pier

Detailed Engineering Design of Chargahawa River Bridge Design Calculation VOLUMN-III

4.0 Design of Pier4.1 Design Data:

Concrete: M20

Reinforcement: Fe415

Basic Permissible Stresses of Concrete as per IRC : 21-2000:

Permissible direct compressive stress, σco = 5.0 N/mm2

Permissible flexural compressive stress, σc = 6.67 N/mm2

Maximum Permissible shear stress, τmax = 1.8 N/mm2

Basic Permissible tensile stress, σt = 0.53 N/mm2

Basic Permissible Stresses of Reinforcing Bars as per IRC : 21-2000:

Permissible Flexural Tensile stress, σst = 200 Mpa

Permissible direct compressive stress, σco = 170 MPa

20

Design of Data:

Modular Ratio, m = 10

Neutral axis depth factor, n = (mσc)/( mσc+σst) = 0.250

Lever arm factor, j = (1-n/3) = 0.917

Moment of resistance coefficient, R= ½ x n x j x σc = 0.765

Unit weight of materials as per IRC : 6-2000:

Concrete (cement-Reinforced) = 2.4 t/m3

Macadam (binder premix) = 2.2 t/m3

Water = 1.0 t/m3

Backfill = 1.6 t/m3

4.2 Geometrical Properties:center to center Span of bearing 18.00 mTotal span length 18.6 mHFL = 1002.15 mLBL = 997.00 mMax. scor depth below HFL = 3.28 mDepth of Superstructure = 1.4 mDepth of Bearing & pad = 0.05 mAve. Velocity of water current = 2.2 m/secPier shape, semi-circular, = 0.66 mRL of max scour lev. = 998.87 mFree board = 0.9 mRL of Pier Cap required = 1003.05 mRL of Pier Cap provided = 1003.05 mRL of foundation top = 997.50 mHt. of HFL from base of pier = 4.65 mHt. upto deck from GL = 7.50 mSize of bearing l = 400 b = 250 mmExpansion joit = 40 mmC/C distance between girder = 2.4 mDiameter of pier, b1 = 1.2 mNumber of pier, n = 3Width of pier cap, b3 = 1.6 m

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Length of pier cap,b4 = 6.6 mb5 = 0.64 mb6 = 0.9 mh1 = 4.80 m

Pier cap Thickness,h2 = 0.75 mh3 = 0.05 m

Calculations of loads and Moments

Due to dead LoadDead load from superstructure = 162.16 tSelf wt. of bearing & expansion joint = 9.01 tTotal DL from superstructure = 171.17 tSelf wt. of pierwt. of pier shaft = 39.09 twt.of pier cap = 19.01 tTotal DL at pier base = 229.27 t

Due to LL Load - unequal loadingClass A wheel loading in longitudinal directionFor maximum load on pier, the arrangement of IRC class A loading will be as shown below:

Max. LL on pier from right side = 2(11.4*20+11.4*18.8+6.8*14.5+6.8*11.5+6.8*9.5+6.8*5.5)/20= 69.01 t

impact factor = 1.188Max LL including Impact = 81.95 tLL / m length of pier = 68.29LL from left side = 2(2.7*17.44+2.7*16.34)/18

= 10.13LL including Impact = 12.03 tLL / m length of pier = 10.03 t

Moment due to unequal loading of LL 22.37 t-mIn transverse directionMax. LL including impact = 93.99 tEccentricity = 0.25 mMoment due to LL = 23.50 t-m

Class AA Track loading in longitudinal direction

Max. LL on pier = 62.13 timpact factor = 1.100Max LL including Impact = 68.34 tLL / m length of pier = 62.13

Moment due to unequal loading of LL 21.87 t-mIn transverse directionEccentricity = 0.35 mMoment due to LL = 23.92 t-m

Due to longitudinal Forces

11.4 11.4 6.8 6.8 6.8 6.82.72.7

PLAN




Due to tractive effort or braking = 14 tThis act at = 1.2 m above deck levelor at a distance from base of pier = 6.75 mMoment due to longitudinal force = 94.5 t-m

Due to resistance in bearing due to temperature

Maximum temperature variation, T = 25oC

Coefficient of thermal expansion, α = 1.17E-05 /m/oCMax. Elongation , δ = 4.68 mmShear Mdulus of bearing material, G = 1.00 MpaDepth of bearing = 50.00 mmLongitudinal force / bearing = 0.936 tThe total resistance offered by bearing = = 2.808 tUnbalanced force at the bearing = 2.81 tThis force acts at the bearing level, i.e. at a distance from pier base = 7.40 m

Total longitudinal force due to Breaking & temperature 16.81 tMoment due to longitudinal force = 158.68 t-mMoment due to temperature on DL only = 20.78 t-m

Due to water currentFor pier parrallel to direction of water current, the intensity of pressure is given by

Ic = 52KV2where, K = 0.66 a constant depending on the geometry of pier

V = 2.2 velocity of current in m/secIc = 166.11 kg/m2

Height of HFL from base of pier = 4.65 mWidth of pier at HFL = 1.1 ma.) water current force parallel to the pier = 1.19 t

And corresponding moment = 3.86 t-mb.) Water current varying at 20 degree

Intensity parallel to the pier = 156.09 kg/m2Intensity perpendicular to the pier = 56.81 kg/m2Force parallel to the pier 1.12 tForce perpendicular to the pier 0.41 tMoment parallel to the pier 3.63 t-mMoment perpendicular to the pier 1.32 t-m

Due to wind forceThe area of superstructure in elevation providing 25% for railing = 31.13 m2a. height upto deck level from ground level = 7.20 mWind pressure at that height (refer. IRC:6-2000), P = 86 kg/m2Wind force on superstructure = 2.68 tWind force against moving load 20.4 m long corresponding to IRC class A @ 300 kg/m

= 6.12 tTotal wind force = 8.80 tb. Minimum force on deck at 450 kg/m 7.47 tc. Minimum force with wind pressure of 225 kg/linear m in the plane of unloaded structure

7.00 tWind force to be considered = maximum of (a,b,c) = 8.89 tMoment parallel to pier due to wind force acting at 1.5m above deck = 64.02 t-mDue to BuoyancyForce due to buoyancy on the pier shaft = 15.78 tDue to seismic forcesA. Along longitudinal direction

Description162.16 6.2519.01 5.17539.09 2.4

Total Load (t) Seismic Load(t) Lever arm (m) Moment (t-m)Superstructure DL 16.22 101.35Pier cap 1.90 9.84Pier Shaft 3.91 9.38




220.26

B. Along Transverse directionMoment due to dead load is taken to be same as that in longitudinal direction. In addition, seismicforces and moments on LL including impact must be considered for transverse condition.

Total LL on pier = 93.99 tSeismic force due to LL = 9.40 tLever arm = 8.15 m

Moment due to seismic force on LL = 76.60 t-m

Total Seismic force = 31.42 tTotal Moment due to seismic force = 197.2 t-m

Summary of Loads and Moments

Long. Tran. Long.220.2693.99 23.92 21.87

16.81 158.68

1.12 0.41 3.63 1.328.89 64.02

15.7831.42 22.03 197.17 120.6

The forces and moments due to seismicity are grater than those due to wind forces. As per standard design practce, seismic forces and moments will be adopted neglecting the effect due to wind forces.

Design of pier shaft sectionThe pier section will be designed for the Case A and the section adequacy will be checked for both thecases. As the moment of inertia of the pier along Y-Y axis is greater than along X-X axis, the designneeds to be done for stresses along the X-X axis only.

Design vertical load = 104.75 tDesign Moment = 60.62 t-mEccentricity = 0.58 m

Diameter of the pier shaft, D = 1.2 mEffective diameter, d = 1109 mm

Check for position of eccentricity D/4<e<1.5D0.28 < 0.58 < 1.80 O.K.

Length of pier, L = 5.55 mEffective length of column, le =1.2xL = 6.66 mRadius of gyration, r = 0.3 mRatio of effective length to radius of gyration = 22.2 > 12

< 50 Short columnMinimum area of steel = 0.8 %Maximum area of steel = 8 %Assume percentage area of steel = 1 %Modular ratio. α = 10Section modulus, W = 0.131 m3Gross area of concrete, Ag = 0.950 m2Net area of concrete, Ac = 0.941 m2

degree skiew

Total 22.03 120.57

Description Vertical LoadHorizontal Force (t) Moment (t-m)

Tran.DL (Superstr.+pier)LL (Unequal loading)Longitudinal force(Braking+Temp)Water current at 20

Wind forcesBuoyancySeismic forces

Case A: DL+LL+LF+WC

Moment (t-m) 160.00 302.44181.87

C: Case A + SFVertical load (t) 204.48 314.25Horizontal Load (t) 17.21 39.24

B: DL+LF+WC+B314.2517.21




X-sectional area of steel, Asc = 9503 mm2

Therefore, direct stress, σco,cal = 0.98 N/mm2

Bending compressive stress, σc, cal = 4.61 N/mm2

Checki) Combined stress = 0.89 < 1.0 O.K.

ii) Resultant tension, σt, cal = 3.64 N/mm2 > 0.53 N/mm2

iii) Resultant compressive stress = 5.59 N/mm2 < 6.67 O.K.iv) Tensile stress in steel = 36.36 N/mm2 < 200 O.K.Provide 20 numbers 25 mm diameter bars giving area of steel= 9817 mm2Spacing of bars = 158 mm O.K.

Design for lateral tie:

Diameter of lateral tie >φL/4 = 6.25 mm

> 8 mmPitch of lateral tie < least lateral dimension of column = 1009 mm

< 12x dia of smallest longitudinal bar = 300 mmProvide 10 mm dia lateral tie @ 150 mm c/c with cross links.

Check for stresses in concrete and steel for case BDesign vertical load = 65.65 tDesign Moment = 53.33 t-mEccentricity = 0.81 m







iii) Resultant compressive stress = 4.71 N/mm2 < 6.67 O.K.iv) Tensile stress in steel = 34.54 N/mm2 < 200 O.K.

Check for stresses in concrete and steel for case CDesign vertical load = 101.67 tDesign Moment = 110.16 t-mEccentricity = 1.08 m







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iii) Resultant compressive stress = 9.41 N/mm2 < 10.01 O.K.iv) Tensile stress in steel = 74.49 N/mm2 < 300 O.K.

Design of pier cap:Calculations of loads and MomentsDue to dead LoadTotal DL from superstructure = 28.53 twt.of pier cap = 9.50 tTotal DL on pier = 38.03 tLoad from each girder = 20.82 t

Due to live LoadTotal live load including impact = 93.99 tLoad from each girder = 29.85 t

Design of pier capAs the bearings are placed centrally over the column and the width of bearing is located within half the depth of cap from any external face of the columns, the cap beam need not be designed for flexure.Min. area of steel required = 2212 mm2Provide 8 nos. of 25 mm φ bars giving area of steel = 3927 mm2at top and 8 nos. of 25 mm φ bars giving area of steel = 3927 mm2at bottom.Equivalent shear force = 50.67 tShear stress, tv = 0.47 N/mm2Percentage area of tension steel = 0.36 %Permissible shear strength of concrete, tc = 0.265 N/mm2Assuming 10 mm φ 8 legged vertical stirrups giving area of steel = 628 mm2Spacing of stirrups = 385 mm c/cProvide 10 mm φ 8 legged vertical stirrups @ 100 mm c/c.Provide 4 nos. 20 mm dia. Bars on each face as a side face bars.

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 =

C0 = 5.00 MPa, permissible direct compressive stress in concrete

A1 = 1.776 m2, dispersed largest concentric area similar to A2

A2 = 0.356 m2, loaded area

Therefore, Α1/Α2 = 4.989 > 2

And, C = 7.07 Mpa

Actual compressive stress = 1.42 < C O.K.

2

10

A

AC ×




Design of pier footing (Piles)Diameter of pier = 1.2 mLength of footing, L = 6.5 m along longitudinal directionWidth of footing, B = 6.5 m along transverse directionDepth of footing = 1.1 mBearing capacity of soil = 0 t/m2Load bearing capacity of pile = 40 tNumbers of piles along longitudinal dirrection = 4 mNumbers of piles along transverse dirrection = 4 mTotal numbers of piles = 16Diameter of piles = 700 mmLength of piles = 16 mSoil Type : Clayey Silt

Determination of depth of fixity of piles:

Unconfined compressive strength, S = 0.46 kg/cm2

Modulus of subgrade reaction, K = 13.2 kg/cm2

Therefore, Lf/d = 6 from figure 1 and 2.

and, depth of fixity,Lf = 4.2 m

Calculation of loadsself weight of footing = 111.54 tBuoyancy force on foundation base = 46.48 t

The forces and moments due to seismicity are grater than those due to wind forces. As per standard design practce, seismic forces and moments will be adopted neglecting the effect due to wind forces.

Horizontal Load (t) 17.21 17.21 38.76

Case A: DL+LL+LF+WC B: DL+LF+WC+B C: Case A + SFVertical load (t) 314.25 262.00 416.56

Moment (t-m) 180.91 160.00 330.49




Case: A

Deducting the bearing capacity of pile cap as a open footing, vertical loads to be taken by piles

= 314.25 t

Loads on each pile due to vertical load = 19.64 t

Axial load on extreme pile due to moment = 7.80 t

Resultant compressive force = 27.44 t < 40.00 t O.K.

Resultant tensile force = 0.00 t < 40.00 t O.K.

Horizontal force on each piles = 1.08 t

Moment on each piles = 4.52 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) = 0.774

Safe stress are,

Permissible direct compressive stress, σco = 3.87 N/mm2

Permissible flexural compressive stress, σco = 5.16 N/mm2

Permissible Flexural Tensile stress, σst = 154.762 Mpa (in steel)

Maximum compressive stress in concrete = 0.88 N/mm2 < 3.869 N/mm2

Provide minimum area of compression steel @ 1.25 %.

Asc = mm2

Provide 15 nos. Of 25 mm φ bars, giving area of steel = 7363 mm2

Check for combined stress:

Percentage area of steel provided, pt = 1.276

p/fck = 0.064

d'/D = 0.089

Pu/fckD2 = 0.052

From chart, Mu/fckD3 = 0.07

and Mu = 48.02 t-m

Working moment, M = 32.01 t-m > 4.52 t-m O.K.

Shear stress due to horizontal force = 0.03 N/mm2

Permissible shear strength of concrete = 0.45 N/mm2 O.K.

Case: C

Axial load on extreme pile due to moment = 14.25 t

Resultant compressive force = 40.28 t < 90.00 t O.K.

Resultant tensile force = 0.00 t < 90.00 t O.K.

Horizontal force on each piles = 2.42 t

Moment on each piles = 10.17 t-m

Pu/fckD2 = 0.062

From chart, Mu/fckD3 = 0.072

and Mu = 61.74 t-m

4811




Working moment, M = 41.16 t-m > 10.17 t-m O.K.

Shear stress due to horizontal force = 0.06 N/mm2

Permissible shear strength of concrete = 0.56 N/mm2 O.K.

Lateral Reinforcement:

Lateral reinforcement should be 0.2% of the gross volume. Using dia. 10 mm ties,

we have the volume of one tie = 145577 mm3

If p be the pitch of the ties in mm, then

Volume of pile per pitch length = 384845.1 p

Hence equating, we get, p = 189 mm

Hence provide dia 10 mm ties at 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 = 322.5 t-m

At face of column = 260.3 t-m

Bending moment due to moment on steam = 90.5 t-m

Maximum bending moment = 413.0 t-m

The effective depth required is given by,

dreq = 912 mm

Adopt 1030 mm effective depth and overall depth = 1100 mm

Area of steel required along width of cap = 21872 mm2

Using 25 mm dia. bars, spacing would be 146 mm

Provede 25 mm dia. Bars @ 110 mm c/c, provided area of steel = 29006

at both top and bottom.

Distribution reinforcement @ 0.12% = 8580 mm2

Provide 25 mm dia bars @ 150 mm c/c, giving area of stee = 21271 mm2.

Shear force = 263.94 t

Shear stress, tv = 0.39 N/mm2

Percentage area of steel, pt = 0.43 %

Permissible shear strength of concrete, tc = 0.278 N/mm2 < tv O.K.

Shear reinforement to be designed..

Assuming 10 mm dia 20 legged vertical tirrups, then Asv = 1571 mm2

And spacing of stirrups, S = 416 mm

Maximum spacing of stirrups, = 218 mm

Provide 10 mm dia 20 legged vertical tirrups @ 160 mm c/c.

Check for punching shear along pier,


bo = 6692 mm

tv = 0.15 N/mm2

Punching shear strength, tc = 0.716 N/mm2 > tv Safe.

Check for punching shear along pile,


bo = 5435 mm

tv = 0.06 N/mm2

Punching shear strength, tc = 0.716 N/mm2 > tv Safe.



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