DESIGN CALCULATION OF SPILLWAY GATE 1 Design data Type Radial gate Max.Flood level 337 m High water level 336 m Low water level 336 m Crest El. 340 m Clear span 14 m

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DESIGN CALCULATION OF SPILLWAY GATE







1

Design data







Type

Radial gate






Max.Flood level


337

m





High water level

336

m





Low water level


336

m





Crest El.


340

m





Clear span


14

m





Height



11

m





Design head
-
H


11

m





Quantity


2

sets





All.stress
-
SS400

1200

kg/cm2





All.stress
-
SM490

1600

kg/cm2





Corr.all.



2

mm























2

Economic radius of radial gate







R =

1.25 * H






=

1.25*11






=

13.75

m





sin β/2=


h/2







R





=


5.5







13.75





=


0.4





β/2 =

23.58

o





β =

47.16

o





< =

0.26

radiant











































3

Arc length of gate







L =

R * (< in radiant)






=

3.14*13.75 *0.26






=

11.31

m











4

Spacing of horizontal girder






(a)

Number of girder
-
N, according to standard for H= 8.5 to 12




=

3

(b)

Water pressure, t/m2


spacing, m

kg/cm2



p0 =

0.000





0.000


a










2.245





b1 =

3.726











p1 =

4.491





4.491


b










6.134





b2 =

3.189











p2 =

7.778





7
.778


c



8.910





b3 =

2.644






9.559





p3 =

10.042





10.042


d








b4 =

1.441

10.521



p4 =

11.000





11.000


e


11.000











5

Vertical stiffeners






5.1

Bending moment on vertical stiffeners







M ab=

p1*b1^2/6






=

4.49*3.73^2/6






=

10.391

tm





=


1,039,086

kg
-
cm












M cb & bc =

(p1*b2^2/12)+(p2*b2^2/12)






=

(4.49*3.19^2/12)+(7.78*3.19^2/12)






=

10.40

tm





=


1,039,761

kg
-
cm












M cd & dc =

(p2*b3^2/12)+(p3*b3^2/12)






=

(7.78*2.64^2/12)+(10.04*2.64^2
/12)






=

10.38

tm





=


1,038,110

kg
-
cm












M de =

(p3*b4^2/2)+(p4*b4^2/3)






=

(10.04*1.44^2/2)+(11*1.44^2/3)






=

10.43

tm





=


1,042,558

kg
-
cm





Max. bending moment
-

Mb max.








Mb max. =


1,042,558

kg
-
cm











5.2

Number of

vertical stiffeners







Spacing, s =


0.60

m





N =


(B/s)+1






=


14/0.6+1






=


24












5.3

Bending stress







σcd =

Mmax

=


1042557.53





Zxl


815.90













=

1042557.53







815.90






=


1,277.80


tf/cm2
















Ma
terial:


JIS G 3101


SS400





H

B

Tw

Tf


I

mm

450

200

9

14


Cor.all.

mm



1




I x1 =



18,195

cm4




Z x1=



816

cm3




Aw =



77.54

cm2




G =



60.87

kg/m


5.4

Weight of vertical stiffener







Gst =

N * L * G






=

24*11.31*60.87






=


16,75
3.18

kgs




6

Horizontal girder






6.1

The end frames shall be fixed at about 0.2 * B distance from each end.







Distance of end frame from each end of the gate, e







b =

0.2 * B






=

0.2 *14






=

2.8

m





=

280

cm





Distance between end frames







e =

B
-

( 2*b )






=

14
-
(2 *2.8)






=

8.4

m





=

840

cm











6.2

Loads acting on horizontal girder






(1)

Beam B







wb = wb1 + wb2







wb1 =

(p0 + 2 * p1) * b1







6






=

(0 + 2 * 4.49) * 3.73


=

5.5
77

tf/m



6






wb2 =

(2 * p1 + p2) * b2







6






=

(2 * 4.49 + 7.78) * 3.19


=

8.908

tf/m



6






wb = 5.577 + 8.908



=

14.485

tf/m





=

144.85

kgf/cm

(2)

Beam C







wc = wc1 + wc2







wc1 =

(p1 + 2 * p2) * b2


=





6







(4.49 + 2 * 7.78)
* 3.19


=

63.930




6



6






=

10.66

tf/m


wc2 =

(2 * p2 + p3) * b3


=





6






=

(2 * 7.78 + 10.04) * 2.64


=

94.23




6



6






=

15.71

t/m
















wc= 10.66 + 15.71



=

26.36

tf/m





=

263.60

tf/cm

(3)

Beam D







wd= wd1 + wd2







wd1 =

(p2 + 2 * p3) * b3







6






=

(7.78 + 2 * 10.04) * 2.64


=

12.278

tf/m



6






wd2 =

(2 * p3 + p4) * b4







6







(2 * 10.04 + 11) * 10.52


=

7.465

tf/m



6






wd = 12.278 + 7.465



=

19.743

tf/m





=

197.43

kg/cm









wmax =



=

263.60

kg/
cm


MBA =

wmax * b ^2

=

263.6 * 280 ^2





2


2






11,356,240.00







2








=


5,678,120.00


kg
-
cm

6.3

Bending stress







σBC =

MBC

=

5678120





Zxl


10454.64






=

11356240







10454.64




Upper girder


=


1,086.24


kgf/cm2


Material:


JIS

G 3101


SS400





H

B

Tw

Tf


I

mm

900

400

25

25


Cor.all.

mm



1











I x1 =



468,368


cm4



Z x1=



10,455


cm3



Aw =



379.50


cm2



G =



297.91


kg/m








6.4

Weight of horizontal girder







Gst =

N * B * G






=

3*14*297.91






=


166,828.20

kgs











7

Skin plate







From the practical safety consideration the thickness of skin plates is of not less than 10 mm.







An allowance of 2 mm thickness is made to allow for corrosion and rusting.







This allowance is included
in the minimum specified thickness of 10 mm.







So that the effective thickness of skin plate is only 8 mm.













7.1

Bending stress







σs =

K * a^2 * p







100 x t^2












7.2

Result of calculation







k =

50





a

b

b/a

p

t



(cm)

(c
m)


(kgf/cm2)

(cm)

σs


60

372.600

6.21

2.245

1.80

(kgf/cm2)


60

318.900

5.315

6.134

2.80

1247.4


60

264.400

4.407

8.910

3.80

1408.4


60

144.100

2.402

10.521

3.80

1110.6



allowable bending stress



SM
-
490

1311.5



Corrosion allowance




0.2

cm


Thickn
ess of skin plate



1

20

mm





2

30

mm





3

40

mm





4

40

mm

7.3

Weight of skin plate, G=W * H * t * g







Width
-
W

Height
-
H

Thickness
-
t

Weight
-
G




(dm)

(dm)

(dm)

(kg)



1

140

37.26

0.2


8,189.75



2

140

31.89

0.3


10,514.13



3

140

26.44

0.4


11,
623.02



4

140

14.41

0.4


6,334.64



5

Total




36,661.54










8

Total weight of moving parts exluding end frame







Weight of vertical stiffener


=


16,753.18

kgs



Weight horizontal girder


=


166,828.20

kgs



Weight of skin plate


=


36,661.54

kgs



Total


=


220,242.92

kgs









9

Friction force






9.1

Friction force on pin







Assume pin pin diameter
-
df



=

600

mm





=

0.3

m


Friction between bronze and steel
-

ff



=

0.25









(a)

Maximum horizontal thrust on pin, P=







P =

B * H^2







2 * n






where n = number of end frame = 2








=

14 *11^2


=

484





2 * 2


4







=

121

tf

(b)

Friction force on pin







Fp =

P * ff

=

121 * 0.25







=

30.3

tf

( c)

Load due to friction
-

Tp








Tp * R = Fp * r






Tp =

Fp * r / R

=

30.25 *0.3 / 13.75




where



=

0.66

tf


R =

Radius of radial gate






Fp =

Friction force on pin






R =

Radius of pin












9.2

Friction force due to seal rubber






Assume width of seal bears water pressure
-

wr



=

50

mm





=

0.05

m


Pressure area of rubber
-

ar =


2*L*wr =

2*11.31*0.05







=

1.13

m2


Average water pressure
-

pw



=

5.578

t/m2


Friction coefisien between rubber and steel
-

fr



=

1.1



Load due to friction of rubber seal
-

Tr







Tr =

pw * ar * fr

=

5.58*1.1
3*1.1







=

6.94

tf

10

Center of gravity







DESCRIPTIONS

WEIGHT

DISTANCE

MOMENT





(kgs)

(mm)

(kgs.
-
mm)



1

gate leaf


220,243


11,311


2,491,162,385



2

end frame


24,575


6,875


168,953,560



3

total


244,818



2,660,115,945



4

center of g
ravity



10,866

from center of pin





























11

Steel wire rope






11.1

Tension on Steel wire rope in normal condition







diameter, d




35

mm


breaking strength, T





kgf


unit weight, Wc




90

kg/m


Taking length of wire rope
-

Lc



=

12.31

m


Tension in wire rope due to its own weight



=

1.11

tf


Total tension in one side of wire rope in normal condition
-

Wc







Wc =

(Gg+Gf+Tr)/2

=

220.24+1.96+6.94/2




where



=


114.57

tf


Gg =

weight of gate leaf


=


220.24

tf


Df =

por
tion of end frame weight at wire rope position






(Dwg.3)



=


1.96

tf








11.2

Tension on wire rope in Emergeny condition






(a)

Related dimensions







span
-

B



=

14

m


Radius
-

R



=

13.75

m


tan A =B/R

=

13.75/14

=

0.98



angle < A



=

44.4
8

o

(i)

Nearest distance from pin to diagonal line of another pin to the edge of gate leaf
-

"X"







X/B=

SIN(A)






X=

sin(A)*B


=

9.81

m


angel < D =



=

45.49

o

(ii)

Distance from intersection of diagonal line and gate center line to pin centerline
-

"Y"







Y/X =

SIN(D)






y =

sin(D)*X



7.00

m

( iii)

C of G =



=


10.87

m





from center of pin



(iv)

Distance from intersection of diagonal line and gate center line to center of gravity
-

"Z"







Z =

10.87
-
7


=


3.87

m

(v)

Nearest distance from c.g to diagonal line of another pin to the edge of gate leaf
-

"p"














p =

sin(D)*X


=

2.758

m


(Dwg.4)













(b)

Moments







Taking moment about vertical plane passing through diagonal line







Gg*p = Rd*X






(i)

Reaction at point d, Rd







Rd =

Gg * p







X






=

110.12*2.76

=

303.68





9.81


9.81







=

30.94

tf






(downward)














(ii)

Taking moment about vertical plane passing through wire rope line P
-
Q












Rd*Cc
-

Gg*Dc = Ra*Bc







Ra

=

reaction at point A













Ra =

(Rd*Cc)
-

(Gg*Dc )







Bc







=

(30.94*0.6)
-

(110.12*6.4)



=


(686.21)




13.4



13.4







=


(51.21)

tf






(upward)









where







Cc =

nearest distance of wire rope from point D



0.6

m


Gg =

Load on c.g/2




110.12

tf


Dc =

nearest distance of wire rope from point c.g



6.4

m


Bc =

nearest distance of wire rope from point A



13.4

m








(iii)

Tension in emergency







Taking moment about vertical plane passing through D, parallel to line
AK







X*(Ra+T')=Gg*(X+Cc)







(Ra*X)+(T'*X)=Gg*(X+Cc)







T'*X=Gg*(X+Cc)
-
(Ra*X)







T'*X=

Gg*(X+Cc)
-
(Ra*X)






T' =

110.12*(9.81+0.6)
-
(
-
51.21*9.81)







9.81








=


1,649.35

=


168.13

tf




9.81



















Added with friction loads



=


175.07

tf


T=


















































12

End frame






12.1

Axial loads on end frame members






(a)

Axial load







wb =

14.49

t/m





wc =

26.36

t/m





wd =

19.74

t/m





w max =

26.36

t/m











( b)

Maximum
Hydrostatic load
-

F







F =

wmax * B /2






=

26.36 *14/2






=

621.52

tf











( c)

Inclined component along the line of end member
-

F'







Radius
-

R



=

13.75

m


Angle of end frame inclination
-

θ



=

11.51

o


Length of end frame
-

L = R /
cos (θ)



=

13.75/
-
0.87






=

15.80

m


F' =

F * L / R


=

621.52*15.8/13.75






=

713.98

tf

(d)

Profil for end frames







Upper girder







Material:


JIS G 3101


SS400





H

B

Tw

Tf


Double
-
I

mm

750

300

30

30


Cor.all.

mm



1




I x1 =


586,564


cm
4




Z x1=



15,726

cm3




Aw =



722.40

cm2




G =



567.08

kg/m









(e)

Bending moment
-

Bm







length of end frame, L' = L
-

(hf+hg)



=


14.45

m


where







L = incline distance of end frame



=


15.80

m


hf

= height of vertical stiffeners web



=


-


m


hg = height of horizontal girders web



=


-


m









Bm =

G * L'^2







8







=

567.08 *(14.45)^2







8







=


118,331







8






=


14,791.40

kgm











(f)

Bending stress
-

Sb







Sb
=

Bm*100

+

F'





z


A





=

14791.4*100

+

713.98*1000





15725.57


722.4





=


1,479,140.40

+

713975.11





15725.57


722.4




=


94.06

+

988.34




=


1,082.40

kg/cm2











(g)

Weight of one element of end frame







Gst =

L' * G






=

14.45*567.08


=


8,191.69

kgs









Total weight
-

Gef = N * Gst



=


24,575.06

kgs


N = number of arm



=


3