1
3.7 DESIGN OF SUB-STRUCTURE:
3.7.1 PIER DESIGN:
Dead Load from super structure:
Dead load from three girders:
49 18.33 3 = 2694.51kN 2695 kN
Cross beams at end:
10.5 24.2 = 88.2 kN
Intermediate cross beam:
10.5 3 4.2 = 132.3 kN
Total un-factored dead load = 2695 +88.2+132.2 = 2915.5 kN 2916 kN
Live Load:
Live load in yy direction = 695.2 kN 700kN
Factored live load in yy = 1050 kN
ex = 150+450/2 = 375 mm = 0.375 m
Live load about xx direction = 413 kN
Factored live load = 1.5 1.1 413 = 682 kN 700 kN
[As the girder-B, reaction passes through the C.G of the pier, no eccentricity is due to that reaction but the reaction due to girder-A will not pass through the C.G of the pier so that the eccentricity and that reaction should be taken. ]
ey = 2.5 m
Mux = 682 2.5 =1705 kN-m
Muy = 1050 0.375 = 394 kN-m
So dead load from the super structure un-factored = 2916 kN
factored load = 2916 1.5 = 4374 kN
Live load in yy direction = 682 kN (factored)
Live load in xx direction = 1050 kN (factored)
ex = 0.375 m, ey = 2.5 m , Mux = 1705 kN-m and Muy = 394kN-m
Stability Check:
Pier size = 9 1 = 9m2
We have provided rubber pad which acts as elastomeric bearing.
eeff length = 1.3 6 = 7.8 m (page-114, IRC : 112 – 2011)
Total ht. of pier=6m
Pier cap is provided assuming thickness 250mm and 20mm projection.
So DL of pier cap=9x1.4.0.25x25=78.75KN.
We have to provide 6 pedestal of size (450x450)mm each having ht. 450mm.
DL of 6 pedestals=6x(0.45x0.45x0.45x0.25)=13.67KN
DL of pier=6x1x9x25+2(π/8x0.52x6x25)=1350+29.45 1380KN
DL due to pier cap, pier, pedestal=1380+78.75+13.67=1472.742=1473 kN
Total DL =1473+2916=4389kN
Factored DL=4389x1.5=6584kN
1. Stress at pier base due to DL=6584/9=731.556kN/m2=0.731556N/mm2
2. Stress due to buoyancy effect
Ht.of water above pier base=5.5m
Submerged vol. of pier=49.5m3
Reaction due to buoyancy effect=49.5x10=495KN
Stress at pier base =495/9=55KN/m2(upward)=0.055N/mm2
3.stress due to eccentric loading due to LL
Stress in XX-dinn=P xx/A+Mx/Z
Max.=1050/9+(1705x6)/1x92
=116.67+129.3
=245.97KN/m2=2.246N/mm2
Min.=1050/9-1705/1x92/6
= 116.67-129.30= -12.63KN/m2
= -0.0126N/mm2
Stress in YY dinn
Max.=Pyy/A+M/Z
=Pyy/A+Muy/Z
=682/9+394x 6/9
=75.77+262.667
=338.45KN/m2=0.338N/mm2
Min.=Pyy/A-M/z
=682/9 –{(394x 6)/9}
=-186.897KN/m2=-0.187n/mm2
4. Stress due to Longitudinal forces
i)Due to breaking
According to IRC 6:2000 the breaking force should be considered 20% of LL
i.e. 0.2x700=140KN.
LA=6+0.45+0.15=6.6m(breaking effect is considered from bearing)
Factored moment=140x1.5x6.6=1386kN/m
Z= bd2/6=9x12/6=1.5m3
So M/Z=140x1.5x6.6/1.5=924KN/m2 =0.924N/mm2
ii)Due to resistance of bearing:
from worst case analysis
Left side span=no LL only DL
Right side span=DL +LL
Left side coefficient=0(for worst analysis)
=>resistance by left side rubber pad is
=coefficient of bearing x DL of left span=0xDL of left span=0
According to AASHTO LRFD code for elastomeric bearing coefficient of resistance lies between 0.02 to 0.04
2)Right span coefficient of bearing=0.04
Dl + LL =4374+1050=5424KN
Resistance of bearing=0.04x5424+=217KN
La from bearing =6.6m
Z=9x1/6=1.5m3
Stress =I m/2=217x6.6/1.5=955Kn/m2 =0.955N/mm2
5.Wind analysis-(IS -875 part III):
Wind design speed (Vb)=K1VbK1K2K3
K1=1.08(referring to T1)
K2=category II (up to 10m) =1
K3=1(topographic factor)
Design wind pressure (Vz)=0.6Vz2=0.6x542=1750KN/m2
For wind expressed area is super structure area only.
Superstructure exposed area
={(18.33x0.25)+(18.33x1)+18.33x(1.4+0.15)}
=57.234m2
Total wind fence=1.75x51.234=89.82kN
Moment at base of pier=89.82xLA
LA=6+(0.45+0.05+1.4+0.95+0.25/2) =7.15m
Moment=89.82x7.15=642.213KN/m=643kNm
Factored wind moment=643x1.2=772kNm
Factored wind load=107.78kN
Wind stress at pier base=772/1x92/6=57.18=0.057N/mm
According to code lead combination IS: 456:200
a)1.5(DL+LL)=1.5(4389+700)=7633.5KN
b)1.5(DL+WL)=1.5(4389+107.89)=6746KN
c)1.2(DL+LL+WL)=1.2(4389+107.89)=5396.27KN
At here LL should not be considered as our wind speed exceeds 130kmph(IRC 6:2000 LL-212.5)
So in our case no wind lead should be considered
6.Water current:
Intensity of water current(P)=52KV2(P-29 IRC-6)
K=0.66(semicircular easeway & cut way)
V=3m/s
P=52KV2=o.52x0.66x32=3.1KN/m2
Factored according to IRC-45:1972
P=3.1x1.4=4.34KN/m2
Area of obstruction=1x5.5=5.5m2
Force due to obstruction=4.43x5.5=23.87KN
It acts at h/3 distance from base ABC is the pressure distribution after water current max. at top & min. at bottom
So moment = (23.87x5.5)/3=87.53/2 88/2=44KNm
According to IRC-45:1972 the dirn is assumed to vary by 20.
Pressure IInl to pier=4.34xcos20=4.37kN/m2
Pressure perpendicular to pier =4.34xsin20=1.48KN/m2
FP=fence in parallel dinN to pier
=4.07x5.5x1=22.39KN
Fn=fence in perpendicular dinn to pier
=1.48x9x5.5=73.26KN
Moment considering max.stress at pier base
M=73.26x5.5/3=268.62/2 kNm=270/2=135KNm
Z=9x12/6=1.5m3
thus = M/Z=268.62/2/1.5=179.08Kn/m2=(0.179N/4)N/mm2
STABILITY ANALYSIS:
At pier base
When dry(N/mm2) when HFL(N/mm2)
Maxm Minm Max m Minm
1)Di stress 0.732 0.732 0.732 0.732
2)buoyance ____ ____ -0.055 -0.055
3)eccentric
loading due to LL
->XX –dirn 0.246 -0.01263 0.246 - 0.01263
->YY –dirn 0.388 -0.187 0.388 -0.187
4)longitudinal Forces
Breaking 0.924 -0.924 0.924 -0.924
Bearing 0.955 -0.955 0.955 - 0.955
5) Water current
_____ ______ 0.09 -0.09
------------------------------------------------------------------------------------------
3.245 -1.347 3.36 -1.58
Max. permissible bearing stress=0.45x35=15.75N/mm2 >3.360N/mm2 (OK)
Checking for 2m above pile cap
DL of pier=9x1x4x25+2(π/8x0.5x0.5x4x25) 920KN
So total DL=2916+920+78.67+13.67=3929KN
Factored DL stress=3929/9=436.56KN/m2=.436N/mm2
Stress due to breaking =m/2=140x1.5x4.6/9x1x1/6=633KN/m2=0.644N/mm2
Resistance force due to bearing (stress)=m/2=217x4.6/1.5=665.47=0.666N/mm2
Above 2 meters above pile cap
When dry when HFL
maxmminm maxm minm
1)DL stress 0.436 0.436 0.436 0.436
2)buoyance ______ ______ -0.055 -0.055
3)eccentric loading
Due to LL
->XX dinn 0.246 -0.01263 0.246 - 0.01263
->YY dinn 0.388 -0.187 0.388 -0.187
4)longitudunaltenus
Breaking 0.644 -0.644 0.644 -0.644
bearing 0.666 -0.666 0.666 -0.666
5)watercurrent _____ _____ 0.09 -0.09
_____________________________________________________
2.38 -1.074 2.414 -1.22
Max. permissible bearing stress=0.45x35=15.75N/mm2 >2.414N/mm2 (OK)
Stability analysis above 4m from pik cap
When dry when HFL
Maxm minm maxmminm
1)DL stress 0.385 0.385 0.385 0.385
2)buoyancy _____ _____ -0.055 -0.055
3)eccentric loading
due to LL
XX dirn 0.246 -0.01263 0.246 -0.01263
YY dirn 0.388 -0.187 0.388 -0.187
4)longitudinalForces
Breaking 0.364 -0.364 0.364 -0.364
Bearing 0.376 -0.376 0.364 -0.364
5) Water current _____ ______ 0.09 -0.09
___________________________________________________
-0.55 1.793 0.78
So max value =1.793N/mm2<0.45x35=15.61N/mm2
REINFORCEMENT IN PIER:
Rectangular column size (1x9)m
Or (1000x9000)mm
M35concrete and Fe 415 steel
Pu (factored) =6584+1050=7634kN
Mux = 682x2.5=1705kNm
Muy=1050x0.375=394kNm
ey=2.5m and ex=0.375m
Checking whether column/pier is biaxially loaded or not.According to IRC -112:2011 (clause – 8.3.2 (3),P-74) LL=8.32
Iy=13x9/12=0.75mm4
z=1x93/12=60.75mm4
breq=iy 12=0.75 12=2.59
hreq=iz 12=60.75 12=210.44
As we know and slenderness ratio = eff. Length/center radius of gyration
In our case 2 radius of gyration Kyand Kx and Kz
y=eff.length/Ky;Ky=iyA=0.7591 = 0.288
So y =1.3x6/0.288=27.02
z=eff. Length/Kz; Kz=ixA= 60.7591 = 2.59
So z = 1.36/2.59 = 3.00; zz = xx
Condition I
y/y=27.002/3.0022=9>2{both should be satisfied for axially loaded so our isBiaxial loading}
z/y=1/9=0.111<2
ey=1.475m
ez or ex =0.375m
for axial compression
Condition – II
(ey/heq)/(ez/beq) 0.2 or (ez/beq)/(ey/heq)0.2
Or (1.475/210.44)/(0.375/2.59) = (ey/heq)/(ez/heq) = 0.0435<0.2
(ez/beq)/(ey/heq)=(0.375/2.59)/(1.475/210.44)=20.65>0.2
Hence condition-II is satisfied
As the condition-I fails it should be designed as Bi-axially loaded column.
Design is done according to Sp-16
Assuming p=1.5
P/fck=1.5/35=0.042857=0.043
Uniaxial bending action about x on z axis
dI=45+20/2+10=65mm
dl/D=65/9000=0.00722
takingdI/D=0.05
Pu/fckbD=7634x103/35x1000x9000=0.0248=0.025
Muy=1050x0.375=394kNm
+ 1386kNm(breaking)
+ 1433kNm (resistance)
+ 135kNm(water current)
So Muy=3349kNm
Mux=1705kNm
Referring to chart-43 of SP-16, following P/fck=0.043
d'/D=0.05and Pu/fckbD=0.072
=>Muy1=0.072x35x1000x90002=204120KNm
dl/D]y axis=65/1000=0.065=0.1
referring to chant-43 of SP-16 following
P/fck=0.043,d1/D=0.1,Pu/fckbD =0.025
Muy1/fck bD2=0.0672
=>Muy1=fck bD20.0672
=>Muy1=0.0672x35x9000x10002=21168KN
Referring to chart 63 of SP-16 and following the values of P=1.5 Fe=415 m=3
Puz/Ag=21
Puz=21x9000x1000=189000KN
Pu/Puz=7634/189000=0.041
Muy/Muy1=3349/21168=10.1
Mux/Mux1=1705/204120=0.0084
Referring to chart -64 Sp-16 and following
Pu/Puz=0.041 and Muy/Muy1=0.16
Mux/Mux1=0.85
The chart value of Mux/Mux1=0.85 .>>>Mux/Mux1=0.0084
So providing min. steel
P/fck =0.8/35=0.0228
Pu/fckbD=7634x103/35x1000x9000=0.0248=0.025
Referring to chart-43 and following d1/D=0.05
P/fck=0.0228 Pu/fckbD=7634000/3510009000=0.025
We get Mux1=0.48x35x1000x90002=136080KNm
d1/D for y axis = 65/1000=0.065=0.1
from chart-44 we get
Muy1/fck bD2=0.04
=>Muy1=0.04x35x9000x10002
=12600KNm
From chart-63, p=0.8%
Puz/Ag=18.2
Puz=18.2x9000x1000=163800KN
Pu/Puz=7634/163800=0.0467
Muy/Muy1 =3349/12600= 0.26
Mux/Mux1=1705/136080=0.0126
From chart -64,
The value Mux/Mux1=0.74>>>0.0126
So we have to providemin. Steel as 0.8% of Ag.
Ag=0.8/100x1000x9000=72000mm2
Providing 28ф bars, no. of bars=72000/{(π/4)x302}=102 bars
Spacing=18000/102=176.47c/c
Providing spacing 170mm c/c, which is lessthan 200mm(ok) .
IRC-112:2011 16.3.1(4) (P-174)
Astp=18000x(π/4)x302/170=74844mm2
According to IRC-112:2011 CL-16.3.1(2)
Min. vertical ref.=0.0024xAc=0.0024x1000x9000=21600mm2
Max. vertical ref.=0.04x1000x9000=360000mm2> Astp
f. SECONDARY REINFORCEMENT IN PIER:
Provide 10 mm # transverse tie at spacing 200mm c/c throughout the section.
Provide 10mm # 4-legged transverse reinforcement @ 200 mm c/c.
Since 1/4th of the main reinforcing bars i.e. 30mm # is 8.5 mm, that's why we have chosen 10mm # bars as lateral and transverse reinforcement.
For reinforcement detailing, refer to Appendix-C
3.7.2 PEDESTAL DESIGN:
Fig-29 (Pedestal on pier and abutment)
DL from super structure = 505 kN.
LL from super structure = 453 kN.
Assuming the size of pedestal
Taking ht. as = 450 mm.
DL of pedestal
Total load = 505 + 453 + 2.3 = 960.3 961 kN
factored load = 1441.5 1442 kN.
Taking rubber paid size as
Pressure or rubber paid
So providing rubber paid
Pressure on loaded area = 9.0125 N/mm2
Permissible bearing stress
According to CL – 34.4 IS : 456:2000
Pressure on loaded area and permissible bearing stress
Permissible bearing stress (ok)
min. size of pedestal to carry out this type of pressure is
(ok)
= 229
According IRC - 112, P - 171 cc - 16.22
or
min reinforcement
Providing 16mm bans
Providing 3 bans in spacing
According to IRC – 112 page - 121
the spacing should not be more than 200 mm c/c
So providing 4-12 mm bars Fe-415 type
(ok)
Spacing
here eff. cover is assumed to be 25 mm.
So clear length = 450 – 2 = 400 mm.
According toRankire formula
(are end is ticked and other end is hinged)
and
According to cl. - 39.3 1S – 456 : 2000 (P-71)
; (ok)
3.7.3 ABUTMENT DESIGN:
Fig-30 (Section of Abutment showing all Dimensions)
Preliminary dimensions
base slab width = 8m
thickness = 1.5m (base slab)
ht. of stem = 6m
top width of stem = 1.2m
bottom width of stem = 1.5m
(angle of shearing resistance) = 35°
(bulk density) = 19 kN/m3 (coarse sand)
e = 0.325m (for dead load)
According to IRC-6 : 2014, LL surchargeIS-456:2000,CL - 214.1.1.3 equivalent ht. of soil for vehicular loading shall be 1.2m.
According IRC-6 : 2014
So
= 9.547
STABILITY ANALYSIS:
Earth pressure:
{
Providing abutment cap
Surcharge due to LL
(assuming class AA tracked vehicle)
Providing 2 wheeled vehicle
So virtual (acc to IRC-6:2014)
According to IRC – 112 : 2014,
Height above base to centre of pressure
(LA for earth pressure)
Live Load Surcharge and approach slab
Horizontal force due to LL surcharge = 56 kN/m
Horizontal force due to approach slab = 18.15 kN/m
The above 2 forces act at a distance act at a distance from base.
Vertical load due to LL surcharge and approach slab
Weight of earth on heel slab:
Vertical load
DL from super structure
Total DL from super structure = 73.5 + 26.25 = 99.75 kN/m.
Table-5 [Stability Analysis of Abutment (Traffic Surcharge + Earth Surcharge)]
Sl no*
Details
Force(kN)
Lever arm(m)
Moment(kNm)
V
H
Mv
MH
1
DL from superstructure
99.75
----
2.075
208
----
2
Active Earth pressure
----
232
4.179
----
960
3
Horizontal force due to LL surcharge & approach slab
----
74.15
9.95/2
----
369
4
Vertical load due to LL surcharge & approach slab
151.5
----
5.5
833.25
-----
5
Self wt (1)
180
----
2.4
432
-----
6
Self wt (2)
22.5
----
1.833
41.2425
-----
7
Self wt (3)
300
----
4
1200
-----
8
Self wt (4)
25.3125
----
2.775
70.24
-----
9
Self wt (5)
802.75
----
5.5
4416
-----
Total
1581.8125
306.15
7200.7325
-----
x =mv-mHv
=7200.7325-13291581.8125 =3.712 m
(ok or safe)
FOS against overturning,
FOS against sliding,
(ok)
Taking another arrangement :
Removing surcharge so that providing LL
In our analysis the max. LL was found out to be 682 kN 700kN
Factored LL =
Putting the LL just outside the abutment i.e. on deckslab.
So equivalent virtual surcharge ht. is zero (0).
Live load surcharge and approach slab.
Horizontal force, due to LL surcharge = 0
Horizontal force due to approach slab
It acts from base of abutment.
Vertical load due to LL surcharge and approach slab
Table-6 [Stability Analysis of Abutment Considering LL Surcharge]
Sl no*
Details
Force(kN)
Lever arm(m)
Moment(kN-m)
V
H
Mv
MH
1
DL from superstructure
99.75
----
2.075
208
----
2
Active Earth pressure
----
232
4.179
----
960
3
Horizontal force due to LL surcharge & approach slab
----
18.15
9.95/2
----
369
4
Vertical force due to LL surcharge & approach slab
37.5
----
5.5
207
-----
5
Self wt (1)
180
----
2.4
432
-----
6
Self wt (2)
22.5
----
1.833
41.2425
-----
7
Self wt (3)
300
----
4
1200
-----
8
Self wt (4)
25.3125
----
2.775
70.24
-----
9
Backfill vertical load
802.75
----
5.5
4416
-----
10
Live load
97.22
----
0
0
-----
Total
1565
250.15
6575
1050.3
FOS against over turning (ok)
FOS against sliding (ok)
Comparing the two condition we get the worst case as traffic load surcharge and earth surcharge.
DESIGN OF STEM :
Design is done by considering surcharge i.e. both traffic surcharge and earth surcharges.
Pressure at the top of the wall
Pressure at the top of the stem
Pressure at the bottom of the stem
Area under the pressure diagram will give force (P).
Fig-31 (Pressure diagram of Stem)
Factored
Providing 25 bars,
Spacing d = 1500 – 70 = 1430 mm c/c,
providing 20 bars
Spacing
providing 16 bars
Spacing = 200 mm c/c.
Astp
This rf. to be provided for one face only i.e. in the back fill side.
According to the code both side to be equally reinforced so increasing rf. =
Assuming the rectangular portion or stem should carry all the loads than acc. to
IS 456-2000 0.8% of Ag=0.8/100×(1000×1200)
Ast required = 9600 mm2 = 10,000 mm2
So rf. to be provided (each side)
providing 28 bars.
Spacing
Providing spacing 120 mm c/c.
Astp =
No. of bars
So providing 9 bars of 28 bars in spacing 111 mm c/c.
Astp= (in one face)
% ofAstp (ok)
CHECK FOR SHEAR IN STEAM:
As calculated shear = 120 kN
Factored shear
From T-19 IS – 456 : 2000
τcp =0.6476 n/mm2
Checking according IRC-112:2011, cl-10.3.2, page-88
The design for shear resistance VRd,c VEd
(max. value)
(super structure load)
= 1460 kN. (DL from super structure)
(so safe) (No. shear ref. reqd).
CHECK FOR LOAD FROM SUPER STRUCTURE:
Assuming the rectangular section will take all the loads.
Loads:
DL from super structure = 1460 kN.
Factored DL = 2190 kN.
LL in xx =
LL in yy = 682 kN.
Fig-32 (Plan of top of Abutment Stem)
As the both should be less than 2 so we have to design as biaxial bending.
(LL bending)
(breaking moment)
+ 135(water current) 150 kNm
(for DL of super structure)
= 2772.25 kNm2800 kNm.
(due to eccentric loading of live load)
+ dead load of dirt wall+ dead load of abutment cap+ 3 pedestals (stem load) (dirt wall) + 93kN (pile cap)
+
=5188 kN (Total factored load by adding above)
Design checking for abutment whether the design reinforcement to be provided will take the load or no additional reinforcement required. Solved referring to SP - 16.
P = 1.5%
referring to chart-43
For YY
,
referring to chart - 44
Referring to chart – 63,
P = 1.5%, Fe-415, M - 35
referring to chart 64 and comparing values
and
We get
So reducing rf. and providing min rf.i.e., P = 0.8%
referring the chart - 43.
= 142884 kNm.
So referring to chart - 44.
P = 0.8%, fck = 35, Fe - 415 and referring to chart - 64.
Referring to chart 64 and comparing values
and
We get
So providing min. reinforcement as 0.8% of Ag
Astreqd=0.8/100(1000×9000)
=72000mm2
So providing min. rf. as 0.8% of Ag.
Ast required
We have provided
Reinforcement required = 72000 – 11288 = 60712 mm2 = 61000 mm2
Spacing
So providing 111 mm c/c spacing for 28 bans.
So % of steel = 1.03%
Spacing =
Referring to chart-43 of SP-16 we get
Referring to chart-44 of SP-16 we get
= 197316 kNm
= 24948 kNm
from chart - 64
p – 1.03%, fck = 35, Fe – 415
,
from chart - 64.
referring to chart - 64,
Pu = 5188 kN
= 204692 kN.
So Pu
So taking
(ok)
Providing 10 - 4 legged stirrups throughout the section with spacing 200 mm c/c (IRC-112:2011)
Providing approach slab of 3.5m length with 12 mm # bars with spacing 150 mm c/c. (IRC – 6 : 2014)
Providing 30 mm # bars with 114 mm c/c spacing throughout the 18m.i.e., through both side of long section. Side face reinforcement.
For reinforcement detailing refer to Appendix-C
DESIGN OF HEEL SLAB:
P = 1582 kN
e = 0.325 mm
Maximum soil pressure
Minimum soil pressure = 149.55 kN/m2
Total downward pressure = wt. of earth + self wt. of heel + surcharge
Fig-33 (Pressure diagram for Heel Slab)
Net pressure
Net pressure
Shear
Moment about
Factored
Providing 30 mm bars,
Astreqd
Providing 30 mm # bars @ 90 mm c/c
Provide30 mm bans with spacing 90 mm c/c.
SHEAR CHECK FOR HEEL SLAB:
Factored
Referring T-19 Zs-456 : 2000
(so shear reinforcement(rf.) required)
Providing 24 legged 10 structures
(min. value)= 300 mm
Providing 4 legged – 10 stirrups with spacing 200 mm c/c throughout the heel slabs. Provide 0.12% of Ag as distribution reinforcement.
Providing 16 bans spacing
So provide 16 bans as spacing 100mm c/c as distribution reinforcement.
DESIGN OF TOE SLAB:
Self-wt. of toe
(Earth Pressure diagram for toe slab)
(Net Pressure Diagram)
Fig-34 (Pressure diagram for toe slab)
Factored
Mu about
Mu (factored)
= 650 mm2
min. steel
from T – 19, τcp= 0.33 N/mm2 >Zv (ok)
No shear ref. required.
But provide 2-legged 10 stirrups with spacing 200 mm c/c.
Increasing the half reinforcement from stem and heel slab to the intersection portion of heel slab and stem.
Spacing of bars
So provide 30 mm # bars with spacing 240 mm c/c spacing in toe.
DISTRIBUTION REINFORCEMENT:
So provide 16 reinforcement with spacing 100 mm in toe slab.
The base slab thickness is increased upto 4.5m as that the abutment can be designed as shallow foundation and stress at heel & toe will be safe.
3.7.4 DESIGN OF DIRT WALL:
(from previous analysis)
Horizontal force due to live load surcharge
(width of wall)
.
It acts above bare of direct wall.
Maximum moment at the bare of the back wall
Horizontal moment due to surcharge
Un-factored moment = 15 + 9 = 24 kN-m.
Factored moment
Our case M35, Fe-415 grade is used.
assuming 70 mm cover
Our column width is 450 mm (fixed)
So the section size of column is fixed to be 450 (width) 500 mm (breadth)
Pu = 0.446 fck Ag – 0.446fckAst + 0.75 fyAst
According to Cl-39.5, IS – 456 : 2000,
10% should be deducted due to eccentricity
Pu = 0.4 fck Ag – 0.4fckAst + 0.75 fyAst
(so theoretically no rf. required).
but according to IRC-45 : 2000 0.8% of Ag should be provided as reinforcement.
Providing 16 bans
Spacing
So providing 16 bans with 200 mm c/c spacing this are providing to increase ductility,
minimum diameter of lateral tic > diameter of longitudinal bar or 5mm whichever more.
So providing 8 2 legged stirrups throughout section pitch
least lateral dimension = 450 mm
16 times diameter of bar = 256 mm
300 mm.
Provide 8 mm- 2-legged stirrups with spacing 250 mm c/c.
3.8 FOUNDATION DESIGN:
3.8.1 DESIGN OF PILE CAP:
Vertical force vu = 10867kN
X - direction moment=Mux=1.5 L.L+1.4 W.C = 1747kN-m
X - direction moment=Mux=1.2(L.L)+1.4×W.C = 799 kN-m
Wind force and live load cannot be considered at a time as per IRC:6-2014
Mux=1747kN-m
Muy=3348 kN-m
P= Vn± Mux dy dy2± Muy dx dx2
dx2=3×[(4.52)2+(4.52)2]=30.375m2
dy2=2×((4.5)2+(4.5)2)=81m2
F1 =108676-1747×4.581+3348×4.52×30.375
= 1811.2-97.056+248=1962.144kN
F2= 1811.2-97.056-248 = 1467kN
F3= 1811.2+248 = 2060kN
F4= 1811.2-248 = 1564kN
F5= 1811.2+248+97.056=2157kN
F6 = 1811.2+97.056-248 = 1660.256kN
F1+F3+F5 = 6180 kN
F5 + F6= 3818 kN
DESIGN REINFORCEMENT FOR PILE CAP:
Mux=3818×4.5=17181kN-m
Muy=6180×4.52=13905kN-m
(dreq)=17181×1060.36×35×0.48×6.2×103×(1-0.416×0.48)
=756.64mm
dprovided=1800-200(pile+ P.C.C)-55-15=1530mm
(Ast req.)y-direction=0.5×3.5415×(1-1-4.6×17181×10635×6200×15302×1530×6200)=32548.206mm2
Providing 30 - mm bars @ 130 mm c/c
(Ast prov.)y – direction=32896.1mm2
(Ast req.)x-direction =0.5×35415×[1-1-4.6×13905×10635×10700×15002×1500×1-10700 ]
=26194.87mm2
Providing 30 - mm 280mm c/c
(Ast)prov x-dir=26507.18mm2;
After considering the shear criteria (explained in next article), revised reinforcement is given by, (Ast prov.)revised = 30mm- @ 110mm c/c=67472.8422 mm2
For reinforcement detailing, refer to Appendix-D
SHEAR FORCE FOR PILE CAP DESIGN:
Shear will be checked at a distance d/2 fromface of column as per IS-2911-part 3
Xx =1.75m, Xy =0 m
d/2=750mm ,(d/2)+150=900mm<1750 mm.
So the full reaction of the piles will be considered as the shear force to be resisted by the cap.
VEd = Vu =6180-361.25= 5820kN-m
As per IRC:112-2011,
VRdc = (0.12K(80ρ1fck)0.33+0.15σcp)×bwd
K=1+200 d=1+2001500=1.365<2.0 (Hence O.K)
Due to no lateral compression pre stressing cp =0 kN/m2
Vmin =0.2925
(VRdc)min =0.2925×10.7×103×1500 =4694.29 kN.
Ρ1=Ascbwd=0.00165<0.02
VRdc =0.12×1.365×80×0.00165×350.33×10700×1500
= 4356.33kN< (vmin)Rdc < VEd
Hence let us provide 30 mm bars @ 110 mm c/c
(Ast prov.)x-direction=97472.8422 mm2
Ρ1 =0.0042
VRdc =5929.60kN> VEd = 5820 kN (OK)
ANCHORAGE LENGTH:
bnet. = αa bAstreq./Astprov. ,(cl-15.2.4.3 of IRC:112-2011)
αa=1, b=k = 30×30=900mm
(Astreq./Astprov.)x-direction = 0.388
(Astreq./Astprov.)y-direction = 0.989
Hence calculated bnet. will be smaller than b. But let's continue the bars of base up to top of the cap having 60 mm cover at top.
3.8.2 DESIGN OF UNDER-REAMED PILE:
3.8.2.1 SOIL DESIGN OF PILE:
Scour depth=3.72m
Pile cap=1.8m
Water current parallel to pier=22.39 kN
Water current to pier=74kN
Maximum load on outer most piles on 4-corners =2157 KN
Фpile =1.2m
Lpile =11.7m
leffect = 0.7Lpile =0.7×11.7=8.19m (table 11.1 of IRC:112-2011)
leff/d =8.191.2=6.825short column.
Qu= ApNcCp+ AaNcCa' + Ca'As' + αCaAs (clause-5.2.3.1 of IS: 2911-part 3) [for cohesive soil]
Qu=π4( Du2 – D2 ) [0.5 DunN + Nqr=1ndr]+π4D2 (0.5DN + dfNq)
+(0.5 πDK tanδ) (d12 + df2 – dn2) [for sandy soil] (Cl-5.2.3 of IS:2911-part3)
Du =3m,D=1.2m.
n =2, =18.87 kN/m3(soil testing report)
N=22.21875=22.22(average of all value)
Nq=17.293(fig-2 ,is 2911-3-1980,page- 15) and dr = 6+10.5 = 16.5 m
Df =11.7m,K=1.75
δ= =25.375°average of all values
d1= 6m, dn=10.5m
(Qu)2ud =39438.41+4602.503+1849.34=45890.253 kN
Providing single under ream at 6m depth, we have
(Qu)1ud = 15359.71+4602.503+4041.44 = 24003.654 kN
Let us provide single under ream
(Qu)compression =9601.462 kN =Qu2.5
(Qu)uplift =Qu3= 8001.218 kN
Due to group action, 10% strength will be reduced of each pile as per IS:2911-3-1980(CL-5.2.8.1)
So (Qu)comp =9601.462×0.9=8641.31=8640 kN
(Qu)uplift=7201.09kN=7200kN
3.8.2.2 ALL FORCES TO BE CARRIED BY EACH PILE:
Vertical force due to super-structure+live load +pier+ pile cap+moment(in both direction)= 2157kN
Though middlepiles will be having lesser load, let us take it as same as that of corner piles.
Horizontal /Lateral Force:
As per IS2911-3-1980,appendix-c,
T=5EIK1,R=4EIK2
Using table- 2 of appendix-c.IS:2911-3-1980,since all the layers are impervious &66%(approx.) of soil is sand/gravel group with in 13.50m, let us chose
K1 =1.245(dense sand &submerged condition) = 1.245 kg/cm2 = 0.1245 N/mm2(category: dense sand in submerged condition)
Un-confined compression=2Cu
Up to 4.5 m below the ground level ,unconfined compression is varying from 0.50-0.64 in bore hole -1 up to 4,5 m below ground level of in all other case it is zero let us take
K2 =7.75kg/cm2lower value(table -3, appendix-c,IS:2911-3-1980)=0.775N/mm2
Modulus of elasticity of concrete Ec =5000fck=5000×35=29580N/mm2
I=π64×d4=1.017876×1011mm4
T=777.556 m
R=249.32m
AS per IS:2911-3-1980,Appendix-c,note (above table-2)
Using flexible piles will be those for which embedded length is 4R Or 4T
In our case, embedded length will be 11.7-(3.77-1.8)=9.78m 4R or 4T
So chart cannot be used as given on appendix-C of IS:2911-3-1980
Calculation Of Water Force:
IRC-6;2014 ,cl-210.7(page-36)will be used .
K=1.25, V =3m/sec
P=52KV2=52×1.25×32=585 kg/m2 = 5.85 kN/m2
The area obstructing the flow in short direction=(4.5+1.2)×1.92=10.944m2
So force FalongW.C =5.85×10.94=64.022 kN
(Fu)short face=1.4×64.012=89.63kN
(Fu)long face =1.4×5.85×sin (20)×10.2×1.92=54.857kN
If the table -1 of appendix –B of IS:2911-3-1980 is extrapolated to the required dia of pile i.e.1.2m ,for single under reaming ,it'll be found that,
Lateral thrust that safely can be carried is 10.8t= 108kN>89.63kN
&>54.857kN
(Hence safe against lateral force)
Also safe load in uplift resistance is given by (using extrapolation),
50.4t=504KN but minimum vertical force in our case is 1467kN>504kNok
As per appendix –B -1-11 of IS:2911-3-1980, the loads lesser than above extrapolated loads need not be designed separately. Since in our case it is lesser, that's why no need of separate design for horizontal forces.
Using Brom's chart (page -274of Foundation Engineering, PHI - publication by P.C. Varghese),
eL=1.929.78=0.1963=0.2 and Lb=9,781.2=8.15
From chart;
HuKpB3γ=30and Kp = 1+sin 1-sin = 1+sin25.3751-sin25.375=2.5
B = (1.2)3 = 1.728 m3 , γ=18.87 kN/m3
So Hu = 301.7282.518.87 = 2445.552 kN
For safe design, taking factor of safety 2.5, Hs = 978.2208 kN >> Fushort face and Fulongface(Safe)
So our piles will be designed as short axially loaded columns with axial force =Pu=2157 kN
3.8.2.3 DESIGN OF REINFORCEMENT FOR PILES:
Let us use IS 456:2000 for design.
Pu=0.4fck Ac+0.67fyAst (cl-39.3, IS456:2000)
Or 2157×103=0.4×35×[π4×1200)2-Ast+ 0.67×415×Ast
Or 2157000=15.83×106-14Ast+278.05Ast
Or 264.05Ast=(-)1.367×107
That means minimum reinforcement will be provided. Since the design is based on IS456:2000,minimum reinforcement is 0.8% of Ag as against 0.4% of Ag as per IS: 2911 (3)
Ast =0.8100×[π4×1200)2=9047.786mm2
Providing 30mm bars, no. of bars=9048π4×302=12.80=13numbers.
ASper clause -16-2-3 of IRC;112-2011,10 mm ф bars will be provided as transverse reinforcement @ 200 mm c/c.
For reinforcement detailing, refer to Appendix-D
4. CONCLUSION:
From our project it's concluded that using limit state method of design, the economy is achieved due to reduction in both reinforcing steel and concrete volume due to reduction in sectional size. Also the limit state of deflection, shear and bending stress are found to be safe as per IRC:112-2011 which is the latest code of practice for designing reinforced and prestressed concrete bridges. The whole structure is found to be stable against sliding and overturning. Besides that, provision of long span decreases the obstruction by increasing the water way.
