Post on 19-Jul-2016
description
MANNING FORMULA
= Q
= V
= A
= B
= D
= Z Q=AxV
= n
= S
= P Duty =
= r
= Vc
1 Acre = sq.m = Ha
1 cubic meter = 1000 Litre
1 litre = 0.001 cu.m
1 litre =0.03529 cu.ft
Duty = 2 l/s/Ha
= 0.07078 cu.ft/s/Ha
= 0.02877 cu.ft/s/Ac
Discharge for 1000 Acs = 28.772358 cu.ft/s
1 cusec = 30 Acs
1000 Acs = 33.3333 Cusecs
= 0.94462 Cu.m/s 35.2876
= 944.62 l/s
STATION B(m) D(m) Z n A(sq.m) P(m) r(m)
Kaudulla Branch canal 1.5 0.9 0 0.015 1.350 3.300 0.409
say Free Board= 0.3 m
0.0004
Cement Mortar 0.018 Excavation earth canals for drains
4048.9292 0.404892921
S
Earth Excavated 0.025 Canals of capacity up to and including 100 cu.ft/s
Concrete- Average 0.015 Canals of capacity over 100 cu.ft/s
Canal Bed slope
Wetted Perimeter
Hydraulic mean radius
Critical Velocity
Type of canal Coeff.(n)Size of Canal
Rock Excavated 0.035
Bed width V = (r^0.6666)x(S^0.5)/n , or V = 1.486x(r^0.6666)x(S^0.5)/n
Full Supply depth Vc =0.5477 x( D^0.64) , or Vc= 0.84x (D^0.64)
Side slope of canal (ss)
Rugosity coefficient for the material Critical Velocity Ratio, CVR =V/Vc = 0.9 to 1.25
Sample canal design
Discharge P =B+2xDx((1+Z^2)^0.5)
Mean Velocity A =(B+DxZ)xD
Mean area r = A/P
Width B=1.5 m
Depth d=0.9 m
FB=0.3 m
Size of trough= 1.5 x 1.2 m
0.02833 cu.ft/s/Ac
Side Slope (ss)
1 on 1.5
1 on 2.0
2.0 on 1
V(sq.m/s)Vc
(sq.m/s)CVR Qd (l/s)
Extent
(Ha)Qr (l/s)
0.735 0.512 1.435 992.000 1000.00 40.000
Excavation earth canals for drains
Canals of capacity up to and including 100 cu.ft/s
Canals of capacity over 100 cu.ft/s
Size of Canal
V = (r^0.6666)x(S^0.5)/n , or V = 1.486x(r^0.6666)x(S^0.5)/n
Vc =0.5477 x( D^0.64) , or Vc= 0.84x (D^0.64)
Critical Velocity Ratio, CVR =V/Vc = 0.9 to 1.25
Sample canal design
P =B+2xDx((1+Z^2)^0.5)
A =(B+DxZ)xD
r = A/P
Refer:BS 8110- Clauses :3.3.1.6, 3.4.12, 3.4.1.6, 3.4.4.4, 3.5.5.2, 3.12.8.4,and 3.4.5.4
tw
D
L
tb
Bedwidth of rough = L
Length of base = Lb
Depth of rough = D
Full supply depth = Df
Weight of wall = Ww
Thickness of wall = tw
Weight of base = Wb
Thickness of base = tb
Weight of concrete = 24 KN/m3
Weight of soil = 2000 Kg/m3
Angle of friction of soil,Ø = 30 degrees
Weight of vehicle = 6.3052 KN/m2 (Not necessary)
Assume the Trough is full of water at critical condition
Consider 1 m -length of Trough
A D
B C
L (m) D (m) Df (m) tw (m)Ww
(KN/m)
Wb
(KN/m)
1.5 1.2 1.28 0.2 5.76 9.12
Water pressure at the bottom of the slab, P1 = 1.9 x 1000 x9.81/1000 = KN/m2
Say the Basic wind speed, V= m/s
And the multiplying factors, S1 = 1.1 S2= 0.7 S3= 1.0
Wind load , wk =0.613 Vs2 N/m
2 and Vs =V.S1S2S3
wk= 0.408 KN/m per m-length
Case:A Flume empty with wind load
Sample Design of Trough
18.64
1.9
33.5
tb (m)
0.2
Lb (m)
Dimension should
be designed
Depth of wind effect Dw= 2.1 m
(+) =
(Wind load) (self weight) (Wind+self weight)
For unit wind load BM at bottom of wall = 1 X Dw2 /2 = 2.2 KNm
For unit dead load BM at mid of slab = 1 XL2/8 = 0.281 KNm
At ulitimate condition, BM at bottom of wall = 1.4 X wk x 2 = KNm
At ultimate condition BM at mid of slab =1.4 X(Wb/L)X1.022
= 2.39 KNm
Resultant BM at mid of slab= 3.65 KNm
A D
B C
1.26 1.26
Case:B
Flume with water and wind
Take partial safety factors as 1.2 for each loading cases
(+) (+) (=)
BM at A, due to water=-[1.2 x(1/2)x1.92x9.81x(1.9/3+0.1)] = KNm
BM at A, due to wind=1.2 x(1/2)x(1.9+0.2)x2.0x0.408 = KNm
Resultant BM at A, due to wind & water = KNm
Design load on slab = 1.2x(0.2x1x1x24)+1.2x1.9x9.81 = KNm
BM at mid of slab =28.127x3.262/8 = KNm
Resultant BM at mid span of slab = KNm
A D
1.02816
3.65
-15.582
1.26
-14.554
28.1268
28.7582
14.2042
-14.554 -14.55404
B C
Case:C Flume only with water
A C
B D
At ultimate condition
Water pressure at the bottom of wall=1.4 X1.9 x9.81= 26.09 KN/m per m- length
BM at bottom of wall =-[(1/2) X 26.09 X1.9 X(0.1+1.9/3) ] =-18.18 KNm
Ultimate load on slab=1.4 X 0.2 X1 X24 +26.09 = 32.81 KN/m
BM at mid of slab= 32.81X 2.862/8 = KNm
Resultant BM at mid of slab = 15.37 KNm
A D
-18.18 -18.18
B C
Case
A
B
C
Take ultimate BM at mid span of slab Mu = KNm
Depth of slab =200 mm
Clear cover to r/f = 40 mm
Dia. Of Tor r/f = 10 mm
Yield strenght of r/f =460 N/mm2
Strength of concrete fcu = 25 N/mm2
Effecive depth d= 155 mm
K= Mu /(b d2).fcu = 0.0256 < 0.156 =K
1
No compression r/f is required
15.37
14.204
33.5466
BM at end of
slab(KNm)
BM at mid of slab
(KNm)
14.2042
15.37
-18.18 15.37
1.26 3.65
-14.55
Z = d [ 0.5 + [0.25-(k/0.9)]0.5
] = 149.49 mm
0.95 d = 147.25 <149.43 hence take Z = 147.25 mm
As =Mu/(0.87fy. Z) = 260.775 mm2
Cross section of 10mm dia. Bar = mm2
Hence No. of bars required =
Use T12@ 250 -B
At the bottom of wall Mu= 18.18 KNm
K= 0.0303 < K1
Z= 149.6 >0.95d mm
Take Z= 147.25 mm
As = 308.49 mm2
No. of bars required = 3.9263
Use T10@ 250 - Inner face of wall
Provide 0.13% of Secondary r/f = 195 mm2 per 1-m length
No.of r/f = 2.4818
Spacing = 402.93
Provide T10 mm dia. Distribution r/f at 300 mm spacing
Check for shear
UDL on slab = 1.4 x0.2 x 1 x 24 +1.6 x 1.9 x 9.81 x1= KN/m per 1-m width
Maximum shear force =36.54*2.86/ 2 = 27.407 KN
Shear stress , v =52.256x1000 /(1000 x 155) = N/mm2
Concrete shear stress, vc =0.79(100As/(bvd))0.33
(400/d)0.25
/rm
As (mm2) bv (mm)
Effective
depth
d(mm)
314.29 1000 155
vc > v, hence shear r/f is not required
Design of wall as beam -simply supported on piers
Wt.of slab = 0.2x2.86 x24/2 = 6.864 KN/m
Wt. of wall = 0.2 x(1.9 +0.2) x1 x 24 = 10.08
Total dead load = 16.944 KN/m
Wt.of water = 1.9 x9.81 x2.86/2 = 26.654 KN/m
Ultimate design load = 1.4 x 16.944 +1.6 x 26.654 = KN/m
Say the length of beam = 6.686 m
KN/m
1.25 0.47
0.17682
rm
66.368
vc (mm2)
36.54
78.5714
3.31896
Max. BM at mid of span = 66.368 x 6.686 2 /8 = KNm
Max. shear force at the end of beam = 66.368 x 6.686/2 = 221.867 KN
Effective depth of beam = 1900+200 -40-10/2 = mm
K =M/bd2 fcu = 0.0176 <0.156
No compression bars are required
Z = 2055 x ( 0.5 +(0.25- 0.0176/9)0.5
) = 2051 mm
0.95 d = 1952.3 < 2051 mm, hence take Z= 1952.25 mm
As = M/(0.87 fy Z) = 474.664 mm2
Use T20 mm bar hence the cross sectional area = mm2
No. of r/f required = 1.5103
Use 2 T20 - B
Asmini = 0.13 x 200 x 2055/100 = 534.3 mm2
Asprovided = 628.57 >534.3 mm2
Max. shear stress at support = 221.87 x 1000/(200x 2055) = 0.540 N/ mm2
100 As /bd =(100 x 628.57/200 x 2055) = mm2
400/d = 400/2055 = 0.1946
Vc =0.79 x (0.1529)0.33
x(0.1946)0.25
/1.25 = N/mm2
0.5Vc = 0.11293466 N/mm2
Vc +0.4 = 0.6259 N/mm2
Hence 0.5Vc < V<Vc +0.4
Max.spacing = 0.75 deff = 1541.3 mm
Spacing of links, Sv < 0.87 fyv Asv /(0.4 bv)
Use T10 links , Asv = 157.143 mm2
Sv < 786.093 mm
Provide T10@400 links
370.851
2055
314.29
0.15294
0.22587
Design of Pier
Refer:BS8110, cl3.8.2.4,3.8.1.6.2,cl3.9.3.7.2,cl3.8.1.3,cl3.8.3.1,cl3.8.4.5,cl3.8.4.3,cl3.12.11.2.3
cl5.2.3.2,cl5.2.3.4,cl5.2.3.3,cl3.4.4.4,cl5.2.7.2.3,cl3.4.5.2,cl3.11.3.2,cl3.12.11.2.7,cl3.4.5.10,cl3,7.6.6
(Dimension should be designed) 0.2
0.95
wk 1.9
0.2
0.95 0.15 2.86
62.698 m MSL 0.35
0.55 0.2
3.66 0.2
0.3
T T
5.0
57.0 m MSL
0.75
0.3 0.3
0.6
1.155
0.955
1.22
1.9
0.2
0.2
0.3
0.75
S S
Actual height of pier, lo =62.698-57.0-0.2 = m
Effective height of pier, le=βlo , where β = 1.3 , hence le = m
For unbraced column , le < 30 x Pt , where Pt = Thickness of pier
Pt > 7.147/30 = m
Say, Pt=600 mm and le/ Pt < > 10 hence the pier is slender
Take the distance between piers = 7.5 m
Weight of flume= (0.2x1.9x2 +(2.86+0.2x2))x7.5x24 = KN
Assume the flume is full of water
Hence the weight of water =7.5x1.9x2.86x9.81 = KN
Say the width of head =1.22m
Hence the weight of pier head =1.22 x0.35x2x0.2X24+1.22x0.2x3.66X24+(3.66X1.22+0.75X0.6)X24/2 KN
= 84.515 KN
Total dead load = 254.16+84.515 = KN
Design load at ultimate condition,N =1.4Gk+1.6Qk =1.4x338.675+1.6x399.81 = KN
Nmax=(0.4xfcu)bd/2=(0.4x25)x750x600/2 = KN
Nmax/2= 2250/2 =1125 KN> 1113.8 KN
Computation of forces due to water current
Force due to water current Pw = KVD2
0.235
5.498
7.147
0.23823
11.9123
338.675
0.15
1113.84
2250
254.16
399.81
Velocity of stream VD=3.232 m/s
Constant K=0.35 , hence Pw= KN/m2
Height of water at stream =HFL- Bed level =61.5 -58.6= 2.9 m
3.656
Pw
2x2.9/3
Force due to water current, fw =3.656x(2.9x0.6 )/2= KN
20% of fw is acting on lateral direction
Hence force on lateral direction= (20/100)x(3.181)= 0.64 KN
Force due to debris= Pd =0.56Vd2 = 0.56x3.232
2= KN/m
2
Assume the average depth of debris=1.2 m
Force due to debris, fD =1.2x0.6x5.85 = KN
Force due to debris on lateral direction =(20/100)x4.212 = KN
1.22
1.9
5.850
4.212
0.8424
3.656
3.181
0.2
0.2
0.3
0.75
61.5 mMSL HFL
fD 0.6
fw 2.3
58.6 mMSL stream Bed 1.93
1.15
P
5.7 mMSL
Y
0.3
b=0.6 m 1.35 X 0.75 X
h=0.75 m
0.3
Y
0.6
Designed ultimate capacity of a section when subjected axial load only,
Nuz=0.45fcuAc +0.87fy Asc
and Madd=Nau=N βaKh =NKh (le/b')2/2000
where, K= (Nuz - N)/(Nuz-Nbal) < or = 1
Take K=1,
and the depth of cross section measured in the plane under consideration, h=1.35 m
Hence, Madd= 1113.8 x 1x1.35x (7.147/0.6)2/2000 = KNm
Self weight of pier =5.0x24x(0.75x0.6+2x0.3x0.3/2) = KN
Self weight of flume = KN
Weight of pier head = KN
Total dead load Gk= KN
254.16
106.67
0.15
84.515
414.28
75.6
Weight of water Qk= KN
Ultimate axial load, N=1.4Gk+1.6Qk =1.4x414.28 +1.6x399.81 =
Application of this load at the edge of the pier is considered to be the critical condition
BM at point P, along Y-Y axis, Minitial= (1.4x(254.16/2)x1.22/2)+(1.6x0.64x(2x2.9/3)+1.15)
+(1.6x0.8424x(1.15+2.9-0.6))
Minitial=
My =Minitial +Madd = 116.306 +106.67 = KNm
MB at point P,along X-X axis due to wind, water current and debris
Mx =1.4x0.408x1.9x7.5x(5+0.3+0.2+0.2+0.95)+1.6x3.181x((2/3)x2.9+1.15)
+1.6x4.212x(1.15+2.3)
Mx = KNm
Say the cover to r/f = 50 mm, hence b'=600-50 =550 mm and h'=750-50=700 mm
My/b' =222.98x106/550 =0.4054 x10
6
Mx/h' =69.82x106/700 =0.0997x10
6< My/b'
N/bhfcu=1219.69x103/(600x750x25) =
β =0.88 +(0.77-0.88)x(0.10842-0.1)/(0.2-0.1) = 0.87
My'=My+β(b'/h')Mx =222.98+0.87x(550/700)x93.07 = KNm
M/bh2 =286.6x10
6/(600x750
2) = N/mm
2
N/bh =1219.69x103/(600x750)= N/mm
2
Since these values are very small nominal r/f can be provided
Hence, 100Asc/bh =0.4, where Asc is the area of steel in compression
Asc=0.4*600*750/100 = mm2
Minimum area required by concrete Ac=N/0.4fcu=1219.96x103/(0.4x25) = mm
2
<750x600= mm2
Ac-required < Ac -provided
Provide r/f for 1% of crushing area, hence Ac = 1x121996/100 = mm2
Diameter of bar=20 mm and cross sectional area = mm2
No of bar required =1219.96/314.286 =
use 4T20@165
Design of Corbal
Self weight of flume = 254.16 KN
2.71042
1800
1219.69
286.60
0.84919
450000
314.28571
121996
KN
116.306 KNm
1219.96
3.88164
399.81
222.98
93.07
0.108417
Weight of water = 399.81 KN
Ultimate load =1.4Gk+1.6Qk =1.4x254.16+1.6x399.81= KN
Self weight of corbal beam =24x[3.66x1.22x0.2 +{3.66x1.22+0.75x0.6}/2] = KN
Ultimate bearing stress=0.4fcu = 0.4x25 = 10 N/mm2
610 av
200
300 300
310
l=1220/2=610 mm
Ultimate design load of beam = KN/m
Reaction at pier =66.368x7.5/2 =
Effective bearing length =(610/2)+100 = mm< 600 mm
Net bearing width = 248.88x103/(405x10) = mm <200 mm (actual bearing width)
Ultimate weight of corbal/area =1.4 x 39.128/(3.66x1.22) = KN/m2
Let the load transmitted at the one third of bearing area from load face
av = (2/3)x(1220/2)-300 = mm
BM at S=(995.5/2)x0.106 +(3.66x0.2x0.31x24)x(0.31/2) +(0.31x0.3/2)x3.66x24x0.31/3 KN
= 54.0277 KNm
h=600 mm, hence d=600-50=550 mm
K=M/bd2fcu =54.028 x10
6/1000x550
2x25 =
Z=550 x[0.5+(0.25-0.0071/0.9)0.5
] = mm
0.95d =0.95x550= 522.5 mm< 545.626 mm
Hence take, Z=522.5 mm
As=M/(0.87fyZ) =54.028x106/(0.87x460x522.5) = mm
2
As-minimum=0.0015bh =0.0015 x1000x600 = mm2
Use T20 mm bars, hence No.of bars required = 900/(πd2/4) =
900
2.864
995.5
39.1277
1220
66.368
248.88 KN
405
61.45185
12.27
545.626
258.38
106.667
0.00714
Use 3T20 at 300 mm spacing
BM at T due to water& flume =(995.5/2)x(0.955/2) = KNm
MB at T due to self weight of corbal =12.27 x1.155x0.235x1.155/2 = KNm
Total BM at T=237.68+2.58 = KNm
K=M/bd2fcu =240.26x10
6/(1000x550
2x25) = <0.156
Hence take Z=0.95d=522.5 mm
As =M/(0.87fyZ) =240.26x106/(0.87x460x522.5) = mm
2
Use 16 mm dia. Bars, No. of bar required =1149/(πd2/4) =
Use 6T16@160
Area of horizontal links As =(Main tension r/f)/2 = (3xπx202/4)/2 = mm
2
Depth of link provided= (2/3)x550 = mm
Use T16@175
Axial load N=1219.69 KN
Shear stress on corbal = 1219.69x103/1220x550 = N/mm
2
Allowable shear stress =0.8 fcu0.5
= >1.818 N/mm2
Design of foundation
0.6
0.3 0.3 2.6
0.3
0.3
0.6 0.6 0.3 0.375 0.3 0.6
3.3
Assume the bearing capacity of soils = KN/m2
Take the width of foundation = 1.8 m
Length of foundation = 2.55 m
Depth of foundation = 0.45 m
Self weight of foundation=2.55x1.8x0.45x24 = KN
Self weight of flume = KN
Weight of water = KN
Self weight of pier = KN
Self weight of pier head = KN
Total dead load on foundation = KN
Imposed load on foundation = KN
0.375
366.7
1.818
4
237.68
0.03177
2.58
240.26
1149.0
5.71235795
471.43
863.657
399.81
399.81
75.6
84.515
254.16
150
49.572
Total service load on foundation = KN
Bearing pressure =1263.47/(1.8x2.55) = KN/m2 >150 KN/m
2 ,not allowable
Required base area =1263.47/150= m2
Say the size of foundation = 2.6x3.3 m2
Hence the bearing pressure =1263.47/(2.6x3.3) = KN/m2 < 150 KN/m
2safe
Ultimate axial load =1.4 Gk +1.6 Qk =1.4*863.657+1.6*399.81 = KN
Moment in X-direction due to water, wind & debris, Mx =93.07 KNm
Moment in Y-direction due to flume load, My=222.98 KNm
B
1.0
A A
0.3 0.3 2.6
0.3
0.3
1.0 0.975 0.3 0.375 0.3 0.975
B
3.3
Stress due to axial load =1848.82/(2.6x3.3) = KN/m2
Bending stress, σx =Mx.y/I =93.07x(3.3/2)/(2.6x3.33/12) = KN/m
2
Maximum resultant moment, σx-x =(215.48+19.72) = KNm
maximum clear cover to r/f = 40 mm
use 16 mm dia. Bar
Hence effective depth = 450-(16/2)-40 = mm
Critical BM at B-B =235.2 x0.675x2.6x0.675/2 = KNm
M/(bd2)= 139.3x10
6/(2600x402
2)=
100 As/bd2 =0.14
As =0.14x2400x402/100 = 1350.7 mm2
As -minimum=0.15 bd %=0.15x2400x402/100= mm2
No of bars required = 1447.2/(πx162/4) =
use 8T16@290
Bending stress σy =My.X/I =222.98x(2.6/2)/(3.3x2.63/12)= KN/m
2
Maximum resultant Moment σy-y =215.48+59.973 = KNm
Critical BM at A-A=275.453x0.3x3.3x0.3/2 = KNm
19.72
235.2
1447.2
139.3
0.460
0.375
402
1848.82
275.453
40.90
7.19
59.973
8.42313
147.25758
275.266
1263.47
215.4802
M/bd2=40.9x10
6/(3300 x402
2) = N/mm
2very small
Hence minimum r/f required =0.15bd% =1447.2 mm2
Use 11T16@290
As-provided =201x11 = 2211 mm2
Check for distribution of r/f
X-direction, c=600 mm -width of column, d-eff.depth
(3c/4)+(9d/4) = 1354.5 > lc= 1000 mm ,distance to the edge of pad
Y-direction,3c/4+9d/4 =3x(750+600)/4+9x402/4 = 1917 > lc=975 mm
Hence the distribution is O/k
Max. spacing between bars =3d=3x402 = mm -O/K
Anchorage length =40φ =40x16 = mm
Distance between column face and of the footing =[3300-(750+600)]/2 & (2600-600)/2
=975 mm & 1000 mm -O/K
Check for vertical line shearX
d
1.5d
1.5d=1.5x402 = mm
2556
1806
603
1206
640
0.07669
X
Stress at section X-X =215.48+59.973x(1.3-0.402)/1.3 = KN/m2
= 0.257 N/mm2
vc=0.79(100As/bvd)0.33
x(400/d)0.25
x1/гm =0.79x(100x2211/3300x402)0.33
x(400/402)0.25
x1/1.25
= N/mm2
safe.
Stress at section Y-Y =215.48+19.72X(1.65-0.402)/1.65 = KN/m2
= 0.23 N/mm2
vc = 0.79x(100x2211/2600x402)0.33x
(400/402)0.25
/1.25 = N/mm2
safe.
Vmax =1848.82x103/(2x(1806+2556))x402 = <0.8fcu0.5
= 4 N/mm2
safe.
Area out side the perimeter =3.3x2.6-2.556x1.806 = m2
Shear force =215.48x3.96 = 853.3 KN
Shear stress =853.3x103/(2x(1806+2556))x402 = < 4 N/mm
2safe0.24331
256.91
18.2884
230.40
19.7114555
0.52717
3.96