radial+gate+with+arm+ +6+m.+++haji+shaher

Design of Weir Type Radial Gates

Project :-

Six ( 6 ) Dispersal Structure On Nari River( Package - 1 )

Client :- Irrigation and Power Department of Balochistan

Consultants :- CAMEOS Consultants.Consulting Engineers, Architects & Planners Islamabad.

Contractor :-

Design Criteria & Computation

For Radial Gates 8.18 m. X 5.08 m.( Haji Shaher )

With Arm 6 m.

DATE OF SUBMISSION : May., 2012

SUN ENGINEERS;Climax abad , G. T. Road Gujranwala.

SUN Engineers Nari River Desig. By: Engr. Liaqat Ali.

G.T. Road Gujranwala Radial Gates Check. By: Engr. M. UsmanWith Arm 6 m. App. By: Capt.(R.) Engr. Nasir

TABLE OF CONTENTS.Page No.

1 - Data 1

2 - Material Used & Stresses 2

3 - Computation of Hydraulic Forces 3

4 - Pin Friction 5

5 - Seal Friction 5

6 - Load on Stiffeners 7

7- Design of Vertical Beams ( Angle ) 8

8 - Design of Main Horizontal Girders 14

9 - Design of Struts ( Radial Arms ) 18

10 - Design of Trunnion Assembly 21

11 - Design of Trunnion Pin 22

12 - Design of Wire Rope 23

13 - Design of Rope Drum Axle 24

14 - Design of Cross Shaft 25

15 - Hoisting Effort 26

16 - Hoisting Bridge. 28

17 - Design of Trunnion Embedded Parts #REF!

18 - Design of Side Embedded Parts 33

19 - Weight of Gate 36

20 - Hoisting M / C

21 -

22 -



Data

1 - Clear Span = 8.180 m. = 26.83 ft.

2 - Max. Water Level = 113.610 m. = 372.64 ft.

3 - 114.110 m. = 374.28 ft.

4 - Trunnion Height = 4.580 m. = 15.022 ft.

5 - Gate Sill Level = 109.530 m. = 359.26 ft.

6 - Gate Top Level = 114.610 m. = 375.92 ft.

7 - Gate Arm Radius = 6.00 m. = 19.680 ft.

8 - Hoist deck Level = 115.610 m. = 379.20 ft.

9 - Hoist deck Height = 6.080 m. = 19.94 ft.

10 - No. of Hoist = 2

Gate Parameters

1- Gate Height 5.08 m. = 16.66 ft.

2- Gate Width 8.16 m. = 26.75 ft.

3- Arc Length 4.71 m. = 15.45 ft.

4- No. of Gates = 11

Effective Head

Head on Upstream 4.08 = 13.38 ft.

Page : 1 of 42

Trunnion Center Level =



Material Used & Stresses

Design criteria & computations for

Radial Gate With Arm - 19.68 ft

19.68 ft

Technical Specifications

1- Structural Steel.

a- For Skin plate, Beams & Stiffeners of Gate.steel".

b- For Embedded parts ( except Seal & track plates & ASTM-A 36 Specification for structural Side guides) steel".

2- ASTM - 276

Specification for stainless steel( bars, bolts, nuts & washer etc.)

3- ASTM - 276 Type 410

Specification for stainless steel( pins & rods)

4- Normal Nuts, Bolts And Washers. ASTM 307Grade A

ALLOWABLE STRESSESFy = 36,000 psi.

= 36 ksi.

Type of StressesLoad Case

Normal Occasional Exceptional a - bending Stress 0.6 0.8 0.95 b - Tension 0.45 0.5 0.6 c - Shear 0.4 0.5 0.6 d - Combined Stress 0.75 0.95 No e - For All Mech. & Electrical Components 0.33 Fy 0.67 Fy 0.90 Fy

or 0.2 Fu or 0.4 Fu or 0.55 FuWhich ever is Less

Job: Design of Radial Gate With Arm -

ASTM-A 36 Specification for structural

Corrosion- Resisting (or Corrosion Resistant) Steel.

Corrosion- Resisting (or Corrosion Resistant) Steel.

Page:- 2 of 42



Computation of Hydraulic Forces

H = Head = EL = 372.64 - 359.26= 13.38 ft.

The Horizontal Components of the Water Thrust isW = ω x B x h ( H - h / 2 ) Ref :1 Eq 4.15 Page 103

The Vertical Components of the Water Thrust is

Ref. 1 Eq. 4.16 Page 103

Where ; Fig. 1

= 62.4= 26.83 ft.= 13.38 ft.= 13.38 ft.

Skin Plate radius ( Measured on the wet surface. )

Difference bettween the elevation of the water level and center of curvature of Skin Plate.

Arc Sine Ds / R

Difference between the elevations of Center of Curvature of the Skin Plate andthe Water Level ( for Weir Gates )

Difference between the elevations of Center of Curvature of the Skin Plate andthe sill.

Note ; - a ) refer to Difference of the elevations and may assume positive or Negative sign , depending on the gate arrangement.

b )positive or Negative sign.

c ) The Direction of the Vertical Components is given by its sign ;Positive : UpwardNegative: Down ward

EL =

Wv = ωB R [ Dm (Cos as - Cos ai + R (ai - as ) / 2 + R ( Sin as Cosas - Sinai Cosai ) / 2 ]

ω = specific weight of water lbs./ ft3.B = span of Side Seals H = Max. Head of Water on SillH = h = Gate Sealing HeightR =

Dm =

as =

a i = Arc Sine Di / R

Ds =

D i =

Dm , Ds , & Di

The angle as & ai are taken in radians and can also assume a

EL 114.11

6.08

EL 109.53

M.W.L.113.61

Gate Top Level =114.61

PIER TOP=115.61

5.08

4.084.58

RADIAL GATE (HAJ I SHAHER)

Page : 3 of 42



The Magnitude and Direction of the resultant Water Thrust are Calculated by

Ref. 1 Eq. 4.19 Page 103

Ref. 1 Eq. 4.20 Page 103In our case ;B = 26.83 ft. R = 19.680 ft.

372.64 ft. - 374.28 ft. = -1.64 ft. 372.64 ft. - 359.26 ft. = 13.38 ft. 372.64 ft. - 359.26 ft. = 13.38 ft.

374.28 ft. - 372.64 ft. = 1.64 ft.

374.28 ft. - 359.26 ft. = 15.02 ft.

Arc Sin Ds / R

= Arc Sin ( 1.64 ¤ 19.680 )== 0.0834 rad.

= Arc Sin ( 15.02 ¤ 19.680 )

0.8685 rad.

= 0.9965 = 0.0833

= 0.6460 = 0.7633Horizontal Component of W (Eq 4.15 ) Page 103

= 62.4 x 26.83 x 13.38 ( 13.38- 13.38 / 2 )

= 149916.61 lbs. = 149.92 KipsVertical Component of W ( Eq 4.16 )

= 62.4 x 26.83 x 19.680 [ -1.64 ( 0.9965- 0.6460 ) + 19.680 { 0.8685 0.0834 ) } / 2+ 19.680 ( 0.0833 x 0.9965 - 0.7633 x 0.646 ) / 2 ]

= 32948.590 [ -0.575 + 7.725 + -4.035 ]= 102624.90 lbs. = 102.625 KipsThe magnitude and direction of the resultant Water thrust are calculated by:

= √ 149.92 102.62 2

= 181.68 Kips

W = √ W2 h + W2

v

β = arc tan W v / W h

Dm =H = h =

Ds =

Di =

α s =

α i = Arc Sin Di / R

Cos as Sin as

Cos ai sin ai

W h

W v = ω B R [Dm (cos a s - cos a i ) + R (a i - a s) / 2 + R ( sin a s cos as - sin a I cos a i ) / 2]

- (

W = √ W2 h + W2

v

2 +

Page : 4 of 42



Direction of W =Arc Tan ( 102.62 )

149.92

0.600 rad.= 34.39 degree

Pin Friction

Fig : 2The friction on supports and hinges is proportional to the water load on the gate and the coefficient of friction of the surface in contact. It is calculated by the equation.

Where Max . Horizontal thrust on each Trunnion pinCoefficient of friction

Let the pin diameter is equal to 6 in. and coefficient offriction between two metals as 0.30 for steel bronze understatic conditions (Table 9-1 Ref 1 page 226) the max horizantalthrust on the pin will be when the gate is fully closed and full head is acting.

Max.Thrust = 5.59 Kips/ft.Force for both Pins per gate = 5.59 x 26.83

= 149.92 KipsForce of Friction = 0.30 x149.92

= 44.97 Kips

Seal Friction

SIDE SEAL FRICTION 2" width of the seal , the water pressure

Total head = 13.38 ftAverage pressure at the bottom of seal / ft length of seal (maximum )

= 62.4 x 13.38 x 1 x 2 / 12 = 139.18 lb / ft. length

The friction factor between rubber and steel is 1.0 in the start

and 0.8 in running condition Ref. 1 Page 228Total seal friction = 1.0 x F x Sector lengthArc length or Sector length = 15.45 ft.Seal friction ( both sides) = 2 x 1 x 139.18 x 15.45 / 1000

4.301 kips

139.18

β =

Fa = µ Fh Eq. 9.1 Ref 1 Page 226

Fh =µ =

Fs =

WBO

W

Wh

Wv

Page : 5 of 42



App.Wt. of Gate =

= 76.58 KN

= 17.21 Kips.

Net Effective Force

Down ward forceWt. of Gate , D = 17.21 kips ( weight of gate )

= 4.301 kips

= 44.97 kips Buoyancy = 2.19 kips

Total = 64.293 kips

Net Effective Force

Down ward forceWt. of Gate , D = 17.21 kips ( weight of gate )

= 49.28 kips

Hoist Capacity = 66.48 kips Adding 20% for misc. = 79.78 kips

Hydrostatic Pressure

Hs = ω x h at the bottom of gateWhere h = Effective head on the gateh = 13.38 ft

ω = 62.4Hs = 62.4 x 13.38 / 1000

= 0.835 Ksf

== 26.8 x 16.66 x 0.835= 373.32 Kips

Hydrodynamic Pressure= 7/8 x a x ω x √ hxy

Where y = Lavel at dynamic force required in this case = H a = Facter of dynamic force = 0.1

= 7 / 8 x 62.4 0.1 13.38= 0.0731 Ksf

0.64(8.182 x 4.082)0.682

Seal Friction , Fs

Pin Friction, Fa

Seal Friction , Fs & Pin Friction, F

lbs / ft3

FHs Gate Area x Hs

Hd

Hd x Ö 2 / 1000

Page : 6 of 42



Load on Stiffeners

Gate Area, 26.83 x 16.66 = 447

F = 181.68 Kips

Pressure , P = F / A181.68 / 447

= 0.406

0.0028 Ksi.

Load on Stiffeners = P x Width of Gate

= 0.406 x 26.83

= 10.90 Kips / ft.

= 0.909 Kips / in.

A = ft.2

W =

Kips / ft.2

Page : 7 of 42



Design of Vertical Beams ( Angle )

Section Angle 100 x 100 x 10

Chossen Section Properties: -

Area = 2.40

Area ( Angle ) = 7.35

I = 3.48

S = 1.22

Wt. = 8.20 lbs./ft.

Loads on vertical stiffeners :-

P = 0.406 ksf

Load on stiffeners, w = P x Width of Gate= 0.406 x 27= 10.90 kips / ft.= 0.91 kips/in.

Moment ( M ) =L = #REF! in

M = 0.91 x #REF! 12

= #REF! kips.-in.

Section Modulus (Reqd,) = #REF! / ( 0.6 x 36 )

= #REF!

Section Modulus ( provided,) = 1

Section Modulus (Reqd,) < Section Modulus (Provided,)

in.2

in.2

in.4

in.3

w L2 / 12

2 /

in.3

in.3

Page : 8 of 42



Check for Flexure :-= #REF! kips.-in.

=

= 0.9 x 0.6 x 36 x 1= 23.65 kips.-in.

> Hence it is O.K

Check for Shear :-

= w L / 2= 0.91 x #REF! / 2= #REF! kips.

= 0.4 x 36 x 2= 34.5960692 kips.

>Hence it is O.K

Check for Deflection :-

Max. Deflection occures is given by =Total Live & Dead Load = 0.91 kips/in.

= 3.486 x 13 x 0.375

Max. Deflection :- = 0.91 x ( #REF!

3.48= #REF! in

Allowable Deflection = L / 750= #REF! / 750= #REF! in

Allowable Deflection > Actual Deflection

M U

j M n j x 0.6 F y x S

j M n

M n M U

VU

VC

VC VU

W L4 / 384 x E I

Value of I for T in.4

) 4

384 x 30 x 10 3 x

Page : 9 of 42



= V T / R Fig. 3

= 1.64 / 19.680 )== 0.08 rad

= Q S / R

= 15.0 / 19.680 )= 0.868 rad

= 0.868 - 0.08 = 0.785 rad

L = R x= 19.680 x 0.785= 15.449 ft.

= 0.1414 x L = 2.185 ft.

= 0.4734 x L = 7.314 ft.

= 0.3852 x L = 5.951 ft.

Total = 15.449 ft.

Sin b1

b 1 Sin - 1 (

Sin b 2

b2 Sin - 1 (

b

b

l 1

l 2

l 3

AV Q

S

T

C

l1

l2

l3

B1

B2

Page : 10 of 42



Support of vertical Beams is made on the main horizontal girders, where the arms are also connected. In more recent designs ,the use of two pairs of arms per gate prevails,whichresults in vertical beams with two supports. The vertical beams may be dimensioned as straight ones , supported on the main H.girders and subjected to a triangular load (Weir gates)

For Weir Gates with two pairs of radial arms the load diagram is as follows

A B 3.89 D

2.185 7.314 5.951

Fig. 4

Fig. Loading on vertical beams of Weir segment gate Where : depth corresponding to lower seal =

= 372.64 ft.

= depth corresponding to support A

= L

373 ( 7.314 += 5.951 ) / 15.45= 319.95 ft.

= depth corresponding to support D

= L

= 372.64 x 5.951 / 15.45

= 143.54 ft.

g = specific weight of water = 62.4

The bending moments at the supports per unit width of span ,are

= g = 62.4

= 62.4 x 2.185 372.64 + 319.95 ) / 6= 52868.34 lbs. - ft. / ft.

ghi ghA

ghD

RA RD

l 1 = l 2 = l 3 =

hi

hA

g hi hA = hi ( l 2 + l 3 ) / L

g hA l 2 + l 3

hD

g hi hD = hi x l 3 / L

g hD l 3

lbs/ft.3

MA/ g l1

2 ( 2 hi + hA ) / 6 lbs. / ft. 3

2 ( 2 x

=

= 62.4 x 5.951 143.54 / 6= 52869.15 lbs. - ft. / ft.

Page : 11 of 42



The bending moment on any point B between the supports and at a distance x from the support D is ;

2 2 Page142

62.4 143.54 x ( X - 7.314 ) + ( 319.95 -2 6 x 7.314

143.54 7.314 52868.34 + 52869.15 ) ∕ 2

= 62.4 { 71.771 524.91 X + 4.02 215.03321 X }+ 52868.746

= 62.4 { 4.02 71.77 739.94 X } + 52868.746

= 250.85 4478.49 46172.45 X + 52868.75

= 250.85 4478.49 46172.45 X + 52868.75 ( A )Deriving the Eq. ( A ) w.r.to X and making it equal to zero

752.55 8956.97 X - 46172.45 = 0

a = 752.55 b = 8956.97c = - 46172.45

X = - 8956.973 ± Ö 8956.97 752.55 x ( - 46172.454 )2 x 752.55

= 3.89 ft. & = -15.79 ft. Which is not Applicable

= 250.85 x ( 3.89 4478.49 x ( 3.89 )2 -

46172.45 x 3.89 + 52868.75= 14721.57 + 67632.92 - 179430.545 + 52868.75

= -44207.31 lbs. - ft. / ft.

= x B / 2= 52868.34 x 27 / 2 = 709239.40 lbs. ft.

= x B / 2= -44207.31 x 27 / 2 = -593049.92 lbs. ft.

= x B / 2= 52869.15 x 27 / 2 = 709250.218 lbs. ft.

Max. Bending Moment = 709250.22 lbs. ft. and S = 1.22Max. Bending Stress = B. M. X 12 / S

= 709250.22 x 12 ∕ 1.22 = 6994500.73 psi

Allowable . Bending Stress = 0.6 x 36,000

MD/ g l 3

2 h D / 62 x

MB/ = g [ X x hD (X - ℓ 2 ) + ( h A - h D ) X ( X 2 - ℓ2 2 ) ] + ( M A' + M D' )

6 ℓ 2

MB/ = { X x

) X ( X 2 - 2 ) } + (

X 2 - X 3 -

X 3 + X 2 -

X 3 + X 2 -

X 3 + X 2 -

X 2 +

2 - 4 x

MB/ )3 +

MA MA/

MB MB/

MD MD/

in3

= 21600 psi

Page : 12 of 42



Fig. 5

properties of chosen section ( T. Fab. )

a = 6 in.

b = 9.449 in.

d = 12.50 in.

h = 11.61 in.

= 0.50 in.

= 0.394 in.

t = 0.375 in.

Area = Fig. 6

= + += 3 + 4.352 + 3.72000744

= 11.07237

8.07 == 12.25

y = =

= 6.1968504

=y = 5.82 in. = 0.1968504

s1

s 2

a x s 1 + h x t + s 1 + b x s 2

A 1 A 2 A 3

in.2

Area ( T ee ) = h x t + b x s 2 = Y 1 S 1 / 2 + h + S 2

A 1 Y 1 + A 2 Y 2 + A 3 Y 3 Y 2 h / 2 + S 2

A 1 + A 2 + A 3

Y 3 S 2 / 2

M'D = 6906.44 lbs.-ft./ ft.

M'B = -5774.92 lbs.-ft./ ft.

D

B

A

M'A = 6906.34 lbs.-ft./ ft.

2.53'

I =

= 291.247

S = I / Y = 50.03

Wt. / ft. == 25.02 lbs / ft.

Page : 13 of 42

a x s 1 3 / 12 + A 1 ( Y - Y 1 ) 2 + t h 3 / 12 + A 2 ( Y - Y 2 )2 +b s 2 3 / 12 + A 3 ( Y - Y 3 ) 2

in.4

in.3

( A1 + A 2 ) x 12 x 0.2836



Design of Main Horizontal Girders

Beem 24 x 84 lbs/ft.Arc length , = L = R θ

= 19.680 ft.

of the skin plate = 44.98 degree

19.680 x 44.98

= 15.45 ft.

Fig. 7In particular case of Weir gates with two pair of arms.

0.1414 x L = 2.185 ft.

0.4734 x L = 7.314 ft.

0.3852 x L = 5.951 ft.

= 15.45 ft.

a = 44.98 x 0.1414 = 6.36

b = 44.98 x 0.4734 = 21.29

g = 44.98 x 0.3852 = 17.33

= 44.98

R = Radius of armsθ = Angle subtended by the wet part

L =

x p / 180

ℓ1 =

ℓ2 =

ℓ3 =

L = Arc length

EL 114.11

6.08

EL 109.53

M.W.L.113.61

Gate Top Level =114.61

PIER TOP=115.61

5.08

4.084.58

RADIAL GATE (HAJ I SHAHER)

Page : 14 of 42



Bending Moment of Girder

As Horizantal girders are rigidly connected with adjacent assembly. ( For Rounded Connections )

L = 26.83 ft. = 322 in.

a = 5.91 ft. = 70.87 in. @ 71 in.

b = 15.0 ft. = 180.2 in. @ 180 in.Bending Moments

= =

=

Ref .1 Design of hydraulic Gate page 134 Case 5.13 (a)

Fig. 8

Where,0.91 kips. / in.

= - q a2 / 16

= 0.91 x ( 71= 286.27 kips. - in.

MC MA MB

- q a2 / 16

q = w =

MC

)2 / 16

(-1)(-1) c

BA

Page : 15 of 42



Deflection =

f c = 0.91 x 180.23 180.23 24 x 70.86614

2370.08

= 0.04528 in.Allowable Deflection. = b / 750

= 180.23 / 750

= 0.24 in.

Maximum Bending Moment = 286.27 kips. - in.

Maximum Bending Stress =

= 286.27 ∕ 196.74= 1.46 ksi

Allowable Bending Stress = 0.6 x 36

= 21.60 ksi

Allowable Stress > Stress AvailableMax. Bending Moment is M = 286.27 kips. - in.

Section Modulus (Reqd.) = M / fb

= 286.27 / 0.6 x 36

= 13.25

Section Modulus (Provided) = 196.74

Section Modulus (Reqd.) < Section Modulus (Provided)

Properties of chosen section

A = 24.65

= 2370.08

= 196.74

Wt. = 84.00 lbs./ft.Check for Flexure

Mμ = 286.27 kips. - in.== 0.9 x 0.6 x 36 x 196.74= 3824.64 kips. - in.

φMn > Mu Ok

fc = q b2 ( 5 b2 - 24 a2 ) / 384 E I

2 ( 5 x 2 - 2 )

384 x 30 x 10 3 x

M ∕ S

in3

in3

in2

I xx in.4

S xx in.3

φM n j x f b x S x

φM n

Page : 16 of 42



properties of chosen section ( W 24x84 lbs/ft. )

a = 9.00 in.

d = 24.00 in.

h = 22.46 in.

= 0.77 in.

t = 0.469 in. Fig. 9

A = 24.65

I = 2370.08

S = 196.74

s1

in.2

in.4

in.3

ht

a

s1

d

Page : 17 of 42



Design of Struts ( Radial Arms )

The lower arms / struts are designed as it has the mainload and head.

Load 1) Load transfered by the Horizontal Girder P = Load reaction at the supports of Horizantal girder

= WL / 2= 10.90 x 26.75 / 2

= 145.84 kips

2) Total force acting on 4 Arms / Struts

181.68 kips Ref : Page 5 / 49

= 181.68 / 4 kips force on each Arm

W = 45.42 kips

Properties 0f chosen section. Fig. 10

A = 15.04

= 471.13

S xx = 66.52

Wt. = 51.15 lbs. / ft.

Total force =

Chosen Section = Beam IPN 360 76.1 kg./m

in2

I xx in.4

in.3

Fn

Fr

W/2

Fn

Fr

FLOW

W/2

W=

W= 16

16

Page : 18 of 42

SUN Engineers Nari River Desig. By: Engr. Liaqat Ali.G.T. Road Gujranwala Radial Gates Check. By: Engr. M. Usman

With Arm 6 m. App. By: Capt.(R.) Engr. Nasir

Buckling Check= 2370.08

= 471.13h = 82.02 in.

K == 2370.08 ⁄ 471.13 x 82.02 ¤ 180.00= 2.292

a = 71.00 in.

b = 180.00 in.

=

= 0.91 x 71.00= 286.27 kips-in

=

= 0.91 / ( 2.292 + 2 ) x ( 180.00 71.00

= 76.00 kips-in.

Section Modulus Reqd. = / 0.75 x 36= 286.27 / 0.75 x 36

= 10.60

= 66.52

Stability Check

= 1.04= 181.68

= 45.42 kips

A = Cross - sectional area = 15.04M = Net Moment = 286.27 kips-in

= Section Modulus (Compression) = 66.52= 1.04 x 45.419 15.04

+ 0.9 x ( 286.27 ∕ 66.52 )

= 7.02

For combined stresses from table at page No 2

0.75 x Fy = 82.02 in.= 0.75 x 36 Radius of gyration = 27.00 Kips. = Ö I / A

= 5.6082.0209974 ¤ 5.60

= 14.65

1.04 For 14.65

Force Area RelationshipForce = 181.68 Kips.

No of Arms = 4Force on each Arm = 181.678 ⁄ 4 = 45.419 kipsAllowable Stress = 27.00 kips

Area = = 45.419 ∕ 27.00

= 1.68

Area provided = 15.04

Area provided > Area Requred

K = I2 / I1 x h / b

I 2 in.4

I 1 in.4

I2 / I1 x h / b

Mβ1 wa2/162 ∕ 16

Mβ2 w / ( k+2 ) x ( b2 / 6 - a2 )

2 ¤ 6 - 2 )

Mβ1

Sx ( Reqd. ) in.3

Sx Provided in.3

ω x F/A + 0.9 M / Wc ≤ σ allowable Compression

ω = buckling coefficient ( From Table 5.7 Page # 125 )F = compression / Axial load ∕ 4

in2

Wc in.3

∕

Slenderness Ratio ,l = Lf / i

Where Lf = Buckling length

f b = i =

l =

Allowable Stress is more than combined stress ω = Reff.# 1, Page 125 table 5.7

P / Fb

in2

in2

Hence Our design is OK.

Page : 19 of 42



Arm. BEAM ( Fab. )

d = 14.17 in.

b = 5.63 in.

h = 12.64 in.

s = 0.77 in.

t = 0.51 in.Fig. #REF!

A = 15.04

I = 471.13

= 474.88

S = 66.52

Wt. / ft. = 51.15 lbs.

in.2

in.4

in.3

ht

s

b

d

Page : 20 of 42



Design of Trunnion Assembly

Fig. #REF!

Properties of Chosen Section (comulatives)

A = 91.26

= 490.58

= 2651.77Material used = ASTM A - 36

= 36 ksi

Load on Trunnion Assembly

Load on trunnion assemblyResultant Load = 181.68 kipsLoad on each side = 181.68 / 2

= 90.84 kipsArea Required = 90.84 / ( 0.4 x 36 )

= 6.31

Area provided = 91.2691.26 - 6 x 1.25 x 2 = 91.26 - 15

= 76.26Fy = 36 ksiArea Provided > Area Required

Check for Shear= 90.84 kips

== 0.9 x 0.4 x 36 x 76.26= 988.33 kips

Vu <

in2

Sxx in3

Ixx in.4

Fy

in2

in.2

Effective area provided =

in.2

Vu

φV c 0.9x0.4 x Fy x A

φ Vc

Page : 21 of 42



Design of Trunnion Pin Let Dia of Pin = 125 mm ( provided. )

h = 522.00 mm effectiveP = 181.68 / 2 = 90.839 kips

Material ASTM A 276 Type 304 or Equivalent

Fy = 65 Ksi Fig. #REF!Allowable Shear Stress = 0.20 Fy

= 0.20 x 65= 13 kips

Required Area = P /= 90.839 ∕ 13

= 6.99

= 6.99

D = √ 6.99

= 2.98 in

= 75.76 mm

Which is less than Dia provided.

σall.

σ all.

in2

π / 4 x D2 in2

x 4 / p

Page : 22 of 42



CALCULATIONS FOR STEEL WIRE ROPE

( i ) Total Load = 64292.97 lbs.( ii ) Load on each rope Drum = 32146.49 lbs.( iii ) Load on each rope = 16073.24 lbs.( iv ) Factor of Safety = 5( v ) Total bearing strength of wire rope = 16073.24 x 5

= 80366.22 lbs.= 1.00 in.

6 x 19 + IWRC construction, The breaking strength of wire rope = 45.70 Tons. ( 1 Ton = 2000 lbs. )

= 91400 lbs.

Hence our Design is safe.

ROPE DRUMS

1- Pitch dia of Rope Drum = 24 in.2- Dia. of Steel Wire Rope . = 1.00 in.3- Maximum travel of Gate = 16.66 ft.4- No. of turns required for Gate = 16.66 x 2 ∕ π x 2.00

= 5.30 ≡ 6.00

5- Total No.of turns ( min ) = 6 + 3 = 9.00 Adopted = 10 turns6- Min Rope Drum Length adopted = 12 in.

Wire Rope dia .

Page : 23 of 42



Design For Rope Drum Axle

Maximum Vertical Load = 64,292.97 lbs.Load on each rope Drum = 32146.486Load on each rope = 16073.243Maximum Torque = 16,073.24 x 24 / 2

= 192878.92 lbs-in.Un-supported length of shaft = 20 in.Maximum Bending Moment = 16073.24 x 24 / 4

= 80366.22 lb-in.Reqd. Section Modulus of shaft = 80366.22 /( 0.6 x 45000)

= 2.977

Z for shaft dia , d = 2.977

d = ( 2.977 = 2.47 in.We are using Shaft of dia = 4 in.

Checking the shaft against combined bending and torsion.Considering Key Way Effect

Z == 12.57

= 12.57

Maximum Stress =

= √ ( 80366.22 192878.92 12.57 x 1.25

= 20784.86 psi.

Checking of the shaft against Torsion only.

t = T / Z x 1.25= 192878.92 12.57 x 1.25 = 19186.02 psi.

Allowable shear stress for SAE 1060 = 60,000 x 0.4 = 24000 psi.

in3

0.196 d3 in3

/ 0.196 )1/3

π d3 / 16

in3

t max = √ M 2 + t 2 / Z x 1.25

) 2 + ( ) 2 /

Torsional Stress = /

Page : 24 of 42



Design of Cross Shaft

Maximum Load On each Rope Drum = 32146.49 lbs.Maximum Load On each Rope = 16073.24 lbs.

= 16073.24 ∕ ( 0.96 x 0.92 ) x 24 / 2= 218386.46

Let Diameter of Cross Shaft. = 4.00 in.

Z == 12.57

= 12.57

T = T / Z = 218386.46 / 12.57

= 17,378.64 psi < 18000 Psi (Allowable)Which is Safe. (Mat. :- SAE 1045 )

We are using Dia 4.00 in. Shaft which is Safe.

Torque (T )

π d3 / 16

in3

Page : 25 of 42



Hoisting Efforts A - Gate Weight , W = 17,209 lbs.

B - Side Seal Friction = 4300.503 lbs.

C - Pin Friction = 44975 lbs.

Hoisting Load = 66,484 lbs.

Hoisting Efforts for lifting Gate. == 66,484 ¤ 2 x 24 ∕ 2= 398,907 lb - in.

=

= Torque on Rope Drum = 398,907 lb - in.

Torque on Hand Wheel (Handle ) ? lb - in.

n = Eff. Of Worm / Worm Gear = 65%

= 0.65

c = Gear Reduction Ratio = 20 / 52 x 1 / 60 x 19 / 47 x 19 / 100

= 0.000492

= 398,907 lb - in.

=

= = 398,907 x 0.000492 ∕ 0.65

= 302.16 lb - in.

P x rWhere

P = Pull Req. at the Hand Wheel.

r = Radius of the Handle = 15 in.T1 = P x 15 = 302.16 lb - in.

P = 20.14 lbsRPM = 30

Lifting Speed of Gate = 30 x 20 / 52 x 1 / 60 x 19 / 47 x 19 / 100 x π x 24 / 2

Manual = 4.496 in / minOpening Time of Gate = 18.00 x 12 / 4.496

= 48.05 Min.

The Eq. Will be T2 T1 x n / c

Where T2

T1 =

T 2

T 2 T1 x n / c

T1 T2 x c / n

T1 =

Page : 26 of 42



Horse Power Required To Lift the Gate.

We have the general formula for H.P.

H.P =33000

H.P = Horse Power Required To Lift the Gate.N = Revolution of Rope Drum per minute = 12 x 2 = 0.318

π x 24

We Require to Lift the gate at a speed of 12 inches per minute So there are 0.315 revolutions of rope Drum in one minute

Where T = 398,907 lb - in.

= 33242.24 lb - ft.

33000

= 0.318 x 33242.2433000

= 2.015 h.p.

Lifting Speed of Gate = 1440 x 23 / 48 x 20 / 52 x 1 / 60 x 19 / 47 x 19 / 100 x π x 24 / 2

Motorized 12.808 in / minOpening Time of Gate = 18.00 x 12 / 12.808

16.87 Min.

2 π N T

Torque on Rope Drum =

H.P = 2 π N T

2 π x

Page : 27 of 42



Hoisting Bridge.

Width of the Platform should be sufficient to accommodate the gear box and enough space for manual operation and maintenance.

1 - Clear Span = 26.8 ft.2 - Let length of the bridge = 33.39 ft.3 - Weight of Gate = 17,208.99 lbs.4 - Width of the bridge = 6.5 ft.5 - Weight of Hoisting m / c. = 1,500 lbs.6 - Dead load i.e. Railing ,Girder ,Channel , = 6140.87 lbs.

Angles,Chequered Plate etc.7 - Width of Pier = 6.56 ft.

8 - = 20359.248 lbs.9 - C / C of foundation Plates = 29.83 ft.10 - Self Weight of Girder = 56.44 lbs./ ft.

Maximum Bending Moment due to UDL ;

= Live load + Dead load = 20359.248 + 6140.871= 26500.12

Load / Girder = 26500.119 ∕ 2 = 13250.06 lbs.Maximum Bending Moment due to UDL. = W L / 8

= 592881.8532 in. - lbs.

Value of S for Beam IPN - 380 = 76.89

= 13250.059294 / 76.88992 7710.79 psi.

Maximum Bending Moment due to Concentrated Load.Concentrated Load. = Weight of Hoisting m / c. + Weight of Gate

= 1,500 + 17,209= 18,709 lbs.

Load / Beam = 18,709 ∕ 2 = 9354.494 lbs.

Maximum Bending Moment due to Concentrated Load. = W L / 4= 9354.494 x 29.83 x 12 / 4

= 837144.9 lbs.- in.

M / S= 837144.9 / 76.89= 10887.58 psi.

B. Stress due to UDL. + B. Stress due to Conc. Load.= 7710.79 + 10887.58= 18598.37 psi.

Maximum Bending Stress is 18598.37 < 21,600 psi , Hence our design is safe.

Live loads @ 105lbs/ft 2

a)

in3

Bending Stress f due to UDL. = M / S

b)

Bending Stress f due to Concentrated Load. =

Total Stress =

Page : 28 of 42



Properties of Chosen Section of Beam .

IPN 380 x 84 Kg. / m

A = 16.59 5.87

= 576.84

= 76.89

d = 14.96 in = 380 mm.

= 5.87 in = 149 mm.

= 0.81 in = 20.5 mm.

= 0.54 in = 13.7 mm.

Wt. = 56.44 lbs. / ft. 0.54 14.96

Check For Flexure.

= 1430026.78 lbs. in.

= 0.81= 0.9 x 21600 x 76.88992= 1494740.00 lbs. - in

>Hence Our Design is Safe. Fig. #REF!

Check for Deflection.

Max. Deflection occures at the Centre of the Beam Which is =

=### x ( 29.8304 + 9354.494 x ( 29.8304

576.84256 576.8426

= 0.457 + 0.517= 0.974 in

Allowable Deflection = L / 350= 29.8304 x 12 / 350= 1.023 in

Allowable Deflection > Actual Deflection Hence Our Design is Safe.

in 2

Ix in 4

Sx in 3

bf

tf

tw

Mu

Mn 0.9 x Fy x Sx

Mn Mu

5 W L3 / 384xEI + P L 3 / 48E I

x 12 ) 3 x 12 )3

384 x 30 x 106 x 48 x 30 x 106 x

Page : 29 of 42



Design of Supports

Max. Load on each side = (Wt. of Gate + seals Friction + Pin Friction +Dead Load + Live Load )/2

= ( 17,209 + 4300.50 + 44975 + 6140.87+ 20359.25 ) / 2= 46492.30 lbs.

Load on each Channel = 46492.30 / 2= 23246.15 lbs.

Maximum Length of Support = 72 in.

Bending Moment. = W L / 8

= 23246.15 x 72 / 8= 209,215.34 lbs.- in.

Allowable Bending Stress = 36,000 x 0.60= 21600 psi

Section Modulus Required. = 209,215.34 / 21600

= 9.69 We are using Channel 240 x 33.2 kg/mFor Which

= 18.31

Hence the Selected Support is Safe.

in.3

Sx in3

Page : 30 of 42



Design of Bottom Embedded parts

Total downward Force = wt. of gate + Frictional force due to Water head= 17.21 + 49.28= 66.48 kips

200 x 8

= 1600 = 186 in

175 x 20

= 3500 = 172 in

12 x 150

= 1800 = 87 in

140 x 12

= 1680 in2 = 6 inY =

{ ( 1600 x 186 ) + ( 3500 x 172 ) + ( 180087 ) + ( 1680 x 6 ) } 1600 + 3500

1800 + 1680= 124.275 mm

Moment of Inertia I =

200 x 8 1600 x 186 - 124.275

+ 175 x 20 3500 x 172 - 124.275

+ 12 x 150 1800 x 87 - 124.275

+ 140 x 12 1680 x 6 - 124.275

I = 43590630.86 4.36E+07

= 24862 Mpa

= 200,000 Mpab = 200 mmS = I / Y

= 43590630.86 ∕ 124.275

= 350759.29

P == Total down ward force

Width of gate

= 66.48 x 4.45 x 3.28 / 26.75= 36.27 KN/m

= 24862 Mpa

= 200,000 Mpa

b = 200 mm. Fig. #REF!

A1 =

in2 Y1

A2 =

in2 Y2

A3 =

in2 Y3

A4 =

Y4

3 / 12 +3 / 12 +3 / 12 +3 / 12 +

mm4 mm4

Ec

Ea

mm3

( k/4Ea x I ) 1/4

Ec

Ea

Page : 31 of 42



β =

K =

K = 0.4 x 24862 x ( 24862 x 200 200,000

x 4.36E+07= 16494.55 Mpa= 2391709.09 psi

β = ( 2E+06 ∕ ( 4 x 200,000 x 4.36E+07

= 0.016

M =

WhereM = 36.27 ∕ ( 4 x 0.016 x 1000 )

0.560 KN - m

Concrete Bearing Pressure

The load is uniformly distributedFormulae x is for wheels but in our case the load is uniformly distributed and hence whole of

Po = 36.27 x 0.016 ∕ ( 2 x 200 )= 0.0015 Mpa= 0.213 psi

This value is very small hence OK

( k / 4Ea x I)1/4

0.4 Ec ( Ec b4 / Ea I ) 1/3 ( Ref. 1 Eq. 6.2c Page # 153 )

4 ∕

) 1/3

)) 1 / 4

mm-1

P / 4βe - β x ( Cos βx - Sin βx )

e ßx ( Cos βx - Sin βx ) = 1

Po = P β / 2b ( Aβ X 1 + Aβ X 2 + Aβ X 3 ) Ref 1 page 154

Aßx = e βx ( Cos βx - Sin βx )

the term ( Aßx1 + Aßx2 + Aßx3 ) = 1

Page : 32 of 42



Design of side Embedded Parts

Fy 36 ksi

Fb 21.6 ksi

The load acting on the gate = 66.484 kips

(Resultant)

The load on one side track = 66.484 / 2

= 33.242 kips

= 147.93 KNThe height of the gate = 16.66 ft. = 5078.70 mm

Load on each side = 147.93 ∕ 5078.70

= 0.029 KN / mm

Modulus of foundationEc = 24862 Mpa ( 4000psi for concrete assumed )

Ea = 200,000 Mpa

b = 1 mmt = 8 mm

I =

= 1 x 8

= 42.67

Figure

Fig. #REF!

Fig. #REF!

b t 3 / 123 ∕ 12

mm4

300 8

27512

12300

12150

Page : 33 of 42



Modulus of footing

K =

K = 0.4 x 24862 x ( 24862 200,000 x 42.67 = 1420.37 Mpa

β = ( 1420.37 ∕ 4 x 200,000 x 42.67

= 0.0803

Effective length = = 0.0803(width) = 39.12 mm

Provided width of side track = 300 mmwidth provided > width required

Hence it is Ok

Max . Bending Moment

= 0.0803 x 300 = 24.10 radians

Cos h 24.10 = 1.456E+10 =

Sin h 24.10 = 1.456E+10 =Cos 24.10 = 0.5082Sin 24.10 = -0.8613

= Max. Bending Moment.

= = 0.029 x { ( 1.4565E+10 - 0.508 ) ∕ ( 4 x 0.080

x ( 14564586066 - -0.8613 )} = 0.09 N - mm

Bending stress, s =

= I / y = 42.67 ∕ 20

= 2.13Y = (12+8)

= 20 mm

Bending stress = 0.09 ∕ 2.13

= 0.0425

= 6.16 psi

0.4 Ec ( Ec b4 / Ea I ) 1/3 ( Ref. 1 Eq. 6.2c Page # 153 )

x 14 / ) 1 / 3

) 1 / 4

mm-1

π / β π ∕

β1

1.513 x 10 11

1.513 x 10 11

M c

P { ( cos h β 1 - Cos β 1 ) / 4 β ( sin h β 1 - Sin β 1 )}

Mc / Sx

Sx

mm3

Mpa

Page : 34 of 42



Concrete Bearing Pressure

Po =

= 147.93 ∕ 0.00

= #DIV/0!

= #DIV/0! psi

#DIV/0! < 4000 psi

Hence Ok

Lifting Eye Pin

Total Hoist load = 146531.78 Kg.

= 322956.05 lbs.

Load / Pin = Hoist Load / 2

322956.05 / 2

161478.02 lbs

Dia of pin used = 1.75 in.

Area of pin = 1.75

= 2.41

Shear Stress induced in pin = 161478.02 ∕ 2.41

= 67134.76 psi

Allowable Shear stress = 0.4 x 65,000

= 26000 psi

Allowable Shear stress > Actual Stress

Hence Dia 1.75 in. Selected is Ok

P / L eft.

Mpa

π / 4 x ( ) 2

in2

Page : 35 of 42


G.T. Road Gujranwala Radial Gates Check. By: Engr. M. Usman


Weight of Gate

Sr.No. Description Dimensions Qty. X - Sect. Unit Wt. Total Wt.Area

lbs. lbs.1 - Skin Plate 9.67 12 0.50 1 2369.44 2369.44

2 - H. Girder 6 12 0.375 9.67 2 196.74 6474.52 12949.05

3 - Gate Arm 9.22 4 #REF! 260.21 1040.83

( UB 309 x 42kg / m )

4 -

5 - V. Stiff. (Angle) #REF! #REF! #REF! 9.67 6 #REF! #REF! #REF!

6 - Top Angle 0.625 8 4 9.67 1 5.75 189.23 189.23

7 - Side Angle 0.625 8 4 4.71 2 5.75 92.17 184.35

8 - Side Backing Pl. 0.375 4 4.71 2 1.50 24.05 36.07

9 - Side Seal 0.50 6 4.71 2 3.00 9.52 19.04

10 - Side Clamping Pl. 0.375 3.50 4.71 2 1.31 21.04 42.08

11 - H. Backing Pl. 0.50 4 26.8304 1 2 182.62 182.62

12 - Bottom Seal 0.50 4.25 26.8304 1 2.125 38.41 38.41

13 - H. Clamping Pl. 0.375 4 26.8304 1 1.50 136.96 136.96

#REF!

14 - Bolts,Nuts,Washers #REF!

Pins.L.E.Pl.& Welds. etc.

Weight of Gate = #REF!

Total Weight = #REF!= #REF! Kips.

in.2

Page : 36 of 42



HOISTING M/C DESIGN

Spur Gear Design. (Refer Machinery Handbook)

= 23m, Module = = 4 mm

Pitch diameter of pinion = = 92 mm

= 3.62 in.

= = 6.35 ina = Addendum = 1/ P = 0.1575 in.b =Dedendum = 1.25/P = 0.1969 in.

= 2 / P = 0.3150 inC = Clearance = 0.25 / P = 0.0394 in.

= (T + 2) P = 3.937 in.

= 3.228 in.

Pinion HP for wearMaterial = A-322, Grade 8620

= 1400

= 0.935Pressure angle = 20 degreeZ = Zone factor = 2

F = Face Width = 1.77 in.N = Rpm = Revolutions per minute = 30 rpmNo. of teeth = 23

K = Pitch factor = = 4.3875p = Diameteral Pitch = 6.35

== 1400 x 0.935 x 2 x 1.77 x 30 x 21 / ( 126,000 x 4.3875 x 6.35 )= 0.911 Hp

Data of Spur Gear. Teeth of Spur Gear = 48

Module ,m = 4 mm

Pitch diameter of pinion = = 192 mm

= 7.559 in.

P = Diameteral pitch = = 6.35 in.a = Addendum = 1/ P = 0.1575 in.b =Dedendum = 1.25/P = 0.1969

= 2 / P = 0.3150 inC = Clearance = 0.25 / P = 0.0394 in.

= (T + 2) P = 7.874 in.

= 7.165 in.

Tp , Teeth of pinion

Tp x m

Dp

P =Diameteral pitch Tp / Dp

hk = Working depth

Do = Outside diameter

DR = Root diameter = (N - 2.50) / P

Sc = Surface stress factor

X c = Speed factor for wear

P 0.8 = ( 6.35 )0.8

Pinion HP for wear Sc Xc Z F N T / 126,000 K P

T p =

Tp x m

D p =

T p / D p

hk = Working depth

Do = Outside diameter

DR = Root diameter = (N - 2.50) / P

of 42


G.T. Road Gujranwala Fixed Wheel Gate. Check. By: Engr. M. UsmanWith Arm 6 m. App. By: Capt.(R.) Engr. Nasir

Gear HP for wear Material :- A 322 Grade 8620

S = 1400 psi.

= 0.935

Z = 2

F = 1.102 in.N = 30 x 23 / 48 = 14.375 rpmT = 48K = 4.39P = 6.35

= = 1,400 x 0.935 x 2 x 1.102 x 14.375 x 48 / (126,000 x 4.39 x 6.35 )

= 0.5671 hp

Horse power for strength Material ;- A 322 Grade 8620

= 17000 Psi

= 0.60Y = Strength factor = 0.70F = Face width = 1.77 inrpm = Revolutions per minute = 30 rpm N = Number of teeth = 23Diameteral pitch = 6.35 in

Horse power for strength (pinion) =

== 1.716 hp

Gear HP for strength

= 17000 psi.

= 0.60Y = 0.59F = 1.102 in.N = 30x 21 / 59 = 14.375 rpmT = 48P = 6.35 in.

Gear HP for strength =

== 0.901 hp

X c

Gear HP for Wear = Sc Xc Z F N T / 126,000 K P

Sb = Bending stress

Xb = Speed factor for strength

Sb Xb Y F N T / ( 126000 x P2 )

17,000 x 0.60 x 0.70 x 1.77 x 30 x 23 / (126,000 x 6.352 )

Sb

Xb

Sb Xb Y F N T / ( 126000 x P2 )

17,000 x 0.60 x 0.59 x 1 x 10.678 x 48 / (126,000 x 6.35 )

of 42



Pinion Hp for wearMaterial :- A 322 Grade 8620

= 1400 psi.

= 0.935Z = 1.83F = 2.56 in.N = 30 x 23 / 48 = 14.375 rpmT = 20K = 4.3875P = 6.35 in.

Pinion Hp for wear = Sc Xc Z F N T / (126,000 K P )1,400 x 0.935 x 1.83 x 2.56 x 14.375 x 20 / (126,000 x 4.3875 x 6.35 )

= 0.502 hp

Gear HP for WearMaterial :- A 322 Grade 8620

= 1400 psi.

= 0.935Z = 1.83F = 2.36 in.N = 30x 23 / 48 x 20 / 52 = 5.529 rpm

T = 52k = 4.3875P = 6.35 in.

Gear HP for Wear = Sc Xc Z F N T / (126,000 K P )= 1,400 x 0.935 x 1.83 x 2.36 x 5.529 x 52 / 126,000 x 4.3875 x 6.35 )= 0.463 hp

Pinion Hp for strengthMaterial :- A 322 Grade 8620

= 17000 psi.

= 0.60Y = 0.575F = 2.56 in.N = 30x 23 / 48 = 14.375 rpmT = 20P = 6.35 in.

Pinion HP for strength =

=0.850 rpm

Sc

X c

Sc

Xc

Sb

Xb

Sb Xb Y F N T / ( 126000 x P2 )

17000 x 0.60 x 0.575 x 2.56 x 14.375 x 20 / (126000 x 6.35 2 )

of 42



Gear HP for strength

= 17000 psi.

= 0.60Y = 0.62F = 2.36 in.N = 30x 23 / 48 x 20/ 52 = 5.529 rpmT = 52

P = 6.35 in.

== 0.845 hp

Worm Gear Design

C = Center Distance = 11.575 in. = 294.005 mmModule = m = 8 mm = 0.315 in.Pressure angle = 20 degreeAxial pitch = p = module / 0.3183

= 0.315 / 0.3183= 0.9895

No of worm starts = 1L = Lead of worm thread = t P

= 1 x 0.9895 = 0.9895 inT = No of worm wheel teeth = 60

D = T P / π= 60 x 0.9895 / 3.1416= 18.898 in

D = Pitch Dia of worm = 2C - D

= 2(11.575) - 18.898= 4.252 in

Lead angle (Worm)= Tan. La = L / π d

L a = 0.9895 / 3.1416 x 4.252

= 0.074= 4.232

Ratio = T / t = 60 / 1 = 60a = Addendum = 0.3183 P

0.3183 (1.484) = 0.3183 x 0.9895= 0.315 in

W = Whole depth = 0.623 x P 0.623 (1.484) = 0.623 x 0.9895

= 0.616 in

Sb

Xb

Gear HP for strength Sb Xb Y F N T / (126,000 x P2)

17,000 x 0.60 x 0.62 x 2.36 x 5.529 x 52 / (126,000 x 6.352 )

Pitch Dia of worm wheel =

Tan-1

Tan-1

of 42



= d + 2a= 4.252 + 2 x 0.315= 4.882 in.

F = Face width, worm wheel = 2.38 P + 0.25= 2.38 x 0.9895 + 0.25= 2.605 in.

F = Actual = 3.7795 in.C = Clearance = 0.2 m Cos La

= 0.20 x 0.315 Cos (4.232)= 0.06282

G = Worm thread length = P ( 4.5 + 0.02 x 60 )

= 0.9895 ( 4.5 + 0.02 x 60 )= 5.640

Actual G = 8 in

Rubbing speed = V =

== 6.176 ft/min

Efficiency = 100 x L / ( L + f π d )= 100 x 0.9895 / (0.9895 + 0.09 π x 4.252 )

45.149 %

Wear load (wheel)Material :- Phosphor Bronze. Sand Cast

S = 700

= 0.85

= 3.465

D = 18.898

=

Wear load (Wheel) == 73629.28 lbs- in.

Wear load (Worm)Material :- Corbon steel, Normalized

S = 900

= 0.72

= 3.465D = 18.898

Wear load (Worm) =

d o = Outside dia of worm

0.262 x n √ d2+(D/R)2

0.262 x 3.828 √ (4.252)2+(18.898 / 60 ) 2

Xw

Fe

0.18 S Xw Fe D1.8

0.18 x 700 x 0.85 x 3.465 x 18.8981.8

Xw

Fe

0.18 S Xw FeD1.8

== 80187.85 lbs. - in.

Page 41 of 42



Strength load (wheel)Material :- Phosphor Bronze. Sand Cast

= 7,000 Psi

= 0.56Q = Arc length of teeth at root

= 3.76 m = 0.315

D = 18.898

= 4.232

Strength load (Wheel) == 0.625 x 7,000 x 0.56 x 3.76 x 0.315 x 18.898 x cos (4.232 )

= 54689.776 lbs. - in.

Strength load (Worm)Material :- Corbon steel, Normalized

= 20,000 Psi

= 0.56Q = 3.76m = 0.315D = 18.898

= 4.232

Strength load (Worm) == 0.625 x 20,000 x 0.56 x 3.76 x 0.315 x 18.898 x 0.9973= 156256.5 lbs. - in.

Horsepower Ratingconsidering the minimum of values i.e. Strength load ( wheel ).

Mn / 63,000 = ( 54689.78 x 30 x 19 / 59 x 20 / 52 x 1 / 60 rpm ) / 63000

2.400 hp

Required Hp = Lift x Load / 33000hp 12.808 33242.24 ) / 33000

= 1.075 hp

References:1- Design of Hydraulic Gate . ( by Paullo C. F. Erbisti )2- Steel Construction Manual. ( by AISC. )

0.18 x 900 x 0.72 x 3.465 x 18.8981.8

Sb

Xs

La

0.625 Sb Xs Q m D cos La

Sb

Xs

La

0.625 Sb Xs Q m D cos La

hp =

= ( in. / min )(1ft / 12in x

3- ( By ERIK OBERG AND FRANKLIN & JONES. )

4- Formulas for Stress and Strain ( By Raymond J. Roark. )

5- Hand Book of Gear Design . ( By Gitin M. Maitra .)

Page 42 of 42

Machinery ,s Hand Book

radial+gate+with+arm+ +6+m.+++haji+shaher

Documents

nari riverradial

allowable

resultant

weir gates

xw fe

bending stress

water thrust

bending moment