Box Culvert Design

1 In side diamentions 4 x 3.05 m

2 Super imposed load N/m3

3 Live load N/m2

4 Wieght of soil N/m2 N/m3

5 Angle of repose Degree

6 Nominal cover top/bottom mm Nominal cover Side 50 mm

6 Cocrete M - N/m3

σ_{cbc N/m}2

7 Steel Fy N/m2

N/m2 1 Solution Genral

For the purpose of design , one metre length of the box is considered. The analysis is done for the following cases.

(I) Live load, dead load and earth prssure acting , with no water pressure from inside.

(II) Live and dead load on top and earth pressure acting from out side, and water pressure acting from inside, with no live load on sides

(III) Dead load and earth pressure acting from out side and water pressure from in side.

Let the thicness of Horizontal slab 400 mm = 0.4 m

Vertical wall thicness 400 mm = 0.4 m

Effective slab span 4 + 0.4 = 4.4 m

Effective Height of wall 3.05 + 0.4 = 3.45 m

2 Case 1 : Dead and live load from out side of while no water pressure from inside. 150 DESIGN OF BOX TYPE CULVERT

12000 45000 18000 30 20 9800 wt. of water 50 wt. of concrete 7 415 25000 13 190 m Out side

σ

_st

water side side

σ

_st

2 Case 1 : Dead and live load from out side of while no water pressure from inside.

Self weight og top slab = 0.4 x 1 x 1 x 25000 = N/m2

Live load and dead load = 45000 + 12000 = N/m2

Total load on top = N/m2

Weight of side wall = 3.45 x 0.4 x 25000 = N/m

∴ Upward soil reaction at base = ( 67000 x 4.4 )+( 2 x 34500

1 - sin 30 1 - 0.5

0.5

1

1 + sin 30 1 + 0.5

1.5

3

∴ Latral pressure due to dead load and live load = Pv x Ka

p = 57000 x 0.333 = N/m2

Latral pressure due to soil Ka x w x h = 0.333 x 18000 h = h

Hence total pressure = 19000 + 6000 h

Latral presure intencity at top = N/m2

Latral pressure intencity at bottom = 19000 + 6000 x

3.45

= N/m2 w = 67000 N/m2 19000 A E B h 3.45 6000 h D F C 39700 w = 82682 N/m2 Fig 1 20700 19000 4.40 Fig 1 show the box culvert

frame ABC D, along with the external loads, Due to symmetry, half of frame (i.e. AEFD ) of box culvert is considered for moment distribution. Since all the members have uniform thickness, and uniform diamentions, the relative stiffness K for AD will be equal to 1 while the relative stiffness for AE and DF will be 1/2.

19000

6000

19000 39700 19000 19000 10000 57000 67000 34500 4.4 )= 82681.8 N/m2 Ka = = = = = 0.33

1 1/2 1+1/2 1+1/2 wL2 67000 x 3.452₌ 12 wL2 82682 x

3.45

2₌ 12

pL2 WL Where W is the total tringular earth pressure.

12 15 19000 x 3.45 2_{+ 20700 x 3.45}

_3.45

15

pL2 WL 12 15 19000 x 3.45 2-_{- 20700 x 3.45}

_3.45

10

The Moment distribution is carried out as illustrate in table

Fixed End Moments

Member

The moment distribution carried out as per table 1 for

case 1

Joint

Member

Distribution factore

Fix end moment A A

Balance Carry over 115575 45330 116810 115575 DA -31165 AD 27059 45330 19000 1.725 -16948 -33897 13132

Distribution factore for AD and DA= = 2/3 Distribution factore for AB and DC= = 1/3

Fix end moments will be as under : =

M_fdc= + = -66456 N - m M_FAB= N-m = 12 12 82010 N - m x MFAD = + MFAD = + 27059 + 82010.03 MFDA = - -12 2 2 MFDA = -12 0.33 82010.03 D DC A DC AB 0.67 -31165 DA = -31165 -12319 -18846 = x AD 0.67 27059 26264 -16948 AB -66456 54096 13132 -66456 54096 67000 0.33 45330 45330 Carry over balance 3.45 m Carry over balance Carry over D D balance 75044 Carry over balance Carry over balance Carry over balance Carry over balance Final moment

For horizontal slab AB, carrying UDL @ N/m2.

Vertical reactionat a and B = 0.5 x x 3.45 = N/m2

Similarly, for the Bottom slab DC carrying U.D.L.loads @ 82682 N/m2

Vertical reaction at D and C = 0.5 x x 4.40 = N

The body diagram for various members, including loading, B.M. And reactions are shown in fig.2 For the vertical member AD, the horizontal reaction at A is found by taking moments at D.Thus

( -ha x 4.40 ) + 45330 - 58023 + 19000 x

4.40

x 4.40 x 1/2 + x x

4.40

x 4.40 x 1/3 -ha x 4.40 + + + From which, ha = Hence , hd =( 19000 + 39700 )x 4.40 - 54096 = N 2 67000 115575 58023 ##### 142067 181900 -6 15 1/2 45330 -45330 -46 -58023 18 -12 -23 75044 20700 1.725 -209 -486 58023 -23 -54 -108 70 628 162 70 162 324 13132 -8755 5649 1459 -1883 -4377 -3766 1459 -973 628 -418 -16948 11299 -4377 2918 -1883 1255 -486 -209 139 -54 67000 36 8 183920 75044 66792 5649 82682 39700 181900 82681.8 181900 Fig 2 18 20700 -12693 54096 58023

x 4.40 2₌ 8 Net B.M. at E = - = x 4.40 2 = 8 Net B.M. at F = - =

For vertical member AD , Simply supported B.M. At mid span x 4.40

8 + 2

3 Case 2 : Dead load and live load from out side and water pressure from inside.

In this case , water pressure having an intensity of zero at A and

9800

x 3.45 = N/m2

w = 67000 N/m2 Itensity = 19000 A E B And = 39700 - 33810 = 5890 D F C 5890 w = 82682 N/m2 3.45 19000 At D, is acting, in addition to the pressure

considered in case 1. The various pressures are marked in fig 3 .The vertical walls will thus be subjected to a net latral pressure of

N/m2 At the Top N/m2 at the bottom Similarly, free B.M. at F = N-m -82681.8182 200090 58023 45330 67000

Free B.M. at mid point E = 162140 N-m

162140 45330 116810 N e t la tr a l p re s s u re d ia g ra m N -m 200090 58023 142067 N-m 19000 20700 13110 N-m Net B.M. = = 51677

Simply supporetd at mid sapn =

71027 = 1/16 x x 4.40 2₌ 2₊ 71027 4.40 39700 33810 -19351 39700 19000 19000 N/m Fig 3 wL2 = 67000 x 4.402₌ 12 wL2 82682 x

4.4

2₌ 12

pL2 WL Where W is the total tringular earth pressure.

12 10 5890 x 4.4 2_{+ 13110 x 4.4}

_4.4

10

pL2 WL 12 15 5890 x 4.4 2_{- 13110 x 4.4}

_4.4

15

Fixed End Moments

Member

The moment distribution carried out as per table 1 for case 1 Joint

Member

Distribution factore 20044

Fix end moment A A

Balance Carry over balance 3.45 Carry over balance 51903 1.725 -4275 -8550 6363 3181 79378 25651 12826 12826 -9544 61460 -9544 -19089 1.725 28633 -38477 133393.3 -17963 22194 -108093 -38477 -76954 57266 28633 61460 0.33 0.67 0.67 0.33 19000 38224 DC D A 115575 115575 DC DA AD AB 133393.3 -17963 22194 -108093 20044 DA AD 61460

The moment distribution is carrired out as illustred in table.

AB 61460 MFDA = - x = 12 2 -17963 N -m = 22194 N-m 12 2 MFDA = - -MFAD = + + MFAD = + Mfdc= = 133393 N - m 12 Fix end moments will be as under : M_FAB=

67000 -108093 N - m 12

balance Carry over balance Carry over balance Carry over balance Carry over balance Final moment

For horizontal slab AB, carrying UDL @ N/m2.

Vertical reactionat a and B = 0.5 x x 3.45 = N/m2

Similarly, for the Bottom slab DC carrying U.D.L.loads @ 82682 N/m2

Vertical reaction at D and C = 0.5 x x 3.45 = N

The body diagram for various members, including loading, B.M. And reactions are shown in fig.3 For the vertical member AD, the horizontal reaction at A is found by taking moments at D.Thus

( -ha x 3.45 ) + 61460 - 79378 + 5890 x

3.45

x 3.45 x 1/2 + x x

3.45

x 3.45 x 2/3 -ha x 3.45 + + + From which, ha = Hence , hd =( 5890 + 19000 )x 3.45 - 20044 = N 2 x 3.45 2₌ 8 22892 22892

Free B.M. at mid point E = 67000 99684 N-m

1/2 13110 -17918 35052.9 52013.9 20044 67000 67000 115575 82682 142626 79378 -79378 61460 -61460 39 -53 -13 -26 35 18 82682 -53 -106 79 39 Fig 4 158 -118 -118 -236 317 158 353 -475 22892 -475 -950 707 353 142626 142626 43637 1425 -1060 -1060 -2121 2850 1425 79378 8 Net B.M. at E = - = x 3.45 2 = 8 Net B.M. at F = - =

For vertical member AD , Simply supported B.M. At mid span x 3.45

8 + 2

4 Case 3 : Dead load and live load on top water pressure from inside no live load on side. in this case, it is assume that there is no latral oressure due to live load . As before .

N/m2

and the bottom slab is subjected to a load w = 67000 N/m2

Itensity = 82682 _N/m2

A E B

1/3 x 12000 = N/m2

1/3 x 18000 = 6000 _N/m2

4000 + 6000 h D F C

Earth pressure intensity at top = 4000 33810w= 82682 N/m 33810 Fig 5

Earth pressure intensity at Bottom= 4000 + 6000 x 3.45 = N/m2 In addition to these, the vertical wall lslab subjectednto water pressure of intensity ZERO at top and N/m2 at Bottom, acting from inside . The lateral pressure on vertical walls Is shown in fig 5 and 6

N/m2 24700 33810 N e t la tr a l p re s s u re d ia g ra m 13110 The top slab is subjected to a load of '=

3.45 67000

Lateral pressure due to dead load = 4000 Lateral pressure due to soil =

Net B.M. = 79378 70419

Hence earth pressure at depth h is =

51903 N-m 4000 4000 2_{= 18516} 61460 = - 18516 = N-m

Simply supporetd at mid sapn = 5890

2₊

1/16 x 13110 x 3.45 Similarly, free B.M. at F = 82682 123015 N -m

123015.043 79378 43637

Free B.M. at mid point E = 99684 N-m

99684 61460 38224 N-m

4000

4.40

wL2 67000 x 4.402₌ 12

wL2 82682 x

4.4

2₌ 12

pL2 WL Where W is the total tringular earth pressure.

12 15 4000 x 4.4 2_{- 13110 x 4.4}

_4.4

15

pL2 WL 6453 - 8460 12 10 4000 x 4.4 2_{- 13110 x 4.4}

_4.4

10

Fixed End Moments

Member

The moment distribution carried out as per table 1 for case 1 Joint

Member

Distribution factore

Fix end moment A A

Balance Carry over balance 3.45 Carry over 31029 15514 15514 -12233 49361 1.725 36700 -46543 64504 -12233 -24467 133393.3 6237 -2007 -108093 -46543 -93087 73400 36700 49361 0.33 0.67 0.67 0.33 4000 112779 D A 115575 115575 DC DA AD AB 67000 49361 133393.3 6237 -2007 -108093 =

The moment distribution is carrired out as illustred in table.

DC DA AD AB 49361 = 6237 N -m 12 2 MFDA = - + MFDA = - x x = -2007 N-m 12 2 M_FAD = + -MFAD = + 12 M_fdc= = 133393 N - m 12

Fix end moments will be as under : M_FAB= = -108093 N - m

Carry over balance Carry over D D balance Carry over balance Carry over balance Carry over balance Carry over balance Final moment

For horizontal slab AB, carrying UDL @ N/m2.

Vertical reactionat a and B = 0.5 x x 3.45 = N

Similarly, for the Bottom slab DC carrying U.D.L.loads @ 82682 N/m2

Vertical reaction at D and C = 0.5 x x 3.45 = N

The body diagram for various members, including loading, B.M. And reactions are shown in fig.6 For the vertical member AD, the horizontal reaction at A is found by taking moments at D.Thus

( ha x 3.45 ) + 49361 - 67280 + 4000 x

3.45

x 3.45 x 1/2 - x x

3.45

x 3.45 x 1/3 -ha x 3.45 + + -From which, ha = Hence , hd =( 13110 x 3.45 )- 4000 x 3.45 - 5833 = 2 x 4.40 2₌ 8

Free B.M. at mid point E = 67000 162140 N-m

1/2 13110 -17919 23805 26007 5833 67000 67000 115575 82682 142626 67280 -67280 49361 -49361 50 -64 -17 -34 43 21 82682 -64 -128 101 50 Fig 4 192 -151 -151 -302 383 192 453 -575 5833 -575 -1149 906 453 142626 142626 132810 1724 -1359 -1359 -2719 3448 1724 13110 67280 1.725 4078 -5171 0 -5171 -10343 8156 4078 67280 15514 -12233 2981.8

Net B.M. at E = - = x 4.40 2

= 8

Net B.M. at F = - =

For vertical member AD , Simply supported B.M. At mid span x 4.40

8 + 2

5 Design of top slab :

Mid section

The top slab is subjected to following values of B.M. and direct force

The section will be design for maximum B.M. = N -m for water side force

σ

_{st =}

=

150 N/mm2 = 25000 N/m3

σ

_{cbc =}

=

7 N/mm2 = 9800 N/mm2

m

=

13 for water side force

x

13 x 7 + 150 =

0.378

K = 0.378

m*c+σst

Case B.M. at ends (A)

116810 wt. of concrete

wt of water

B.M. at Center (E) Direct force (ha)

116810 45330 54096 k= m*c = 13 7 38224 (i) 49361 112779 20044 5833 (II) (II) 61460

Simply supporetd at mid sapn = 4000

2₊ 1/16 x 13110 4.40 64504 N-m = N-m N-m x 2_{= 6183.1} Net B.M. = 67280 49361 = 58321 + 6183 Similarly, free B.M. at F = 82682 200090 N -m 200090 67280 132810 162140 49361 112779 13 x 7 + 150 j=1-k/3

=

1 - 0.378 / 3 = J = R=1/2xc x j x k

=

0.5 x 7 x 0.87 x 0.378 = R = =

400

mm so effective thicknesss =

350

mm Mr = R . B .D2

=

1.155 x 1000 x 350 2 = > O.K. = 2545 mm2 150 x 0.874 x 350 3.14xdia2 3.14 x 20 x 20 4 x100 4

Spacing of Bars = Ax1000/Ast 314 x 1000 / 2545 = 123 say = mm

mm Φ Bars @ mm c/c

1000 x 314 / 120 = 2617 mm2 Bend half bars up near support at distance of L/5 = 4.40 / 5 = 0.90 m

0.1 x( 400 - 100 %

450 - 100

Ast = 0.21 x 400 x 10 = 858 mm2 area on each face= # mm2

3.14xdia2 3.14 x 8 x 8

4 x100 4

Spacing of Bars = 50 x 1000 / 429 = 117 say = mm

Hence Provided

8

mm c/c on each face

Section at supports

:-Maximum B.M.= N-m. There is direct compression of N also. But it effect is not considered because the slab is actually reinforced both at top and bottom .

Since steel is at top

σ

st= 190 N/mm2 concrete M 20

k = 0.324 J = 0.892 R = 1.011

190 x 0.892 x 350 Provide over all thickness

141516561 116810000

120

120

= 0.21 314 mm2 mm Φ bars A = = Ast = BMx100/σstxjxD= 116810000 using 20 Ax1000/Ast =

20

using 8 mm Φ bars

0.874

0.874

1.155

1.155 =

Area of distributionn steel = 0.3 -Acual Ast provided

Hence Provided =

0.378

K = 0.378

m*c+σst

k= =

120

mm2

110

110

mm Φ Bars @ = = 50 = 1037 mm2 = = A 61460 54096 ∴ Ast = 61460000

Area available from the bars bentup from the middle section = / 2 = 1308 mm2 1037 < 1308

6 Design of bottom slab:

The bottom slab has the following value of B.M. and direct force.

The section will be design for maximum B.M. = N -m for water side force

σ

_{st =}

=

150 N/mm2 = 25000 N/m3

σ

_{cbc =}

=

7 N/mm2 = 9800 N/mm2

m

=

13 for water side force

x 13 x 7 + 150 j=1-k/3

=

1 - 0.378 / 3 = J = R=1/2xc x j x k

=

0.5 x 7 x 0.87 x 0.378 = R = 1000 x 1.155

410

mm so that d = mm = 3010 mm2 150 x 0.874 x 360 3.14xdia2 3.14 x 20 x 20

Provide thickness of bottom slab D=

2617 2617

Hence these bars will serve the purpose. However, provide 8 mm dia. Additional bars @ 200 mm c/c

Case B.M. at Center (F) B.M. at ends (D) Direct force (ha)

(i) 142067 58023 75044 (II) 43637 79378 22892 (II) 132810 67280 2982 142067 wt. of concrete wt of water k= m*c = 13 7 =

0.378

K = 0.378

m*c+σst

0.874

0.874

1.155

1.155 401 mm

D

=

360

Ast = BMx100/σstxjxD= 142067000 ∴ ∴ ∴ ∴ _d

₌

142067000 = 351 mm 3.14xdia2 3.14 x 20 x 20 4 x100 4

Spacing of Bars = Ax1000/Ast 314 x 1000 / 3010 = 104 say = mm

mm Φ Bars @ mm c/c

1000 x 314 / 100 = 3140 mm2 Bend half bars up near support at distance of L/5 = 4.40 / 5 = 0.90 m

0.1 x( 410 - 100 450 - 100

A_st = 0.21 x 410 x 10 = 867 mm2

area on each face= mm2

3.14xdia2 3.14 x 8 x 8

4 x100 4

Spacing of Bars = 50 x 1000 / 434 = 116 say = mm

Hence Provided

8

mm c/c on each face

Section at supports

:-Maximum B.M.= N-m. There is direct compression of N also. But it effect is not considered because the slab is actually reinforced both at top and bottom .

Since steel is at top

σ

st= 190 N/mm2 concrete M 20

k = 0.324 J = 0.892 R = 1.011

190 x 0.892 x 360

Area available from the bars bentup from the middle section = / 2 = 1570 mm2

Additional reinforcemet required = -269 mm2

3.14xdia2 3.14 x 8 x 8

4 x100 4

Spacing of Bars = 50 x 1000 / -269 = -187 say = mm

Hence Provided

8

mm Φ Bars @

-180

mm c/c throught out the slab, at its bottom.

mm2 Ax1000/Ast =

-180

using 8 mm bars A = = = 50

20

100

using 20 mm bars A = = 0.21 = = 314 mm2

100

% 434 using 8 mm bars A = = = 50 ∴ Ast = 79378000 = 1301 mm2 79378 Acual Ast provided

Area of distributionn steel = 0.3 -Hence Provided

3140

Hence these bars will serve the purpose. However, provide 8 mm dia. Additional bars @ 200 mm c/c mm2

110

110

Ax1000/Ast = mm Φ Bars @ 75044 1570 < 1301

7 Design of side wall:

The side wall has the following value of B.M. and direct force.

The section will be design for maximum B.M. = N -m, and direct force = N x

proposed thickness of side wall '= 400 mm ∴ e / D 437 / 400 = 1.09 < 1.5 OK thickness of side wall is OK

20

mm Φ bars @

300

mm c/c provided on both faces, as shown in fig xxx . With cover of 50 mm = 400 mm

1000 3.14 x 20 x 20

300 4

The depth of N.A. is computed from following expression: b n D - dt - n + (m - 1)Asc 1 (n - dc)(D - dt- dc) 3 3 n D b n + (m -1) Asc n - dc D - dt - n 2 n n 1000 n n 1047 2 3 n or 400 - 50 - + n - x -100 437 x x 12 Let us reinforce the section with

mm Eccentricity = 79378 1000 181900 = 142626 79378 181900 (i) -19351 58023 181900 (II) 51903 142626

Case B.M. at Center (F) B.M. at ends (D) Direct force (ha)

Asc = Ast = 79378 67280 (II) 64504 = x 1047 mm2 - m Ast = e + - dt and D 50 2 3 n 1000 n+ 1047 1047 2 n n n 3 n 12560 13607 n n n n n multiply by n 175000 n2 - 167 n3 + n -500 n2 + n- - + 13607 n 175000 n2 - 167 n3 + n -293500 n2 + n --118500 n2 - n - = 167 n3 -711 n2 + n - = n3

Solwing this trial and error we get, n = mm

12 x 1047 13 x 1047 x ( 400 - 50 - ##### ) or -100 x 12 x - 13 x + 150 500 n 350 - -1256000 84623 18607954 14103833 -3101325667 x 3164125667 -+ 15359833 -587 -1256000 -62800000 = 12560 -1256000 628000 628000 + 12560 4762333 500 n + n --62800000 -437 x 350 -- 4762333 175000 n+ 500 167 n2 x n -400 - 50 - + 12 x x n - 50 = 587 = n -1256000 -62800000 n - 50 400 - 50 -- n 50 ∴ _c' = ( 500 x 112.86 + ##### -112.86 ( 50 ) - 112.86

112.86

0 13607 = 50 + n x x

or 56432 + 111.3 x 62.86 - 120.6 x = m c' 13 x 5.22 n =

142.6

N/mm2 <

190

N/mm2 O.K. = ∴ c' 181900 34840 34840 237.14 112.86 )

Stress in steel is less than permissiable Hence section is O.K. 112.86 x ( 400 - 50

-=

5.22

<

7

Stress is less than permissiable

Also stress in steel t = (D-dc-n) =