Abutment 6w44y

Generated by Foxit PDF Creator © Foxit Software http://www.foxitsoftware.com Design For evaluation Detailed Engineering Design of Chargahawa River Bridge Calculation only. VOLUMN-III

B.

Design of Substructure

1.0 Design Data: 1.1. Materials and its Properties: Concrete: Reinforcement:

M20 Fe415

Basic Permissible Stresses of Concrete as per IRC : 21-2000: Permissible direct compressive stress, σco = Permissible flexural compressive stress, σc = Maximum Permissible shear stress, τmax

5.0 Mpa 6.67 Mpa

=

1.8 N/mm2

Basic Permissible Stresses of Reinforcing Bars as per IRC : 21-2000: Permissible Flexural Tensile stress, σst = 200 N/mm2 Permissible direct compressive stress, σco

=

170 N/mm2

Design of Data: Modular Ratio = Neutral axis depth factor, n = (mσc)/( mσc+σst) = Lever arm factor, j = (1-n/3) = Moment of resistance coefficient, R = ½ x n x j x σc = Unit weight of materials as per IRC : 6-2000: Concrete (cement-Reinforced) = Macadam (binder premix) = Water = Backfill = 1.2. Geometrical Properties: Effective Span of Bridge Deck = Total length of each Span = Angle of internal friction of backfill Approach slab Length, L Width, B Depth, D Size of bearing Length Width Thickness High Flood Level, HFL Mean scour depth from HFL Lowest Bed Level Bottom level of abutment Clearance above HFL Top level of abutment Depth of superstructure Total height of abutment h1 = 4.59 m h2 = 0.26 m

Unique Engineering Consultancy (P) Ltd

10.98 0.268 0.911 0.814

2.4

t/m3

2.2

t/m4

1.0

t/m5

1.92

t/m6

18.0 m 18.60 m

=

29 degree

=

3.5 m

=

7m

=

0.26 m

=

400 mm

=

250 mm

=

50 mm

=

1002.15 m

=

3.28 m

=

997 m

=

999.600 m

=

0.9 m

=

1003.00 m

=

1.4 m

=

Fig.20

4.85 m

b1 = b2 =

1m 0m

Page 1


h3 = h4 = h5 = h6 =

b3 = b4 = b5 = b6 = b7 =

1.45 m 0.4 m 3.00 m 0.05 m

Cap overhang, Min. length of cap = Length of abutment shaft =

0.25 m 0.55 m 0.9 m 0.35 m 0.15 m

m 6.9 m 7.2

1.3. Calculation of weight and C.G. of the abutment stem Elements Area, Ai (m2) P1 3.00

Weight (t)

Xi (m)

Yi (m)

AiXi

AiYi

49.68

0.5

1.50

1.50

4.50

P2

0.00

0.00

1.00

1.00

0.00

0.00

P3 P4

0.46 0.3625 3.82

7.95 6.00 63.63

0.58 0.125

3.20 4.12

0.26 0.05

1.47 1.50 7.47

C.G. from back of abutment, x = e= Moment about back face =

1.81 0.473 from bottom of abutment, y = 0.125 m 30.13 t-m

1.954

1.4. Calculation of Loads and Moments Due to Dead Load = 81.08 t Dead load from superstructure, = 4.05 t Weight of bearings, expansion t etc. Total dead load from superstructure on each abutment, PDL = 85.14 t = 0.6 m Distance of bearing center from back of abutment Eccentricity of DL & LL from superstructure acting through bearing, e=

0.125 m

Moment due to DL of superstructure about CG of abutment =

10.64 t-m

Due to Live Load Live load from superstructure: Due to IRC Class A wheel load, in longitudinal direction, 11.4 t

11.4 t

6.8 t

6.8 t

6.8 t

6.8 t

Fig. 21 Max. LL on abutment from right side impact factor Max LL including Impact, PLL

= (11.4*18+11.4*16.8+6.8*12.5+6.8*9.5+6.8*6.5+6.8*3.5)/18 = 68.26 t = 1.188 = 81.06 t

Moment due to eccentric load

=

10.13 t-m

In transverse direction, eccentricity of load Moment due to live load

= =

0.25 m 20.26 t-m


Page 2


Due to IRC Class AA Tracked Vehicle, in longitudinal direction, 70 t

Fig. 22 Max. LL on abutment from right side impact factor Max LL including Impact, PLL

= 70x(20-1.8)/20 = = =

63.70 t 1.10 70.07 t


=

8.76 t-m


= =

0.35 m 24.52 t-m

Due to IRC Class AA Wheeled Vehicle, in longitudinal direction, 20 t

20 t

Fig. 23

Max. LL on abutment from right side impact factor Max LL including Impact, PLL

= 20+20x14.8/20 = = =

34.80 t 1.188 41.33 t


=

5.17 t-m


= =

0.35 m 14.46 t-m

Due to Earth Pressure = 0.5γs×h1

×tan2(45o-φ/2)×t

2

Horizontal force due to earth pressure

= Which acts at a distance from abutment base = 0.42xh1 = Magnitude of surcharge, q = 1.2xγs = 2

48.42 t 1.93 m 2.304 t/m2

Horizontal force due to surcharge = qxh1xtan (45-φ/2)xt =

25.32 t/m2

Which acts at a distance from abutment base = h1/2

2.295 m

=

Weight of backfill behind the abutment resting on foundation = Moment due to earth pressure about abutment base

=

0.0 t 151.45 t-m

Due to Temperature Variation Maximum temperature variation, T = 25 Coefficient of thermal expansion, α = 1.17E-05 Max. Elongation , δ = 4.68 Shear Mdulus of bearing material, G = 1.00 Depth of bearing = 50.00 Longitudinal force / bearing = 1.053 The total resistance offered by bearing = = 3.159 Unbalanced force at the bearing = This force acts at the bearing level, i.e. at a distance from pier base = Moment due to temperature variation = 10.90

o

C /m/oC mm Mpa mm t t 3.16 t 3.45 m t-m

Due to Braking Effect Effect due to class A loading = 0.2x70


=

14 t

Page 3


This acts at distance above deck level

=

1.2 m

And at a distance from abutment base

=

6.05 m

Moment due to braking about abutment base

=

84.70 t-m

Due to Seismic force Description Total Load Seismic Lever arm (m) (t) Load (t) Superstructure DL 85.14 8.51 4.15 Abutment shaft P1 49.68 4.97 1.50 Abutment shaft P2 0.00 0.00 1.00 Abutment shaft P3 7.95 0.79 3.00 Abutment shaft P4 6.00 0.60 4.12 Approach slab 15.288 1.53 4.72 Backfill 0 0.00 2.29 Total 164.06 16.41

Moment (t-m) 35.33 7.45 0.00 2.38 2.48 7.22 0.00 54.86

Summary of Loads and Moments The summary of loads and moments are tabulated below. The transverse forces and moments are not considered, since it would not be critical due to high moment of inertia of abutment. Therefore stresses are checked in longitudinal direction only. Description Superstructure DL LL including Impact Self wt. of abutment Braking effect Temperature Earth Pressure Sub Total Seismic Force Total

Vertical load (t) 85.14 81.06 63.63

Horizontal force (t)

Moment (t-m) 10.64 10.13 0.00 84.70 10.90 151.45 267.83 54.86 322.69

14 3.16 73.74 90.90 16.41 107.31

229.82 229.82

As per standard design practice, design of abutment would be carried out for case A, and checked for case B, as given below. Case Vertical load (t) Horizontal force (t) Moment (t-m)

A: (N+T) 229.82 90.90 267.83

B: (N+T+S) 229.82 107.31 322.69

Design of abutment stem section Abutment section will be designed for case A and the section adequacy will be checked for case B. Design vertical load = 229.82 t Design moment = 267.83 t-m Depth of section required =

M Rb

Clear cover to reinforcement = Effective depth provided =

=

691 mm

75

mm

910

mm

σ Unique Engineering Consultancy (P) Ltd

st

<

910

O.K.

M = × j× d Page 4


Area of steel required for tension, Ast = 38 Provide mm2 and spacing

nos. of

16 Provide area of steel, Ast =

mm φ bars @

σ

25

182

st

M = × j× d

16160 mm2

mm φ bars giving area of steel =

18653

mm c/c. 150

mm c/c at the front side of the abutment, giving

9651 mm2. Provided nos of bars =

90.90 t Design horizontal force = Corresponding shear stress, τv = 0.14 N/mm2 0.30 % Percentage area of tension steel, 100Ast/bd = Shear strength increment factor = 1+5P/(Agxfck) = 1.083 < 1.5 > τv Design shear strength of concrete, τc = 0.26 O.K.

48

O.K.

Check for shear at the bottom of the abutment cap. 17.16 t Total horizontal force at the bottom of the cap = = 1000 mm Depth of abutment = 893 mm Effective depth, d < t Corresponding shear stress, τv = 0.03 O.K. c Tension reinforcement required at different level along the height of abutments. At

1.93 m above base of abutment Load Type Lever arm (m) Moment Load (t) DL of Superstructure 85.14 0.125 10.64 Live load (LL) 81.06 0.125 10.13 Breaking load 14 4.12 57.68 Temperature load 3.16 1.52 4.80 Earth Pressure 16.26 1.12 18.17 Surcharge 14.67 1.33 19.51 Total 120.94 Overall depth of abutment = 1000 mm Effective depth of abutment = 893 mm Area of tension reinforcement req. = 7436 mm2 Provide 23 numbers 25 mm φ bars and 0 numbers 11290 mm2 20 mm φ bars, giving area of steel = Actual point of curtailment = 3.08 m from base of abutment. NO CURTAILMENT IS DONE

Rear side Reinforcement Provide 16 mm φ bars @ 150 mm c/c at rear side of the abutment. Check for stresses for case B Bending Moment = 322.69 t-m Stress in steel = 208.75 N/mm2 < 300 O.K. Horizontal force = 107.31 t 0.17 N/mm2 < 0.45 Corresponding shear stress =

O.K.

Design of abutment cap: Vertical reaction due to


Page 5


Dead load, PDL/3

=

28.38 t

Live load including impact, PLL/3 = 27.02 t = 55.40 t Total vertical reaction Providing minimum area of steel @ 1% assuming cap thickness 225 mm, area of steel required /m of cap = 2250 mm2 (for both top and bottom layer). Provide 16 mm φ bars @ 150 mm c/c, at both top and bottom layers in both ways, giving area of steel, Ast = 2680 mm2. As per IRC:78-2000, clause 710.10.4 the two layers of mesh reinforcement, one at 20 mm from top and the other at 100 mm from top of abutment cap each consisting of 8 mm φ bars @ 100 mm c/c in both directions, shall be provided directly under the bearings. Check for bearing stresses The allowable bearing pressure with nearly uniform distribution on the loaded area of a footing or base under a bearing or column shall be given by following equation, C=

C

0

×

A1 A2

C0 = 6.25 MPa, permissible direct compressive stress in concrete 0.52 m2, dispersed largest concentric area similar to A2 A1 = 0.1 m2, loaded area A2 = Therefore, Α1/Α2 = 5.2 >2 8.84 Mpa And, C = Actual compressive stress =

5.54

< C

O.K.

Design of Backwall: 2 2 o Horizontal force due to earth pressure = 0.5γs×(h3- h2) ×tan (45 -φ/2) = 0.4998 m Which acts at a distance from backwall base = Magnitude of surcharge, q = 2.304 t/m2 Horizontal force due to surcharge = q×(h3-h2)×tan2(45o-φ/2) = Which acts at a distance from backwall base = 0.595 m Self weight of backwall /m width = 0.714 t/m Live load on back wall considering class AA loading = 10 t This acts at a distance from backwall toe = 0.125 m

0.47 t/m

0.95 t/m

Moment due to earth pressure about backwall base= 0.80 t-m/m Moment due to self weight and LL = 1.34 t-m/m Total Moment = 2.14 t-m/m Effective depth of backwall = 194 mm Area of steel required, Ast = 606 mm2 Provide 12 mm φ bars @ 150 mm c/c, giving area of steel, Ast = 753 mm2. And provide 10 mm φ bars @ 200 mm c/c as distribution bar.

Design of Return wall: Return wall will be monolithic with the backwall. They are ed together through 30 cm x 30 cm fillets. The load acting on the wing walls would be earth pressure and is designed to withstand a live load equivalent to surcharge of 1.2 m height of earth fill according to IRC:78-2000.


Page 6


h1 = h2 = h3 = h4 = b1 =

1.65 m 1.2 m 1.30 m 0.4 m

2.25 m Thickness of return wall assumed =

250 mm

Average value of earth pressure = 0.5γs×(h1/2+h2/2+1.2) = 1.84 t Acting at a distance from wing wall bottom Moment due to earth pressure Depth required Effective depth provided Area of steel required

=

×tan (45 -φ/2)

2

2

o

= = =

0.7125 m 3.68 t-m/m 178 mm 194 mm

=

1369 mm2

<

Provided O.K.

Provide 16 mm f bars @ 150 mm c/c, giving area of steel, Ast = 1910 mm φ bars @ 150 mm2. And provide 12 mm c/c as distribution bar in vertical direction. In the fillet at the t provide nominal reinforcement of 10 mm φ bars @ 200 mm c/c. The wing wall would be properly anchored to the abutment, backwall and cap. Design of Approach Slab: The approach slab is resting over the abutment and the other end ed by the soil underneath. It should be designed on the basis of elastic base theory, which is complicated. Hence the dimensions and reinforcement is provided as per the standard design practice. Self weight /m width 0.624 = t/m Weight of pavement 0.176 = t/m Maximum live load 11.44 = t/m2 Total udl t/m = 12.24 Assuming maximum cavitation of 1.2m behind the abutment, Maximum bending moment = 22.03 kN-m/m Effective depth of slab required = 164 mm Effective depth provided = 212 mm Area of steel required = 751 mm2 mm φ bars @ Provide 16 150 mm c/c both top and bottom, in both direction Provided area of steel, Ast = 1340 mm2. Maximum shear force = 73.4 kN/m Shear stress 0.35 = N/mm2 Percentage area of steel = 0.63 % Permissible shear stress = N/mm2 > actual shear stress O.K. 0.365 Design of RCC Stoppers: Base width of sttoper, b1 = Top width of sttoper, b2 = Height above abutment cap, h1 = Length of sttoper, l =

0.4 0.3 0.45 0.8

m m m m

Horizontal force acting on the stopper above pier cap @ 10% of vertical load= 7.94 t 11.92 Friction resistance = t However as additional factor of safety, assuming an extra lateral force in excess of frictional resistance taking αR = 0.20 in worst case, Fs = 15.89 t 3.97 t Difference =


Page 7


Moment at the base of stopper = 1.79 t-m 166 mm Effective depth required = 342 mm Effective depth provided = 287 mm2 Area of steel required = Provide 12 mm f bars @ 150 mm c/c. Area of steel provided = 716 mm2 = N/mm2 Max. Shear stress, tv 0.15 % Percentage area of steel provided, p = 0.26 N/mm2 > tv Permissible minimum shear strength of concrete =0.226 Provide nominal shear reinforcement of 8 mm φ bars @ 100 mm c/c. Design of abutment footing (Piles) 4.2 Geometrical Properties: High Flood Level, HFL = Maximum depth of scour from HFL = Elevation of maximum scour level (MSL) = Max. Water depth in the river = Water depth near the abutment = Proportionate scour depth near the abutment Elevation of MSL near abutment = Lowest Bed Level = Elevation of top of foundation base = Bottom level of foundation = h1 h2 h3 h4 h5

4

= = = = =

4.59 0.26 1.45 0.4 3.00

=

O.K.

1002.2 m

=

3.28 m

=

998.87 m

=

3.28 m

=

3.280 m

=

3.280 m

=

998.87 m

=

997 m

=

999.600 m

=

998.500 m

m m m m m

CL of Bearing

3

HFL

1

MSL 2

LBL

h6 = HFL above MSL, h7 = HFL above min. bed level h8 =

0.05 m 3.28 m 5.15 m

MSL below min. bed level, h9 = Top of foundation base below MSL, h10 = Depth of pile cap, h11 =

-1.87 m -0.73 m 1.1 m


Page 8

Generated by Foxit PDF Creator © Foxit Software http://www.foxitsoftware.com Design For evaluation Detailed Engineering Design of Chargahawa River Bridge Calculation only. VOLUMN-III b1 = b3 =

1.0 m 0.25 m

b4 =

0.55 m

b5 =

0.9 m

b6 =

0.35 m

b7 b8 b9 b10 b11 b12

= = = = = =

Foundation base length, b13 =

0.15 0.4 3.5 3.1 1 3.1

m m m m m m

7.2 m

Abutment base length Foundation projection beyond abutment stem Total width of foundation Angle of internal friction of soil Safe bearing capacity of soil Load bearing capacity of pile = Numbers of piles along longitudinal dirrection = Numbers of piles along transverse dirrection = Total numbers of piles = Diameter of piles = Length of piles = Soil Type : Clayey Silt

Determination of depth of fixity of piles: Unconfined compressive strength, S = Modulus of subgrade reaction, K = Lf/d Therefore, and, depth of fixity, Lf

6.9 0.15 7.2 29 0

m m m

degree t/m2

40 t 4 4

m m

16 700 mm 16 m

0.46 kg/cm2 13.2 kg/cm2 =

7

=

4.9

from figure 1 and 2. m

Calculation of vertical load on foundation base Due to Dead Load Superstructure 85.14 t = Backwall 6.00 t = Abutment cap 7.95 t = Abutment shaft 49.68 t = Soil behind abutment 196.70 t = Approach slab 15.29 t = Soil in front of abutment = 0.00 t Total = 360.76 t Due to Live Load Maximum live load including impact = 81.06 t Due to self weight of foundation base =

Loss in weight due to Buoyancy Depth of abutment shaft below HFL Abutment width at HFL Loss in weight of abutment shaft Loss in weight of soil behind abutment Loss in weight of soil in front of abutment


136.86 t

=

2.55 m

=

1.000 m

=

17.59 t

=

56.92 t

=

0.00 t

Page 9


Total buoyant force on abutment

=

78.06 t

Buoyant force on foundation base

=

57.02 t

=

166.19 t

Calculation of Moments at Foundation Base Level Due to superstructure DL+LL Total superstructure DL+LL Eccentricity of load about pile cap transverse axis Corresponding moment Superstructure DL Moment at foundation toe due to DL

= = = =

0m 0.00 t-m 85.14 t 0.00 t-m

Due to Abutment + Soil filling DL A. Maximum Load Condition i) Due to abutment body = 63.63 t Load = -0.13 m Lever arm = -7.95 t-m Moment ii) Due to Backfill and approach slab = 196.70 t Load = -2.1 m Lever arm = -413.07 t-m Moment iii) Due to soil fill in front of abutment = 0.00 t Load = 2m Lever arm = 0.00 t-m Moment Total moment due to abutment + soil filling DL =

-421.03

t-m

B. No LL + Buoyancy Condition i) Due to abutment body = 85.14 t Load = -0.13 m Lever arm = -10.64 t-m Moment ii) Due to Backfill and approach slab = 133.93 t Load = -2.1 m Lever arm Moment = -281.25 t-m iii) Due to soil fill in front of abutment = 0.00 Load = 1.55 Lever arm = 0.00 Moment Total moment due to abutment + soil filling DL =

-291.89

t-m

Due to Earth Pressure Earth pressure Lever arm Surcharge Lever arm Total Moment


= = =

48.42 t 3.03 m 25.32 t

=

3.395 m

=

232.57 t-m

Page 10


Due to Longitudinal forces: a) Due to tractive effort or braking Which acts at a distance above foundation bottom Moment due to tractive effort or braking about foundation base

b) Force due to resistance in bearings due to temperature Which acts at a distance above foundation bottom Moment due to tractive effort or braking about foundation base

= = = = = =

14 6.05 84.70 3.16 3.4 10.90

Due to Seismic Forces: A. Along longitudinal direction for maximum load condition Load (t) Description Seismic Load (t) Lever arm 85.14 8.51 4.15 Superstructure DL Abutment shaft P1 49.68 4.97 1.50 Abutment shaft P2 0.00 0.00 1.00 Abutment shaft P3 7.95 0.79 3.00 Abutment shaft P4 6.00 0.60 4.12 Approach slab 15.29 1.53 4.72 Backfill 196.70 19.67 2.1 Total 360.76 36.08 B. Case of buoyancy and no live load

t m t-m t m t-m

Moment (t-m) 35.33 7.45 0.00 2.38 2.48 7.22 41.31 96.17

The seismic load will be lesser than that for N+T case due to submerged part of the sub structure which will be under buoyancy. Description

Superstructure DL Abutment shaft P1 above HFL Abutment shaft P1 below HFL Abutment shaft P2 above HFL Abutment shaft P2 below HFL Abutment Cap P3 Abutment backwall P4 Approach slab Backfill above HFL Backfill below HFL Total

Load (t) 85.14 7.45 24.63 3.73 0.00 7.95

6.26 15.29 109.28 79.68 339.41

Seismic Load (t)

8.51 0.75 2.46 0.37 0.00 0.79 0.63 1.53 10.93 7.97 33.94

Lever arm (m)

4.15 3.875 2.375 3.80 #DIV/0! 3 4.125 4.72 4.67 2.375

Moment (t-m) 35.33 2.89 5.85 1.42 #DIV/0! 2.38 2.58 7.22 51.03 18.92 #DIV/0!

Summary of Moments and Forces Vertical Loads Case I Case II Case III Max. Buoyancy+No LL Case I + Seismicity Load 360.76 DL from superstructure + abutment 282.70 360.76 136.86 DL of foundation base 79.83 136.86 81.06 Live load 81.06 578.67 Total 362.53 578.67 Horizontal Forces Description

Description

Earth Pressure Tractive effort or braking


Case I Case II Case III Max. Buoyancy+No LL Case I + Seismicity Load 73.74 73.74 73.74 14 14

Page 11


3.16 90.90

Temperature effect Seismic Force Total

3.16

3.16 40.41 131.31

76.90

Resulting Moments Description Superstructure DL+LL Abutment +Soil filling DL Earth pressure Tractive effort or braking Temperature effect Seismic force Total

Case I Max. -291.89 -421.03 232.57 84.70 10.90 -384.75

Case II Buoyancy+No LL 0.00 -291.89 232.57 -59.32

Case III Case I + Seismicity -291.89 -421.03 232.57 84.70 10.90 171.28 -213.47

Summary of loads Cases

Unit T T t-m

Vertical Horizontal Moment

Case I Case II Max. Load Buoyancy+No LL 80.37 50.35 12.62 10.68 -53.44 -8.24

Case III Case I + Seismicity 80.37 18.24 -29.65

Case: I Deducting the bearing capacity of pile cap as a open footing, vertical loads to be taken by piles = 80.37 t Loads on each pile due to vertical load

= =

5.02 t 12.62 t

Axial load on extreme pile due to moment Axial load on second row of pile due to moment = = Resultant compressive force on extreme pile Resultant tensile force on extreme pile = Resultant compressive force on second row of pile= Coefficient of friction between concrete and clay = = Friction resistance of cap Horizontal force on each pile = Moment on each piles =

4.02 t 21.67 t

<

40.00

t O.K.

0.00 t

<

40.00

t O.K.

17.34 t 0.20 0.00 t 11.17 t 54.72 t-m

Design of piles: Length of pile, L = 16 m Diameter of pile, D = 0.7 m L/D = 22.86 > 12 Hence the pile is designed as long column. Least radious of gyration, r = 17.5 cm Reduction coefficient for reduction of stress values is given as, cr

= (1.25-L/48B) =

Safe stress are, Permissible direct compressive stress, σco = Permissible flexural compressive stress, σco = Permissible Flexural Tensile stress, σst = Maximum compressive stress in concrete =


0.774

3.87 5.16 154.76 1.50

N/mm2 N/mm2 Mpa

(in steel)

N/mm2 <

3.869

N/mm2

Page 12


Provide minimum area of compression steel @ Asc

=

4811

15 nos. of Provide Check for combined stress: Percentage area of steel provided, pt p/fck d'/D

25

1.25

=

2.400

=

0.120

=

0.091 0.088

Pu/fckD2

=

Mu/fckD3

=

Mu

=

82.32 t-m

M = Working moment, Shear stress due to horizontal force = Permissible shear strength of concrete =

54.88 t-m

From chart, and

>

54.72

t-m O.K.

0.29 N/mm2 0.45 N/mm2 O.K.

=

27.49 t

=

65.07 t <

90.00 t O.K.

0.00 t <

90.00 t O.K.

= 6.18 t 30.27 t-m

Pu/fckD2

=

0.100

Mu/fckD3

=

0.106

Mu

=

72.72 t-m

M = Working moment, Shear stress due to horizontal force = Permissible shear strength of concrete =

48.48 t-m

and

7363

0.12

Case: C Axial load on extreme pile due to moment Resultant compressive force Resultant tensile force Horizontal force on each piles = Moment on each piles = From chart,

%.

mm2 mm φ bars, giving area of steel =

>

30.27

t-m O.K.

0.16 N/mm2 0.56 N/mm2

O.K.

Lateral Reinforcement: Lateral reinforcement should be 0.2% of the gross volume. Using dia. 10 mm ties, 145577 mm3 we have the volume of one tie = If p be the pitch of the ties in mm, then Volume of pile per pitch length =

384845 p

Hence equating, we get, p

Hence provide dia

10

= mm ties at

189 mm 160

mm c/c in the main body of the pile.

Design of Pile Cap: Maximum bending moment in the pile cap due to vertical load, At center of cap = 312.91 t-m 375.77 t-m At face of abutment = = 233.3 t-m Bending moment due to moment on steam 609.0 t-m Maximum bending moment = The effective depth required is given by, dreq =

999

mm

Adopt 1030 mm effective depth and overall depth = 1100 mm 32466 mm2 Area of steel required along width of cap = Using 25 mm dia. bars, spacing would be 113 mm Provede 25 mm dia bars @ 120 mm c/c, provided area = 33469 mm2 at both top and bottom. Distribution reinforcement @ 0.12% = 9900 mm2 20 mm dia bars @ 150 mm c/c, giving area of stee = 15708 mm2. Provide


Page 13


413.23 t Shear force = Shear stress, tv = 0.53 N/mm2 Percentage area of steel, pt = Permissible shear strength of concrete, tc = Shear reinforement is required. Assuming 10 mm dia 20 And spacing of stirrups, S = 163

mm

Maximum spacing of stirrups, Provide 10 mm dia

189 mm legged vertical tirrups @

= 4

0.43

%

0.28

N/mm2

<

tv

legged vertical tirrups, then Asv =

160

1571 mm2

mm c/c.

Check for punching shear along pier, 269.49 t Shear force = 20.32 m bo = tv =

0.13 N/mm2

Punching shear strength, tc = 0.716 Check for punching shear along pile, 57.69 t Shear force = bo = 5435.0 m tv =

N/mm2

>

tv

Safe.

N/mm2

>

tv

Safe.

0.10 N/mm2

Punching shear strength, tc =


0.72

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