = 520 x 1035 = 538.20 kN

538.20 103

f c = at level of steel = -8.9760000

ES

= 68.97 = -53.82 MPa

f Pe (aft. loss) = 1035.00 – 53.82 = 981.00 MPa

f cTop,Bot (aft. loss)

F= e

AG

981520= = -8.50 MPa

60000

Example - Pre-tensioned member – eccentric tendonA pre-tensioned beam of size 200 mm x 300 mm is eccentrically pre-stressed with 520 mm2 wires

anchored to bulkheads with a f i = 1035 MPa. The cgs is 100 mm above the bottom of the beam.

Assuming n = 6, compute the stresses in concrete and steel immediately after transfer due to pre-stressonly.

Fi = 1035 x 520 = 538.20 kN

e

AG

= 150 – 50

= 200 x 300

= 50 mm

= 60000 mm2


SCE Dept of Civil

Ac

c

200 3003

I =12

= 450 x106 mm4

F F e2

f i i at level of steelAG I

f c = 538.20103

60000

538.20103 5050

450106= -11.96 MPa

ES nf c

ES = 611.96 = 71.76 MPa

f Pe (aft. loss) = 1035.00 – 71.76 = 963.24 MPa

Fe = 983.24 x 520 = 500.88 kN


Fe= AG

Fe ey

I

3 3

= 500.8810

500.8810 50

15060000 450106

= 8.348 8.348

f top = 8.348 8.348 = 0 MPa

f bot = 8.348 8.348 = -16.70 MPa

f c at level of steel could also be approximated to

Ff i

G

In that case

500.88103

f c = = 8.34860000


SCE Dept of Civil

ES nf c

ES = 6 x 8.348 = 53.82 MPa

f Pe (aft. loss) = 1035.00 – 53.82 = 981.18 MPa

Fe = 981.18 x 520 = 510.21 kN


3 3

=510.2110

510.2110 50

15060000 450106

f top = 8.50 8.50 = 0 MPa

f bot = 8.50 8.50 = -17.0 MPa

Which show that the approximate method is fast and quite accurate.

Example - Post-tensioned member – eccentric tendonA post-tensioned beam of size 200 mm x 300 mm is eccentrically pre-stressed with 520 mm2 wires

stressed to a f i = 1035 MPa. The cgs is 75 mm above the bottom of the beam. Immediately after

transfer the stress reduces by 5% owing to anchorage and other losses. The size of the duct is 50 mm x

75 mm. Compute the stresses in concrete and steel immediately after transfer due to pre-stress only.

Fi = 1035 x 520 = 538.20 kN

Fe = 0.95 x Fi

= 0.95 x 538.20 = 511.29 kN

e = 150 – 50 = 50 mm

AG = 200 x 300 = 60000 mm2

200 3003

I =12

= 450 x106 mm4


SCE Dept of Civil


Fe= AG

Fe ey

I

3 3

= 511.2910

511.2910 75

15060000 450106

= 8.52 12.78

f top = 8.52 12.78 = 4.26 MPa

f bot = 8.52 12.78 = -21.30 MPa

Stress in concrete due to pre-stress & loadsStresses in concrete produced by external bending moment, whether due to the beam’s self-weight orapplied load is:

f M

yc I

The resulting stress in concrete due to both the pre-stress and loads is:

f F

Fey

Myc A I I

Example - Post-tensioned member with loadsA post-tensioned beam of size 300 mm x 600 mm and le = 12 m is pre-stressed with1575 kN which

eventually reduces to 1350 kN due to losses. The cgs is 175 mm above the bottom of the beam. Thebeam carries two live loads of 45 kN each in addition to its self-weight of 4.5 kN/m. Compute theextreme fiber stresses at mid-span for (a) initial condition with full pre-stress and no live load and (b)pre-stress after losses with full live load.

Fi = 1575 kN


SCE Dept of Civil

Fe = 1350 kN

e

AG

= 300 – 175

= 300 x 600

= 125 mm

= 180000 mm2

3006003

I =12

= 5400 x106 mm4

MG =4.5103 122

8= 81 kN-m

ML = 45 4.5 = 202.5 kN-m

MT = 81 + 202.5 = 283.50 kN-m

Initial condition

F F e Mf i i y yc A I I

3 3 6

= 157510

157510 125

300 8110

300180000 5400106 5400106

= 8.75 10.94 4.5

f cTop = 8.75 10.94 4.5 = -2.31 MPa

f cBot = 8.75 10.94 4.5 = -15.19 MPa

Final condition

F F e Mf e e y yc A I I

3 3 6

=135010

135010 125

300 283.510

300180000 5400106 5400106

= 7.5 9.38 15.75

f cTop = -13.87 MPa


SCE Dept of Civil

f cBot = -1.13 MPa

Example - Post-tensioned member with loadsSolve Ex 6.2.1 by locating the center of pressure C for concrete section.

Fe = 1350 kN

MT = 81 + 202.5 = 283.50 kN-m

283.50106

a =1350103

= 210 mm

e = 210 – 125 = 85 mm

C = Fe = 1350 kN

f C

Ceyc A I

3 3

= 135010

135010 85

300180000 5400106

= 7.5 6.37

f cTop = 7.56.37 = -13.87 MPa

f cBot = 7.5 6.37 = -1.13 MPa

Stress in steel due to loadsIn RCC members, the lever arm between the resultant compression and tension remains almostconstant but the tension in steel increases almost proportionately with increasing moment till yielding.


SCE Dept of Civil

In pre-stress concrete resistance to external bending moments is furnished by a lengthening of the leverarm between the resisting forces C and T which remain relatively unchanged in magnitude.

After cracking, the stress in pre-stressing steel increases rapidly with moment.

The following sketch explains the variations of the stress in pre-stressing steel

The variations are shown for bonded and un bonded tendons.

f p with increasing load.

After the pre-stress is transferred while the member is supported at the ends, the stress will tend to

increase from the value after losses f po due to the moment under self weight. Simultaneously the stress

will tend to drop due to the time dependent losses such as creep, shrinkage and relaxation. The

effective pre-stress after time dependent losses is denoted as f pe .


SCE Dept of Civil

Due to the moment under service loads, the stress in the pre-stressing steel will slightly increase from

f pe . The increase is more at the section of maximum moment in a bonded tendon as compared to the

increase in average stress for an un bonded tendon.

The stress in a bonded tendon is not uniform along the length. Usually the increase in stress is neglectedin the calculations under service loads. If the loads are further increased, the stress increases slightly tillcracking.

After cracking, there is a jump of the stress in the pre-stressing steel. Beyond that, the stress increases

rapidly with moment till the ultimate load. At ultimate, the stress is f pu .

Similar to the observation for pre-cracking, the average stress in an un bonded tendon is less than thestress at the section of maximum moment for a bonded tendon.

As discussed above, at the section of maximum moment, the stresses in the un bonded tendonincreases more slowly than that for bonded tendon. This is because any strain in an un bonded tendonwill be distributed throughout its entire length.

If MR is the resultant moment in at a cross-section of a bonded beam and the beam deflects downwards,there is an increase in steel stress due to this bending given by

f s nf c nM R yI

Let M be the moment at any given point of an un bonded beam, f c the stress in concrete at a section,

f M

yc I

If c is the strain in concrete in that section

c f

Ec

M

yEc I

Then , the total strain along the cable is,


SCE Dept of Civil

E I

M dx

c

ydx

M

L Ec

ydxIL

is the average strain

The stress in steel f s

E s Mf s E s

n

L Ec

M

ydxIL

ydxL I

Example – stress in steelA post-tensioned beam of span le = 12 m and size 300 mm x 600 mm, carries a superimposed load of 11kN/m in addition to its own weight of 4.5 kN/m. The initial pre-stress in steel is 950 MPa and reduces to820 MPa after all losses and assuming no bending in beam. The cable of are 1600 mm2 is parabolic. N=6.Compute the stress in steel at mid-span assuming (a) the steel is bonded by grouting and (b) the steel isun bonded and entirely free to slip.

Bonded tendon:

Fi = 1600 x 950 = 1520 kN

Fe = 1600 x 820 = 1312 kN


SCE Dept of Civil

s

e at mid span = 125 mm

A = 300 x 600 = 180000

3006003

I =12

= 5400 x106 mm4

wt ws wG

= 4.5 + 11.0 = 15.50 kN/m

w l 2

M t eT 8

15.5103 12 2

=8

= 279 kN-m

Moment due to Fe

= 1312 x 125 = -164 kN-m

MR = 279 – 164 = 115 kN-m

At level of steel

f M R yc I

115106

= 1255400106

= 2.66 MPa

Increase in stress in steel

f s = nf c

= 6 x 2.66 = 15.97 MPa

Resultant f s = 820 + 16 = 836 MPa

Un bonded beam

f n M

L Iydx


SCE Dept of Civil

2

2

2

L

from the BMD and y diagram

M M 1 x

o L

2

y y 1x

o L 2

2

2

f nM y 1

x dxs LI2

o o l

2

f 8n M o yo where

nM o yo is the stress at mid span of a bonded beam = 16 MPa.s 15 I I

f 8

16 = 8.53 MPas 15

Resultant f s = 820 + 8.53 = 828.53 MPa

Cracking momentMoments producing first crack in a pre-stressed concrete beam, assuming cracks start when tensile

stress in the extreme fiber of concrete reaches its modulus of rupture, f cr ,

f cr 0.7 f ck cl 5.2.2

Therefore cracks appear when

F F e Mf e e y

y orcr A I I

F I f IM F e e cr

e Ay y


SCE Dept of Civil

ExampleFor the problem in 6.3.1 compute the total dead and live load that can be carried by the beam for (a)

zero tensile stress at bottom fiber and (2) cracking in the bottom fiber assuming

1350 kN.

f cr 4.2MPa and Fe =

To obtain zero stress in the bottom fiber, the center of pressure must be located at the top kern point

a = (e+kt)

= 125 + 100 = 225 mm

M = Fe x a

w 8M

Tl 2

= 1350000 x 225 = 303.75 kN-m

8303.75106

=12000 2

= 16.87 kN/m

For cracking moment, additional moment

f cr I=

y

4.25400106

=300

= 75.6 kN-m

M = 303.75 + 75.60 = 379.35 kN-m

wT =8379.35106

12000 2= 21.07 kN/m


SCE Dept of Civil

At transfer: At working load

Fi Fi e M GTop: A

Z

Z f tt (1)

Fi Fi e M GBot: A

Z

Z f ct (2)

Fe Fee M G M LTop: f cw (3)A Z Z Z

Fe Fee M G M LBot: f tw(4)A Z Z Z

2.2 Elastic Design for flexure

Derivations

t t t t t

b b b b b

TakingFe Fi


SCE Dept of Civil

t f b f

tt ct

t

b

F F e M F F e M i i f G (1) i i f G (2)

A Z t Z t A Z b Z b

Fi Fi e M G M L Fi Fi e

M G M L

A Z t Z t Z t

f cw(3)

A Z b Z b Z b

f tw(4)

M G M G M L M G

M G M L f tt Z t

Z t Z t

f cw(3) f ct Z b

Z b Z b

f tw(4)

M G 1nM L

Z t

f cw f tt f tr (3)M G 1nM L

Z b

f tw f ct f cr (4)

M 1 nM M 1 nMZ G L (5)

tr

Z G L (6)cr

Max. pre-stressing force is limited by

1. Tension at top - f tt during transfer – Eq.1

2. Min. Comp stress at bottom - f tw during working load – Eq.4

Therefore from Eq.1 and Eq.2 and taking

F F ef i i

t A Z

F F ef i i

b A Z

f t f tt

M G (7) form Eq.1

Z t

f b M G Z b

M L

Z b

f tw(8) from Eq.4

M M

f 1 f G L (8)b tw

Z b


SCE Dept of Civil

t

b

bt

b

A

A

Also, since:

f Fi

Fi e(a)A Z t

F F ef i i (b)

A Z b

Fi e f t

Fi ZA

t

(a)

f Fi fb A

t Fi

Z t (b)A Zb

f Fi Zb Z t f

Z t b A Z Z

F f Z f Zi b b t t (9)A Z b Z t

f F 1

e (a )t i Z t

f F 1

e (b )b i Zb

f t Zb Ae Z t

fb Z t Ae Zb

Z Z f f Ae b t b t (10)

f t Z t fb Zb

Remember, in these equations:


SCE Dept of Civil

i

b

b

i

f t f tt

M G

Z t

1 M M f b f tw

G L Z b

When MG is large as will be the case for long span and/or heavy girders, the computed eccentricity efrom Eq.10, may fall below the bottom of the beam. In that case, the e available is worked out and Fi isincreased suitably.

Fi for know eccentricity e working load is:

F F e 1 M M f i i f G L from Eq.4b A Z

twZ b

From the first part of the above equation

F F ef i i

A Z b

Z Ae F b

AZ b

f AZ F b b (11)

Z b Ae 2.3 Permissible stresses for flexure member

Steel – Cl 8.5.1Steel stress for pre-tensioned tendons immediately after transfer or post-tensioned tendons afteranchorage is:

f pi 0.87 f pu

Where f pi = Maximum initial pre-stress, and f pu = Ultimate tensile stress in tendon.

Concrete in compression – Cl 22.8.2.1, 22.8.1.1Concrete stress after transfer and before losses in extreme fiber

Compression = 0.54 fck to 0.37 fck (for M30 to M60) for post-tension

= 0.51 fck to 0.44 fck (or M40 to M60) for pre-tension


SCE Dept of Civil

Concrete stress at service loads after transfer and after losses in extreme fiber

Compression = 0.41 fck to 0.35 fck (for M30 to M60) for post-tension

= 0.34 fck to 0.27 fck (or M40 to M60) for pre-tension

Concrete in tension – Cl 22.7.1Concrete stress after transfer and before losses in extreme fiber

1. For Type 1 members, Tension = 0.2. For Type 2 members, Tension = 3.0 MPa to 4.5 MPa3. For Type 3 members, Tension = 4.1 MPa to 4.8 MPa

Concrete stress at service loads after transfer and after losses in extreme fiber

Tension = same as at transfer before losses

Example

Depth not restricted - beamDesign a post-tensioned beam of le = 12 m to carry a live load of 12 kN/m throughout its length. The

width of beam b = 250 mm. f ct f cw 17MPa and f tt f tw 1.4MPa . = 0.85.

Assume depth of beam = h mm

A = 250h mm2


SCE Dept of Civil

f

0.25

h 22412

MG = 1000

8= 0.108h kN-m

ML =1212 2

8= 216 kN-m

Min Z is governed by Zb. From Eq.4

f cr

f cr

f tw f ct

= 1.4 0.8517

= 15.85 MPa

M 1 nMZ G L (6)b f cr

0.108h10.85106 216106

=15.85

106 216 0.0162h=

15.85

Z b also =250h 2

6

From which

h = 580 mm

A = 250 x 580 = 145x103 mm2

Zt = Zb = Z =2505802

6= 14x106 mm3

MG = 62.64 kN-m

f t f tt

M G

Z t

= 1.462.64106

14106= 5.87 MPa

1 M M f b tw

G L Z b


SCE Dept of Civil

=

6

1 62.64 216106 1.4

6= -21.76 MPa

0.85 1410


=21.765.8714106

2 14106

= 7.945

Fi = 7.945 x 145 x103 = 1152 kN


f t Z t fb Zb

14141012 21.76 5.87=

14106 5.87 21.76= 24.3436x10

e = 167.89 mm

Depth restricted - slabA post-tensioned concrete bridge slab of le = 10 m is 380 mm thick. It is stressed with parallel cables

stressed to 360 kN each. wL = 25 kN/m2. Losses are 20%.

spacing of cable at mid-span.

f tt f tw 0.7MPa . Calculate the emax and

= 0.80


SCE Dept of Civil

1

f

A = 1000 X 380 = 380000 mm2

10003803

I =12

= 4572.66 x106 mm4

Zt = Zb = Z =4572.6610 6

380

= 24.07 x106 mm3

2

ws = 1 x 0.38 x 24 = 9.12 kN/m

MG =9.1210 2

8= 114 kN-m

ML =25102

8= 312.50 kN-m

M G

Z

114106

=24.07 106

= 4.74

M L

Z

312.5106

=24.07106

= 12.98

At mid-span

f t f tt

M G

Z t

= 0.7 4.74 = 5.44 MPa

1 M M f b tw

G L Z b

= 0.7 4.7412.980.8

= -21.275 MPa


=21.2755.4424.07106

2 24.07106

= 7.9175


SCE Dept of Civil

6

tw

Fi = 7.9175 x 380 x103 = 3008.65 kN


f t Z t fb Zb

24.0724.071012 21.275 5.44=

24.07106 5.44 21.275= 40.6081 x10

e mid-span = 106.86 mm

At support

MG = ML = 0

e at support = 68.944

A more complicated solution is:

At mid-span, the stress at top and bottom at transfer and working load are respectively.

Fi A

Fi e Z t

M G

Z t

f tt (1)

Fe Fe e M G M L A Z b Z b Z b

f tw(4)

ie. Fi A

Fi e Z b

M G Z b

M L

Z b

f tw(4)

Multiplying Eq.1 by and adding it to Eq.4 above, and remembering Zt = Zb = Z,

2Fi 1 M G

M L f f ttA Z Z

Fi 20.8 1 0.8 4.74 12.98 0.7 0.80.7 from whichA

Fi = 3009 kN

Likewise, multiplying Eq.1 by and subtracting it from Eq.4 above,


SCE Dept of Civil

tw 2Fi e 1M G

M L f f ttZ Z Z

Fi e 20.8 1 0.8 4.74 12.98 0.7 0.80.7A

e at mid-span = 106.358 mm

At support

MG = ML = 0

Fi A

Fi e

Z t

0 f tt (1)

6 6

300910

300910 e

0.7380000 24.07106

e at support = 68.944

Spacing of cables

Fi = 3009 kN

Force per cable = 360 kN

No of cables =3009103

3601039 Nos

Spacing =1000

9= 112 mm c/c

Depth not restricted - slabA post-tensioned concrete one-way bridge slab of le = 10 m is stressed with parallel cables stressed to

500 kN each. wL = 25 kN/m2. Losses are 20%. f ct f cw 15MPa and f tt f tw 0 .


SCE Dept of Civil

= 0.80

Assume depth of slab = h mm

Width of slab = 1000 mm

A = 1000h mm2

1

h 22410

MG = 1000

8= 0.3 kN-m

ML =25102

8= 312.5 kN-m

Min Z is governed by Zb. From Eq.4

f cr

f cr

f tw f ct

= 00.815

= 12 MPa

M 1 nMZ G L (6)b f cr

0.3h1 0.8106 312.5106

=12

106 312.5 0.06h=

12


SCE Dept of Civil

f

=

6

Z b also =100h 2

6

From which

h = 410 mm

A = 1000 x 410 = 410000 mm2

Zt = Zb = Z =1000410 2

6= 28.02x106 mm3

MG = 123 kN-m

f t f tt

M G

Z t

123106

= 028.02106

= 4.39 MPa

1 M M f b tw

G L Z b

1 123 312.50106 0

6= -91.43 MPa

0.8 28.0210


19.43 4.39=

2= 7.52

Fi = 7.52 x 410000 = 3083.20 kN


f t Z t fb Zb

28.02106 19.43 4.39=

4.3919.43= 443.774 x10

e mid-span = 108.24 mm


SCE Dept of Civil

No of cables =3084 103

500 1037 Nos

Spacing =1000

7= 143 mm c/c

Depth restricted - beamA pre-tensioned simply supported beam of size 80 mm x 120 mm and le = 3 m caries two 4 kN loads at

third points along the span. Losses are 20%. f tt 0, f tw 1.4MPa . Design the beam with 3mm wires

for f i 1400 MPa each.

= 0.80

A = 80 X 120 = 9600 mm2

801203

I =12

= 11.52 x106 mm4

Zt = Zb = Z =11.5210 6

120

= 0.192 x106 mm3

2

ws = 0.08 x 0.12 x 24 = 0.23 kN/m

MG =0.23 32

8= 0.2592 kN-m

ML = 41 = 4.0 kN-m


SCE Dept of Civil

1

f

6

M G

Z

0.2592106

=0.192106

= 1.35

M L

Z

4.0106

=0.192 106

= 20.83

At mid-span

f t f tt

M G

Z t

= 01.35 = 1.35 MPa

1 M M f b tw

G L Z b

= 1.4 1.35 20.830.8

= -25.975 MPa


=25.9751.35

2= 12.3125

Fi = 12.3125 x 9600 = 118.20 kN


f t Z t fb Zb

0.192106 25.975 1.35=

1.35 25.975= 0.213x10

e = 22.193 mm


32

=4

= 7.07 mm2


SCE Dept of Civil

f

=

6

f i in one wire = 7.07 x 1400 = 9.896 kN

No of cables =118.20 103

9.896 10312 Nos

Spacing =1000

9= 112 mm c/c

below bottom in heavy girder - beamAn unsymmetrical I section has the following sectional property: h = 1000 mm, A = 345 000 mm2, Zt = 95x106 mm3, Zb = 75 x 106 mm3, cgc = 440 mm from top, MG = 1012 kN-m, ML = 450 kN-m. Design the

section if f ct f cw 15MPa and f tt f tw 0 . = 0.85

f t f tt

M G

Z t

= 01012x106

95106= 10.65 MPa

1 M M f b tw

G L Z b

1

1012 450106

0 6

= -22.93 MPa0.85 7510


22.93 7510.6595=

75 95= 4.1647

Fi = 4.1647 x 345000 = 1436.82 kN


f t Z t fb Zb

75951012 22.9310.65=10.6595 22.93 75106

= 361.339 x10

e = 1047.36 mm


SCE Dept of Civil

i

eavil = yb – cover

= (1000-440) – 100 = 460 mm

For this eavil, the Fi required is:

f AZ F b b (11)

Z b Ae

Fi =22.93 34500075106

75106 345000460= 2538.78 kN


SCE Dept of Civil

2.4 End block

Bursting forceA portion of a pre-stressed member surrounding the anchorage is the end block. Through the length ofthe end block, pre-stress is transferred from concentrated areas to become linearly distributed fiberstresses at the end of the block. The theoretical length of this block, called the lead length is not morethan the height of the beam.

But the stress distribution within this block is rather complicate.

The larger transverse dimension of the end zone is represented as yo. The corresponding dimension ofthe bearing plate is represented as ypo. For analysis, the end zone is divided into a local zone and ageneral zone.


SCE Dept of Civil

The local zone is the region behind the bearing plate and is subjected to high bearing stress and internalstresses. The behavior of the local zone is influenced by the anchorage device and the additionalconfining spiral reinforcement.

The general zone is the end zone region which is subjected to spalling of concrete. The zone isstrengthened by end zone reinforcement.

The transverse stress (σt) at the CGC varies along the length of the end zone. It is compressive for adistance 0.1yo from the end and tensile thereafter, which drops down to zero at a distance yo from theend.

The transverse tensile stress is known as splitting tensile stress. The resultant of the tensile stress in atransverse direction is known as the bursting force (Fbst).


SCE Dept of Civil

Besides the bursting force there is spalling forces in the general zone.

Fbst for an individual square end zone loaded by a symmetrically placed square bearing plate according

to Cl 18.6.2.2 is,

y po Fbst PK 0.32 0.3

yo


SCE Dept of Civil

Where, PK = pre-stress in the tendon;

dimension of the end zone.

y po = length of a side of bearing plate; yo = transverse

It can be observed that with the increase in size of the bearing plate the bursting force

Fbst reduces.

End Zone reinforcementTransverse reinforcement - end zone reinforcement or anchorage zone

reinforcement or bursting link - is provided in each principle direction based on the value of Fbst. Thereinforcement is distributed within a length from 0.1yo to yo from an end of the member.

The amount of end zone reinforcement in each direction Ast is:

FA bst

st f s

The parameter represents the fraction of the transverse dimension covered by the

bearing plate.

The stress in the transverse reinforcement, f s = 0.87fy.

When the cover is less than 50 mm, f s = a value corresponding to a strain of 0.001.

The end zone reinforcement is provided in several forms, some of which are proprietary of theconstruction firms. The forms are closed stirrups, mats or links with loops.


SCE Dept of Civil

Bearing plate & End blockDesign the bearing plate and the end zone reinforcement for the following bonded

post-tensioned beam. The strength of concrete at transfer is 50 MPa. A pre-stressing force of 1055 kN isapplied by a single tendon. There is no eccentricity of the tendon at the ends.


SCE Dept of Civil

Bearing Plate

Assume area of bearing plate to be 200 mm x 300 mm

Pf K

br Apun

PK = 1055 kN

Apun = 200 x 300 = 60000 mm2

f br

1055103

=60000

= 17.58 MPa

Abr

f 0.48 f

= 400 x 600 = 240000 mm2

Abrbr,all ci Apun

= 0.4850240000

60000= 48 MPa

0.8 f ci = 40 MPa

fbr fbr,all 40MPa


SCE Dept of Civil

End Block

In vertical direction


yo

1055

.32 0.3300

= 179.35 kN= 0 600

In horizontal direction


yo

1055

.32 0.3200

= 179.35 kN= 0 400

Ast =Fbst

0.87 f y

179.35103

=0.87250

= 824.60 mm2

Provide 10 mm 2L stirrups in both directions as Fbst is same in those

Aw = 10 2

4= 78.54 mm2

No of stirrups =

2 rd

824.60

2 78.54

1 rd

= 6 Nos

Provide3

Ast from 0.1 yo = 60 mm to 0.5 yo = 300 mm and3

Ast from 0.5 yo = 300 mm to yo = 600

mm, both vertically and horizontal.


SCE Dept of Civil

Chapter III

CIRCULAR PRESTRESSING

Design of prestressed concrete tanks – Pipes.

3.1 Design Procedure for circular tanks Computations

1. Minimum wall thickness

2. Circumferential Prestress

3. Vertical Prestress.

Estimate

1) Maximum, ring tension Nd

2) Bending Moment Mw

3) Minimum wall thickness = Ndηfct – fmin.w

Minimum cover 35mm

4) Circumferential Prestress

fc = Nd + fmin.w N/mm2

ηt η

5) Spacing of wires

As = Cross sectional area of wire coinding, mm2

Wt = average radial Pressure of wires at transfer at a given section N/mm2

D = Diameter of the tank, mm

S = Spacing of wires at the given section mm

fs – Stress in wires at transfer, N/mm2

t – Thickness of the tank wall, mm

fc – compressive stress in concrete, N/mm2

Hoop compression due toprestressing

= wt . D2


SCE Dept of Civil

Equating wt. D = fs As2 S1

Wt = 2 fs AssD

Nd – hoop tension due to hydrostatic working pressure, Ww

Nt – hoop compression due to radial pressure of wires, wt

Then Nt = Nd wtWw

Also Nt = t fc

Spacing of the wire winding

S = 2 Nd . fs. As mmWw fc. Dt

Mt = Mw wtWw

Where Mt = Vertical moment due to the prestress at transfer.

Mw = Vertical moment due to hydrostatic pressure.

The compressive prestress required

Fc = fmin. W + Mwη ηz

When the tank is empty

fc = fmin. W + Mtη Z

Vertical prestressing force is required

P = fc. Ac

(Note: Vertical Prestressing force = 30% of hoop compression.]

1. A cylindrical prestressed concrete water tank of internal diameter 30m is required to storewater over a depth of 7.5m. The permissible compressive stress in concrete at transfer is 13

N/mm2 and the minimum compressive stress under working presuure is 1 N/mm2. The loss ratiois 0.75. Wires of 5mm diameter with an initial stress of 1000N/mm2 are available for


SCE Dept of Civil

circumferential winding and freyssinet cables made up of 12 wires of 8mm diameter stressed to1200N/mm2 are to be used for vertical prestressing. Design the tank walls assuming the base asfixed. The cube strength of concrete is 40N/mm2.Solution:

From table 16.1Assume t = 150mm

H2 = 7.52 = 12.5Dt 30 x 0.15

Ww = wH = 10 x 7.5 = 75kN/m2 = 0.075N/mm2

From table 16.2 & 16.3

Maximum ring tension Nd = (coefficient) wHR kN/m

= 0.64 x 10 x 7.5 x 15

= 720kN/m = 720N/mm.

Moment in tank wall for the fixed base condition = (coefficient) Wh3 kNm/m

= 0.01 x 10 x 7.53

= 42.5 kNm/m

= 42500Nmm/mm

Minimum wall thickness = t = Ndηfct –fmin.w

= 720 = 82.3mm0.75 x 13 – 1

Net thickness available (allowing for vertical cables of diameter 30mm) is (150 – 30) = 120mm

fc = Nd + fmin.wηt η

= 720 + 1 = 9.42N/mm2

0.75 x 120 0.75


SCE Dept of Civil

Spacing of circumferential wire winding at base.

S = 2 Nd fs. AsWw fc. Dt

= 2 x 720 x 1000 x π/4)(5)2

0.075 x 9.4 x 30 x 103 x 120

= 11.4mm

Number of wires / metre = 87

Ring tension Nd at 0.1 H (0.75m) from top

Nd = (coeff) wHR kN/m

= 0.097 X 10 X 7.5 X 15 = 109 kN/m = 109N/mm

fs = 2 x 109 x 1000 x 200.075 x 2.5 x 30 x 103 x 120

= 64mm

Number of wires / metre t the top of tank = 16

Vertical moment = Mw wtWw

Wt = 2 fs As = 2 X 1000 X 20 = 0.117 N/mm2

Sd 11.4 x 30 x 103

Mt = 42500 0.117 = 67,000 Nmm / mm = 67 x 106 Nmm / m.0.075

Considering one metre length of tank

Along the circumferential

Z = 100 x 1502 = 375 x 104mm3

6

fc = fmin. w + Mtη Z

= 1 + 67X 106 = 19.20N/mm2

0.75 375 X 104

Since this stress exceeds the permissible value of fct = 13N/mm2, the thickness of the tank wall of base isincreased to 200mm.


SCE Dept of Civil

Z = 1000 x1002 = 666 x104 mm3

6

fc = 1 + 67 x 106 = 12N/mm2

0.75 666 x 104

Vertical Prestressing force = fc = A = 12 x 1000 x 2001000

= 2400kN

Using 8mm diameter (12Nos.) Freyssinet cables

Force / cable = π/4 x 82 x 12 x 1200 = 720kN1000

Spacing = 1000 x 720 = 800mm2400

App. Vertical Prestress = 0.3 fc

= 0.3 x 9.4 = 2.82 N/mm2

Vertical prestressing force = 2.82 x 1000 x 2001000

= 564kN

Ultimate tensile force = 87 x 20 x 1500 = 2610kN1000

Load factor = 2610 / 720 = 3.6

Direct tensile strength of concrete = 0.267 √40

= 1.7N/mm2

Cracking load = 1000 x 200 0.75 x 9.4 + 1.71000

= 1760Kn

F.S against cracking = 1760 / 720 = 2.45

Nominal reinft. 0.2 percent circumferential & longitudinal directions

8mm ф @ 300mm spacing on both faces at a cover of 20mm.


SCE Dept of Civil

A

Ac

3.2 Circular pre-stressing

Circumferential pre-stressCircumferential pre-stressing is done to resist hoop tension in circular structures, like water-tanks andpipes. Essentially each horizontal slice of the wall forma a ring subjected to uniform internal pressure.This ring may be considered as a pre-stresses concrete member under tension.

Considering one half of a thin cylindrical slice of a tank as a free-body: under the action of pre-stress Fi insteel, the total compression C in the concrete equals Fi. The C-line coincides with the cgs line, which is aconcordant cable linearly transformed.

Due to pre-stress, initially after transfer of pre-stress,

Ff i , A = Area of concretec C

c

Ff e

c

which after losses in pre-stress reduces to


SCE Dept of Civil

Ac

c

When internal liquid pressure is acting at working load stage, the internal pressure intensity,

f pR

T

where

p = internal pressure intensity

R = internal radius of the vessel

AT = transformed area = Ac nAp

AP = area of steel

The resultant stress in concrete at working load due to internal pressure ‘p’ and pre-stress Fe is

F pRf e (1)

Ac AT

In Eq.1, if it is assumed that hoop tension is entirely carried by the effective pre-stress, Fe pR .

And since At Ac ,

concrete.

f c is always negative, implying that there is always a residual compressive stress in

Design method 1Ap = area of steel

Ac = area of concrete

f ct = permissible stress at transfer in concrete


SCE Dept of Civil

f

f

A A

f cw = permissible stress at working load in concrete

Fi = initial pre-stress

Fe = effective pre-stress after losses

m = factor of safety

En s

Ec

Fi Ap f i

Fe Ap f e

At transfer

A Fi (2)c

ct

At working load

Fe

Ac

pR

AT

f cw (3) where At Ac nAp

Assuming that hoop tension is entirely carried by the effective pre-stress, Fe pR ,

A pR

p f e

Fi Ap f i

FA i from Eq.2c

ct

F pRf e

from Eq.3cw

c T


SCE Dept of Civil

c

Design method 2If both f ct and f cw are to be kept in concrete, which may be the case when a tensile stress f cw =

cracking stress, may be allowed, and if a factor of safety ‘m’ is required, then Eq.2 and Eq.3 can becombined together into the following from.

Ap = area of steel

Ac = area of concrete

f ct = permissible stress at transfer in concrete

f cw = permissible stress at working load in concrete

Fi = initial pre-stress

Fe = effective pre-stress after losses

m = factor of safety

En s

Ec

Fi Ap f i

Fe Ap f e

At transfer

F f i ApA i f ct f ct

At Ac nAp Ap

f(n i )

f ct

At working load

F mpR e

f cwAc AT

f e Ap mpR f cw

Ap

f i

f ct

Ap n

f i f ct


SCE Dept of Civil

ct

i ct

Ap f ct

f e

f i

f cw

mpR

nf ct f i

f ct

f e f ct f i f cw mpR

Ap

f i

nf ct f i

f ct

AmpR

p nf f i f e f ct f i f cw fct f i

mpR

f ct f i f f f n ct 1 f f e f i

cw f i f ct

mpR

f f f e f i

cw 1 n ct f ct f i

3.3 Design of pipesPre-stressed concrete pipes are suitable when the internal pressure is within 0.5 to 2.0

Mpa. There are two types of pre-stressed concrete pipes:

1. Cylinder type which has a steel cylinder core, over which the concrete is cast and pre-stressed.2. Non-cylinder type which is made of pre-stressed concrete only.

IS:784-2001 Pre-stressed Concrete Pipes (Including Specials), provides guidelines for the design of pre-stressed concrete pipes with the internal diameter ranging from 200 mm to 2500 mm. The pipes aredesigned to withstand the combined effect of internal pressure and external loads. The minimum gradeof concrete in the core should be M40 for non-cylinder type pipes.

The pipes are manufactured either by,

1. Centrifugal method: In the centrifugal method the mould is subjected to spinning till theconcrete is compacted to a uniform thickness throughout the length of the pipe.

2. Vertical casting method: In the vertical casting method, concrete is poured in layers up to aspecified height.

After adequate curing of concrete, first the longitudinal wires are pre-stressed. Subsequently, thecircumferential pre-stressing is done by the wire wound around the core in a helical form. The wire iswound using a counter weight or a die. Finally a coat of concrete or rich cement mortar is applied overthe wire to prevent from corrosion. For cylinder type pipes, first the steel cylinder is fabricated andtested. Then the concrete is cast around it.


SCE Dept of Civil

5

f

Example 1 – non-cylinderDesign a non-cylinder pre-stressed pipe for the following specifications: R = 300 mm, p = 1.05 MPa, f i =

1000 MPa, f e = 800 MPa, f ct = -14 MPa, f cw =-0.7 MPa, Es = 2.1x10 MPa, Ec = 0.35 x 105 MPa and 2.5

mm wires are used. And what would be the internal pressure ‘p’ required to balance the pre-stress attransfer before losses to maintain a stress of -0.7 in concrete?

Method 1:


A pR

p f e

Ap =1.05103 300

800= 394 mm2

Fi Ap f i

Fi

FA i

= 3941000 = 394 kN

cct

Ac =394103

14= 28143 mm2

Taking a 1000 mm height of the pipe

28142t = = 29 mm

1000

30 mm

Ac = 301000 = 30000 mm2

Checking for final stress

At Ac nAp

At = 30000 6394 = 32364 mm2


SCE Dept of Civil

A A

F pRf e cw

c T

f cw

394800=

30000

1.05103 300

32364= -0.77 MPa

Since f cw is slightly more than specified -0.7 MPa, another trail could be made in the design.

Aw = 2.52

4= 4.91 mm2

ApNo of wires =

Aw

394=

4.9181 wires

Spacing =1000

81= 12 mm

Method 2:

Ap mpR

f f f e f i

cw 1 n ct f ct f i

Ap =

1.05103 300

0.7 14

= 388 mm2

8001000

1 614

1000

Fi = 3881000 = 388 kN

Ac =388103

14= 27715 mm2

27715t = = 28 mm

1000

Ac = 281000 = 30000 mm2


SCE Dept of Civil

5


At = 28000 6388 = 30328 mm2

f cw

388800=

28000

1.05103 300

30328= -0.70 MPa

At transfer before losses,

f cw

3881000=

28000

p103 300

30328= -0.7 MPa

p = 1.33 MPa

Example 2 – non-cylinderDesign a non-cylinder pre-stressed pipe for the following specifications: R = 800 mm, p = 1 MPa, f i =

1000 MPa, f e = 800 MPa, f ct = -12 MPa, f cw = 0, Es = 2.1x10 MPa, Ec = 0.35 x 105 MPa and 5 mm wires

are used. If cracking stress is +2 MPa, what is the F.S against cracking?

Ap mpR

f f f e f i

cw 1 n ct f ct f i

Ap =

1103 800

0 12

= 933 mm2

800 1000

1 612

1000

Fi = 9331000 = 933 kN

Ac =933103

12= 77750 mm2

77750t = = 78 mm

1000

Checking for stresses

Ac = 781000 = 78000 mm2

At = 78000 6933 = 83598 mm2


SCE Dept of Civil

f cw

933800=

78000

1103 800 = 0

83598

If cracking stress is allowed, f cw = 2 MPa

Ap mpR

f f f e f i

cw 1 n ct f ct f i

m1103 800

933 = 2 12 8001000

1 612

1000

m = 1.2

3.4 Design of circular water tanksConcrete liquid retaining structures must be impervious. Hence, their design is based on no in crackingin concrete. Circular pre-stressed liquid retaining structures, are stressed to avoid tension in concrete.

Pre-stressed concrete liquid retaining structures require low maintenance and resist seismic forcessatisfactory.

Circular pre-stressed concrete tanks are used in water treatment, water distribution, storm watermanagement, large industrial tanks, bulk storage tanks and for storing liquefied natural gas (LNG).

The construction of the circular tanks is in the following sequence. First, the concrete core is cast andcured. The surface is prepared by sand or hydro blasting. Next, the

circumferential pre-stressing is applied by strand wrapping machine. Shotcrete is

applied to provide a coat of concrete over the pre-stressing strands.

IS:3370-1967 (1-4) Code of Practice for Concrete Structures for the Storage of Liquids providesguidelines for the analysis and design of liquid storage tanks. The four sections of the code are titled asfollows:

Part 1: General Requirement.

Part 2: Reinforced Concrete Structures.

Part 3: Pre-stressed Concrete Structures.


SCE Dept of Civil

Part 4: Design Tables.

In IS:3370-1967 (3), the design requirements for pre-stressed tanks are mentioned. A few of them are:

1. The computed stress in the concrete and steel, during transfer, handling and construction, andunder working loads, should be within the permissible values as specified in IS:1343-1980.

2. The liquid retaining face should be checked against cracking with a load factor of 1.2.3. The ultimate load at failure should not be less than twice the working load.4. When the tank is full, there should be compression in the concrete at all points of at least 0.7

N/mm2. When the tank is empty, there should not be tensile stress greater than 1.0 N/mm2.Thus, the tank should be analyzed both for the full and empty conditions.

5. There should be provisions to allow for elastic distortion of the structure during pre-stressing.Any restraint that may lead to the reduction of the pre-stressing force should be considered.

6. The cover requirement is as follows. The minimum cover to the pre-stressing wires should be 35mm on the liquid face. For faces away from the liquid, the cover requirements are as perIS:1343-1980.

The general equations from Eq 1 to Eq 3, would serve well for the design of circular pre-stressedliquid retaining structure.

Example 1Determine the area of steel required per meter height of a circular pre-stressed water tank with aninside diameter of 18 m and a height of 6 m water pressure. Compute the thickness of concrete

required.

f i = 1034 MPa, f e = 827 MPa, f ct = -5.17, MPa and n = 10.

Design for the following two cases:

1. Assume that the entire hoop-tension is carried by the effective pre-stress.2. For a load factor of 1.25, producing zero stress in concrete. f ct = -5.17, f cw =0.

Case 1:

p =6101000

106(on an area of 1m x 1m) = 0.06 MPa


A pR

p f e


SCE Dept of Civil

f

A A

Ap =0.06103 9000

827= 653 mm2

Fi Ap f i

Fi

FA i

= 6531034 = 675 kN

cct

Ac =675103

5.17= 130600 mm2

Taking a 1000 mm height of the pipe

t =130600

= 130.60 mm1000

140 mm

Ac = 1401000 = 140000 mm2


At Ac nAp

At = 140000 10 653

F pRf e

= 146530 mm2

cwc T

f cw

653827=

140000

0.06103 9000

146530= -0.172 MPa

Case 2:

Ap mpR

f f f e f i

cw 1 n ct f ct f i


SCE Dept of Civil

Ap =

1.250.06103 9000= 778 mm2

0 5.17 827 1034

1105.17

1034

Fi = 7781034 = 805 kN

Ac =805103

5.17= 156 x103 mm2

156103

t =1000

= 156 mm

165 mm

Ac = 1651000 = 165000 mm2


At = 165000 10 778 = 172780 mm2

f cw

778827=

165000

0.06103 9000

172780= -0.77 MPa

If we had provided the actual


SCE Dept of Civil

Chapter IV

COMPOSITE CONSTRUCTION

Analysis for stresses – Estimate for deflections – Flexural and shear strength of composite members.

4.1 INTRODUCTIONMany applications of prestressed concrete involve the combination of precast prestressed concretebeams and in situ reinforced concrete slabs. Some examples of such composite construction are shownin Fig. 10.1. An in situ infill between precast beams is shown in Fig. 10.1(a) while an in situ topping isshown in Fig. 10.1(b). The former type of construction is often used in bridges, while the latter iscommon in building construction. The beams are designed to act alone under their own weight plus theweight of the wet concrete of the slab. Once the concrete in the slab has hardened and provided thatthere is adequate horizontal shear connection between them, the slab and beam behave as a compositesection under design load. The beams act as permanent formwork for the slab, which provides thecompression flange of the composite section. The section size of the beam can thus be kept to aminimum, since a compression flange is only required at the soffit at transfer. This leads to the use ofinverted T-, or ‘top-hat’, sections.

4.2 SERVICEABILITY LIMIT STATEThe stress distributions in the various regions of the composite member are shown in Fig. 10.2(a)–(d).The stress distribution in Fig. 10.2(a) is due to the self weight of the beam, with the maximumcompressive stress at the lower extreme fibre. Once the slab is in place, the stress distribution in thebeam is modified to that shown in Fig. 10.2(b), where the bending moment at the section, Md is thatdue to the combined self weight of the beam and slab.

Once the concrete in the slab has hardened and the imposed load acts on the composite section, theadditional stress distribution is shown in Fig. 10.2(c). This is determined by ordinary bending theory, butusing the composite section properties.The final stress distribution is shown Figure


SCE Dept of Civil

Stress distribution within a composite section.

The floor slab shown in Fig. 10.3 comprises precast pretensioned beams and an in situ concrete slab. Ifthe span of the beams is 5 m and the imposed load is 5 kN/m2 (including finishes), determine the stress

distributions at the various load stages. Assume all long-term losses have occurred before the beams areerected and that the net force in each wire is 19.4 kN. Section properties of the beams:Ac=1.13×105 mm2Ic=7.5×108 mm4Zt=Zb=6×106 mm3.Eccentricity of the wires=125−40=85 mm.(i) Self weight of the beams=0.113×24=2.7 kN/m.Mo=(2.7×52)/8=8.4 kNm.Total prestress force after all losses have occurred is given byßPo=6×19.4=116.4 kN.The stress distribution in the beams is thus given by

(ii) The weight of the slab is supported by the beams acting alone, so that Md=8.4+0.075×0.6×24×52/8=11.8 kNm.The stress distribution within the beams is now given by(iii) The imposed load of 5 kN/m2 is supported by the composite section and the


SCE Dept of Civil

section properties of this are now required. To find the neutral axis of the composite section, takingmoments about the soffit of the beams gives(1.13×105+75×600)y=(1.13×105×125+75×600×288)∴y=171 mm.Icomp=7.5×108+1.13×105 (171–125)2+(753×600)/12+(75×600)/(288–171)2=1.63×109 mm4.The imposed load bending moment, (Mdes−Md)=0.6×5×52/8=9.4 kNm.The stress distribution within the composite section under this extra bending momentis given by

The maximum compressive stress occurs at the upper fibres of the beams, but is significantly lower thanthe level of stress had the beam carried the total imposed load alone. This explains the advantage ofinverted T-sections in composite construction, where only a small compression flange is required forbending moments Mo and Md, the

Stress distribution for composite section in Example 10.1 (N/mm2): (a) beam; (b)beam and slab; (c)beam and slab and imposed load.

compression flange for bending moment Mdes being provided by the slab. The maximum compressivestress in the slab is much lower than in the beam and, for this reason, in many composite structures alower grade of concrete is used for the in situ portion. The modulus of elasticity for this concrete islower than that for the beam and this effect can be taken into account in finding the composite sectionproperties by using an approximate modular ratio of 0.8.The in situ slab in Example 10.1 lies above the composite section neutral axis and, therefore, the slab isin compression over its full depth under the total design load. However, for composite sections asshown in Fig. 10.1(a) the in situ portion of the section extends well below the neutral axis, so that thelower region is in tension. If the tensile strength of this concrete is exceeded then the composite sectionproperties must be determined on the basis of the in situ section having cracked below the neutral axis.


SCE Dept of Civil

4.3 ULTIMATE STRENGTHThe basic principles for the analysis of prestressed concrete sections at the ultimate limit state offlexural strength described in Chapter 5 are also applicable to composite sections. For the section shownin Fig. 10.5(a), it may be assumed initially that, at the ultimate limit state, the neutral axis lies within theslab and the section may then be treated effectively as a rectangular beam. The position of the neutralaxis should later be checked to see whether it does, indeed, fall within the slab. For the section shown inFig. 10.5(b), the position of the neutral axis may be determined on the assumption that the section isrectangular,

but the different strengths of the concrete in the slab and beam regions of the compression zone shouldbe taken into account.

4.4 HORIZONTAL SHEARThe composite behaviour of the precast beam and in situ slab is only effective if the horizontal shearstresses at the interface between the two regions can be resisted. For shallow members, such as thatshown in Fig. 10.3, there is usually no mechanical key between the two types of concrete, and reliance ismade on the friction developed between the contact surfaces. For deeper sections, mechanical shearconnectors in the form of links projecting from the beam are used, which provide a much better shearconnection. The determination of the horizontal shear resistance is based on the ultimate limit state,and if this condition is satisfied it may be assumed that satisfactory horizontal shear resistance isprovided at the serviceability limit state. A simply supported composite section carrying a uniformlydistributed load is shown in Fig. 10.8(a) and the free-body diagram for half the length of the in situ slabis shown in Fig. 10.8(b). At the simply supported end there must be zero force in the slab, while themaximum force occurs at the midspan. The distribution of shear forces on the underside of the slab isalso shown in Fig. 10.8(b), being zero at midspan and reaching a maximum at the support. Thisbehaviour is similar to that in an elastic beam, where the vertical and horizontal shear stresses increasetowards the support for a uniformly distributed load.


SCE Dept of Civil

The following expression is given in Part 1–3 of EC2 for the horizontal shear stress, where ß is the ratioof the longitudinal force in the slab to the total longitudinal force, given by Msd/z, both calculated for agiven section; Vsd is the transverse ultimate shear force; z is the lever arm; and bj is the width of theinterface.

The design shear resistance for horizontal joints with vertical shear reinforcement is given byτRdj=kTτRd+μσN+0.87 fykϱ μ≤0.33 vfck,where kT is a coefficient with kT=0 if the joint is subjected to tension;τRd is the basic design shear strength from Table

Horizontal shear: (a) composite section; (b) free-body diagram for in situ slab.

DIFFERENTIAL MOVEMENTSThe fact that the slab of a composite member is usually cast at a much later stage than the beam meansthat most of the time-dependent effects of shrinkage of the slab take place with the section actingcompositely. Most of the shrinkage of the beam will already have occurred by the time the slab is inplace, and the movement due to the shrinkage of the slab will induce stresses throughout the whole ofthe composite section. The water content of the slab concrete is often higher than that of the beam,since a lower strength is required, and this aggravates the problem of differential shrinkage. These extrastresses, which occur even under zero applied load, are not insignificant and should be considered indesign. Both the slab and beam undergo creep deformations under load and, although some of thecreep deformations in the beam may have taken place before casting of the slab, the level ofcompressive stress is higher in the beam, and so the creep deformations are larger.

Load-deflection curve for composite section in Example 10.4.


SCE Dept of Civil

Differential movements.

composite section which tend to reduce those set up by differential shrinkage. A problem which isencountered, particularly in connection with bridge decks, is that of varying temperature across acomposite section, although this may still be a problem in composite members used as roof structures.The hotter upper surface tends to expand more than the cooler lower surface and stresses are inducedthroughout the composite section.

A method for determining the stresses due to differential shrinkage will now be outlined, and this can beadapted to find the stresses due to differential creep and temperature movements. Consider acomposite member as shown in Fig. 10.13, where the slab is shown to have a free shrinkage movementof δsh relative to the beam. In reality this movement is restrained by the shear forces which are set upbetween the slab and beam, putting the slab into tension and the beam into compression. Themagnitude of the tensile force in the slab is given byT=εshAc,slabEc,slab,where Ac,slab and Ec,slab are the cross-sectional area and modulus of elasticity of the slab, respectively,and εsh is the free shrinkage strain of the slab concrete.The compressive force in the beam must be numerically equal to this tensile force. In addition to thedirect stresses described above, bending stresses are also introduced by restraint of the free differentialshrinkage. In order to determine these stresses, the free bodies of the slab and beam are considered, asshown in Fig. 10.14. Initially, the slab can be regarded as having a force T applied through its centroid, sothat its length is equal to that of the beam. There must be no net external force on the compositemember due to differential shrinkage alone, so a pair of equal and opposite compressive forces must beapplied to maintain equilibrium. However, these compressive forces act on the composite section andinduce a bending moment at the ends of the member of


SCE Dept of Civil

Internal stress resultants due to differential movements.

Stresses due to differential movements.


SCE Dept of Civil

Chapter V5.1 PRESTRESSED CONCRETE BRIDGES

General aspects – pretensioned prestressed bridge decks – Post tensioned prestressed bridge decks –Principle of design only.


SCE Dept of Civil


SCE Dept of Civil

PRE STRESSED CONCRETE VII/IV CIVIL ENGINEERING

1 P.JAGATEESH 2015-16

UNIT-I

INTRODUCTION-THEORY AND BEHAVIOUR PART A

1. What are the advantages of PSC construction

In case of fully prestressed member, which are free from tensile stresses

under working loads, the cross section is more efficiently utilized when

compared with a reinforced concrete section which is cracked under working

loads.

The flexural member is stiffer under working loads than a reinforced

concrete member of the same length.

2. Define Pre tensioning and Post tensioning

Pre tensioning: A method of Pre stressing concrete in which the tendons are

tensioned before the concrete is placed. In this method, the prestress is

imparted to concrete by bond between steel and concrete.

Post tensioning: A method of pre stressing concrete by tensioning the

tendons against hardened concrete. In this method, the prestress is imparted

to concrete by bearing.

3. What is the need for the use of high strength concrete and tensile steel in Pre

stressed concrete?

High strength concrete is necessary for prestress concrete as the material

offers highly resistance in tension, shear bond and bearing. In the zone of

anchorage the bearing stresses being hired, high strength concrete is

invariably preferred to minimizing the cost. High strength concrete is less

liable to shrinkage cracks and has lighter modulus of elasticity and smaller

ultimate creep strain resulting in a smaller loss of prestress in steel. The use

of high strength concrete results in a reduction in a cross sectional

dimensions of prestress concrete structural element with a reduced dead

weight of the material longer span become technically and economically

practicable.

Tensile strength of high tensile steel is in the range of 1400 to 2000 N/mm2

and if initially stress upto 1400 N/mm2 their will be still large stress in the high

tensile reinforcement after making deduction for loss of prestress. Therefore high

tensile steel is made for prestress concrete.

4. Define Kern Distance.

Kern is the core area of the section in which if the load applied tension

will not be induced in the section

Kt = Zb/A, Kb = Zt/A,

If the load applied K Compresser will be the maximum at the top most fiber

and zero stress will be at the bottom most fiber. If the load applied at Kb

compressive stress will be the maximum at the bottom most fiber and zero

stress will be at the top most fiber.



5. What is Relaxation of steel?

When a high tensile steel wire is stretch and maintained at a constant strain

the initially force in the wire does not remain constant but decrease with

time. The decrease of stress in steel at constant strain is termed relaxation of

steel.

6. What is concordant prestressing?

Pre stressing of members in which the cable follow a concordant profile. In

case of statically indeterminate structures. It does not cause any changes in

support reaction.

7. Define bonded and non bonded prestressing concrete.

Bonded prestressing: Concrete in which prestress is imparted to

concrete through bond between the tendons and surrounding concrete.

Pre tensioned members belong to this group.

Non-bonded prestressing: A method of construction in which the tendons

are not bonded to the surrounding concrete. The tendons may be placed in

ducts formed in the concrete members or they may be placed outside the

concrete section.

8. Define Axial prestressing

Members in which the entire cross-section of concrete has a uniform

compressive prestress. In this type of prestressing, the centroid, of the

tendons coincides with that of the concrete section.

9. Define Prestressed concrete.

It is basically concrete in which internal stresses of a suitable magnitude

and distribution are introduced so that the stresses resulting from external

loads (or) counteracted to a desire degree in reinforced concrete member

the prestress is commonly introduced by tensioning the steel reinforcement

10. Define anchorage.

A device generally used to enable the tendon to impart and maintain

prestress to the concrete is called anchorage. e.g. Fressinet, BBRV

systems,etc.,



PART-B

1. a) What are the advantages of Prestressed Concrete

In case of fully prestressed member, which are free from tensile stresses

under working loads.

The cross section is more effectively utilized when compared with a

reinforced concrete section which is cracked under working loads.

Within certain limits, a permanent will be counteracted by increasing the

eccentricity of the prestressing force in a prestressed structural elements, thus

effecting saving in the use of materials.

Prestressed concrete members possess improved resistance to shearing forces,

due to the effect of compressive prestress, which reduces the principal tensile

stress.

The use of high strength concrete and steel in prestressed members

results in lighter and slender members than is possible with reinforced

concrete.

It is free from cracks, contributes to the improved durability of the structure

under aggressive environmental conditions.

The economy of prestressed concrete is well established for long span

structures.

A prestressed concrete flexural member is stiffer under working loads

than a reinforced concrete member of the same depth.



b) Describe briefly Fressinet system of post tensioning

PRINCIPLES OF POST-TENSIONING:

In post-tensioning, the concrete units are first cast by incorporating ducts

or grooves to house the tendons. When the concrete attains sufficient strength, the high-

tensile wires are tensioned by means of jack bearing on the end face of the member and

anchorags by wedges or nuts.

FREYSSINET SYSTEM OF POST TENSIONING:

The Freyssinet system of post-tensioning anchorages which was

developed in 1939.

The Freyssinet anchorage system, which is widely used in Europe and

India, consists of a cylinder with a conical interior through which the high-tensile wires

pass and against the walls of which the wires are wedged by a conical plug lined

longitudinally with grooves to house the wires. The main advantages of the Freyssinet

system is that a large number of wires or strands can be simultaneously tensioned using

the double-acting hydraulic jack.

2. a) Discuss about the importance of control of deflections and the factors

influencing the deflection of PSC beams

Importance of control of deflection:

The structural concrete members shall designed to have adequate stiffness to limit

deflections, which may adversely affect the strength or serviceability of the

structure at working loads.

Suitable control on deflection is very essential for the following reasons:

Excessive, sagging of principal structural members is not only unsightly,

but at times, also renders the floor unsuitable for the intended use.

Large deflections under dynamic effects and under the influence of

variable loads may cause discomfort to the users.

Excessive deflections are likely to cause damage to finishes, partitions

and associated structures.

FACTORS INFLUENCING DEFLECTIONS:

The deflections of prestressed concrete members are influenced byy the following

salient factors:

Imposed load and self weight

Magnitude of the prestressing force

Cable profile

Second moment of area of cross section

Modulus of elasticity of concrete

Shrinkage, creep and relaxation of steel stress

Span of the member

Fixity conditions



b) Describe the various types of losses in prestress. What steps may be taken to reduce

these losses

LOSS DUE TO ELASTIC DEFORMATION OF CONCRETE:

The loss of prestress due to elastic deformation of concrete depends on the

modular ratio and the average stress in concrete at the level of steel.

If fc= prestress in concrete at the level of steel.

Es= modulus of elasticity of steel.

Ec= modulus of elasticity of concrete.

αe= Es/ Ec = modular ratio.

Strain in concrete at the level of steel = (fc/ Ec)

Stress in steel corresponding to this strain = (fc/ Ec) Es

Loss of stress in steel = αe fc

If the initial stress in steel is known, the percentage loss of stress due to the

elastic deformation of concrete can be computed.

LOSS DUE TO SHRINKAGE OF CONCRETE:

The shrinkage of concrete in prestressed members results in a shortening

of tensioned wires and hence contributes to the loss of stress. The

shrinkage of concrete is influenced by the type of cement and aggregates

and the methowd of curing used of high-strength concrete with low water

cement ratios result in a reduction in shrinkage and consequent loss of

prestress.

According IS1343 for the loss of prestress due to the shrinkage of

concrete

Єcs = total residual shrinkage strain having values of 300x106

for pre

tensioning and [200x106/log10(t+2)]

Where, t = age of concrete at transfer in days.

The loss of stress in steel due to the shrinkage of concrete is estimated as,

Loss of stress = Єcs x Es

LOSS DUE TO CREEP OF CONCRETE:

The sustained prestress in the concrete of a prestressed member results in

creep of concrete which effectively reduces the stress in high-tensile steel.

The loss of stress in steel due to creep of concrete can be estimated if the

magnitude of ultimate creep strain or creep coefficient is known.

ULTIMATE CREEP STRAIN METHOD:

If Єcc = ultimate creep strain for a sustained unit stress

fc = Compressive stress in concrete at the level of steel.

Es = modulus of elasticity of steel.

Loss of stress in steel due to creep of concrete = Єcc fc Es

CREEP COEFFICIENT METHOD:



If = creep coefficient

Єc = creep strain

Єe = elastic strain

αe = modular ratio

fc = stress in concrete


Ec = modulus of elasticity of concrete.

Creep coefficient( ) = (Єc/ Єe)

Loss of stress in steel = fc αe

LOSS DUE TO RELAXATION OF STRESS INN STEEL:

Most of the code provides for the loss of stress due to relaxation of steel as

a percentage of the initial stress in steel. The Indian standard code

recommends a value varying from 0 to 90 N/mm2

for stress in wire

varying from 0.5 fup to 0.8 fup .

LOSS OF STRESS DUE TO FRICTION:

On tensioning the curved tendons, loss of stress occurs in the post-

tensioned members due to friction between the tendons and the

surrounding concrete ducts. The magnitude of this loss is of the following

types:

(a) Loss of stress due to the curvature effects, which depends upon

the tendon from or alignment which generally follows a curved profile along the

length of the beam.

(b) Loss of stress effect, which depends upon the local deviation

in the alignment of the cable. The wobble or wave effect is the result of accidental

or unavoidable misalignment, since ducts or sheaths cannot be perfectly located to

follow predetermined profile throughout the length of the beam.

Px = Poe-(µα+ kx)

LOSS DUE TO ANCHORAGE SLIP:

In most post-tensioned system, when the cable is tensioned and the jack is

released to transfer prestress to concrete, the friction wedges, employed to

grip the wires, slip over a small distance before the wires are firmly

housed between the wedges. The magnitude of slip depends upon the type

of wedge and the stress in the wire.

∆ = (PL/AEs)

Where ∆ = slip of anchorage, mm

L = length of the cable,mm

A = cross sectional area of the cable, mm2


P = Prestressed force in the cable.



3.A prestressed concrete beam of section 120 mm wide by 300 mm deep is

used over an effective span of 6 m to support a uniformly distributed load

of 4 kN/m, which includes the self-weigbt of the beam. The beam is

prestressed by a straight cable carrying a force of 180 kN and located at an

eccenlricity of 50mm. Determine the location of the thrust-line in the

beam and plot its position at quarter and central span sections.

P=180kN

E=50mm

A=36000mm2

z=1800000mm3

Stresses due to prestressing force:

P/A(180X10^3/36X10^3) = +5 N/mm2

Pe/Z = (180 x 103 x 50) /(18x10^5) = +5 N/mm^2

Bending moment at the centre of the span= (0.125 x 4 x 62) =

18 kN m

Bending stresses at top and bottom=(18x10^6/18x10^5) =+ 10 N/mm^2

Resultant stresses at the central section :

At top =(5-5+10)=10 N/mm^2

At bottom =(5+5-10)=0 N/mm^2

Shift of pressure –lne from cable –line(M/P)=( (180 x 106) /(18x10^4)=100mm

Bending moment at quarter span section=(3/32)qL2= (3/32)x4x62

=13.5kNm

Bending stresses at top and bottom =(13.5x10^6/18x10^5)=7.5 N/mm^2

Resultant stresss at quarter span section:

At top =(5-5+7.5)=7.5 N/mm^2

At bottom =(5+5-7.5)=2.5 N/mm^2

Shift of pressure –lne from cable –line(M/P)=( (13.5x 106) /(18x10^4)=75mm

PRESTRESSED CONCRETE IV/VII CIVIL ENGINEERING


PRESTRESSED CONCRETE STRUCTURES

UNIT-II

DESIGN CONCEPTS

PART-A

1. What is meant by end block in a post tensioned member?

The zone between the end of the beam and the section where only longitudinal

stress exists is generally referred to as the anchorage zone or end block.

2. List any two applications of partial prestressing.

Used in large diameter concrete pipes

Used in railway sleepers

Water tanks

Precast concrete piles to counter tensile stress during transport and erection.

used in bridges construction

3. What is meant by partial prestressing?

The degree of prestress applied to concrete in which tensile stresses to a limited

degree are permitted in concrete under working load. In this case, in addition to

tensioned steel, a considerable proportion of untensioned reinforcement is

generally used to limit the width of cracks developed under service load.

4. Define degree of prestressing

A measure of the magnitude of the prestressing force related to the resultant

stress occurring in the structural member at working load.

5. Define Bursting tension.

The effect of transverse tensile stress is to develop a zone of bursting tension in a

direction perpendicular to the anchorage force resulting in horizontal cracking.

6. Define Proof stress

The tensile stress in steel which produces a residual strain of 0.2 percent of the

original gauge length on unloading.

7. Define cracking load.

The load on the structural element corresponding to the first visible crack.

8. Define Debonding.

Prevention of bond between the steel wire and the surrounding concrete.

9. Write formula for Moment of resistance in BIS code.

Mu = Apb Aps (d-dn)

10. What are the types of flexural failure?

Fracture of steel in tension

Failure of under-reinforced section

Failure of over-reinforced section

Other modes of failure



PART-B

1. a) What is meant by partial prestressing? Discuss the advantages and disadvantages when

partial prestressing is done

PARTIAL PRESTRESSING:

The degree of prestress applied to concrete in which tensile stresses to a

limited degree are permitted in concrete under working load. In this case,

in addition to tensioned steel, a considerable proportion of untensioned

reinforcement is generally used to limit the width of cracks developed

under service load.

ADVANTAGES:

Limited tensile stresses are permitted in concrete under service

loads with controls on the maximum width of cracks and

depending upon the type of prestressing and environmental

condition.

Untensioned reinforcement is required in the cross-section of a

prestresseed member for various reasons, such as to resist the

differential shrinkage, temperature effects and handling stresses.

Hence this reinforcement can cater for the serviceability

requirements, such as control of cracking, and partially for the

ultimate limit state of collapse which can result in considerable

reduction in the costlier high tensile steel.

Saving in the cost of overall structure.

DISADVANTAGES:

The excessive upward deflections, especially in bridge structure

where dead loads from a major portion of the total service loads,

and these deflections may increase with time of creep.



(b) Explain about the types of flexure failure occurs in prestressed concrete section



2. (a) Explain concept of limit states, partial safety factor.

Partial safety factors, are therefore used for each limit state being reached.

The values of partial safety loads recommended in the British, Indian

and American codes.

IS code:

Load combination Limit state of collapse Limit state of serviceability

DL LL WL DL LL WL

DL+LL 1.5 1.5 - 1.0 1.0 -

DL+WL 1.5 - 1.5 1.0 - 1.0

DL+LL+WL 1.2 1.2 1.2 1.0 0.8 0.8

Partial safety factor for materials has a values which depends on the

important of limit states being materials to which is applies difference between

strength of materials when tested and when incorporated in construction during the

service life.

(b) Discuss difference in load deflection of under prestressed, partially prestressed and

fully prestressed.

The load deflection characteristics of a typical prestressed concrete members

and discussed below:

If the beam is sufficient loaded, tensile stresses is develop in the soffit and

when this exceed the tensile strength of concrete, cracks are likely to develop in the

member.



The load deflection curve is approximately linear upto the stage of visible

cracking, but beyond this stage the deflection increase at a faster rate due to the

reduced stiffness of the beam.

In the port- cracking of the beam of beam is parallel to that of reinforced

concrete member.

The deflection of cracked structural member, may be estimated by the unilinear or

bilinear method recommended by the ECC.

In the unilinear method, the deflection will be,

a= βL2M/ Ec Ir

where a = Max deflection

L = Effective span

M = Max moment

Ec = Modulus of elasticity of concrete

Ir = IInd

commend of area.

In the bilinear method, the moment curvature is approximately by second straight

line.

The instantaneous deflection in the post cracking stage is obtained as the sum

of deflection upto cracking load based on gross section and beyond the cracking

load considering the cracked section.

Hence deflection are estimated by

a= βL2

{(Mcr/ EcIc)+((M-Mc)/0.85Ecfck)}



3. The end block of a post-tensioned PSC beam, 300 x 300 mm is subjected to a

concentric anchorage force of 832.8 kN by a Freyssinet anchorage of area

11720 mm2. Design and detail the anchorage reinforcement for the end

block.(NOV-DEC 2009)

Prestressing Concrete,P= 832800 N

Average compressive stress, fc= (832800/300x300) =9.3 N/mm2

2ypo = (π/4xd2)

(1/2) = √(11720x4/π = 123mm

2yo = 300/2 = 150mm

Ypo/yo = (123/300)=0.41

Fc = P/A = 832.8/(300x300) = 9.25N/mm2

Tensile stress Fv(max) =fc(0.98 – 0.825 ypo/yo) = 9.3(0.98 – 0.825x0.41) = 6 N/mm2

Bursting tension Fbst = p(0.48 – 0.4 ypo/yo)

= 832800(0.48-0.4x0.41)

= 264000N

Using 10mm diameter mild steel links with yield stress of 260

N/mm2

Ast = Fbst/0.87fy = (260x103)/(0.87x260) = 706.85mm

2

Number of reqd =(264000/(0.87x260x79))= 15

The reinforcement is to be arranged in the zone 0.2 yo=(0.2x150)

=30mm

PRESTRESSED CONCRETE VII/IV CIVIL ENGINEERING


UNIT III

Circular prestressing

Part A

1. Sketch the loop reinforcement, hair-pin bars in end blocks.(NOV-DEC 2009)

2. Sketch the correct arrangement of sheet cage in anchorage zone.(NOV-DEC 2009)



3. Define two stage constructions.(NOV-DEC 2012)

One-stage construction: Construct and initialize the object in one stage, all

with the constructor.

Two-stage construction: Construct and initialize the object in two separate

stages.

The constructor creates the object and an initialization function initializes it.

4. Write any two general failures of prestressed concrete tanks.(NOV-DEC 2012)

deformation of the pre-cast concrete units during construction

Manufacturing inaccuracies led to out of tolerance units being delivered to the

site under investigation and may have affected the ability to achieve a good

seal.

5. Mention the importance of shrinkage in composite construction?

(NOV-DEC 2010)

The time dependent behavior of composite prestressed concrete

beams depends upon the presence of differential shrinkage and creep of the

concretes of web and deck, in addition to other parameters, such as relaxation

of steel, presence

of untensioned steel, and compression steel etc.



Part B

1. Explain the effect of varying the ratio of depth anchorage to the depth of end block

on the distribution of bursting tension. (8) (NOV-DEC 2012)

Bursting tensile forces

a) The bursting tensile forces in the end blocks, or regions of bonded post-

tensioned members, should be assessed on the basis of the tendon jacking load.

For unbonded members, the bursting tensile forces should be assessed on the basis

of the tendon jacking load or the load in the tendon at the limit state of collapse,

whichever is greater ( see Appendix B ).

The bursting tensile force, Fbst existing in an individual square end block loaded

by a symmetrically placed square anchorage or bearing plate, may be derived from

the equation below:

b) The force Fbst will be distributed in a region extending from 0.1 yo to yo from

the loaded face of the end block. Reinforcement provided to sustain the bursting

tensile force may be

assumed to be acting at its design strength (0.87 times

characteristic strength of reinforcement) except that the stress should be limited to

a value corresponding

to a strain of 0.001 when the concrete cover to the

reinforcement is less than 50 mm.

c) In rectangular end

blocks, the bursting tensile forces in the two principal

directions should be assessed on the basis of 18.6.2.2. When circular anchorage or

bearing plates are used, the side of the equivalent square area should be used.



Where groups of anchorages or bearing plates occur, the end blocks should be

divided into a series of symmetrically loaded prisms and each prism treated in the

above manner. For designing end blocks having a cross-section different in shape

from that of the general cross-section of the beam, reference should be made to

specialist literature.

d) Compliance with the requirements of (a), (b) and (c) will generally ensure that

bursting tensile forces along the load axis are provided for. Alternative methods of

design which make allowance for the tensile strength of the concrete may be used,

in which case reference should be made to specialist literature.

e) Consideration should also be given to the spalling tensile stresses that occur in

end blocks where the anchorage or bearing plates are highly eccentric; these reach

a maximum at the loaded face.

2.(i) Explain the junctions of tank wall and base slab with neat sketch. (8)

(NOV- DEC 2012)

Joint in the concrete introduced for convenience in construction at which

special measures are taken to achieve subsequent continuity without provision for

further relative movement, is called a construction joint. A typical application is

between successive lifts in a reservoir.



The position and arrangement of all construction joints should be predetermined

by the engineer. Consideration should be given to limiting the number of such

joints and to keeping them free from possibility of percolations in a similar

manner to contraction joints.

A gap temporarily left between the concrete of adjoining parts of a structure which

after a suitable interval and before the structure is put into use, is filled with mortar

or concrete either completely ( Fig. 5A) or as provided below, with the inclusion

of suitable jointing materials ( Fig. 5B and SC). In the former case the width of the

gap should be sufficient to allow the sides to be prepared before filling.

Where measures are taken for example, by the inclusion of suitable jointing

materials to maintain the water tightness of the concrete subsequent to the filling

of the joint, this type of joint may be regarded as being equivalent to a contraction

joint ( partial or complete ) as defined above.



3. (a) What are the different types of joints used between the slab of prestressed

concrete tank

Joints shall be categorized as below:

a) Movetnent Joints - There are three categories of movement joints:

contraction joint - A movement joint with a deliberate discontinuity but no initial

gap between the concrete on either side of the joint, the joint being intended to

accommodate contraction of the concrete ( see Fig. 1 ).

A distinction should be made between a complete contraction joint (see Fig. 1A )

in which both concrete and reinforcing steel are interrupted, and a partial

contraction joint (. see Fig. 1B ) in which only the concrete is interrupted, the

reinforcing steel running through.

Expansion joint - A movement joint with complete discontinuity in both

reinforcement and concrete and intended to accommodate either expansion or

contraction of the structure (see Pig. 2).

In general, such a joint requires the provision of an initial gap between the

adjoining parts of a structure which by closing or opening accommodates the

expansion or contraction of the structure. Design of the joint so as to incorporate

sliding surfaces, is not, however, precluded and may sometimes be advantageous.



b) Construction Joint-A joint in the concrete introduced for convenience in

construction at which special measures are taken to achieve subsequent continuity

without provision for further relative movement, is called a construction joint. A

typical application is between successive lifts in a reservoir.







c) Temporary Open Joints - A gap temporarily left between the concrete of

adjoining parts of a structure which after a suitable interval and before the

structure is put into use, is filled with mortar or concrete either completely ( Fig.

5A) or as provided below, with the inclusion of suitable jointing materials ( Fig.

5B and SC). In the former case the width of the gap should be sufficient to allow

the sides to be prepared before filling.







(b) Design the circular tank (only procedure).(NOV-DEC 2010) .(NOV-DEC

2010)

in the construction of concrete structures for the storage of liquids, the

imperviousness of concrete is an important basic requirement. Hence, the design

of such construction is based on avoidance of cracking in the concrete. The

structures are prestressed to avoid tension in the concrete. In addition, prestressed

concrete tanks require low maintenance. The resistance to seismic forces is also

satisfactory.

Prestressed concrete tanks are used in water treatment and distribution systems,

waste water collection and treatment system and storm water management. Other

applications are liquefied natural gas (LNG) containment structures, large

industrial process tanks and bulk storage tanks. The construction of the tanks is in

the following sequence. First, the concrete core is cast and cured. The surface is

prepared by sand or hydro blasting. Next, the circumferential prestressing is

applied by strand wrapping machine. Shotcrete is applied to provide a coat of

concrete over the prestressing strands.

Analysis

The analysis of liquid storage tanks can be done by IS:3370 - 1967, Part 4, or by

the finite element method. The Code provides coefficients for bending moment,

shear and hoop tension (for cylindrical tanks), which were developed from the

theory of plates and shells. In Part 4, both rectangular and cylindrical tanks are

covered. Since circular prestressing is applicable to cylindrical tanks, only this

type of tank is covered in this module.

The following types of boundary conditions are considered in the analysis of the

cylindrical wall.

a) For base: fixed or hinged

b) For top: free or hinged or framed.

For base



Fixed: When the wall is built continuous with its footing, then the base can be

considered to be fixed as the first approximation.

Hinged: If the sub grade is susceptible to settlement, then a hinged base is a

conservative assumption. Since the actual rotational restraint from the footing is

somewhere in between fixed and hinged, a hinged base can be assumed.

The base can be made sliding with appropriate polyvinyl chloride (PVC) water-

stops for liquid tightness.

For top

Free: The top of the wall is considered free when there is no restraint in expansion.

Hinged: When the top is connected to the roof slab by dowels for shear transfer,

the boundary condition can be considered to be hinged.

Framed: When the top of the wall and the roof slab are made continuous with

moment transfer, the top is considered to be framed. The hydrostatic pressure on

the wall increases linearly from the top to the bottom of the liquid of maximum

possible depth. If the vapour pressure in the free board is negligible, then the

pressure at the top is zero. Else, it is added to the pressure of the liquid throughout

the depth. The forces generated in the tank due to circumferential prestress are

opposite in nature to that due to hydrostatic pressure. If the tank is built

underground, then the earth pressure needs to be considered. The hoop tension in

the wall, generated due to a triangular hydrostatic pressure is given as follows.

The hoop tension in the wall, generated due to a triangular hydrostatic pressure is

given as follows.

T = CT w H Ri (9-6.15)

The bending moment in the vertical direction is given as follows.

M = CM w H3 (9-6.16)

The shear at the base is given by the following expression.

V = CV w H2 (9-6.17)

In the previous equations, the notations used are as follows.

CT = coefficient for hoop tension



CM = coefficient for bending moment

CV = coefficient for shear

w = unit weight of liquid

H = height of the liquid

Ri = inner radius of the wall.

The values of the coefficients are tabulated in IS:3370 - 1967, Part 4, for various

values of H2/Dt, at different depths of the liquid. D and t represent the inner

diameter and the thickness of the wall, respectively. The typical variations of CT

and CM with depth, for two sets of boundary conditions are illustrated.

The roof can be made of a dome supported at the edges on the cylindrical wall.

Else, the roof can be a flat slab supported on columns along with the edges.

IS:3370 - 1967, Part 4, provides coefficients for the analysis of the floor and roof

slabs.

Design

IS:3370 - 1967, Part 3, provides design requirements for prestressed tanks. A few

of them are mentioned.

1) The computed stress in the concrete and steel, during transfer, handling and

construction, and under working loads, should be within the permissible values as

specified in IS:1343 - 1980.

2) The liquid retaining face should be checked against cracking with a load factor

of 1.2. σCL/σWL ≥ 1.2 (9-6.18)

Here,

σCL = stress under cracking load

σWL = stress under working load.

Values of limiting tensile strength of concrete for estimating the cracking load are

Specified in the Code.

3) The ultimate load at failure should not be less than twice the working load.

4) When the tank is full, there should be compression in the concrete at all points

of at least 0.7 N/mm2. When the tank is empty, there should not be tensile stress



greater than 1.0 N/mm2. Thus, the tank should be analysed both for the full and

empty conditions.

5) There should be provisions to allow for elastic distortion of the structure during

prestressing. Any restraint that may lead to the reduction of the prestressing force,

should be considered.

4. (a) What are the design considerations of prestressed concrete poles? (4)

The pre stressed concrete pole for power transmission line are generally designed

as member with uniform prestress since they are subjected to bending moment of

equal magnitude in opposite directions. The poles are generally designed for

following critical load conditions,

1. Bending due to wind load on the cable and on the exposed face.

2. Combined bending and torsion due to eccentric snapping of wire.

3. Maximum torsion due to skew snapping of wires.

4. Bending due to failure of all the wires on one side of the pole.

5. Handling and erection stresses.

(b) What are the advantages of partially prestressed concrete poles?

Resistance to corrosion in humid and temperature climate and to erosion in

desert areas.

Freeze thaw resistance in cold region.

Easy handling due to less weight than other poles

Fire resisting, particularly grassing and pushing fire near ground line.

Easily installed in drilled holes in ground with or without concrete fill.

Lighter because of reduced cross section when compared with reinforced

concrete poles.

Clean and neat in appearance and requiring negligible maintenance for a

number of years, thus ideal suited for urban installation.



UNIT III

Circular prestressing

Part A

1. Sketch the loop reinforcement, hair-pin bars in end blocks.(NOV-DEC 2009)

2. Sketch the correct arrangement of sheet cage in anchorage zone.(NOV-DEC 2009)



3. Define two stage constructions.(NOV-DEC 2012)

One-stage construction: Construct and initialize the object in one stage, all

with the constructor.

Two-stage construction: Construct and initialize the object in two separate

stages.

The constructor creates the object and an initialization function initializes it.

4. Write any two general failures of prestressed concrete tanks.(NOV-DEC 2012)

deformation of the pre-cast concrete units during construction

Manufacturing inaccuracies led to out of tolerance units being delivered to the

site under investigation and may have affected the ability to achieve a good

seal.

5. Mention the importance of shrinkage in composite construction?

(NOV-DEC 2010)

The time dependent behavior of composite prestressed concrete

beams depends upon the presence of differential shrinkage and creep of the

concretes of web and deck, in addition to other parameters, such as relaxation

of steel, presence

of untensioned steel, and compression steel etc.



Part B

1. Explain the effect of varying the ratio of depth anchorage to the depth of end block

on the distribution of bursting tension. (8) (NOV-DEC 2012)

Bursting tensile forces

a) The bursting tensile forces in the end blocks, or regions of bonded post-

tensioned members, should be assessed on the basis of the tendon jacking load.

For unbonded members, the bursting tensile forces should be assessed on the basis

of the tendon jacking load or the load in the tendon at the limit state of collapse,

whichever is greater ( see Appendix B ).

The bursting tensile force, Fbst existing in an individual square end block loaded

by a symmetrically placed square anchorage or bearing plate, may be derived from

the equation below:

b) The force Fbst will be distributed in a region extending from 0.1 yo to yo from

the loaded face of the end block. Reinforcement provided to sustain the bursting

tensile force may be

assumed to be acting at its design strength (0.87 times

characteristic strength of reinforcement) except that the stress should be limited to

a value corresponding

to a strain of 0.001 when the concrete cover to the

reinforcement is less than 50 mm.

c) In rectangular end

blocks, the bursting tensile forces in the two principal

directions should be assessed on the basis of 18.6.2.2. When circular anchorage or

bearing plates are used, the side of the equivalent square area should be used.



Where groups of anchorages or bearing plates occur, the end blocks should be

divided into a series of symmetrically loaded prisms and each prism treated in the

above manner. For designing end blocks having a cross-section different in shape

from that of the general cross-section of the beam, reference should be made to

specialist literature.

d) Compliance with the requirements of (a), (b) and (c) will generally ensure that

bursting tensile forces along the load axis are provided for. Alternative methods of

design which make allowance for the tensile strength of the concrete may be used,

in which case reference should be made to specialist literature.

e) Consideration should also be given to the spalling tensile stresses that occur in

end blocks where the anchorage or bearing plates are highly eccentric; these reach

a maximum at the loaded face.

2.(i) Explain the junctions of tank wall and base slab with neat sketch. (8)

(NOV- DEC 2012)

Joint in the concrete introduced for convenience in construction at which

special measures are taken to achieve subsequent continuity without provision for

further relative movement, is called a construction joint. A typical application is

between successive lifts in a reservoir.







A gap temporarily left between the concrete of adjoining parts of a structure which

after a suitable interval and before the structure is put into use, is filled with mortar

or concrete either completely ( Fig. 5A) or as provided below, with the inclusion

of suitable jointing materials ( Fig. 5B and SC). In the former case the width of the

gap should be sufficient to allow the sides to be prepared before filling.







3. (a) What are the different types of joints used between the slab of prestressed

concrete tank

Joints shall be categorized as below:

a) Movetnent Joints - There are three categories of movement joints:

contraction joint - A movement joint with a deliberate discontinuity but no initial

gap between the concrete on either side of the joint, the joint being intended to

accommodate contraction of the concrete ( see Fig. 1 ).

A distinction should be made between a complete contraction joint (see Fig. 1A )

in which both concrete and reinforcing steel are interrupted, and a partial

contraction joint (. see Fig. 1B ) in which only the concrete is interrupted, the

reinforcing steel running through.

Expansion joint - A movement joint with complete discontinuity in both

reinforcement and concrete and intended to accommodate either expansion or

contraction of the structure (see Pig. 2).

In general, such a joint requires the provision of an initial gap between the

adjoining parts of a structure which by closing or opening accommodates the

expansion or contraction of the structure. Design of the joint so as to incorporate

sliding surfaces, is not, however, precluded and may sometimes be advantageous.



b) Construction Joint-A joint in the concrete introduced for convenience in

construction at which special measures are taken to achieve subsequent continuity

without provision for further relative movement, is called a construction joint. A

typical application is between successive lifts in a reservoir.







c) Temporary Open Joints - A gap temporarily left between the concrete of

adjoining parts of a structure which after a suitable interval and before the

structure is put into use, is filled with mortar or concrete either completely ( Fig.

5A) or as provided below, with the inclusion of suitable jointing materials ( Fig.

5B and SC). In the former case the width of the gap should be sufficient to allow

the sides to be prepared before filling.







(b) Design the circular tank (only procedure).(NOV-DEC 2010) .(NOV-DEC

2010)

in the construction of concrete structures for the storage of liquids, the

imperviousness of concrete is an important basic requirement. Hence, the design

of such construction is based on avoidance of cracking in the concrete. The

structures are prestressed to avoid tension in the concrete. In addition, prestressed

concrete tanks require low maintenance. The resistance to seismic forces is also

satisfactory.

Prestressed concrete tanks are used in water treatment and distribution systems,

waste water collection and treatment system and storm water management. Other

applications are liquefied natural gas (LNG) containment structures, large

industrial process tanks and bulk storage tanks. The construction of the tanks is in

the following sequence. First, the concrete core is cast and cured. The surface is

prepared by sand or hydro blasting. Next, the circumferential prestressing is

applied by strand wrapping machine. Shotcrete is applied to provide a coat of

concrete over the prestressing strands.

Analysis

The analysis of liquid storage tanks can be done by IS:3370 - 1967, Part 4, or by

the finite element method. The Code provides coefficients for bending moment,

shear and hoop tension (for cylindrical tanks), which were developed from the

theory of plates and shells. In Part 4, both rectangular and cylindrical tanks are

covered. Since circular prestressing is applicable to cylindrical tanks, only this

type of tank is covered in this module.

The following types of boundary conditions are considered in the analysis of the

cylindrical wall.

a) For base: fixed or hinged

b) For top: free or hinged or framed.



For base

Fixed: When the wall is built continuous with its footing, then the base can be

considered to be fixed as the first approximation.

Hinged: If the sub grade is susceptible to settlement, then a hinged base is a

conservative assumption. Since the actual rotational restraint from the footing is

somewhere in between fixed and hinged, a hinged base can be assumed.

The base can be made sliding with appropriate polyvinyl chloride (PVC) water-

stops for liquid tightness.

For top

Free: The top of the wall is considered free when there is no restraint in expansion.

Hinged: When the top is connected to the roof slab by dowels for shear transfer,

the boundary condition can be considered to be hinged.

Framed: When the top of the wall and the roof slab are made continuous with

moment transfer, the top is considered to be framed. The hydrostatic pressure on

the wall increases linearly from the top to the bottom of the liquid of maximum

possible depth. If the vapour pressure in the free board is negligible, then the

pressure at the top is zero. Else, it is added to the pressure of the liquid throughout

the depth. The forces generated in the tank due to circumferential prestress are

opposite in nature to that due to hydrostatic pressure. If the tank is built

underground, then the earth pressure needs to be considered. The hoop tension in

the wall, generated due to a triangular hydrostatic pressure is given as follows.

The hoop tension in the wall, generated due to a triangular hydrostatic pressure is

given as follows.

T = CT w H Ri (9-6.15)

The bending moment in the vertical direction is given as follows.

M = CM w H3 (9-6.16)

The shear at the base is given by the following expression.

V = CV w H2 (9-6.17)

In the previous equations, the notations used are as follows.

CT = coefficient for hoop tension



CM = coefficient for bending moment

CV = coefficient for shear

w = unit weight of liquid

H = height of the liquid

Ri = inner radius of the wall.

The values of the coefficients are tabulated in IS:3370 - 1967, Part 4, for various

values of H2/Dt, at different depths of the liquid. D and t represent the inner

diameter and the thickness of the wall, respectively. The typical variations of CT

and CM with depth, for two sets of boundary conditions are illustrated.

The roof can be made of a dome supported at the edges on the cylindrical wall.

Else, the roof can be a flat slab supported on columns along with the edges.

IS:3370 - 1967, Part 4, provides coefficients for the analysis of the floor and roof

slabs.

Design

IS:3370 - 1967, Part 3, provides design requirements for prestressed tanks. A few

of them are mentioned.

1) The computed stress in the concrete and steel, during transfer, handling and

construction, and under working loads, should be within the permissible values as

specified in IS:1343 - 1980.

2) The liquid retaining face should be checked against cracking with a load factor

of 1.2. σCL/σWL ≥ 1.2 (9-6.18)

Here,

σCL = stress under cracking load

σWL = stress under working load.

Values of limiting tensile strength of concrete for estimating the cracking load are

Specified in the Code.

3) The ultimate load at failure should not be less than twice the working load.



4) When the tank is full, there should be compression in the concrete at all points

of at least 0.7 N/mm2. When the tank is empty, there should not be tensile stress

greater than 1.0 N/mm2. Thus, the tank should be analysed both for the full and

empty conditions.

5) There should be provisions to allow for elastic distortion of the structure during

prestressing. Any restraint that may lead to the reduction of the prestressing force,

should be considered.

4. (a) What are the design considerations of prestressed concrete poles? (4)

The pre stressed concrete pole for power transmission line are generally designed

as member with uniform prestress since they are subjected to bending moment of

equal magnitude in opposite directions. The poles are generally designed for

following critical load conditions,

1. Bending due to wind load on the cable and on the exposed face.

2. Combined bending and torsion due to eccentric snapping of wire.

3. Maximum torsion due to skew snapping of wires.

4. Bending due to failure of all the wires on one side of the pole.

5. Handling and erection stresses.

(b) What are the advantages of partially prestressed concrete poles?

Resistance to corrosion in humid and temperature climate and to erosion in

desert areas.

Freeze thaw resistance in cold region.

Easy handling due to less weight than other poles

Fire resisting, particularly grassing and pushing fire near ground line.

Easily installed in drilled holes in ground with or without concrete fill.

Lighter because of reduced cross section when compared with reinforced

concrete poles.

Clean and neat in appearance and requiring negligible maintenance for a

number of years, thus ideal suited for urban installation.


1P.JAGATEESH 2015-16

UNIT V

PRE-STRESSED CONCRETE BRIDGES

Part A

1. what is main advantage of prestressed concrete bridge deck.

High-strength concrete and high-tensile steel, besides being economical,make for slender sections, which are aesthetically superior.

Prestressed concrete bridges can be designed as class I type structures without any tensilestresses under service loads, thus resulting in a crack-free structure.

In comparison with steel bridges, prestressed concrete bridges require verylittlemaintenance.

Prestressed concrete is ideally suited for composite bridge construction in which precastprestressed girders support the cast in situ slab deck. This type of con• struction is verypopular since it involves minimum disruption of traffic.

2.Typical types Of Pre-Tensioned Prestressed Concrete Bridges .

a. Voided slabb. Single teec. Box beamsd. Double teee. Aasho-type girders with slab(U.S.A)f. Y-tube standard beams with slab

3.what is main advantage of prestressed concrete bridge deck.


Segmental construction is ideally suited for post –tensioning work. Post-tensioning facilities the use of curved slabs hich improve the shear resistance of

girders.

4.Typical types Of Post-Tensioned Prestressed Concrete Bridges .

a. Solid slab (10-15m)b. Hollow slab(15-25m)c. Tee beams(20-40m)d. Box girders, two cell(30-70m)e. Box girders ,trapezoidal (30-80m)



Part-B

1. what is main advantage of prestressed concrete bridge deck.


Prestressed concrete bridges can be designed as class I type structures without any tensilestresses under service loads, thus resulting in a crack-free structure.

In comparison with steel bridges, prestressed concrete bridges require verylittlemaintenance.

Prestressed concrete is ideally suited for composite bridge construction in which precastprestressed girders support the cast in situ slab deck. This type of con• struction is verypopular since it involves minimum disruption of traffic.

Post-tensioned prestressed concrete finds extensive applications in long-spancontinuous girder bridges of variable cross-section. Not only does it make forsleek. structures, but it also effects considerable saving in the overall cost ofconstruction.

In recent years, partially prestressed concrete (type-3 structure) bas been pre•ferred for bridge construction, because it offers considerable economy in the useof costly high-tensile steel in the girder.



2.Typical types Of Pre-Tensioned Prestressed Concrete Bridges

(a) Voided slab (b) Single tee

(c) Box beams (d) Double tee

',;,,·,:,(;

. ·~:.~.,_;,..:Ji.,.'

UNIT II - DESIGN CONCEPTS - WordPress.com · UNIT 1 INTRODUCTION – THEORY AND BEHAVIOUR Basic...

Documents

Transcript of UNIT II - DESIGN CONCEPTS - WordPress.com · UNIT 1 INTRODUCTION – THEORY AND BEHAVIOUR Basic...