Strengthening and Maintenance of

Reinforced concert Structures

Introduction :-

Although there is enough safety and design factors in the Egyptian

code of practice ECOP and in other codes of practice, defects in

building a raises from poor man workshop, poor quality of materials

used in structural process or errors in design and in estimating of

material / Soil strength.

Studying the causes of these defects may lead to eliminating

theses problems.

Types of defects in Structures :-

a) Defects in the whole structure :

a-1) Serviceability defects :

Settlement

هبوط للمبنىSlope or tilting

الميل أو االلتواء Horizontal

displacement

ازاحة أفقية

Vertical

Displacement

إزاحة رأسية

Vibration

اهتزازات المصانع –الكبارى

Noticeable

cracks

ملحوظة شروخ

Insufficient

Isolation

عزل غير كافى

a-2) Safety defects :-

Full collapse Partial collapse Unbalanced

عدم إتزان case إنهيار جزئى إنهيار كامل b) Defects in Structural elements :-

b-1) Deformations and deflections : يمخر ل والتالتشك

Deflection in members

Subjected to bending

ل فى العناصر المعرضة لالنحناء التشك

Buckling in members

Subjected to compression

االنبعاج فى العناصر المعرضة للضغط

b-2) Cracks in Reinforced concrete elements :

Efflorescence

يحالتنقيح والتمل And stains

Cracks Spilling of

cover

تساقط الغطاء الخرسانى

Surface

Erosion

تآكل السطح

Corrosion

الصدأ Aggregate

Alkali

reaction

التفاعل القلوى للركام

1) Can be produced due to alkali – aggregate reaction forming a

white gel.

Efflorescence is a white powder stain produced by the leaching

of lime form Portland cement.

Discoloration can also result from metals inserted into the

concrete, or from corrosion products dripping into the surface.

2) Cracks are either active or dormant, active cracks widen deepen

thought the concrete, Dormant cracks remain unchanged. Some

dormant cracks such as those caused by shrinkage during the

curing process pose no danger, but if left unprepared they can

provide convenient channels for moisture penetration causing

corrosion for instance structural cracks results from temporary or

continued overloads, uneven foundation settling, defects in

structural design, they are active if over load settlement are

continued. They are dormant if the temporary overloads have

been removed, or if the differential settlement have been

removed.

Map cracks may result from alkali agg. Reaction.

3) Spilling may result when reinforcing bars corrode, creating high

stresses with concrete, thus chunks of conc. pop off from the

surface. It may result also from improper conditions.

4) Erosion is the weathering of concrete surface by wind, rain,

snow, salts. Also, it can be caused by mechanical action of

water.

5) Corrosion is the rusting of reinforcing bars in concrete. The

protection of steel bars can be damaged either by carbonation,

which occurs when carbon dioxide in the air reacts chemically

with cement paste and reduces the alkalinity of concrete or by

chloride ions from salts in moisture to produce an electrolyte

that corrodes the bars. Corrosion produces rust which occupies

more space than the original metal causing expansion forces.

Lecture NO. 2:-

Reasons for strengthening of structures :

A) Types of structural elements to be strengthened:

(A-1) Slabs.

(A-2) Beams.

(A-3) Columns.

(A-4) Foundations.

“The main reasons for strengthening any structural element may

differ according to the cause of the problem”.

تتوقف أسباب تدعيم أى عنصر إنشائى على سبب المشكلة التى أدت إلى التدعيم.

B) Main Reasons :-

a) Need for increasing capacity of structure:

“Like building an additional floor in the same building”.

دة والقواعد على سبيل الرغبة فى زيادة عدد أدوار فى المبنى وبالتالى البد من تدعيم األعم المثال ال الحصر.

b) Defects in the existing building :

حدوث عيوب فى العناصر اإلنشائية للمبنى. These defects may be caused by:

(B-1) Material weakness: ضعف فى المادة اإلنشائية

This may be affected by aggressive environment or salts

and other harmful.

This weakness may be presented in both steel (corrosion of

steel) & in concrete (Alkali aggregate reaction).

(B-2) In adequate cross sections for design :

Corrosion for instance head to decreading the cross section of steel ( 07 % of As-actual As ) for instance.

Or sulphate attack in concrete may lead to decreasing the cross section

of concrete itself.

يحدث تقليل فى القطاع اإلنشائى وذلك إمكا فكى حديكد التسكليح أو فكى القطكاع الخرسكانى فنجكد % مكن المسكاحة األصكلية 07أنه نتيجة لصدأ الحديد أصبحت المساحة الفعالكة للحديكد حكوالى

)على سبيل المثال( كما أن القطاع الخرسانى قد تحدث فيه أضرار نتيجة لألمالح.

(B-3) In adequate geometry of sections in order to redesign for lateral

loads .

قككد تكككون عككزوم الحسككاءة لككبعا العناصككر اإلنشككائية كاألعمككدة علككى سككبيل المثككال غيككر كافيككة لمقاومككة األحمككال الجانبيككة ممككا يتطلككب زيككادة القطككاع اإلنشككائى بنسككبة معينككة لتحقيكك الحسككاءة

لجانبية. الكلية للمنشأ لمقاومة األحمال ا

Methods for testing Existing materials :- طرق اختبار المواد لتدعيمها

Destructive Tests : االختبارات المتلفة

Before any strengtheing operation, a complete survey for the elements

strength should be made to know the existing capacity of the building

and the amount of materials (both steel & concerete) to be added to

surve for the new design case.

قبكككل إجكككراء أى أعمكككال تكككدعيم علكككى المبنكككى يجكككب عمكككل مسكككح شكككامل للمبنكككى الحكككالى وقكككدرة مستقبلية له. العناصر اإلنشائية له على تحمل األحمال الواقعة عليه أو األحمال ال

Destructive tests: like core-test depends on testing an actual part of the

structural element to illustrate its actual strength.

يعتبككر اختبككار القلككب الخرسككانى هككو االختبككار المتلككف الككذى عككن طريقككه يككتم اسككتقطاع جككزء مككن ط إليجاد مقاومة الضغط للمنشأ. العنصر اإلنشائى واختباره فى الضغ

None Destructive Tests: االختبارات غير المتلفة

These tests are not that harmfull to the structural elements such as

schmiddet rammer & pulse velocity test.

ل اختبككار القلككب الخرسككانى وتشككمل مطرقككة شككميدت والموجككات هككذه االختبككارات غيككر ضككارة مثكك فو الصوتية.

The main important remark about schmiddet hammer is that it

deviates 20% than the actual strength.

قد تكون قراءات شميدت همر بعيدة عن القراءات الحقيقية )المقاومة الحقيقيكة( للمنشكأ وتتكراوح %. 07بة الخطأ تل بين نس

Analysis Methods :-

After determining the actual properties of materials and gathering

drawings of the structure and modifieng any changes from the actual

phase of the structure, a complete analysis of the building should be

made.

القطاعكات الحقيقيكة( وبعكد تجميكع رسكومات –بعد تحديكد خكواا المكواد اإلنشكائية )المقاومكات المنشأ وتعديل أى إضافات من أرا الواقع قد تكون تمت على المنشأ يجب البدء فى عمليكة

تحليل إنشائى.

The analysis process could be achieved either by using simple

methods or finite element methods to know the actual response of the

building and to determine shear forces, normal forces, and bending

moments for structural elements.

قككد تكككون طريقككة التحليككل اإلنشككائى بسككيطة أو قككد تكككون عككن طريكك اسككتخدام طريقككة العناصككر واالنحناء والقوى المحورية للقطاعات التى نرغب فى تدعيمها. المحددة لتجديد قوى القا

For the case of changing the usage of the building a new analysis

process is to be conducted in order to calculate the additional sections

to be added to the existing structure.

امية المبنكككى أو تحميكككل بأحمكككال إضكككافية يجكككب عمكككل تحليكككل إنشكككائى فكككى حالكككة تغييكككر اسكككتخدللوضع القديم وتحليل إنشائى آخر للوضع الجديكد بحيكث نحصكل علكى اإلضكافة المطلوبكة فكى

المنشأ من ناحية القطاعات )حديد أو خرسانة(.

Methods for strengthening :-

a) Concrete jackets.

b) Steel sections

(sections like, I , T , U ,

, )

c) (Combined sections of concrete jackets & stel sections).

d) Fiber Reinforced polymers (FRP) (foundations, columns, beam,

slabs, walls).

Strenthening of slabs

Using steel beams

L.L (Existing) = 3 KN/M2

L.L (Target) = 5 KN/M2

Wu (old) = 142 KN/M2

= 018*25+2+3=14.25 KN/M2

Broad flange B.I Standard SI

2 steel beams 1 steel beam

conc. = 25 KN/m3 & Floor cover = 2.0 KN/m

2

Wu = (0.72* conc+2+5)*1.5 = 17.25KN/M2

We can treat the cont. beam as a simple beams and hence.

MKNML

Mu .4.198

)3(1725

8

22

Calculate Moment capacity:-

TCFx 0.0

s

y

cu

fAsbaf

***45.0

15.1

360*7961000**)25(45.0 a

(a=22.16mm)

mma

yct 92.1432

16.225.15)

2(5.15

bafcuFCxs

fyAsFT cT **45.0,

CTCap

yCorTM *)()(

mmNMCap

.36.3586292.143*)6.249182()(

(N.mm *10-06

Kn.m)

mKNM

cap.86.35

)(

Mu < Mcap Safe

Therefore there no need to strengthen the old slab in its new position.

And the beam must be designed.

Design of steel Beam : (Non – contacted)

Wu = 14.25 KN/M2 , Wu = 17.25 KN/m

2

Mcap = 35.86 KN.M.

Load on steel beam =

(3.0*17.25)=5175KN/M

51.75 KN/M

MKNMu BS .97.3168

)7(75.51 2

).(

Assume S.I.B. No. 34:

From tables :

Ix = 15700 cm4

= 157*106 mm

4

)(10*157

))(2/340(*).(10*97.316.6

6

mm

mmmmN

I

YMCT

2/21.343 mmNCT

For steel 44, 2/160 mmNallowable

“We can take 2 S.I.B. with increading the depth somewhat”

Take 2 S.i.B No. 35

Ix=19610*104 mm

4

22

6

6

/160/1452*10*1.196

)2/360(*10*97.316mmNmmNC

T

“Safe”

Existing slab Grout

Design of Steel Beam : (Contacted) :

“In this procedure, the steel and concrete work together as a composite

sectrion”.

Assume 25IB “No 36”

Ix =2 x 19610*104 mm

4

A= 2*97.1*102 mm

2

More accurate method”

Assume (ST 52)

F = 210 N/mm2

366

10*51.1210

10*97.316mm

F

MZ req

=1.51*103cm

3

From SIB table 1.51*103/2=754.7cm

3

Take 2 SIB “32”.

Ix =2*12510*104=250.2*10

6mm

4

A=2*77.8*10

2=15.56*10

3mm

4

Callculate Beff = smaller of 16 ts + bo = 16*180+286=3160

or 14862865

6000

50 b

L

or (L – L) = 3000 mm

Beff = 1486 mm

Calculate N.A :-

No A Y Ay

A1 310*7.26

10

1486*180

450 12*106

A2 15.56*103

180 2.8*106

42.26*103 14.8*10

6

mmA

Ayy 2.350

10*26.42

10*8.143

6

Calculate I comp :

“n = Modular ratio”

I comp = I1 + I2

23

1 )2.350450(*10

180*148612/

10

)180(*)1486(

nnI

=72.2*106+266.4*10

6=338.6*10

6 mm

4

I2 =250.2*106+266.4*10

6+(15.56*10

3)(450-180)

2=1384.5*10

6 mm

4

I comp=338.6*106+1384.5*10

6=1723.1*10

6 mm

4

Calculate C T :-

copm

cc

I

MY Yc = 189.8 mm & YT = 350.2 mm

22

6

6

/5.9/4.310/10*1.1723

8.189*10*97.316mmnmmN

O.K.

NI

MY

comp

TT 2104.64

10*1.1723

2.350*10*97.3166

6

“safe”

Design of Shear connectors :

I

MYshshear

6

6

10*1.1723

8.9*10*97.316

=1.8 N/mm2

Qshear =1.8 KNNmm

N mmmm 5.47110*6.4711000*262* 3

2

“Assume using 4 dowels in cross section & assume using 4

dowels/m'”

Total No of dowels to cover this area = 4*4=16 dowel.

force / dowel = 471.6 KN/16 = 29.5 KN/dowe.

Use 16mm : MNA

Pdowel /6.146

)8(

10*5.29/

2

3

126210*6.06.0 fsall

“Not Safe”

Use 18mm : 126/9.115/ 2 mmNdowel

O.K.

Remark :

This No. of dowels was calculated at wide spam, it can be reduced left

& right.

Check the safety of conc. Beams :-

A shear=285.4*11.5*2

= 6564.2mm2

Shear of Shear of steel beam :

2

3

/8.152.6564

10*5.103mmNweb

q all = 0.35fs =0.35*210 =73.5N/mm2

3.152

6*75.51uQ 155.25 KN

Q working= 5.1035.1

Qu 155.25 KN

1.5*qall=110.25N/mm2

allweb Safe in shear.

For R.F.C Beam :-

4.1*)6.03(*25.0*153

mmmm

KNWu

+ (0.4*0.6*25KN)*1.4

Wu=126+8.4=21KN/m’

3*)2/3.155(8

2

WL

M u

MKN.45.3272

3*3.155

8

)6(21 2

Mu = 327.45 KN/m

Data from prev. Fcu = 25 N/mm2, FY = 360 N/mm

2

Assume s, s' > y

Nfy

AsTu

32 10*47615.1

360*

2

224

15.1*

Cu=0.45 (fc)*a*b=0.45*25*a*400

Nfy

AsCs

s

32 10*3.15915.1

360**

2

18*2)*(

C =T

4500 a +159.3*103 = 476*10

3

4500a=316.7*103 a=70.4mm

Cu=4500*70.4=316800N

mma

C 888.0

4.70

8.0

Check comp. & Tension RFT for strain :-

88

003.0

2588

'

s

0021.0' s

0018.010*200

3603

s

y

yE

f .

ys ' O.K

ys ' also

Calculate capaicity Moment :-

)2

600()25600(a

CCM uusCAP

)2

4.70600(316800)575(10*3.159 3

66 10*9.17810 CAPM

mmNMCAP .10*5.270 6

MCAP=270.5 KN.M Mu = 327.45

Unsafe

& need strengthening

Strengthening of Beam using

Concrete jacket

Assume that As1 & As2 are yielded:

C=0.45fcu*a*600=0.45*25*a*a600

C=6750 a

15.1

360*10*52.1

15.1* 3

11 fy

AsT

NT 3

2 10*65.951

Fx = 0 C = T1 + T2

6750 a =475.8*103+951.65*10

3

a = 211.47 mm

Check of yielding :

3.2644753.264

003.0 1

s

0024.01 s

0018.010*2

3605

E

f y

y ys 1

ys 2

Get M capacity :

capMTT )2

47.211675()

2

47.211475( 21

capMTT )27.569()27.369( 21

475.8*103(369.27)+951.65*10

3(569.27)=Mcap

Mcap=717.44*106

Mcap=717.44 KN.m > Mu

Safe

Check of Shear

Wu=O.W + wall = 27.3 KN/m

Wu= 4.1*)6.03(*025.0*3

15 mm

KN4.1*)

325*7.0*6.0(

m

KN

Wu=27.3KN/m

Qu=159.60KN

qu= db

Qu

.

Qu=2

2

3

/38.0700*600

)(10*60.159mmN

mm

N

qcu=0.24c

cuf

(ECC 4-18)

qcu=0.242/98.0

5.1

25mmN

qcu >qu “No need to use stirrups”

(If it needs more stirrups)

s

y

su

f

q

sb

Ast

. (4-22)

Where qsu = qu – 0.5 qcu (4-21)

“Tied Stirrups”

Strengthening of concrete beam using steel Beams

(Non-composite)

For R.C.B

Mu=327.45 & Mcap=270.5KN.

Mu – Mcap = 327.45 – 270.5 = 56.95 KN.m

1.2)5.3*6.0()6.0(

/210

10*95.56 6

yball

all

fff

mmNf

MZ

=271.19*103 mm

3

Z = 271.19 cm3

From S.I.B tables

Use No. 24 o.w. = 36.2 kg/m

= 362 N/m

= 0.362 kn/m O.K.

Check of shear of steel beam : Aweb = (240-(13.2*2))*8.7

Aweb = 1860mm2

1860

Qweb

Assume 8

2WlM

mKNw /7.1236

8*95.56

KNL

WQ 382

6*7.12

2*

23

/4.201860

10*38mmNweb

qall;;=0.35fy=0.35(350)=122.5N/mm2

allq O. K.

KNQ 382

6*7.12max

shear on dowels KN)33.6(6

38

Design of steel dowels :

Qmax=38KN

Shear on dowels KN)33.6(6

38

Shear stress on dowels2

2

3

/4.31

2

16

10*33.6mmN

qall (for bolts) = 0.27fy

= 0.27fy = 0.27(350)=94.5N/mm

qu < qall O.K.

Strengthening of slabs

using FRP systems

The following slabs [(ts=20cm) & Top R.F.T=5 10]

Is transformed into storage area with a (L.L=10KN/m2).

It is required to: check the capacity of the slab in both directions & to

calculate the ultimate moment of the slab in both directions and hence

strengthen the slab with FRP- Epoxy systems if it neds to the FRP –

has the following chacterstics :

Type FRP plies

Thickness

ultimate tensile strength

(N/mm2)

Rupture strain (mm/mm)

Modulus of Elasticity

(N/mm2)

0.13 (MM)

2000

0.01

2.1*105

Calculate Ultimate Moments due to proposed New load :

Wu=[(0.2*25+2)+10)]*1.5=25.5KN/m2

R = 1.4 from table (6-3) = 0.797 ; β = 0.203

(Wu + ) = 20.32 , (wu * β)=5.2

mKNw

M u

shortu .5.63

8

25*32.20

8

)5)(*( 2

mKNw

M u

longu .85.31

8

)49(2.5

8

)7)(*( 2

Calculate Moment capcity :

Short direction

0

xF

Cc=0.45*25*a*1000=11250 7.392*2

10*5'

2

As

2.12293215.1

'* fy

AsCs 8.549*2

10*7'

2

As

3.17211115.1

* fy

AsT

11250a + 122932.2=172111.3 a = 4.37mm

EM = 0

Cs*(150)+Cc* capM )2

37.4175(

122932.2(150)+11250(4.37)(172.8)=Mcap

Es=0.09>0.018 o.k. 5.63.94.26 MumKNMshort

cap

Long direction

11250a + 122932.2=122932.2 a = 0

122932.2(150)+=Mcap

short

long

cap MumKNM .44.18

It is needed to be strengthened in 2 directigons.

Slab is protected (Internal) & carbon Epoxy

Ce=0.95

ffu=2000*0.95=1900N/mm2

Efu=0.01*0.95=0.0095mm/mm

)/(4000 2mmNfcuEc (2-1)

= 4000 23 /10*2225 mmN

Calculate n’s :

1.910*22

10*23

5

Ec

Esns & 5.9

10*22

10*1.23

5

Ec

Efn f

For short Direction:

Z=N.A depth at workign leads

)200(**)5.2'*()1(2

*)(1000 ZAnzAsnz

z ffs

-ns*As*(175-z)=0

500Z2 + (8.1)392.7*(z – 2.5) – 9.5*(4x0.26x150)(200-z)

-9.1*549.8(175-z)=0

500Z2 + 3180.9z – 7952.2 – 296400 + 1482Z

-875556.5 + 5003.2Z = 0

500Z2 + 9666.1Z – 1.8*10

6 = 0

Z2 + 19.33Z – 2360 = 0

2

06.9985.17

2

)2360(4)33.19(33.19 2

Z Z = 40.60mm

Calculate Icr :

Icr= 1/3 (1000*40.6)3+(8.1)(392.7)(40.6-25)

2+9.1(549.8)(175-40.6)

2

+9.5(4*0.26*150)(200-40t).

Icr=22.3*106+0.774*10

6+90.4*10

6+37.66*10

6

Icr=151.13*106mm

4

Calculate MD.L:

Working dead load = (0.w+F.C)*x

=(2.0*25+2)*0.797=5.58KN/m

MD.L(working)=5.58(5)2/8=17.4KN.m

calculate bi: (Strain at the conc. Substrate at the time of FRP

instullation).

EEIcr

Myc

63

6

.

10*13.151*10*22

)6.40200(*10*4.17

*

)6.40200(

cr

LDbi

IEc

M

00083.0bi

calculate Km : (factor to account for debonding)

n Ef tf=2*2.1*105*0.13=54600 < 180000

9.0360000

160

1

ff

fu

m

tnE

EK

1.75 0.9 Km = 0.9

pages (25-4)

Calculate fe:

Assume C=40mm

Cch

bife 003.0

fe=0.011 km Efu

0.900*0.0095

fe=0.011 0.009

Calculate s&fs&ffe :

chcd

s bife

s = 0.0084

s

fsEs

fs = 2*10

5*0.0084

fs=1658.8N/mm2

Ef=ffe/fec ffe=Ef. fe = 2.1*05*0.009

Ffe=1890N/mm2

0

xF Es=0.0084>0.005

Ductile failure, c=1.5, s=1.15

0.45*25*a*1000+392.7*184/1.15=105300

4.1

1890*)150*13.0*4(

15.1

360*8.549

A=30.2 C=37.8MM

Iterate unitl Ci=Cf

C = 24.5mm

Check capacity :

C = 24.5mm, a=0.8c=19.6

fe =0.009, fs=fy=360 N/mm2

s’ =0.00008, fs'=-5.7 N/mm2

s = 0.00855 ffe=1890 Nmm2

s > 0.005 c =1.5 , s = 1.15, f = 1.4

0

xF

Cc=0.45*25*19.6*1000=220500

4.194615.1

7.5*7.392

15.1

''*'

fsAsCs

3.17211115.1

360*8.549

15.1*

fsAsTs

210600)4.1

1890)(150*26.0*4(

4.1*

fe

ff

fAT

0M

Ma

Csa

Ta

Ts f )2

25(')2

200()2

175(*

172111.3(165.2)+210600(190.2)+1946.4(15.2)=M

M=68.52KN.m < (Mu=63.5)

O.K. “Safe”

Remarks :

“Increasing (1) layer may increase capacity by 40%”.

“2 layers of FRP are 2.5 times the existing steel”.

Calculate stresses in R.F.T due to working (oads)

Ms=O.W+F.C.+L.L.= mKNMu

/175.1

5.25

5.1

Ms=63.5/1.5=42.33KN.m

Eszddz

fsAszdzdEAzdzdAsEs

EszdzdfEAEMsf

ffff

ffbi

ss

))(3

(''))(3/())(3/(

))(3/((.

556 10*2*)6.40175(*)3

6.40200(10*1.2*150*26.0*4*00083.010*33.2 A

25510*74.010*97.010*27.3

10*27.1121212

15

0.8*fy=0.8*360=288

fs=255<0.8fy O.K.

Calculate stresses in FRP under working loads :

fufbi

f

sssf fEEzd

zdf

Es

Eff 55.0,,

5

5

5

10*2*00083.06.40175

6.40200

10*2

10*1.24.316

=151.6<0.55(1900)

=151.6<1045 O.K.

“We can Repeat the same procedure for strengthening long side”.

Strengthening of R.F.C. columns

Concrete Jacket CFRP laminate

Example :-

A central column that carries a load of “3000” KN, the column is

short, the column . Carries 5 storey’s . The owner of the building is

inteding to increase No. of floars to 8 floars (Column height =3.0m).

It is required to :

a) Check the capacity and RFT details.

b) Strengthen the column with conc. Jacket.

c) Use CFRP “Sika” to strengthen it.

(use full confinement laminates with the following data):

Table (FRP-Sika) data :

Thickness (mm) 0.13

Ult. Tne. Strength (N/mm2) 2000

Repture strain 0.01

Modulus of Elasticity

(N/mm2) 2.1*1

Cover = 2.5cm = 25mm

Main R.F.T= 16 16

Stirrups 8 (spiral)

Stirrups pitch = 5cm = 50mm”

a) Check capacity of column:

0.35 fcu Ax + 0.67 fy Asc + 1.38 fyp Vsp

Pu = lesser of

0.4 Fcu Ac + .76 fy Asc

ECC ( 4 - 2 – b & 4 – 12 – c )

Where: Ax = core Area = 2

2

2375832

550mm

Fyp = 240 N / mm2 (stirrups), Ac =

2

2

3.2827432

600mm

Asc = 2

2

1.2012

16mm

Vsp = p

DkAsp. , Asp =

2

2

27.502

8mm

Δk = 550 mm

P = 50 mm

21.173750

)550)(27.50(mmVsp

Pu = 0.35 ( 25 ) ( 237583 ) + 0.67 ( 360 ) * 16 * 201.1 + 1.38 (240)

(1737.1) - ( 4 – 12 – b )

= 3430.3 KN

Pu = 0.4 ( 25 ) ( 282743.3 ) + 0.76 * ( 360 ) (201.1) * 16 = 2915.4 KN

(4 – 12 – c)

Take lesser value = 2915.4 KN < 3000 KN

Check Min % of stirrup ( < 0.25 Vc )

No of stirrup in meter = mm

mm

50

1000 = 20 stirrup

% st = 20*(50.27)*(2T(550/2)=1737206

0.25Vc3.7068581000*

2

550

100

25.02

b) Using Concrete Jacket :

Pu/floar floorKN /6005

3000

Pu (8 flowers)=4800KN

Pu(on conc. Jacket)=4800 – 2882.5 = 1917.5

1917.5 = 0.4(25) (AC) + 0.76(360) 2827)(100

25.1 AC

in spiral columns As=bigger of

2851)(100

20.1

2827100

1

AC

AC

AC*100

25.1)360(76.0)25(4.05.1917*103

AC = 142883.8mm2 = 1428.8cm

2 d’=73.6 = 75cm

43.1590*2

60

2

7522

.

jacketconcA

2

. 9.19)43.1590(100

25.1cmAs jacketconc

No of bars 16101.201

100*9.19

c) USING CFRP laminates;

According to FRP code (Eq. 4-20-b)

Pu=0.4fcuc Ac+0.76fyAsc where

25.15.2875.9125.2 11

cucu

cucucf

f

f

fff

(Eq (4-21))

full confinement & circular :

Use : )224.....(..........2

1 xf

EEMf

feff

nn

D

ntfM f )10*87.0(600

13.0**44 3

xf = 1.3 (4-3-2-3-1-1)

Efe=0.004 0.75 Efu

Efu=CE*Efu (3-5)

Internal & protected col. CE=0.95(carbor& Epox).

Efu=0.95*0.01=0.0095

Efe=0.0040.75(0.0095)

=0.0040.007

Ef=2.1*105N/mm

2

3.1*2

004.0*10*1.2*)10*87.0( 53

1

nf

nf 281.01

2.1)25

)281.0(5.7)

25

281.0(875.9125.2[25

nnfcuc

Assume:

if n = 1 fcuc = 27.31

if n = 2 fcuc = 29.48

if n = 3 fcuc = 31.51

if n = 4 fcuc = 33.44

if n = 6 fcuc = 37.14

Pu=0.4(fcuc*282743.3)+0.76*360*201.1*12

4800*103=113097.3 fcuc+660251.5

fcuc=36.60N/mm2<37.14

use 6 layer of CFRP.

(If geometry of column is periority we have to use CFRP; increase Jus

2cm in diameter).

(If the geometry is not essential conider the more economical solution.

Conc. Jacket or CFRP

use spirals (pitch = 50mm, 8).

Check 0.25% Vc = Vst.

990.22102

7002*)27.50(*20

STV

5.3976071000*)10*43.1590(100

25.0%25.0 2 cV

cst VV %25.0 O.K.

Design of Dowels : Loads to be carried by Dowel (Swhear loads) = 19/7.5KN

Use 12 dowel (use 6/m' of column)

Total No of dowels = 4*6*3=72

force/dowel .63.2672

5.1917KN

fy=360N/mm2 fall=0.35(360)=126N/mm

2

alldowel fmmNf

2

2/235

2

12

1000*63.26

use (6 dowel in each section).

No of dowels =8*6*3=144

force/dowel KN3.13144

5.1917

allu ff

6.117

2

12

10*3.132

3

O.K.