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Document No.: Doc WRD 14 (496)C

Title: GUIDELINES FOR DESIGN OF BRANCHING IN PENSTOCKS FOR HYDROELECTRIC PROJECTS

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For official use onlyDoc. WRD 14 (496)C

September 2009

BUREAU OF INDIAN STANDARDS

DraftIndian Standard

GUIDELINES FOR DESIGN OF BRANCHING IN PENSTOCKS FOR HYDRO

ELECTRIC PROJECTS

ICS No. 27.140; 93.160

(Not to be reproduced without the permission of BIS or used as standard)

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Last date for receipt of comments is 20-11-2009

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FOREWORD

(Formal clauses to be added later)

Different types of branching like bifurcation, trifurcation etc are used in penstocks carrying water

from surge tanks or reservoirs to the power houses. This standards covers guidelines for design of

branching in Penstocks for Hydro Electric Projects.

For the purpose of deciding whether a particular requirement of this standard is complied with, the

final value, observed or calculated. Expressing the result of a test or analysis, shall be rounded off

in accordance with IS 2:1960 Rules for rounding off numerical values (revised). The number of

significant places retained in the rounded off value should be the same as that of the specified valuein this standard.


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1. SCOPE

This standard covers guidelines for design of branching in Penstocks for Hydro Electric Projects. It

covers Types of Branching, Requirement for Design, Types of Reinforcements, Analysis of Wyes,

Analytical Design of Internal Sickle Plate Type Bifurcation and Design of Spherical Branch.

2. TERMINOLOGY

Definition for some of the key terms used in the Code are given below for ready reference:

2.1 Branch: A penstock is generally called a branch when the flow of water is to be divided into

two or more branches, or when two or more flows are to be gathered to a main pipe.

2.2 Wye Branch: It is a branch to diverge a main pipe to two branch pipes, attaching stiffening

girders on their intersection lines.

2.3 Symmetrical and Unsymmetrical Branch: In a symmetrical branch, as shown in the Fig.1,

angles a and b are equal. Otherwise, the branch is unsymmetrical (see Fig. 12).

2.4 Equibranch: When branches have similar diameter, the branch is known as equibranch.

2.5 Tie Rod: A tie rod is the structural member placed inside the pipe for support, at the

branching. The same is illustrated in Fig. 2.

2.6 Spherical Branch: It is a type to connect main pipe and branch pipes to a spherical shell

through reinforcing rings (see fig. 12).

2.7 Sickle Plate: It is a crescent shaped rib inside the branch pipe, to give strength at the joint.

The Sickle Plate, shaped as an internal horse-shoe girder, is also called splitter plate. It is provided

at the intersection of the two branches for resistance against the forces being developed there.

2.8 Reinforcement: It is the support provided to counter the unbalanced forces acting on

unsupported areas at the branching junction.

3. TYPES OF BRANCHING

3.1 Geometrically, there are several types of branching possible, such as bifurcation,

trifurcation etc. However, in practical applications, generally a bifurcation is employed.

3.2 The Wye Branching is the one in which the main pipe diverges into two branch pipes. In the

Wye branching, the following categories are available:

a)

Wyes with sharp transition.b) Wyes with conical transition.

c) Wyes with tie rods.

d) Wyes with sickle.

3.3 In case of branching into 3 or more branches, the following types are available:

a) a type in which the main is pipe directly trifurcated

b) manifold type in which pipes are branched in the same direction in succession from a straight

main pipe, and


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c) a type to combine bifurcation.

These are illustrated in the Fig.3.

3.4 Another type of branching used in Hydro Electric Projects is the Spherical branching in

which the main pipe is connected to the branch pipe through a spherical shell and having

reinforcing rings. In India, so far the application of Spherical branching has been rather restricted.

4. REQUIREMENT FOR DESIGN OF BRANCHING

For the safe design of a branching, the following factors should be considered:

a) Hydraulics of Water Flow through the branch

b) External Pressure

c) Internal Pressure

4.1 Hydraulic Considerations for Flow of Water through the Branch

4.1.1 While designing the branch, the following hydraulic considerations should be taken care of,during pressure flow conditions:

a) Head loss due to branch should be small.

b) The total head loss in the penstock before and after the branching should be small.

c) Turbulent and secondary flows should not be allowed to be generated.

d) For equibranch, head loss in each pipe should be of similar value.

e) In case the flow rate in one branch pipe changes, a large vortex or a hydraulic pulsation should

not take place in the flow of the other branch pipes.

4.1.2 Design Considerations

In order to achieve the above conditions, the following design considerations are recommended:

a) In case of equibranch, the angle of branching should be kept symmetrical about the main pipe

axis, i.e., symmetry should be ensured in case of equibranch. In the context of Fig. 1, angle a

and angle b should normally be equal.

b) The head loss coefficient due to branching varies to a large extent due to the distribution ratio

of the flow. It has been observed that the head loss coefficient is minimal when the angle

between the equi-diverging branches is between 45 to 600, and it rises sharply with increase of

angle beyond this limit. Therefore, in the context of Fig.1, the sum of angle a and angle b

should normally be between 45 to 600.

c) Sharp changes in cross sectional area of a passage should be avoided as far as possible.

d) While deciding on the type of support for a branching, the fact that any hindrance to the flow

leads to significant increase in head loss, should be taken in to view.

e) Right angle branches and cylindrical outlets or inlets should be avoided where hydraulic

efficiency is important or cavitation cannot be allowed.

f) While selecting Tie Rod type of support at the branching, the fact that the head loss

coefficient increases largely due to presence of a tie rod, should also be considered. Tie Rod is

an obstruction to the flow in the penstock as shown in Fig. 1A.


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g) Similarly, the head loss coefficient largely increases due to presence of obstruction to the flow

in the penstock, caused by sickle in Sickle type branching.

h) The velocity of flow in the branches should be selected so that the Reynolds Number (Re) is

greater than 104.

i) The use of conical connections with side-wall angles, , equal to 6-8 degrees, reduces

hydraulic losses to about one third of those resulting from use of cylindrical connections.

Therefore, in practical applications, appropriate conical angles should be implemented.

4.1.3 Additional Hydraulic Considerations for Spherical Branch

Apart from the overall hydraulic considerations given above, for the spherical branch the following

additional aspects should be considered suitably:

a) The ratio of sphere diameter to main pipe diameter should not be kept very high in order to

limit the head loss coefficient. When the flow distribution ratio of a branch pipe becomes

high, i.e., when the % of flow in one branch is much higher than the other branch, the head

loss increases rapidly. However, from construction point of view, it is not desirable to employ

a spherical branch having too small a diameter compared with the main pipe. Therefore, the

ratio of sphere diameter to a pipe diameter of 1.3 to 1.6 should generally be used.

b) It is normally preferable to install a flow regulating plate inside. However, this method is

insufficient when the ratio of sphere diameter to a main pipe diameter is larger than 1.6.

c) While designing the diameter of the spherical branching, it is to be kept in view that the head

loss in spherical wye increases rapidly with increasing diameter of wye.

4.1.4 Loss Coefficient for Bifurcation

a) The head loss due to branch, H, can be expressed in the following equation:

g

v

2H

20

=

where

v0 = mean velocity of flow in the main pipe,

= head loss coefficient.

Values of are influenced by the branch angle, change in the sectional area, distribution ratio

of the flow to each branch pipe, and the Reynolds Number. An estimation of the head loss

coefficient for different branch angles can be made from Fig. 4, while influence of Reynolds

number of main pipe over the head loss coefficient, in case of conical wye having equal

distribution amongst the branches, is given at Fig. 5.

b) The head loss coefficient for conical wyes and manifolds, with various types of transitions,

and with/ without tie rods is given in Fig. 6, wherein Open branch refers to branch where nogate is provided, while Closed branch refers to branching having gates for regulating flow

through the branch. An estimate of the head loss in spherical wye with increasing diameter

may be made from Fig. 7.

4.1.5 Loss Coefficient for a Trifurcation

A trifurcation is illustrated in Fig. 12. The loss coefficient for a trifurcation can be given as

tsCoefficienLossEntry3

2

32

2

2 ++= SinQ

QSin

Q

Q

mm


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where

Qm = Discharge through main pipe,

Q2 andQ3 = Discharges through the branches

2 and3 = Horizontal angle of take off (see Fig. 8)

However, the losses would be higher than that calculated by the above formula, in case when one ormore branches are closed.

5. TYPES OF REINFORCEMENTS

5.1 To compensate for the openings for the branching made in the normal circular section, some

reinforcement should be provided to take care of the unbalanced forces acting on unsupported areas

resulted by these junctions. These reinforcements can be either inside the pipes (internal) or

provided externally at the junction (external).

5.2 External Reinforcements

Based on the extent of unsupported area, internal water pressure, ratio of main and branch pipeareas, clearance restriction for fabrication etc., various types of external reinforcements are

possible. Generally one or more exterior girders welded with tie rods or ring girder or a

combination of these is provided. The various types of external reinforcement are:

a) one plate

b) two plate reinforcement, and

c) three plate reinforcement.

The placement of reinforcements in case of one plate, two plate and three plate reinforcements is

shown in Fig. 9.

The exterior girder, also called as yoke, is of horseshoe shape which is welded to the periphery of

the junction of the pipes and finally welded to a tie rod or a ring girder provided at the beginning of

the bifurcation. The section of the girder may be T shaped attached to the penstock surface. Some

portion of the penstock steel liner also is assumed to act monolithically as a flange of the yoke

girder as in the case of stiffener rings (see Fig. 10).

5.3 Internal Reinforcements

Another type of reinforcement is to provide an internal horse-shoe girder called splitter or sickle

plate. It generally consists of crescent shaped rib inside the branch pipe and it is designed in such a

way that rib is directly subjected to tension and has the same magnitude as the stress in shell section

of pipes adjacent to it. Apart from being structurally strong, this type is more economical becauseof smaller external dimensions taking lesser space and enables large branch pipes to be fabricated,

transported, stress relieved and erected as a single unit (see Fig.11).

5.4 Spherical Branch

It is a type to connect main pipe and branch pipes to a spherical shell through reinforcing rings.

This type is normally treated as an axis symmetrical shell, and it is possible to decrease the local

bending stress of a spherical shell connected to a reinforcing ring if its cross section is properly

selected. Therefore, the plate thickness is comparatively less. The arrangement of a spherical

branch, vis--vis a wye branch is shown at Figure 12.


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5.5 Selection of type of Branch

On account of higher plate thickness in case of wye branch, normally it is preferred for low Design

Head. For comparatively higher discharge values or heads, Spherical branch is normally preferred.

However, no strict rules are available and the selection of type of branch is generally left to the

discretion of the designer.

6. ANALYSIS OF WYES

6.1 As mentioned above, at the junction where the main pipe diverges into two branch pipes, the

attaching stiffening girders on their intersection lines can be carried out internally or externally.

6.2 The forces on a bifurcation with external reinforcement, are shown in Fig.13. Some typical

examples of reinforcements for non-wye branching are illustrated in Fig. 14, while examples for

wye branching are given in Fig. 15.

6.3 The method of stress analysis used for branch outlets and wyes is approximate, with

simplifying assumptions. The reinforcement is proportioned to carry the entire unbalanced load as

indicated by the loading diagrams in the figures above. The total load carried by the reinforcementis equal to the product of the internal pressure and the unsupported area projected to the plane of

fitting. A portion of the pipe shell is considered to be acting monolithically with the girders as in

the case of stiffener rings.

6.4 Allowable Stresses

For branching, it is prudent to use lower allowable stresses as compared to the penstock, on account

of limitation of the designers to determine critical stresses in these complex structures with the

same degree of accuracy as is possible for the penstock. Consequently, these allowable stresses are

based on one half times the minimum specified yield stress or one-fourth times the minimum

specified tensile strength, whichever is less.

6.5 Analysis of External Reinforcement

6.5.1 In case of simple curved reinforcing plate, it is assumed that the plate is acting as a plane ring

and the loads in both directions are uniformly distributed and the plate is circular.

6.5.2 When the external girders are used in combination with tie rods or ring girders, the analysis

becomes statically indeterminate. For analysis, the deflection of the girder at the junctions of the tie

rods or ring girder under the load in both directions are evolved and equated to the deflections of

the tie rods or ring girders.

6.5.3 The loads considered for external reinforcement are shown in the Fig.13. As seen from the

figure, the yoke is considered as an elliptical cantilevered beam. It is assumed to be loaded byvertical forces varying linearly from zero at X=0 to P (r1 cos1 + r2 cos2) at X=L and by the forces

V1 and V2 due to tie rod at 0 and C and by the moment M (see Fig. 15).

6.5.4 Analytical Design of External Reinforcement

The method of stress analysis used for branch outlets and wyes is approximate. Simplifying

assumptions are made in the analysis that yield results of efficient accuracy for practical design

purposes. The reinforcement is proportioned to carry the entire unbalanced loads. The total load

carried by the reinforcement is equal to the product of the internal pressure and the unsupported


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area projected to the plane of the fitting. A portion of the pipe shell is considered to be acting

monolithically with the girders, similar to the stiffener rings.

The external stiffener(s) (see Fig. 9) are analyzed as a C-clamp with a portion of the pipe shell

considered as an equivalent flange. The width (W) of the equivalent flange may be obtained from

the formula:

56.1 RtaW +=

where

a = thickness of girder

R = radius of the main pipe

t = shell thickness.

The distribution of the design load in the case of one plate external reinforcement is shown in Fig.

15 (b). The increase in the bending stress, if the radius of curvature at the crotch is small, can be

evaluated using a correction factor in the bending formula for straight beams.

If the external girder is used in conjunction with the ring girder [see Fig. 15(c)], the same is

statically indeterminate. To simplify analysis, the ring girders are assumed to be free at the common

intersection point, and loaded with the triangularly distributed design load and the unknown shear

load concentrated at the intersection point. The deflection of each girder is calculated with the

direction of the unknown shear force assumed. The deflection of all intersecting members are

equated and the shear forces calculated, and the direct and bending stresses at any point along the

girders and ring may be determined. With the elongation known, the stress in the tie rod can also be

determined.

The sample computation sheet, with Fig. 13 as reference, illustrates the steps taken in the analysis

of a typical external reinforcement analytically. The same is given at Annex 2.

6.5.5 The typical design of one plate and two plate external reinforcement using Nomograph

As an alternative to the complex calculations involved in analytical method, a simplified graphical

method can be used, which has simplified the process of design of external reinforcement. The

steps involved in use of Nomograph for design of external reinforcement are illustrated below.

While wye depth dw and base depth db refer to two plate design, dw and db refer to one plate

design:

Step 1: Lay a straight edge across the nomograph through the appropriate points on the pipe

diameter and internal pressure scales. Read off the depth of plate (d) from its scale. This reading is

the crotch depth (see Fig. 1) for 2.54 cm thick plate for a two plate 90 degree wye branch pipe.

Step 2a: Based on the deflection angle, use the N factor curve (Fig. 17) to get the factors that,

when multiplied by the depth of the plate found in Step 1, will give the wye depth d w and base

depth db for the new wye branch.

Step 2b: If the wye branch has unequal diameter pipe, the larger diameter pipe will have been

used in steps 1 and 2a, and these results should be multiplied by the Q factors found on the single

plate stiffener curves (Fig.18) to give wye depth dw and base depth db. While Qw is to be

multiplied with dw to get dw, Qb is to be multiplied with db to get db. These factors vary with the

ratio of radiusof small pipe to the radius of the large pipe.


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Step 3: If the wye depth, dw found so far is greater than 30 times the thickness of the plate (2.54

cm) then wye depth dw and base depth db should be converted to conform to a greater thickness t,

by use of the general equation:

360917.0

11

=

t

tdd

whered1 = existing depth of plate

t1 = existing thickness of plate

d = new depth of plate

t = new thickness of plate selected

= deflection angle of the wye branch.

Step 4: To find the top depth, dt (for two plate design) or dt (for one plate design), use Fig. 19.

This dimension gives the top and bottom depths of plate at 90 deg from the crotch depths (see Fig.

20).

Step 5: The interior curves follow the cut of the pipe, but the outside crotch radius in both

crotches should equal dt plus the radius of the pipe for two plate design, or, in the single plate

design, dt plus the radius of the smaller pipe. Tangents connected between these curves complete

the outer shape.

The important depths of the reinforcement plates, dw, db and dt (see Fig. 20) can be found from the

nomograph. If a curved exterior is desired, a radius equal to the inside pipe radius plus d t can be

used, both for the outside curve of the wye section and for the side curve of the base section.

6.6 Three Plate Design

In the case of three plate external reinforcement, as shown in Fig. 9, the function of the third plateis to act like a clamp in holding down the deflection of the two main plates. In doing so, it accepts

part of the stresses of the other plates and permits a more economical design. This decrease in the

depths of the two main plates, is small enough to make it practical simply to add a third plate to a

two plate design. The two factors that dictate the use of a third plate are diameter of pipe and

internal pressure. When both of these are above the limits i.e., nominal diameter of 1500 mm and

design pressure of 2 N/ mm2, a ring plate should be used advantageously. If either of these limits is

exceeded, the designer may elect to use a third plate.

The size should be dictated by the top depth (dt). Since the other two plates are flush with the inside

surface of the pipe, the shell plate thickness, plus clearance, should be subtracted from the top

depth. This dimension should be constant throughout, and the plate should be placed at right angles

to the axis of the pipe, giving it a half ring shape. Its thickness should be equal the smaller of the

main plates.

7. DESIGN OF INTERNAL SICKLE PLATE TYPE BIFURCATION

7.1 The design of internal sickle plate is based on the principle that the stresses which occur in

the line of intersection of two parts of pipes are transmitted to a strengthening collar which lies in

the plane of intersection.


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7.2 The design of strengthening collar involves firstly, the determination of resultant of all forces

from the beginning of the collar to the point in question, determined according to size, direction and

position for various points along the line of intersection. The cross-section of the strengthening

collar which passes though this point and is at right angles to the resultant, is made symmetrical to

this resultant and proportional to its size. This results for the strengthening collar in a body of

constant strength, which is only subjected to normal stresses, i.e., with an absolute minimum of

material requirements.

7.3 Considering the requirement of equal thickness for ease of fabrication purpose, the collar is

obtained in the shape of sickle, which lies symmetrical to the line of intersection at the crown right

inside the branch piece and at its apexes.

7.4 For designing the strengthening collar, the size, position and direction of the resultant forces

which are transmitted at various points from the pipe walls at the line of intersection to the

strengthening collar, should be known.

7.5 Fig. 21 to Fig. 24 show half of the sickle shaped strengthening rib and also the resultant

forces. The coordinates of the intersection curve are obtained as follows:

7.5.1 The coordinates of the intersection curve (x and y coordinates) are obtained as below. The

Notations are explained below and also in the Fig 21 to Fig 24:

R1 Radius of the main pipe

- Half angle of bifurcation

Cone angle

- angle varied from 0at vertical to 90at horizontal, in steps of 2.5

sin

sinrx = ; cosry = , where r is given by

cotsintan1

Rr 1

+=

Similarly, z is given by

cotsintan1

cotsinRz 1+

=

Therefore2

1

)cotsintan1(

dcotcosRdz

+=

Therefore)cotsintan1(sin

sinRx 1

+=

and)cotsintan1(

cosRy 1

+=

Length of the Sickle Plate:

Projection of the intersection curve on the horizontal plane is obtained by putting 2/= in valueof x above.

)cottan1(sin

Rx 12/

+= which gives the length of the sickle plate.

7.5.2 The pipe walls transmit forces at the point of intersection from both sides on to the

strengthening collar which lies in the plane of intersection AB (see Fig. 22 - 24). On account of

their symmetry the resultant of these forces must always fall in the plane of the intersection. It is

assumed that the wall of the pipe is so thin that they can be considered as membranes and therefore


11/38 11

possesses no resistance against bending. Hence, they inflict only tractive and shearing forces on the

strengthening structure in the case of internal pressure.

7.5.3 With an internal pressure p, the forces per unit length in a cylindrical membrane are:

in circumferential direction =cos

pr

in axial direction =cos2

pr

7.5.4 In an element of the line of intersection of length dl the following forces are inflicted on the

strengthening from one side:

(a)As a result of the circumferential stresses :

dzcos

P1

pr= (see Fig. 24)

But since 21)cotsintan1(

dcotcosRdz

+=

3

2

11

)cotsintan1(cos

dcotcosRP

+=

(b)As a result of axial stresses:

drcos2

T1pr

= (see Fig. 22)

2

2

11

)cotsintan1(cos2

dRT

+

=p

7.5.5 The forces Pl and Tl acting from both sides on the elemental length dl can be resolved into

vertical and horizontal components in the plane of strengthening collar. The resolution of forces is

shown in Fig. 23. The forces in vertical direction are given by

d

p-

p]

cosK

sinR

cosK

cossincotR2[dV

2

1

2

1

3

2

1=

The forces in the horizontal direction are given by:

dp-p ]cosK

)cos(cosRcosK

sincoscotR2[dH2

1121

3

221 +=

where cotsintan1K +=

=

x

y

cotRtan

1

1

1;

=

cot

sintan 11

Integration of this equation from 0 to gives the forces:Resulting vertical force is =

0dVV


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Resulting horizontal force is =

0dHH

7.5.6 For the resultant of all forces which act in the elliptical arc EF (see Fig. 21) on the

strengthening collar is finally obtained as below:

22VR +=

=

y

x1tan

=

H

Vtan 1

7.5.7 As seen from the geometry of the triangle RVH in Fig. 21, when the branches are cylindrical

(i.e., R1 = R2), = and hence the resultant R will be perpendicular to the line OF (see Fig. 21). Inthe case of conical branches, > and therefore R has components giving rise to a normal stress

fn and shear stress qt on the plane passing through O.

The principal stresses, to be kept within permissible limits, are

ntnt f2

1q4f

2

1f

22+

+=

and ntnn f2

1q4f

2

1f

22

+=

To determine the position of the resultant R, moments of the forces with respect to O are calculated

as below:

Moment of Vertical Forces =

0dVxMV and =

0dHyMH

The integration of V, H, MV and MH can be done numerically using Simpsons Rule or other such

numerical techniques.

Total Moment = MV + MH

The distance of the resultant from O is given byR

Ml = .

For ,2/ = 2/

2/

2/R

Ml

=

7.5.8 Theoretical width of sickle plate at its crown, i.e., 2/ = , is given by the followingexpression:

2/l-ax2

B=

but)cottan1(sin

Rxax 12/

+==

However for practical consideration, since additional width is required to be provided to project

outside the intersection, the actual width is enhanced by a factor Ax where Ax ranges from 30 to

50mm.


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7.5.9 Width of the plate at any other section is then calculated in proportion to the value of R , i.e.,

2/R

RBb

=

The thickness of sickle should be such that the principal stresses are within acceptable limits.

Theoretical estimate of the sickle thickness S may be taken as

3BR

tR2S

1

2/

p

=

7.6 A procedure for design of internal sickle plate using analytical method given above, isenclosed at Annex 1 in a tabular form.

8. DESIGN OF SPHERICAL BRANCH

8.1 When a model consisting of a sphere and a cylinder having cave cover as Fig. 25 is

considered:

2

pr=HC

where

p = internal pressure

r = radius of the connecting main branch.

The horizontal component of a spheres membrane tensile force pr/2 is:

2

prcos

2

pa== HS

where

a = radius of the sphere and angle

= angle from vertical to the point where the cylindrical main branch meets the sphere

Thus, the horizontal direction force is kept balanced with HC = HS

8.2 On the other hand, a reinforcing ring is attached so as to resist the vertical component v of a

spheres membrane tensile force as illustrated in Fig. 26. The tensile force T1 generated in areinforcing ring by the internal pressure acting on the reinforcing rings breadth b is

T1 = prb

where b = width of the reinforcing ring.

The tensile force T2 generated in a reinforcing ring by the vertical component of a spheresmembrane is

cosa2

102 prT = ,

where

r0 = radius of the C.G of the reinforcing ring

= angle shown in Figure 26.


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8.3 Supposing only the membrane tensile force acting on a sphere, i.e. a sphere under membrane

stress condition, the cross sectional area Sof reinforcing ring is:

+=+

= )cosar0.5(brS00

21

o

rr

PTT

where r0 = Tensile stress of reinforcing ring.

The above formula includes the stress of a reinforcing ring r0 but it is possible to determine the

formula which does not include r0 from a displacement boundary condition as follows:

The radial force V acting on a reinforcing ring is:

)cos2

a+b(p=cosa

2 p

pbV +=

The displacement rof a reinforcing ring in radial direction is :

=

=

cos

2

sin 222 abp

SE

aV

S

rr

where

E = modulus of elasticity

S = cross area of the reinforcing ring.

The displacement S, of a sphere in radial direction at connecting point with reinforcing ring is :

( ) sin12

2

=Eh

pas

where h = spheres plate thickness.

r= S is to be essential in order that a sphere is under the membrane stress condition, which gives

+

= sinhcos

2

ab

1

2S

The spheres plate thickness h is

so

pa

2h =

where so = spheres membrane tensile stress

While, a cylinders plate thickness t is to be determined only by the tangential stress powith no consideration given to the reinforcing rings effect :

po

pr

2t =

If thickness of cylindert is to be determined so that the radial displacement c of cylinder may beequal to the radial displacement r of a reinforcing ring, it is not required to consider the effect of

a cylinder of a reinforcing ring and the sphere, and the sphere can be kept under a membrane

stress.


15/38 15

=

11

'

2

Et

pr

c

( ) sin12

2

==EL

pasr

8.4 With c = S, a sin = r,= 0.3, and the corrosion allowance is expressed with , the

cylinders plate thickness required is expressed with the following equation :

+= ha

rt 43.2'

The range to increase t to t is determined from the following equation:

( )3.1

2 '' tr

b where b is the new width of the reinforcing ring.

8.5 Practically, it is seldom to satisfy the above equations in computation of S, t and h, sobending moment and a shearing force at each point should be calculated by a statically

indeterminate calculation method etc., to attempt a strict solution.

8.6 When considering an actual use of steel penstocks, concave covers do not exist, and thus theaxial forces HS and HC in Fig. 25 do not act. As a practical solution in this case, there is a concept

to solve a structure model shown in Fig.27 with an assumption that a pipe is embedded in concrete

and the pipes axial displacement is restrained and fixed at a certain distance point from a

reinforcing ring. Internal pressures acting on the branch are explained above, but the external

pressure should be examined.

9. DESIGN OF BRANCHES BY NUMERICAL TECHNIQUES

The Wye piece, designed based on the above mentioned criterion is an indeterminate structure. It is

necessary that stress concentration occurring at the various intersection points should be ascertained

and suitable strengthening measures should be carried out, if found necessary. For this either

physical model studies / photo elastic studies or mathematical modeling deploying methods asFinite Element etc. with appropriate boundary and loading conditions should be deployed.


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ANNEX A

(Clause7.6 )

PRACTICAL PROCEDURE FOR DESIGN OF INTERNAL SICKLE PLAT

Data to be provided:

R1 Radius of the main pipe

- angle of cone

half angle of intersection

t thickness of sickle plate (assumed to be checked through stresses)

Table 1

1 2 3 4 5 6 7

Alpha(deg)

Alpha(Radians)

cotsintan1

Rr 1

+=

sin

sinrx =

cosry =

=

y

x1tan Radians)tan(sintan 11

=

Radians=1

0

2.55.0

..

..

90

Table 2

OUTER & INNER PROFILE OF THE SICKEL PLATE

1 2 3 4 5

Alpha (deg) xo

yo

x y

= x + h sin = y + h cos =

0 (b

2.5

5.0

..

..

90


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17

Table 3

CALCULATIONS OF VERTICAL FORCE

ALPHA = X dV/dX dV avg dV

cosK

sin

cosK

cossincot22

1

3- =[ dV/dX (R1)

2] x [2(

susequent

current)] = 0.5(dV

prev.+dV

curren

0

2.5

5.0

..

90

Table 4

CALCULATIONS OF HORIZONTAL FORCE

ALPHA = X dH/dX dH avg dH

cosK

)cos(cos

cosK

sincoscot22

11

3

2 +- =[ dH/dX (R1)

2] x [2(

subequent

current)] = 0.5(dH

prev.+

0

2.5

5.0

7.5

..

90


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18

Table 5

CALCULATION OF VERTICAL AND HORIZONTAL MOMENT

1 2 3 4 5

ALPHA

=X dM V /dX dMV avg dMV M V

sincosK

sinsin

sincosK

cossincot23

1

4

2

- =[dMv /dX] x [2(susequent current)] = 0.5(dMvprev.+dMvcurrent) = avg dM

0

2.5

5.0

7.5

..

90

7 8 9 10 11 12

dMH avg dMH M H M ResultantResultant

Moment

=[ dMH /dX ] x [2(susequent

current

)] = 0.5(dMHprev.

+dMHcurrent

) = avg dMH =MH+MV = R/36 = (R1)3M/36

Co

B =2 (x at alpha 90 * 1000 M/R at alpha 90)


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19

TABLE 6

CALCULATION OF THE VALUES OF PRINCIPAL TENSILE AND COMPRESSIV

1 2 3 4 5

Alpha Gamma Theta Theta - Gamma Normal Stress

Tan-1

(V/H) From Table 1 Col.3 - Col2Resultant (from col.11 of table 5 above) *

cos (col.4)

Resulta

0

2.5

5.0

7.5

..

90

7 8 9

Fn Qt Pt

(Col. 5)*100000/ (thickness in cm * b

from table 5 * 1000)

(Col. 6)*100000/ (thickness in cm * b

from table 5 * 1000)

(Col. 7) + sqrt(col.72

+ col.82)

2


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ANNEX B

(Clause6.5.4 )

SAMPLE COMPUTATION SHEET FOR ANALYTICAL DESIGN OF EXTERNAL

REINFORCEMENT (SEE FIG. 13 AND FIG. 13-a)

NOTATIONS:

1. AR Cross sectional area of circular ring

2. AU Cross sectional area of U beam of section considered

3. E Modulus of elasticity of steel

4. IR Moment of inertia of circular ring

5. IU Moment of inertia of U-beam at section considered

6. IUC Moment of inertia of U-beam at centre of element s

7. MU Bending Moment of U beam at section considered

8. IUC Bending Moment of U beam at at centre of element s due to reaction

between circular ring and U beam, and unbalanced pressure9. R Radius of main pipe

10. R1 Radius of branch pipes at point M

11. R2 Radius of centroidal axis of circular rings

12. R3 Radius of curvature of inside of U beam at point considered

13. T Total vertical shear in U-beam at section considered

14. Y Reaction between inner circular ring and external U-beam girder.

15. b1 Effective width of inner ring (equivalent flange)

16. b2 Effective width of U-beam (equivalent flange)

17. c Distance from centroidal axis to extreme inner fiber of U beam

18. c Distance from centroidal axis to extreme outer fiber of U beam

19. e Distance from centroidal axis of extreme inner fiber of inner ring

20. e' Distance from centroidal axis of extreme outer fiber of inner ring

21. ki, ko Ratio of actual stress in inner or outer fiber, respectively, to stress

computed by flexure formula for straight beam

22. mc Bending moment of U-beam at center of element s due to unit load at C.

23. p Internal pressure

24. q Constant for steel

25. Poissons ratio for steel

26. x Horizontal distance from centre line of circular ring

27. Angle of cross section of U beam with vertical

28. c Deflection of U beam at point C

Note: positive moments produce tension of inner fiber

+ sign of tension

- sign of compression


21/38 21

ANNEX C

(Fig. 13)

FORMULAE

A. Calculation of Y

1. Deflection c of U beam at point C

a. Moments and Vertical Shear(for a and L see Fig. 13) :

MUC = Yx for x from C to K T = -Y

YxaxpRL

UC = cos)(3

1M

2

1

for x from K to MYaxpR

L= cos)(

1T

2

1

YxLaxpRUC = cos)3

2(M 1

for x from M to N YLpR = cosT 1

mc = x For x from C to N

b. DeflectionUC

cUCc

EIsmM

=

1 2 3 4 5 6

x MUC S MUC x S IUC Col.4/ Col.5

Ec =

2. Deflection c of circular ring:

a. Hoop Stress S1 due to internal pressure = ( )

+ qd

ApR

R

2

12

RTq

1285= . Find the value of qwith this formula.

b. Radial deflection c at point C:

Due to S1 Ec1 = (+) S1R . Find Ec1 based on this formula.

Due to Triangular loadR

CI

pbRR222 27E 4+0.0=

Due to YR

CI

YR23 57E 4+0.0=

3. Solve for Y : Ec = Ec = Ec1 + Ec2 + Ec3

a. Stress in U Beam:

1 2 3 4 5 6

x MU c c k1 IU


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7 8 9 10 11 12

Bending Stress Combined stress

uI

ckMU

1

uI

kocMU

' T

UA

T sin Inside Outside

b. Stress in Circular Ring:

Point MR e e IR Bending Stress

R

Ri

I

eM=S

R

Ro

I

eM 'S =

C -0.07074pbR22 0.3183YR2 - +

E -0.065pbR22

+ 0.0353YR2

F +0.6959pbR22

+ 0.1817YR2 + -

Point Direct Stress Combined Stress

Inside Outside

C S1

E

( ) +=+= YpbRAR 354.01768.01

S1

F ( ) +=+= YpbRAR

54.05.01

S1

4. Determination of Values for ki and ko:

Case I ( See Fig. 13-a(a)-considering tie rod and inner circular ring)

A= b1t + t1h; h1 = r-R3 ; h2 = R6 r

4

5

1

3

4

1

r

R

RLogt

R

RLogb

A

ee +

=

=

c

I

AeR

h ui

3

1k ;

=

'

5

2kc

I

AeR

h uo

Case II ( See Fig. 13-a(b)-considering tie rod, inner circular ring and external girder))

A= b1t + t1h + b2t2; h1 = r-R3 ; h2 = R6 r

7

62

4

71

3

41

r

R

RLogb

R

RLogt

R

RLogb

A

eee ++=

= c

I

AeR

h ui

3

1

k ;

= '5

2

k c

I

AeR

h uo

5. Determination of C, Rmax, Rmin and R3 pertaining to elliptical intersection of U-beam:

See Fig. 13-a (c). F and F1 are the foci of the ellipse. R is the Radius of curvature at any point P.

r1 = a +cx;r2 = a -cx;r1 + r2 = 2a;

a

ba 22c'

= ;

b

a2

maxR = ;a

b2

minR = ;( ) 2

3

213R

ab

rr=


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