Design Of Abutment Bridge 51b2a
This document was ed by and they confirmed that they have the permission to share it. If you are author or own the copyright of this book, please report to us by using this report form. Report 2z6p3t
Overview 5o1f4z
& View Design Of Abutment Bridge as PDF for free.
More details 6z3438
- Words: 8,723
- Pages: 51
Substructure Kondre (right) Dimension of slab and beams (brought from slab sheet)
Check for correctness of data
Length of single span = Carriageway width, Bc = Overall width of Bridge, Bo =
35.000 m 7.500 m 10.700 m
No. of main girder = 5 Dimensions of the longitudinal girder is taken as follows: Length (L si ) = 35.000 m Breadth (B si )
=
0.550 m
Thickness(Tsi) Distance betn girders =
=
1.680 m 2.375 m
No. of cross beams = 15 For the design purposes the dimensions of the cross beam is taken as follows: Length (L si ) = 2.375 m Breadth (B si ) = 0.140 m Thickness(Tsi) = 0.300 m Distance betn cross beams = 2.500 m For the design purposes the dimensions of the internal slab is taken as follows: Length (L si ) = 2.500 m Hence, effec.span in long.dir. = Breadth (B si ) = 2.375 m Hence, effec.span in transv.dir. = Thickness(Tsi) = 0.230 m Avg. pavement cover over Slab(T sp ) = 0.075 m kerb/footpath breadth = 0.225 m at bottom and 0.225 m at top height = 1.000 m For design purpose dimension of cantilever slab is taken as Length (L si ) = 1.00 m Breadth (B si ) = 0.600 m from c/c of outer girder Breadth (B si ) = 0.325 m from outer face of outer main girder Breadth of carriageway = 1.000 m from c/c of outer girder 1.016 m from outer face of outer main girder Thickness(Tsi) = 0.230 m at the end Thickness(Tsi) = 0.230 m at outer face of main girder Avg. pavement cover over Slab(T sp ) = 0.08 m
2.2
2.360 m 1.825 m
DESIGN OF RIGHT ABUTMENT
The freebody diagram of the R.C.C Abutment is shown below:
i=0
0.200 0.40
CG of Stem from inner vertical face:
Pa
d
14.454
0.23
0.600
A 1=
3.324
A2=
2.674
M.A.= A1+A2 =
1.327 5.998
CG =
0.221
b = 90 A Dimensions taken for the design of abutment: Retaining wall Length = 9.500 Breadth = 0.600 Height = 14.454 Ballast wall Length = 9.500 Breadth = 0.400 Breadth of ret. wall = 0.200 Height = 1.000 Return wall No. = 2 Breadth = 0.200
m m at bottom
m m m m m
m
Bottom level of base = Road level = 0.23 m at top 1.25 CG of retain. wall from point A = 0.221 m
height of superstr. plus bearing without pav.cover Height of Return wall H1=
Page 1
0.7
0.000 5.000 ######
Substructure Kondre (right) Length = Total height of abutment = Approach slab Length = Breadth = Thickness = Foundation is open foundation Length = Width Height of footing = Volume = 2.2.1
6.000 m
Height of Return wall H2= 15.454 m
5.000 m 3.000 m 0.200 m
16.354 -11.354
6.000 m 2.750 m 0.900
0.4
Heel Toe
1.000 m 1.000 m
W =
356.40 KN
0.750 0.750
1.250
14.850 m3
CALCULATION OF LOAD
STRESS DUE TO DEAD LOAD S.N
Elements
Unit
Dead Load of Superstructure 1 Hand rails 6.368 2 Railing posts 3 Wearing course 4 Kerbs 5 Deck slab i) rectangle ii)trapezoid 6 Main Girder i) fillets ii) webs 7 Internal Cross Girder i) fillets ii) webs 8 End Cross Girder i) fillets ii) webs
Dead Load of Abutment retaining wall return walls ballast wall
2
rm cum cum cum cum cum cum cum cum cum cum cum
9.500 6.000 9.500
No.
L(m)
B(m)
H(m) Volume Unit Wt
4 35.60 18 0.15 0.15 1.00 1 35.60 7.50 0.08 2 35.60 0.23 1.00 1 35.60 2.93 0.23 2 35.60 0.33 0.23 8 35.60 0.30 0.15 5 35.60 0.60 1.45 0 1.83 0.15 0.15 0 1.83 0.14 0.07 2 1.83 0.15 0.15 2 1.83 0.14 0.07 Total Dead load of the superstructure Total DL coming to one abutment =
0.415 0.200 0.400
14.454 0.55 1.00
56.98 1.32 3.80
142.40 0.41 20.03 16.02 23.95 5.32 6.41 154.86 0.00 0.00 0.04 0.04
24 1367.64 24 31.68 24 91.20 1490.52
STRESS DUE TO LIVE LOAD Live load due to class A loading: Maximum load on the given abutment is produced when the train of load is positioned as shown in the figure: Axle Load,KN 27 27 114 114 68 68 68 68 xi , m 35.0 33.8 29.5 26.5 23.5 20.5 yi 1.0 1.0 0.8 0.8 0.7 0.6 load, KN 114.0 110.1 57.3 51.5 45.7 39.8 Maximum total live load = 418.4 KN Load due to impact = 418.37714 x 0.110 = 45.919 KN Live load including impact = 418.4 + 45.919 = 464.3 KN Live load due to class AA wheeled loading: Axle Load,KN xi , m yi load, KN Maximum total live load = Load due to impact = Live load including impact =
210.28571 210.3
Live load due to class AA tracked loading: Maximum total live load = Load due to impact = Live load including impact =
664 664.0
200 19.0 0.54 108.6
200 17.8 0.51 101.71 210.3 KN x 0.18 = + 37.851429 =
664.0 KN x +
Page 2
0.1 = 66.4 =
37.851 KN 248.1 KN
66.4 KN 730.4 KN
0.06 24 22 24 24 24 24 24 24 24 24 24
Total Wt e from c/c 8.54 9.72 440.55 384.48 574.80 127.73 153.79 ###### 0.00 0.00 0.99 0.86 ##### 2709 KN
0.000 m 2.208 m 0.007 m
Substructure Kondre (right) Out of above three loadings class AA tracked loading gives maximum load to abutment. Maximum live load to abutment = 730.4 KN CALCULATIONOF EARTH PRESSURE FOR RETAINING WALL Horizontal earth pressure due to backfil Corresponding to
F=
30 ° of the backfill material
g =
18 KN/m3
Ka =
g'
(
0.2973
=
8
Coulomb's theory applicable
Ep1 = Ko x g h /2 = 639.1 KN/m h= 15.454 m Horizontal component of earth pressure = 639.1 cos 20 which acts at 0.42 x h = 6.49 m from the top of footing base. Vertical component of earth pressure = 639.1 x which acts at 0.221 m inward from the CG of wall.
KN/m3
)
2
Total Horizontal Thrust, E1 = Ep1 x L = Total Vertical Thrust, E2 = Ep2 x L =
= sin 20
600.5 KN/m =
218.570 KN/m
5704.9 KN 2076.4 KN
WEIGHT OF BACKFILL BEHIND THE ABUTMENT Wb =
9.1
x
1.000
x
15.5
x
18
=
2531.37 KN
STRESS DUE TO SEISMIC FORCES Horizontal Forces due to seismicity are computed using the formula as shown in IS Code 1893 or IRC - 6 F eq = where
a= b= l=
axbx l xG 0.08 for the zone V of seismic intensity 1.2 for medium soil with isolated footing 1 importance factor
Thus seismic coefficient is taken as
0.096
a.
Seismic force due to Superstructure height of action of the force
= = =
b.
Seismic force due to Abutment act at cg of abutment
= =
c.
Seismic force due to Back fill acts at cg
= =
0.096 14.45 + 15.319 m
x 2709.05 0.025 +
0.096 x 7.727 m 0.096 x 7.727 m
2531.4 =
= 260.1 KN 1.68 / 2 =
1491 =
143.1 KN
243.01 KN
The critical case will be when seismic force acts in the direction of traffic. But according to IRC 6-1966 (222.7) "The seismic force due to live load shall not be considered when acting in the direction of traffic,..". STRESS DUE TO WATER CURRENT Horizontal Thrust due to water current is ignored, because it usually contributes to the stability of the wall and the load case with seismic condition does not concide with HFL.
STRESS DUE TO LONGITUDINAL FORCES Due to Tractive Effort or Braking Force Braking Force due to Class AA tracked load = @20% of live load/2 = = height of the line of action of the brake from the road level = height of the bearing level from the top of base =
Due to Resistance in Bearing (Temperature)
Page 3
1/2 to one abutment 55.4 KN 1.2 m 14.454 m
Substructure Kondre (right) Severe moment is produced when the LL is within span. Assume frictional coeff. = = Frictional resistance at free bearing
=
773.9 KN
Frictional resistance of fixed bearing
=
687.9 KN
Unbalanced frictional force = Line of action of the force from the bottom of the abutment =
0.225 in free bearing 0.2 in other bearing
86.0 KN 14.454 m
STRESS DUE TO WIND LOAD Exposed height of structure
= depth of the beams + thickness of slab + + height of kerb = = 2.68 m
Exposed area contributing to wind pressure =
93.8 m²
Avg. height of superstructure from the bed = intensity of wind load velocity of wind The design is to be conducted for one span only loaded with class A train of vehicle
= =
Length of Class A train
=
20.4 m
a.
Wind Force on Surface (SS)
=
0.73 x
b.
Wind Force on Moving vehicle Class A train acts at 1.5 m above road level
=
20.4 x =
Maximum load is obtained in exposed area = For lateral direction wind load acts at
6.000 m 0.73 KN/m² 107 km/hr
93.8 =
68.474 KN
3 = m
61.2 KN
-9.854 129.7 KN
= =
129.7 KN 15.454 m
SUMMARY OF LOADS FOR DESIGN OF ABUTMENTS S.No.
Load
Vertical (KN)
Horizontal (KN) Arm Later. (m)
Moment (KN) Longit. Later.
Longit. dir
1 1.1 1.2
Superstructure Dead load of SS(D) Live Load of SS (L) including impact Substructure DL of abutment(E) Retaining wall Return wall - Top section Ballast wall Backfill (BF) Earth pressure (EP) (EP1)
2
3 4 4.1 4.2 4.3 4.4 4.5 4.6
2709.1 438.0
0.48 1297.0 0.48 209.7
1367.6 31.7 91.2 2531 35062.779
Due to overburden(EP2) Frontal ive soil mass (EPp) Buoyancy (B) Longitudinal force Tractive force (T) Due to temperature (F) Wind Load Centrifugal force (CF) Thrust due to water current (WC) Seismic forces (FS) on superstructure on abutment on live load from backfill
Taking Rankin's earth pressure coefficient.
0.00 -2.21 -0.01 0.00 5704.9
6.49 ######
55.40 85.99
130.03 71.54 121.51
-69.9 -0.7 0.0
129.67
14.45 800.8 14.45 1242.8 15.45 2003.982
260.07 143.09 42.05 243.01 243.011
15.32 3984.0 3983.994 7.73 1105.7 1105.654 16.95 712.8818 7.727 1877.7 1877.746
260.07 143.09
Load combination Group II (Normal+Temp. condition) D + L + E + Ep1+ Ep2 + Ep3+ T + F + CF 4638
Page 4
5846
129.7
#####
2004.0
Substructure Kondre (right) Load combination Group IV (for Seismic dry condition) D + L + E + Ep1+ Ep2 + T + F + CF + FS
7492
6492
688.2
Note: Moment is taken at the critical section at the bottom of the ABUTMENT 2.2.2
STRUCTURAL DESIGN OF ABUTMENT RETAINING WALL
Design of the returning wall will be done based on normal loading condition, but checked on maximum loading Design parameters Vertical force = 4638 KN (Under combination group IV the permissible increase of stress is Long. Moment = 40508 KN-m 50%, hence Combination I governs the design longitudinal moment) Substructures are designed using M20 concrete and Fe415 steel reinforcement
Conccrete M20
Parameters for RCC design sck = 20 N/mm2
Steel
sy =
415 N/mm2
Page 5
#####
7680.3
Substructure Kondre (right) Approximate Area of Steel reinforcement required is calculated considering singly reinforced beam theory Ast = Mu/ (0.87xσy x 0.7984 x d) = Provide Required spacing =
234060 mm2
25 mm dia. bars 0.9 mm c/c
Ast = Provide at
Area of reinforcement provided, Ast =
490.625 mm2 100 mm c/c
45873.438 mm2 0.8047971 % >
r=
Decrease spacing !
0.2 %
At the frontal face, the design moment is greatly reduced because of lack of earth pressure. Therefore on this side provide only 50 % of the main reinforcement Ast =
234060 x0.5
=
Provide Required spacing =
117030 mm2
20 mm dia. bars 20 1.2 mm c/c
Ast = Provide at
Area of reinforcement provided, Ast =
314 mm2 100 mm c/c
Decrease specing !
29359 mm2
Same spacing of reinforcement is provided on the side faces of the abutment. Distribution re-bar Providing 0.12% of cross sectional area = Provide Required spacing =
0.0012 x
12 mm dia. bars 314.0 mm c/c
Check for stresses
Ast = Provide at
600 x 1000=
113.04 mm2 250 mm c/c
(For seismic condition) 7492018 ######### 6336.850 9500 600 512.5 75 75
mm mm mm mm mm mm
at both faces.
ok
Parameters for RCC design Conccrete M20
Normal Thrust , W = Bending Moment = Ecentricity, e = breadth, b = Total depth , D = Effective depth, d = Compressive steel cover, dc = Tensile steel cover, dt =
720 mm2
N N-mm > m=
Ac = At =
Steel
sck = sy =
20 N/mm2 415 N/mm2
128.125 mm 0.0
29359 mm2 45873.44 mm2
Ast = Mu/ (0.87xσy x 0.7984 x d) =178050.265 mm2/m
x=
0.87*σy*Ast 0.36*σck*b
8928 mm
150 mm say
Check for shear stress: Critical secion for shear in case of abutment stem is considered at distance d = of base slab. Horizontal force due to soil pressure = 5559.80 KN Breaking force = 55.40 KN Longitudinal force due to temperature effect = 85.99 KN Shear force V = 5701.18 KN Shear stress tc = V/(b d) =
A
0.744 N/mm2
Permissible shear stress without shear reinforcement tcmax =2.8
Mpa for M20 concrete
Where, As = Area of Steel provided in mm2 b = width of abutment vertical wall and d = thickness of abutment wall tc = basic values given for different p% of Steel provided = From Table 19 of IS 456-200 The design of ballest wall and return wall will be as in left side abutment. 2.1.3
512.5 mm from the junction
DESIGN OF BALLAST WALL
Page 6
p=
4.05%
0.28 Mpa for M20 concrete
<
2 0.744 N/mm
Substructure Kondre (right)
1.000 m
RCC balast wall (as shown in the figure) monolithic to the abutment has been designed. 0.40 m
CALCULATION OF DEAD LOAD STRESSES Dead load from the approach slab (triangular portion) = = Dead load of the Ballast wall
36 KN 3.7894737 KN/m
=
91.2 KN 9.6 KN/m
= Earth pressure from the back fill
= 21 KN/m3 0.2973
r= ca =
ca x r x h2 /2 =
3.12 KN/m
LIVE LOAD STRESS CALCULATION Class AA (tr) load component on wall , LL =
Add impact to live load @ , I =
291.667 30.702 3.070 0.15
= = = e=
0.1
KN KN/m KN/m m from axis of ballast wall
Horizontal Load due to braking of train of vehicle class A vehicle, T =(0.2 ) x(27+114*2) = (Front vehicle only considered)
=
51.0 KN 5.4 KN/m
Compaction load is maximum at the depth of 1.2m from the top. P comp = 25.2 KN/m Maximum shear force on section I-I H = T or Pcopm+ Ep = V = DL + LL + I =
28.3 KN/m 47.2 KN/m
Moment given due to the forces on the critical section I - I at the base of the ballast wall M = ( LL + I) x e + T x h + Ep x h /3 = = 17.92 KN-m/m Area of vertical steel reinforcement required Ast = Mu/ (0.87xσy x 0.7984 x d) =165.746387 mm2/m Provide
12 mm dia bars @ Area of steel provided = Provide 12 mm dia bars @ Area of steel provided = Horizontal links are provided of
200 c/c
for the continuation of bar from retaining wall 565.2 mm2/m 200 c/c on the opposite face 565.2 mm2/m 12 mm dia. bars @ 200 c/c
Stresses at the section, Shear stress,tmax =
0.083 N/mm2
Permissible shear stress without shear reinforcement tcmax = 2.8 2
Where, As = Area of Steel provided in mm
2.1.4
p=
tc = basic values given for different p% of Steel provided = From Table 19 of IS 456-200 DESIGN OF RETURN WALLS
Page 7
N/mm2
0.28%
0.28 0.38 Mpa for M20 concrete
>
2 0.083 N/mm ok
Substructure Kondre (right) These will be monolithic with the back wall. They are ed together through 300 mm x 300 mm fillets. The loads acting on the wing wall (return wall) would be earth pressure and horizontal force to kerb as specified in the code. STRESS DUE TO VERTICAL LOADS All the stresses are calculated for the critical section I-I at the bottom face of the abutment Dead load of the returnwall = 6.000 x Vert. LL from Class A load = Add Impact due to LL @ 0.25=
168.72 KN 42.18 KN
56.24 14.06
0.550
x
0.200
x 24 = =
15.84 KN 2.64 KN/m
KN/m KN/m
STRESSES DUE TO HORIZONTAL FORCES Horizontal Thrust due to DL+LL+IL=q*L*ka acts at h/2 0.28 m
= 24.070 KN/m from the top of foundation
Moment due to hor. component of vertical loads =
433.2636 KN-m/r.m ka x rs x h^2/2
Earth pressure from the backfill =
=
Moment due to earth pressure = Momt due to later. force =
16.998 KN-m 7.5 KN/m for the curb as per code = 135.000
Total horz. uniformly distributed load = Total Moment = 433.26 + depth required = where b = ( Provide overall depth =
17.00 + M /( 0.138σck b ) 700 + 650 mm
Ast = Mu/ (0.87xσy x 0.7984 x d) =
24.07
+ 135.000 =
= 400 ) / 2 =
0.94 KN/m
KN-m
0.94 +
7.5 =
32.51 KN/m
585.26 620.925 mm 550 mm
3530.97 mm2/m
Tensile steel re-bar equired = 3530.97 mm2 Provide 25 mm dia. bars Required spacing = 139.0 mm c/c
Ast = Provide at
2.2.3 DESIGN OF FOUNDATION Foundation is designed as open foundation of the rectangular shape Length = 6.00 m Breadth = 2.75 m heel toe width of abutment Height = 0.9 m Concrete cover 75 mm DEAD LOAD due to foundation =
490.87 mm2 150 mm c/c
Decrease specing !
1.000 m 1.000 m 0.600 m
356.4 KN
CALCULATIONOF EARTH PRESSURE FOR FOUNDATION Horizontal earth pressure due to backfill Ka = 0.2973 As calculated earlier for abutment stem.
( Coulomb's theory applicable
Ep1 = Ka x g h2 /2 = 715.66 KN/m h = 16.354 m Horizontal component of earth pressure = 715.7 x cos 20 which acts at 0.42 x h = 6.87 m from the bottom of footing base.
=
)
672.499 KN/m
Pressure due to surcharge: As approach slab has been provided no surcharge load is to be taken as per code. WEIGHT OF BACKFILL BEHIND THE ABUTMENT Wb = Wf =
6.000 6.000 acts at
x x 0.8
1.000 1.000 m
x 15.454 x 0 from centre of abutment
Seismic force due to Foundation act at cg of abutment Seismic force due to back fill act at cg of backfill SUMMARY OF LOADS FOR DESIGN OF FOUNDATION
x x
= = = =
Page 8
18.00 = 18.00 =
0.096 x 0.45 m 0.096 x 8.627 m
1669.0 KN 0.0 KN
356.4 =
34.214 KN
1669.0 = 160.23 KN from the bottom of foundation base
Substructure Kondre (right) S.No.
Load
Vertical (KN)
Horiz (KN)
1 1.1 1.2
2
4.1 4.2 4.3
M(o) KN-m
Superstructure Dead load of SS(D) Live Load on SS including impact (L)
2709.05 438.0
0.8 2224.79 0.8 359.705
Substructure DL of abutment(E) Retaining wall Return wall Ballast wall
1367.64 31.68 91.20
1.3 1777.93 3.5 111.118 1.3 118.56
1669.03
2.10 3504.97
Backfill (BF) Backfill (FR) Earth pressure (EP) Triangular dry(EP1)
3 4
L. arm M[r] (m) KN-m
6388.7
Due to overburden(EP2) Frontal ive soil mass (EPp) DL of foundation Buoyancy (B) Longitudinal force Breaking (breaking) force (T) Due to temperature (F) Wind Load
-6.9
-43882
0.00 356.40
Centrifugal force (CF) Thrust due to water current (WC) Seismic forces (FS) on superstructure on abutment on live load on foundation from backfill Load combination Group I (N+T condition) D + L + I + E + Ep1+ Ep2 +BF+ T + F + CF Load combination Group IV (for Seismic dry condition) D + L + E + Ep1+ Ep2 +BF+ T + F + CF + FS
0 1.4
490.05
55.40 85.99
-15.4 -15.4
-850.6 -1320
130.03 71.54
260.1 143.1
-15.3 -7.7
17.11 80.11
34.2 160.2
-0.5 -8.6
-1992 -552.8 0 -7.698 -691.1
6663.0
6530
8587.12 -46053
6961.8
7128
8587 -48446
4.4 4.5 4.6
Note: Moment is taken
about the toe for check against overturning . 403.82
For Normal condition P= 6663 KN Mnet = Mr-Mo= -37465.918 x = Mnet/P= -5.6229806 e = b/2-x'= 6.9979806 > b/6, hence tesion Pmax=P/A*(1+6*e/b) = 6569.444 KN/m2 Pmin=P/A*(1-6*e/b) = -5761.807 KN/m2 Where, A=b x d = 16.5 m2 CHECK AGAINST OVERTURNING Case: Normal Factor of Safety against overturning =Mres/Mov= Case: Seismic Factor of Safety against overturning =Mres/Mov=
For Seismic condition P= 6962 KN Mnet = Mr-Mo= -39859 x = Mnet/P= -5.7254 e = b/2-x'= 7.10 Pmax=P/A*(1+6*e/b) = Pmin=P/A*(1-6*e/b) =
KN-m m > b/6, hence tesion 6958.3 KN/m2 -6114 KN/m2
8587.1227
/
46053 =
0.2
> 2.0 Safe
8587.1227
/
48446 =
0.2
>1.5 Safe
CHECK AGAINST SLIDING Case : Normal The minimum vertical force will be when there is no live load and the front backfill is washed out = ###### KN The resisting force againest sliding = (vertical force) x (coefficient of friction b/w concrete and soil) = 6225.00 x 0.45 = 2801.25 KN The sliding force is the horizontal forces = 6474.72 KN Hence, Factor of Safety against sliding =Pres/Psl= 2801.25 / 6474.7 = 0.433 >1.5 Hence Safe (f= 0.45 soil/concrete) Case: Seismic The minimum vertical force will be when there is no live load and the front backfill is washed out = ###### The resisting force againest sliding = (vertical force) x (coefficient of friction b/w concrete and soil) = 6523.80 x 0.45 = 2935.71 KN
Page 9
Substructure Kondre (right) The sliding force is the horizontal forces = Hence, Factor of Safety against sliding =Pres/Psl=
7072.32 2935.71
KN /
7072.3 =
0.415 >1.25 Hence Safe
STRUCTURAL DESIGN OF FOUNDATION
6569.44 KN/m2
Wt of toe slab =
21.6 KN/m2
D
F
G
1412.739
C B #####
A
Design of toe slab:
E
-1277.72
1.000 m
0.600 m 2.75 m
1.000 m
Pressure distribution diagram
####
6547.84 Moment about EF =
concrete grede : Steel grade :
2414.47 KN-m d= M /( 0.138σck b ) 564.0 mm Adopt D = 600 mm d= 500 mm Ast = Mu/ (0.87xσy x 0.7984 x d)16752 = mm2 Using 16 mm dia bar, spacing = Adopt 16 mm dia bar, spacing =
M 20 Fe 415
2414.47
Minimum reinforc. as per code = 1000 mm2 12.0 mm 150 mm c/c Area of re-bar provided = 1339.733 mm2
Check for shear stress:
Shear stress =
V/(b d) =
b=
6.00 m
1391
15776.00 KN
500 3969
Shear force at critical section V =
d =
6548
In toe critical section for shear is considered at a distance from abutment stem.
5.259 N/mm2
Permissible shear stress without shear reinforcement tcmax =2.8 Mpa for M20 concrete Where, k1 = 1.14 - 0.7 d >= 0.5 ( d in m ) Where, As = Area of Steel provided in mm2 b = width of abutment vertical wall and d = thickness of abutment wall p=
0.5 m
0.037% 0.04 2 0.280 N/mm2 < 5.259 N/mm Not ok. Revise section or provide shear re-bar
tc = basic values given for different p% of Steel provided = From Table 19 of IS 456-200
Design of heel slab: Upward pressure on the heel slab varies from -5761.8073 Kn/m2 to -1277.72 Kn/m2 In addition to the upward pressure the heel slab is subjected to downward pressure due to the self weight of heel slab and weight of earth fill on heel slab: Down pressure due to self weight = 14.4 KN/m2 Down pressure due to backfill = 278.172 KN/m2 292.572 KN/m2 1570 6054 Net down pressure = Moment about DG= d= M /( 0.138σck b ) Adopt
6054 1570 2279.841 908.862
KN/m2 KN/m2 KN-m mm
max min
D= 600 mm d= 500 mm Ast = Mu/ (0.87xσy x 0.7984 x d) = 15818 mm2 Using 16 mm dia bar, specing = Adopt 16 mm dia bar, specing =
Page 10
1.000 m
Minimum reinforc. as per code = 1000 mm2 13 mm 150 mm c/c Area of re-bar provided = 1339.7 mm2
Substructure Kondre (right) Distribution re-bar 0.12/100 x
900 Using Adopt
=
1080 mm2/runnig meter 12 mm dia bar, specing = 209.3 mm 12 mm dia bar, specing = 200 mm c/c
at top and bottom
Check for shear stress: Since the load on the heel is predominantly downward, it induces tensile reactio at its junction with stem. Hence the critical section for shear for the heel slab is considered at the face of . Shear force V = 22874.002 KN b= 6.00 m Shear stress = V/(b d) = 7.625 N/mm2 Permissible shear stress without shear reinforcement tcmax =2.8 Where, k1 = 1.14 - 0.7 d >= 0.5 ( d in m )
N/mm2
Where, As = Area of Steel provided in mm2 b = width of abutment vertical wall and d = thickness of abutment wall tc = basic values given for different p% of Steel provided = From Table 19 of IS 456-200
Mpa for M20 concrete
p= 0.28 N/mm2
Page 11
0.037%
<
0.04 2 7.6247 N/mm
Not ok. Revise section or p
Substructure Kondre (right)
Page 12
Substructure Kondre (right)
i
i Pa
Weight of soil is not to be considered in Coulumb's Theroy.
d 90-b
Page 13
Ws
Pa i
Substructure Kondre (right)
d 90-b
Weight of soil is not to be considered in Coulumb's Theroy.
i 0 h= a= Coul. Rankin
Pa i
b
b Coulomb
F 30
b 90
d 20
30
a Let us find out which theory is applicable:
60
1.7321 m <
0.75 0.9397 2.6845 1 0.75 0.5
Hor pressure from Rankin's theory = Horizontal thrust = 6806.5 acts at
Rankin Weight of soil is also to be considered in Rankin's Theroy.
1.5
15.454 m hence Coulomb's theory applicable Ka= 0.2973 0.3333
716.48 KN/m 5.15 m from the top of foundation footing.
Page 14
Substructure Kondre (right)
Page 15
Substructure Kondre (right)
Page 16
Substructure Kondre (right)
Decrease specing !
Page 17
Substructure Kondre (right)
Page 18
Substructure Kondre (right)
Page 19
Substructure Kondre (right)
> b/6, hence tesion
Page 20
Substructure Kondre (right)
Page 21
Substructure Kondre (right)
Not ok. Revise section or provide shear re-bar
Page 22
Dimension of slab and beams (brought from slab sheet) Length of single span = Carriageway width, Bc = Overall width of Bridge, Bo =
Check for correctness of data 8.000 m 3.000 m 6.000 m
No. of main girder = 3 Dimensions of the longitudinal girder is taken as follows: Length (L si ) = 8.000 m Breadth (B si ) Thickness(Tsi) Distance betn girders =
=
0.600 m
=
0.750 m 2.400 m
No. of cross beams = 2 For the design purposes the dimensions of the cross beam is taken as follows: Length (L si ) = 3.000 m Breadth (B si ) = 0.600 m Thickness(Tsi) = 0.900 m Distance betn cross beams = 8.000 m For the design purposes the dimensions of the internal slab is taken as follows: Length (L si ) = 8.600 m Hence, effec.span in long.dir. = Breadth (B si ) = 3.600 m Hence, effec.span in transv.dir. = Thickness(Tsi) = 0.200 m Avg. pavement cover over Slab(T sp ) = 0.070 m kerb/footpath breadth = 1.000 m at bottom and 0.225 m at top height = 0.500 m For design purpose dimension of cantilever slab is taken as Length (L si ) = 1.00 m Breadth (B si ) = 0.600 m from c/c of outer girder Breadth (B si ) = 0.300 m from outer face of outer main girder Breadth of carriageway = 2.400 m from c/c of outer girder 0.300 m from outer face of outer main girder Thickness(Tsi) = 0.200 m at the end Thickness(Tsi) = 0.200 m at outer face of main girder Avg. pavement cover over Slab(T sp ) = 0.07 m
2.2
DESIGN OF RIGHT ABUTMENT
The freebody diagram of the R.C.C Abutment is shown below:
0.20
i=0 Pa
0.60 3.100
d
CG of Stem from inner vertical face: A 1=
1.860
A2=
0.233
M.A.=
0.709
3.100
A1+A2 =
0.750 A
b = 90
CG =
Dimensions taken for the design of abutment: Retaining wall Length = 6.000 Breadth = 0.750 Height = 3.100 Ballast wall Length = 6.000 Breadth = 0.200 Height = 1.000 Return wall No. = 2 Breadth = 0.200 Length = 6.000 Total height of abutment = Approach slab Length = Breadth = Thickness = Foundation is open foundation Length = Width Height of footing = Volume =
m m at bottom
m m m m
m m
Bottom level of base = Road level = 0.60 m at top 1.25 CG of retain. wall from point A = 0.339 m
height of superstr. plus bearing without pav.cover Height of Return wall H1= Height of Return wall H2= 4.100 m
5.000 m 3.000 m 0.200 m
0.7 0.4
5.000 0.000
6.000 m 2.750 m 0.900
2.093
0.339
Heel Toe
1.000 m 1.000 m
W =
356.40 KN
0.750 0.750
1.250
14.850 m3
2.2.1 CALCULATION OF LOAD STRESS DUE TO DEAD LOAD S.N Elements
Unit
Dead Load of Superstructure 1 Hand rails 6.368 2 Railing posts 3 Wearing course 4 Kerbs 5 Deck slab i) rectangle ii)trapezoid 6 Main Girder i) fillets ii) webs 7 Internal Cross Girder i) fillets ii) webs 8 End Cross Girder i) fillets ii) webs
rm cum cum cum cum cum cum cum cum cum cum cum
No.
L(m)
B(m)
H(m) Volume
4 8.60 5 0.15 0.15 1 8.60 3.00 2 8.60 0.61 1 8.60 3.00 2 8.60 0.30 4 8.60 0.30 3 8.60 0.60 0 2.40 0.15 0 2.40 0.60 2 2.40 0.15 2 2.40 0.60 Total Dead load of the superstructure Total DL coming to one abutment =
1.00 0.07 0.50 0.20 0.20 0.15 0.55 0.15 0.70 0.15 0.70
34.40 0.11 1.81 5.27 5.16 1.03 0.77 8.51 0.00 0.00 0.05 2.02
Dead Load of Abutment retaining wall return walls ballast wall
2
6.000 6.000 6.000
0.675 0.200 0.200
3.100 0.55 1.00
12.56 1.32 1.20
24 24 24
301.32 31.68 28.80 361.80
STRESS DUE TO LIVE LOAD Live load due to class A loading: Maximum load on the given abutment is produced when the train of load is positioned as shown in the figure: Axle Load,KN 27 27 114 114 68 68 68 68 xi , m 8.0 6.8 2.5 -0.5 -3.5 -6.5 yi 1.0 0.9 0.3 -0.1 -0.4 -0.8 load, KN 114.0 96.9 21.3 -4.3 -29.8 -55.3 Maximum total live load = 142.9 KN Load due to impact = 142.9 x 0.321 = 45.932 KN Live load including impact = 142.9 + 45.932 = 188.8 KN Live load due to class AA wheeled loading: Axle Load,KN xi , m yi load, KN Maximum total live load = Load due to impact = Live load including impact = Live load due to class AA tracked loading: Maximum total live load = Load due to impact = Live load including impact =
200 19.0 2.38 475.0 920 920.0
200 17.8 2.23 445.00 920.0 KN x +
0.18 = 165.6 =
165.6 KN 1085.6 KN
542.5 542.5
542.5 KN x +
0.1 = 54.25 =
54.25 KN 596.8 KN
Out of above three loadings class AA tracked loading gives maximum load to abutment. Maximum live load to abutment = 1085.6 KN CALCULATIONOF EARTH PRESSURE FOR RETAINING WALL Horizontal earth pressure due to backfil Corresponding to
F=
30 ° of the backfill material
g =
18 KN/m3
Ka =
0.2973
g'
(
=
8
Coulomb's theory applicable
Ep1 = Ko x g h /2 = 45.0 KN/m h= 4.100 m Horizontal component of earth pressure = 45.0 cos 20 = which acts at 0.42 x h = 1.72 m from the top of footing base. Vertical component of earth pressure = 45.0 x sin 20 which acts at 0.339 m inward from the CG of wall.
KN/m3
)
2
Total Horizontal Thrust, E1 = Ep1 x L =
253.6 KN
42.3 =
Total Vertical Thrust, E2 = Ep2 x L =
92.3 KN
WEIGHT OF BACKFILL BEHIND THE ABUTMENT Wb =
5.6
x
1.000
x
4.1
x
18
=
413.28
STRESS DUE TO SEISMIC FORCES Horizontal Forces due to seismicity are computed using the formula as shown in IS Code 1893 or IRC - 6 F eq = wherea = b=
l=
axbx l xG 0.08 for the zone V of seismic intensity 1.2 for medium soil with isolated footing 1 importance factor
Thus seismic coefficient is taken as
0.096
a.
Seismic force due to Superstructure height of action of the force
= = =
b.
Seismic force due to Abutment act at cg of abutment
= =
c.
Seismic force due to Back fill acts at cg
= =
0.096 3.10 + 3.5 m
x 296.058 0.025 +
0.096 x 2.05 m 0.096 x 2.05 m
362
413.28 =
The critical case will be when seismic force acts in the direction of traffic. But according to IRC 6-1966 (222.7) "The seismic force due to li load shall not be considered when acting in the direction of traffic,..". STRESS DUE TO WATER CURRENT Horizontal Thrust due to water current is ignored, because it usually contributes to the stability of the wall and the load case with seismic condition does not concide with HFL.
STRESS DUE TO LONGITUDINAL FORCES Due to Tractive Effort or Braking Force Braking Force due to Class AA tracked load = @20% of live load/2 = = height of the line of action of the brake from the road level = height of the bearing level from the top of base =
55.4 KN 1.2 3.100 m
Due to Resistance in Bearing (Temperature) Severe moment is produced when the LL is within span. Assume frictional coeff. =
0.225
= Frictional resistance at free bearing
=
310.9 KN
Frictional resistance of fixed bearing
=
276.3 KN
Unbalanced frictional force = Line of action of the force from the bottom of the abutment =
34.5 KN
0.2
3.100 m
STRESS DUE TO WIND LOAD Exposed height of structure
= depth of the beams + thickness of slab + + height of kerb = = 1.25 m Exposed area contributing to wind pressure = Avg. height of superstructure from the bed = intensity of wind load = velocity of wind = The design is to be conducted for one span only loaded with class A train of vehicle Length of Class A train
=
20.4 m
a.
Wind Force on Surface (SS)
=
0.73 x
b.
Wind Force on Moving vehicle Class A train acts at 1.5 m above road level
=
20.4 x =
Maximum load is obtained in exposed area = For lateral direction wind load = acts at =
10 m² 6.000 m 0.73 KN/m² 107 km/hr
10 =
1.500
3= m
68.5 KN 68.5 KN 4.100 m
SUMMARY OF LOADS FOR DESIGN OF ABUTMENTS S.N o.
Load
Vertical (KN)
Horizontal (KN) Longit. dir
1
2
Superstructure 1.1 Dead load of SS(D) 1.2 Live Load of SS (L) including impact Substructure DL of abutment(E) Retaining wall Return wall - Top section Ballast wall Backfill (BF) Earth pressure (EP) (EP1) Due to overburden(EP2)
296.1 438.0
301.3 31.7 28.8 413 413.526
Taking Rankin's earth pressure coefficient.
253.6
Later.
3 4 4.1 4.2 4.3 4.4 4.5 4.6
Taking Rankin's earth pressure coefficient.
Frontal ive soil mass (EPp) Buoyancy (B) Longitudinal force Tractive force (T) Due to temperature (F) Wind Load Centrifugal force (CF) Thrust due to water current (WC) Seismic forces (FS) on superstructure on abutment on live load from backfill
55.40 34.54 68.50
14.21 17.37
28.42 34.73
28.42 34.73 42.05 39.675 39.6749
19.84
Load combination Group II (Normal+Temp. condition) D + L + E + Ep1+ Ep2 + Ep3+ T + F + CF Load combination Group IV (for Seismic dry condition) D + L + E + Ep1+ Ep2 + T + F + CF + FS
1096
344
68.5
1561
446
144.9
Note: Moment is taken at the critical section at the bottom of the ABUTMENT 2.2.2 STRUCTURAL DESIGN OF ABUTMENT RETAINING WALL Design of the returning wall will be done based on normal loading condition, but checked on maximum loading Design parameters Vertical force = 1096 KN (Under combination group IV the permissible increase of stress is Long. Moment = 753 KN-m 50%, hence Combination I governs the design longitudinal moment) Substructures are designed using M20 concrete and Fe415 steel reinforcement
Conccrete M20
Parameters for RCC design sck = 20 N/mm2
sy = Steel 415 N/mm2 Approximate Area of Steel reinforcement required is calculated considering singly reinforced beam theory Ast = Mu/ (0.87xσy x 0.7984 x d) = Provide Required spacing =
3332 mm2
25 mm dia. bars 88.3 mm c/c
Area of reinforcement provided, Ast = r=
Ast = Provide at
490.625 mm2 100 mm c/c
28701.563 mm2 0.6378125 % >
At the frontal face, the design moment is greatly reduced because of lack of earth pressure. Therefore on this side provide only 50 % of the main reinforcement Ast =
3332 x0.5
=
1666 mm2
0.2 %
Provide Required spacing =
20 mm dia. bars 20 113.1 mm c/c
Ast = Provide at
Area of reinforcement provided, Ast =
314 mm2 100 mm c/c
18369 mm2
Same spacing of reinforcement is provided on the side faces of the abutment. Distribution re-bar Providing 0.12% of cross sectional area = Provide Required spacing =
0.0012 x
12 mm dia. bars 251.2 mm c/c
Check for stresses
Ast = Provide at
750 x 1000=
113.04 mm2 250 mm c/c
(For seismic condition)
Parameters for RCC design Conccrete M20
Normal Thrust , W = Bending Moment = Ecentricity, e = breadth, b = Total depth , D = Effective depth, d = Compressive steel cover, dc = Tensile steel cover, dt =
1560553 ######### 643.948 6000 750 662.5 75 75
mm mm mm mm mm mm
900
N N-mm > m=
Ac = At =
Steel
sck = sy =
165.625 mm 0.0
18369 mm2 28701.56 mm2
Ast = Mu/ (0.87xσy x 0.7984 x d)3768.77108 = mm2/m
x=
0.87*σy*Ast 0.36*σck*b
189 mm
150 mm say
Check for shear stress: Critical secion for shear in case of abutment stem is considered at distance d = of base slab. Horizontal force due to soil pressure = 173.25 KN Breaking force = 55.40 KN Longitudinal force due to temperature effect = 34.54 KN Shear force V = 263.19 KN Shear stress tc = V/(b d) =
662.5 mm from the junction
A
0.125 N/mm2
Permissible shear stress without shear reinforcement tcmax =2.8
Mpa for M20 concrete
2
Where, As = Area of Steel provided in mm b = width of abutment vertical wall and d = thickness of abutment wall tc = basic values given for different p% of Steel provided = From Table 19 of IS 456-200 The design of ballest wall and return wall will be as in left side abutment.
p=
0.21%
0.28 Mpa for M20 concrete
2.1.3 DESIGN OF BALLAST WALL
1.000 m
RCC balast wall (as shown in the figure) monolithic to the abutment has been designed. 0.20 m
CALCULATION OF DEAD LOAD STRESSES Dead load from the approach slab (triangular portion) = = Dead load of the Ballast wall
36 KN 6 KN/m
=
28.8 KN 4.8 KN/m
= Earth pressure from the back fill
= 21 KN/m3 0.2973
r= ca =
ca x r x h2 /2 =
3.12 KN/m
LIVE LOAD STRESS CALCULATION Class AA (tr) load component on wall , LL =
Add impact to live load @ , I =
0.1
= = = e=
291.667 48.611 4.861 0.15
KN KN/m KN/m m from axis of ballast wall
Horizontal Load due to braking of train of vehicle class A vehicle, T =(0.2 ) x(27+114*2) = (Front vehicle only considered) Compaction load is maximum at the depth of 1.2m from the top. P comp = 25.2 KN/m Maximum shear force on section I-I H = T or Pcopm+ Ep = V = DL + LL + I =
28.3 KN/m 64.3 KN/m
Moment given due to the forces on the critical section I - I at the base of the ballast wall M = ( LL + I) x e + T x h + Ep x h /3 = = 27.76 KN-m/m Area of vertical steel reinforcement required
=
Ast = Mu/ (0.87xσy x 0.7984 x d)256.816378 = mm2/m Provide 12 mm dia bars @ Area of steel provided = Provide 12 mm dia bars @ Area of steel provided = Horizontal links are provided of
200 c/c
for the continuation of bar from retaining wall 565.2 mm2/m 200 c/c on the opposite face 565.2 mm2/m 12 mm dia. bars @ 200 c/c
Stresses at the section, Shear stress,tmax =
0.202 N/mm2
Permissible shear stress without shear reinforcement tcmax = 2.8 Where, As = Area of Steel provided in mm2
p=
N/mm2
0.57%
0.57
tc = basic values given for different p% of Steel provided = From Table 19 of IS 456-200 2.1.4 DESIGN OF RETURN WALLS
0.51 Mpa for M20 concrete
These will be monolithic with the back wall. They are ed together through 300 mm x 300 mm fillets. The loads acting on the wing wall (return wall) would be earth pressure and horizontal force to kerb as specified in the code. STRESS DUE TO VERTICAL LOADS All the stresses are calculated for the critical section I-I at the bottom face of the abutment Dead load of the returnwall = 6.000 x 0.550 Vert. LL from Class A load = Add Impact due to LL @ 0.25=
168.72 KN 42.18 KN
56.24 14.06
x
0.200
KN/m KN/m
STRESSES DUE TO HORIZONTAL FORCES Horizontal Thrust due to DL+LL+IL=q*L*ka acts at h/2 0.28 m Moment due to hor. component of vertical loads =
433.2636 KN-m/r.m ka x rs x h^2/2
Earth pressure from the backfill = Moment due to earth pressure = Momt due to later. force =
= 24.070 KN/m from the top of foundation
16.998 KN-m 7.5 KN/m for the curb as per code = 135.000
=
KN-m
Total horz. uniformly distributed load = 24.07 + 0.94 + Total Moment = 433.26 + 17.00 + 135.000 = 585.26 depth required = M /( 0.138σck b ) = 620.925 mm where b = ( 700 + 400 ) / 2 = 550 mm Provide overall depth = 650 mm Ast = Mu/ (0.87xσy x 0.7984 x d) = 3530.97 mm2/m
0.94
7.5
Tensile steel re-bar equired = 3530.97 mm2 Provide 25 mm dia. bars Required spacing = 139.0 mm c/c
Ast = Provide at
2.2.3 DESIGN OF FOUNDATION Foundation is designed as open foundation of the rectangular shape Length = 6.00 m Breadth = 2.75 m heel toe width of abutment Height = 0.9 m Concrete cover 75 mm DEAD LOAD due to foundation =
490.87 mm2 150 mm c/c
1.000 m 1.000 m 0.750 m
356.4 KN
CALCULATIONOF EARTH PRESSURE FOR FOUNDATION Horizontal earth pressure due to backfill Ka = 0.2973 As calculated earlier for abutment stem.
( Coulomb's theory applicable
Ep1 = Ka x g h2 /2 = 66.90 KN/m h= 5.000 m Horizontal component of earth pressure = 66.9 x cos 20 which acts at 0.42 x h = 2.10 m from the bottom of footing base.
=
Pressure due to surcharge: As approach slab has been provided no surcharge load is to be taken as per code. WEIGHT OF BACKFILL BEHIND THE ABUTMENT Wb = Wf =
6.000 6.000 acts at
x x 0.875
1.000 1.000 m
x 4.100 x 0 from centre of abutment
Seismic force due to Foundation act at cg of abutment Seismic force due to back fill act at cg of backfill SUMMARY OF LOADS FOR DESIGN OF FOUNDATION S.No . Load
x x
= = = =
Vertical (KN)
18.00 = 18.00 =
442.8 0.0
0.096 x 0.45 m 0.096 x 2.95 m
356.4
Horiz (KN)
1
2
Superstructure 1.1 Dead load of SS(D) 1.2 Live Load on SS including impact (L) Substructure DL of abutment(E) Retaining wall Return wall Ballast wall Backfill Backfill
(BF) (FR)
442.8 from the bottom of foundation ba
L. arm (m)
296.06 438.0
1.2 1.2
301.32 31.68 28.80
1.4 3.7 1.4
442.80
2.25
Earth pressure (EP) Triangular dry(EP1)
377.2
Due to overburden(EP2) Frontal ive soil mass (EPp) DL of foundation 3 Buoyancy (B) 4 Longitudinal force 4.1 Breaking (breaking) force (T) 4.2 Due to temperature (F) 4.3 Wind Load
0.00 356.40
4.4 Centrifugal force (CF) 4.5 Thrust due to water current (WC) 4.6 Seismic forces (FS) on superstructure on abutment on live load on foundation from backfill Load combination Group I (N+T condition) D + L + I + E + Ep1+ Ep2 +BF+ T + F + CF Load combination Group IV (for Seismic dry condition) D + L + E + Ep1+ Ep2 +BF+ T + F + CF + FS Note: Moment is taken For Normal condition P= 1895 Mnet = Mr-Mo= x = Mnet/P= e = b/2-x'= Pmax=P/A*(1+6*e/b) = Pmin=P/A*(1-6*e/b) = Where, A=b x d =
-2.1
1.4
55.40 34.54
-4.0 -4.0
14.21 17.37
28.4 34.7
-3.5 -2.1
17.11 21.25
34.2 42.5
-0.5 -3.0
1895.1
467
1965.0
607
about the toe for check against overturning . For Seismic condition P= 1965 Mnet = Mr-Mo= x = Mnet/P= e = b/2-x'= Pmax=P/A*(1+6*e/b) = Pmin=P/A*(1-6*e/b) =
KN 1797.12344 0.94832107 0.42667893 ok 221.772 KN/m2 7.932 KN/m2 16.5 m2
CHECK AGAINST OVERTURNING Case: Normal Factor of Safety against overturning =Mres/Mov= Case: Seismic Factor of Safety against overturning =Mres/Mov=
2948.9419
/
1151.8 =
2948.9419
/
1086 =
CHECK AGAINST SLIDING Case : Normal The minimum vertical force will be when there is no live load and the front backfill is washed out = The resisting force againest sliding = (vertical force) x (coefficient of friction b/w concrete and soil) = 1457.06 x 0.45 = 655.676 The sliding force is the horizontal forces = 411.71 KN Hence, Factor of Safety against sliding =Pres/Psl= 655.68 / 411.7 = (f= 0.45 soil/concrete) Case: Seismic The minimum vertical force will be when there is no live load and the front backfill is washed out =
The resisting force againest sliding = (vertical force) x (coefficient of friction b/w concrete and soil) = 1527.00 x 0.45 = 687.149 The sliding force is the horizontal forces = 551.59 KN Hence, Factor of Safety against sliding =Pres/Psl= 687.15 / 551.6 = STRUCTURAL DESIGN OF FOUNDATION
Wt of toe slab =
21.6 KN/m2
221.77 KN/m2
A
Design of toe slab:
E
D
F
G
144.012
85.692
1.000 m
0.750 m 2.75 m
Pressure distribution diagram
####
200.17 Moment about EF =
concrete grede : Steel grade :
87.13 KN-m d= M /( 0.138σck b ) 107.1 mm Adopt D = 600 mm d= 500 mm Ast = Mu/ (0.87xσy x 0.7984 x d) =604 mm2 Using 16 mm dia bar, spacing = Adopt 16 mm dia bar, spacing =
Minimum reinforc. as per code = 201.1 mm 150 mm c/c Area of re-bar provided =
Check for shear stress:
Shear stress =
V/(b d) =
542.20 KN
b=
500
6.00 m
161.3
Shear force at critical section V =
d =
#####
In toe critical section for shear is considered at a distance from abutment stem.
2
0.181 N/mm
Permissible shear stress without shear reinforcement tcmax =2.8 Mpa for M20 concrete Where, k1 = 1.14 - 0.7 d >= 0.5 ( d in m ) Where, As = Area of Steel provided in mm2 b = width of abutment vertical wall and d = thickness of abutment wall p= tc = basic values given for different p% of Steel provided = From Table 19 of IS 456-200
0.5 m
0.280 N/mm2 ok. Shear re-bar is not necessary.
0.037%
>
Design of heel slab: Upward pressure on the heel slab varies from 7.93215698 Kn/m2 to 85.69 Kn/m2 In addition to the upward pressure the heel slab is subjected to downward pressure due to the self weight of heel slab and weight of earth fill on heel slab: Down pressure due to self weight = 14.4 KN/m2 Down pressure due to backfill = 73.8 KN/m2 88.2 KN/m2 3 Net down pressure = Moment about DG= d= M /( 0.138σck b ) Adopt
80 3 27.174 99.225
D= d= Ast = Mu/ (0.87xσy x 0.7984 x d) = Using 16 Adopt 16 Distribution re-bar 0.12/100 x 900 = Using 12 Adopt 12
KN/m2 KN/m2 KN-m mm
max min
600 mm 500 mm 189 mm2 mm dia bar, specing = mm dia bar, specing =
1.000 m
Minimum reinforc. as per code = 201 mm 150 mm c/c Area of re-bar provided =
1080 mm2/runnig meter mm dia bar, specing = 209.3 mm mm dia bar, specing = 200 mm c/c
at top and bottom
Check for shear stress: Since the load on the heel is predominantly downward, it induces tensile reactio at its junction with stem. Hence the critical section for shear for the heel slab is considered at the face of . Shear force V = 248.3274 KN b= 6.00 m Shear stress = V/(b d) = 0.083 N/mm2 Permissible shear stress without shear reinforcement tcmax =2.8 Where, k1 = 1.14 - 0.7 d >= 0.5 ( d in m )
N/mm2
Where, As = Area of Steel provided in mm2 b = width of abutment vertical wall and d = thickness of abutment wall tc = basic values given for different p% of Steel provided = From Table 19 of IS 456-200
Mpa for M20 concrete
p=
0.037%
0.28 N/mm2
>
k for correctness of data
8.000 m 3.000 m
0.000 5.000 ######
Unit Wt
Total Wt e from c/c
0.06 2.06 24 2.70 22 39.73 24 126.42 24 123.84 24 24.77 24 18.58 24 204.34 24 0.00 24 0.00 24 1.30 24 48.38 592.12 296.1 KN
0.000 m 2.338 m 0.238 m
i Pa
i 0 h= a= KN/m
Coul. Rankin
Ws
d 90-b
Weight of soil is not to be considered in Coulumb's Theroy.
b
b Coulomb
F 30
b 90
d 20
30
Let us find out which theory is applicable:
60
1.7321 m <
0.75 0.9397 2.6845 1 0.75 0.5
1.5
4.100 m hence Ka= 0.2973 0.3333
15.384 KN/m Hor pressure from Rankin's theory = Horizontal thrust = 302.6 acts at
50.43 KN/m 1.37 m from the top of foundation footing.
KN
= 28.42 KN 0.75 / 2 =
=
34.73 KN
39.675 KN
6 (222.7) "The seismic force due to live
1/2 to one abutment m
in free bearing
in other bearing
7.3 KN
61.2 KN
Arm (m)
Moment (KN) Longit. Later. 0.16 0.16
47.7 70.6
0.00 -2.34 -0.24 0.00
-74.1 -6.8 0.0
1.72
436.7
3.10 3.10 4.10
3.50 2.05 5.60 2.05
171.7 107.1 280.85
99.5 99.47549 71.2 71.20224 235.4688 81.3 81.3335
752.9
280.9
1004.9
487.5
le increase of stress is gn longitudinal
Decrease spacing !
ok
mm2
at both faces.
ok
Parameters for RCC design 20 N/mm2 415 N/mm2
mm from the junction
>
2 0.125 N/mm
51.0 KN 8.5 KN/m
rom retaining wall
>
2 0.202 N/mm ok
specified in the code.
x 24 = =
15.84 KN 2.64 KN/m
KN/m
=
32.51 KN/m
Decrease specing !
)
62.8613 KN/m
KN KN
=
34.214 KN
= 42.509 KN from the bottom of foundation base
M[r]
M(o)
KN-m
KN-m
359.382 531.683
414.315 117.612 39.6 996.3
-792.1 0 490.05
-221.6 -138.2
-49.74 -35.6 0 -7.698 -62.7 2948.94
-1152
2949
-1086
114.85
eismic condition KN 1862.99 0.94809 0.43 =P/A*(1+6*e/b) = =P/A*(1-6*e/b) =
KN-m m ok 230.02 KN/m2 8.1638 KN/m2
2.6
> 2.0 Safe
2.7
>1.5 Safe
###### KN KN 1.593 >1.5 Hence Safe soil/concrete) ######
KN 1.246 >1.25 Hence Safe
C
7.9
B
1.000 m
M 20 Fe 415
87.1259
rc. as per code =
1000 mm2
Area of re-bar provided = 1339.733
122.4
mm2
0.04 2 0.181 N/mm
80
1000 mm2
of re-bar provided =
1339.7 mm2
0.04 2 0.0828 N/mm
ok
i Pa i Rankin Weight of soil is also to be considered in Rankin's Theroy. a theory is applicable: Coulomb's theory applicable
ndation footing.
Check for correctness of data
Length of single span = Carriageway width, Bc = Overall width of Bridge, Bo =
35.000 m 7.500 m 10.700 m
No. of main girder = 5 Dimensions of the longitudinal girder is taken as follows: Length (L si ) = 35.000 m Breadth (B si )
=
0.550 m
Thickness(Tsi) Distance betn girders =
=
1.680 m 2.375 m
No. of cross beams = 15 For the design purposes the dimensions of the cross beam is taken as follows: Length (L si ) = 2.375 m Breadth (B si ) = 0.140 m Thickness(Tsi) = 0.300 m Distance betn cross beams = 2.500 m For the design purposes the dimensions of the internal slab is taken as follows: Length (L si ) = 2.500 m Hence, effec.span in long.dir. = Breadth (B si ) = 2.375 m Hence, effec.span in transv.dir. = Thickness(Tsi) = 0.230 m Avg. pavement cover over Slab(T sp ) = 0.075 m kerb/footpath breadth = 0.225 m at bottom and 0.225 m at top height = 1.000 m For design purpose dimension of cantilever slab is taken as Length (L si ) = 1.00 m Breadth (B si ) = 0.600 m from c/c of outer girder Breadth (B si ) = 0.325 m from outer face of outer main girder Breadth of carriageway = 1.000 m from c/c of outer girder 1.016 m from outer face of outer main girder Thickness(Tsi) = 0.230 m at the end Thickness(Tsi) = 0.230 m at outer face of main girder Avg. pavement cover over Slab(T sp ) = 0.08 m
2.2
2.360 m 1.825 m
DESIGN OF RIGHT ABUTMENT
The freebody diagram of the R.C.C Abutment is shown below:
i=0
0.200 0.40
CG of Stem from inner vertical face:
Pa
d
14.454
0.23
0.600
A 1=
3.324
A2=
2.674
M.A.= A1+A2 =
1.327 5.998
CG =
0.221
b = 90 A Dimensions taken for the design of abutment: Retaining wall Length = 9.500 Breadth = 0.600 Height = 14.454 Ballast wall Length = 9.500 Breadth = 0.400 Breadth of ret. wall = 0.200 Height = 1.000 Return wall No. = 2 Breadth = 0.200
m m at bottom
m m m m m
m
Bottom level of base = Road level = 0.23 m at top 1.25 CG of retain. wall from point A = 0.221 m
height of superstr. plus bearing without pav.cover Height of Return wall H1=
Page 1
0.7
0.000 5.000 ######
Substructure Kondre (right) Length = Total height of abutment = Approach slab Length = Breadth = Thickness = Foundation is open foundation Length = Width Height of footing = Volume = 2.2.1
6.000 m
Height of Return wall H2= 15.454 m
5.000 m 3.000 m 0.200 m
16.354 -11.354
6.000 m 2.750 m 0.900
0.4
Heel Toe
1.000 m 1.000 m
W =
356.40 KN
0.750 0.750
1.250
14.850 m3
CALCULATION OF LOAD
STRESS DUE TO DEAD LOAD S.N
Elements
Unit
Dead Load of Superstructure 1 Hand rails 6.368 2 Railing posts 3 Wearing course 4 Kerbs 5 Deck slab i) rectangle ii)trapezoid 6 Main Girder i) fillets ii) webs 7 Internal Cross Girder i) fillets ii) webs 8 End Cross Girder i) fillets ii) webs
Dead Load of Abutment retaining wall return walls ballast wall
2
rm cum cum cum cum cum cum cum cum cum cum cum
9.500 6.000 9.500
No.
L(m)
B(m)
H(m) Volume Unit Wt
4 35.60 18 0.15 0.15 1.00 1 35.60 7.50 0.08 2 35.60 0.23 1.00 1 35.60 2.93 0.23 2 35.60 0.33 0.23 8 35.60 0.30 0.15 5 35.60 0.60 1.45 0 1.83 0.15 0.15 0 1.83 0.14 0.07 2 1.83 0.15 0.15 2 1.83 0.14 0.07 Total Dead load of the superstructure Total DL coming to one abutment =
0.415 0.200 0.400
14.454 0.55 1.00
56.98 1.32 3.80
142.40 0.41 20.03 16.02 23.95 5.32 6.41 154.86 0.00 0.00 0.04 0.04
24 1367.64 24 31.68 24 91.20 1490.52
STRESS DUE TO LIVE LOAD Live load due to class A loading: Maximum load on the given abutment is produced when the train of load is positioned as shown in the figure: Axle Load,KN 27 27 114 114 68 68 68 68 xi , m 35.0 33.8 29.5 26.5 23.5 20.5 yi 1.0 1.0 0.8 0.8 0.7 0.6 load, KN 114.0 110.1 57.3 51.5 45.7 39.8 Maximum total live load = 418.4 KN Load due to impact = 418.37714 x 0.110 = 45.919 KN Live load including impact = 418.4 + 45.919 = 464.3 KN Live load due to class AA wheeled loading: Axle Load,KN xi , m yi load, KN Maximum total live load = Load due to impact = Live load including impact =
210.28571 210.3
Live load due to class AA tracked loading: Maximum total live load = Load due to impact = Live load including impact =
664 664.0
200 19.0 0.54 108.6
200 17.8 0.51 101.71 210.3 KN x 0.18 = + 37.851429 =
664.0 KN x +
Page 2
0.1 = 66.4 =
37.851 KN 248.1 KN
66.4 KN 730.4 KN
0.06 24 22 24 24 24 24 24 24 24 24 24
Total Wt e from c/c 8.54 9.72 440.55 384.48 574.80 127.73 153.79 ###### 0.00 0.00 0.99 0.86 ##### 2709 KN
0.000 m 2.208 m 0.007 m
Substructure Kondre (right) Out of above three loadings class AA tracked loading gives maximum load to abutment. Maximum live load to abutment = 730.4 KN CALCULATIONOF EARTH PRESSURE FOR RETAINING WALL Horizontal earth pressure due to backfil Corresponding to
F=
30 ° of the backfill material
g =
18 KN/m3
Ka =
g'
(
0.2973
=
8
Coulomb's theory applicable
Ep1 = Ko x g h /2 = 639.1 KN/m h= 15.454 m Horizontal component of earth pressure = 639.1 cos 20 which acts at 0.42 x h = 6.49 m from the top of footing base. Vertical component of earth pressure = 639.1 x which acts at 0.221 m inward from the CG of wall.
KN/m3
)
2
Total Horizontal Thrust, E1 = Ep1 x L = Total Vertical Thrust, E2 = Ep2 x L =
= sin 20
600.5 KN/m =
218.570 KN/m
5704.9 KN 2076.4 KN
WEIGHT OF BACKFILL BEHIND THE ABUTMENT Wb =
9.1
x
1.000
x
15.5
x
18
=
2531.37 KN
STRESS DUE TO SEISMIC FORCES Horizontal Forces due to seismicity are computed using the formula as shown in IS Code 1893 or IRC - 6 F eq = where
a= b= l=
axbx l xG 0.08 for the zone V of seismic intensity 1.2 for medium soil with isolated footing 1 importance factor
Thus seismic coefficient is taken as
0.096
a.
Seismic force due to Superstructure height of action of the force
= = =
b.
Seismic force due to Abutment act at cg of abutment
= =
c.
Seismic force due to Back fill acts at cg
= =
0.096 14.45 + 15.319 m
x 2709.05 0.025 +
0.096 x 7.727 m 0.096 x 7.727 m
2531.4 =
= 260.1 KN 1.68 / 2 =
1491 =
143.1 KN
243.01 KN
The critical case will be when seismic force acts in the direction of traffic. But according to IRC 6-1966 (222.7) "The seismic force due to live load shall not be considered when acting in the direction of traffic,..". STRESS DUE TO WATER CURRENT Horizontal Thrust due to water current is ignored, because it usually contributes to the stability of the wall and the load case with seismic condition does not concide with HFL.
STRESS DUE TO LONGITUDINAL FORCES Due to Tractive Effort or Braking Force Braking Force due to Class AA tracked load = @20% of live load/2 = = height of the line of action of the brake from the road level = height of the bearing level from the top of base =
Due to Resistance in Bearing (Temperature)
Page 3
1/2 to one abutment 55.4 KN 1.2 m 14.454 m
Substructure Kondre (right) Severe moment is produced when the LL is within span. Assume frictional coeff. = = Frictional resistance at free bearing
=
773.9 KN
Frictional resistance of fixed bearing
=
687.9 KN
Unbalanced frictional force = Line of action of the force from the bottom of the abutment =
0.225 in free bearing 0.2 in other bearing
86.0 KN 14.454 m
STRESS DUE TO WIND LOAD Exposed height of structure
= depth of the beams + thickness of slab + + height of kerb = = 2.68 m
Exposed area contributing to wind pressure =
93.8 m²
Avg. height of superstructure from the bed = intensity of wind load velocity of wind The design is to be conducted for one span only loaded with class A train of vehicle
= =
Length of Class A train
=
20.4 m
a.
Wind Force on Surface (SS)
=
0.73 x
b.
Wind Force on Moving vehicle Class A train acts at 1.5 m above road level
=
20.4 x =
Maximum load is obtained in exposed area = For lateral direction wind load acts at
6.000 m 0.73 KN/m² 107 km/hr
93.8 =
68.474 KN
3 = m
61.2 KN
-9.854 129.7 KN
= =
129.7 KN 15.454 m
SUMMARY OF LOADS FOR DESIGN OF ABUTMENTS S.No.
Load
Vertical (KN)
Horizontal (KN) Arm Later. (m)
Moment (KN) Longit. Later.
Longit. dir
1 1.1 1.2
Superstructure Dead load of SS(D) Live Load of SS (L) including impact Substructure DL of abutment(E) Retaining wall Return wall - Top section Ballast wall Backfill (BF) Earth pressure (EP) (EP1)
2
3 4 4.1 4.2 4.3 4.4 4.5 4.6
2709.1 438.0
0.48 1297.0 0.48 209.7
1367.6 31.7 91.2 2531 35062.779
Due to overburden(EP2) Frontal ive soil mass (EPp) Buoyancy (B) Longitudinal force Tractive force (T) Due to temperature (F) Wind Load Centrifugal force (CF) Thrust due to water current (WC) Seismic forces (FS) on superstructure on abutment on live load from backfill
Taking Rankin's earth pressure coefficient.
0.00 -2.21 -0.01 0.00 5704.9
6.49 ######
55.40 85.99
130.03 71.54 121.51
-69.9 -0.7 0.0
129.67
14.45 800.8 14.45 1242.8 15.45 2003.982
260.07 143.09 42.05 243.01 243.011
15.32 3984.0 3983.994 7.73 1105.7 1105.654 16.95 712.8818 7.727 1877.7 1877.746
260.07 143.09
Load combination Group II (Normal+Temp. condition) D + L + E + Ep1+ Ep2 + Ep3+ T + F + CF 4638
Page 4
5846
129.7
#####
2004.0
Substructure Kondre (right) Load combination Group IV (for Seismic dry condition) D + L + E + Ep1+ Ep2 + T + F + CF + FS
7492
6492
688.2
Note: Moment is taken at the critical section at the bottom of the ABUTMENT 2.2.2
STRUCTURAL DESIGN OF ABUTMENT RETAINING WALL
Design of the returning wall will be done based on normal loading condition, but checked on maximum loading Design parameters Vertical force = 4638 KN (Under combination group IV the permissible increase of stress is Long. Moment = 40508 KN-m 50%, hence Combination I governs the design longitudinal moment) Substructures are designed using M20 concrete and Fe415 steel reinforcement
Conccrete M20
Parameters for RCC design sck = 20 N/mm2
Steel
sy =
415 N/mm2
Page 5
#####
7680.3
Substructure Kondre (right) Approximate Area of Steel reinforcement required is calculated considering singly reinforced beam theory Ast = Mu/ (0.87xσy x 0.7984 x d) = Provide Required spacing =
234060 mm2
25 mm dia. bars 0.9 mm c/c
Ast = Provide at
Area of reinforcement provided, Ast =
490.625 mm2 100 mm c/c
45873.438 mm2 0.8047971 % >
r=
Decrease spacing !
0.2 %
At the frontal face, the design moment is greatly reduced because of lack of earth pressure. Therefore on this side provide only 50 % of the main reinforcement Ast =
234060 x0.5
=
Provide Required spacing =
117030 mm2
20 mm dia. bars 20 1.2 mm c/c
Ast = Provide at
Area of reinforcement provided, Ast =
314 mm2 100 mm c/c
Decrease specing !
29359 mm2
Same spacing of reinforcement is provided on the side faces of the abutment. Distribution re-bar Providing 0.12% of cross sectional area = Provide Required spacing =
0.0012 x
12 mm dia. bars 314.0 mm c/c
Check for stresses
Ast = Provide at
600 x 1000=
113.04 mm2 250 mm c/c
(For seismic condition) 7492018 ######### 6336.850 9500 600 512.5 75 75
mm mm mm mm mm mm
at both faces.
ok
Parameters for RCC design Conccrete M20
Normal Thrust , W = Bending Moment = Ecentricity, e = breadth, b = Total depth , D = Effective depth, d = Compressive steel cover, dc = Tensile steel cover, dt =
720 mm2
N N-mm > m=
Ac = At =
Steel
sck = sy =
20 N/mm2 415 N/mm2
128.125 mm 0.0
29359 mm2 45873.44 mm2
Ast = Mu/ (0.87xσy x 0.7984 x d) =178050.265 mm2/m
x=
0.87*σy*Ast 0.36*σck*b
8928 mm
150 mm say
Check for shear stress: Critical secion for shear in case of abutment stem is considered at distance d = of base slab. Horizontal force due to soil pressure = 5559.80 KN Breaking force = 55.40 KN Longitudinal force due to temperature effect = 85.99 KN Shear force V = 5701.18 KN Shear stress tc = V/(b d) =
A
0.744 N/mm2
Permissible shear stress without shear reinforcement tcmax =2.8
Mpa for M20 concrete
Where, As = Area of Steel provided in mm2 b = width of abutment vertical wall and d = thickness of abutment wall tc = basic values given for different p% of Steel provided = From Table 19 of IS 456-200 The design of ballest wall and return wall will be as in left side abutment. 2.1.3
512.5 mm from the junction
DESIGN OF BALLAST WALL
Page 6
p=
4.05%
0.28 Mpa for M20 concrete
<
2 0.744 N/mm
Substructure Kondre (right)
1.000 m
RCC balast wall (as shown in the figure) monolithic to the abutment has been designed. 0.40 m
CALCULATION OF DEAD LOAD STRESSES Dead load from the approach slab (triangular portion) = = Dead load of the Ballast wall
36 KN 3.7894737 KN/m
=
91.2 KN 9.6 KN/m
= Earth pressure from the back fill
= 21 KN/m3 0.2973
r= ca =
ca x r x h2 /2 =
3.12 KN/m
LIVE LOAD STRESS CALCULATION Class AA (tr) load component on wall , LL =
Add impact to live load @ , I =
291.667 30.702 3.070 0.15
= = = e=
0.1
KN KN/m KN/m m from axis of ballast wall
Horizontal Load due to braking of train of vehicle class A vehicle, T =(0.2 ) x(27+114*2) = (Front vehicle only considered)
=
51.0 KN 5.4 KN/m
Compaction load is maximum at the depth of 1.2m from the top. P comp = 25.2 KN/m Maximum shear force on section I-I H = T or Pcopm+ Ep = V = DL + LL + I =
28.3 KN/m 47.2 KN/m
Moment given due to the forces on the critical section I - I at the base of the ballast wall M = ( LL + I) x e + T x h + Ep x h /3 = = 17.92 KN-m/m Area of vertical steel reinforcement required Ast = Mu/ (0.87xσy x 0.7984 x d) =165.746387 mm2/m Provide
12 mm dia bars @ Area of steel provided = Provide 12 mm dia bars @ Area of steel provided = Horizontal links are provided of
200 c/c
for the continuation of bar from retaining wall 565.2 mm2/m 200 c/c on the opposite face 565.2 mm2/m 12 mm dia. bars @ 200 c/c
Stresses at the section, Shear stress,tmax =
0.083 N/mm2
Permissible shear stress without shear reinforcement tcmax = 2.8 2
Where, As = Area of Steel provided in mm
2.1.4
p=
tc = basic values given for different p% of Steel provided = From Table 19 of IS 456-200 DESIGN OF RETURN WALLS
Page 7
N/mm2
0.28%
0.28 0.38 Mpa for M20 concrete
>
2 0.083 N/mm ok
Substructure Kondre (right) These will be monolithic with the back wall. They are ed together through 300 mm x 300 mm fillets. The loads acting on the wing wall (return wall) would be earth pressure and horizontal force to kerb as specified in the code. STRESS DUE TO VERTICAL LOADS All the stresses are calculated for the critical section I-I at the bottom face of the abutment Dead load of the returnwall = 6.000 x Vert. LL from Class A load = Add Impact due to LL @ 0.25=
168.72 KN 42.18 KN
56.24 14.06
0.550
x
0.200
x 24 = =
15.84 KN 2.64 KN/m
KN/m KN/m
STRESSES DUE TO HORIZONTAL FORCES Horizontal Thrust due to DL+LL+IL=q*L*ka acts at h/2 0.28 m
= 24.070 KN/m from the top of foundation
Moment due to hor. component of vertical loads =
433.2636 KN-m/r.m ka x rs x h^2/2
Earth pressure from the backfill =
=
Moment due to earth pressure = Momt due to later. force =
16.998 KN-m 7.5 KN/m for the curb as per code = 135.000
Total horz. uniformly distributed load = Total Moment = 433.26 + depth required = where b = ( Provide overall depth =
17.00 + M /( 0.138σck b ) 700 + 650 mm
Ast = Mu/ (0.87xσy x 0.7984 x d) =
24.07
+ 135.000 =
= 400 ) / 2 =
0.94 KN/m
KN-m
0.94 +
7.5 =
32.51 KN/m
585.26 620.925 mm 550 mm
3530.97 mm2/m
Tensile steel re-bar equired = 3530.97 mm2 Provide 25 mm dia. bars Required spacing = 139.0 mm c/c
Ast = Provide at
2.2.3 DESIGN OF FOUNDATION Foundation is designed as open foundation of the rectangular shape Length = 6.00 m Breadth = 2.75 m heel toe width of abutment Height = 0.9 m Concrete cover 75 mm DEAD LOAD due to foundation =
490.87 mm2 150 mm c/c
Decrease specing !
1.000 m 1.000 m 0.600 m
356.4 KN
CALCULATIONOF EARTH PRESSURE FOR FOUNDATION Horizontal earth pressure due to backfill Ka = 0.2973 As calculated earlier for abutment stem.
( Coulomb's theory applicable
Ep1 = Ka x g h2 /2 = 715.66 KN/m h = 16.354 m Horizontal component of earth pressure = 715.7 x cos 20 which acts at 0.42 x h = 6.87 m from the bottom of footing base.
=
)
672.499 KN/m
Pressure due to surcharge: As approach slab has been provided no surcharge load is to be taken as per code. WEIGHT OF BACKFILL BEHIND THE ABUTMENT Wb = Wf =
6.000 6.000 acts at
x x 0.8
1.000 1.000 m
x 15.454 x 0 from centre of abutment
Seismic force due to Foundation act at cg of abutment Seismic force due to back fill act at cg of backfill SUMMARY OF LOADS FOR DESIGN OF FOUNDATION
x x
= = = =
Page 8
18.00 = 18.00 =
0.096 x 0.45 m 0.096 x 8.627 m
1669.0 KN 0.0 KN
356.4 =
34.214 KN
1669.0 = 160.23 KN from the bottom of foundation base
Substructure Kondre (right) S.No.
Load
Vertical (KN)
Horiz (KN)
1 1.1 1.2
2
4.1 4.2 4.3
M(o) KN-m
Superstructure Dead load of SS(D) Live Load on SS including impact (L)
2709.05 438.0
0.8 2224.79 0.8 359.705
Substructure DL of abutment(E) Retaining wall Return wall Ballast wall
1367.64 31.68 91.20
1.3 1777.93 3.5 111.118 1.3 118.56
1669.03
2.10 3504.97
Backfill (BF) Backfill (FR) Earth pressure (EP) Triangular dry(EP1)
3 4
L. arm M[r] (m) KN-m
6388.7
Due to overburden(EP2) Frontal ive soil mass (EPp) DL of foundation Buoyancy (B) Longitudinal force Breaking (breaking) force (T) Due to temperature (F) Wind Load
-6.9
-43882
0.00 356.40
Centrifugal force (CF) Thrust due to water current (WC) Seismic forces (FS) on superstructure on abutment on live load on foundation from backfill Load combination Group I (N+T condition) D + L + I + E + Ep1+ Ep2 +BF+ T + F + CF Load combination Group IV (for Seismic dry condition) D + L + E + Ep1+ Ep2 +BF+ T + F + CF + FS
0 1.4
490.05
55.40 85.99
-15.4 -15.4
-850.6 -1320
130.03 71.54
260.1 143.1
-15.3 -7.7
17.11 80.11
34.2 160.2
-0.5 -8.6
-1992 -552.8 0 -7.698 -691.1
6663.0
6530
8587.12 -46053
6961.8
7128
8587 -48446
4.4 4.5 4.6
Note: Moment is taken
about the toe for check against overturning . 403.82
For Normal condition P= 6663 KN Mnet = Mr-Mo= -37465.918 x = Mnet/P= -5.6229806 e = b/2-x'= 6.9979806 > b/6, hence tesion Pmax=P/A*(1+6*e/b) = 6569.444 KN/m2 Pmin=P/A*(1-6*e/b) = -5761.807 KN/m2 Where, A=b x d = 16.5 m2 CHECK AGAINST OVERTURNING Case: Normal Factor of Safety against overturning =Mres/Mov= Case: Seismic Factor of Safety against overturning =Mres/Mov=
For Seismic condition P= 6962 KN Mnet = Mr-Mo= -39859 x = Mnet/P= -5.7254 e = b/2-x'= 7.10 Pmax=P/A*(1+6*e/b) = Pmin=P/A*(1-6*e/b) =
KN-m m > b/6, hence tesion 6958.3 KN/m2 -6114 KN/m2
8587.1227
/
46053 =
0.2
> 2.0 Safe
8587.1227
/
48446 =
0.2
>1.5 Safe
CHECK AGAINST SLIDING Case : Normal The minimum vertical force will be when there is no live load and the front backfill is washed out = ###### KN The resisting force againest sliding = (vertical force) x (coefficient of friction b/w concrete and soil) = 6225.00 x 0.45 = 2801.25 KN The sliding force is the horizontal forces = 6474.72 KN Hence, Factor of Safety against sliding =Pres/Psl= 2801.25 / 6474.7 = 0.433 >1.5 Hence Safe (f= 0.45 soil/concrete) Case: Seismic The minimum vertical force will be when there is no live load and the front backfill is washed out = ###### The resisting force againest sliding = (vertical force) x (coefficient of friction b/w concrete and soil) = 6523.80 x 0.45 = 2935.71 KN
Page 9
Substructure Kondre (right) The sliding force is the horizontal forces = Hence, Factor of Safety against sliding =Pres/Psl=
7072.32 2935.71
KN /
7072.3 =
0.415 >1.25 Hence Safe
STRUCTURAL DESIGN OF FOUNDATION
6569.44 KN/m2
Wt of toe slab =
21.6 KN/m2
D
F
G
1412.739
C B #####
A
Design of toe slab:
E
-1277.72
1.000 m
0.600 m 2.75 m
1.000 m
Pressure distribution diagram
####
6547.84 Moment about EF =
concrete grede : Steel grade :
2414.47 KN-m d= M /( 0.138σck b ) 564.0 mm Adopt D = 600 mm d= 500 mm Ast = Mu/ (0.87xσy x 0.7984 x d)16752 = mm2 Using 16 mm dia bar, spacing = Adopt 16 mm dia bar, spacing =
M 20 Fe 415
2414.47
Minimum reinforc. as per code = 1000 mm2 12.0 mm 150 mm c/c Area of re-bar provided = 1339.733 mm2
Check for shear stress:
Shear stress =
V/(b d) =
b=
6.00 m
1391
15776.00 KN
500 3969
Shear force at critical section V =
d =
6548
In toe critical section for shear is considered at a distance from abutment stem.
5.259 N/mm2
Permissible shear stress without shear reinforcement tcmax =2.8 Mpa for M20 concrete Where, k1 = 1.14 - 0.7 d >= 0.5 ( d in m ) Where, As = Area of Steel provided in mm2 b = width of abutment vertical wall and d = thickness of abutment wall p=
0.5 m
0.037% 0.04 2 0.280 N/mm2 < 5.259 N/mm Not ok. Revise section or provide shear re-bar
tc = basic values given for different p% of Steel provided = From Table 19 of IS 456-200
Design of heel slab: Upward pressure on the heel slab varies from -5761.8073 Kn/m2 to -1277.72 Kn/m2 In addition to the upward pressure the heel slab is subjected to downward pressure due to the self weight of heel slab and weight of earth fill on heel slab: Down pressure due to self weight = 14.4 KN/m2 Down pressure due to backfill = 278.172 KN/m2 292.572 KN/m2 1570 6054 Net down pressure = Moment about DG= d= M /( 0.138σck b ) Adopt
6054 1570 2279.841 908.862
KN/m2 KN/m2 KN-m mm
max min
D= 600 mm d= 500 mm Ast = Mu/ (0.87xσy x 0.7984 x d) = 15818 mm2 Using 16 mm dia bar, specing = Adopt 16 mm dia bar, specing =
Page 10
1.000 m
Minimum reinforc. as per code = 1000 mm2 13 mm 150 mm c/c Area of re-bar provided = 1339.7 mm2
Substructure Kondre (right) Distribution re-bar 0.12/100 x
900 Using Adopt
=
1080 mm2/runnig meter 12 mm dia bar, specing = 209.3 mm 12 mm dia bar, specing = 200 mm c/c
at top and bottom
Check for shear stress: Since the load on the heel is predominantly downward, it induces tensile reactio at its junction with stem. Hence the critical section for shear for the heel slab is considered at the face of . Shear force V = 22874.002 KN b= 6.00 m Shear stress = V/(b d) = 7.625 N/mm2 Permissible shear stress without shear reinforcement tcmax =2.8 Where, k1 = 1.14 - 0.7 d >= 0.5 ( d in m )
N/mm2
Where, As = Area of Steel provided in mm2 b = width of abutment vertical wall and d = thickness of abutment wall tc = basic values given for different p% of Steel provided = From Table 19 of IS 456-200
Mpa for M20 concrete
p= 0.28 N/mm2
Page 11
0.037%
<
0.04 2 7.6247 N/mm
Not ok. Revise section or p
Substructure Kondre (right)
Page 12
Substructure Kondre (right)
i
i Pa
Weight of soil is not to be considered in Coulumb's Theroy.
d 90-b
Page 13
Ws
Pa i
Substructure Kondre (right)
d 90-b
Weight of soil is not to be considered in Coulumb's Theroy.
i 0 h= a= Coul. Rankin
Pa i
b
b Coulomb
F 30
b 90
d 20
30
a Let us find out which theory is applicable:
60
1.7321 m <
0.75 0.9397 2.6845 1 0.75 0.5
Hor pressure from Rankin's theory = Horizontal thrust = 6806.5 acts at
Rankin Weight of soil is also to be considered in Rankin's Theroy.
1.5
15.454 m hence Coulomb's theory applicable Ka= 0.2973 0.3333
716.48 KN/m 5.15 m from the top of foundation footing.
Page 14
Substructure Kondre (right)
Page 15
Substructure Kondre (right)
Page 16
Substructure Kondre (right)
Decrease specing !
Page 17
Substructure Kondre (right)
Page 18
Substructure Kondre (right)
Page 19
Substructure Kondre (right)
> b/6, hence tesion
Page 20
Substructure Kondre (right)
Page 21
Substructure Kondre (right)
Not ok. Revise section or provide shear re-bar
Page 22
Dimension of slab and beams (brought from slab sheet) Length of single span = Carriageway width, Bc = Overall width of Bridge, Bo =
Check for correctness of data 8.000 m 3.000 m 6.000 m
No. of main girder = 3 Dimensions of the longitudinal girder is taken as follows: Length (L si ) = 8.000 m Breadth (B si ) Thickness(Tsi) Distance betn girders =
=
0.600 m
=
0.750 m 2.400 m
No. of cross beams = 2 For the design purposes the dimensions of the cross beam is taken as follows: Length (L si ) = 3.000 m Breadth (B si ) = 0.600 m Thickness(Tsi) = 0.900 m Distance betn cross beams = 8.000 m For the design purposes the dimensions of the internal slab is taken as follows: Length (L si ) = 8.600 m Hence, effec.span in long.dir. = Breadth (B si ) = 3.600 m Hence, effec.span in transv.dir. = Thickness(Tsi) = 0.200 m Avg. pavement cover over Slab(T sp ) = 0.070 m kerb/footpath breadth = 1.000 m at bottom and 0.225 m at top height = 0.500 m For design purpose dimension of cantilever slab is taken as Length (L si ) = 1.00 m Breadth (B si ) = 0.600 m from c/c of outer girder Breadth (B si ) = 0.300 m from outer face of outer main girder Breadth of carriageway = 2.400 m from c/c of outer girder 0.300 m from outer face of outer main girder Thickness(Tsi) = 0.200 m at the end Thickness(Tsi) = 0.200 m at outer face of main girder Avg. pavement cover over Slab(T sp ) = 0.07 m
2.2
DESIGN OF RIGHT ABUTMENT
The freebody diagram of the R.C.C Abutment is shown below:
0.20
i=0 Pa
0.60 3.100
d
CG of Stem from inner vertical face: A 1=
1.860
A2=
0.233
M.A.=
0.709
3.100
A1+A2 =
0.750 A
b = 90
CG =
Dimensions taken for the design of abutment: Retaining wall Length = 6.000 Breadth = 0.750 Height = 3.100 Ballast wall Length = 6.000 Breadth = 0.200 Height = 1.000 Return wall No. = 2 Breadth = 0.200 Length = 6.000 Total height of abutment = Approach slab Length = Breadth = Thickness = Foundation is open foundation Length = Width Height of footing = Volume =
m m at bottom
m m m m
m m
Bottom level of base = Road level = 0.60 m at top 1.25 CG of retain. wall from point A = 0.339 m
height of superstr. plus bearing without pav.cover Height of Return wall H1= Height of Return wall H2= 4.100 m
5.000 m 3.000 m 0.200 m
0.7 0.4
5.000 0.000
6.000 m 2.750 m 0.900
2.093
0.339
Heel Toe
1.000 m 1.000 m
W =
356.40 KN
0.750 0.750
1.250
14.850 m3
2.2.1 CALCULATION OF LOAD STRESS DUE TO DEAD LOAD S.N Elements
Unit
Dead Load of Superstructure 1 Hand rails 6.368 2 Railing posts 3 Wearing course 4 Kerbs 5 Deck slab i) rectangle ii)trapezoid 6 Main Girder i) fillets ii) webs 7 Internal Cross Girder i) fillets ii) webs 8 End Cross Girder i) fillets ii) webs
rm cum cum cum cum cum cum cum cum cum cum cum
No.
L(m)
B(m)
H(m) Volume
4 8.60 5 0.15 0.15 1 8.60 3.00 2 8.60 0.61 1 8.60 3.00 2 8.60 0.30 4 8.60 0.30 3 8.60 0.60 0 2.40 0.15 0 2.40 0.60 2 2.40 0.15 2 2.40 0.60 Total Dead load of the superstructure Total DL coming to one abutment =
1.00 0.07 0.50 0.20 0.20 0.15 0.55 0.15 0.70 0.15 0.70
34.40 0.11 1.81 5.27 5.16 1.03 0.77 8.51 0.00 0.00 0.05 2.02
Dead Load of Abutment retaining wall return walls ballast wall
2
6.000 6.000 6.000
0.675 0.200 0.200
3.100 0.55 1.00
12.56 1.32 1.20
24 24 24
301.32 31.68 28.80 361.80
STRESS DUE TO LIVE LOAD Live load due to class A loading: Maximum load on the given abutment is produced when the train of load is positioned as shown in the figure: Axle Load,KN 27 27 114 114 68 68 68 68 xi , m 8.0 6.8 2.5 -0.5 -3.5 -6.5 yi 1.0 0.9 0.3 -0.1 -0.4 -0.8 load, KN 114.0 96.9 21.3 -4.3 -29.8 -55.3 Maximum total live load = 142.9 KN Load due to impact = 142.9 x 0.321 = 45.932 KN Live load including impact = 142.9 + 45.932 = 188.8 KN Live load due to class AA wheeled loading: Axle Load,KN xi , m yi load, KN Maximum total live load = Load due to impact = Live load including impact = Live load due to class AA tracked loading: Maximum total live load = Load due to impact = Live load including impact =
200 19.0 2.38 475.0 920 920.0
200 17.8 2.23 445.00 920.0 KN x +
0.18 = 165.6 =
165.6 KN 1085.6 KN
542.5 542.5
542.5 KN x +
0.1 = 54.25 =
54.25 KN 596.8 KN
Out of above three loadings class AA tracked loading gives maximum load to abutment. Maximum live load to abutment = 1085.6 KN CALCULATIONOF EARTH PRESSURE FOR RETAINING WALL Horizontal earth pressure due to backfil Corresponding to
F=
30 ° of the backfill material
g =
18 KN/m3
Ka =
0.2973
g'
(
=
8
Coulomb's theory applicable
Ep1 = Ko x g h /2 = 45.0 KN/m h= 4.100 m Horizontal component of earth pressure = 45.0 cos 20 = which acts at 0.42 x h = 1.72 m from the top of footing base. Vertical component of earth pressure = 45.0 x sin 20 which acts at 0.339 m inward from the CG of wall.
KN/m3
)
2
Total Horizontal Thrust, E1 = Ep1 x L =
253.6 KN
42.3 =
Total Vertical Thrust, E2 = Ep2 x L =
92.3 KN
WEIGHT OF BACKFILL BEHIND THE ABUTMENT Wb =
5.6
x
1.000
x
4.1
x
18
=
413.28
STRESS DUE TO SEISMIC FORCES Horizontal Forces due to seismicity are computed using the formula as shown in IS Code 1893 or IRC - 6 F eq = wherea = b=
l=
axbx l xG 0.08 for the zone V of seismic intensity 1.2 for medium soil with isolated footing 1 importance factor
Thus seismic coefficient is taken as
0.096
a.
Seismic force due to Superstructure height of action of the force
= = =
b.
Seismic force due to Abutment act at cg of abutment
= =
c.
Seismic force due to Back fill acts at cg
= =
0.096 3.10 + 3.5 m
x 296.058 0.025 +
0.096 x 2.05 m 0.096 x 2.05 m
362
413.28 =
The critical case will be when seismic force acts in the direction of traffic. But according to IRC 6-1966 (222.7) "The seismic force due to li load shall not be considered when acting in the direction of traffic,..". STRESS DUE TO WATER CURRENT Horizontal Thrust due to water current is ignored, because it usually contributes to the stability of the wall and the load case with seismic condition does not concide with HFL.
STRESS DUE TO LONGITUDINAL FORCES Due to Tractive Effort or Braking Force Braking Force due to Class AA tracked load = @20% of live load/2 = = height of the line of action of the brake from the road level = height of the bearing level from the top of base =
55.4 KN 1.2 3.100 m
Due to Resistance in Bearing (Temperature) Severe moment is produced when the LL is within span. Assume frictional coeff. =
0.225
= Frictional resistance at free bearing
=
310.9 KN
Frictional resistance of fixed bearing
=
276.3 KN
Unbalanced frictional force = Line of action of the force from the bottom of the abutment =
34.5 KN
0.2
3.100 m
STRESS DUE TO WIND LOAD Exposed height of structure
= depth of the beams + thickness of slab + + height of kerb = = 1.25 m Exposed area contributing to wind pressure = Avg. height of superstructure from the bed = intensity of wind load = velocity of wind = The design is to be conducted for one span only loaded with class A train of vehicle Length of Class A train
=
20.4 m
a.
Wind Force on Surface (SS)
=
0.73 x
b.
Wind Force on Moving vehicle Class A train acts at 1.5 m above road level
=
20.4 x =
Maximum load is obtained in exposed area = For lateral direction wind load = acts at =
10 m² 6.000 m 0.73 KN/m² 107 km/hr
10 =
1.500
3= m
68.5 KN 68.5 KN 4.100 m
SUMMARY OF LOADS FOR DESIGN OF ABUTMENTS S.N o.
Load
Vertical (KN)
Horizontal (KN) Longit. dir
1
2
Superstructure 1.1 Dead load of SS(D) 1.2 Live Load of SS (L) including impact Substructure DL of abutment(E) Retaining wall Return wall - Top section Ballast wall Backfill (BF) Earth pressure (EP) (EP1) Due to overburden(EP2)
296.1 438.0
301.3 31.7 28.8 413 413.526
Taking Rankin's earth pressure coefficient.
253.6
Later.
3 4 4.1 4.2 4.3 4.4 4.5 4.6
Taking Rankin's earth pressure coefficient.
Frontal ive soil mass (EPp) Buoyancy (B) Longitudinal force Tractive force (T) Due to temperature (F) Wind Load Centrifugal force (CF) Thrust due to water current (WC) Seismic forces (FS) on superstructure on abutment on live load from backfill
55.40 34.54 68.50
14.21 17.37
28.42 34.73
28.42 34.73 42.05 39.675 39.6749
19.84
Load combination Group II (Normal+Temp. condition) D + L + E + Ep1+ Ep2 + Ep3+ T + F + CF Load combination Group IV (for Seismic dry condition) D + L + E + Ep1+ Ep2 + T + F + CF + FS
1096
344
68.5
1561
446
144.9
Note: Moment is taken at the critical section at the bottom of the ABUTMENT 2.2.2 STRUCTURAL DESIGN OF ABUTMENT RETAINING WALL Design of the returning wall will be done based on normal loading condition, but checked on maximum loading Design parameters Vertical force = 1096 KN (Under combination group IV the permissible increase of stress is Long. Moment = 753 KN-m 50%, hence Combination I governs the design longitudinal moment) Substructures are designed using M20 concrete and Fe415 steel reinforcement
Conccrete M20
Parameters for RCC design sck = 20 N/mm2
sy = Steel 415 N/mm2 Approximate Area of Steel reinforcement required is calculated considering singly reinforced beam theory Ast = Mu/ (0.87xσy x 0.7984 x d) = Provide Required spacing =
3332 mm2
25 mm dia. bars 88.3 mm c/c
Area of reinforcement provided, Ast = r=
Ast = Provide at
490.625 mm2 100 mm c/c
28701.563 mm2 0.6378125 % >
At the frontal face, the design moment is greatly reduced because of lack of earth pressure. Therefore on this side provide only 50 % of the main reinforcement Ast =
3332 x0.5
=
1666 mm2
0.2 %
Provide Required spacing =
20 mm dia. bars 20 113.1 mm c/c
Ast = Provide at
Area of reinforcement provided, Ast =
314 mm2 100 mm c/c
18369 mm2
Same spacing of reinforcement is provided on the side faces of the abutment. Distribution re-bar Providing 0.12% of cross sectional area = Provide Required spacing =
0.0012 x
12 mm dia. bars 251.2 mm c/c
Check for stresses
Ast = Provide at
750 x 1000=
113.04 mm2 250 mm c/c
(For seismic condition)
Parameters for RCC design Conccrete M20
Normal Thrust , W = Bending Moment = Ecentricity, e = breadth, b = Total depth , D = Effective depth, d = Compressive steel cover, dc = Tensile steel cover, dt =
1560553 ######### 643.948 6000 750 662.5 75 75
mm mm mm mm mm mm
900
N N-mm > m=
Ac = At =
Steel
sck = sy =
165.625 mm 0.0
18369 mm2 28701.56 mm2
Ast = Mu/ (0.87xσy x 0.7984 x d)3768.77108 = mm2/m
x=
0.87*σy*Ast 0.36*σck*b
189 mm
150 mm say
Check for shear stress: Critical secion for shear in case of abutment stem is considered at distance d = of base slab. Horizontal force due to soil pressure = 173.25 KN Breaking force = 55.40 KN Longitudinal force due to temperature effect = 34.54 KN Shear force V = 263.19 KN Shear stress tc = V/(b d) =
662.5 mm from the junction
A
0.125 N/mm2
Permissible shear stress without shear reinforcement tcmax =2.8
Mpa for M20 concrete
2
Where, As = Area of Steel provided in mm b = width of abutment vertical wall and d = thickness of abutment wall tc = basic values given for different p% of Steel provided = From Table 19 of IS 456-200 The design of ballest wall and return wall will be as in left side abutment.
p=
0.21%
0.28 Mpa for M20 concrete
2.1.3 DESIGN OF BALLAST WALL
1.000 m
RCC balast wall (as shown in the figure) monolithic to the abutment has been designed. 0.20 m
CALCULATION OF DEAD LOAD STRESSES Dead load from the approach slab (triangular portion) = = Dead load of the Ballast wall
36 KN 6 KN/m
=
28.8 KN 4.8 KN/m
= Earth pressure from the back fill
= 21 KN/m3 0.2973
r= ca =
ca x r x h2 /2 =
3.12 KN/m
LIVE LOAD STRESS CALCULATION Class AA (tr) load component on wall , LL =
Add impact to live load @ , I =
0.1
= = = e=
291.667 48.611 4.861 0.15
KN KN/m KN/m m from axis of ballast wall
Horizontal Load due to braking of train of vehicle class A vehicle, T =(0.2 ) x(27+114*2) = (Front vehicle only considered) Compaction load is maximum at the depth of 1.2m from the top. P comp = 25.2 KN/m Maximum shear force on section I-I H = T or Pcopm+ Ep = V = DL + LL + I =
28.3 KN/m 64.3 KN/m
Moment given due to the forces on the critical section I - I at the base of the ballast wall M = ( LL + I) x e + T x h + Ep x h /3 = = 27.76 KN-m/m Area of vertical steel reinforcement required
=
Ast = Mu/ (0.87xσy x 0.7984 x d)256.816378 = mm2/m Provide 12 mm dia bars @ Area of steel provided = Provide 12 mm dia bars @ Area of steel provided = Horizontal links are provided of
200 c/c
for the continuation of bar from retaining wall 565.2 mm2/m 200 c/c on the opposite face 565.2 mm2/m 12 mm dia. bars @ 200 c/c
Stresses at the section, Shear stress,tmax =
0.202 N/mm2
Permissible shear stress without shear reinforcement tcmax = 2.8 Where, As = Area of Steel provided in mm2
p=
N/mm2
0.57%
0.57
tc = basic values given for different p% of Steel provided = From Table 19 of IS 456-200 2.1.4 DESIGN OF RETURN WALLS
0.51 Mpa for M20 concrete
These will be monolithic with the back wall. They are ed together through 300 mm x 300 mm fillets. The loads acting on the wing wall (return wall) would be earth pressure and horizontal force to kerb as specified in the code. STRESS DUE TO VERTICAL LOADS All the stresses are calculated for the critical section I-I at the bottom face of the abutment Dead load of the returnwall = 6.000 x 0.550 Vert. LL from Class A load = Add Impact due to LL @ 0.25=
168.72 KN 42.18 KN
56.24 14.06
x
0.200
KN/m KN/m
STRESSES DUE TO HORIZONTAL FORCES Horizontal Thrust due to DL+LL+IL=q*L*ka acts at h/2 0.28 m Moment due to hor. component of vertical loads =
433.2636 KN-m/r.m ka x rs x h^2/2
Earth pressure from the backfill = Moment due to earth pressure = Momt due to later. force =
= 24.070 KN/m from the top of foundation
16.998 KN-m 7.5 KN/m for the curb as per code = 135.000
=
KN-m
Total horz. uniformly distributed load = 24.07 + 0.94 + Total Moment = 433.26 + 17.00 + 135.000 = 585.26 depth required = M /( 0.138σck b ) = 620.925 mm where b = ( 700 + 400 ) / 2 = 550 mm Provide overall depth = 650 mm Ast = Mu/ (0.87xσy x 0.7984 x d) = 3530.97 mm2/m
0.94
7.5
Tensile steel re-bar equired = 3530.97 mm2 Provide 25 mm dia. bars Required spacing = 139.0 mm c/c
Ast = Provide at
2.2.3 DESIGN OF FOUNDATION Foundation is designed as open foundation of the rectangular shape Length = 6.00 m Breadth = 2.75 m heel toe width of abutment Height = 0.9 m Concrete cover 75 mm DEAD LOAD due to foundation =
490.87 mm2 150 mm c/c
1.000 m 1.000 m 0.750 m
356.4 KN
CALCULATIONOF EARTH PRESSURE FOR FOUNDATION Horizontal earth pressure due to backfill Ka = 0.2973 As calculated earlier for abutment stem.
( Coulomb's theory applicable
Ep1 = Ka x g h2 /2 = 66.90 KN/m h= 5.000 m Horizontal component of earth pressure = 66.9 x cos 20 which acts at 0.42 x h = 2.10 m from the bottom of footing base.
=
Pressure due to surcharge: As approach slab has been provided no surcharge load is to be taken as per code. WEIGHT OF BACKFILL BEHIND THE ABUTMENT Wb = Wf =
6.000 6.000 acts at
x x 0.875
1.000 1.000 m
x 4.100 x 0 from centre of abutment
Seismic force due to Foundation act at cg of abutment Seismic force due to back fill act at cg of backfill SUMMARY OF LOADS FOR DESIGN OF FOUNDATION S.No . Load
x x
= = = =
Vertical (KN)
18.00 = 18.00 =
442.8 0.0
0.096 x 0.45 m 0.096 x 2.95 m
356.4
Horiz (KN)
1
2
Superstructure 1.1 Dead load of SS(D) 1.2 Live Load on SS including impact (L) Substructure DL of abutment(E) Retaining wall Return wall Ballast wall Backfill Backfill
(BF) (FR)
442.8 from the bottom of foundation ba
L. arm (m)
296.06 438.0
1.2 1.2
301.32 31.68 28.80
1.4 3.7 1.4
442.80
2.25
Earth pressure (EP) Triangular dry(EP1)
377.2
Due to overburden(EP2) Frontal ive soil mass (EPp) DL of foundation 3 Buoyancy (B) 4 Longitudinal force 4.1 Breaking (breaking) force (T) 4.2 Due to temperature (F) 4.3 Wind Load
0.00 356.40
4.4 Centrifugal force (CF) 4.5 Thrust due to water current (WC) 4.6 Seismic forces (FS) on superstructure on abutment on live load on foundation from backfill Load combination Group I (N+T condition) D + L + I + E + Ep1+ Ep2 +BF+ T + F + CF Load combination Group IV (for Seismic dry condition) D + L + E + Ep1+ Ep2 +BF+ T + F + CF + FS Note: Moment is taken For Normal condition P= 1895 Mnet = Mr-Mo= x = Mnet/P= e = b/2-x'= Pmax=P/A*(1+6*e/b) = Pmin=P/A*(1-6*e/b) = Where, A=b x d =
-2.1
1.4
55.40 34.54
-4.0 -4.0
14.21 17.37
28.4 34.7
-3.5 -2.1
17.11 21.25
34.2 42.5
-0.5 -3.0
1895.1
467
1965.0
607
about the toe for check against overturning . For Seismic condition P= 1965 Mnet = Mr-Mo= x = Mnet/P= e = b/2-x'= Pmax=P/A*(1+6*e/b) = Pmin=P/A*(1-6*e/b) =
KN 1797.12344 0.94832107 0.42667893 ok 221.772 KN/m2 7.932 KN/m2 16.5 m2
CHECK AGAINST OVERTURNING Case: Normal Factor of Safety against overturning =Mres/Mov= Case: Seismic Factor of Safety against overturning =Mres/Mov=
2948.9419
/
1151.8 =
2948.9419
/
1086 =
CHECK AGAINST SLIDING Case : Normal The minimum vertical force will be when there is no live load and the front backfill is washed out = The resisting force againest sliding = (vertical force) x (coefficient of friction b/w concrete and soil) = 1457.06 x 0.45 = 655.676 The sliding force is the horizontal forces = 411.71 KN Hence, Factor of Safety against sliding =Pres/Psl= 655.68 / 411.7 = (f= 0.45 soil/concrete) Case: Seismic The minimum vertical force will be when there is no live load and the front backfill is washed out =
The resisting force againest sliding = (vertical force) x (coefficient of friction b/w concrete and soil) = 1527.00 x 0.45 = 687.149 The sliding force is the horizontal forces = 551.59 KN Hence, Factor of Safety against sliding =Pres/Psl= 687.15 / 551.6 = STRUCTURAL DESIGN OF FOUNDATION
Wt of toe slab =
21.6 KN/m2
221.77 KN/m2
A
Design of toe slab:
E
D
F
G
144.012
85.692
1.000 m
0.750 m 2.75 m
Pressure distribution diagram
####
200.17 Moment about EF =
concrete grede : Steel grade :
87.13 KN-m d= M /( 0.138σck b ) 107.1 mm Adopt D = 600 mm d= 500 mm Ast = Mu/ (0.87xσy x 0.7984 x d) =604 mm2 Using 16 mm dia bar, spacing = Adopt 16 mm dia bar, spacing =
Minimum reinforc. as per code = 201.1 mm 150 mm c/c Area of re-bar provided =
Check for shear stress:
Shear stress =
V/(b d) =
542.20 KN
b=
500
6.00 m
161.3
Shear force at critical section V =
d =
#####
In toe critical section for shear is considered at a distance from abutment stem.
2
0.181 N/mm
Permissible shear stress without shear reinforcement tcmax =2.8 Mpa for M20 concrete Where, k1 = 1.14 - 0.7 d >= 0.5 ( d in m ) Where, As = Area of Steel provided in mm2 b = width of abutment vertical wall and d = thickness of abutment wall p= tc = basic values given for different p% of Steel provided = From Table 19 of IS 456-200
0.5 m
0.280 N/mm2 ok. Shear re-bar is not necessary.
0.037%
>
Design of heel slab: Upward pressure on the heel slab varies from 7.93215698 Kn/m2 to 85.69 Kn/m2 In addition to the upward pressure the heel slab is subjected to downward pressure due to the self weight of heel slab and weight of earth fill on heel slab: Down pressure due to self weight = 14.4 KN/m2 Down pressure due to backfill = 73.8 KN/m2 88.2 KN/m2 3 Net down pressure = Moment about DG= d= M /( 0.138σck b ) Adopt
80 3 27.174 99.225
D= d= Ast = Mu/ (0.87xσy x 0.7984 x d) = Using 16 Adopt 16 Distribution re-bar 0.12/100 x 900 = Using 12 Adopt 12
KN/m2 KN/m2 KN-m mm
max min
600 mm 500 mm 189 mm2 mm dia bar, specing = mm dia bar, specing =
1.000 m
Minimum reinforc. as per code = 201 mm 150 mm c/c Area of re-bar provided =
1080 mm2/runnig meter mm dia bar, specing = 209.3 mm mm dia bar, specing = 200 mm c/c
at top and bottom
Check for shear stress: Since the load on the heel is predominantly downward, it induces tensile reactio at its junction with stem. Hence the critical section for shear for the heel slab is considered at the face of . Shear force V = 248.3274 KN b= 6.00 m Shear stress = V/(b d) = 0.083 N/mm2 Permissible shear stress without shear reinforcement tcmax =2.8 Where, k1 = 1.14 - 0.7 d >= 0.5 ( d in m )
N/mm2
Where, As = Area of Steel provided in mm2 b = width of abutment vertical wall and d = thickness of abutment wall tc = basic values given for different p% of Steel provided = From Table 19 of IS 456-200
Mpa for M20 concrete
p=
0.037%
0.28 N/mm2
>
k for correctness of data
8.000 m 3.000 m
0.000 5.000 ######
Unit Wt
Total Wt e from c/c
0.06 2.06 24 2.70 22 39.73 24 126.42 24 123.84 24 24.77 24 18.58 24 204.34 24 0.00 24 0.00 24 1.30 24 48.38 592.12 296.1 KN
0.000 m 2.338 m 0.238 m
i Pa
i 0 h= a= KN/m
Coul. Rankin
Ws
d 90-b
Weight of soil is not to be considered in Coulumb's Theroy.
b
b Coulomb
F 30
b 90
d 20
30
Let us find out which theory is applicable:
60
1.7321 m <
0.75 0.9397 2.6845 1 0.75 0.5
1.5
4.100 m hence Ka= 0.2973 0.3333
15.384 KN/m Hor pressure from Rankin's theory = Horizontal thrust = 302.6 acts at
50.43 KN/m 1.37 m from the top of foundation footing.
KN
= 28.42 KN 0.75 / 2 =
=
34.73 KN
39.675 KN
6 (222.7) "The seismic force due to live
1/2 to one abutment m
in free bearing
in other bearing
7.3 KN
61.2 KN
Arm (m)
Moment (KN) Longit. Later. 0.16 0.16
47.7 70.6
0.00 -2.34 -0.24 0.00
-74.1 -6.8 0.0
1.72
436.7
3.10 3.10 4.10
3.50 2.05 5.60 2.05
171.7 107.1 280.85
99.5 99.47549 71.2 71.20224 235.4688 81.3 81.3335
752.9
280.9
1004.9
487.5
le increase of stress is gn longitudinal
Decrease spacing !
ok
mm2
at both faces.
ok
Parameters for RCC design 20 N/mm2 415 N/mm2
mm from the junction
>
2 0.125 N/mm
51.0 KN 8.5 KN/m
rom retaining wall
>
2 0.202 N/mm ok
specified in the code.
x 24 = =
15.84 KN 2.64 KN/m
KN/m
=
32.51 KN/m
Decrease specing !
)
62.8613 KN/m
KN KN
=
34.214 KN
= 42.509 KN from the bottom of foundation base
M[r]
M(o)
KN-m
KN-m
359.382 531.683
414.315 117.612 39.6 996.3
-792.1 0 490.05
-221.6 -138.2
-49.74 -35.6 0 -7.698 -62.7 2948.94
-1152
2949
-1086
114.85
eismic condition KN 1862.99 0.94809 0.43 =P/A*(1+6*e/b) = =P/A*(1-6*e/b) =
KN-m m ok 230.02 KN/m2 8.1638 KN/m2
2.6
> 2.0 Safe
2.7
>1.5 Safe
###### KN KN 1.593 >1.5 Hence Safe soil/concrete) ######
KN 1.246 >1.25 Hence Safe
C
7.9
B
1.000 m
M 20 Fe 415
87.1259
rc. as per code =
1000 mm2
Area of re-bar provided = 1339.733
122.4
mm2
0.04 2 0.181 N/mm
80
1000 mm2
of re-bar provided =
1339.7 mm2
0.04 2 0.0828 N/mm
ok
i Pa i Rankin Weight of soil is also to be considered in Rankin's Theroy. a theory is applicable: Coulomb's theory applicable
ndation footing.
Related Documents c2h70
Design Of Abutment Bridge 51b2a
April 2020 31
Bridge Abutment Design Example 5x3j4m
October 2019 114
Abutment Design 1c3u49
April 2020 26
Design Calculation Of Abutment 1z4o1g
April 2020 31
Bridge Abutment Pier Design As Per Irc 38314r
February 2021 0
Design Of Abutment-rob (1) 70331f
April 2020 25More Documents from "Prakash Singh Rawal" 1s16b
Design Of Abutment Bridge 51b2a
April 2020 31
Irc Class Vehicle Load.xls 2ux6w
July 2020 0
Steel Industry Financial Analysis 1n6i20
January 2022 0
Staff Selection Commission Exam Papers 2i1q1x
December 2019 89
Important Physics Book.pdf 592f5
November 2019 76
April 2020 22