United Nations Office for Project Services Afghanistan Rural Rehabilitation Programme Programme
UNOPS-KABUL ngineering section
PARACTICAL PARACTICAL DESIGN OF CULVERT CULVERT & BRIDGES R!"!" Slab "ulvert R!"!" Bridge# R!"!" $oot Bridge# R!"!" % Steel "om&osite Bridge Sus&ension Bridge
Pre&ared B'( ng! Sa' ed Khan Ahmad)ai UNOPS-Kabul
Design of culvert 10/Aug/2002
1
R!"!" "UL*R+ "UL*R+ ,ata "ollected from Project site( Cross Section of culvert site. Clear Span = 7.50m Wash Wa sh slope = 0.02 1n 100 m !ive !oa" = #s20 = 20 m.ton Assume(
$hic%ness of Sla& $hic%ness of 'earing coat Clear Wi"th Wi"th of culvert
=50cm =7cm =5.0m
,ata from R!"!" boos( fc (Compressive Strength of Concrete ) fs (+iel" or proof strength of steel) Specific unit 'eight of -.C.C Allo'a&le strength of concrete fc Allo'a&le strength of steel fs
=210%g/S*m =2,00%g/S*m =200%g/S*m =0.210=,%g/S*m =0.52,00=100%g/S*m
,esign Process(
#+"raulic Calculations
ig.1
13$otal 'ette" cross section area 23$otal 'ette" perimeter 63#+"raulic ra"ius
A= 24.245 S*m =2.,4m -=A/=24.245/2.,4=1.054m
3elocit+ 3elocit+ of 'ater in 'ash stream Where
=1/nS 0.5- 2/6 S=Slope of 'ash &e" n=coefficient n=coefficient of roughness=0.06 =1/0.060.021/21.0542/6=5.,,m/sec
53 Discharge 43 Scour "epth
8=A
8=24.2455.,,=15.cum/sec D.scour=0.75(8/f)1/6
2
Where f=lace+ silt factor=0. D. scour=0.75(15./0.) scour=0.75(15./0.)1/6=2.46m &elo' the #..! 9a:.Scour "epth=1.50D.scour=1.50 "epth=1.50D.scour=1.502.64=6.m 2.64=6.m
73 Average Average 9a:..! = (1.20;0.7;2.1 (1.20;0.7 ;2.1;2.66;2.6 ;2.66;2.66;0.,7 6;0.,7)/4=1.40 )/4=1.40m m $herefore Scour "epth un"er river &e"=6.31.40=2.65m ertical ertical Clearance a&ove from #..!=40cm
Slab ,esign of culvert(
$herefore
,ead load(
fig.2 Self 'eight of sla& = 0.502.0 = 1.20 $on/m $on/m Weight We ight of 'earing coat = 0.072.10 = 0.17$on/m * = $otal "ea" loa" = 1.67$on/m
,ead load bending moment( .9. Dea" loa" = *!2/,= 1.6,.102/,=11.07 $on3m
fig.2
Live load bending moment(
fig.6 .9. live. loa"=!eff./
Where =!oa" on each rear tire of vehicle/< <=1.22;0.04!eff.=1.22;0.04,.10=1.704m !oa" on each rear tire of vehicle =7.25 ton =7.25/1.704=.2 ton/m .9. live .loa"= !efc./= .25,.10/=,.404 $on3m $on3m >mpact actor? > f =15/!efc.;6,=0.66 >
[email protected] ? $herefore >mpact 9oment=,.4040.66=2.4 ton3m $otal $otal .9.= .9. Dea" loa" ; .9. live loa" ;.9. >mp.=11.05;,.404;2.,=22.50 =11.05;,.404;2.,=22.50 ton3m r=fs/fc=100/,=14.70 r=fs/fc=100/,=14.70 %=n/n ;r=10/10;14.70=0.67 fc=210%g/S*m275%g//S*m fc=210%g/S*m275%g//S *m therefore n=10
6
B=13%/6=130.67/6=0.,74 Sla& thic%ness " min.=(2$..9./f s%B&)1/2=(222.50105/,0.670.,74100)1/2=0.45cm D real=503(532.50/2)=6.75cm? Where $hic%. of sla&=50cm? rotection. la+er=5cm?
" steel &ar =25mm to &e use"
Area of Steel bars in .//cm /f slab( As=$otal .9./f sB"=22.50105/1000.,746.75=1.6cm2 "ia. 25mm in 100cm of sla&
Longitudinal or distribution steel bars in .//cm of slab(
As=0.55/(! effect .)1/2 As main steel &ars When the main steel &ars is parallel to the !ine of traffic As=1.20/(! effect.)1/2 As main s.&? When the main steel &ars is right angle to the !ine of traffic As=0.55/(,.10)1/2A main steel &ars =0.201.6=,.64S*m ? #ence provi"e , "ia.12 16cm center to center
,esign of "urb or Kerb(
ig.
Dea" loa"
3Weight of %er&=0.0;0.50/2 0.752.0=0.,1 ton/m
!ive loa" #s20=20 ton ? !oa" on each rear tire=7.25ton .9. "ea" loa" =*!eff.2/,=0.,1,.102/,=4.4 ton3m .9. live loa" = 0.107.25,.10=5.,7 ton3m $otal .9.=4.4;5.,7=12.51 ton3m
Area of steel for curb( As=$..9./fsB"=12.51105/1000.,7475=16.40 S*m "ia. 25mm in 50cm ? also provi"e 2 "ia 25mm in the top of cur&.
"hec for shear( ,ead Load shear(
ig.5 Shear orce 9=0? - A,.103*,.12/2
- A=(1.67,.12/2)/,.10=5.5 ton ? - A=- =5.5 t
Live load shear force(
ig.4 9=0? - A,.1037.256.,5
- A=10.70 ton
$otal Shear orce (8)=5.5=10.70=14.15 ton Allo'a&le shear stress =0.06210=4.60 %g/S*m -eal shear stress =8/&B"=14.15106/1000.,756.75=.21 %g/S*m -eal shear stress =.21%gs*m4.60 %g/ S*m ? #ence safe against shear force
"hec for bond shear stress( -eal &on" shear stress &on"=8/∑0B" ? Where ∑0= of un&ent steel &ars erimeter of &ar #ere rom steel &ars ? 6 &ars is un&ent in 100 cm ∑0 = 6pai2.25 = 66.12.25 = 26.55 -eal &on" stress = 14.1510 6/26.550.,76.75 = 1,%g/S*m Allo'a&le shear stress = 0.1f c = 0.1210 = 21%g/S*m $herefore -eal shear stress=1,%g/S*mAllo'a&le shear stress=21%g/S*m #ence safe against &on" shear stress
,imensions of Abutment and 0ing 1alls2 for culvert)( 9a:. ressure on soil ton/S*m
Dimension
20 20 20 0 0 0
A C A C
alue in m for height of meter 1.50
2.0
2.50
6.00
6.50
0.75 0.20 0.5 0.60 0.20 0.15
0.0 0.60 0.5 0.5 0.60 0.15
1.20 0.5 0.5 0.55 0.5 0.15
1.45 0.45 0.5 0.75 0.45 0.15
2.40 1.04 0.5 0.0 0.,5 0.15
0ing 0alls(
13!ength of 'ing 'all 1.5032.0m times the height of a&utment !=(1.50 or 2.0)# 23 $he 'i"th of the 'ing 'all at the &ase shoul" &e from 0.65# to 0.0# 'here # is height of 'ing 'all.
5
R!"!" BR3,4 Selection of Bridge site(
13 23 63 3 53 43
Suita&le foun"ation Straight line irm an" 'ell "efine" &an% Stream line flo' 9inimum 'i"th -ight angle crossing
Bridge and "ulvert (
13 Span of &ri"ge form 6 to 7.5 m? ma"e form -.C.C sla&. 23 Span of &ri"ge 7.5 to 12 m? ma"e form -.C.C &eam an" sla&. 63 Span of &ri"ge more than 12 to 2, m? ma"e form iron &eam an" -.C.C pre3cast sla&. 3 Span more than 2, to 0 m? ma"e form -.C.C pre3cast &eam. ,ia&hragm(
!ocation of "iaphragm 13 Ep to 12 m span? t'o "iaphragm in each en" of &eam. 23 Span from 12 to 15 m? 6 "iaphragms? t'o in each en" an" one in the mi""le of span. 63 Span more than 15 m ? after each (span/6) one "iaphragm. ,e&th of +ee beam(
,e&th of Rectangular beam(
13 or light loa"ing span/15 to 20. 23 or me"ium loa"ing span/12 to 15. 63 or heav+ loa"ing span/12
13 or span up to 10 m Span/20 23 or span longer than 10 m Span/12
*ertical clearance above 5!$!L(
Discharge cum/sec
9in .vertical clearance (mm)
elo' 0.60 cum/sec 0.60 up to 6.0 cum/sec 6.10 to 60.00 cum/sec 61.00 to 600.00 cum/sec 601.00 to 6000.00 cum/sec Fver 6000.00 cum/sec
150 50 400 00 1200 1500
5'draulic "alculation(
Cross section of &ri"ge site #+"raulic calculation for &ri"ge the same as for culvert ? (see the h+"raulic calculation for culvert)
4
,esign of R!"!" Bridge(
<:ample
Clear span accor"ing to the site Wi"th of the &ri"ge 'ith cur&s Wi"th of cur& Wi"th of &eam Depth of &eam = (span/17) !ive loa" $hic%ness of sla& (assume) $hic%ness of ' earing coat
= 1m = 5m = 0.40m = 0.0m = 0.,0m = #s20 = 0.20m = 0.07m
,esign of slab(
Weight of sla& Weight of 'earing coat $otal 'eight (*)
= 0.202.0= 0., ton = 0.072.20 = 0.15 ton = 0.,;0.15 = 0.46 ton
.9. for "ea" loa"
= *l2/, = 0.461.072/10 = 0.076 ton3m
ig.7 .9. for live loa" = (0.,01.4l;1/14) 20 = (0.,01.41.07;1/14) 7.25 = 0., ton3m >mpact factor
= 15/span;6, = 15/1;6, = 0.60
.9. >mpact
= 0.600., = 0.2 ton3m
$otal .9.
= 0.072;0.,;0.2 = 1.64 ton3m
Area of steel bars(
fcG = 100 %g/cm2 9in."epth of sla& -eal "epth of sla& $herefore real "epth
fsG= 100 %g/cm2 B = 0.,4 % = 0.2 0.50 = (2.9/fcG % B &) = (21.64105/1000.20.,4100)0.50 = ,.70 cm = (2035;1.40/2) = 1.20 cm use "ia of 14 mm &ar 'ith 5cm protection la+er in 2 faces of sla& = 1.20 cm @ 9in "epth = ,.70 cm F.H.
As = .9/fsG B " = 1.64 105/100 0.,4 1.20 = ,.14 cm2 Spacing of steel &ars in 1m of sla& = 1002.00/,.14 = 2.4 cm sa+ 20 cm c/c Ese 5 "ia 14 mm &ars 20 cm center to center (As = 10.0, cm 2) Distri&ution or longitu"inal steel &ars As = (120/(!) 0.5 ) /100 = (120/1.070.5)/100 = 1.14 @ 0.47
7
therefore
As = 0.47 10 .0, = 4.76 cm 2 Spacing = 100 1.16/4.76 = 17 cm center to center? Ese 7 "ia 12mm 17 cm center to center.
,esign of 3nterior beam(
= 1;.04 = 1.40 m
Dea" loa" Weight of sla& on &eam Weight of 'earing coat Self loa" of &eam
= 1.7 0.20 2.0 = 0.765 m.ton = 1.7 0.07 2.20 = 0.224 m.ton = 0.,0 0.0 2.50 = 0.,0 m.ton
$otal loa" (*) B! 6oment of dead load(
= 1.74 9.$on .9 = * span2/, = 1.74 1.42/, = 4., $on3m
B! 6oment for live load(
Distri&ution of Wheel loa" on longitu"inal &eam = Span1/1., for one line of traffic. Where Span1 is center to center "istance &et'een t'o &eams? an" 1., is constant $herefore Distri&. 'heel loa" on long. &eam = 1.7/1., = 0.72 <:erte" loa" from rear tire of vehicle = 7.25 0.72 = 5.67 $on <:erte" loa" from front tire of vehicle = 6.425/2 = 1.,2 0.72 = 1.65 $on Where 6.425 ton is loa" on front 'heel of vehicle .
ig., I
,
$otal 9r = 0 = 3-l1.40 .; 5.6,10., ; 1.652.6 ; 5.6,4.5 = 4.46, ton3m $hrefore -l = 4.46, ton3m. +2/,.01 = 4.5/1.40 +6/,.01 = 2.6/1.40
+2 = 6.415 +6 = 1.2,
+1/4.5 = 6.74/1.40
+1 = 1.47
9a:..9oment = 6.4155.6, ; 1.475.6, ;1.2,1.65 = 60.605 ton3m >mpact factor
= 15/1.40;6, = 0.2,5
>mpact 9oment = 0.2,5 9a:..9. = 0.2,560.605 = ,.467 ton3m $otal .9. in the mi""le of span = 9 "ea" loa";9a:..9; 9.impact = 4.,5;60.605;,.467 = ,5.,67 ton3m Shear stress( Shear of dead load(
* = 1.74 ton/m - A = 8ma: = 1.741.40/2 = 12.,, ton
ig.10 Shear of live load(
ig.11 9A = 3- 1.40; 7.251.40; 5.6,10.65; 1.64.10 = 0 - = 11.424 8ma: = - = 11.424 ton -A = 2.6 ton Shear of 3m&act (
>mpact factor = 0.2,5
8ma:. imp. = 0.245 8ma:.live loa" = 0.2,511.424 = 6.61 ton
+otal shear force 7 12.,, ; 11.424 ; 6.61 = 27.7, ton
As = .9/f sG("eff.3t/2) B = ,5.74 105/100(,4320/2) 0.,4 = 7.742 cm 2 Where (t ) is thic%ness of sla& AS = 7.742 therefore 10 "ia 64 (As = 101.,0 cm2)
"hec the section of beam(
= As/&" = 101.,0/17 ,4 =0.0052 n = 0.0052 10 = 0.052 t/" = 20/,4 =0.262 2 6 1/2pn = 1/2 0.005210 = .40 B = 434t/";2(t/") ;(t/") (1/2pn) B = 0.,25 436t/"
=
% = np ;1/2(t/") 2 = 10 0.0052; K(20/,4)2 = 0.2, np ; t/" 100.0052;20/,4 $herefore As 1 = 9/fsGB " = ,5.74 10 5/100 0.,25 ,4 = ,4.67cm2 As = 101.,0 As1 = ,4.67 cm2 therefore the section is F.H Area of steel bars for shear force(
Shear stress = = 8 ma:/&B" = 27.7,106/0 ,4 0.,25 = .7 %g/cm2 fc = 250 %g/cm2
Allo'a&le shear stress = 0.06 250 = 7.50 9ar% of concrete = 250 %g/cm 2 S = 2.2 (coefficient of safet+)
$.c= $ensile shear strength of concrete = 20.0
Allo'a&le shear stress () = $c/S = 20/2.2 = .0 = .7 @ = .0 rom other han" = .7 $c= 20? &eam.
$herefore? nee" for shear steel &ars $here fore no nee" to change the section of the
Lo' "etermine the "istance from 'here shear steel &ars re*uire" !: = (3/)Span/2 = (.73.0/.0)1.4/2 = 0.54m = 54 cm -esultant shear stress- = 0.50(;) ! : = 0.50(.7;.0) 0 54 = 2115 %g - = 2125%g = 21.15 ton We 'ill use stirrup "ia 10 mm in the spacing of 10 cm ? in the "istance of 54 cm Lo of stirrups (Ls) = 54/10 = 5.4 ? sa+ 4 stirrups. m = 2 'here (m) is num&er of stirrup arms. Shear force to support &+ stirrups $s = L s m 8 p/ S 'here 2 8 p = plasticit+ limit of steel &ars = 2500 %g/cm ? an" S=2.2 (coeff. of safet+) $s = (42) 2500/2.2 = 16464.64 %g = 16.464 ton Shear stress to support &+ stirrups s = mAs.s8 p/ Spacing& =20.7,52500/100= .,16 %g/cm 2 Shear stress to support &+ main &ent steel &ars $m. &ar = 0.,0 -3$s = 0.,21.153 16.464= 6.2, ton $herefore from the ta&le of steel &ars strength the "ia of 64 mm has strength =1,,7 %g/cm2 hence no nee" to &ent the main steel &ars? ho'ever one main &ar 'ill &e &ent up.
Lote
>f the "epth of the &eam e:cee" 40cm s%in reinforcement on &oth faces o& 'e& or &eam in the form of !ongitu"inal &ars (min 12 mm &ars spacing not more than 20 cm shoul" &e provi"e". Such reinforcement on each face shoul" &e at lest 0.05M of cross section area.
10
ig 12
11
,esign of masonr' abutment for bridge( Normal scour de&th(
Dscour = 0.76(8/f)1/6
'here 8 is ma: .river "ischarge or *uantit+ of river flo' (f ) is lac+s silt factor
$a&le of Silt factor(f) +'&e of bed soil er+ fine soil ine silt 9e"ium silt 9e"ium san" Coarse san" Stan"ar" silt #eav+ san"
f 0.0 0.40 0.,5 1.25 1.50 1.00 2.00
Si)e of &articles mm 0.052 0.120 0.266 0.505 0.725 0.626 1.20
When the 'i"th of stream a&stracte" &+ a&utment or pier than the scour "epth Dscour = D (W/! a&st.)2/6 ? 'here W is normal 'i"th of river an" !a&st. is length of a&stracte" 'ater 'a+. Dscour .ma:. = (1.50 to 2.0) D scour Design <:ample Data
$+pe of river &e" soil 9e"ium silt -iver "ischarge (8) = A = 600.00 cum/sec Assume = 2.5 m/sec? an" Cross section Area (A) = 120.00 m 2 Supper structure $3 eam an" sla& one lane &ri"ge of clear span 1m !oa"ing #S20 ac%fill Nravel 'ith angle of repose O = 00 ? ' = 1.,0 ton/m 6 Angle of internal friction of soil on masonr+ P= Q =20 0 Angle of a&utment slop 'ith groun" level R = 10,.140 Angle of &ac%fill soil 'ith groun" level = s = 0
Solution(
Lormal scour "epth = 0.76(8/f)1/6 = 0.76(600/0.,5)1/6 = 6.6 m. 9a:. Scour "epth = 1.5 6.6 = 5.01 m &elo' the average height of floo" level. (see scour "epth calculation in culvert). ,ead load from su&&er struture(
!oa" from &eam = (0.,0 0.0 1/2 2.0) = 21.50 ton !oa" from sla& = (0.20 5.00 1/2 2.0)1 = 14.,0 ton !oa" from cur& = (0.55 0.25 1/2 2.0)2 = .42 ton !oa" from 'earing coat = (0.07 6.,0 1/2 2.20)1 = .0 ton $otal Dea" loa" = 7.02 ton !ive loa" from t'o rear tire of vehicle $otal "ea" loa" an" live loa" !oa" per m 'i"th of a&utment
12
= 15 ton = 42.02 ton = 42.02/5 = 12.0 ton/m
Assume reliminar+ "imension of a&utment
ig.16 Wi"th of a&utment in foun"ation = 250 cm ? #eight of a&utment $op 'i"th of a&utment = 120 cm
= 600 cm
Self 'eight of a&utment per m length W= ' 1 ; '2 ; '6 ; ' ; '5 ; earth = = (0.40 6.0) 2.20 ; (0.40 .0) 2.20 ; (K 1.60 6.0) 2.20 ; (0.70 0.0) 2.2 ; (2.50 2.50) 2.20 = 6.4; 5.,0, ; .2 ; 0.414 ; 16.75 ; 0.42= 2.6, ton/m. +otal vertical forces &er meter length of abutment 7 .8!9/9 : 8;!<= 7 9.!>= ton?m Longitudinal forces(
a. orce "ue to attractive effort 7 0.20 #S20 loa" = 0.20 20 = .0 ton orce per meter length of a&utment = .0 /5 = 0.,0 ton &. orce "ue to temperature variation = 0.15 total loa" per meter 'i"th of a&utment = 0.15 12.0 = 1.,41 ton. $otal longitu"inal forces = 0.,0 ; 1.,41 = 2.44 ton? this force act at the &earing level. arth &ressure ( 2 P7 @ 1 h8 2 cos sin 2 C DEE 7 K 1.,0. (cos 10,.14 sin(10,.1430) 2 sin 2 : FEE.?8 : 2 sin 2D:FE.?8 sin 2D-GE (sin(10,.14;20))1/2 ; sin (0;20)1/2 sin(0)1/2
= 6.0,4 ton/m #oriQontal component of () = 6.0,4 Cos(10,.1430) = 2.62 ton/m ertical component of () = 6.0,4 Sin(10,.1430) = 0.42 ton/m #eight a&ove &ase of center of pressure = 0.2 .0 = 1.,5m
ig1
16
7
9oment of vertical forces from the a&utment toe
Source A&utment Structure Do Do Do Do Super structure
orce s+m&ol '1 '2 '6 ' '5 's.s earth.v.
$otal ver.orce 6.4 5.,0, .2 0.414 16.75 12.0 0.42 1.7
Arm (m) 0.60 0.0 1.46 0.65 1.25 0.65 1.20;2.40/6=2.5
Sta&.9oment ton3m 1.1,, 5.227 7.007 0.215 17.1,7 .61 2.05 67.57
-esultant force location from point ($oe) Arm = Sta&. 9oment/vertical forces = 67.57/1.7 = 0., m
9oment of #oriQontal forces from the a&utment toe
Source Structure
orce S+m&ol Sf earth #.
$otal orce 2.44 2.62 5.52
Arm (m) 6.0 1.,5
Fverturning .9. 7., 5.2 16.0
-esultant force location from A line Arm = 16.0/5.52 = 2.67 m "hec for overturning and Sliding 6oment (
Overturning(
S.overturning = Sta&.9oment/Fverturning 9oment = 67.57/16.0 = 2.,0 @ 1.50 F.H
Sliding(
S.sli"ing = .orces f / #.orces = 1.70 0.40 / 5.52 = .7 S. sli"ing = .7 @@ 1. #ence a&utment is safe against sli"ing . "hec location of Resultant (
(#. Arm) ; (3. T) = 0 =@ (5.52 2.67) ; ( 3 1.70 T) =@ T = 16.0/1.70 = 0.621 m a = Arm J T = 0., J 0.621= 0.5, m e = /2 J a = 2.50/23 0.5, = 1.2530.5, = 0.47 m $otal "o'n'ar" forces = 1.7 ton/m
1
ig15
ig 14
15
R!"!" $OO+ BR3,4 Data given Clear span of &ri"ge #eight of &ri"ge from river &e" -iver &e" elevation #igh floo" level Clear 'i"th of &ri"ge !ive loa" earing capacit+ of foun"ation soil Assume sla& thic%ness
1, m .6m .6, 7.66 2.50 m Camel 'ith loa" 62 ton/s*m 15 cm
,esign &rocess(
#eight ("epth) of &eam (span/15 to 20) for light loa" 1,/15 = 1.20 m Wi"th of the &eam 2/5 "epth = 2/5 1.20 = 0., m? assume 0.0 m Slab design (
Dea" loa" Dea" loa" of sla& 0.15 200 = 640 %g/m !ive loa" Weight of camel 'ith loa" 140 ser = 140 7 = 1?120 %g !ive loa" = 1120/1.5 = 74.70 %g/m 2 ? !ive loa" = 750 %g/m 2 Where (1.5 m) "istance &et'een t'o legs of camel.
ig 17 $otal loa" (*) = Dea" loa" ; !ive loa" = 640 ; 750 = 1?110 %g/m 2 !effect. = 1.70 1.05 = 1.7,5m .9ma:. = * ! eff.2/, = 1?110 1.7,5 2/, = 2 %g3m Area of Steel in one meter of span = A s= .9/f s U" = 2 10 2/100 0.,74 15 = 210 2/1,64 As = 2.06 cm 2 #ence provi"e 5 "ia ,mm 20cm center to center (A s = 2.51cm 2 @ 2.1cm 2) ,esign of beam(
Dea" loa" from sla& 1?110 span/2 = 1?1101.70/2 = %g/m
14
Dea" loa" of &eam &h200 = 0.0 1.20200 = 40 %g/m Dea" loa" from han" rail 60%g/m $otal loa" (*) = .0; 40.0; 60.0 = 1?6.0 %g/m
ig 1,
17
"O6POS3+ 2S+L % "ON"R+ E BR3,4 ,esign of com&osite bridge 1ith reinforcement concrete slab and steel &late girders to cover 8< m clear s&an !
Niven Data
Clear span 26 m Wi"th of &ri"ge 5m oot path 0.0 m Concrete 93200 fc 2100 %g/cm2 f+ 2,00 %g/cm2 !ive loa" #s20 $he a&utments of &ri"ge are alrea"+ e:isting .
ig1 ,S3N4 O$ R"" SLAB(
a. Dea" !oa" Dea" loa" of sla& Dea" of asphalt $otal "ea" loa" (*)
0.20 2. ; 0.0 0.25 2.0 = 0.72 ton/m 2 0.07 2.20 = 0.15 ton/m 2 0.,7 ton/m 2
Dea" !oa" 9oment .9"ea" loa" = *!2/, = 0.,7 1.57 2/, = 0.27 ton3m &. !ive !oa" V !ive !oa" 9oment
(1.4! ; 1) !oa" on rear t +re 14 !ive !oa" 9oment = 0.,0(1.41.57;1) 7.25 = 1.60 ton3m 14 c. >mpact 9oment 0.60 1.60 = 0.0 ton3m = 0.,0
$otal 9oment = 0.27 ; 1.60 ; 0.0 = 1.7 ton3m
1,
Area of steel bars in one 2mE slab(
fc fsteel
0.502,00 = 100 %g/cm 2 0.02100 = , %g/cm 2
r = f s/fc = 100/, = 14.70 n = 10 % = n/n;r = 10/10;14.70 = 0.675
U = 13 %/6 = 13 0.675/6 = 0.,75
"min. = (29/fc %U&) 1/2 = (21.710 5/,0.6750.,75100)1/2 = 11., cm
#ence provi"e 4 "ia.1 mm (As = .2cm 2)
Spacing = 1.56100/.2 = 14.4 cm = 14 cm center to center. ,esign of 3nterior steel 23 E beam(
Dea" loa" Wt of (>) &eam ;Wt of "iaphragms ;Wt of angle >ron Wt of -CC sla& = (0.00.027.,5) ;(0.00.607.,5)2 ; (120.005)/26 ; (0.400.017.,5) ; ;(0.150.0177.,5)2 ; 1.540.202. ; 0.07 2.20 = 1.60 ton/m2 9"ea" loa" = *!2/, = 1.602.20 2/, = 5.17 ton3m
! eff.= 1.0526.0 = 2.20
!ive loa" 9oment Coeff.of loa" "istri&ution = Span/1.4, = 1.57/1.4, = 0.65
ig.20 !oa" from the rear tire = 7.250.65= 4.7, ton/S*m !oa" from the front tire = 1.,10.65 = 1.42ton/S*m - a2.20 = 1.4215.42 ;4.7,11.67 ; 1.427.12 = 15.0/2.20 = ,.04 ton 9ma: = ,.0412.,6 J 4.7,.25 = 7.55 ton3m >mpact factor = 15/2.20 = 0.21
1
>mpact 9oment = 0.27.55 = 17., ton3m $otal 9oment = 5.17 ; 7.55; 17., = 1,7.75 ton3m Selection of Steel Section (
9oment of >nertia for composite section (Steel I concrete) Lote $o 'or% together the -CC I steel ? the span of -CC sla& &et'een t'o > &eam 'ill "ecrease 10 times ? therefore span/10 = 157/10 = 15.70 $otal Area of > &eam an" -CC ? A = A1;A2 = (15.7020) ; ((02) ; (60)2) = 61 ; 20 = 76cm 2 Staticall' 6oment = A1+1 ; A2+2 = 6111, ; 20 = 57462 cm 2 "enter of gravit' of com&osite section (
= 9st:3: /A = 57462/76 = 7,.52 cm
ig 21 6oment Of 3nertia or Second 6oment ( A! 9oment of >nertia of steel a&out the : 1 A:is
>:1 = & h 6/12(&h6/12;6072) ; (&h6/12;6072) >:1 = 206/12;(606/12;2450,0);(606/12;2450,0) >:1 = 121500;24520;24520 = 451?,0cm B! 9oment >nertia of Concrete section a&out the : 2 a:is
>:2 = &h6/12 = 15.70206/12 = 1044.47 cm "! 9oment >nertia of composite section a&out the : 0 section
>:o = >:2 ;A2 2.,2 ; >:1 ; A1 2.52 >:o = 1044.47;61,4.070 ; 451,0 ; 20,71.60 = 1601665.1, cm
20
ig22 "oefficient of Strength (
W &ottom= >:o/center of gravit+ from the top = 1601665.1,/7,.52 = 14576.60 cm 6 Wtop = >:o/C.F.N from &ottom = 1601665.1,/1., = 44,06.47 cm 6 Stress(
Stress &ottom X & = $otal 9 loa"/W & = 1,?775?000/14576.60 = 1162.,5 %g/cm 2 Stress top X t = $.9oment/Wt = 1,?775?000/44,06.47 = 2,1.05 %g/cm 2 Allo'a&le stress for steel = f steel = 2,00/2 = 100 %g/cm 2 therefore Strtess & = 1162.,5 %g/cm 2 100 %g/cm 2 Stress t = 2,1.05 %g/cm2 100 %g/cm 2
FH FH
,esign of fillet 1elding 2 Design of suita&le 'el"e" connection &et'een the 'e& an" the lange) (
SiQe of late Nir"er h = 0 cm ? lange siQe 'i"e= 60 cm ? $= cm Clear Span of &ri"ge 26 m Design rocess #oriQontal shear per cm length of plate gir"er 81 = /> (A) ? = 1Span/2 1 = (Dea" loa" ; !ive loa") per (m)length of gir"er 1 = 1.60 ; 4.7, = ,.0, ton/m? = ,0,026/2 = 2?20 %g >:1 = 451.,0 cm ? (center of gravit+) = cm? A = 120 = 5?,,0 cm 6
A = 60 =120cm
81 = 2?20/451?,0(5?,,0) = ,6,.015 %g/cm ertical loa"e" lange (Comp. lange) 1 = ,.0, ton/m = ,0,0 %g/m = ,0.,0 %g/cm -esultant Shear (8) 8 = (812 ; 1)1/2 = (,6,.015 2 ; ,0.,0 2)1/2 = ,1.0 %g/cm Wel"e" shall &e applie" on &oth si"es of the We& plate for top lange as 'ell as for &ottom lange. Wel" coul" &e continuous.
"ontinuous 1eld (
!et (S) &e the siQe of 'el"
21
Strength of 'el" per cm length = 2s% * ? * = ermissi&le stress = 1100 %g
Where % = coeff. = 0.7
S = 8/2% * = ,1.0/20.701100 = 0.5 cm = 5.0 mm S = 5. mm , mm (minimum thic%.of 'el") F.H ut for more safet+ 'e 'ill ta%e 10 mm therefore
S = 1cm
,esign the Butt 1elding 2for Nir"er) (
Strength of the Uoint = ! t t Where ! =
ig26
22
FH
SESFL ->DN< Data Data collecte" fro &ri"ge site
Clear span of &ri"ge Sno' loa" !ive loa"
270 eet 25 l&s/ft2 60 l&s/ft 2
Si)e of various 1ooden com&onents of bridge(
1. $ransom 2. earer 6. lan%s . ost 5. ac% Sta+ 4. Diagonals
7. Nuar" -ails
(6inch5inch/1) 4foot 1 (2Y Y/1) 6Z34Y (1.5Y/12) 1Z Z6.5 (2Y6Y/1) Z2 (2Y6Y/1) Z34Y2 (2Y6Y/1) 5Z34Y (2Y6Y/1)6Z34Y2 $F$A!
= 0.425 cuft = 0.77, cuft = 1.75 cuft = 0.666 cuft = 0.675 cuft = 0.1 cuft = 0.22 cuft = 5.04 Cuft
Weight of 'oo" 50 l&s/cuft $otal 'eight of 'oo" 505.07 = 256.50 !&s Weight per running oot of ri"ge = 256.50 l&s/6.5 = 72.6 l&s/ft
ig 2 "alculation the length of sus&enders (
Suspen"ers are provi"e" at (! 1 = 6Z34Y) intervals . Dip = Span/15 = 270/15 = 1, feet
(Central "ip of the main ca&les)
>nverte" Dip = Span/40 = 270/40 = Z34Y !ength of least suspen"er (assume) = 2 feet #eight of $o'er = Dip ; >nv.Dip ; !ength of least suspen. = 1,Z ; Z34Y ; 2Z = 2Z34Y !ength of the suspen"ers can &e calculate" &+ the follo'ing formula !ength of suspen"ers = ("/! 2):2
26
Where " = "ip
! = Span of &ri"ge
: = Distance from least suspen"er
$herefore $otal length of suspen"er = "1:2/!2 ; "2:2/!2 ; 2Z = /!2("1 ; "2) :2 ; 2 "2 >nverte" "ip : = 6.5 ft 1 = /2702 (1,;.5) 6.5 2 ; 2 = 2.015 ft : = 7.00 ftt 2 = /2702(1,;.5)7.02 ; 2 = 2.0405 ft : = 10.50 ft 6 = 0.1641 ; 2 = 2.1641 ft : = 1.00 ft = 0.22 ; 2 = 2.22 ft : = 17Z34Y 5 = 0.67, ; 2 = 2.67, ft : = 21.00ft 4 = 0.5 ; 2 = 2.5 ft : = 2.50 ft 7 = 0.71 ; 2 = 2.71 ft : = 2,.00 ft , = 0.47 ; 2 = 2.47 ft : = 61.50 ft = 1.225 ; 2 = 6.225 ft : = 65.00 ft 10 = 1.5126 ; 2 = 6.5126 ft : = 6,.50 ft 11 = 1.,2 ; 2 = 6.,2 ft : = 2.00 ft 12 = 2.177, ; 2 = .177, ft : = 5.50 ft 16 = 2.555 ; 2 = .555 ft : = .00 ft 1 = 2.42 ; 2 = .42 ft : = 52.50 ft 15 = 6.02, ; 2 = 5.02, ft : = 54.00 f 14 = 6.,714 ; 2 = 5.,714 ft : = 5.50 ft 17 = .6707 ; 2 = 4.6707 ft : = 46.00 ft 1, = .0 ; 2 = 4.0 ft : = 44.50 ft 1 = 5.54 ; 2 = 7.54 ft : = 70.00 ft 20 = 4.0 ; 2 = ,.0 ft : = 76.50 ft 21 = 4.44 ; 2 = ,.44 ft : = 77.00 ft 22 = 7.61, ; 2 = .61, ft : = ,0.50 ft 26 = ,.0006 ; 2 = 10.00 ft : = ,.00 ft 2 = ,.7111 ; 2 = 10.7111 ft : = ,7.50 ft 25 = .522 ; 2 = 11.522 ft : = 1.00 ft 24 = 10.2265 ; 2 = 12.2265 ft : = .50 ft 27 = 11.025 ; 2 = 16.025 ft : = ,.00 ft 2, = 11.54, ; 2 = 16.,54,ft : = 101.50 ft 2 = 12.71,, ; 2 = 1.71,, ft : = 105.00 ft 60 = 16.4111 ; 2 = 15.4111 ft : = 10,.50 ft 61 = 1.5664 ; 2 = 14.5664 ft : = 112.00 ft 62 = 15.,4 ; 2 = 17.,4 ft : = 115.50 ft 66 = 14.4 ; 2 = 1,.4 ft : = 11.00 ft 6 = 17.,27 ; 2 = 1.,27 ft : = 122.50 ft 65 = 1,.5242 ; 2 = 20.5242 ft : = 124.00 ft 64 = 1.40 ; 2 = 21.40 ft : = 12.50 ft 67 = 20.700 ; 2 = 22.700 ft : = 166.00 ft 6, = 21.,6,6 ; 2 = 26.,6,6 ft : = 164.50 ft 6 = 26.002 ; 2 = 25.002, ft $otal length of suspen"ers one si"e half = 6,,.4655 eet $otal length of suspen"er ca&le ( for entire &ri"ge) = 6,,.4655 ; = 155,.52 eet $otal num&er of suspen"ers = Span/! 1 2 = (270/6.5) 2 = 15 Average length of each suspen"er = 155,.52/15 = 10.120 feet >ncrease length &+ 20M = (10.120) 1.20 = 12.1 feet Weight of suspen"er ca&le = 1.0 l&s/ft
2
$'o suspen"ers ca&le in "esign panel ? therefore Wt = 2112.1 = 2.2,, l&s/ft Weight of suspen"ers ca&le per running foot of &ri"ge = 2.2,,/ = 4.072 l&s/ft 0eight of 6ain cable(
Self 'eight of main ca&le = 2 l&s/ft !L CA!< ! = Span(1;,/6("/span)2) ! = 270(1; ,/6(1,/270) 2) = 276.20 eet $hree ca&les on either si"e ? therefore Weight of 9ain ca&les per running foot of &ri"ge = si"esLo of ca&les Self 't of ca&le = = 262 = 12.0 l&s/ft Sno' 'eight = 25 'i"th = 25 6.5 = ,7.50 l&s/ft !ive !oa" = 606.50 = 105 l&s/ft +otal load on bridge ( +otal Load = ( Wt of 'oo"en "ec% ; Wt of 9ain ca&le ; Wt of suspen"ers ; Sno' ; Slings)1.1 ; ;!ive loa" =(72.6 ;12 ;4.072 ;,7.50 ;,)1.1 ;105 = 60.402 l&s/feet = 0.604 Hips/ft
ig 25
25
,S34N O$ 6A3N "ABL ( Uniforml' distributed load 21E 7 /!
Using (
*ertical +hrust (
= ' Span/2 = 0.604 270/2 = 1.74 %ips
+ension in cable (
$ = ((#) 2 ; () 2)1/2 = ((154.765) 2 ; (1.74) 2)1/2 = 142.212 %ips
Dia (1) inch ca&le (1)'ires each (0.2) inch "ia. <:tra high strength gra"e
rea%ing !oa" = 10?500 l&s
$a%ing factor of safet+ = 6.2
Wor%ing !oa" = 10?500/6.2 = 62?454.25 l&s
Lum&er of ca&les re*uire" for entire &ri"ge = 142?212/62454.25 = .47 for &oth si"e of &ri"ge. $herefore We $a%e (6) Ca&les of (1) inch on each si"e. See anne: Lo 1(ca&le ta&le) ,esign of Anchor bloc for 6ain cable (
ig 24 Span/ = 270/ = 47.50 eet? [ = 1.60 = 10 ?55Z? 56Y
tan [ = #. of $o'er/47.50 = 1,/47.5 = 0.2447 R = 1,03 ( ; 0) = 75.04,4 0 = 70?Z?7Y
ig 27 $A = $ Sin 1.610 = 142.212 Sin 1.61 0 = 1.75 %ips $ = $ Cos 1.61 0 = 142.212 Cos 1.61 = 154.765 %ips =\-
= orce of Sli"ing
Assuming that the total tension in ca&le can cause s li"ing an" is increase" &+ 50M $herefore = 142.212 1.50 = 26.61, %ips
24
\ = Co3efficient of friction = 0.40 - = 'eight of loc% 26.61, = 0.40 - = 26.61,/0.4 = 05.560 Hips olume of &loc% re*uire" = $otal Wt/Enit Wt of &loc% = 05.560/0.15 = 2706.566 cft SiQe of loc% = 17 feet : 17 feet : 10 feet? Wt = 0.15 = 66.5 %ips
= 2,0 %ips
earing ressure = Wt of &loc%/Contact Area = 66.50/1717 = 1.50 %ips/ft 2 2.2%ips/ft 2 #ence F.H "hec for Overturning (
Self 'eight of &loc% = 17 17 10 0.15 = 66.50 %ips Fverturning 9oment $ 10/2 ; $A17/2 =154.7655 ; 1.75,.50 =116,.66 %3ft -esisting 9oment = 66.5017/2 = 64,.75 %3ft (.F.S) actor of Safet+ against Fverturning = -esist. 9oment/Fvert. 9oment .F.S = 64,.75/116,.66 = 6.265? #ence F.H ,esign of 1ind-gu' cable for 1ind (
Calculation of area in "irect contact 'ith 'in" 1. 2. 6. . 5. 4. 7.
$ransom earer lan%s ost ac%sta+ Diagonal Nuar" -ail
(65/1) 1 = 0.10 sft ( inch/12) 6Z34Y 1 =1.1447 Sft (1.5 inch/12)6Z34Y 1 = 0.675 Sft (6/12) Z 1 = 1.0 sft (6/12) Z1 = 1.0 Sft (6/12) 5Z34Y1= 2.75 Sft (6/12) 6Z34Y1 = 0.,75 Sft $otal Area = 7.666 Sft
$otal Area of Woo" in contact 'ith 'in" = 7.666 sft $otal Area of ca&le in contact 'ith 'in" 9ain ca&le "ia (1 inch) 1/12 6Z34Y 6 = 0.,75 sft Suspen"er Ca&le "ia.(1/2 inch) 0.5/12 11.55 = 0.,1 sft $otal area = 1.654 sft $otal Area of &ri"ge in contact 'ith 'in" = 7.666 ; 1.654 = ,.4, sft $a%ing 'in" pressure = 20 l&s/sft $otal pressure = 20 ,.4,6 =176.7 l&s $otal loa" per running foot = 176.7/6.5 = .45 l&s/foot = 0.05 %ips/foot Selecting 'in" tie ca&le "ip " = Span/15 = 270/15 = 1, feet $ag R =1,/165 = 0.1666
R = 7.54 0
Sa+. R = 7.4 0
5ori)ontal +hrust (
# = W! 2/," = 0.05270 2/,1, = 25.6125 %ips
*ertical +hrust (
= W!/2 = 0.05270/2 = 4.75 %ips
+ 7 ((#2 ; 2))1/2 = ((25.6125) 2 ; (4.75) 2)1/2 = 24.20 %ips = 24?200 l&s
27
Using( !ocall+ ma"e ] inch "iameter ca&le 1 'ires stran"s each of 0.2 inch "iameter.
<:tra high strength gra"e 'ith &rea%ing strength = 5,?600 l&s .F.S = 2
Wor%ing strength = 5,?600/2 = 2?250 l&s
Lum&er of ca&les re*uire" = 24?200/2150 = 0.0 $a%ing (1) ca&le "ia ] inch on each si"e. ,esign of Anchor bloc for 1in-gu' cable (
- = Weight of &loc%? = Sli"ing force = 24.20 %ips Sli"ing increasing &+ 50M 1.5 = 24.2 1.5 = 6.60 %ips 6.60 = 0.4 -
\ = 0.4
-= 45.50 %ips
ig 2, olume of Anchor &loc% re*uire" for 'in" ca&le = Wt of &loc%/Enit Wt of &loc% = = 45.5/0.15 = 67 cft SiQe of Anchor &loc% for 'in" ca&le = Z : Z : 4Z earing ressure = Wt of loc%/Contact Area = 45.50/ = 0.,1 %/ft 2 2.2 H/ft 2 #ence O!K . ,esign of Sus&ender for 1in-gu' cable (
!oa" on one Suspen"er = !oa" from Woo"en Dec% ; Sno' !oa" ; !ive loa" = = 72.6 6.5/2 ; ,7.5 6.5/2 ; 105 6.5/2 = 46.46 l&s Using (
Dia (1/2) inch 'ire rope ? 1:7 (0.145) #igh Strength gra"e 'ith &rea%ing strength = 1,.,00 l&s .F.S = Wor%ing Strength = 1,?,00/ = 700 l&s Lum&er of ca&le re*uire" for Suspen"er = !oa"/'or%ing Str. = 46.60/700 = 0.1 $a%ing 1 "ia K inch ca&le for suspen"er.
,esign of 1ind ties (
$otal $hrust on &ri"ge 0.05 6.5 = 0.175 %ips = 175 l&s Esing "ia K inch ca&le 'ith 7 stran"s each "ia 0.145 inch . #igh strength gra"e = 1,?,00 l&s rea%ing strength 1,?,00 l&s .F.S = Wor%ing strength = 1,?,00/ = 700 l&s Lum&er of ca&les re*uire" = 175/700 = 0.067 $a%ing 1 "ia K inch ca&le for 'in" ties ,esign of column (
2,
$angent = "ip/span = 1,/270 = 0.2447
= 1.6 0
# = $ Cos 1.6 = 142.4/2 Cos 1.6 = 7,.75 %ips Where $ = $ension in main ca&le. = $ Sin 1.6 = 142.4/2 Sin 1.6 = 21.0 %ips !oa" on column = 221 = 2 %ips >ncreasing &+ 70M? u = 21.70 = 71. %ips Concrete Column u = (Stress Concrete Acc) ; (Stress Steel As) u = 1000260 ; A s 1,000 9in As = 0.01 A gross of Concrete = 0.012 60 9in .Asteel = 7.20 inch 2 Esing Dia. K inch &ars.?
Lum&er of &ars -e*uire" = A steel /A &ar = 7.2/0.2 = 64 &ars
ig 2 ,esign of combine footing (
!oa" on the top of column = 71.0 %ips !oa" on footing = !oa" on column ; Self loa" of column !oa" on ooting =71.0 ; (1,2/12 60/12) 0.15=71.0 ; 16.50 = ,.0 Sa+ ,5.00 $otal loa" on ooting = ,5 ; ,5 = 170 %ips Self 'eight of ooting = 10M of the total loa" = 170 0.10 = 17.00 %ips $F$A! !FAD FL FF$>LN 170.0 ; 17.0 1,7.00 %ips Area of the &ase of footing sla& 1,7/1.0 = 10 Sft Where 1.0 %/sft is allo'a&le soil pressure. Area ReIuired (
A = Z 12Z =10, sft @ 10 sft Let up'ar" pressure of the soil = 170/ 12 = 1.57 %/sft $he arrangement of the columns 'ith their respective position shall &e as s ho'n in igure.
ig 60 $he loa" per foot run of the footing = 1.57 = 1.147 %/ft
2
Shear force and Bending 6oment (
Shear force to the left of the column (A) = 1.147 6.25 = 4.06 %ips Where 6.25 = "istance from the left si"e toe of the footing to column center. Shear force to the right of column (A) = !oa" on footing J 4.06 = ,53 4.06 = 6,.57 %ips Shear force Bust to the left of column S. = 6,.573 1.147 5.5 = 6,.42 %ips Shear force Bust to the right of the column () S. = ,53 6,.42 = 4.06, %ips
ig 61 Bending 6oment ( Bending 6oment at column A % B 7 1.147 6.25 2/2 = 7.1, %3ft
$he 9a:. &en"ing 9oment 'ill &e in the center of &oth columns at the "ist ance of Distance = 6.25 ; 2.75 = 4 ft from either en"s. 9a:. . 9oment = ,5 2.75 J 1.147 4 2/2 = 266.75 J 255.004 = 321.254 %3ft
ig 62 ,e&th of the foundation (
A. rom punching shear consi"eration $otal punching force aroun" the perimeter of column (A) =!oa" on footing J Ep'ar" loa" & ! = ,5.03 1.57 2 ft 2.5 ft = 77.16 %ips Where & an" ! are the "imension of column &= 2inch = 2ft I ! = 60 inch = 2.5 ft !et (") &e the "epth of the footing un"er the column. $otal resisting force against punching &"S p Where & = Fne "imension of column " =
60
9a:. en"ing 9oment = 7.1, %3ft Depth of ooting = (.9/8) 1/2
Where is 'i"th of footing = ft = 10, inch
" = (7.1,12000/1,510,) 1/2 = 4.47 inch A"opt an overall "epth = 12 inch
As = 9/f s U " =7.1, 12?000/20?000 0.,72 = ,0?24,/154?40 = 5.47 inch rovi"e 1/2 inch "ia &ar . 61 "ia K inch 6.50 inch center to center (61 "ia 12mm cm c/c) Area of +ransverse steel bars ( J 2 "ia 0.6 inch (10 mm) 6.50 inch
ig 66
61
2
62
