Box Culvert Design

DESIGN OF BOX TYPE CULVERT 1 2 3 4 5 6 6

In side diamentions Super imposed load Live load Wieght of soil Angle of repose Nominal cover top/bottom Cocrete

7 Steel 1 Solution

Mσcbc Fy

4 x 12000 45000 18000 30 50 20 7 415

3.05 m N/m3 N/m2 N/m2 Degree mm

wt. of water Nominal cover Side wt. of concrete m Out side σst water side side σst

N/m2

N/m3

9800 50 25000 13 190 150

mm N/m3 N/m2 N/m2

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 Vertical wall thicness Effective slab span Effective Height of wall

400 400 4 3.05

mm mm + 0.4 + 0.4

= = = =

0.4 0.4 4.4 3.45

m m m m

2 Case 1 : Dead and live load from out side of while no water pressure from inside. 10000 Self weight og top slab = 0.4 x 1x 1 x 25000 = 57000 Live load and dead load = 45000 + 12000 = 67000 Total load on top = 34500 Weight of side wall = 3.45 x 0.4 x 25000 = ∴ Upward soil reaction at base = ( 67000 x 4.4 )+( 2 x 34500 )= 82681.8 4.4 Ka =

1 - sin 30 1 + sin 30

=

11+

0.5 0.5

1 3

0.5 = 1.5

=

Latral pressure due to dead load and live load = Pv x Ka 19000 p = 57000 x 0.333 = Latral pressure due to soil Ka x w x h = 0.333 x 18000 h = 6000 Hence total pressure = 19000 + 6000 h 19000 N/m2 Latral presure intencity at top =

N/m2 N/m2 N/m2 N/m N/m2

= 0.33

∴

Latral pressure intencity at bottom

= 19000 +

6000 x w=

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.

N/m2 h

39700

3.45 =

N/m2

67000 N/m2 19000

19000 A

E

B

4.40

h 19000

3.45

6000 h D

F

C 19000

39700 w

=

82682 N/m2

Fig 1

20700

Distribution factore for AD and DA=

1 = 1+1/2

Fix end moments will be as under :

MFAB=

wL2 12

Mfdc= MFAD = + MFAD = + MFDA = MFDA = -

pL2 12 19000

+ x 12

pL2 12 19000 x 12

WL 15 3.45 WL 15 3.45

2/3

+

Distribution factore for AB and DC=

=

3.45 2=

1/3

-66456 N - m

82682 x wL2 3.45 2= 82010 N - m = 12 12 Where W is the total tringular earth pressure.

2

+ 20700 x 2

2-

67000 x 12

1/2 = 1+1/2

- 20700 x 2

3.45

3.45

x

3.45 = 15

x

3.45 = 10

27059

-18846

N-m

-12319

=

-31165

The Moment distribution is carried out as illustrate in table

Fixed End Moments Member

DC 82010.03

DA -31165

AD 27059

AB -66456

The moment distribution carried out as per table 1 for case 1 D A Joint DC DA AD AB Member 0.33 0.67 0.67 0.33 Distribution factore -66456 82010.03 -31165 27059 Fix end moment -16948 -33897 13132 26264 Balance 13132 -16948 Carry over 5649 -4377 -8755 11299 balance 5649 -4377 Carry over -1883 -3766 2918 1459 balance 1459 -1883 Carry over -973 1255 -486 628 balance 628 -486 Carry over -418 -209 324 162 balance -209 162 Carry over 139 70 -54 -108 balance -54 70 Carry over -23 36 18 -46 balance 18 -23 Carry over 8 -6 -12 15 balance 58023 -58023 45330 -45330 Final moment

45330

67000

115575

115575

54096 19000

45330

116810

A

A

45330

1.725 ##### 3.45 m

58023

1.725

20700 D 39700

D 75044

58023

181900

142067

181900

75044

82682

Fig 2

67000 N/m2. For horizontal slab AB, carrying UDL @ 67000 x 3.45 = 115575 N/m2 Vertical reactionat a and B = 0.5 x 2 Similarly, for the Bottom slab DC carrying U.D.L.loads @ 82682 N/m 82681.8 x 4.40 = 181900 Vertical reaction at D and C = 0.5 x 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 1/2 20700 + x x 4.40 x 4.40 x 1/3 -12693 183920 + 66792 -ha x 4.40 + + From which, ha = 54096 Hence , hd =( 19000 + 39700 )x 4.40 - 54096 = 75044 N 2

45330

54096

x 4.40 2= 162140 N-m 8 162140 45330 116810 N-m Net B.M. at E = = 82681.8182 x 4.40 2= Similarly, free B.M. at F = 200090 N -m 8 200090 58023 142067 N-m Net B.M. at F = = For vertical member AD , Simply supported B.M. At mid span 19000 x 4.40 2+ 1/16 x 20700 x 4.40 2= 71027 Simply supporetd at mid sapn = 8 58023 + 45330 - 71027 = -19351 N-m Net B.M. = = 51677 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 = 33810 N/m2 At D, is acting, in addition to the pressure considered in case 1. The various pressures are w= 67000 N/m2 marked in fig 3 .The vertical walls will thus be 19000 19000 19000 subjected to a net latral pressure of 2 13110 Itensity = 19000 N/m At the Top A E B And = 39700 33810 4.40 = 5890 N/m2 at the bottom 3.45

F

D

C

39700

39700 w

2

Fix end moments will be as under :

wL 12

MFAB=

MFAD = +

pL2 12 5890

+ x 12

WL 10 4.4

82682 N/m2

=

= 67000 x 12

Fig 3 4.40 2=

-108093 N - m

82682 x wL2 4.4 2= 133393 N - m = 12 12 Where W is the total tringular earth pressure.

Mfdc= MFAD = +

5890

Net latral pressure diagram

67000

Free B.M. at mid point E =

2

+ 13110 x 2

4.4

x

4.4 10

WL pL2 12 15 5890 x 4.4 2- 13110 x 4.4 4.4 MFDA = x 12 2 15 The moment distribution is carrired out as illustred in table.

=

22194

N-m

=

-17963 N -m

MFDA = -

Fixed End Moments Member

DC 133393.3

DA -17963

AD 22194

AB -108093

The moment distribution carried out as per table 1 for case 1 D A Joint DC DA AD AB Member 0.33 0.67 0.67 0.33 Distribution factore 133393.3 -17963 22194 -108093 Fix end moment -38477 -76954 57266 28633 Balance 28633 -38477 Carry over -9544 -19089 25651 12826 balance 12826 -9544 Carry over -4275 -8550 6363 3181 balance

61460

67000

61460

20044

115575

19000

115575 61460

38224

20044

A

A

61460

1.725 51903 3.45

79378

1.725

-1060

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

-2121 1425 -950 353 -236 158 -106 39 -26 -79378

-475 -118 -53 -13 79378

2850 -1060 707 -475 317 -118 79 -53 35 61460

1425

22892

353

43637

79378

142626

142626

22892

158 82682

39

Fig 4

18 -61460

67000 N/m2. For horizontal slab AB, carrying UDL @ 67000 x 3.45 = 115575 N/m2 Vertical reactionat a and B = 0.5 x 2 Similarly, for the Bottom slab DC carrying U.D.L.loads @ 82682 N/m 82682 x 3.45 = 142626 Vertical reaction at D and C = 0.5 x 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 1/2 13110 + x x 3.45 x 3.45 x 2/3 -17918 35052.9 + 52013.9 -ha x 3.45 + + From which, ha = 20044 Hence , hd =( 5890 + 19000 )x 3.45 - 20044 = 22892 N 2 67000 x 3.45 2= Free B.M. at mid point E = 99684 N-m 8 99684 61460 38224 N-m Net B.M. at E = = 82682 x 3.45 2= Similarly, free B.M. at F = 123015 N -m 8 123015.043 79378 43637 N-m Net B.M. at F = = For vertical member AD , Simply supported B.M. At mid span 5890 x 3.45 2+ 1/16 x 13110 x 3.45 2= 18516 Simply supporetd at mid sapn = 8 79378 + 61460 Net B.M. = = 70419 - 18516 = 51903 N-m 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 . The top slab is subjected to a load of '= 67000 N/m2 w=

67000 N/m2

4000

4000 A

N/m

2

N/m

2

E

4000

B

4.40 3.45

Hence earth pressure at depth h is = 4000

+

6000

h

D

F

C

24700

Earth pressure intensity at top

=

4000

N/m2

24700 33810

w=

Net latral pressure diagram

and the bottom slab is subjected to a load Itensity = 82682 N/m2 Lateral pressure due to dead load = 12000 = 4000 1/3 x Lateral pressure due to soil = 1/3 x 18000 = 6000

13110

82682 N/m 33810

Fig 5 24700 N/m2 Earth pressure intensity at Bottom= 4000 + 6000 x 3.45 = 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

33810

Fix end moments will be as under :

wL2 12

MFAB=

MFAD = +

pL2 12 4000

x 12

WL 15 4.4

67000 x 12

4.40 2=

-108093 N - m

82682 x wL2 4.4 2= 133393 N - m = 12 12 Where W is the total tringular earth pressure.

Mfdc= MFAD = +

=

2

- 13110 x 2

4.4

x

4.4 15

WL 6453 8460 pL2 + 12 10 4.4 4000 x 4.4 2- 13110 x 4.4 MFDA = x 12 2 10 The moment distribution is carrired out as illustred in table.

=

-2007

N-m

=

6237

N -m

MFDA = -

Fixed End Moments Member

DC 133393.3

DA 6237

AD -2007

AB -108093

The moment distribution carried out as per table 1 for case 1 D A Joint DC DA AD AB Member 0.33 0.67 0.67 0.33 Distribution factore 133393.3 6237 -2007 -108093 Fix end moment -46543 -93087 73400 36700 Balance 36700 -46543 Carry over -12233 -24467 31029 15514 balance 15514 -12233 Carry over -5171 -10343 8156 4078 balance 4078 -5171 Carry over -1359 -2719 3448 1724 balance 1724 -1359 Carry over -575 -1149 906 453 balance 453 -575 Carry over -151 -302 383 192 balance 192 -151 Carry over -64 -128 101 50 balance 50 -64 Carry over -17 -34 43 21 balance 67280 -67280 49361 -49361 Final moment

49361

67000

49361

= 115575

115575 49361

4000 A 49361

112779 A 1.725

64504 3.45

67280 0D 13110

1.725

D 67280

142626 5833

142626

82682 Fig 4

67000 N/m2. For horizontal slab AB, carrying UDL @ 67000 x 3.45 = 115575 N Vertical reactionat a and B = 0.5 x 2 Similarly, for the Bottom slab DC carrying U.D.L.loads @ 82682 N/m 82682 x 3.45 = 142626 Vertical reaction at D and C = 0.5 x 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 1/2 13110 x x 3.45 x 3.45 x 1/3 -17919 23805 26007 -ha x 3.45 + + 5833 From which, ha = Hence , hd =( 13110 x 3.45 )- 4000 x 3.45 - 5833 = 2981.8

Free B.M. at mid point E =

67000

2 x 8

4.40

2

=

162140

N-m

132810

49361 112779 N-m = 2 x 4.40 = Similarly, free B.M. at F = 200090 N -m 8 200090 67280 132810 N-m Net B.M. at F = = For vertical member AD , Simply supported B.M. At mid span 4000 x 4.40 2+ 1/16 x 13110 x 4.40 2= 6183.1 Simply supporetd at mid sapn = 8 67280 + 49361 Net B.M. = = 58321 + 6183 = 64504 N-m 2 Net B.M. at E =

162140 82682

5 Design of top slab :

Case (i) (II) (II)

Mid section The top slab is subjected to following values of B.M. and direct force B.M. at Center (E) B.M. at ends (A) Direct force (ha) 116810 45330 54096 38224 61460 20044 112779 49361 5833

116810 N -m The section will be design for maximum B.M. = for water side force wt. of concrete = 25000 N/m3 = 150 N/mm2 σst = 2 wt of water = 9800 N/mm2 7 N/mm = σcbc = for water side force m = 13 m*c 13 x 7 = 0.378 K = 0.378 k= = 13 x 7 + 150 m*c+σst 0.874 j=1-k/3 = 1 0.378 / 3 = 0.874 J = 1.155 R=1/2xc x j x k x 7 x 0.87 x 0.378 = 1.155 R = = 0.5 Provide over all thickness = 400 mm so effective thicknesss = 350 mm 141516561 116810000 1000 x 350 2= > O.K. Mr = R . B .D2 = 1.155 x 116810000 = 2545 mm2 BMx100/σstxjxD= Ast = 150 x 0.874 x 350 3.14 x 20 x 20 3.14xdia2 A = = using 20 mm Φ bars = 314 mm2 4 x100 4 Spacing of Bars = Ax1000/Ast 314 x 1000 / 2545 = 123 say = 120 mm Hence Provided 20 120 mm c/c mm Φ Bars @ Acual Ast provided 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 % Area of distributionn steel = 0.3 = 0.21 450 100 Ast = 0.21 x 400 x 10 = 858 mm2 area on each face= # mm2 2 3.14 x 8 x 8 3.14xdia using 8 mm Φ bars A = = = 50 mm2 4 x100 4 Spacing of Bars = Ax1000/Ast = 50 x 1000 / 429 = 117 say = 110 mm mm Φ Bars @ 110 mm c/c on each face Hence Provided 8 Section at supports :61460 N-m. There is direct compression of 54096 N also. Maximum B.M.= But it effect is not considered because the slab is actually reinforced both at top and bottom . σst = Since steel is at top 190 N/mm2 concrete M 20 k = 0.324 J = 0.892 R = 1.011 61460000 = = 1037 ∴ Ast mm2 190 x 0.892 x 350

Area

2617 / available from the bars bentup from the middle section = 2 = 1308 mm2 1037 < 1308 Hence these bars will serve the purpose. However, provide 8 mm dia. Additional bars @ 200 mm c/c

6 Design of bottom slab: The bottom slab has the following value of B.M. and direct force. 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 The section will be design for maximum B.M. = for water side force wt. of concrete = = 150 N/mm2 σst = 2 wt of water = 7 = N/mm σcbc = m = 13 m*c 13 x 7 k= = = 13 x 7 + 150 m*c+σst j=1-k/3 = 1 0.378 / 3 = R=1/2xc x j x k x 7 x 0.87 x 0.378 = = 0.5 142067000 = 351 mm ∴ d = 1000 x 1.155 Provide thickness of bottom slab D= 410 mm so that d = 142067000 = BMx100/σstxjxD= Ast = 150 x 0.874 x 360 3.14xdia2 = 4 x100 x 1000 / 3010 =

142067 N -m 25000 N/m3 9800 N/mm2 for water side force

0.378

K

=

0.378

0.874 1.155

J R

= =

0.874 1.155

D

=

401

mm

360

mm 3010 mm2

20 x 20 = 314 mm2 4 Spacing of Bars = Ax1000/Ast 314 104 say = 100 mm Hence Provided 20 100 mm c/c mm Φ Bars @ Acual Ast provided 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 = 0.21 % Area of distributionn steel = 0.3 450 100 Ast 434 mm2 = 0.21 x 410 x 10 = 867 mm2 area on each face= 2 3.14 x 8 x 8 3.14xdia using 8 mm bars A = = = 50 mm2 4 x100 4 Spacing of Bars = Ax1000/Ast = 50 x 1000 / 434 = 116 say = 110 mm mm Φ Bars @ Hence Provided 110 mm c/c on each face 8 Section at supports :79378 N-m. There is direct compression of 75044 N also. Maximum B.M.= But it effect is not considered because the slab is actually reinforced both at top and bottom . σst = Since steel is at top 190 N/mm2 concrete M 20 k = 0.324 J = 0.892 R = 1.011 79378000 = = 1301 ∴ Ast mm2 190 x 0.892 x 360 3140 / Area available from the bars bentup from the middle section = 2 = 1570 mm2 using

20

1301 <

mm bars

1570

A

=

Hence these bars will serve the purpose. However, provide 8 mm dia. Additional bars @ 200 mm c/c -269 mm2 3.14 x 8 x 8 3.14xdia A = = = 50 mm2 4 x100 4 50 x 1000 / -269 = -187 say = -180 mm -180 mm c/c throught out the slab, at its bottom.

Additional reinforcemet required 2

using 8

3.14 x

mm bars

Spacing of Bars = Ax1000/Ast = Hence Provided 8 mm Φ Bars @

=

7 Design of side wall: The side wall has the following value of B.M. and direct force. B.M. at Center (F) B.M. at ends (D) Direct force (ha) -19351 58023 181900 51903 79378 142626 64504 67280 142626

Case (i) (II) (II)

The section will be design for maximum B.M. = 79378 Eccentricity = proposed thickness of side wall '=

79378

N -m, 1000

x 181900

and direct force =

400 mm ∴ e/D thickness of side wall is OK

Let us reinforce the section with

437

mm

437 /

400 =

1.09 <

N

1.5 OK

300 mm c/c provided on both faces, as shown and D = mm 400 mm 1000 3.14 x 20 x 20 Asc = Ast = x = 1047 mm2 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) n 3 3 D = e+ - dt D - dt - n 2 b n + (m -1) Asc n - dc - m Ast n n

in fig xxx . With cover of

1000 n 2 1000 n+ 2

20

181900

=

mm Φ bars @

50

50

+

12

x

n

-

50

-

+

n

-

50

x

n

-

50

-

n2

+ -1256000

1047 x n

12

x

500 n

350

-

n 3

500 n+

12560 n

x

175000 n

-

167

500 n

+

12560

multiply by n 175000 n2 500 n2

+

167 n3 12560

+ -1256000 n n- 628000 -

175000 n2 293500 n2

+

167 n3 15359833

+ n

-118500 n2

-

14103833

n

-

-3101325667

= 167 n3

n =

-

18607954

= n3

1047 x n

84623 -711 n2 + Solwing this trial and error we get, n = (

500

x x(

-

112.86 + 400 -

13

n 350

n 4762333

-

-1256000 n -

1047 x 400 n

x

-100

50

-n

= 437 + 150 -

n

-62800000

-

n

-

-1256000

13607 x n

628000

x

n

50

-

∴ c'

-

n 3

400

or

n -

=

587

+ 13607

-62800000 4762333 + 13607 n

-62800000 3164125667

=

=

587

0

112.86 mm 12 x 1047 ( ##### 112.86 50 -

##### )

50

)-

13 x 1047 112.86

or ∴ c' = Also stress in steel t

= =

56432 + 181900 34840 m c' n

111.3 x

5.22

= (D-dc-n)

142.6 N/mm2

62.86 -

=

<

237.14 =

120.6 x

7

N/mm2

13 x 5.22 x( 112.86

400

34840

Stress is less than permissiable -

50

< 190 N/mm2 O.K. Stress in steel is less than permissiable Hence section is O.K.

- 112.86

)