rc pier-f(box & girder).xls

7/28/2019 RC pier-f(Box & Girder).xls

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SUR RIVER BRIDGE RC PIER STABILITY-1 of 36

NOTE:- DO NOT CHANGE VALUES IN BLA CK

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 MPafs = 140 MPa

Es = 200,000 MPa

Modular ratio 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)= 5.0 kg/cm

Allowable Bearing Capacity (1.25*s)= 6.25 kg/cm2

1.2. REFERENCES: -ERA BRIDGE DESIGN MANUAL 2002

-AASHTO STANDARD SPECIFICATION FOR HIGHWAY BRIDGES,199

1.3.DESIGN METHOD: LRFD

2) LOADING

2.1. Dead Loads

2.1.1. From Superstructure

CLN

exterior girder interior girder

LEFT GIRDER SUPSTR.

RIGHT BOXGIRDER

SUPSTR.20 mtr. span 40 mtr. span

Exterior Interior Exterior Interior

X-sectional

(kN/m) 31.85 30.98 37.22 34.94

Diaphrams-

middle(No.) 2 2 0 0

REINFORCED CONCRETE COULMN PIER DESIGN FOR FOUR GIRDER

SUPERSTRUCTURE

TRANSPORT CONSTRUCTION DESIGN S. Co. BRIDGE AND STRUCTURES DIVISION


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(kN) 3.26 6.52 0.00 0.00

ends(kN) 1.63 3.26 2.35 4.70

SupportReaction(kN) 331.35 327.33 756.06 712.24

2.1.2. Selfweight

0.2

L F

0.4

W

A

G

W

H

C E

D

B J

A B C D E F

1.00 7.00 0.80 0.80 11.62 1.10

G H J L W0.9 0.60 3.50 7.40 1.00

Pier Cap w1(kN/m)=(A*L*F+2*0.20*0.40*F)*24/L= 17.07

Bracing w2(kN/m)=C*H*24= 11.52

Pier Column P(kN)=P/4*W2*E*24= 219.01

Summary of Dead Loads

331.35 327.33 327.33 331.35 (Left Supstr.)

756.06 712.24 712.24 756.06 (Right Supstr.)

c a a a c

17.07

Enter

values for

dimensions



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where

aLt.= 2.2 m c= 0.400

aRt.= 2.21 m c= 0.385

11.52

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*11 55 KN

LONGITUDINAL ARRANGEMENT TRANSVERSE ARRANGEMENT

2.2.1 Dynamic Load Allowance

Section 3.13, the vehicular dynamic load allowance IM

IM = 33% Therefore 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= 0.5

Rp2 Rp1

14 - 30ft

(4.267- 9.144m)

6ft

(1.80m)

Longitudinal Moment due to

Unbalanced Superstructure



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Rp1= 1.895 P

Rp2= 0.204 P

Axial Load: Rp=Rp1+Rp2= 2.099 P

Moment about CLN of pier: MCL= 0.423 P

case 2: Maximum Bending Moment

P P P/4

4.267

Rp1

Rp1= 2.092 P

Axial Load: Rp=Rp1= 2.092 P


2.2.2 Transverse Load Distribution For T- Girder

In designing sidewalks, slabs and supporting members, a wheel load located on the sidewalk shall be 1 fo

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

P P TRWW= 8.92 m

1.8 SW= 0.8 m

a= 2.20 m

d1= 0.055 m

RE RI d2= 0.455 md1 a - d1 d2 c= 1.16 m

no.of girders= 4

The distribution coefficient to the exterior girder for shear bw= 0.47 m

REX1 (shear) = 1/a*P*(a+ 1.232 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 giv



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REX2 (shear) = (0.60+de/3 0.908 per lane de=c-bw/2= 0.925 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 1.817 P

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

REX (shear) = 1.817 P

Interior Girder:


The distribution of live load per lane for shear in interior girder is determined according to the formulas giv

RINT 1 (shear) = 0.36 + 0.649 where 1100


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

RA=RP*RA= 2.577 P = 229.35 kN

RB=RP*RB= 2.717 P = 241.84 kN

Case-II: Two or more design lanes are loaded

RA= 1.817 P RB= 1.538 P RC= 1.538 P RD=

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

RA=RP*RA= 3.813 P = 339.33 kN

RB=RP*RB= 3.227 P = 287.21 kN

RC=RP*RC= 3.227 P = 287.21 kN

RD=RP*RD= 3.813 P = 339.33 kN


RA

=RP*R

A=

3.800 P = 338.24 kNRB=RP*RB= 3.217 P = 286.29 kN

RC=RP*RC= 3.217 P = 286.29 kN

RD=RP*RD= 3.800 P = 338.24 kN

2.2.3 Transverse Load Distribution (Box Girder)

In designing sidewalks, slabs and supporting members, a wheel load located on the sidewalk shall be 1 fo

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

Exterior Girder:


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

P P



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1.8 a'= 1.8 m

a= 2.21 m

d1= 0.345 m

RE RI d2= 0.755 m

d1 a' - d1 d2

The distribution coefficient to the exterior girder for shear

REX1 (shear) = 1/a*m*P*( 1.797 P m= 1.2

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 giv

REX2 (shear) = (0.64+de/3 0.683 per lane de=c-bw/2= 1.02 m

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

REX2 (shear) = (0.64+de/3 1.366 P

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

REX (shear) = 1.797 P

Interior Girder:


The distribution of live load per lane for shear in interior girder is determined according to the formulas giv

RINT 1 (shear) = (S/2900)0.6 0.636 where 1800


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Interior Girder

Case-1: One Design lane loaded where 2100


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RA= 1.797 P RB= 1.124 P RC= 1.124 P RD=

a) For maximum Axial load case (from longitudinal arrangement case 1)RA=RP*RA= 3.772 P = 335.71 kN

RB=RP*RB= 2.359 P = 209.95 kN

RC=RP*RC= 2.359 P = 209.95 kN

RD=RP*RD= 3.772 P = 335.71 kN


RA=RP*RA= 3.760 P = 334.63 kN

RB=RP*RB= 2.351 P = 209.27 kN

RC=RP*RC= 2.351 P = 209.27 kN

RD=RP*RD= 3.760 P = 334.63 kN

b)Design Tandem

A) Longitudinal Arrangement

case 1: Maximum Axial Load on pier

P P

1.2

x x= 0.5

Rp2 Rp1

Rp1

= 0.983 P

Rp2= 1.000 P

Axial Load: Rp=Rp1+Rp2= 1.983 P


Moment about CLN of pier: MCL= 0.385 kN

case 2: Maximum Bending Moment

P P

1.2

Rp1

Rp1= 1.970 P

Axial Load: Rp=Rp1= 1.970 P


Moment about CLN of pier: MCL= 43.84 kN

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

Case-I: One Design lane loaded (T-Girder)




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RA=RP*RA= 134.33 kN

RB=RP*RB= 140.77 kN


RA=RP*RA= 133.49 kN

RB=RP*RB= 140.77 kN

Case-II: Two or more design lanes are loaded (T-Girder)


RA=RP*RA= 198.11 kN

RB=RP*RB= 167.68 kN

RC=RP*RC= 167.68 kN

RD=RP*RD= 198.11 kN


RA=RP*RA= 196.87 kN

RB=RP*RB= 166.64 kN

RC=RP*RD= 166.64 kN

RD=RP*RD= 196.87 kN

2.3. Wind Loads

Wind Load onstructure - 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 = 69.63 2.75 191.47

Railing = 9.3 3.45 32.09

Posts = 2.805 3.03 8.4981.73 232.04

FWT= 199.42 kN

Line of Action (about Pier Cap bottom) = 2.84 m

b) Longitudinal Direction W=12lb/ft2

= 0.586 kN/m2

FWL= 47.89 kN

= 2.84 m

2.3.2. Wind Load on Live Load - WL

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

Length of exposed surface= 30.8 mFWLT= 45.892 kN (6ft above the deck surface)

(about pier cap bottom) = 4.58 m

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

Length of exposed surface= 30.8 m

FWLL= 18.357 kN (6ft above the deck surface)


Area (m )



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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/m

Bracing = 1.564 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= 46.30 kN (6ft above the deck surface)


2.3.6 Seismic Force Effects,EQ

Earthquake zones: EBCS Zone -4

Site Coefficient: Type I = 1

Acceleration coefficient(A): = 0.1

The horizontal seismic force is the product of the site coefficient, the acceleration coefficient an

605.82 KNFooting = 470.40 KN

Sum Wp 1076.22 KN

Horizontal earthquake force FH = site coeff.*A* Wp = 107.62 KN

This force is transferred to the substructure at joints

The proportion of this load at the two levels is as in the following:

At bracing level =40% of FH= 43.05 KN

At pier cap level =60% of FH= 64.57 KN

3) STABILITY ANALYSIS

Assume a combined footing with dimensions(m):Width W = 3.5 Length L = 7 Depth D = 0.8

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

Case-I: One Design lane loaded

i) Dead Load

Superstructure

Pier cap,

Bracing & Columns =



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Left = 1,317.36 kN

Right = 2936.58 kN

4253.94

Substructure

605.82 kN

Footing = 470.4 kN

1076.22Total = 5330.16 kN

ii) Live Load (T-Girder)

Case I: a) RA+RB= 472.71 3/2aavg*RA+1/2aavg*RB= 1028.50 MCL = 75.23

b) RA+RB= 471.19 " 1025.19 " 93.09

Case II: a)RA+B+C+D= 1253.08 0.00 2*MCL = 150.46

b)RA+B+C+D= 1249.05 " 0.00 " 186.18

ii) Breaking/Longitudinal Force, BR

FLF= 46.30 kN MLF= 787.09 kNm/m

iii)Stream current Force,WA

FTF= 2.87 kN MTF= 4.03 kNm/m

A) Stability against OVERTURNING:

MREST.= 8922.97 kN/m

MDRIV.= MLL+MBR+MWA= 941.58 kN/m

S.F.= 9.48 > 2.0 OK!B) Stability against BEARING PRESSURE:

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

Ptot(kN)= 5802.86 5913.30 6583.23 6579.21

MT(kN/m)= 1241.32 1380.77 0.00 0.00

ML(kN/m)= 75.23 93.09 150.46 186.18

eT=MT/Ptot= 0.214 0.234 0.000 0.000

eL=ML/Ptot= 0.013 0.016 0.023 0.028

285.54 296.18 279.23 281.57< sall OK!

C) Stability against SLIDING:

FTDRIV=FTWA= 2.87

FLDRIV=FLBR= 46.30

FDRIV=(FTDRIV +FLDRIV )= 46.39

FRESIST=SVtan f=SV*0.7= 3731.11

S.F.=FREST./FDRIV.= 80.43 > 1.5 OK!

MLONGT. (kN/m)

Case II:

RC)=

P (kN)

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

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

Case I:

Pier cap,

Bracing & Columns =

MTRANS.(kN/m) about CL



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3.2. CHECK FOR STRENGTHII =DL+WS+WA

a) Dead Load DL

PDL= 5330.16 kN

MTDL= 0 kN/m

MLDL= 404.81 kN/m

c) Wind Load on Structures W

superstructure

FTW= 199.42 kN MTW= 3042.79 kN/m

FLW= 47.89 kN MLW= 730.77 "

substructure

FTW= 45.43 kN MTW= 300.27 kN/m

FLW= 65.53 kN MLW= 524.38 "

Total

FTW= 244.85 kN MTW= 3343.06 kN/m

FLW= 113.42 kN MLW= 1255.15 "

d) Wind Load on Live Load WL

FTWL= 45.89 kN MTWL= 780.12 kN/m

FLWL= 18.36 kN MLWL= 312.05 "

e)Stream current Force,WA



MREST.= 8922.97 kNm/m

MDRIV.= MLL+MWS+MWA= 1259.18 kNm/m

S.F.= 7.09 > 2.0 OK!

B) Stability against BEARING PRESSURE:

Ptot=PDL= 5330.16



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MT=MTDL+MTWS+MTWA= 3347.08

ML=MLDL+MLWS+MLWA= 1659.96

eT=MT/Ptot= 0.628 OK!

eL=ML/Ptot= 0.311 OK!

smax= 450.81

sall OK!C) Stability against SLIDING:

FTDRIV=FTWS+FTWA= 247.72

FLDRIV=FLWS= 113.42

FDRIV=(FTDRIV2+FLDRIV

2)= 272.45



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


a) Dead Load DL

PDL= 5330.16 kN

MTDL= 0 kN/m

MLDL= 404.81 kN/m

b) Live Load LLPLLI MTLL MLLL

Case I: 472.71 1241.32 75.23

583.14 1380.77 93.09

Case II: 1253.08 0.00 150.46

1249.05 0.00 186.18

c) Wind Load on Structures WS

superstructure

FTW= 45.43 kN MTW= 3042.79 kN/m

FLW= 47.89 kN MLW= 730.77 "

substructure

FTW= 45.43 kN MTW= 300.27 kN/m

FLW= 65.53 kN MLW= 524.38 "

TotalFTW= 244.85 kN MTW= 3343.06 kN/m

FLW= 113.42 kN MLW= 1255.15 "

d) Wind Load on Live Load, WL

FTWL= 45.89 kN MTWL= 780.12 kN/m

FLWL= 18.36 kN MLWL= 312.05 "



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e)Breaking/Longitudinal Force,BR

FTF= 45.89 kN MTF= 780.12 kNm/m

f)Stream current Force,WA




MDRIV.= MLL+MWS+MWA= 3005.60 kNm/m

S.F.= 2.97 > 2.0 OK!


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

Ptot=PDL+PLLI= 5913.30 6583.23

MT=MTDL+MTLL+MTBR+MTWL+MTWS+MTWA= 6288.09 4907.32

ML=MLDL+MLLL+MLBR+MLWL+MLWS+MLWA= 2065.10 1846.14

eT=MT/Ptot= 1.063 0.745 OK!

eL=ML/Ptot= 0.349 0.280 OK!

smax= 605.85 569.56

> sall OK!


FTDRIV=FTBR+FTWL+FTWS+FTWA= 339.50

FLDRIV=FLWS+FLWL= 131.78


2)= 364.18






MDRIV.= MBR+MWA= 3005.60 kNm/m

S.F.= 2.97 > 2.0 OK!




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FTDRIV=FTBR+FTWL+FTWS+FTWA= 339.50

FLDRIV=FLWS+FLWL= 131.78


2)= 364.18



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


a) Dead Load DL

PDL= 5330.16 kN

MTDL= 0 kN/m

MLDL= 404.81 kN/m

b) Live Load LL

PLLI MTLL MLLL

Case I: 472.71 1241.32 75.23

583.14 1380.77 93.09

Case II: 1253.08 0.00 150.46

1249.05 0.00 186.18

c)Stream current Force,WA


d)Seismic Force Effects,EQ

At bracing level = FTF= 43.05 kN MTF= 250.09 kNm/m

At pier cap level = FTF= 64.57 kN MTF= 750.28 kNm/m

At bracing level = FLF= 43.05 kN MLF= 250.09 kNm/m

At pier cap level = FLF= 64.57 kN MLF= 750.28 kNm/m



MDRIV.= MLL+MEQ+MWA= 2385.16 kNm/m

S.F.= 3.74 > 2.0 OK!


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

Ptot=PDL+PLLI= 5913.30 6583.23



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MT=MTDL+MTLL+MTEQ+MTWA= 2385.16 1004.40

ML=MLDL+MLLL+MLEQ+MLWA= 1498.27 1591.36

eT=MT/Ptot= 0.403 0.153 OK!

eL=ML/Ptot= 0.253 0.242 OK!

smax= 460.67 394.66

> sall OK!C) Stability against SLIDING:

FTDRIV=FTEQ+FTWA= 110.49

FLDRIV=FLEQ+FLWA= 107.62


2)= 154.24






MDRIV.= MLL+MEQ+MWA= 1004.40 kNm/m

S.F.= 8.88 > 2.0 OK!


FTDRIV=FTEQ+FTWA= 110.49

FLDRIV=FLEQ+FLWA= 107.62


2)= 154.24


S.F.=FREST.

/FDRIV.

= 29.88>

1.5 OK!



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8 EDITION.



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m

m

2.2

405 KNm

76 36

1.9 1.8



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ot from the face of the rail.

-8)

en in Table 13-8.



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n in Table 13-7.



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1.817 P

ot from the face of the rail.

-8)



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en in Table 13-8.

n in Table 13-7.

13-4)



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0



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1.797 P

Case-I: One Design lane loaded (Box Girder)




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RA=RP*RA= 195.99 kN

RB=RP*RB= 68.92 kN


RA=RP*RA= 194.77 kN

RB=RP*RB= 194.77 kN

Case-II: Two or more design lanes are loaded(Box Girder)


RA=RP*RA= 195.99 kN

RB=RP*RB= 122.57 kN

RC=RP*RC= 122.57 kN

RD=RP*RD= 195.99 kN


RA=RP*RA= 194.77 kN

RB=RP*RB= 121.81 kN

RC=RP*RD= 121.81 kN

RD=RP*RD= 194.77 kN



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d the permanent loads



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ii) Live Load (Box-Girder)

Case I: a) RA+RB= 454.50 1241.32 MCL = 75.23

b) RA+RB= 583.14 " 1380.77 " 93.09

Case II: )RA+B+C+D= 1091.30 0.00 2*MCL = 150.46

b)RA+B+C+D= 1087.80 " 0.00 " 186.18

MLONGT. (kN/m)

RB=

RD)+1/2aavg*(RB-RC)

P (kN) MTRANS.(kN/m) about CL



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SUR RIVER BRIDGE RC PIER

SUMMARY OF LOADS

DL(Lt.Sup.) 331.35 327.33 327.33 331.35 MDL L 404.80625 KNM

DL(Rt.Sup.) 756.06 712.24 712.24 756.06

LL(HL-93)F

LLRtF

LLCtF

LLCtF

LLLtF

WLt=

45.89F

WLl=

18.36FWt= 199.42 FWl= 47.89

0.4 a a= 2.2

26.40

FEQT 64.57 MLONGT FEQL 64.57

MTRANS

2.35 2.35

6.3095

11.52

FEQT 43.05 FEQL 43.05

FWl= 1.96

1.56

1.96

FWt= 1.96 FWt= 1.96 5.81

FWAt= 1.02 FWAt= 1.02

1.4 4.6 1.4

HS20-44/ case I: a) 335.71 195.99 248.51 140.77 0.00 0.00 1241.32 432.16 75.2

Design

Tandem b) 334.63 194.77 248.51 194.77 0.00 0.00 1380.77 429.47 93.0

case II: a) 339.33 198.11 287.21 167.68 339.33 198.11 0.00 0.00 150.4

b) 338.24 196.87 286.29 166.64 338.24 196.87 0.00 0.00 186.1

MLONFRt(kN) FCt (kN) FLt (kN) MTRANS.(kN/m)

TRANSPORT CONSTRUCTION DESIGN S. Co. BR


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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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AWASH-TENDAHO RC PIER FOOTING DESIGN-32 of 36

FOOTING DESIGN

Loading Case Summary

Comb Fz(kN) Mx(kNm) My(kNm) Fz(kN) Mx(kNm) My(kNm)I 2909.597 -31.84 326.71 3462.465 -53.19 326.71

II 2544.063 -739.51 -706.005 3827.999 -762.18 -706.005

Allowable Bearing Capacity = 5 kg/cm2

= 750 kN/m2

Design Constants

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

f'c= 24.00 Mpa

fc=(0.4*f'c) = 9.6 (0.4*f'c) =

Reinforcement steel:

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

fy = 420 MPa

fs = 165 MPaEs = 200,000 MPa

Grade 300 steel: For rebars less than diam. 20

fy = 300 MPa

fs = 140 MPa

Es = 200,000 MPa

FyR x R FyL

4.6

z

f(m)= 1.00 D=x 0.8

B

C

x

W=

y 3.5

A D

B= 7.0

Self Weight W (kN)= 470.4

Location of Resultant R & Eccentricity e

Comb I II

x(m)= 2.486 2.528

Right leg Left Leg



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ex(m)= 0.186 0.228 B/6

ey(m)= -0.103 0.222 W/6

smax= 274.766 439.902 sall

Design Load = 439.90 kN/m2

ex= 0.228 m ey= 0.222 m V=Fz1+Fz2+W= 6842.462 kN

sA=V/A(1-6*ex/B+6*ez/W)= 330.99 kN/m

sB=V/A(1-6*ex/B-6*ez/W)= 118.42 "

sC=V/A(1+6*ex/B-6*ez/W)= 227.58 "

sD=V/A(1+6*ex/B+6*ez/W)= 440.15 "

Depth Checking

1. Punching Shear (Art. 8.15.5.6.2)

v=V/bo*d bo=p(f+d)

design shear stress v < vcp=0.166f'c'

f'c= 3000 psi 24.00 Mpa

vcp= 0.81 N/mm 813.23 kN/m

Vact=Pcol-Psoil= 2781.97 kN where Pcol= 3827.999 kN

trial d= 0.74 m

Vres=vcp*bo*d= 3289.61 Vres Vact OK!

davil=D-F/2-CLR CR= 0.74 m davil d OK!

2. Wide Beam Shear

a. In the Short Direction

s1=P/A(1 + 6*ey/W)= 359.06 kN/m d

s2=P/A(1 - 6*ey/W)= 161.10 "

trial d= 0.74 m

s3= 330.22 s1 s3 s2

Vact= 1230.37 kN

bc -ratio of long side to short side of concentrated load= 2.0

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

vcb=0.166(1+2/bc)f'c0.332f'c= 1.63 N/mm 1626.46 kN/m

Vres=vcb*B*d= 8425.069 kN Vres Vact OK!



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b. In the Long Direction

FyR= 2909.597 FyL= 3827.999

1.2 MxR= -739.51 MxL= -762.18

q2 q3

q4 q1

q1=s1*W=P/A(1 + 6*ex/B)*W= 1088.19 kN/m q3= 793.39 kN/m

q2=s2*W=P/A(1 - 6*ex/B)*W= 732.40 " q4= 1027.199

2193.23

915.47

1.2 2.19 2.41 1.2 SFD

-1634.77

-1994.12

895.41

BMD

-549.28 -1746.74

-1288.79 -2508.92

Vmax= 2193.23 kN corresponding triangle leg = 2.41

Vact= 1064.76 kN trial d = 0.74 m

Vres=vcb*W*d= 4212.53 kN Vres Vact OK!

davil d OK!

3. Flexure

a. In the Short Direction

Using F = 20 mm andClear cover = 50 mm

davil=D-F/2-CLR CR= 0.74 m



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beff=coln width+2(1/2*d)= 1.74 m

Loading

FyR= 2909.597 MyR= -706.005 FyL= 3827.999 MyL= -706.005

Right Leftey(m)= -0.2426 -0.1844 1.25

smax=Fy/beff*W(1 + 6*ey/W)= 279.03 429.84

smin=Fy/beff*W(1 - 6*ey/W)= 676.50 827.31

qmax=smax*beff= 747.92 kN/m qmax q1= 994.91 qminqmin=smin*beff= 1439.51 " 747.92 1439.51

My-y= 905.92 kNm

Reinforcement

Design moment, Mu = 905.92 KNm/m

As = Mu / ( fy (d - a/2 ) ) where Mu= 905.92 KNm/m

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

b= 1740 mm

Assume a 45 mm fy= 420.00 N/mm2

As = Mu / ( fy 3,340 mm2 fc'= 21.00 N/mm2

a = As*fy / ( 0. 45 mm D= 800 mm

diam= 20 mm

cover = 50 mm

d= 740 mm

Required As = 3340 mm2/m

Spacing = 164 mmUse F 20 mm @ 164 mm for width= beff

Provide As = 1,919.7 mm2/m

Use As,min=0.002*1000*d= 1480 mm /m S = 212.27 mm

Use F 20 mm @ 212 mm for the restcm

Top :- temperature & shrinkage

As=1/8 in2/ft=2.65 cm

2/m

Using F =12 mm S= 42.64 cm

Use F 12 mm @ 400 mm



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b. In the Long Direction

Bottom :-

Mmax= 2508.92 kNm

ReinforcementDesign moment, Mu = 2508.92 KNm/m

As = Mu / ( fy (d - a/2 ) ) where Mu= 2508.92 KNm/m

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

b= 3500 mm

Assume a 63 mm fy= 300.00 N/mm2

As = Mu / ( fy 13,079 mm2 fc'= 21.00 N/mm2

a = As*fy / ( 0. 63 mm D= 800 mm

diam= 16 mm

cover = 50 mm

d= 742 mm

Required As = 13079 mm2/m

Spacing = 54 mmUse F 16 mm @ 54 mm

Provide As = 3,736.7 mm2/m

rc pier-f(box & girder).xls

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