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Avg height Avg lengthcounterfort 1 16.283 m 9.415 mcounterfort 2 16.507 m 9.728 mside wall 1 16.006 m 9.794 mside wall 2 16.786 m 9.817 m
Avg height is calculated taking the average of maximum and minimum height of abutment at median and outer edge respectively and then doing the necessary deduction. Average length is the length that
average height would go, to result area equivalent to the actual geometry.
abutment cap maximum dimension = 2.781 m
abutment cap minimum dimension = 1.474 m
active earth coefficient for normal case = 0.279
active earth coefficient for seismic case = 0.368
unit weight of concrete (substructure) = 2.40 t/m3
unit weight of soil = 1.80 t/m3
braking force = 10 t
maximum centrifugal force(normal) = 18 t
maximum centrifugal force(seismic) = 9 t
transverse moment (normal) = 344 t-m
transverse moment (seismic) = 172 t-m
coefficient of friction = 0.5
Allowable soil bearing capacity(seismic)= 56.25 t/m2 (= 1.25 X 45)
STABILITY CALCULATIONSface of dirtwall
0.574
1
0.441
face of stem wall
2 0.754
0.35
1.881
VERTICAL FORCES
weight calculations W (t) Lever arm from end wall1 wt of slab 9.276 x 11.225 x 0.3 x 2.4 = 74.969 4.6382 dirt wall 0.5 x (2.256 + 3.036) x 11.225 x 0.35 x 2.4 = 29.754 9.4513 abut cap 0.5 x (2.781+1.474) x 11.225 x 0.6 x 2.4 = 34.389 10.3734 stem 0.5 x (0.85 + 0.45) x 17.636 x 11.379 x 2.4 = 313.061 10.5195 toe 0.5 x (1.8+0.853) x 13.412 x 2.3 x2.4 = 98.095 12.7926 heel slab 0.5 x (9.337 + 10.644) x 11.225 x 0.85 x 2.4 = 228.772 5.0027 partition wall 0.3 x 8.925 x 16.396 x 2.4 = 105.360696 5.1128 counterfort-1 0.3 x 16.283 x 9.415 x 2.4 = 110.379 5.0089 counterfort-2 0.3 x 16.507 x 9.728 x 2.4 = 115.618 5.164
10 side wall-1 0.85 x 16.006 x 9.794 x 2.4 = 319.796 5.19711 side wall-2 0.85 x 16.786 x 9.817 x 2.4 = 336.168 5.20912 end wall 0.3 x 11.225 x 16.396 x 2.4 = 132.512 0.15013 earth fill -1 41.585 x 16.396 x 1.8 = 1227.29 2.63114 earth fill-2 4.9595 x 8.92 x 16.786 x 1.8 = 1094.909 7.74215 DL + SIDL 170 = 170 10.51916 LL 80 = 80 10.519
For calculating moments about toe for stability and for calculating area and section modulus's of the foundation plan, considering the trapezoid to be an equivalent rectangle for simplicity.
Check for shear :shear force at distance d from face of stem= 18.11 tbending moment at d = 5.35 t-meffective depth available at d= 1.005 mrelief in shear force = (M*tan(beta)/d) = 2.42 t
Check for shear :shear force at distance d from face of stem= 21.39 tbending moment at d = 6.33 t-meffective depth available at d= 1.005 mrelief in shear force = (M*tan(beta)/d) = 2.86 t
Therefore wall overall thickness can vary from 0.80 m at bottom to0.45 m at top
STEEL REQUIREMENT BY SHEAR
Maximum Shear that can occur due to unit load= 3.03 t/m
Inside the box compartments, the earth is confined and hence pressure acting is static, so "active earth pressure at rest" will act.
for given soil, angle of repose = 30 °earth pressure coefficient = 0.279unit density of soil = 1.80 t/m3support thickness = 0.30 m
For soil fill inside the box
S. No. Depth (m) Vcr (t/m) τc (N/mm2)
1 16.786 8.43 18 0.800 0.219 0.221
2 13.000 6.53 14 0.760 0.179 0.224
3 10.000 5.02 10 0.660 0.158 0.215
4 6.500 3.26 7 0.560 0.121 0.227
5 3.000 1.51 3 0.450 0.070 0.246
At every section τv < τc , hence nominal shear shear reinforcement will doProvide 2 legged shear stirrup :
bar dia : 10 φ Asv = 78.5 mm2spacing: @ 200 c/c
Pressure (t/m2)
d provided
(m)τv
(N/mm2)
Design of Partition wall
This member is designed as a tension member with tensile force equal to the maximum reaction acting at the juncture of partition wall and the side walls
Maximum reaction force = 3.030 t/m + 2.750 t/m per 1.00 t/m2 load = 5.780 t/m
Minimumreinforcement provided = 0.2% X bd= 600.00 mm2
Provide 12 150 753.98 mm2
Depth (m)
Pressure (t/m2)
Tension T(t/m)
Reqd Ast (mm2/m)
On each face (mm2/m)
bar dia (mm)
Spacing (mm)
In the end wall there is earth pressure acting from both sides hence net pressure is effectively zero. Thus minimum reinforcement would suffice.
Design Of COUNTERFORT WALLSINNER COUNTERFORTS (T-BEAMS)
influence zone for the counterfort walls= 3.412 m
(this is the width from which each counterfort receives the eatrh pressure)
unit weight of soil = 1.80 t/m3 Fst = 24000 t/m2earth pressure coeffiecnt = 0.279 j = 0.891
surcharge height = 1.20 mmaximum soil height = 16.786 m
pressure due to soil at specified depth = 0.502 X H (t/m2) (varying)pressure due to surcharge at specified depth = 0.603 t/m2 (constant)
pressure diagram will be somewhat like this
0.603 t/m2 2.06 t/m2
X 3.4125 = 16.786 m
9.033 t/m2 30.83 t/m2
D = 9.662 m max bar dia = 32 φd = 9.596 m clear cover = 50 mmb= 0.300 m
S. No. Depth (m)STEEL REQUIREMENT(mm2)
Provide φ nos
1 16.786 30.83 1641 7999.044 32 9.00
2 13.000 24.34 1337 6513.915 32 7.00
3 10.000 19.20 1095 5337.109 32 6.00
4 6.500 13.20 813 3964.169 32 4.00
5 3.000 7.20 532 2591.229 32 3.00
Pressure at this depth
(t/m2)
Moment (t-m/m) Reqd
(mm2)
Two layers of 32φ from the base to height of 6.786 m and lap with 25φ therafter extending to top
Vu (t) τv (N/mm2)
16.786 30.83 t/m2 138.02 0.48 0.24 68.93
13.000 24.34 t/m2 85.80 0.30 0.24 16.71
10.000 19.20 t/m2 53.15 0.18 0.22 0
6.500 13.20 t/m2 24.80 0.09 0.22 0
3.000 7.20 t/m2 6.95 0.02 0.22 0
Provide 2 legged stirrup 8 φ = 100.531 mm2
Spacing = 335.87868 mm
Provide 8φ 2 legged stirrup at a spacing of 200 c/c(horizontal ties)
Average downward pressure = 16.91 t/m2
Total area of reinforcement required to resist this tension = 704.59 mm2/m
Provide 2 legged 8φ = 100.53 mm2
spacing = 100.53 x 1000 / 704.59 = 142.679
Provide 8φ 2 legged stirrup at a spacing of 120 c/c(vertical ties)
Depth (m)
Pressure at this depth
(t/m2)
τc (N/mm2)
Vs = (τv-τc)bd (t)
To secure connection between counterfort and heel slab, counterfort needs to be designed for tension arising out of the net outward pressure of heel slab.
OUTER COUNTERFORTS (L BEAMS)influence zone for the counterfort walls= 1.775 m
(this is the width from which each counterfort receives the eatrh pressure)
unit weight of soil = 1.80 t/m3 Fst = 24000 t/m2earth pressure coeffiecnt = 0.279 j = 0.891
surcharge height = 1.20 mmaximum soil height = 16.786 m
pressure due to soil at specified depth = 0.502 X H (t/m2) (varying)pressure due to surcharge at specified depth = 0.603 t/m2 (constant)
pressure diagram will be somewhat like this
0.603 t/m2 1.07 t/m2
X 1.775 = 16.786 m
9.033 t/m2 16.04 t/m2
D = 9.662 m max bar dia = 32 φd = 9.596 m clear cover = 50 mmb= 0.850 m
S. No. Depth (m)STEEL REQUIREMENT(mm2)
Provide φ nos
1 16.786 16.04 854 4160.876 32 5.00
2 13.000 12.66 695 3388.114 32 5.00
3 10.000 9.99 570 2775.782 32 3.00
4 6.500 6.87 423 2061.395 32 3.00
Pressure at this depth
(t/m2)
Moment (t-m/m) Reqd
(mm2)
5 3.000 3.75 276 1347.008 32 3.00
Two layers of 32φ from the base to height of 6.786 m and lap with 28φ therafter extending to top
Vu (t) τv (N/mm2)
16.786 16.04 t/m2 71.80 0.09 0.2
13.000 12.66 t/m2 44.63 0.05 0.2
10.000 9.99 t/m2 27.65 0.03 0.2
6.500 6.87 t/m2 12.90 0.02 0.2
3.000 3.75 t/m2 3.61 0.00 0.2
τv < τc at all positions, so provide nominal shear reinforcement :
Provide 2 legged stirrup 8 φ = 100.531 mm2
Spacing = 200 mm (horizontal ties)
Average downward pressure = 16.91 t/m2
Total area of reinforcement required to resist this tension = 704.59 mm2/m
Provide 2 legged 8φ = 100.53 mm2
spacing = 100.53 x 1000 / 704.59 = 142.679
Provide 8φ 2 legged stirrup at a spacing of 120 c/c(vertical ties)
Depth (m)
Pressure at this depth
(t/m2)
τc (N/mm2)
To secure connection between counterfort and heel slab, counterfort needs to be designed for tension arising out of the net outward pressure of heel slab.
3413
850
9728
300
modular ratio m = 280/(3σcbc) for M35, σcbc = 11.67= 8
total load per track including impact = 1.25*35 = 43.75 t
effective load on span= 43.75*3.712/3.712 = 43.75 t
moment along short span= (m1+µ*m2)*43.75 = 5.57 t-m
moment along long span= (m2+µ*m1)*43.75 = 2.89 t-m
Live load bending moment due to IRC Class AA wheeled vehicle =
Y
2.6375
4 1 2 3X 3.75 t 6.25 t 6.25 t 3.75 t X
2.6375 8 5 6 73.75 t 6.25 t 6.25 t 3.75 t
1.775 Y 1.775
The Class AA wheeled vehicle is placed as shown to produce the severest moments. The front axle is placed along the centre line with 6.25t wheel at centre of panel. Maximum moments in the short span and long span directions are computed for individual wheel loads taken in the order shown
B = 3.550 mL = 5.275 m
Bending Moment due to wheel 1 = 6.25 t
tyre contact dimensions= 300 X 150 mm
u= sqrt((0.3+2*0.056)^2+0.3^2)= 0.5097
v= sqrt((0.15+2*0.056)^2+0.3^2)= 0.3984
u/B = 0.144 v/L = 0.076 K = 0.70
m1 = 0.221 Total load allowing for 25 % impact = 7.813 tm2 = 0.190
Moment along short span = 1.950 t-mMoment along long span = 1.744 t-m
Bending Moment due to wheel 2 = 6.25 t
Y
X 6.25 t 6.25 t X
0.3 0.85 0.85 0.3Y
2.3
Intensity of loading = (6.25*1.25)/(0.5097*0.3984) = 38.47 t/m2
Consider the loaded area of 0.150 X 2.300
u = sqrt((2.3+2*0.056)^2+0.3^2) = 2.4306v = sqrt((0.15+2*0.056)^2+0.3^2) = 0.3983
u/B = 0.685 v/L = 0.076 K = 0.70
Here wheel load is placed unsymmetrical to YY axis. But Pigeuds Curves are derived for symmetrical loading. Hence we place an equal dummy load symmetrical about the YY axis and consider the whole loading area. Then we deduct the area beyond the actual loaded area. Half of the resulting value is taken as the moment due to actual loading.
m1 = 0.112m2 = 0.119
Moment along short span = 4.578 t-mMoment along long span = 4.788 t-m
Now consider the area beyond the actual loading = 1.7 X 0.15
u = sqrt((1.7+2*0.056)^2+0.3^2) = 1.8367v = sqrt((0.15+2*0.056)^2+0.3^2) = 0.3983
u/B = 0.517 v/L = 0.076 K = 0.70
m1 = 0.132m2 = 0.14
Moment along short span = 3.517 t-mMoment along long span = 3.674 t-m
Net moment along short span = 0.531 t-mNet moment along long span = 0.557 t-m
Bending Moment due to wheel 3 = 3.75 t
By similar procedure as for previous case, we getB.M along short span = 0.052 t-mB.M along long span = 0.121 t-m
Bending Moment due to wheel 4 = 3.75 t
By similar procedure as for previous case, we getB.M along short span = 0.520 t-mB.M along long span = 0.607 t-m
Bending Moment due to wheel 5 = 6.25 t
B.M along short span = 0.823 t-mB.M along long span = 0.195 t-m
Bending Moment due to wheel 6 = 6.25 t
In this case loading is eccentric w.r.t XX axis. By similar procedure as for previous case but with load area extended w.r.t. XX axis, we get
B.M along short span = 0.823 t-mB.M along long span = 0.195 t-m
Bending Moment due to wheel 7 = 3.75 t
By similar procedure as for previous case, we getB.M along short span = 0.486 t-mB.M along long span = 0.115 t-m
Bending Moment due to wheel 8 = 3.75 t
By similar procedure as for previous case, we getB.M along short span = 0.486 t-mB.M along long span = 0.115 t-m
total bending moment along short span = 5.671 t-m
total bending moment along long span = 3.649 t-m
To allow for continuity, the computed momnts are multiplied by a factor of 0.8
Design Bending Moment=along short span= 5.16 t-malong long span= 3.24 t-m
Grade of conc. 35
Grade of steel 500
Dia of bar used 16
Q for concrete grade used 170 t/m2
k value for concrete 0.327
j value for concrete 0.891
Permissible stress in steel 24000 t/m2
Permissible stress in concrete 1167 t/m2
cover 50 mm
effective depth required= 0.174 mprovided deff = 0.242 m OK
In this case loading is eccentric with respect to both XX and YY axes. A strict simulation would be very complicated and laborious. For A reasonable approximation, eccentricity w.r.t. only XX axis is considered and calculations made as for previous case.
main reinforcement required= 998 mm2So provide ϕ 16 @ 150 c/c 1340 mm2 OK
longitudinal reinforcement required = 625 mm2So provide ϕ 12 @ 150 c/c 754 mm2 OK
Design of Abutment Cap
Taking thickness equal to 225 mm for calculation of steel requirement.
Y
X
Along X direction
0.225 m
2.781 m
Steel reqd = 1% of cross section = 6257.25 mm2
steel on both top and bottom face = 1/2 of steel reqd= 3128.625 mm2
Provide ϕ 25 @ 150 c/c = 3272.492 mm2
Along Y direction
0.225 m
1.000 m
Steel reqd = 1% of cross section 2250 mm2
steel on both top and bottom face = 1/2 of steel red= 1125 mm2
1)Earth Pressure due to surcharge equivalent to 1.2m of earthfill
= 0.603 t/m2
2)Earth Pressure due to backfill of earth =
= 1.525 t/m2
Bending moment at the base of dirtwall due to earth pressure (1)
= 2.777 t-m/m
Bending moment at the base of dirtwall due to earth pressure (1)
= 2.342 t-m/m
Total bending moment at the base of dirtwall = 5.120 t-m/mdue to earth pressure
Calculation of force and moment due to the effect of braking :(cosidering 40t bogie loading)
1.7 10t 10.85 10t
2.79m Dirt Wall3.036
1.750 2.79 3.036
Effective width = 7.576 m
Braking force, 0.2*20 = 4 t (Only two wheels can come on dirt wall at a time.)Braking force = 4 t (Impact factor can't be included with braking force)
Braking force per metre width = 0.53 t
Bending moment at the base of dirtwall due to effect of braking 2.24 t-m/m
g =
ka =
width of dirtwall =
height of dirtwall,h =
ka *g*1.2
ka *g*h
ka *g*1.2*h2/2
ka *g*h3/6
Max. wheel load is from 40t boggie load. So, considering this case only. Wheel Loads shown below are placed on dirt wall (not on RCC Solid Slab) along carriage-way. So, dispersion is taken directly at 45
degrees.
Therefore total bending moment at the base of the dirtwall 7.36 t-m/m
Basic Design Data:
Grade of conc. 35Grade of steel 500Dia of bar used 16Permissible stress in concrete 1167 t/m2Permissible stress in steel ###m , Modulur ratio 10 K value for concrete 0.327j value for concrete 0.891Q for concrete 170.08
Max. moment in dirtwall (t-m) 7.36
Effective depth required (mm) 208
Effective depth provided (mm) 292 SAFE
1178
Provide Vertical longitudinal reinforcement:
Use f16 @ 150 c/c
(On Approach Side face) = 1340 Thus OK
Min. Reinforcent in vertical direction (On outer face) = 525 For 1m lengthUse f16 @ 200 c/c
1005 Thus OK
Ast required (mm2)
Ast provided (mm2)
Therefore providing 20 f @ 130 c/c on the approach side and 12 f @ 200 c/c (Min. % reinforcement) on the outer side in the vertical direction . Also providing 12 f @ 200 c/c on both faces in the horizontal
direction .
DESIGN OF DIRTWALL I Normal Case with Live load
In the design of dirtwall the total height cosidered has been calculated taking into account a slope of 2.5% provided in the carriageway.
For Backfill Soil :a = 90.0 degree
0.35 b = 0.0 degree3.036 f = 30.0 degree
1 2 d = 20.0 degreeKah = 0.279
1.525 t/m2 0.603 t/m2 g = 1.8
Earth Pressure diagram
Height of dirtwall = 3.036 m
1 Earth Pressure due to surcharge equivalent to 1.2m of earthfill = 0.603
2 Earth Pressure due to backfill of earth = 1.525
Bending moment at the base of dirtwall due to earth pressure (1) = 2.78 t-m/mBending moment at the base of dirtwall due to earth pressure (2) = 2.95 t-m/m
Total bending moment at the base of dirtwall = 5.73 t-m/m
Calculation of force and moment due to the effect of braking :(cosidering 40t bogie loading)
1.710t 10t
3.04 2.79m
1.700 2.79 3.036
Effective width = 7.526 mBraking force, 0.2*20 = 4.0 tBraking force per metre width = 0.53 tBending moment at the base of dirtwall due to effect of braking = 1.61 t-m/m
Therefore total bending moment at the base of the dirtwall = 7.34 t-m/m
t/m3
t/m2
t/m2
II Seismic CaseIn the design of dirtwall the total height cosidered has been calculated taking into account a slope of 2.5% provided in the carriageway.
For Backfill Soil :a = 90.0 degree
0.30 b = 0.0 degree3.036 f = 33.0 degree
1 2 d = 22.0 degreeCah = 0.134
0.730 t/m2 0.288 t/m2 g = 1.8
Earth Pressure diagram
Height of dirtwall = 3.036 m
1 Earth Pressure due to surcharge equivalent to 1.2m of earthfill = 0.288
2 Earth Pressure due to backfill of earth = 0.7303 Horizondal force due to seismic = 0.231 t/m
Bending moment at the base of dirtwall due to earth pressure (1) = 1.76 t-m/mBending moment at the base of dirtwall due to earth pressure (2) = 1.41 t-m/mBending moment at the base of dirtwall due to seismic horizondal force (3) = 0.35 t-m/m
Total bending moment at the base of dirtwall = 3.52 t-m/mTherefore Normal case govern for the design
CALCULATION OF DESIGN PARAMETERS
Grade of concrete = M 35Grade of steel = Fe 500
Permissible stresses:
sst = 24480
sbc = 1190
Basic Design Parameters:
k = 0.327j = 0.891
Q = 173.41
Required effective depth =(7.34/173.41)^0.5 = 0.206 m