strengthening and maintenance of reinforced concert...
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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.
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