UT Design (14.11.05)

8/11/2019 UT Design (14.11.05)

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DESIGN OF DECK SLAB FOR BRIDGE

Design of RCC Deck Slab for 5.0m clear span :

Clear Span = 5.00 m

Assumed thickness of slab = 400 mm

Clear cover = 35 mm

Main reinforcement = 16mm Φ HYSD barsEffective depth available = 400 - 35 - 16/2

= 357 mm

The single lane road bridge is designed for IRC class 'A' loading.

The RCC Deck Slab is proposed in M - 20 grade.

Allowable stresses : [ RCC M-20]

σcbc = 7 N/mm2

σst = 190 N/mm2

m = 13.33

k = 0.3872

j = 0.8709

Q = 1.1802

Unit weight of water = 10 KN/m3

Unit weight of concrete = 25 KN/m3

Effective span will be least of the following :

i) C/C of Support = 5.0 + 0.5 [bearing]= 5.5 m

ii) Clear Span +Effective depth = 5.00 + 0.357

= 5.357 m

Therefore Effective span = 5.357 m

Clear Carriage way width = 4.25 m

Width of slab = 4.25 + 2 x 0.225 + 2 x0.2 - 2 x 0.09

= 4.92 m

[As per Clause 211.2 of IRC 6 - Pg.22]

Impactor factor = 4.5/(6.0+Effective span)

= 0.396 m

Bending Moment due to Dead Load :

Weight of Slab & Wearing Coat = (0.4 + 0.075) x 25

= 11.875 kN

Bending Moment [Wl2

/8] = 11.875 x 5.357^2/8= 42.60 kN - m

Bending Moment due to Live Load : [As per IRC code 21 -2000, section 305.16.2]

Bef = α x a x (1-(a/Lo)) + bw

Where Bef = the effective width of slab on which the road acts.

Lo = the effective span.

A =

bw =

α =where b is the width of the slab.

[b/Lo] = [4.92/5.357]

= 0.918

For [b/Lo] of For [α] value

0.900 2.360

1.000 2.480

Difference 0.100 Difference 0.120

α = 2.36+[0.018x0.12]/0.10

= 2.382

the distance of the centre of gravity of the concentrated load from

the nearest support.

the breadth of concentration area of the load i.e the dimension of the

type/ track contact area over the road surface of the slab in a

direction at right angles to the span plus twice the thickness of

wearing coat or surface finish above the structural slab and

A constant having the following values depending upon the ration

b/Lo

Road Bridge Design Page1

8/11/2019 UT Design (14.11.05)


Dispersion width under 'C' [11.40 t]


= 2.382 x 2.379 x [1-2.379/5.357] +0.65

= 3.80 m

> 1.8 m

Hence dispersion width overlap.

Therefore, combined width of dispersion = 1.8 + [0.225+0.15+(0.5/2)]+3.8/2

= 4.325 m

Intensity of load = Load (impact factor)/combined dispersion width

= 3.68 t

= 36.8 kN

Dispersion width under 'D'


= 2.382 x 1.778 x [1-1.778/5.357] +0.65

= 3.48 m

> 1.8 m


Therefore, combined width of dispersion = 1.8 + 0.625+3.48/2

= 4.165 m

Intensity of load = Load (impact factor)/combined dispersion width

= 3.82 t

= 38.2 kN

For maximum bending moment the position of load system is as shown below:

3.677

2.379 0.3 0.3 0.6

A C D B

5.357

Taking moments about right support 'B'R A x 5.357 = 3.677 x 2.978 + 3.829 x 1.778

R A = 3.31 t

= 33.1 kN

Max.Bending Moment due to Live Load

= R A x 2.379

= 78.74 kN - m

Therefore totalB.M = Livel Load B.M + Dead Load B.M

= 121.34 kN - m

Effective depth required (d) = sqrt(B.M/Qb)

= sqrt((121.34x 10^6)/(1.1802 x 1000)

= 320.64 mm

Effective depth available = 320.64 mm

< 357.00 mm

Hence assumed depth of slab 400mm is O.K

Hence Safe.

1.778

2.679

3.829

For Maximum bending moment the two wheel loads of 11.40 t kept such that the resultant of two loads

and one of the loads should be kept equidistant from the centre of the span. The position of loads is

shown in the sketch.


8/11/2019 UT Design (14.11.05)


Main Reinforcement Calculations:

Area of steel required (Ast) = BM/σst x j x d

= 2054.01 mm2

Therefore provide 20mm dia @

spacing = 314 x 1000/2054.01

= 152.8715 mm

Say 150 m c/c Area of steel provided = 2093.33 mm

2

Distribution steel: [ As per clause 305.18.1 of IRC code 21 - 2000, section III]

B.M considered for distribution steel.

= 0.3 x Live Load B.M + 0.2 Dead Load B.M

= 32.14 kN-m.

Area of steel [Ast] = 32.14 x 10^6/(190 x 0.8709 x 357)

= 544.10 mm2

But min. area to be provided is 0.12% of gross cross sectional area

= (0.12 x 400 x 1000)/100

= 48.00 mm2

Therefore provide 10mm dia bars @

Spacing = 78.53 x 1000/48

= 144.33 mm

say 140 mm c/c

Ast provided = 560.93 mm2

Check for shear:

Maximum shear due to dead load = WL/2

= 29.69 kN

Maximum shear due to live load, as per clause 305.16.5 of IRC 21 - 2000

Longitudinal dispersion = 0.25+2x(0.4+0.075)

= 1.2 m

Therefore the load of 11.4t may be kept at 1.2/2 = 0.6m from the support for maximum shear.

11.4 t 11.4 t

0.6

C D

A B

Dispersion under Load 'C':


= 2.382 x 0.6 x [1-0.6/5.357] +0.65

= 1.92 m

> 1.8 m


= 2.425 + 1.92/2

= 3.385 m

Intensity of Load = Load (Impact factor)/ combined dispersion width

= [11.4 x (1.396)/3.385

= 4.70 t

200

150 1800

225 250

225+150+1800+250 = 2425

1.2

5.357

Therefore, combined width of

dispersion


8/11/2019 UT Design (14.11.05)


Dispersion under Load 'D':


= 2.382 x [0.6+1.2] x [1-(0.6+1.2)/5.357] +0.65

= 3.50 m

> 1.8 m


= 2.425 + 3.5/2

= 4.175 m

Intensity of Load = Load (Impact factor)/ combined dispersion width

= [11.4 x (1.396)/4.175

= 3.81 t

4.702 t 3.813 t

0.6 1.2

A B

Taking moments about 'B'

R A x 5.357 = (4.702 x 4.757) + (3.813 x 3.557)

R A = 6.71 t

= 67.07 kN

Total Shear = 67.07+29.69

= 96.76 kN

Shear stress = 96.76 x 1000/[1000 x 357]

= 0.27 N/mm2

As per clause 304.7.2 of IRC section - III, the maximum permissible shear stress

Tmax = 0.07 f ck

= 0.07 x 20

= 1.4 N/mm2

> 0.27 N/mm2

T<Tc where Tc = permissible shear stress

Tc = k1 x k2 x Tco

k1 = 1.14 - 0.7d > 0.5 [d in m]

k2 = 0.5 + 0.25p > 1Tco = basic values for different grades of concrete = 0.28 Mpa

= Longitudinal reinforcement ratio = As/bd

k1 = 1.14 - 0.7 x 35.7/[100]

= 0.89 > 0.5

k2 = 0.5+0.25 x [22.34/2] x 1/[100 x 35.7] > 1

= 0.50078 < 1

Hence take k2 = 1

Tc = 0.89 x 1 x 0.28

= 0.249 N/mm2

< 0.27 N/mm2

Hence Tc < T , shear reinforcement shall be provided to cater to ful l shear

V = Asw x [sigma]s x d x [sina+cosa]/spacing

= 0.5 x 2235 x 190 x 357 x [1.414] /180

= 595451.31 N= 595.45 kN

Tbd = 0.249 x 1000 x 357

= 88.89 kN

Balance shear = 96.76 - 88.89

= 7.87 kN

< 595.45 kN

Hence shear

5.357

Therefore, combined width of

dispersion


8/11/2019 UT Design (14.11.05)


DESIGN

The design consist of 1) Design of slab spanning between the Kerb

2) Design of Kerb as beam.

Design of slab:

Loads:

a) Self weight of slab = 0.1 x1 x 24

= 2.4 kN/m

b) Weight of wearing coat = 0.075x24

= 1.8 kN/m

c) Live Load as per bridge code

[as per para 209.1 of IRC 6:1966] = 4 kN

= 8.2 kN/m

Width of slab in between kerbs = 1.50 m

Thickness of slab = 0.10 m

Size of Kerbs = 150 mm x 625 mm

Effective width of slab = 1.5+0.15/2+0.15/2

= 1.65 m

Max.B.M M = Wl2/8

= 2.791 kN-m

M15 grade and HYSD bars

Stresses:

σcbc = 7 N/mm2

σst = 190 N/mm2

m = 13.33

k = 0.332

j = 0.889

Q = 1.034

Unit weight of water = 10 KN/m3

Unit weight of concrete = 24 KN/m3

Effective depth required (d) = sqrt(M/Qxb)= 51.96 mm

Available effective depth = 0.10 - 0.025 - 0.006

0.069 m

> 0.0520 m( required)

Hence OK

Ast required Ast = M/σst x j x d

= 239.35 mm2

Spacing of 12mm ø bars @

Spacing = Area of 1 bar x 1000/ total area

= 472.11 mm

However the spacing should not exceed 3 x d i.e 3 x d

= 207.00 mm

Hence provide 12mm ø @200mm c/c

Ast provided = Area of 1 bar x 1000/spacing

= 565 mm

2

ΣO =

Distribution steel :

0.12% of the section [as per para 25.6.21 of IS : 456 - 2000]

= 0.12/100 x 100 x 1000

120.00 mm2

Spacing = 78.5 x 1000/120

= 654.1667 mm c/c

Spacing should not exceed = 5 x d

= 5 x 69

= 345.00 mm

Provide 10mm ø bars @ 300mm clc

DESIGN OF FOOT BRIDGE

Total load

Design of Foot Bridge Page5

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Check for Shear :

Shear force [V] = 8.2 x 1.5/2

= 6.15 kN

Shear stress [ĩ V] = V/bd

= 6.15x 1000/1000x 69

= 0.089 N/mm2

Percentage of steel per metre run of slab[100 x As/(b x d)] = 100 x 565/(1000 x 69)

= 0.409 %

= kĩ c

= 1.3 x 0.265

= 0.345 N/mm2

> ĩ c

Hence no shear reinforcement required.

Design of beam:

Loads:

a) Self weight of slab = 0.5 x1.5 x 0.1 x 24= 1.80 kN/m

b) Weight of wearing coat = 0.5 x 1.5 x 0.075 x 24

= 1.35 kN/m

c) self weight of Kerb = 0.15 x 0.625 x 24

= 2.25 kN/m

d) Hand Railing = 0.358 kN/m

e) Live load as per bridge code = 0.5 x 4 x 1.5

= 3.00 kN/m

Total Load (W) = 8.758 kN/m

Clear span = 5.00 m providing 16 mm ø bars.

Effective depth = 625 - 25 - 16/2

= 592.00 mm

Effective span (l) :

i) Clear span + bearing = 5.0 + 0.45

= 5.45 m

ii) Clear span + effective depth = 5.0+0.592

= 5.592 m

Adopt least of the two, = 5.45 m

Maximum Bending Moment = Wl2/8

= 32.517 kN - m

Effective depth required = sqrt(32.517x10^6/1.034 x 150)

= 457.97 mm

< 592.00 mm Provided

Ast required = M/(σst xjxd)

= 325.07 mm2

Provide 2 Nos of 16mm ø bars @600mm clc

Spacing = 201 x 1000/325.07

= 618.33 mm

Area of steel provided = 201 x 1000/600

= 335.00 mm2

> 325.07 mm3

required

Hence OK

Minimum reinforcement:

Minimum reinforcement as per clause 26.5.1.1 of IS:456 - 2000

is As/bd = 0.85/f y

Ast = 0.85 x 150 x592 /415

= 192.02 mm2

As per B-5.2.1.1, the permissible shear

stress in conrete

From table 23 of IS: 456:2000 the permissible shear stress [ without shear reinforcement in slab] for 0.409% steel is 0.265 N/mm2


8/11/2019 UT Design (14.11.05)


Design of Kerb for horizontal thrust of 7.5 kN/ run:

Horizontal load per metre run = 7.5 kN/m

Bending Moment due to this force = 7.5 x 0.625

= 4.6875 kN-m

Effective depth available = 150 - 25 - 16/2

= 117.00 mm

Ast required = 4.6875/(190 x 0.889 x 117)

= 237.11 mm2

Spacing of stirrups = [1000 x 50.26]/237.11

= 211.97 mm2

Provide 8mm dia bars as vertical links @200mm clc.

Side face reinforcement:

[As per IS: 456 - 2000 para 26.5.1.3 ] = 0.1% of beam area

= 0.1/100 x 625 x 150

= 93.75 mm2


Provide 8mm dia bars @500mm c/c

Check for Shear:

Shear force [V] = 8.758 x 5/2

= 21.90 kN

Shear stress [ĩ V] = V/bd

= 21.90 x 1000/(150 x 592)

= 0.247 N/mm2

Percentage steel requirement = 100 x 335/(150 x 592)

= 0.377 %

= 0.26 N/mm2

> 0.247 N/mm2

Hence Ok

From table 23 of IS:456 - 2000 the permissible shear

stress without shear reinforcement


8/11/2019 UT Design (14.11.05)


I

S.No Description Unit Particulars

1 Discharge Cumecs

2 Bed width m

3 Full Supply Level m

4 Bed fall --

5 Velocity m/sec

6 Free board m

7 Value of 'n' --

8 Side slopes (inner/outer)

9 Top width of banks (L/R) m

10 Canal bed level m

11 Full Supply depth m

12 Top of bank level m

II Particulars of Stream are :

a) Catchment area Km2

16.87

b) Max.flood discharge (as per

Dicken's formula) MFD Cumecs 139.02

c) Observed MFL at the Crossing

d) Stream bed level at the Crossing

e) Bed fall of stream from

400m.U/S to 1700m D/S

III Hydraulic Calculations:

Catchment area = 16.87 KM2

MFD as per Dicken's formula(Q) = CM3/4

( taking the value of C = 16.7 since the catchment area of stream is in between 2.5 to 75KM2)

Q = 139.012 Cumecs

Say 139.02 Cumecs

IV Vent Way:

Adopt 5 vents of 5m clear span for stream flow.

The C.B.L at the site of crossing = 18.26

Thickness of trough (assumed) = 0.36 m

The bottom of trough =

Barrel width at entry =

Barrel width at exit =

Floor level :

DESIGN OF BUTTRESS TYPE U.T

Hydraulic Particulars of Canal at the site of Crossing are:

Keeping in view the larger M.F.D and length of structure in canal flow direction a B.T.U.T is proposed.

A down ward gradient of 1 in 223 is given in the barrel floor (150mm fall in the entire length of barrel).

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The floor level at the entry of barrel (at the end of u/s transition) =

The floor level at the centre =

The floor level at the starting of d/s transition or at the end of barrel =

The d/s stream transition level =

V Transitions :

a) U/S transition:Length of transition =

Length of transition upto the toe of drop wall =

Width of transition over drop wall =

b) D/S transition:

Length of transition =

Width of transition =

Length of transition upto the end of cut off wall =

VI Area of Flow

=

Depth of barrel at the centre =

Barrel width at the bottom =

Barrel at xx m depth =

Area of flow at the centre of barrel =

Velocity in the barrel =

Assume the depth of flow in the barrel as twice the differencebetween OMFL and floor level

8/11/2019 UT Design (14.11.05)


I0.3

a) Bottom slab:- (Designed as simply supported)

0.75 free board

1.8

0.42

25

Clear span = 5.00 m

Adopt M20 grade

From IS 3370 (part -II) - 1965 & IS - 456- 2000

σcbc = 7 N/mm2

σst = 190 N/mm2

m = 13.33

k = 0.3321

j = 0.8893

Q = 1.0338

Unit weight of water = 10 KN/m3Unit weight of concrete = 25 KN/m

3

Unit weight of plain concrete = 24 KN/m3

Assume an overall thickness of slab = 420 mm

Using 20mm dia bars with 30m clear cover

effective depth = 420 - 20/2 -30

= 380 mm

effective span = Clear span + (end bearings/2)

or Clear span + effective depth

effective span = which ever is less

= 5.50 m

or 5.38 m

Hence effective span = 5.38 m

Loads per m span:Water Load = (1.8+0.75)x10

= 25.50 KN/m

Self weight of slab = (0.42+0.04)x25

= 11.5 KN/m

Total Loa = 37.00 KN/m

Max.BM = Wl2/8

= 133.87 KN-m

= sqrt(M/Qb)

= 359.85 mm

< 380.00 mm

Hence assumed depth is OK

Design of trough slab

effective depth

required (d)

40mm sealing coat

Design of Trough Slab page10

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Area of steel required:

Ast = M/σst x jxd

= 2084.96 mm2

Provide 20mm dia Hysd bars @

Spacing = 150.67 mm Say 110 mm c/c


Distribution steel:

Ast = 672 mm2

Provide 16mm dia bars @

Spacing = 299.11

Say 250 mm c/c

Ast provided = 804 cm2

Hence OK

Check for shear :

Shear force at the face of supports (V) = Wl/2

= 92.50 KN

Shear stress = V/bd

ĩc = 0.24 N/mm2

% of reinforcement 100 As/bd = 0.7516

ĩc max = 0.268 N/mm2

ĩc < ĩc max

Hence safe

As per IS: 3370 - part - II, para 7.1.1 & 7.1.1.1 the minimum reiforcement thickness of section above 450mm is

0.16% of gross area

From table 23 of IS 456 - 2000 the allowable shear stress without shear reinforcement for 0.7516% of tensile

steel


8/11/2019 UT Design (14.11.05)


II Design of side slab

Design of top segment:

Weight of water = (2.55 - 1.2 )x10 KN/m

= 13.5 KN/m

The thickness of slab at 1.2 m will be 360 mm using 20mm ø bars with 30mm clear cover,

effective depth of slab = 360 - 30 - 20/2

= 320 mm

effective span = 5.320 m

or 5.50 m

Hence effective span = 5.320 m

Using 20mm dia bars with 30mm clear cover the effective depth of slab

Wt. of water = (2.55 - 1.2 )x10

= 13.50 KN/m

Self wt. of slab = 0.55 x 1 x 25

= 13.75 KN/m

Total = 27.25 KN/m

Max. B.M = WL2/8

= 96.41 KN - m

Effective depth required = sqrt((145.73x10^6)/1.1802x1000)

= 305.37 mm

< 320 mm provided

Hence OK

Area of steel required :

Ast = 1783.017 mm2

Provide 20mm dia bars @Spacing = 176.1901

Say 130 mm c/c


Hence OK

Distirbution steel:

Ast = 512.00 mm2

Provide 16mm dia bars @

Spacing = 392.58 mm

Say 380.00 mm


Hence OK

Check for shear :

Shear force at the face of supports (V) = 68.125 KN

Shear stress (Tc) = 0.2129 N/mm2

percentage of reinforcement = 0.755

from the table 23 Tc max = 0.28

Tc < Tc max

The thickness of slab is uniformly reduced from420mm at bottom to 300mm at top. In the first or bottom most

segment the thickness and reinforcement are same as bottom slab. For the remaining segments thickness ofslab and the reinforcement will be changed according to the requirement.

The side slab is designed in 2 segments from bottom. The two segments are 0 to 1.2m height and 1.2m to

2.55m height.


8/11/2019 UT Design (14.11.05)


DESIGN OF RETURN WALLS

= kN

= kN

CALCULATION OF STRESSES IN CONCRETE :

+ C.B.L

W5 W1

W3

+

B

+

Taking moments about A (Stresses in concrete)

W1 x x

W2 x x

W3 x x x

W4 x x xW5 x x x

Pv x x

Ph x x

Lever arm = M / V = / = m

Eccentricity = - /2 = m

Permissible 'e' = = > HENCE OK

= x ( 1 + 6 ) = t/sq.m.

Min. Stress = x ( 1 - 6 ) = t/sq.m.

Taking moments about B (Stresses on soil)

1.200 1.200

1.200 1.200

58.48 x 0.139 14.864

1.20 / 6 0.200 0.139

Max. comp. Stress 58.48 x 0.139 82.607

26.966 58.48 0.461

0.461 1.20 0.139

0.968 84.591

∑ V 58.483

∑ M 26.966

0.134 5.856 21.0 50.113

0.179

0.0384 5.856 21.0 4.723 ---

0.145 2.420 21.0 0.5 3.68 0.048

0.897 11.848

0.145 2.420 24.0 0.5 4.21 0.097 0.408

0.455 2.420 24.0 0.5 13.21

0.445 0.962

0.60 2.420 21.0 30.49 0.445 13.569

ForceL.A. Moment

V H

0.600 0.150 24.0 2.160

1.2000.6

W7

1.800 15.240

Load Particulars

W4

0.30 0.145 0.600 0.455 15.840

0.6

18.260

W6 W2

unit wt of concrete 24.000

unit wt of earth 21.000

0.15

8/11/2019 UT Design (14.11.05)


W1 x x

W2 x x

W3 x x x

W4 x x xW5 x x x

W6 x x

W7 x x

Pv x x

Ph x x




= x ( 1 + 6 ) = t/sq.m.

Min. Stress = x ( 1 - 6 ) = t/sq.m.102.961

0.00 0.000

0.00 0.000

73.37 x 0.049

1.80 / 6 0.300 0.049

Max. comp. Stress 73.37 x 0.049 122.453

670.079 73.37 1.749

1.749 1.800 0.049

∑ V 73.366

∑ M 670.079

--- ---

0.134 9.120 21.0 - 50.113 0.968 96.618

0.0384 9.120 21.0 7.355

2.287

1.80 0.60 24.0 25.920 0.900 23.328

0.30 2.420 21.0 15.246

8.422 0.39724.0 0.50

0.150

3.3410.145 2.420 21.0 0.50 7.369 0.348 2.5670.145 2.420

0.745 25.962

0.455 2.420 24.0 0.50 13.213 1.197 15.812

0.60 2.420 24.0 34.85

MomentV H

0.60 0.150 24.0 2.160 0.745 1.610

Load ParticularsForce (kN)

L.A.

8/11/2019 UT Design (14.11.05)


DESIGN OF HORIZONTAL WING WALLS

= kN

= kN

CALCULATION OF STRESSES IN

+

+ Road level

W1

+

++

W3

+

Taking moments about A (Stresses in concrete)

W1 x x

W2 x x

W3 x x xW4 x x x

W5 x x x

Pv x x

Ph x x




= x ( 1 + 6 ) = t/sq.m.

Min. Stress = x ( 1 - 6 ) = t/sq.m.

2.500 2.500

2.500 2.500

230.82 x 0.158 57.317

2.50 / 6 0.417 0.158

Max. comp. Stress 230.82 x 0.158 127.338

251.980 230.82 1.092

1.092 2.50 0.158

1.688 84.591

∑ V 230.819

∑ M 251.98

0.134 17.808 21.0 50.113

39.243

0.0384 17.808 21.0 14.361 ---

1.63 4.220 21.0 0.5 72.23 0.543

2.320 8.9451.63 4.070 24.0 0.5 79.61 1.087 86.5090.27 1.360 21.0 0.5 3.86

1.93 4.169

0.60 4.070 24.0 58.61 1.93 113.114

ForceL.A. Moment

V H

0.600 0.150 24.0 2.160

0.600 0.27

3.400 15.240

Load Particulars

W4

17.200

15.99015.840

2.5000.6

W7

0.30 1.630

20.285 0.6

20.060

W5

W6 W2

unit wt of concrete 24.000

unit wt of earth 21.000

0.15 W1

8/11/2019 UT Design (14.11.05)


Taking moments about B (Stresses on soil)

W1 x x

W2 x x

W3 x x xW4 x x x

W5 x x

W6 x x

W7 x x

Pv x x

Ph x x




= x ( 1 + 6 ) = t/sq.m.

Min. Stress = x ( 1 - 6 ) = t/sq.m.102.961

3.40 3.400

3.40 3.400

383.20 x 0.049

3.40 / 6 0.567 0.049

Max. comp. Stress 383.20 x 0.049 122.453

670.079 383.20 1.749

1.749 3.400 0.049

∑ V 383.204

∑ M 670.079

---

0.134 17.808 21.0 - 50.113 1.928 96.618

0.0384 17.808 21.0 14.361

0.567

3.40 0.60 24.0 48.960 1.700 83.232

0.60 0.30 21.0 3.780

159.218 1.38724.0 0.50

0.150

220.783

1.63 4.22 21.0 144.451 0.843 121.821

1.63 4.070

2.23 130.696

0.27 1.360 24.0 0.50 4.406 2.620 11.545

0.60 4.070 24.0 58.61

MomentV H

0.60 0.150 24.0 2.160 2.23 4.817

Load ParticularsForce (kN)

L.A.

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unit wt of concrete = 24.000 kN/m3

unit wt of earth = 21.000 kN/m3

unit wt of concrete = 25.000 kN/m3

The stability for the entire pier is examined for two different cases1) Canal full and stream full and

2) Canal empty but with crowd load and stream full

Case I Canal full and stream full

I. Determination of Longitudinal Moments:

Moments are reckoned about 'A' on the upstream end.

0.3 3.825

0.3

18.26

0.420 0.04

CBL 18.26

17.96

17.92

17.5017.20

25 mm dia Holding

Down Bolts at 2000 c/c

1

10

15.99

15.84

15.24

1000

PIER UNDER CANAL TROUGH

23.980

24.100

24.560

DESIGN OF PIER

The pier is subject to both longitudinal and transverse moments due to the Direct Loads, Live Loads on

water currents which determines the maximum and minimum stresses induced in the pier.

25.000

1.5 :1

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a) Weight of bottom slab = [23.98+24.56] x 0.42 x 6

= 1529.01 kN

b) Weight of side slabs = 2 x [5.21+5.927] x [0.3+

= 603.56 kN

c) Weight of sealing coat = 0.04 x [24.1+23.98] x 6.

= 138.47 kN

d) Weight of earth cushion in the trough = 0.3 x [24.1+25] x 6.00 x

= 927.99 kN

e) Weight of water in the trough = [25+32.65] x 6.00 x 25/2

= 4323.75 kN

f) Weight of bed block under canal trough = 0.3 x 24.56 x 24

= 176.83 kN

g) Weight of rectangular portion of pier under road bridge = 5.01 x 1 x 3.485 x 24

= 370.94 kN

h)= 1.035 x 3.56 x 1 x 24

= 88.43 kN

i) Weight of triangular portion under canal trough = 2 x 4.51 x [20.06-17.50]

= 352.86 kN

j)= 2 x 4.51 x 0.3 x 1 x 24

= 64.944 kN

k)= 1.31 x [20.06-17.2] x 1 x

= 111.93 kN

l) Weight of Road Bridge: =

i) Slab : = 4.92 x 6 x 0.4 x 25

ii) Wearing coat : = 4.25 x 6 x 0.075 x 24

Weight of rectangular portion of pier between Canal trough

and foot bridge

Weight of rectangular portion [below triangular portion]under canal trou h

Weight of rectangular portion of pier between Road Bridge

and Canal Trough

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iii) Kerbs : = 2 x 6 x 0.225 x 0.225 x

iV) Hand rails : = 2 x 6 x 120/1000

m) Weight of pier below foot bridge = 1.8 x 1 x [20.96-17.2] x

= 162.43 kN

n) Weight of pier in batter portion = 42.885 x [17.20 - 15.84]

= 1590.13 kN

o) Weight of foot bridge :

i) Slab: = 1.8 x 0.1 x 6 x 25

ii) Wearing coat : = 1.5 x 6 x 0.075 x 24

iii) Kerbs : = 2 x 0.15 x 0.225 x 6 x 2

iV) Beams : = 2 x 0.15 x 0.225 x 6 x 2

v) Haunches : = 2 x 0.5 x 0.1 x 0.1 x 6 x

vi) Crowd load : = 400 x 1.5 x 6/1000

Taking moments about 'A'

8/11/2019 UT Design (14.11.05)


3.825 0.3

20.81

5.21

5.967

17.50

RCL 20.760

20.68520.285

19.985

17.20

15.99

15.84

15.24

1000

PIER UNDER ROAD BRIDGE

1272

the road, forces due to wind and

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.00 x 25/2

.42] x 6 x 25/[2 x 2]

0 x24/2

21/2

x 1 x 24/2

24

= 295.20 kN

= 45.90 kN

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= 15.19 kN

= 1.44 kN

4

x [1+1.272] x 24/2

= 27.00 kN

= 16.20 kN

= 10.13 kN

= 10.13 kN

= 1.50 kN

= 3.60 kN

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I. Design of Abutment under canal trough:

[ with 50% of Hydrostatic Pressure]

Reaction of canal trough on to the abutment

= 1/2 x [(6x0.42 x 25) + (6 x 0.04 x 24) + ( 6 x 2.55 x 10)]

= 110.88 kN

18.2617.96

17.92

17.50

0.5

600

1730

1000

270 15.84

3000

15.24

3900

CALCULATIONS OF STRESSES IN CONCRETE:

Case - I with 50% hydrostatic pressure:

Taking moments about 'A', we have

Vertical (V) Horizontal (H)

1 R As calculated 110.88 - 2.23 247.26

2 W1 0.5 x 0.3 x 25 3.75 - 2.23 8.36

3 W2 0.5 x 1.36 x 24 16.32 - 2.23 36.39

4 W3 0.5 x 0.27 x 1.36 x 24 4.41 - 2.86 12.60

5 W4 0.5 x 0.46 x 24 5.52 - 1.98 10.93

6 W5 0.5 x 2.12 x 24 19.92 - 1.98 39.447 W6 0.5 x2.12 x 1.73 x 24 44.01 - 1.153 50.74

8 W7 2.23 x 2.55 x 10 56.86 - 1.115 63.40

9 W8 2.23x 0.3 x [21 - 10] 4.05 - 1.115 4.51

10 W9 0.5 x 1.73 x 2.12 x 11 20.17 - 0.577 11.64

11 Pv0.0384 x [3.11 - 0.3 ] x 11 13.58 - - -

12 Ph0.134 x [3.11 - 0.3 ] x 11 - 8.36 0.85 7.09

13 P1 0.5 x 2.35 x 24 - 28.20 2.97 83.70

14 P20.5 x 0.5 x 2.12 - 1.12 1.12 1.26

Σ V 299.473 Σ M 577.34

L.A = Σ M/ΣV

= 577.3/299.473 = 1.928

Ecentricity = 1.928 - 3/2 = 0.428

Allowable (e) 3/6 = 0.5

> 0.428

Hence no tension

Max and Min stresses = 325.997/3.0 x [1± (6 x 0.482)/3.0]

[Compressive] 110.62 kN/m2 (Max)

3.91 kN/m2 (Min)

CALCULATIONS OF STRESSES IN SOIL:


Vertical (V) Horizontal (H)Sl.No Load Description

Sl.No Load Description

Lever arm

m

17.20

Lever Arm

(m)

Moment

kN-m

Force (kN) Moment

kN-m

Force (kN)

Design of Abutments page23

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1 R As calculated 110.88 - 2.53 280.53

2 W1 0.5 x 0.3 x 25 3.75 - 2.53 9.49

3 W2 0.5 x 1.36 x 24 16.32 - 2.53 41.29

4 W3 0.5 x 0.27 x 1.36 x 24 4.41 - 3.16 13.92

5 W4 0.5 x 0.46 x 24 5.52 - 2.28 12.59

6 W5 0.5 x 1.66 x 24 19.92 - 2.28 45.42

7 W6 0.5 x 2.12 x 1.73 x 24 44.01 - 1.453 63.95

8 W7 2.23 x 2.55 x 10 56.87 - 1.415 80.469 W8 2.23x 0.3 x [21 - 10] 7.36 - 1.415 10.41

10 W9 0.5 x 1.73 x 2.12 x 11 20.17 - 0.877 17.69

11 W10 0.3 x 2.42 x11 + 0.3 x 2.8 x 10 18.81 - 0.15 2.82

12 W11 3.9 x 0.6 x 24 56.16 - 1.95 109.51

13 PV0.0384 x [3.02 - 0.3 ] x 11 3.81 - - -

14 Ph 0.134 x [3.022 - 0.3

2] x 11 - 14.12 1.088 15.37

15 P1 0.5 x 2.72 x 2.8 x 10 - 38.08 1.36 51.79

16 P2 0.5 x 0.5 x 2.722

- 1.85 1.85 3.42

ΣV 367.99 ΣM 758.66


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L .A = 2.06

Ecentricity = 2.07 - 3.9/2

= 0.12 m

Allowable (e) 3.9/6.00 = 0.65 m

> 0.12 m

Hence no tension

Max and Min Stresses = 400.97/3.9 x [1± (6 x 0.12)/3.9][Compressive] = 121.79 kN/mm

2(Max)

= 83.83 kN/mm2(Min)

Case - IIStability calculation considering saturated earth :

CALCULATION OF STRESSES IN CONCRETE :

Taking moments about 'A' , we have


1 R As calculated 110.88 - 2.23 247.26

2 W1 0.5 x 0.3 x 25 3.75 - 2.23 8.36

3 W2 0.5 x 2.05 x 24 24.6 - 2.23 54.86

4 W3 0.5 x 0.4 x 2.05 x 24 9.84 - 2.86 28.14

5 W4 0.5 x 0.46 x 24 5.52 - 1.98 10.93

6 W5 0.5 x 2.35 x 24 28.2 - 1.98 55.84

7 W6 0.5 x 2.81 x 1.6 x 24 53.95 - 1.153 62.21

8 W7 2.1 x 2.55 x 10 53.55 - 1.115 59.71

9 W8 2.1x 0.3 x 21 13.23 - 1.115 14.75

10 W9 0.5 x 1.6 x 2.81 x 21 47.21 - 0.577 27.24

11 PV0.0384 x [3.11

2 - 0.3

2] x 21 7.73 - - -

12 Ph0.134 x [3.11

2 - 0.3

2] x 21 - 26.96 0.848 22.87

ΣV 358.46 ΣM 592.16

L . A = 1.65 m

Ecentricity = 1.65 - 3/2

= 0.15 m

Allowable (e) 3.0/6 = 0.5 m> 0.15 m

Hence no tension

Max and Min Stresses = 3.0 x [1± (6 x 0.15)/3.0]

[Compressive] = 155.33 kN/mm2(Max)

= 83.64 kN/mm2(Min)

Force (kN) Lever arm

m

Moment

kN - mSl.No Load Description


8/11/2019 UT Design (14.11.05)


CALCULATION OF STRESSES IN SOIL :

Take moments about 'B'


1 R As calculated 110.88 - 2.53 280.532 W1 0.5 x 0.3 x 25 3.75 - 2.53 9.49

3 W2 0.5 x 2.05 x 24 24.6 - 2.53 62.24

4 W3 0.5 x 0.4 x 2.05 x 24 9.84 - 3.16 31.09

5 W4 0.5 x 0.46 x 24 5.52 - 2.28 12.59

6 W5 0.5 x 2.35 x 24 28.2 - 2.28 64.30

7 W6 0.5 x 2.81 x 1.6 x 24 53.95 - 3.16 170.49

8 W7 2.1 x 2.55 x 10 53.55 - 2.28 122.09

9 W8 2.1x 0.3 x [21 - 10] 6.93 - 2.28 15.80

10 W9 0.5 x 1.6 x 2.81 x 21 47.21 - 1.453 68.59

11 W10 0.3 x 3.11 x11 + 0.3 x 2.85 x 10 18.81 - 0.15 2.82

12 W11 3.9 x 0.6 x 24 56.16 - 1.95 109.51

13 PV 0.0384 x [3.11 - 0.3 ] x 21 7.73 - - -

14 Ph 0.134 x [3.112 - 0.3

2] x 21 - 26.96 1.088 29.34

ΣV 427.13 ΣM 978.87

L .A = 2.29


= 0.34 m

Allowable (e) 3.9/6.00 = 0.65 m

> 0.34 m

Hence no tension

Max and Min Stresses = 427.13/3.9 x [1± (6 x 0.34)/3.9]


= 49.03 kN/mm2(Min)

Lever arm

( m)

Force (kN)DescriptionLoadSl.No Moment

kN-m


8/11/2019 UT Design (14.11.05)


II. DESIGN OF ABUTMENT UNDER ROAD BRIDGE

.Reaction for clear span of 5.00m

a) Weight of slab = 6 x 4.92 x 0.4 x 25

= 295.20 kN

b) Weight of wearing coat = 6 x 4.25 x 0.075 x 24

= 45.9 kN

c) Weight of kerbs = 2 x 6 x 0.2875 x4.25 x 25= 366.56 kN

d) Weigth of Hand Rails =

(assuming 120 kg/m = 2 x 6 x 0.15

= 1.80 kN

Total Load = 524.95 kN

a) Dead load reaction = 663.56/2

= 262.48 kN/m

b) Change in live load reaction due to braking effect. The braking force will be acting at a height of 1.20m above Roa

Level arm = 1.2 + 0.475

= 1.675 m

Change in vertical direction = [20/100] x [11.4+11.4+2.7] x [1.675/5.357]

= 1.595 t

= 15.95 kN


R A x 5.357 = 11.4 x 5.357 + 11.4 x 4.157 + 2.7 x 0.957

R A = 111.044/5.357

= 20.729 t

= 207.29 kN

Live Load reaction = 207.29 kN

c) Impact load reaction = 207.29 x 0.396 x [X]

Where [X] is the impact factor multiplying factor which depends upon the height of abutment.

Height of abutment below bed block = 17.2 - 15.15

2.05 m

As the height of the abutment below the bed block is more than 3.0m, hence the impact multying factor is zero.

Therefore there is no change in Live load reaction.

Total Reaction = 485.72 kN

Length of abutment = 1.92 mReaction per m run [R] = 252.98 kN/m

As the span is less than 8m, no frictional forces due to temperature variation are considered.

Horizontal force due to braking effect 'Bk'

Bk = [20/1000] x [11.4+11.4+2.7]

= 0.51 t

Braking force felt by each bearing = 0.51/2

= 0.26 kN

Force per m run = 0.26/4.92

= 0.052 t

= 0.52 kN


8/11/2019 UT Design (14.11.05)


CALCULATION OF STRESSES IN CONCRETE:

Taking moments about 'A', i.e., + 17.20



1 R As calculated 252.98 - 2.645 669.13

2 W1 0.5 x 0.3 x 25 3.75 - 2.645 9.923 W2 0.5 x2.985 x24 35.82 - 2.645 94.74

4 W3 0.5 x 3.685 x 24 44.22 - 2.145 94.85

5 W4 0.5 x 1.895 x 3.685 x 24 83.80 - 1.263 105.84

6 W5 0.5 x 1.895 x 3.685 x 21 73.32 - 0.632 46.34

7 Pv0.0384 x 3.685 x 21 10.95 - - -

8 Ph0.134 x 3.685 x 21 - 38.21 1.424 54.41

9 Bt As calculated - 0.52 3.085 1.60

ΣV 504.84 ΣM 1076.83

L .A = 2.13 m

Ecentricity = 2.13 - 3.095/2

= 0.743 m

Allowable (e) 3.095/6.00 = 0.516 m

> 0.743 m

Hence no tensionMax and Min Stresses = 427.13/3.9 x [1± (6 x 0.34)/3.9]


= 49.03 kN/mm2(Min)

Taking moments about 'A', i.e., + 15.84



1 R As calculated 252.98 - 3.48 880.36

2 W1 0.5 x 0.3 x 25 3.75 - 3.48 13.05

3 W2 0.5 x4.345 x24 52.14 - 3.48 181.45

4 W3 0.5 x 0.27 x 1.36 x 24 4.41 - 3.82 16.83

5 W4 0.5 x 5.12 x 24 61.44 - 2.98 183.09

6 W5 0.5 x 2.73 x 5.12 x 24 167.73 - 1.82 305.27

7 W6 0.5 x 2.73 x 5.12 x 21 146.76 0.91 133.56

8 Pv0.0384 x 5.12 x 21 21.14 - - -

9 Ph 0.134 x 5.12 x 21 - 73.77 1.968 145.17

10 Bt As calculated - 0.52 4.145 2.16

ΣV 710.35 ΣM 1860.94

Force (kN) Lever Arm

m

Moment

kN - m

Lever Arm

m

Moment

kN - m

Force (kN)


8/11/2019 UT Design (14.11.05)


L .A = 2.62 m

Ecentricity = 2.62 - 4/2

= 0.620 m

Allowable (e) 4/6.00 = 0.667 m

> 0.620 m

Hence no tension

Max and Min Stresses = 710.35/4 x [1± (6 x 0.62)/4.0]

[Compressive] = 342.68 kN/mm2 (Max)

= 12.50 kN/mm2(Min)

CALCULATION OF STRESSES ON SOIL

Taking moments about 'C'



1 R As calculated 252.98 - 3.78 956.26

2 W1 0.5 x 0.3 x 25 3.75 - 3.78 14.18

3 W2 0.5 x4.345 x24 52.14 - 3.78 197.09

4 W3 0.5 x 0.27 x 1.36 x 24 4.41 - 4.12 18.15

5 W4 0.5 x 5.12 x 24 61.44 - 3.28 201.52

6 W5 0.5 x 2.73 x 5.12 x 24 167.73 - 2.12 355.59

7 W6 0.5 x 2.73 x 5.12 x 21 146.76 1.21 177.59

8 W7 0.3 x 5.12 x 21 32.26 0.15 4.84

9 W8 4.9 x 0.6 x 24 70.56 2.45 172.87

10 Pv0.0384 x 5.12 x 21 21.14 - - -

11 Ph0.134 x 5.12 x 21 - 73.77 1.968 145.17

12 Bt As calculated - 0.52 4.145 2.16

ΣV 813.17 ΣM 2245.41

L .A = 2.76 m


= 0.310 m

Allowable (e) 4.9/6.00 = 0.817 m

> 0.310 m

Hence no tension



= 102.96 kN/mm

2

(Min)

Force (kN) Lever Arm

m

Moment

kN - m


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8/11/2019 UT Design (14.11.05)


L .A = 2.06 m


= 0.110 m

Allowable (e) 3.9/6.00 = 0.650 m

> 0.110 m

Hence no tension


[Compressive] = 141.60 kN/mm2 (Max)

= 100.61 kN/mm2(Min)

UT Design (14.11.05)

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