DESIGN CALCULATION OF SPILLWAY GATE 1 Design data

Type Radial gateMax.Flood level 337 mHigh water level 336 mLow water level 336 mCrest El. 340 mClear span 14 mHeight 11 mDesign head-H 11 mQuantity 2 setsAll.stress-SS400 1200 kg/cm2All.stress-SM490 1600 kg/cm2Corr.all. 2 mm

2 Economic radius of radial gateR = 1.25 * H= 1.25*11= 13.75 msin β/2= h/2

R= 5.5

13.75= 0.4β/2 = 23.58 oβ = 47.16 o< = 0.26 radiant

3 Arc length of gateL = 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/cm2p0 = 0.000 0.000

a2.245

b1 = 3.726

p1 = 4.491 4.491b

6.134b2 = 3.189

p2 = 7.778 7.778c 8.910

b3 = 2.6449.559

p3 = 10.042 10.042d

b4 = 1.441 10.521p4 = 11.000 11.000

e 11.000

5 Vertical stiffeners5.1 Bending moment on vertical stiffeners

M ab= p1*b1^2/6= 4.49*3.73^2/6

β

= 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-cmMax. bending moment- Mb max. Mb max. = 1,042,558 kg-cm

5.2 Number of vertical stiffenersSpacing, s = 0.60 mN = (B/s)+1 = 14/0.6+1 = 24

5.3 Bending stressσcd = Mmax = 1042557.53

Zxl 815.90

= 1042557.53815.90

= 1,277.80 tf/cm2

Material: JIS G 3101 SS400H B Tw Tf

I mm 450 200 9 14Cor.all. mm 1

I x1 = 18,195 cm4Z x1= 816 cm3Aw = 77.54 cm2G = 60.87 kg/m

5.4 Weight of vertical stiffenerGst = N * L * G= 24*11.31*60.87= 16,753.18 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, eb = 0.2 * B= 0.2 *14= 2.8 m= 280 cmDistance 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 + wb2wb1 = (p0 + 2 * p1) * b1

6= (0 + 2 * 4.49) * 3.73 = 5.577 tf/m

6wb2 = (2 * p1 + p2) * b2

6= (2 * 4.49 + 7.78) * 3.19 = 8.908 tf/m

6wb = 5.577 + 8.908 = 14.485 tf/m

= 144.85 kgf/cm(2) Beam C

wc = wc1 + wc2wc1 = (p1 + 2 * p2) * b2 =

6(4.49 + 2 * 7.78) * 3.19 = 63.9306 6

= 10.66 tf/mwc2 = (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 Dwd= wd1 + wd2wd1 = (p2 + 2 * p3) * b3

6= (7.78 + 2 * 10.04) * 2.64 = 12.278 tf/m

6

wd2 = (2 * p3 + p4) * b46(2 * 10.04 + 11) * 10.52 = 7.465 tf/m6

wd = 12.278 + 7.465 = 19.743 tf/m= 197.43 kg/cm

wmax = = 263.60 kg/cmMBA = wmax * b ^2 = 263.6 * 280 ^2

2 2 11,356,240.00 2

= 5,678,120.00 kg-cm6.3 Bending stress

σBC = MBC = 5678120Zxl 10454.64

= 1135624010454.64

Upper girder = 1,086.24 kgf/cm2Material: JIS G 3101 SS400

H B Tw TfI mm 900 400 25 25Cor.all. mm 1

I x1 = 468,368 cm4Z x1= 10,455 cm3Aw = 379.50 cm2G = 297.91 kg/m

6.4 Weight of horizontal girderGst = N * B * G= 3*14*297.91= 166,828.20 kgs

7 Skin plateFrom 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 stressσs = K * a^2 * p

100 x t^2

7.2 Result of calculation

k = 50a b b/a p t(cm) (cm) (kgf/cm2) (cm) σs60 372.600 6.21 2.245 1.80 (kgf/cm2)60 318.900 5.315 6.134 2.80 1247.460 264.400 4.407 8.910 3.80 1408.460 144.100 2.402 10.521 3.80 1110.6

allowable bending stress SM-490 1311.5Corrosion allowance 0.2 cm Thickness of skin plate 1 20 mm

2 30 mm 3 40 mm4 40 mm

7.3 Weight of skin plate, G=W * H * t * gWidth-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 frameWeight of vertical stiffener = 16,753.18 kgsWeight horizontal girder = 166,828.20 kgsWeight of skin plate = 36,661.54 kgsTotal = 220,242.92 kgs

9 Friction force9.1 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 * nwhere n = number of end frame = 2 = 14 *11^2 = 484

2 * 2 4= 121 tf

(b) Friction force on pinFp = P * ff = 121 * 0.25

= 30.3 tf( c) Load due to friction- Tp

Tp * R = Fp * rTp = Fp * r / R = 30.25 *0.3 / 13.75where = 0.66 tf

R = Radius of radial gateFp = Friction force on pinR = Radius of pin

9.2 Friction force due to seal rubberAssume width of seal bears water pressure - wr = 50 mm

= 0.05 mPressure area of rubber - ar = 2*L*wr = 2*11.31*0.05

= 1.13 m2Average water pressure - pw = 5.578 t/m2Friction coefisien between rubber and steel- fr = 1.1Load due to friction of rubber seal - TrTr = pw * ar * fr = 5.58*1.13*1.1

= 6.94 tf10 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 gravity 10,866 from center of pin

11 Steel wire rope11.1 Tension on Steel wire rope in normal condition

diameter, d 35 mmbreaking strength, T kgfunit weight, Wc 90 kg/mTaking length of wire rope - Lc = 12.31 mTension in wire rope due to its own weight = 1.11 tfTotal tension in one side of wire rope in normal condition - WcWc = (Gg+Gf+Tr)/2 = 220.24+1.96+6.94/2where = 114.57 tfGg = weight of gate leaf = 220.24 tfDf = portion of end frame weight at wire rope position(Dwg.3) = 1.96 tf

11.2 Tension on wire rope in Emergeny condition

center of gravity

(a) Related dimensionsspan - B = 14 mRadius - R = 13.75 mtan A =B/R = 13.75/14 = 0.98angle < A = 44.48 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 mangel < 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 mfrom 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) MomentsTaking moment about vertical plane passing through diagonal line

Gg*p = Rd*X(i) Reaction at point d, Rd

Rd = Gg * pX

= 110.12*2.76 = 303.689.81 9.81

= 30.94 tf(downward)

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

Rd*Cc - Gg*Dc = Ra*BcRa = 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 mGg = Load on c.g/2 110.12 tfDc = nearest distance of wire rope from point c.g 6.4 mBc = nearest distance of wire rope from point A 13.4 m

(iii) Tension in emergencyTaking 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 tfT=

A

B C

D

Clear span,B b

Dwg.7 - Basic arrangement of upper structure

"X"

90o

"Y"

O

"Z"

"p"

12 End frame12.1 Axial loads on end frame members (a) Axial load

wb = 14.49 t/mwc = 26.36 t/mwd = 19.74 t/mw max = 26.36 t/m

( b) Maximum Hydrostatic load - FF = wmax * B /2= 26.36 *14/2= 621.52 tf

( c) Inclined component along the line of end member - F'Radius - R = 13.75 mAngle of end frame inclination - θ = 11.51 oLength of end frame - L = R / cos (θ) = 13.75/-0.87

= 15.80 mF' = F * L / R = 621.52*15.8/13.75

= 713.98 tf(d) Profil for end frames

Upper girderMaterial: JIS G 3101 SS400

H B Tw TfDouble-I mm 750 300 30 30Cor.all. mm 1

I x1 = 586,564 cm4Z x1= 15,726 cm3Aw = 722.40 cm2G = 567.08 kg/m

(e) Bending moment - Bmlength of end frame, L' = L - (hf+hg) = 14.45 mwhereL = incline distance of end frame = 15.80 mhf = height of vertical stiffeners web = - mhg = height of horizontal girders web = - m

Bm = G * L'^28

= 567.08 *(14.45)^28

= 118,331 8

= 14,791.40 kgm

(f) Bending stress - SbSb = 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 frameGst = L' * G= 14.45*567.08 = 8,191.69 kgs

Total weight- Gef = N * Gst = 24,575.06 kgsN = number of arm = 3