Design of Box Culvert

DESIGN OF SLAB CULVERT 1.0

DATA

Thickness of slab

D

=

500.00 mm

Carriageway width

Cw

=

7.50 mm

Footpath width

Fp

=

1.00 mm

Clear span



=

!.00 m

$c

=

#0.00 mm

$idth of end s"pport%bearing

=

0.&0 m

Concrete 'rade

=

()5

*teel 'rade

=

Fe&15

Thickness of wearing co"rse

+ermissible stress

bc

=

#., -% -%s.mm

+ermissible stress

st

=

)00 -% -%s.mm

(od"lar ratio

m

=

10

/

=

0.0



=

1.10

a1

=

,.!0 m

=

0.#5 m 1.)0 m

ie load dimension ie load dimension

b1

ie load dimension

b)

=

Total lie load

$1

=

Clear coer to reinforcement

2.0

700 k-

=

,0 mm

2nit weight of concrete

8c

=

)& -%c"m

2nit weight of wearing co"rse

8wc

=

)) -%c"m

3erall thickness of slab

D

=

500.00 mm

4ffectie thickness of slab

d

=

&!0.00 mm

i Clear span 6 4ffectie depth

=

!.&! m

ii C%C of 4nd *"pports

=

!.&0 m

=

!.&0 m

CALCULATIN ING G EFFECTIVE SPAN

4ffectie span is lesser than of

*o 4ffectie *pan

3.0 3. 0

BEN END DIN ING G MOM OME ENT BY PE PERM RMAN ANE ENT LOA OAD DS

l

= $eight of wearing co"rse

$)

= 9$c%1000:8wc =

1.7! k-%s.mm

=

1,.7! k-%s.mm

*ay

=

1&.00 k-%s.mm

;ending moment for permanent loads

= w:l)%#

Total load

$

(1

4.0 4. 0

1) k-%s.mm

=

71.!# k-
BEND BE NDIN ING G MO MOME MENT NT BY VE VEHI HICL CLE E LO LOAD AD / LI LIVE VE LO LOAD AD

For 5m span impact factor

=

)5

For m span impact factor

=

10

*o for !.&0 m span >mpact Factor ength of load

a1

=

1.75 

=

,.!0 m

= a16):99D6$c%1000

ength of load incl"ding &5 0 dispersal ld

=

be

= B : ? : 91<9?% 6 b w

?

=

&.7! m

4ffectie width of slab perpendic"lar to span

+lacing the load symmetrically on the span Distance from centre of end s"pport to centre of load $idth of slab

= ;

l%) ,.)0 m

= Cw 6 9) : F p =

.50 m

;%l

=

1.&#& m

bw

= b169):9$c%1000 =

1.01 m

From Table of >@C )1A)000 For ;%l = 1.&#& for simply s"pported slab B = *o 4ffectie $idth of oad

).#& be

= B : ? : 91<9?% 6 b w =

$idth of load with &5 0 dispersal

$d

= 9) : 9be%) 6 9) : 9b 1%) =

Total ie oad incl"ding impact

T

5.55 m

7.!0& m

= w1 : 99f act%10061

= ;ending (oment for ie oad

()

%) : 9l% 9l%) ) < 99 99  = 992 : ld%) =

Design ;ending (oment

5.0

(

),.1! k-%s.mm

110.7 k-
= (16() =

1#).&7 k-
SHEAR FORCE BY LIVE LOAD

4ffectie span

l

=

!.&0 m

ength of load incl. &5 0 dispersal

ld

=

&.7! m

To get maim"m *hear Force at s"pport let "s place the load coinciding the start point of the aboe lengths. ?

From >@C )1A)000

4ffectie width of load

= ld%) =

).,# m

;%l

=

1.&#

B

=

).#&

bw

=

1.01 m

be

= B : ? : 91<9?%l 6 b w =

$idth of load with &5 0 dispersal

wd

5.)! m

= 99):be%) 6 99):b 1%) 6 b ) =

ie oad per 2nit rea

2

7.,1 m

= T%9ld:wd =

*hear force by lie load

E1

)&.11 m

= G2:ld:):9b16):9$ c%1000 =

*hear force by dead load

E)

7).07 k-

= $ : l%) =

Total Design *hear

E

= E1 6 E) =

6.0

&&.#0 k-

11!.#7 k-

STRUCTURAL DESIGN OF SLAB

@e"ired effectie depth

d

= srt99( : 10 !%9I:b =

&07.) mm

=

)&##.!&1 s.mm

) = 99∏:d %&%st:1000

2sing )0mm dia bars *pacing

=

1)!.)) mm

+roide +roide )0 mm dia bars J 1)0 mm c%c as main steel. = 0.):(160.,:()

;ending (oment for Distrib"tion *teel

=

&7.57 k-
=

&#.00 k-
4ffectie depth

=

&&& mm

@e"ired steel

=

2sing 1)mm dia bars *pacing

=

!00.!00!01 s.mm 1##.,1 mm

+roide +roide 1) mm dia bars J 150 mm c%c as distrib"tion steel.

7.0

SHEAR REINFORCEMENTS

Design *hear

T

, = 9E:10 %9b:deff 

= B1

0.)5 -%s.mm

= 1.1& < 0.7 : 9d eff %1000 =

0.#1# K 0.50

;ending "p alternate bars proide )0 mm bars J )&0 mm c%c. st proided

=

st

= B)

0.)# 

= 0.560.)5:st =

+ermissible shear stress

1,0#.!& s.mm

0.571 L 1.0

= B1:B):TMC0 =

0.,, K 0.)5 -%*.

b)

 : ld%) : 9l d%&

2

6):9D%1000H%l

o.k.

-ot o.k

m o.k

C!"#$%$ G#&'$

TMCo

M15

0.)#

M20

0.,&

M25

0.&0

M30

0.&5

M35

0.50

M40

0.50