revised design

PROJECT : PROPOSED ROAD OVER BRIDGE AT BURNPUR COMPLEX OF IISCO LTD.

CLIENT: IISCO,BURNPUR, WEST BENGAL

CONSTRUCTOR: HINDUSTHAN STEELWORKS CONSTRUCTION LIMITED

KOLKATA

CONSULTANT: N P DESIGNERS, 14, K.N. SEN ROAD,BLOCK-B,FLAT-2C,

KOLKATA-700 042

TITLE: DESIGN CALCULATION OF BRIDGE

1

Basic Design Parameter

1. Span (center to center of Pier/Abutment) - 18.0 M/21.0 M2. Finished Road level - 139.066 M3. Top of Rail level - 130.9 M4. Av, Ground level - 130.0 M.5. Height of Pier (below bed block) - 6.20 M6. Loading - IRC Class A two lane or Class 70 R Single lane whichever produces severe effect.7. Grade of Concrete - M-35 for pile, pile cap& superstructure and M-30 for others. 8. Grade of steel - Fe 500 as per IS:17869. Environmental Exposure Condition - Moderate10. Type of Superstructure - R.C.C. T-Beam with R.C.C. Deck Slab.

GENERAL

1. Type of Bridge – R.C.C. Deck supported by R.C.C. Girder as superstructure and on R.C.C. substructure over R.C.C. Bored cast in situ pile foundation.

2. Loading – Two lanes of IRC – class – A or single lane of class 70 R loading. 3. No. of span – 4 spans one of 21.0 m and three of 18.0 m each C/C of Pier/Abutment.4. Foundation system – R.C.C. Bored cast in situ pile foundation of 1000 mm dia and with M.S. Liner up to

soft rock level strata if required. 5. Superstructure system – R.C.C. girder with R.C.C. Deck of Total depth 2166 mm/1866 mm including

R.C.C. deck slab.6. Substructure – R.C.C. wall type pier/abutment .7. Depth of foundation – Top of Pile Cap at 0.5 m below existing ground level.

STRUCTURAL DESIGN DATA

1. Clear carriage way – 7500 mm2. Footpath on each side – 1475 mm on each side3. Width of railing – 275 mm on each side extreme end and 450 mm wide crash barrier in between

carriageway and footpath.4. Effective span – 18.0m/21.0 m c/c of pier/abutment.5. C/c of girder across traffic – 2650 mm6. Wearing Course – 65 mm thick mastic asphalt.7. Expansion joint – As per MOST or Exjomet. 8. Drainage spout – PVC pipe and MS frame with MOST details.9. Total depth of deck slab and girder – 2166 mm/1866 mm10. Total width of bridge – 12000 mm

Reference books and codes

The following books and standards have been considered for designing the bridge.

1. IRC Codes

2

DESIGN OF PIER

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Design of Superstructure : The design and drawing of superstructure has been considered from the standard

designs of MOST.

Design of Pier:

Design of Foundation under PierLoad Calculation over pier pile:-

Lev.139.066 M75 thk Wearing course 40 Exp Gap

1866 2166

21.0 m Span 18.0 m Span 300 300

300. 300 700 1500 1000 300 300

6100 900

G.L. 130.0 M 500

1500

Length of pier cap = 12 MHeight of pier from pile cap level to formation level = 9.93 mOverall depth of pile cap = 1.5 mCompilation of loads and moments about top and bottom of pile cap

Calculation of Loads

(A) Dead Loads:

(1) Dead Loads of Superstructure:

Wt. from 21.0 m spanC/c of pier from P2 to P3 = 21.0 m

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So, overall length = 21.0– 0.040 = 20.96 m


(a) Wt. of crash barrier = {0.45 x 0.15 + ½ (0.45 + 0.275) x 0.25 + ½ (0.275 + 0.175) x 0.5} x 2.5 x 2 sides = 1.360 T/m(b) Wt. of P.C.C. under footpath = 1.25 x 0.4 x 2.4 x 2 sides = 2.400 T/m(c) Wt. of railing base = 0.55 x 0.45 x 2.5 x 2 sides = 1.238 T/m(d) Wt. of railing = 0.170 x 0.175 x 3 x 2.5 x 2 sides = 0.447 T/m(e) Wt of rail post = 0.275 x 0.25 x 1.1 x 2.5 x 2 sides/1.5(spacing) = 0.252 T/mWt. of services neglected as we have considered full of P.C.C. inside the service duct to be on higher side)(f) Wt. of wearing course = 7.5m x 0.065 x 2.5 = 1.219 T/m(g) Wt. of deck slab = [12.0 x 0.2 + ½ x 10.9 x 0.136 + 2 x ½ x 1.3125 x 0.10] x 2.5 = 8.182 T/m(h) Wt of R.C.C. girder = 4 x [0.6 x 0.25 + 2 x ½ x 0.15 x 0.15 + 1.58 x 0.325 + 2 x ½ x 0.3 x 0.15] x2.5 x 1.14 = 8.334 T/m Total = 23.432 T/m say 23.5 T/mSo, reaction on each side Pier = {(20.96 x 23.5)/ 2} T = 246.28 T say 250 T

Wt. from 18.0 m spanC/c of pier from P2 to P3 = 18.0 m



(a) Wt. of crash barrier = {0.45 x 0.15 + ½ (0.45 + 0.275) x 0.25 + ½ (0.275 + 0.175) x 0.5} x 2.5 x 2 sides = 1.360 T/m(b) Wt. of P.C.C. under footpath = 1.25 x 0.4 x 2.4 x 2 sides = 2.400 T/m(c) Wt. of railing base = 0.55 x 0.45 x 2.5 x 2 sides = 1.238 T/m(d) Wt. of railing = 0.170 x 0.175 x 3 x 2.5 x 2 sides = 0.447 T/m(e) Wt of rail post = 0.275 x 0.25 x 1.1 x 2.5 x 2 sides/1.5(spacing) = 0.252 T/mWt. of services neglected as we have considered full of P.C.C. inside the service duct to be on higher side)(f) Wt. of wearing course = 7.5m x 0.065 x 2.5 = 1.219 T/m(g) Wt. of deck slab = [12.0 x 0.2 + ½ x 10.9 x 0.136 + 2 x ½ x 1.3125 x 0.10] x 2.5 = 8.182 T/m(h) Wt of R.C.C. girder = 4 x [0.6 x 0.25 + 2 x ½ x 0.15 x 0.15 + 1.28 x 0.325 + 2 x ½ x 0.3 x 0.15] x2.5 x 1.14 = 7.222 T/m Total = 22.320 T/m say 22.5 T/mSo, reaction on each side Pier = {(17.96 x 22.5)/ 2} T = 202.05 T say 205 TWt. of superstructure on the pier = 250 + 205 = 455 T (2) D.L. of substructure:

i) Pedestals = 4 x (0.75 x 1.30 x 0.525 – 0.75 x 0.63 x 0.3) x 2.5 = 3.702 T say = 4.0 Tii) Bed block = [1.60 x 0.70+ ½ x (1.6 + 0.9) x 0.30] x 12.0 x 2.5 = 44.85 T say = 45.0 Tiii) Pier shaft = 0.90 x 1/2(5.2+9.0)x6.10 = 104.31 T say = 105.0 T

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154.0 T iv) Pile cap = 6.5 x 4.3 x 1.5 x 2.5 = 104.82 T say 105.0 Tv) W.t of earth on pile cap = [6.5 x4.3 – (0.90 x 6.2)] x 0.5 x 1.8 = 20.14 T say 21.0 T

280.0 T

(3) Moments

Moment about C.L. of pier shaft, due to eccentricity of D.L. of superstructure,= (250 x 0.30 – 205 x 0.30) = 13.50 T-m say 14 T-m

(B) Live Load :-(1) Footpath Live load:a) From 21.0 m span:Intensity of footpath live load as per IRC – 6, for substructure design P = 500 – (40L – 300)/ 9 = 500 – (400 x 20.4 – 300)/ 9 = 442.67 kg/m2 say 450 kg/m2

i) For both side load case total load = 2 x 1.475 x 0.45 x 20.96 = 27.83 T say 28 T Reaction on pier =28/2 = 14.0 Tii) For one side load case total load = 1.475 x 0.45 x 20.96 = 13.913 T say 14 T Reaction on pier =14/2 = 7.0 T Moment across traffic = 7.0 x 4.9375=34.57 T-m. Say 35 T-m.

b) From 18.0 m span:Intensity of footpath live load as per IRC – 6, for substructure design,P = 500 – (40 x L – 300)/ 9 = 500 – (40 x 17.4 – 300)/ 9 = 456 kg/m2 say 460 kg/m2

ii) For both side load case total load = 2 x 1.475 x 0.46 x 17.96 = 24.38 T say 25 T Reaction on pier =25/2 = 12.50 Tii) For one side load case total load = 1.475 x 0.46 x 17.96 = 12.19 T say 13 T Reaction on pier =13/2 = 6.50 T Moment across traffic = 6.50 x 4.9375=32.09 T.m. Say 32 T.m.

Reaction on pier:

i) For both side loaded case = 14 + 12.5 = 26.50 T say 28 Ti) For both side loaded case = 7.0 + 6.5 = 13.5 T Say 14 T Moment across traffic = 35 + 32 = 67 T-m.

(2) Vehicular Live load:(i) Class 70 R wheeled:

8 T 12 T 12 T 17 T 17 T 17 T 17 T

3960 1520 2130 1370 3050 1370

13275

300 300A B C D

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300 20400 600 17400 300

Max. pier reaction will occur when the 2nd 17T load will be at end off span AB.Support reaction at B,RB = (17/ 20.40) x (20.70 + 17.65 +16.28) + (12/ 20.40) (14.15 + 12.63) + 8 x (8.67/20.40) = 45.53+15.76 + 3.40 = 64.69 TRC = 17 x (16.33/17.40) = 15.96 T Total reaction at pier = 64.69+15.96 = 80.65 T say 81 TLongitudinal force in the spans = 0.2 x 100 = 20 TForce on the pier considering fixed type pot bearing = 10 T

(ii) Class – A 2 lane:

6.8 T 6.8 T 6.8 T 6.8 T 11.4 T 11.4 T 2.7 T 2.7 T

3000 3000 3000 4300 1200 3200 1100

6200

A B C D

300 20400 300 300 17400 300

For getting maxm reaction one 11.4 T load is kept at end of long span.Max. reaction at B, for one lane loads,RB = (11.4/ 20.04) (20.70 + 19.50) + (6.8/ 20.4) (15.02 + 12.20 + 9.20 + 6.20) = 22.47 + 14.27 = 36.74 TRC = (2.7/17.44) x (14.75 + 13.4) = 4.33 TSo, for one lane loading, max. reaction on pier = (36.74 + 4.33) = 41.07 T For two lane loading Reaction = 2 x 41.07 = 82.14T Say 83TLongitudinal force due to braking = 0.2 x 55.4 = 11.08 TForce on the pier considering elastomeric bearing = 11.08 /2 = 5.40T

Final Live Load Effect on Piers As per code for this bridge there will be either i) One lane of IRC- class 70R load or ii) Two lanes of class A loading.

Case -i - Class 70 R Reaction on pier = 81 T Longitudinal force = 10 T

Case ii - 2 Lanes of class – AReaction on pier = 83 T

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Longitudinal force =5.40T

From the above two load cases, it is clear that class 70R wheeled load is critical is critical.Moment at top of pile cap level = 10 x 7.70T-m = 77.0 T-m Moment at bottom of pile cap = 10 x 9.20 T-m = 92.0. T-m Vertical reaction on pier = 81 TMoment due to eccentricity along traffic At top of pile cap level = (64.69 x 0.30 – 15.96 x 0.30) = 14.62 T-m say 15T-m Horizontal force at bearing level = 10 T

Eccentricity of load (in across direction):-

Eccentricity of 70R live load = (7500/2) – {1200 + (2790/2)} = 1155 mm Moment due to track eccentricity, perpendicular to traffic = 81 x 1.155 = 93.56 T- m say 94 T- m

Summary of Live load: (Footpath Load and Vehicular Load)

Vertical load on pier = 81 + 14 = 95TMoment along traffic,

a) At top of Pile cap level = 15 T-mb) At bot. of Pile cap level = 15 T-m

Moment across traffic = 67 + 94 = 161 T-mLongitudinal force:Horizontal Force = 10 TMoment due to longitudinal force,

a. At top of pile cap level = 77 T-mb. At bot. of pile cap level = 92T-m

Shear Deformation of Bearing

Shear deformation of each bearing (assumed) = 0.5 TTotal Nos of bearings on one Pier Cap = 8So, total force = 8 x 0.5 T = 4.0 TMoment at top of pile cap level = 4 x 7.70 T-m = 30.60 say 31 T-mMoment at bottom of pile cap =4 x 9.20 T-m = 36.80. T-m say 37 T-m

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Check for Seismic forces

Horizontal seismic coefficient (h) for zone III,= .I.0 = 1.0 x 1.5 x 0.04 = 0.06

Part of Structure LeverArm up to top of Pile Cap(M)

Moment at top of pile cap (T-m)

LeverArm up to bottom of

pile cap(M)

Moment at bottom

of pile cap(T-m)

(A) Seismic Forces due to Dead loads:- Horz. Force(i) D.L. of Super structure = 0.06 x 455 = 27.30 T

7.70 210.21 9.2 251.16

(ii) D.L. of Bed block = 45.0 x 0.06 = 2.70 T 7.2 19.44 8.7 23.49

(iii) D.L. of Pier shaft = 0.06 x 105 = 6.30 T Total =36.30 T

3.05 19.22 4.55 28.67

(iv) D.L. of Pile cap = 0.06 x 105 = 6.30 T Total = 42.60 T

Total=248.87

0.75 4.73 Total=30\8.05

(B) Seismic Force due to Live loads:-

Horz. Force p = 0.06 x (1/2 x 95) = 2.85 T Total = 45.45T say 46 T

7.70 21.95Total = 270.82

9.2 26.22 Total = 334.27

Now let us consider two different cases of Seismic Force(1) Seismic force along traffic (No seismic effect on L.L.)At Top of Pile CapTotal Seismic force horizontal = 36.30 T say 37 TTotal Seismic force vertical = ½ x 37.0 = 18.50 T say 19 TMoment at top of Pile cap = 248.87T-m say 246 T-mAt Bottom of Pile CapTotal Seismic force horizontal = 42.60 T say 43 TTotal Seismic force vertical = ½ x 42.60 = 21.30 T say 22 TMoment at bottom of Pile cap = 308.05 T-m say 309 T-m

(2) Seismic force across traffic (Seismic force on 50% of L.L.)Total horizontal Seismic force = 46TTotal vertical Seismic force = 46/2 = 23 TMoment at top of Pile cap = 270.82T-m say 271 T-mMoment at bottom of Pile cap = 334.27 T-m say 335 T-mSince the dimension of foundation is higher across the traffic direction, seismic along traffic will be governing.

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Summary of Loads and Moments(Non-seismic Case)

Case – I- All loadsA) Longitudinal loads:

Sl. No.

Type of loads At bottom of Abutment At bottom of FootingVertical

loadHorizontal load

Moment Vertical load

Horizontal load

Moment

1. D.L. of super structure 455 T - 14 T-m 455 T - 14 T-m

2. D.L. of substructure with wt. of Pile cap and earth on Pile cap (back fill)

154 T - 280T -

3. Live load of Superstructure

95 T - 15 T-m 95 T - 15 T- m

4. Longitudinal force - 10 T 77 T-m - 10 T 92 T-m

5. Shear Deformation of Bearing

4.0 T 31 T-m 4.0 T 37 T-m

Total 704 T 14.0 T 137 T-m 830 T 14.0 T 158 T-m

B) Transverse load:

Due to eccentricity of L.L. moment at bottom of foundation = 161 T-m

Summary of Loads and Moments(Seismic Case)Case – II- Load Case I + Seismic force along TrafficA) Longitudinal loads:

Sl. No.


loadHorizontal load


Horizontal load

Moment

1. D.L. of super structure 455 T - 14 T-m 455 T - 14 T-m


154 T - 280 T -

3. Live load of Superstructure

95 T - 15 T-m 95 T - 15 T- m

4. Longitudinal force - 10 T 77 T-m - 10 T 92 T-m

5. Shear Deformation of Bearing

4.0 T 31 T-m 4.0 T 37 T-m

6. Seismic forces 19 T 37 T 249 T-m 22 T 43 T 309 T-m

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Total 723 T 51.0 T 386 T-m 852 T 57.0 T 467 T-m

B) Transverse load:Due to eccentricity of L.L. moment at bottom of foundation = 161 T-m

Load on Piles Let us use 4 nos. of 1000 mm dia Piles in two rows at 3.0 m on along traffic direction and 5.2 m on across direction spacing. 4300

2600 n = 4

Traffic ZAL = 4 x 1.5 = 6.0 ZAC = 4 x 2.6 = 10.4

2600

1500 1500For Load Case-I (non-seismic)Max / Min. Vertical loads on outer line pile = (MAL/ZAL) ± (MAC/ZAC) = (830/4) ± (158/6) (161/10.4)

= 207.50 ± 26.33 ± 15.48 249.31 T = < 165.69T

Longitudinal horizontal force on each pile = 14/4= 3.5 TFor Load Case-II (seismic)Max / Min. Vertical loads on outer line pile = (MAL/ZAL) ± (MAC/ZAC) = (852/4) ± (467/6) (161/10.40)

= 213 ± 77.84 ± 15.48 306.32 T = < 119.68 T

Longitudinal horizontal force on each pile = 57/4= 14.25 TRef: Clause No. 706.1.2 of IRC:78-2000, Permissible increase in base pressure i.e. capacity of pile is 25 %Since our load in seismic case is only (306.32/249.31 = 1.228) 22.8 % higher hence seismic case load is not governing.

Section Design of Pile:(Foe Load Bearing Capacity see page 34)Let us try with 1000 pile with M-35 grade concrete and Fe-500 grade steel.Ref.:- Appendix C of IS:2911 (Part I/Sec 2)T = 5(EI/K1) E for M-25 grade cement = 500025 = 29580.4 N/mm2 = 2,95,804kg/cm2

I = /64 (100)4 = 4908738.521 cm4

Since the top 15.0 m shaft of pile below pile cap goes into slag equivalent to medium sand Layers , K1 = 0.525T = 5{(2,95,804 x 4908738.521)/0.525} = 307.86 cm say 308 cmL1 = 0 for fixed head pile.

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L1/T = 0 Lf/T = 2.15 Lf = 662.2 cm = 6.63 m

For Load Case-I (non-seismic) Moment at fixity level = 3.50 x {(0 + 6.63)/2} x 0.825 T-m = 9.58 T-m So, we will design the piles for 250 T axial load and 9.58 T-m moment. Area of cross section of pile A = (/4) x 1002 = 7853.98 cm2

Section Modulus of pile section Z = (/32) x 1003 = 98174.77 cm3

41.59 Kg/cm2

Stress in pile section = P/A M/Z = 250 x 103 /7853.98 9.58 x 105 /98174.77 = 31.83 9.76 = < 22.07 Kg/cm2

Total section remains in compression and nominal reinforcement will suffice.

For Load Case-II (seismic) Moment at fixity level = 14.25 x {(0 + 6.63)/2} x 0.825 T-m = 38.98 T-m So, we will design the piles for 307 T axial load and 38.98 T-m moment. The design of pile section is carried out by a computer programme with input and output details as below.Input DataP,M 307 T 38.98 T-mR1,R2 50 cm 0 cmNO1,NO2 10 Nos 0 nosCC1,CC2 5 cm 0 cmBDIA1,BDIA2 2.0 cm 0 cmMODR 10START,VAR,ERROR 1 0.1 0.5Output Data

FINAL RESULTS............. FOR BETA= 174.861400 DEGREES FOR DIST= 99.799050CM............. AEFF= 8166.940000 EEFF= 7.332245E-03 IEFF= 5179447.000000 Distance found = 50.302910 Dist to be matched = 49.806380 COMP. STRESS DEVELOPED....................... Direct stress = 37.835460kg/Cm^2 Bending stress= 37.602110kg/cm^2 TotalStress = 75.437580kg/cm^2 TENSILE STRESS DEVELOPED........................... Total stress = 47.352370 kg/cm^2

Min Ast = (0.4/100) x (/4) x 1002 = 31.42 cm2

Provide 10-20 (31.42 cm2) vertical bars all through as minimum and 8 ties @ 200 c/c and 16 stiffener bars @ 1500 mm c/c.Deflection in pile

Max deflection in pile head (non seismic case) = = Q (L1 + Lf)3/(12 EI)= 3.50 x 1000 x (0 + 663)3/(12 x 2,95,804 x 4908738.521) = 0.055 cm = 0.55 mm < 10 mm (permissible)

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Max deflection in pile head (seismic case) = = Q (L1 + Lf)3/(12 EI)= 14.25 x 1000 x (0 + 663)3/(12 x 2,95,804 x 1.333x 4908738.521) = 0.179 cm = 1.79 mm < 10 mm (permissible)

Design of Pile CapLoad on individual Piles

4300

1 2

Traffic

3 4

3000

n = 4ZAL = 4 x 1.5 = 6.0ZAC = 4 x 2.6 = 10.40

Load on Pile marked (1) = 207.50 – 26.33 + 15.48 = 196.65 TLoad on Pile marked (2) = 207.50 + 26.33 + 15.48 = 249.31 TLoad on Pile marked (3) = 207.50 – 26.33 – 15.48 = 165.69 TLoad on pile marked (4) = 207.50 + 26.33 – 15.48 = 218.35 T

Total force on pile marked 2, & 4 = 249.31 + 218.35 = 467.66 T Wt. of pile cap = 6.5 x 1.5 x 2.5 T/m = 24. 37 T/m

1

2

1050 900 1412.5 6500

3 4

Critical section650 1500 1500 650 For shear

1500 1500

Moment at face of pier shaft, M = 467.66 x (1.5–0.45) – 24.37 x 1.72/2 = 491.04-35.21 = 455.83 T-m say 456T-mLength of pile cap = 6500mm; Try with M-35 grade of concrete.d = √{(486 x 105)/ (17 x 650)} cm = 64.24 cm [ Q value for M-35 Concrete = 1.70 MPa = 17 Kg/cm2] Provide 1500 mm depth,

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d = 1500 – 50 – 25 – (25/ 2) = 1412.5 mmAst = (456 x 105)/ (2400 x 0.89 x 141.25) mm2 = 151.14 cm2

Min main reinf. = (0.2/100) x 141.25 x 650 cm2 = 183.63 cm2

Min. distribution reinf = (0.12/100) x 141.25 x 100 = 16.95 cm2/mProvide 38-25 (186.53 cm2) along traffic at bottom as main reinforcement and 16 @ 115 c/c as distribution at bottom. Provide 16 @ 115 mm c/c both ways at top.ShearOne way ShearMax. S.F. at a distance of d = 1412.5 mm away from face of pile shaftThe central line of pile is 1050 mm away from face of abutment wall. so, shear is not governing So, no shear reinforcement necessary.Punching Shear or two way shear 650

d/2 = 706.25 500 650

Critical line of shear

Worst Punching shear will occur at corner pile to have least resisting periphery.

Max pile load = 249.31 T. Available length of resistance = 650 + 650 + 2 x x (500 + 706.25)/4 = 1300 + 1894.77 = 3194.77 mmEffective Depth = 1412.5 mmPunching Shear Stress = 249.31 x 1000/(319.477 x 141.25) = 5.52 Kg/ cm2

Allowable stress of shear in punching = 0.16 x √35 MPa = 0.946 MPa = 9.46 Kg/ cm2 > 5.52 kg/ cm2

Design of Pier:

(A) loads and Moments in non-seismic case:Total vertical load on Pier bottom section P = 704 TTotal moment at Pier bottom along Traffic = 137 T-m Total moment across Traffic =161 T-m

Let us use 6200 mm long 900 mm wide wall 6200 900

A = 620 x 90 = 55800 cm2

ZAL = 6200 x 902/6 = 8,37,000 cm3

ZAc = 90 x 6202/6 = 5,766,000 cm3.Stress in extreme fibre = P MAL MAC

A ZAL ZAc

= (732 x 103/ 55800) (137 x 105/837000) (161 x 105/5,766,000) = 12.62 16.37 2.80 = 31.79 Kg/cm2

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< = - 6.55 Kg/cm2

So, nominal reinforcement will suffice Ast = 0.3 x 55800/100 cm2 = 167.4 cm2

Provide 80 Nos – 20 (251.20 cm2) of which 36 nos on each long side portion and 4 nos on each short side portion.Provide 8 ties and links (every alternate 3 rd bar) @ 200 mm c/c.

N.B. : We have calculated the actual moment of inertia and vis-à-vis the stresses in concrete and steel in the following page.

Actual stress analysis of Pier body 60 780 60

60 6200 225 9136

X X

36-20@174c/c(approx)

4-20 4-20@150c/c(approx) 900A comb = Ac + A(m-1) Ast = 620x 90 + (10-1) x 80 x 3.14 = 55800 + 2260.00 = 58060.80 cm2

Iyy = 620 x 903/12 + (10-1) x 3.14 [ 172 x 392 + 2 x 7.52 + 2 x 22.52 ] = 37665000 + 3126601 = 40791601 cm4

Ixx = 90 x 6203/12 + (10-1) x 3.147 [12 x 3042 + 4 x (286.62 + 269.22 + 251.82 + 234.42 + 2172 + 199.62 + 182.22

+ 164.82 + 147.42 + 1302 + 112.62 + 95.22 + 77.82 + 60.42 + 432 + 25.62 + 822 )]

= 1787460000 + 88003612 = 1875463612 cm4

Stress in extreme fibre of concrete in non seismic case,

= 704 x 103/58060.80 137 x 105 x 45/40791601 161 x 105 x 310/1875463612

= 12.13 15.12 2.66

= < 29.91 kg/cm2 60

- 5.81 kg/cm2 2.945

5.81/x = 29.91/(90 – x) x 19.777

or 90/x – 1 = 5.148 or, x = 14.64 cm.

So, fc’ = (14.64 – 6)/14.64 x 5.81 kg/cm2 = - 3.43 kg/cm2

So, stress is steel in tension zone, = m fc’ = 10 x 3.43 = 34.3 kg/cm2

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(B) Loads and Moments in seismic case:Total vertical load on Pier bottom section P = 723 TTotal moment at Pier bottom along Traffic = 386 T-m Total moment across Traffic =161 T-m

Stress in extreme fibre of concrete in seismic case (considering 33.33% increase of permissible stress),

= 723 x 103/(1.33 x 58060.8) 386 x 105 x 45/(1.33 x 40791601) 161 x 105 x 310/(1.33 x 1875463612)

= 6.50 31.94 2.0 90

= < 43.28 kg/cm2 60

- 24.60 kg/cm2 15.53

24.60/x = 43.28/(90 – x) x 28.53

or 90/x – 1 = 1.76 or, x = 32.62 cm.

So, fc’ = (32.62 – 6/32.62) x 24.60 kg/cm2 = - 20.08 kg/cm2

So, stress is steel in tension zone, = m fc’ = 10 x (-20.08) = - 200.8 kg/cm2

Check of Pier Section for one span dislodge condition

To get worst effect let us consider one 18.0 m span is dislodged

So, Moment along the traffic will be 250 x 0.3 = 75 T-m.In this case the traffic will not be in the span and hence moment due to the longitudinal force will be absent.Now the value of the moment due to longitudinal force is 78 T-m > 75.0 T-m and hence this case will not be governing.

DESIGN OF PIER CAP : (GRADE OF CONCRETE-M-25)

The Pier cap is 150 x 11000 x 1200 thk.The bearing point is well within the pier shaft along and across traffic and hence no eccentricity is induced and nominal reinforcement is to be provided.Ast (shorter span) on two faces = (1/ 100) x 225 x 12000 mm2 = 27000 mm2

Provide 70-16 on top and bottom. (28140 mm2)Ast (longer span) on two faces = (1/ 100) x 225 x 1500 mm2 = 3375 mm2

Provide 10-16 at top and bottom (4020) and 3-12 closed bars.

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DESIGN OF

ABUTMENT

17

Design of Superstructure : The design and drawing of superstructure has been considered from the standard

designs of MOST.

Design of Abutment:

Top of Deck 139.066M 2501866

500 1916 1050

2961000 1000

9.366 m 280 720

6204. 130.0 M 1150 2000 1150 300 129.70 M top of pile cap Pile cap 2000 1500

DESIGN OF FOUNDATION UNDER ABUTMENT Length of abutment cap = 12 M x Length of Abutment = 10 MHeight of pier from pile cap level to formation level = 9.93 mOverall depth of pile cap = 1.5 m; Angle of internal friction of back fill material = 40Compilation of loads and moments about top and bottom of pile cap.

Calculation of Loads

(A) Dead Loads:(1) Dead Loads of Superstructure:Wt. from 18.0 m span

C/c of pier Mkd P2 to inner face of dirt wall at Abutment A1 = 18.0 m


(1) Dead Loads of Superstructure:(a) Wt. of crash barrier = {0.45 x 0.15 + ½ (0.45 + 0.275) x 0.25 + ½ (0.275 + 0.175) x 0.5} x 2.5 x 2 sides = 1.360 T/m(b) Wt. of P.C.C. under footpath = 1.25 x 0.4 x 2.4 x 2 sides = 2.400 T/m(c) Wt. of railing base = 0.55 x 0.45 x 2.5 x 2 sides = 1.238 T/m(d) Wt. of railing = 0.170 x 0.175 x 3 x 2.5 x 2 sides = 0.447 T/m(e) Wt of rail post = 0.275 x 0.25 x 1.1 x 2.5 x 2 sides/1.5(spacing) = 0.252 T/mWt. of services neglected as we have considered full of P.C.C.

18

inside the service duct to be on higher side)(f) Wt. of wearing course = 7.5m x 0.065 x 2.5 = 1.219 T/m(g) Wt. of deck slab = [12.0 x 0.2 + ½ x 10.9 x 0.136 + 2 x ½ x 1.3125 x 0.10] x 2.5 = 8.182 T/m(h) Wt of R.C.C. girder = 4 x [0.6 x 0.25 + 2 x ½ x 0.15 x 0.15 + 1.28 x 0.325 + 2 x ½ x 0.3 x 0.15] x2.5 x 1.14 = 7.222 T/m Total = 22.320 T/m say 22.5 T/mSo, reaction on each side Support = {(17.94 x 22.5)/ 2} T = 201.83 T say 205 T

(2) D.L. of substructure:

i) Dirt wall = 1.916 x 0.4 x 12.0 x 2.5 = 23.00 Tii) Bed block = 1.0 x 1.0 x 12.0 x 2.5 = 30.00 Tiii) Abutment shaft = (1.0 x 6.20 + ½ x 1.0 x 6.20) x 10.0 x 2.5

= 155 + 77.50 = 232.50 T 285.5 T say 286 T

iv) Wt. of cantilever return wall a) Return wall over pile cap = 2 x 9.116 x 0.5 x 1.15 x 2.5 = 26.21 T

311.71 T say 312 Tv) Pile cap = 4.3 x 10.3 x 1.5 x 2.5 = 166.09 T say 167 Tvi) Wt. of earth over pile cap = 1.15 m x (10.3 – 0.50 x 2) x 9.116 x 1.8 = 175.50 T say 176 T

(back side) vii) Wt. of earth over footing (front side) = 0.3m x 10.0 x 1.15 x 1.8 T = 6.40 T say 7T

Total = 659.70 T say 660T

(3) Moments

Moment about C.L. of abutment shaft, due to eccentricity of various D.L. (except earth on back side as it is self supporting).= -205 x 0.28 – 23 x 0.8 – 30 x 0.8 – 30 x 0.5 – 155 x 0.5 + 77.5 x 0.334= -57.40 – 18.40 – 15.0 – 77.50 + 25.89= -142.41 T-m say -142 T-mMoment about C.L. of Pile cap= -142.41 – 26.21 x 1.575 – 175.50 x 1.575 + 6.4 x 1.575= -142.41 – 41.28 – 276.42 + 10.08= -450.03 T-m say – 450 T-m

(B) Live Load :-(1) Footpath Live load:(B) Live Load :-(1) Footpath Live load:

Intensity of footpath live load as per IRC – 6, for substructure design,P = 500 – (40 x 17.38 – 300)/9 = 500 – 43.91 = 456.09 Kg/m2 say 460 Kg/m2

Reaction on abutment:

ii) For one side loaded case = 0.46 x 1.475 x 17.94 x 2 = 6.09 T say 6.5 Tiii) For both side loaded case = 2 x 6.5 = 13.0 T

Moment across traffic due to one side loaded case, = 6.09 x {(7.5/ 2) + 1.475/2+0.45 } = 30.07 T-m19

(2) Vehicular Live load:

(i) Class 70 R wheeled: 8 T 12T 12T 17T 17T 17T 17T

3960 1520 2130 1370 3050 1370

42601.0 1.018

A B

280 17380 280

Max. support reaction at B,RB = (17/ 17.38) x (17.66 + 16.29 +13.24 + 11.87) + (12/ 17.38) (9.74 + 8.22) + {(8 x 4.26)/ 17.38} = 57.77 + 12.40 + 12.40 + 1.96 = 72.13 T say 73 TLongitudinal force in the span due to braking = 0.2 x 100 T = 20 TForce on the abutment = 10 T

(ii) Class – A 2 lane:6.8 T 6.8 T 6.8 T 6.8 T 11.4 T 11.4 T

3000 3000 3000 4300 1200

17100

A B

280 17380 280

For getting maxm reaction the small wheel loads of 2.7 T is kept beyond the span.Max. reaction at B, for one lane loads,RB = (11.4/ 17.38) (17.66 + 16.46) + (6.8/ 17.38) (12.16 + 9.16 + 6.16 + 3.16) = 22.38 + 11.99 = 34.37 TSo, for 2 lane loading, max. reaction = 2 x 34.37 T = 68.74 T say 69 TLongitudinal force due to braking = 0.2 x (6.8 x 4 + 2 x 11.4) = 10 TForce on the abutment = 10/2 = 5.0 TFrom the above two load cases, it is clear that class 70R wheeled load is critical.Vertical reaction on abutment = 73 T

20

Moment due to eccentricity along traffic At top of pile cap level = 73 x 0.28 = 20.44 T-m say 21 T-mAt bottom of pile cap level = 21 T-mHorizontal force at bearing level = 10 TMoment at top of pile cap level = 10 x 7.5 T-m = 75 T-mMoment at bottom of pile cap level = 10 x 9.0 T-m = 90 T-m

Eccentricity of load:-

Eccentricity of live load = (7500/2) – {1200 + (2790/2)} = 1155 mm Moment due to track eccentricity, perpendicular to traffic = 73 x 1.155 = 84.32 T-m say 85 T-m

Summary of Live load: (Footpath Load and Vehicular Load)

Vehicular load on abutment = 73 + 7 = 80 TMoment along traffic, due to eccentricity of load,

c) At top of Pile cap level = 21 T-md) At bot. of Pile cap level = 21 T-m

Moment along traffic, due to longitudinal force,a) At top of Pile cap level = 75 T-mb) At bot. of Pile cap level = 90 T-m

Longitudinal Force = 10 TMoment across traffic = 85 + 30 = 115 T-m

(C) Earth pressure :-

Let us assume = 40Ka = {cos2 ( - )}/ cos2 cos ( + ) [1 + {sin ( + ) sin ( - 1)/ cos ( + ) cos ( - 1)}]2

= 2/3 = 26.66>22.5hence = 22.5, i = 0, = 0 Ka = cos2 (40 - 0)/[ cos20 cos (0 + 22.5)] x [1 + sin (40 + 22.5) sin (40 – 0)/{ cos (0 + 22.5) cos (0 – 0)}]2

= 0.19922 say 0.20Active E/P up to top of pile cap = (1/ 2) x Ka x x H2 x Lt = (1/2) x 0.20 x 1.8 x 9.1152 x 10.0 T = 149.55 T say 150 T

Moment due to active E/P at top of pile cap = 149.55 x 0.42 x 9.115 = 572.53 T-m say 573 T-mActive earth pressure upto bottom of pile cap = (1/2) x 0.20 x 1.8 x 10.6152 x 10.0 T = 202.82 T say 203 TMoment due to active E/P at bottom of pile cap = 202.82 x 0.42 x 10.615 = 904.24 T-m say 905 T-m

D) Surcharge pressure 21

a) Horizontal: Ka. . h = 0.30 x 1.8 x 1.2 = 0.648 T/m2

Ka..h = 0.2 x 1.8 x 1.2 =0.432 T/m2

9.115m

1.5m

Total force due to live load surcharge at top of pile cap level = 0.432 x 9.115 x 10.0 = 39.38 T say 40 TMoment at top of pile cap level = 39.38 x (9.115/2) = 179.48 T-m say 180 T-mTotal force due to L.L. surcharge at bottom of pile cap level = 0.432 x 10.615 x 10.0 = 45.86 T say 46 TMoment at bottom of pile cap level = 45.86 x (10.615/ 2) = 243.40 T-m say 244 T-m

b) (Vertical):-For L.L. only:

The L.L. Surcharge force will create an extra vertical load at the back of the abutment which will increase the vertical load but this load will produce relieving moment to the abutment pile group system. The width of loading will be 1.15 m length will be clear distance between the two return walls = 1.15 m Total load on the pile cap = 1.2 x 1.8 x 1.15 x 9.0 T = 22.36 T say 23 TRelieving moment about C.L. of pile cap = 22.36 x 1.575 = 35.22 T-m say 35 T-m

Shear Deformation ofBearing:- Force (for 4 nos of bearing) = 4 x 0.50 = 2.0 T Moment at top of pile cap = 2.0 x 7.50 = 15.0 T-m At bottom of pile cap = 2.0 x 9.0 = 18 T-m

Check for Seismic forcesHorizontal seismic coefficient (h) for zone III, = .I.0 = 1.0 x 1.5 x 0.04 = 0.06

Part of Structure Lever Moment at Lever Moment at

22

Arm up to top of Pile

Cap(m)

top of pile cap (T-m)

Arm up to bottom of pilecap(m)

bottom of pile cap

(T-m) (A) Seismic Forces due to Dead loads:- Horizontal. Force(i) D.L. of Super structure = 0.06 x 205 = 12.30 T

7.50 92.25 9.0 110.70

(ii) D.L. of Dirt wall = 0.06 x23.00 = 1.38 T 8.162 m 11.27 9.662 13.34

(iii) D.L. of Bed block = 30x 0.06 = 1.08 T 6.704 12.07 8.204 14.77

(iv) D.L. of Abutment shaft = 0.06 x 155 = 9.30 T 0.06 x 77.50 = 4.65 T 29.43 T

3.1022.068

28.85 19.24 163.68

4.6023.568

21.4016.60

(v) D.L of Return wall over pile cap cap = 0.06 x 26.21 = 1.58 T

4.558 7.20 6.058 9.58

(vi) D.L. of earth over pile cap (back) = 0.06 x 175.50= 10.53 T

4.558 48.00 6.058 63.80

(vii) D.L.of earth over pile cap (front) = 0.06 x 6.40 = 0.39 T 0.15 0.06 1.65 0.65

(viii) D.L. of pile cap = 0.06 x 167 = 10.02 T Total = 51.95 T say 52 T

218.94 say 219

0.75 7.52 258.36 say 259

(B) Seismic Force due to Live loads:-

Horz. Force p = 0.06 x (1/2 x 80) = 2.40 T 54.35 T say 55 T

7.5 18.0236.94 say 237

9.0 21.60279.96 say 280

Now let us consider two different cases of Seismic Force(1) Seismic force along traffic (No seismic effect on L.L.)At Top of Pile CapTotal Seismic force horizontal = 29.43 T say 30 TTotal Seismic force vertical = ½ x 30 = 15 TMoment at top of Pile cap = 163.68 T-m say 164 T-mAt Bottom of Pile CapTotal Seismic force horizontal = 52 TTotal Seismic force vertical = ½ x 52 = 26 TMoment at bottom of Pile cap = 259 T-m(2) Seismic force across traffic (Seismic force on 50% of L.L.)Total horizontal Seismic force = 55 TTotal vertical Seismic force = 55/2 = 27.5 T say 28 TMoment at bottom of Pile cap = 280 T-mSince the dimension of foundation is much higher across the traffic direction, seismic along traffic will be governing.Summary of Loads and Moments

Case – I- All loads

23

A) Longitudinal loads:

Sl. No.


loadHorizontal load


Horizontal load

Moment

1. D.L. of super structure 205 T - -142 T-m

205 T - -450 T-m


286 T - 660 T -

3. Live load of superstructure

80 T -21 T-m 80 T -21 T-m

4. Longitudinal Force 10 T 75 T-m 10 T 90 T-m5. Earth pressure - 150 T 573 T-m - 203 T 905

T-m6. Horizontal surcharge

a) D.L.b) L.L.

- 40 T 180 T-m - 46 T 244 T-m

7. Vertical surcharge a) D.L.b) L.L.

--

--

--

23 T - -35 T-m

8. Shear deformation of bearing

2.0 T 15 T-m 2.0 T 18 T-m

Total 571 T 202 T 680 T-m

968 T 261 T 751T-m

B) Transverse load:


Case – II- Load Case I + Seismic force along TrafficA) Longitudinal loads:

24

Sl. No.


loadHorizontal load


Horizontal load

Moment

1. D.L. of super structure 205 T - -142 T-m

205 T - -450 T-m


286 T - 660 T -

3. Live load of superstructure

80 T -21 T-m 80 T -21 T-m

4. Longitudinal Force 10 T 75 T-m 10 T 90 T-m5. Earth pressure - 150 T 573 T-m - 203 T 905

T-m6. Horizontal surcharge

c) D.L.d) L.L.

- 40 T 180 T-m - 46 T 244 T-m

7. Vertical surcharge c) D.L.d) L.L.

--

--

--

23 T - -35 T-m

8. Shear deformation of bearing

2.0 T 15 T-m 2.0 T 18 T-m

9. Seismic forces 15 T 30 T 164 T-m 26 T 52 T 259 T-mTotal 571 T 232 T 844 T-m 994 T 313 T 1010 T-m

B) Transverse load:


Load on Piles

Let us use 8 nos. of 1000 mm dia Piles in two rows at 3.0 m spacing on along traffic direction and 5.65 m on across traffic directions.

3000

3000 n = 8.0ZAL = 8 x 1.5 = 12 IAC = 4 x (4.52+1.52) = 90

3000

ZAC (outer) = 90/4.5 = 20.0

ZAC (inner) = 90/1.5 = 60 1500 1500

For Load Case-I (non-seismic case)Max / Min. Vertical loads on outer line pile = (MAL/ZAL) ± (MAC/ZAC) = (968/8) ± (751/12) (115/20)

= 121.00 ± 62.59 ± 5.75

25

189.39 T = < 52.66 T

Vertical loads (Max/Min) on Intermediate piles = P/n (MAL/ZAL) ± (MAC/ZAC) = (968/8) ± (751/12) (115/60)

= 121.00 ± 62.59 ± 1.92

185.51 T = <

56.49 TLongitudinal horizontal force on each pile = 261/8 = 32.63 T

For Load Case-II (seismic case)Vertical loads (Max/Min) on Corner piles = P/n (MAL/ZAL) ± (MAC/ZAC) = (994/8) ± (1010/12) (115/20)

= 124.25 ± 84.17 ± 5.75 214.17 T

= < 34.33 T

Vertical loads (Max/Min) on Intermediate piles = P/n (MAL/ZAL) ± (MAC/ZAC) = (994/8) ± (1010/12) (115/60)

= 124.25 ± 84.17 ± 1.92

210.34 T = <

38.16 T

Longitudinal horizontal force on each pile =313/8 = 39.13 TRef: Clause No. 706.1.2 of IRC:78-2000, Permissible increase in base pressure i.e. capacity of pile is 25 %Since our load in seismic case is only (214.17/189.39 = 1.13) 13 % higher hence seismic case load is not governing.

Section Design of Pile:(Foe Load Bearing Capacity see page 34)Let us try with 1000 pile with M-35 grade concrete and Fe-500 grade steel.Ref.:- Appendix C of IS:2911 (Part I/Sec 2)T = 5(EI/K1) E for M-25 grade cement = 500025 = 29580.4 N/mm2 = 2,95,804kg/cm2

I = /64 (100)4 = 4908738.521 cm4

Since the top 15.0 m shaft of pile below pile cap goes into slag equivalent to medium sand Layers , K1 = 0.525T = 5{(2,95,804 x 4908738.521)/0.525} = 307.86 cm say 308 cmL1 = 0 for fixed head pile. L1/T = 0 Lf/T = 2.15 Lf = 662.2 cm = 6.63 m

For Load Case-I (non-seismic) Moment at fixity level = 32.63 x {(0 + 6.63)/2} x 0.825 T-m = 89.24 T-m So, we will design the piles for 190 T axial load and 89.24 T-m moment. The design of pile section is carried out by a computer programme with input and output details as below.

26

Input Data

P,M 190 T 89.24 T-m

R1,R2 50 cm 0 cm

NO1,NO2 22 Nos 0 nos CC1,CC2 5 cm 0 cm

BDIA1,BDIA2 3.2 cm 0 cm

MODR 10

START,VAR,ERROR 1 0.1 0.5

Output Data FINAL RESULTS............. FOR BETA= 80.212100 DEGREES FOR DIST= 41.499940CM............. AEFF= 4778.060000 EEFF= 16.690250 IEFF= 2573655.000000 Distance found = 8.645795 Dist to be matched = 8.190189 COMP. STRESS DEVELOPED....................... Direct stress = 23.440480kg/Cm^2 Bending stress= 90.309390kg/cm^2 TotalStress = 113.749900kg/cm^2 < 11.67 MPa =116.7 Kg/cm2

TENSILE STRESS DEVELOPED........................... Total stress = -1394.762000 kg/cm^2RUN TIME = 0 Min 0 SecTRIALS = 406

For Load Case-II (seismic) Moment at fixity level = 39.13 x {(0 + 6.63)/2} x 0.825 T-m = 107.02 T-m So, we will design the piles for 215 T axial load and 107.02 T-m moment. The design of pile section is carried out by a computer programme with input and output details as below.

Input Data

P, M 214 T 107.02 T-m

27

R1, R2 50 cm 0 cm

NO1, NO2 22 nos 0 nos

CC1, CC2 5 cm 0 cm

BDIA1, BDIA2 3.2 cm 0 cm

MODR 10

START, VAR, ERROR 1 0.1 0.5

Output Data

FINAL RESULTS............. FOR BETA= 81.025120 DEGREES FOR DIST= 42.199930CM............. AEFF= 4847.122000 EEFF= 16.568570 IEFF= 2579046.000000 Distance found = 9.246328 Dist to be matched = 8.768496 COMP. STRESS DEVELOPED....................... Direct stress = 30.946200kg/Cm^2 Bending stress= 111.890400kg/cm^2 Total Stress = 142.836600kg/cm^2 <1.333 x11.67 MPa =15.56 MPa = 155.6 Kg/cm2 TENSILE STRESS DEVELOPED........................... Total stress = -1697.604000 kg/cm^2RUN TIME = 0 Min 0 SecTRIALS = 413

Min Ast @ 0.4 % = (0.4/100) x (/4) x 1002 = 31.415 cm2 Provide 22 – 32 (144.72 cm2) at top 11.0 m and 11 – 20 (34.54 cm2) at bottom and 8 ties @ 200 mm c/c and 16 stiffener bars @ 1500 mm c/c.

Deflection in pile

Max deflection in pile head (non seismic case) = = Q (L1 + Lf)3/(12 EI)= 32.63 x 1000 x (0 + 6633)/(12 x 2,95,804 x 4908738.521) = 0.546 cm = 5.46 mm < 10 mm (permissible)Max deflection in pile head (seismic case) = = Q (L1 + Lf)3/(12 EI)= 39.13 x 1000 x (0 + 663)3/(12 x 2,95,804 x 1.333 x 4908738.521) = 0.491 cm = 4.91 mm < 10 mm (permissible)

DESIGN OF PILE CAP

28

Load on individual Piles in non-seismic case 3000

1 5 3000

n = 8 2 6 ZAL =10.0

Traffic 3000 ZAC (outer)= 12.0 ZAC (inner) = 60.0 3 7 3000

4 8

Load on Pile marked (1) = 121.00 – 62.59 + 5.75 = 64.16 TLoad on Pile marked (2) = 121.00 – 62.59 + 1.92 = 60.33 TLoad on Pile marked (3) = 121.00 – 62.59 - 1.92 = 56.49 TLoad on pile marked (4) = 121.00 – 62.59 - 5.75 = 52.66 TLoad on pile marked (5) = 121.00 + 62.59 + 5.75 = 189.34 TLoad on pile marked (6) = 121.00 + 62.59 + 1.92 = 185.51 TLoad on pile marked (7) = 121.00 + 62.59 - 1.92 = 181.67 TLoad on pile marked (8) = 121.00 + 62.59 - 5.75 = 177.84 T

Total force on pile marked 5, 6, 7 & 8 = 189.34 + 185.51 + 181.67 + 177.84 = 734.36 T Wt. of pile cap = 10.3 x 1.5 x 2.5 T/m = 38.63 T/m

1 5 2 6 10300 3 7

4 8

650 2000

1500 1500 650 PLAN OF PILE CAP

29

1412.5

500 2000 1150

Critical section For shear(outside)

SECTION OF PILE CAP

Moment at face of abutment shaft, M = 734.36 x (1.5 – 1.0) – 38.63 x 1.152/2 = 67.18 – 25.55 = 341.63 T-m say 342 T-m Width of pile cap = 1030 cmd = √342 x 105/ (11.05 x 1030) = 54.82 cmProvide 1500 mm depth, d = 1500 – 50 – 25 – (20/2) = 1415 mmAst = (342 x 105)/ (2400 x 0.90 x 141.5) cm2 = 111.90 mm2

Min main reinf. = (0.2/100) x 141.5 x 1030 cm2 = 291.49 cm2

Min. distribution reinf = (0.12/100) x 141.5 x 100 = 16.98 mm2/mProvide 60-25 (294.60 cm2) along traffic bottom as main reinforcement and 16 @ 115c/c as distribution at bottom. Provide 16 @ 115 mm c/c both ways at top.

Shear

One way ShearMax. S.F. at a distance of d = 1412.5 mm away from face of abutment shaft whch is beyond the structure and hence one way shear is not critical.

Punching Shear or two way shear 650

d/2 = 706.25 500 650

Critical line of shear

Worst Punching shear will occur at corner pile to have least resisting periphery.

Max pile load in non seismic case = 190 T. Available length of resistance = 650 + 650 + 2 x x (500 + 706.25)/4 = 1300 + 1894.77 = 3194.77 mmEffective Depth = 1412.5 mmPunching Shear Stress = 190 x 1000/(319.477 x 141.25) = 4.21 Kg/ cm2

30

Allowable stress of shear in punching = 0.16 x √35 MPa = 0.946 MPa = 9.46 Kg/ cm2

Design of Abutment

(A) Vertical loadsTotal vertical load on abutment bottom section P = 571 TTotal moment on abutment bottom section M = 680 T-mLet us assume a section of 10000 x 2000 mm and M-25 concrete

Pu = 1.5 P = 857 T Mu = 1.5 M = 1020 T-mPu / fck bd = 857 x 104/(25 x 10000 x 2000) = 0.0171 say 0.017 Mu/ fck bd2 = 1020 x 107/(25 x 10000 x 20002) = 0.010 Min Ast = (0.3/100) x 10000 x 2000 = 60,000 mm2.Provide 192 – 20 equally spaced vertical bars 90 + 90 = 180 nos on long side and 6 + 6 = 12 nos on short sides

and 8 ties @ 200 c/c and links at every alternate third bar.

DESIGN OF ABUTMENT CAP

The abutment cap is 1000 x 10000 x 1000 thk.The bearing point is well within the abutment shaft and hence no eccentricity is induced and nominal reinforcement is to be provided.Ast (shorter span) on two faces = (1/ 100) x 225 x 10000 mm2 = 22500 mm2

Provide 56-16 on top and bottom. (22512 mm2)Ast (longer span) on two faces = (1/ 100) x 225 x 1000 mm2 = 2250 mm2

Provide 8-16 at top and bottom (3216 mm2) and 2-12 close bars.

Design of Dirt wallDepth of dirt wall = 1916 mm.The dirt wall is subjected to active earth pressure and live load surcharges.(a) Earth pressure

Ka = 0.20Horizontal Force due to active earth pressure = ½ x Ka. H2 = ½ x 0.20 x 1.8 x (1.916)2 T-m/m = 0.66 T/mMoment at bottom of wall = 0.66 x (0.42 x 1.916) = 0.532 say 0.54 T-m/m

(b) Live load surcharge

1916

0.432 T/m2 Total horizontal force due to LL surcharge = 0.432 x 1.916 T/m = 0.83 T/mMoment at bottom of dirt wall = 0.83 x (1.916/2) = 0.80 T-m/m

31

Design Moment

Total moment at base of wall,M = 0.54 + 0.80 = 1.34 T-m/mSelf wt. of dirt wall P = 0.4 x 1.916 x 1.0 x 2.5 T/m = 1.92 T/m Pu = 1.5 P = 2.88 TMu = 1.5 M = 2.01 T- m/nPu/ fck bd = 2.88 x 104/(25 x1000 x 400) = 0.0029Mu/ fck bd2 = 2.01 x 107/(25 x 1000 x 4002 ) = 0.005 So, min Reinforcement will sufficeAst = 0.3 x 400 x 1000 = 1200 mm2/m.Provide 12 @ 150 mm c/c both face vertical steel (1506 mm2) and 10 @ 200 mm c/c both face as horizontal steel

Load Bearing Capacity of Pile 32

[Ref: Clause- 9 of Appendix-5 of IRC:78-2000 and Appendix – B, IS: 2911 (Part – I/Sec – 2)]

Data : - Dia of pile –1000 mm.

Cut-off Level of Pile 2.0 m from EGL

Termination Level 20.5 m from EGL

The pile is resting on solid rock layer and is socketed for a length of 0.5 m within rock the layer.

From bottom of pile cap to next 15.0 m is slag and then there is a clay layer of 3.0 m thick after which rock

layer is encountered.

Total ultimate capacity, Qu = Qub+Qus

Existing/Finished G.L.

2.0 m

I 17.0 m

15.0 m

Slag

3.0 m C= 4.5 T/m2 II 3.0m

(Clay) =0.4

0.5 m Rock Layer III 0.50 m

Qub = Ultimate end bearing capacity,

= Ksp x qc x df x Ab (for Rock layer)

[ Ksp = An empirical Coefficient ranges from 0.1 to 0.4; Let us take it as 0.15;

qc = Average uniaxial compressive strength of rock

= 500 to 1000 kg/cm2 (As per revised soil report done by CMERI); Let us take 500 kg/cm2

df = depth factor = 1 + 0.4 x (Length of socket / Dia of socket) = 1 + 0.4 x (0.5/1.0) = 1.20

Ab = Area of Cross Section = (/4) x 1002 =7853.98 cm2 ]

= 0.15 x 500 x 1.2 x 7853.98 = 706858 Kg = 706.85 T

Qus = Ultimate skin friction capacity, (for soil layer as per IS: 2911)

= åK.Pdi tan. Asi (for cohesion less soil) + c As (for cohesive soil)

33

Qus = Ultimate skin friction capacity, (for rock layer as per IRC: 78)

= As x qs

In our present case neglecting the skin resistance of slag layer, we take skin resistance of 3.0 m clay

Layer and 0.5 m rock socket.

For Cohesive (Clay) Layer

layer-I , Qus = 0.4 x 4.5 x ( x 1.0 x 3.0) = 16.96 T

For Rock Socket Layer

For layer II, Qus =( x 1.0 x 0.5) x 200 = 314.15 T [ qs = shear along socket = 50 kg/cm2 for

Normal rock and 20 kg/cm2 for weather rock;

Let us take qs = 20 kg/cm2 = 200 T/m2 ]

So, Total Skin friction capacity,

Qus = 16.96 + 314.15 = 331.11 T

So, Ultimate capacity, Qu = 706.85 + 331.11 = 1037.96 T

Allowing a factor of safety of 3.0

Max. Compression capacity Qsafe = 1037.96/3.0 T = 345.98 T say 345 T(However the safe load bearing capacity of pile is to be ascertained by vertical load test as per the soil report)

34

revised design

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