Design of Hydraulic Gate
Design of Hydraulic Gate
Design of Hydraulic Gate
Page 1 to 20
~2~
REGULATOR CUM ROAD BRIDGE AT CH 19.115KM OF DABU KHAL IN PS CANNING
Canal data
1. Design discharge= 56.4cumec
2. Crest level of sluice= (-)1M
3. Full discharge level= 0.5M
4. High flood level= 2.395M
5. Bed level= (-)2.5M
6. Bed width= 19M
River data
1. Tidal lockage period= 8hr
2. Low tide level= (-)0.19M
Bridge data
1. Carriage way width= 3.5M
2. Proposed road top level at centre= 3.905M
3. Minimum vertical clearance from HFL to the soffit of bridge slab= 0.9M
~3~
=
(3.395 2.1)2
_______________________
(3.395)2 - ((3.395 2.1)2
5.18
= 0.5259
________________
9.849
2 x 3.395
y1 =
________________________
[ (1 + 0.5259)
1.5
- (1 1 + 0.5259)1.5 ]
_________________
3 4 + 0.5259
= 1.0639 [ 1.8849 0.3814 ]
= 1.5996
y2
=
y3 =
=
y4 =
=
2.2656
1.0639 [ (3.5259)1.5 - 4.0144 ]
2.7729
1.0639 [ (4.5259)1.5 - 6.6207 ]
3.2
l1 = 3.395 3.2 = 0.195 M
l2 = 3.395 2.7729 0.195 = 0.4271M = 427 mm.
l3 = 3.395 2.2656 0.195 - 0.427 =
l5
l2 = 430 mm.
l4
l3 = 500 mm.
l4 = 670 mm.
l3
C
l2
B
l1
A
l5 = 305 mm.
~4~
Ah (for HFL)
max
= 5407626 N.mm.
Bending Stress :
= M/Z
Allowable bending stress for wet and inaccessible condition of structural steel
permissible = 100 N/mm2
Section modules of horizontal girder, Z = M/permissible = 5407626 N/mm
100 N/mm2
= 54076.26 mm3
~5~
=
=
n (H - h) 2
------------------H2 - (H h)2
2(3.395 2.1) 2
---------------------------------(3.395) 2 - (3.395 2.1) 2
3.3541
= -------------9.849
y
= 0.3406
2H
= ----------------- [ (k + p)3/2 - (k + + - 1)
---------3 n +
3/2
2 x 3.395
y1
= -----------------------------
[ (1 + 0.3406)2.5 - ( 0.3406)1.5 ]
---------------------
2 + 0.3406
= 1.4794 [ 1.3534 ]
= 2 m.
xx
R1
= 68100 mm3
P1
P1
= 10869.6 N
Reactiona at roller position
R2
R1
R2 21739.2 N
P1
Bending moments
MA = 10.869.6 N x (400 195) mm. = - 2228.268 x 103 N mm.
MB
10.869.6 N x 430 + 21739.2 x (430 + 195 400) mm. = + 217.392 x 103 N mm.
MD
= -
MC
= -
~6~
M Max
max =
--------------
3206.532 x 10 N mm
---------------------------- =
3
68100 mm
47.08 N/mm2
__F________
2 h permissible
21739.2
2 x 125 x 75
(hence OK)
N/mm2
F) Skin Plate
Bending stress in flat plate
K
________ x
p a2
____________
100
where
s2
Let plate thickness = 8 mm, considering 1.5 mm corrosion allowance effective thickness =6.5 mm.
Panel B C
3y
4y
a=430
4x
b=1800
g =
1000 kg
_________
m3
m
x 9.8 ____
s2
~7~
Pressure at the plate centre (p) = 3.395 0.195 (0.43/2)
= 2.985 m. of H20
= 9.8 x 10-6 x 2985 N/mm2
= 29.253 x 10-3 N/mm2
From table 2, page 13 of IS 4622/03
k2x = 25, k2y = 7.5 , k3x = 50, k4y = 34.3
2y =
= 9.6
3x =
N/mm2
= 64 N/mm2
(hence OK)
3y =
(0.3 x 64)
4y =
= 19.2 N/mm2
(6.5)2 mm2
= 43.9 N/mm2
4x =
Combined Stress
________________________
= x 2 + y 2 - x y + 3Fxy 2
__________________________
=
[ 32 2 + 9.6 2 - 32 x 9.6 ]
= 23.44 N/mm2
_________________________
= (64) 2 + (19.2) 2 + 64 x 19.2
= 56.38 N/mm2
________________________________
= (43.9) 2 + (39.51) 2 43.9 x 39.51
= 41.87 N/mm2
2,
3,
(hence OK)
~8~
Panel C D
3y
4y
a=500
4x
b=1800
p
gh ;
24.66 x 10
-3
N/mm2
2y = (36.479 x 7.5/25)
= 20.94
N/mm2
3x = (36.479 x 50/25)
= 72.958
N/mm2
3y = (0.3 x 72.958)
= 21.88 N/mm2
4y = (36.479 x 34.3/25)
= 50.05
N/mm2
________________________
= 36.5 2 + 11 2 - 36.5 x 11
=
37.64 N/mm2
___________________________________
= (72.95) 2 + (21.88) 2 72.95 x 21.88
= 64.84 N/mm2
(hence OK)
~9~
________________________
= (50) 2 + (15) 2 15 x 50
= 44.44 N/mm2
2,
3,
(hence OK)
Panel D - E
b=1800
a=670
4y
4x
3y
3x
Pressure at the panel
centre = 9.8 x 10
-6
= 18.959 x 10
-3
h = 1.5996 + (0.67/2)
N/mm2
= 1.9346 m.
2x
k2x
= ------ x
100
pa2
-------s2
18.959 10
k2x x
k2x x 2.01
25 x 2.0
50 N/mm2
2y =
-3
N/mm2
x (670
---------------------------------------------------------------------100 x (6.5)2
k2y x 2.0
8 x 2
16 N/mm2
3x
50 x 2
3y
=
=
100 N/mm2
0.3 x 100
4y
=
=
30 N/mm2
34.3 x 2
4x
= 68.6 N/mm2
= 0.3 x 68.6 =
20.58 N/mm2
all)
~ 10 ~
Combined Stresses
____________________
= 50 2 + 16 2 - 50 x 16
=
44.22 N/mm2
_________________________
= (100) 2 + (30) 2 100 x 30
= 88.88 N/mm2
_______________________________
=
=
Panel A - B
b=1800
15x
a=195
13x
13y
14x
12x
14y
12y
11x
12x
15x
~ 11 ~
11y
12x
12y
13x
13y
14x
14y
= 62 x 0.2908 = 18 N/mm2
15x
G) Roller Dimension :
Material : Cast steel grade 280-520 of IS 1030/1998 with bush bearing
Roller dia. = 179mm, width = 60mm
Crown rad. = 850 mm
Bush ID =50mm, thickness=6mm, width= 60mm
Bush shall be force fit in the wheel
Wheel pin shall be running fit in the bushing.
Wheel load (normal condition) = 21739.2 N
when one wheel is not connecting (worst condition)
86956.82
Max. Wheel load = ------------3
= 28985.6 N
YS = 245 N/mm2
TS = 590 N/mm2
= 73.5 N/mm2
= 110 N/mm2
~ 12 ~
3/2
dmin =
32 x1018164.69
------------------3.14 x 98
= 47.3 mm.
J) Bronze Bushing
Bearing presume
= P/ld
29090.42 N
------------60 x 50
= 9.69 N/mm2
< 15 MPa
(hence OK)
I)
~ 13 ~
Proposed regulator cum road bridge at ch 19.115km of Dabu Khal in PS Canning
= 7.85 kg/cm3
Bronze ...
= 8.80 kg/cm3
Natural rubber
= 0.93 kg/cm3
MC 150 x 75
Total : (237.38 + 17.27 + 55.02+ 120.96 + 42.74 + 14.78 + 9.7 + 1.1) kg.
= 498.64 kg.
Flat bar for seal :
( 50 x 6)
Bar (100 x 8) :
~ 14 ~
2.
F = P/R (f
86956.82
= ----------------- [ 0.2 x 25 + 1.0 ]
89.5
= 5829.5 N (starting)
86956.82
= ----------------- [ 0.15 x 25 + 1 ]
89.5
= 4615 N (running)
3.
L=length of seal
P = av. pressure at seal centre.
Side Seal :
a = 32 , c= 28, L = 2100mm, p = 2.345 M of H2O
FR = Lp/2 [ a2/c + c ]
-3
N/mm2
3.11 N
~ 15 ~
Fs1 = f R
Top Seal
9.81
(3.395 2.1) x 1000 x --------- N/mm2
10-6
= 12.7 x 10-3 N/mm2
p=
= 0.738 N
4.
~ 16 ~
Design Hoist load = 1.5 Ton
A) ROPE SELECTION :
Single fall rope
Load on wire rope (each side) = 1500/2 kg. = 750 kg.
Factor of safety for normal operating condition = 6
Design load = 750 x 6 = 4500 kg. = 45 KN
We select round steel core wire rope of 6 x 37 construction 12 mm. dia.
rope of 1960 grade.
Breaking force of 12 mm did rope (IS 2266/02, Table No.3) = 90 KN
B) ROPE DRUM
Pitch dia of drum = 20 x 12 = 240 mm.
Lift of gate = 2.5 m.
No. of turns required = 2500/ x 240 = 3.31 = 4 (say)
Provide (4 + 2 + 1) = 7 nos turns. ,
say 4 mm.
1 x 7500 N
= --------------------------14 mm x 40 N/mm2
= 13.39 mm.
(hence OK)
= 845740 mm3
~ 17 ~
b =
Bending Stress
Mb
58.5 x 104
----------- = ------------------ = 0.6917 N/mm2
Z
845740
F = tooth load
q= constant
For Z = 16, q = 3.75 (IS6938, table 4)
For Z = 48, q = 2.8+(2.9-2.8)x2 /10 = 2.82
PCD of G1 =48x7 = 336 mm
PCD of P1 = 16x7 = 112 mm
Tooth load of gear G1, F = 947368.42/168 = 5639 N
Tooth load of Pinion p1, F = 5639 N
Tooth bending stress
For gear ,
For pinion ,
=
=
.
.
40.56MPa
53.94MPa
< 21 N/mm2
(hence OK)
~ 18 ~
D) Torque Shaft/floating Shaft :
Length = 500mm
Material FS C45, class 4, (IS 2004), hot rolled & normalised
TS= 620MPa, YS= 320MPa
Permissible shear stress,
,
Twisting moment Mt = 332409.97 N.mm
Minimum shaft dia., d
min
= (16 Mt /
Fpermissible )1/3
1/3
16 x 332409.97
--------------------3.14 x 96
26 mm.
(hence OK)
E) Drum shaft :
Length = 386 mm
Material FS C45, class 4, (IS 2004), hot rolled & normalised
TS= 620MPa, YS= 320MPa
Permissible bending stress,
P2
128
165
R1
R2
P1=390+5639= 6029 N
P2 = 7500+ 400 = 7900 N
Reactions, R1 = 7954 N , R2 = 5975 N
Moment at P1 = 7654 x 93 =739722 Nmm
Moment at P2 = 5975 x 165 = 985875 N mm
Bending stress
32 M
---------- d3
~ 19 ~
Take shaft dia. d = 50 mm
b =
32 x 985875
-------------------- = 80.37 MPa < 124 MPa (hence OK)
x 503
= 26.59 N/mm2
947368.42
Crushing Stress = ---------------------50/2 x 50 x 5
= 53.18 N/mm2
~ 20 ~
Material:
GEAR(G1)
PINION(P1)
Teeth:
48
16
Module:
PCD:
336
112
Width:
56
56
Addendum:
Dedendum:
8.75
8.75
Hub dia:
100
Hub length:
56
4. Drum shaft: Material: C45, IS2004, hot rolled & normalised, dia.=50mm, L=386mm
5. Floating shaft: Material: C45, IS2004, hot rolled & normalised, dia.=40mm, L=500mm
6. Wire rope: Round steel core type, 6x37 construction, 12mm, 1960grade, IS2266(table:3)
7. Flange coupling: Cast steel, flange OD160, Thickness 20, Hub OD 90, Length 82, 04 nos
M16 bolt at 125, as per drawing & IS 6196.
8. Manual drive handle: MS, crank length=350mm