rc pier taba

7/30/2019 RC Pier Taba

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RC PIER

NOTE:- DO NOT CHANGE VALUES IN BLACK

1) DESIGN DATA AND SPECIFICATION

1.1.MATERIAL PROPERTIES:

Concrete :- Grade C-30 concrete ( section 9.3)

fc'= = 24 MPa fc' cylinder )

fc=0.4*fc' = 9.6 MPa

Ec=4800sqrt(fc') 23,515 MPa

Reinforcement steel:

Grade 420 steel: For rebars diam. 20mm and above

fy = 420 MPa

fs = 165 MPa

Es = 200,000 MPa

Grade 300 steel: For rebars less than diam. 20

fy = 300 MPa

fs = 140 MPa

Es = 200,000 MPa

Modular rati Ec / Es = 8.51 Use n = 9

Live Loading: (1) Design Truck : AASHTO HS20 - 44 live load+ 25% increment

(2) Design Tandem

Bearing Capacity(s)= 3.5 kg/cm2

Allowable Bearing Capacity (1.5*s)= 5.25 kg/cm2

1.2. REFERENCES: -ERA BRIDGE DESIGN MANUAL 2002

-AASHTO STANDARD SPECIFICATION FOR HIGHWAY BRIDGES,1998 EDITION.

1.3.DESIGN METHOD: LRFD

2) LOADING

2.1. Dead Loads2.1.1. From Superstructure

CLN

exterior gi rder interior girder

LEFT GIRDER SUPSTR. RIGHT GIRDER SUPSTR.12 mtr. span 12 mtr. span

Exterior Interior Exterior Interior

X-sectional

(kN/m) 27.86 20.94 27.86 20.94

Diaphrams-

middle(No.) 1 1 1 1

(kN) 4.86 4.86 4.86 4.86 Final level = 2410.76 mends(kN) 2.43 2.43 2.43 2.43 Found. level 2398.12 m

Support Wearing surf. 0.10 mReaction(kN) 178.99 135.74 178.99 135.74 Girder depth 0.95 m

REINFORCED CONCRETE COULMN PIER DESIGN FOR FIVE GIRDER

SUPERSTRUCTURE


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RC PIER

2.1.2. Selfweight

0.25

L F

0.5

W

A

G

W

H

C E

D

B J

A B C D E F

1.00 9.00 0.00 1.00 9.59 1.00

G H J L W

1.1 0.00 3.00 9.35 1.00

Pier Cap w1(kN/m)=(A*L*F+2*0.25*0.50*F)*24/L= 24.64

Bracing w2(kN/m)=C*H*24= 0

Pier Column P(kN)=P/4*W2*E*24= 180.77

Summary of Dead Loads

178.99 135.74 135.74 135.74 178.99 (Left Supstr.)

178.99 135.74 135.74 135.74 178.99 (Right Supstr.)

c a a a a c

24.64

where

aLt.= 2.00 m c= 0.675 m

aRt.= 2.00 m c= 0.675 m

0

Enter

values for

dimensions


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RC PIER

2.2 Live Loads

a) Design Truck Load : HS 20-44 + 25% increment

P/4 P P P P14ft

(4.267m)

P = wheel load = 1.25(71.20KN)= 89 KN

LONGITUDINAL ARRANGEMENT TRANSVERSE ARRANGEMENT

b) Design Tandem

P P P P

1.20m 1.80m

P = wheel load = 1/2*110KN = 55 KN

LONGITUDINAL ARRANGEMENT TRANSVERSE ARRANGEMENT

2.2.1 Dynamic Load Allowance

Section 3.13, the vehicular dynamic load allowance IM

IM = 33% There fore IM 33%

The live loads shall be factored by 1+IM/100 = 1.33

A) Longitudinal Arrangement

case 1: Maximum Axial Load on pier

P/4 P P

4.267

x x= 0.46Rp2 Rp1

Rp1= 1.659 P

Rp2= 0.174 P

Axial Load: Rp=Rp1+Rp2= 1.833 P

Moment about CLN of pier: MCL= 0.341 P

case 2: Maximum Bending Moment

P P P/4

4.267

Rp1

Rp1= 1.738 P

Axial Load: Rp=Rp1= 1.738 P


2.2.2 Transverse Load Distribution

In designing sidewalks, slabs and supporting members, a wheel load located on the sidewalk shall be 1 foot from the face of the rail.

Distribution Factor for Shear (Sec. 13.4: Table 13-7 & 13-8)

Exterior Girder:

Case-1: One Design lane loaded

The lever rule is applied assuming that the slab is simply supported over the longitudinal beams (Table 13-8)

14 - 30ft

(4.267- 9.144m)

6ft

(1.80m)

0.16

50

0.2

0

0.2

0

0.25

0.20

0.

40

0.9

0

0.15

d1

1.830.30

d2

0.2

5


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RC PIER

P P TRWW= 10.3 m

1.83 SW= 1.5 m

a= 2.00 m

d1= 0.600 m

RA RB d2= 0.770 md1 a - d1 d2 c= 1.15 m

no.of girders= 5

The distribution coefficient to the exterior girder for shear bw= 0.38 mREX1 (shear) = 1/a*P*(a+d1+d2) 1.685 P

Case-2: Two or more design lanes loaded

The distribution of live load per lane for shear in exterior girder is determined according to the formulas given in Table 13-8.

REX2 (shear) = (0.60+de/3 0.920 per lane de=c-bw/2= 0.96 m

This factor is for one lane load which is equivalent to two lines of wheels, and thus multiplied by 2

REX2 (shear) = (0.60+de/3 1.840 P

There fore, REX (shear) in exterior girder is maximum of the above two values, REX1 or REX2

REX (shear) = 1.840 P

Interior Girder:

Case-1: One Design lane loaded

The distribution of live load per lane for shear in interior girder is determined according to the formulas given in Table 13-7.

RINT 1 (shear) = 0.36 + 0.623 where 1100


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RC PIER

b)Design Tandem

A) Longitudinal Arrangement

case 1: Maximum Axial Load on pier

P P

1.2

x x= 0.46

Rp2 Rp1

Rp1= 0.941 P

Rp2= 1.000 P

Axial Load: Rp=Rp1+Rp2= 1.941 P


Moment about CLN of pier: MCL= 1.212 kN

case 2: Maximum Bending Moment

P P1.2

Rp1

Rp1= 1.904 P

Axial Load: Rp=Rp1= 1.904 P


Moment about CLN of pier: MCL= 38.97 kN

B) Transverse Arrangement (Distribution factors are same as Truck Load)

Case-I: One Design lane loaded

a) For maximum Axial load case (from longitudinal arrangement case 1)RA=RP/2*RA= 89.93 kN

RB=RP/2*RB= 65.26 kN

b) For maximum Moment case (from longitudinal arrangement case 2)

RA=RP/2*RA= 88.23 kN


Case-II: Two or more design lanes are loaded

a) For maximum Axial load case (from longitudinal arrangement case 1)



RC=RP/2*RC= 76.92 kN

RD=RP/2*RD= 76.92 kNRE=RP/2*RE= 98.20 kN

b) For maximum Moment case (from longitudinal arrangement case 2)



RC=RP/2*RD= 75.46 kN

RD=RP/2*RD= 75.46 kN

RE=RP/2*RE= 96.34 kN

2.3. Wind LoadsWind Load on structure - WS

2.3.1. Wind Load on superstructure - W

a) Transverse Direction W=50lb/ft2

= 2.44 kN/m2

Arm (m) AiYi (m )

Girder web & Curb = 21.84 2.68 58.53

Railing = 3.9 3.38 13.18

Posts = 1.32 2.96 3.90

27.06 75.61

FWT= 66.03 kN

Line of Action (about Pier Cap bottom) = 2.79 m

Area (m2)


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RC PIER

2.3.2. Wind Load on Live Load - WL

a) Transverse Direction: WL=100lb/ft= 1.49 kN/m

Length of exposed surface= 12.8 m

FWLT= 19.072 kN (6ft above the deck surface)

(about pier cap bottom) = 4.51 m

b) Longitudinal Direction: WL=40lb/ft = 0.596 kN/m

Length of exposed surface= 12.8 m

FWLL= 7.629 kN (6ft above the deck surface)


2.3.3. Wind Load on substructure

a) Transverse Direction: W=40lb/ft2= 1.955 kN/m

2

per linear meter = 1.955 kN/m

b) Longitudinal Direction: W=40lb/ft

Pier cap = 1.955 kN/mBracing = 0 kN/m

Columns = 1.955 kN/m

2.3.4. Forces of Stream Current,WA v = 5.6 ft/sec2, k=2/3

P = kv2= 20.907 lb/ft

2= 1.02 kN/m

2

per linear meter = 1.02 kN/m

2.3.5. Breaking/Longitudinal Force,BR

Taken 5% of the live load in all lanes

(lane load w=9.3kN/m plus the concentrated load P=81.72kN for moment)

LF= 20.26 kN (6ft above the deck surface)


2.3.6 Seismic Force Effects,EQ

Earthquake zones: EBCS Zone -1

Site Coefficient: Type I = 1

Acceleration coefficient(A): = 0.06

The horizontal seismic force is the product of the site coefficient, the acceleration coefficient and the permanent loads

772.70 KN

Footing = 648.00 KN

Sum Wp 1420.70 KN

Horizontal earthquake force FH = site coeff.*A* Wp = 85.24 KN

This force is transferred to the substructure at jointsThe proportion of this load at the two levels is as in the following:

At bracing level = 0.00 KN

At pier cap level = 85.24 KN

3) STABILITY ANALYSIS

Assume a combined footing with dimensions(m):

Width W = 3 Length L = 9 Depth D = 1

3.1. CHECK FOR STRENGTHI =DL+(LL+I)+BR+WA


i) Dead Load

Superstructure

Left = 765.18 kN

Right = 765.18 kN

1530.35

Substructure

Pi

Pier cap,

Bracing & Columns =


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RC PIER

Case I: a) RA+RB= 478.08 aavg*RA = 549.62 MCL = 60.79

b) RA+RB= 453.41 " 521.27 " 71.15

Case II: a)RA+B+C+D+E 1305.34 a*(RA-RE)= 0.00 2*MCL = 121.57

b)RA+B+C+D+E 1238.00 " 0.00 " 142.30

ii) Breaking/Longitudinal Force, BR

FLF= 20.26 kN MLF= 305.96 kNm/m

iii)Stream current Force,WA

FTF= 1.83 kN MTF= 1.64 kNm/m

A) Stability against OVERTURNING:

MREST.= 4426.58 kN/mMDRIV.= MLL+MBR+MWA= 828.87 kN/m

S.F.= 5.34 > 2.0 OK!

B) Stability against BEARING PRESSURE:

case (a) case (b) case (a) case (b)

Ptot(kN)= 3429.13 3404.46 4256.40 4189.05

MT(kN/m)= 549.62 521.27 0.00 0.00

ML(kN/m)= 60.79 71.15 121.57 142.30

eT=MT/Ptot= 0.160 0.153 0.000 0.000

eL=ML/Ptot= 0.018 0.021 0.029 0.034

145.08 144.23 166.65 165.69< sall OK!

C) Stability against SLIDING:

FTDRIV=FTWA= 1.83

FLDRIV=FLBR= 20.26

FDRIV=(FTDRIV +FLDRIV )= 20.34

FRESIST=SVtan f=SV*0.7= 2065.74

S.F.=FREST./FDRIV.= 101.54 > 1.5 OK!



MREST.= 3454.81 kNm/m

MDRIV.= MBR+MWA= 307.60 kNm/m

S.F.= 11.23 > 2.0 OK!

3.2. CHECK FOR STRENGTHIII =DL+WS+WA

a) Dead Load DL

PDL= 2951.05 kN

MTDL= 0 kN/m

MLDL= 0.00 kN/m

c) Wind Load on Structures W

superstructure

FTW= 66.03 kN MTW= 883.72 kN/m

FLW= 15.86 kN MLW= 212.24 "

smax (kN/m2) =(Ptot/A)*

(1+6*eT/L+6*eL/W)=

Case I: Case II:


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RC PIER

FTWL= 19.07 kN MTWL= 287.99 kN/m

FLWL= 7.63 kN MLWL= 115.19 "

e)Stream current Force,WA




MDRIV.= MLL+MWS+MWA= 633.89 kNm/m

S.F.= 6.98 > 2.0 OK!


Ptot=PDL= 2951.05

MT=MTDL+MTWS+MTWA= 1102.66

ML=MLDL+MLWS+MLWA= 632.25

eT=MT/Ptot= 0.374 OK!

eL=ML/Ptot= 0.214 OK!

smax= 183.36

sall OK!C) Stability against SLIDING:

FTDRIV=FTWS+FTWA= 105.35FLDRIV=FLWS= 71.63

FDRIV=(FTDRIV2+FLDRIV

2)= 127.40


S.F.=FREST./FDRIV.= 16.21 > 1.5 OK!

3.3. CHECK FOR STRENGTHV =DL+(LL+I)+BR+WS+WL+WA


a) Dead Load DL

PDL= 2951.05 kN

MTDL= 0 kN/m

MLDL= 0.00 kN/m

b) Live Load LL

PLLI MTLL MLLL

Case I: 478.08 549.62 60.79

453.41 521.27 71.15

Case II: 1305.34 0.00 121.57

1238.00 0.00 142.30

c) Wind Load on Structures WS

superstructureFTW= 37.50 kN MTW= 883.72 kN/m

FLW= 15.86 kN MLW= 212.24 "

substructure

FTW= 37.50 kN MTW= 217.29 kN/m

FLW= 55.78 kN MLW= 420.01 "

Total


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RC PIER

f)Stream current Force,WA




MDRIV.= MLL+MWS+MWA= 1352.17 kNm/m

S.F.= 3.27 > 2.0 OK!


(b)case I: (b)case II:

Ptot=PDL+PLLI= 3404.46 4189.05

MT=MTDL+MTLL+MTBR+MTWL+MTWS+MTWA= 2199.90 1678.63

ML=MLDL+MLLL+MLBR+MLWL+MLWS+MLWA= 818.60 774.55

eT=MT/Ptot= 0.646 0.401 OK!

eL=ML/Ptot= 0.240 0.185 OK!

smax= 241.05 253.97

> sall OK!C) Stability against SLIDING:

FTDRIV=FTBR+FTWL+FTWS+FTWA= 143.50

FLDRIV=FLWS+FLWL= 79.26


2)= 163.93






MDRIV.= MBR+MWA= 1352.17 kNm/m

S.F.= 3.27 > 2.0 OK!


FTDRIV=FTBR+FTWL+FTWS+FTWA= 143.50

FLDRIV=FLWS+FLWL= 79.26


2)= 163.93



3.4. CHECK FOR EXTREME EVENTI =DL+(LL+I)+WA+EQ


a) Dead Load DL

PDL= 2951.05 kN

MTDL= 0 kN/m

MLDL= 0.00 kN/m


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RC PIER

At bracing level = FTF= 0.00 kN MTF= 0.00 kNm/m

At pier cap level = FTF= 85.24 kN MTF= 817.47 kNm/m

At bracing level = FLF= 0.00 kN MLF= 0.00 kNm/m

At pier cap level = FLF= 85.24 kN MLF= 817.47 kNm/m



MDRIV.= MLL+MEQ+MWA= 1340.38 kNm/m

S.F.= 3.30 > 2.0 OK!


(b)case I: (b)case II:

Ptot=PDL+PLLI= 3404.46 4189.05

MT=MTDL+MTLL+MTEQ+MTWA= 1340.38 819.12

ML=MLDL+MLLL+MLEQ+MLWA= 888.62 959.78

eT=MT/Ptot= 0.394 0.196 OK!

eL=ML/Ptot= 0.261 0.229 OK!

smax= 247.32 239.52

> sall OK!C) Stability against SLIDING:

FTDRIV=FTEQ+FTWA= 87.07

FLDRIV=FLEQ+FLWA= 85.24







MDRIV.= MLL+MEQ+MWA= 819.12 kNm/mS.F.= 5.40 > 2.0 OK!


FTDRIV=FTEQ+FTWA= 87.07

FLDRIV=FLEQ+FLWA= 85.24





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RC PIER


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RC PIER LOAD

SUMMARY OF LOADS

DL(Lt.Sup.) 178.99 135.74 135.74 135.74 178.99

DL(Rt.Sup.) 178.99 135.74 135.74 135.74 178.99

LL(HL-93) FLLRt FLLCt FLLCt FLLCt FLLLt FWLt= 19.07 FWLl= 7.63 FLF=

FWt= 66.03 FWl= 15.86

0.675 a a a= 2 a

24.00 2.79

FEQT= 85.24 MLONGT FEQL= 85.24

MTRANS

2.73 2.73

5.295

0

FEQT= 0.00 FEQL= 0.00

FWl= 1.96

0.00 (bracing)

1.96 (pier cap)FWt= 1.96 FWt= 1.96 4.80

FWAt= 1.02 FWAt= 1.02

1.6 6.15 1.6

HL-93/ case I: a) 274.81 89.93 203.26 65.26 0.00 0.00 549.62 179.86 60.79 1.21

Design

Tandem b) 260.63 88.23 192.78 65.26 0.00 0.00 521.27 176.45 71.15 38.97

case II: a) 300.09 98.20 235.05 76.92 235.05 76.92 0.00 0.00 121.57 2.42

b) 284.61 96.34 222.93 75.46 222.93 75.46 0.00 0.00 142.30 77.95

MLONGT. (kN/m)FRt(kN) FCt (kN) FLt (kN) MTRANS.(kN/m)

CDSCo-CWCE


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Pier section capacity analysis

Ultimate facored loads from load combinations(from STAAD)

Mux= 361.51

Muy= 119.38

Nominal moment strength of the section in the direction of the x- and y- axis(from STAAD)

Mnx= 8253.79

Mny= 1179.11

= 0.65

Mux + Muy < 1.0

Mnx Mny

0.223 < 1.0 OK!


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CONTENTS

STABILITY OF THE WHOLE STRUCTURE

LOADING SUMMARY

ANALYSIS RESULT

BEAM & COLUMN SECTION DESIGN RESULT

FOOTING DESIGN


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RC PIER

RC FOOTING DESIGN FOR COLUMN PIER

Loading Case Summary

Group Fy(kN) Mx(kNm) Mz(kNm) Fy(kN) Mx(kNm) Mz(kNm) Fy(kN) Mx(kN Mz(kNm)

I 1620 67.55 97.65 1030 69.42 34.08 1550 67.6 -36.38

II 1690 67.55 96.59 1130 69.42 35.49 1640 67.6 -34.65

III 1030 344.71 13.4 660.74 355.76 -15.17 1000 345 -80.75

V 1710 111.61 53.72 1080 119.64 -17.32 1670 112 -99.62E.EVE. I 1100 351.76 -73.18 668.84 361.51 -119.37 922.1 352 -156.99

Bearing Capacity = 3.5 kg/cm2

= 350 kN/m2

Allowable Bearing Capacity = 5.25 kg/cm2

= 525 kN/m2

Design ConstantsConcrete :- Grade C-30 concrete ( section 9.3)

f'c= = 24.00 Mpa

fc= (0.4*f'c) = 9.60 "


Grade 420 steel: For rebars diam. 20mm and above Mpa

fy = 420 MPa "

fs = 165 MPa

Es = 200,000 MPa


fy = 300 MPa

fs = 140 MPa

Es = 200,000 MPa kN/m2

FyR x R FyL

6.15

3.075y

f(m)= 1.00 D=x 1

B C

x

W=

z 3.0

A D

B= 9.00

Self Weight W (kN)= 648

Location of Resultant R & Eccentricity e

Group I II III V E.EVE. I

x(m)= 3.046 3.062 3.010 3.033 2.742

ex(m)= 0.029 0.013 0.065 0.042 0.333 < B/6 OK!

ez(m)= 0.049 0.046 0.388 0.077 0.396 < W/6 OK!

smax= 200.460 208.129 225.077 223.538 249.022 < sall OK!

Design Load = 249.02 kN/m2

ex= 0.013 m ez= 0.046 m V=Fy1+Fy2+Fy3+W= 5108 kN

sA=V/A(1-6*ex/B+6*ez/W)= 204.95 kN/msB=V/A(1-6*ex/B-6*ez/W)= 170.14 "

sC=V/A(1+6*ex/B-6*ez/W)= 173.42 "

sD=V/A(1+6*ex/B+6*ez/W)= 208.23 "

Depth Determination

1. Punching Shear (Art. 8.15.5.6.2)

Right leg Center Leg Left Leg


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RC PIER

2. Wide Beam Shear

a. In the Short Direction

s1=P/A(1 + 6*ez/W)= 149.99 kN/m ds2=P/A(1 - 6*ez/W)= 180.38 "

trial d= 0.934 m

s3= 150.66 s1 s3 s2

Vact= 89.29 kN

bc -ratio of long side to short side of concentrated load= 3.0

0.166(1+2/bc)= 0.277 < 0.332 take 0.277

vcb=0.166(1+2/bc)f'c0.332f'c= 1.36 N/mm 1355.38 kN/m

Vres=vcb*B*d= 11393.36 kN Vres Vact OK!

b. In the Long Direction

FyR= 1690 FyC= 1130 FyL= 1640

1.425 MzR= 96.59 MzC= 35.49 MzL= 34.65

q2 q3 q5

q4 q1

q1=s1*W=P/A(1 + 6*ex/B)*W= 499.85 kN/m q3= 492.62 kN/m

q2=s2*W=P/A(1 - 6*ex/B)*W= 491.26 " q4= 498.4903

q5= 495.5556

a= 860 g= 600

b= 697 h= 144

c= 707 i= 338

d= 707 j= 149

e= 666 k= 538

f= 828

Vmax= 860.00 kN corresponding triangle leg = 1.79 (from STAAD)

SFD & BMD(max) Diagram ordinates(from STAAD)


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RC PIER

3. Flexure

a. In the Short Direction

Bottom :-

Using F = 32 mm andClear cover = 50 mm

davil=D-F/2-CLR CR= 0.934 m

beff=coln width+2(1/2*d)= 1.93 m

Loading

FyR= 1690 MxR= 67.55 FyC= 1640 MxC= 69.42 FyL= 1640 MxL= 67.55

Right Center Left

ez(m)= 0.0400 0.0423 0.0412 1

smax=Fy/beff*W(1 + 6*ez/W)= 314.56 306.59 305.95smin=Fy/beff*W(1 - 6*ez/W)= 267.99 10.95 259.38

qmax=smax*beff= 608.37 kN/m qmax q1= 578.34 qminqmin=smin*beff= 518.30 " 608.37 518.30

Mx-x= 299.18 kNm

The moment for design is therefore,

Mfact. = 299.18 kNm/m

The cracking moment given by :

Ig =bh3/12 = 2.5E+11 mm

4

Mcr =frIg/yt = 1543.18 KNm/m yt= 500 mm

fr = 0.63*sqrt(fc') = 3.086357076 MPa

The amount of reinforcement shall be adequate to develop a factored flexural resistance at least equal to the

lesser of :

- 1.2 times the cracking strength determined on the basis of elastic stress distribution and the

modulus of rupture, fr, of the concrete.

Besides the area of reinforcement shall not be less than the minimum shrinkage and temprature reinforcement.

Reinforcement

Mu = 1851.81 KNm/m

Design moment, Mu = 1851.81 KNm/m

As = Mu / ( fy (d - a/2 ) ) where Mu= 1851.81 KNm/m

a = As*fy / ( 0.85 * fc' b ) = 0.90

b= 3000 mm

Assume a 36 mm fy= 420.00 N/mm2

As = Mu / ( fy ( 5,302 mm2 fc'= 24.00 N/mm2

a = As*fy / ( 0.8 36 mm D= 1000 mm

diam= 16 mm

cover = 50 mm

d= 942 mm

Required As = 5302 mm2/m

Spacing = 114 mmUse F 16 mm @ 114 mm for width= beff

Provide As = 5,301.9 mm2/m

According to section 5.10.8, the area of reinforcement for walls and footings in each direction shall not be less than :

As = 0.0015Ag

Use As,min=0.0015*b*d= 4239 mm /m S = 142.29 mm

Use F 16 mm @ 142 mm for the rest


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RC PIER

b. In the Long Direction

Bottom :-

Mmax = 600.00 kNm/m


Ig =bh3/12 = 2.5E+11 mm

4


fr = 0.63*sqrt(fc') = 3.086357076 MPa

The amount of reinforcement shall be adequate to develop a factored flexural resistance at least equal to the

lesser of :




ReinforcementMu = 1851.81 KNm/m



a = As*fy / ( 0.85 * fc' b ) = 0.90

b= 3000 mm


As = Mu / ( fy ( 5,313 mm2 fc'= 24.00 N/mm2

a = As*fy / ( 0.8 36 mm D= 1000 mm

diam= 20 mm

cover = 50 mm

d= 940 mm


Spacing = 177 mmUse F 20 mm 177 mm for width= beff

Provide As = 5,313.4 mm2/m

According to section 5.10.8, the area of reinforcement for walls and footings in each direction shall not be less than :

As = 0.0015Ag

Use As,min=0.0015*b*d= 4230 mm /m S = 222.81 mm

Use F 20 mm @ 222 mm

Use F 20 mm @ 177 mmProvide As = 5,313.4 mm2/m

rmax = 0.75*rb = 0.75*0.85*b1*(fc'/fy)*600/(600+fy) b1= 0.85= 0.0182 fc' = 24 N/mm2

rprovided = As/bd = 0.0019


19/20

RC PIER PIER CAP DESIGN-19 of 20

1) DESIGN DATA

1.1.MATERIAL PROPERTIES:

Concrete :- Grade C-30 concrete ( section 9.3)

fc'= = 24 MPa fc' cylinder )

fc=0.4*fc' = 9.6 MPa

Ec=4800sqrt(fc') 23,515 MPa


Grade 420 steel: For rebars diam. 20mm and abovefy = 420 MPa

fs = 165 MPa

Es = 200,000 MPa


fy = 300 MPa

fs = 140 MPa

Es = 200,000 MPa

Modular ra Ec / Es = 8.51 Use n = 9

2) DESIGN FOR FLEXURE

Design Loads:

(from STAAD)

Mu(max-ve)= 772.94 kNm

Mu(max+ve)= 25.69 kNm

Pier Cap Section:

b= 1000 h

h= 1000

b


Ig =bh3/12 = 83,333,333,333.33 mm

4


fr = 0.63*sqrt(fc') = 3.086357076 MPaThe amount of reinforcement shall be adequate to develop a factored flexural resistance at least equal to the

lesser of :




Reinforcement

Mu = 772.94 KNm/m



a = As*fy / ( 0.85 * fc' b ) = 0.90b= 1000 mm


As = Mu / ( fy 2,205 mm2 fc'= 24.00 N/mm2

a = As*fy / ( 0. 45 mm D= 1000 mm

diam= 20 mm

cover = 40 mm

d= 950 mm


Use 8 F 20 both for top and bottom

rmax = 0.75*rb 0.75*0.85*b1*(fc'/fy)*600/(600+fy) b1= 0.85= 0.0182 fc' = 24 N/mm2

rprovided =As/bd = 0.0023


20/20

RC PIER PIER CAP DESIGN-20 of 20

3) DESIGN FOR SHEAR

Shear strength

Design of cross sections subject to shear shall be based on

Vu < = Vn = (Vc + Vs ) = 0.900

where Vu = factored shear forces at the section

Vn = the nominal shear strength, determined as the lesser of Vn = Vc + Vs or Vn=0.25*fc'*bv*dv

Vc = the nominal shear strength provided by the concrete, determined by:

Vc = 0.083*b*SQRT(fc')*bv*dv, where b= 2.00

Vs = the nominal shear strength provided by the shear reinforcement

Vs = Av*fv*d / s

Shear strength provided by concrete

Shear stress provided by concrete

vc = 0.083*b*sqrt(fc') = 0.813 (MPa )

The shear strength carried by concrete,Vc=vc*bw*d= 0.813 bw*d (N)

Shear strength provided by shear reinforcement

Where the factored shear force, Vu, exceeds shear strength Vc, shear reinforcement shall be provided to satisfy

the equation, Vu < = Vn = ( Vc + Vs )

When shear reinforcement perpendicular to the axis of the member is used

Vs = Vu/ - Vc = Av*fy*d / s

There fore, spacing of shear reinforcement, s, is :

S = Av*fy*d / Vs = Av*fy*d/(Vu/-Vc)=Av*fy*d/(Vu/ - 0.813*bw*d )Minimum shear reinforcement ( Eq.12.34 )

Av = 0.083*sqrt(fc')*bw*S / fy fc' = 24 MPa

Smax = Av*fy / ( 0.083*sqrt( fc ')*bw ) fy = 300 MPa

Maximum spacing of shear reinforcement Av =diam.12 226 mm2

If Vu < 0.10*fc' * bw * d, S

rc pier taba

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