Box Culvert

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1

Box Culverts

Design Example

Design a reinforced concrete box culvert under a drain for the following data: Properties of the drain:

1 2

300 , 8 , 15 , 3.5 ,

side slopes 2 :1 , 75 and bed level 201

n Q cumec W m W m d m H V b m m        Properties of channel: 1 1 1

60 , 30 ,side slopes 2 :1 , water table in the region 200 , 2 , _b 5.5 , _b 3.5 , freeboard above channel water surface 0.4 .

Q cumec b m H V m

d m h m f m m

   

Bed reduced level (B.R.L.) = 200 m 3

2.4 /

c T m

 

Design the same transition type for both inlet and outlet. Draw the details of your design.

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2 Solution

Drain waterway

For the drain check the waterway

min 4.83 4.83 300 83.65 P Q m   

Applied wetted parameter _. 75 2 3.52



3.5 2



2 90.65 83.65 O.K. app P m m        Channel waterway

For the channel,



30 2 260



2 0.8823

V   m s

  

Maximum channel fluming is 40%, 0.4 30 12  m take 12.05m Let the channel waterway be reduced from 30 m to 12.05 m. Take 4 vents 2.75 1.9 meach

Figure 2: Box culvert vents.

60 2.87 3 OK 4 2.75 1.9 V   m s m s   2.87 0.66 1 OK 9.81 1.9 r F     Length of expansion 30 12.05 3 26.925 2 m     Length of contraction 30 12.05 2 17.95 2 m     1.9 m 2.75 m 12.75 m 12.05 m

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3 Pucca Floor

Length of u.s. pucca floor 17.95 8.975 say 9

2 m m

 

Length of u.s. pucca floor 3 26.925 20.20

4 m    Design of Transitions a. Expansion transition





30 , 12.05 , 26.925 c f f x f c c f c f f B B L B L B x B B B m B m L m       542.2 45 x B x   x (m) 0 5 10 15 20 26.925 Bx (m) 12.05 13.56 15.49 18.07 21.69 30.00 b. Contraction transition 30 , 12.05 , 17.95 c f f B  m B  m L  m 361.49 30 x B x   x (m) 0 5 10 15 17.95 Bx (m) 12.05 14.46 18.07 24.10 30.00 Uplift Pressures a. Seepage pressure Seepage head



201 3.5



2004.5 m

Total seepage path









1

3

0.98 2 1.9 0.35 0.35 1 33.95 20.2 22.61m

         

Depth of earth at inside edge0.6 75 0.005  0.98 m 1. At bottom of barrels 1.96 2.6 4.56 4.5 18.05 3.59 22.61 L h m      

Uplift at the base of the barrels 2 2 3.59 2.58 6.17 say 6.2 t m t m    2. At d.s. end of barrels 4.5 6.73 1.34 22.61 h    m

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4

3. At 5 m from d.s. end of floor 4.5 5

0.33 22.61 3

h   _{ } m

 

0.66 22.61 3

h  _ _ m

 

0.99 22.61 3

h  _ _ m

 

Figure 3: Seepage head at different points under the pucca floor. b. Static head

1. At bottom of barrels floor

Elevation of bottom of barrels floor201



0.98 0.35 1.9 0.35  



197.4 m Static head200 197.4 2.58 m

Weight of earth, water and concrete 3.5 00.98 2



0.35 0.35



2.4 0.35 1.9 5 2.4 12.75 7.76m of water 6.2mO.K.             2. At d.s. end of barrels

El. of end of barrels 201



0.6 0.35 1.9

 

75 33.95



0.005 197.61m

    _   _



Assume floor thickness 3.0 m

El. of lower point 197.61 3 194.61m   Static head200 194.61 5.39 

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Min. floor thickness 6.72 3.05

2.2 m

  this is larger than the assumed 3.0 m, Assume thickness3.1m, therefore,

El. of lower point197.61 3.1 194.51m  Static head200 194.51 5.49 m 

Total head5.49 1.33 6.82 m

Min. floor thickness 6.82 3.1 assumed thickness O.K.

2.2 m

  

3. At 5 m from d.s. end of pucca floor Assume thickness2.5 m

El. of lower point197.61 2.5 195.11m  Static head200 195.11 4.895 m  Total head4.895 0.33 5.22 m Min. floor thickness 5.22 2.37

2.2 m

  , this is smaller than the assumed 2.5 m thickness, Assume thickness2.3m

Total head4.69 0.33 5.02 m

Min. floor thickness 5.02 2.28 2.3 O.K.

2.2 m m

  

El. of lower point197.61 2.5 195.11m  Static head200 195.11 4.895 m  Total head4.895 0.66 5.55 m

Min. floor thickness 5.55 2.52 2.5 not O.K.

2.2 m m

  

Assume thickness2.6 m

El. of lower point 197.61 2.6 195.01m   Static head200 195.01 4.99 m 

Total head4.99 0.66 5.65m

2.2 m m

  

El. of lower point 197.61 3.0 194.61m   Static head200 194.61 5.39 m 

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6

Min. floor thickness 6.38 2.9 3.0 not O.K., revise

2.2 m m

  

Assume thickness2.9 m

Total head5.29 0.99 6.28m

2.2 m m

  

Figure 4: Floor thickness at different points.

Upstream pucca floor

Length of seepage path









1

3 0.6 2 0.35 1.9 0.35 1 23.55 9 14.65 m           Seepage head204.5 200 4.5m 1. U.S. end of barrels

23.55 0.6 2 2.6 11.65 3 4.5 3 0.92 14.65 L m h         Assume thickness = 0.8 m Total head0.8 0.92 1.72 m 

2.2 m m

  

2. At 5 m from u.s. end of floor 4.5 5

0.51 14.65 3

h    m

Assume floor thickness = 0.6 m Total head0.6 0.51 1.11m 

2.2 m m

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7

PROFILE

PLAN

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Structural Design

Number of barrels = 4

Size of barrels = 2.75 m×1.9 m

Bank level = 205.5 m

Drain high flood level = 204.5 m

Uplift at base of barrel = 6.2 t/m2 Unit weight of dry earth = 1.6 t/m3 Unit weight of saturated earth = 2.0 t/m3 Unit weight of submerged earth = 1.0 t/m3 Angle of internal friction in all conditions = 30º

Depth of earth cover = 5.5 m

Figure 6: A cross-section of the proposed box culvert. Dimensions are in meters.

Design

Depth of dry earth cover205.5 204.5 1m  Depth of saturated earth204.5 200 4.5m

Weight of dry and saturated earth 



1 1.6 4.5 2  



10.6t m3 Weight of top slab0.35 2.4 0.84t m2

Weight on top slab including its own weight10.6 0.84 11.44t m  2

Weight of barrels per meter length



12.75 2 5 1.9  



0.35 2.4 29.4t m Total dead load/meter of barrels29.48 10.6 12.75 164.55t m  

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Net vertical load acting on foundation164.55 79.05 85.5t m Pressure on foundation soil 85.5 6.70 2

12.75 t m

 

Pressure acting on the base slab= soil reaction + uplift6.7 6.2 12.9t m  2 Net upward pressure on the base slab 2

2 12.9 0.84 12.06 say 12.1 t m t m    Earth Pressure 1 sin 30 1 1 sin 30 3 p C      

Pressure at point (a)C_p  _d 1 C_p_s



204.5 199.825



_w



204.5 199.825



6.770t m2 Pressure at point (n) 6.77 1 1 2.25 1 2.25 9.77 2

3 t m

      

Figure 7: Loading on the culvert barrels.

Distribution Factors

At joints a, e, f and n

For ab, mn, ed and fg, distribution factor 2.25 2.25 0.42 2.25 3.1 5.35

  

 For an, ef, distribution factor 3.1 0.58

5.35   At joints b, c, d, g, h and m For ba 2.25 3.1 3.1 6.975, D.F. 0.3 3.1 I I       For bc 2.25 3.1 3.1 6.975, D.F. 0.3 3.1 I I       For bm 2.25 3.1 3.1 9.61, D.F. 0.4 2.25 I I      

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10 Fixing Moments

 

2 2 11.44 3.1 9.16 12 12.1 3.1 9.69 12 F ab F nm M t m M t m       

















2 2 2 2 6.77 2.25 3 2.25 2.86 0.506 3.37 12 30 6.77 2.25 3 2.25 2.86 0.76 3.62 12 20 F an F na M t m M t m              

Figure 8: Fixed end moments in t∙m.

Table 1: Finding moments using moment distribution method.

Joint m n a b c h Member mh mb mn nm na an ab ba bm bc cb hm D.F. 0.3 0.4 0.3 0.42 0.58 0.58 0.42 0.3 0.4 0.3 F.E.M. 9.69 - -9.69 9.69 -3.62 3.37 -9.16 9.16 - -9.16 9.16 -9.69 Balance -2.549 -3.521 3.358 2.432 C.O. -1.275 1.679 -1.760 1.216 Balance 0.382 0.510 0.382 -0.705 -0.974 1.021 0.739 -0.365 -0.486 -0.365 C.O. -0.243 -0.353 0.191 0.510 -0.487 -0.182 0.370 0.255 -0.182 0.191 Balance 0.179 0.238 0.179 -0.295 -0.407 0.388 0.281 -0.187 -0.250 -0.187 C.O. -0.125 -0.147 0.089 0.194 -0.203 -0.094 0.141 0.119 -0.094 0.089 Balance 0.082 0.109 0.082 -0.119 -0.164 0.172 0.125 -0.078 -0.104 -0.078 C.O. -0.052 -0.060 0.041 0.086 -0.082 -0.039 0.062 0.054 -0.039 0.041 Balance 0.033 0.045 0.033 -0.053 -0.074 0.070 0.051 -0.035 -0.047 -0.035 Moment 10.37 0.48 -10.85 6.29 -6.29 5.85 -5.85 10.28 -0.46 -9.83 8.84 -9.37

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Figure 9: Design centerline moments in t∙m.

Design Moments

Span ab, de

At face: Sagging moment

2 11.44 3.1 11.44 0.17 0.17 2.84 2 2 t m        Fixing moment 5.85 2.93



10.28 5.85



10.04 3.1 t m     

Net fixing moment10.01 2.84 7.2t m

At centre: Sagging moment

2 11.44 3.1 13.74 8 t m     Fixing moment 5.85 10.28 8.065 2 t m    

Net sagging moment13.74 8.065 5.68t m

Span bc, cd

2 11.44 3.1 11.44 0.17 0.17 2.84 2 2 t m        Fixing moment 8.84 2.93



9.82 8.84



9.77 3.1 t m     

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12 Span nm, gf

2 12.1 3.1 12.1 0.17 0.17 3.02 2 2 t m        Fixing moment 6.29 2.93



10.85 6.29



10.6 3.1 t m     

Net fixing moment 10.6 3.02  7.58t m

Span mh, hg

At face: Sagging moment3.02t m

Fixing moment 9.37 2.93



10.37 9.37



10.32

3.1 t m

    

At centre: Sagging moment14.54t m

Fixing moment 9.37 10.37 9.87

2 t m



  

Span an, ef

a. Due to rectangular portion

2 6.77 2.25 6.77 0.17 0.17 1.197 2 2 t m       

b. Due to triangular portion 3 2.25 2 2.25 2.25 2 3 3 2.25 3 n n R R       Sagging moment 2 3 2.25 2.77 0.17 0.17 0.343 3 2 t m       

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Fixing moment 5.85 2.08



6.29 5.85



6.26

2.25 t m

    

At centre: Sagging moment14.54t m

a. Due to rectangular portion 6.77 2.25

4.28

8 t m



 

b. Due to triangular portion

3 2.25 2.25 1.5 1.13 1.13 0.95 6 2 2 3 t m        

Total sagging moment4.28 0.95 5.23t m

Fixing moment at centre 5.85 6.29 6.07

2 t m



  

Net sagging moment5.23 6.07  0.84t m

Reinforcement

Use thickness of slab= 35 cm 32.5

e

d  cm

Span ab, bc, cd, de

At face (-ve steel) 5 2 6 7 7.2 10 21.54 1200 32.5 t A    cm   Use Ø16 mm bars, A_s 2cm2 Spacing 100 9.3 21.54 2 cm   Use Ø16 mm bars @ 9 cm c/c At centre (+ve steel)

5 2 6 7 5.68 10 16.99 1200 32.5 t A    cm   Spacing 100 11.77 16.99 2 cm  

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14

Use Ø16 mm bars @ 11.5 cm c/c

Refer to Table 2 for the reinforcement of the rest of the members.

Table 2: Steel Reinforcement.

Member Fixing moment t∙m

(-ve moment) Sagging moment t∙m (+ve moment) ab,de 7.20 Ø16 mm bars @ 9 cm c/c 5.68 Ø16 mm bars @ 11.5 cm c/c bc, cd 6.93 Ø16 mm bars @ 9 cm c/c 4.41 Ø16 mm bars @ 11.5 cm c/c nm, gf 7.58 Ø16 mm bars @ 8.5 cm c/c 5.97 Ø16 mm bars @ 11 cm c/c mh, hg 7.30 Ø16 mm bars @ 8.5 cm c/c 4.67 Ø16 mm bars @ 11 cm c/c an, ef 4.72 Ø16 mm bars @ 14 cm c/c -0.84 Ø12 mm bars @ 25 cm c/c bm, dg 0.48 Ø12 mm bars @ 25 cm c/c -0.47 Ø12 mm bars @ 25 cm c/c Maximum moment7.58t m 5 7.58 10 28.74 32.5 O.K. 100 9.18 d    cm  cm 