Slender structures

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TECHNISCHEWIIVERSITEIT Laboratorium voor

00pshydromeChaflI

Archief MekeiWeg 2, 2628

CD Deift

T&.:O15.7868fl015'81

WEGEMT School

Trondheixn, 1991 January 21 - 25

basic

equations and some special effects

Marinus van HÖ15t

Professor Mechanical Engineering

Delf t University of Technology

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p= -pgZ

where

p pressure

p density of the liquid

g gravitati9nal acceleration Z coordinate of the point P

A small area dA through the point P experienCes a force

di=pndA

where:

d' force vector on dA

n = inward normal vector on dA

It is assumed that the inner surfaçe of the area is not exposed. to the liquid. The magnitude of F is indepefldent of the orièntatiOn of dA.

(2.1)

(2.2)

WEGEMT slender structures jan'91

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i Summary

This section deals with an introduction to the basic equations that describe the static and dynamic behaviour of submerged slender

prismatic bodies In a more practical sense this means the equations for marine risers and submarine pipelines during thé lay operation.

2. HydrostatiCs

2.. 1 Basics

In principle hydrostatics is very simple and does not offer unsolved problems The hydrostatic pressure in a homogeneous body of

liquid in the earth's gravitational field depends on the vertical distance from the point under consideration to the liquid surface only.

Take an XYZ coordinate system (Fig

2.1)

in space. The XY plane lies

in the liquid surface and the Z-axis points upwards. In a submerged point P(O,O,-Z) the presure is

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WÊGEMT slender structures jan9i

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With the. help of the integral proposition of the veçtor calculus it can be proved that

ffpffc1i=fffVP

dV

(2.5)

2.3 Buoyancy force distribution

The Archimedian law is a proposition on the integral of the

hydrostatic pressure It does not say anything about the distribution of the forces The distribution of the hydrostatic forces have to be obtained front expression

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For example see Fig 2 2 The solid vertical cylinder with a cross

sectional area A, is partly submerged over a length i The weight of the cylinder is smaller than the weight of the displaced water

Vertical equilibrium is obtained by a support at the top of the cylinder. The buoyancy force applies at the bottom of the çylinder

and is not axially distributed as the weight The pressure distribution

iS

shown in Fig 2 2a and the resulting internal

axial force distribution (due to hydrostatics only) is shown in Fig 2 2b Fig 2 2c presents the internal axial force distribution

including the effect of the weight of the cylinder an4 the support reaCtiQfl.

2.2

Archlmediafl law

A body submerged in a liquid buoyancy force B that

equals

ff=

This

relatiot is derived wit1i the B

B.

B2

ï=ffpffdA

the = experiences weight o o pgV principle

an upward force, the of the displacèd liquid,

stated n 2.1.

r:

(2.3)

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or:

3 Buckling due tò hydrostatic forces

Assume a long, slender, prismatic, steel rod suspended from the earth's surface in a deep drilled hole filled with drilling mud See Fig 3 1 The hydrostatic force is a compression force at

the lower ènd of the rod. The normal force distribution is linear inÇ:

N)=-p0gA1+p2gAC

(3.1)

where:

p0 density of the liquid in the hole

p5 density of steel

A cross sectional area of the rod

1 submerged length of the rod

ç vertical distance from the lower end of the rod

Ç is solved for N(Ç) O from:

p8gA(p0gA1

p0gAl

_poi

p3gA p2

Example

p0 20Ó0 kg/rn2, heavy drilling mud

p5 7800 kg/ma, steel 1 1000 m

then

2000 l000

_264m

7800

-So 264 ni of the rod is under compression. Note that the cross

sectional area has disappeared from the expression The cross

sectional area can be very small, for example 1 cm2 - iO rn2 The hydrostatic force on the lower end of the rod is then F

= p0

g A i =

2000 9 81 l0 1000 1962 N, working on a rod with a cross section of 1 cm2, 264 in length under compression This is a situation that

buckling of the rod is expected However, buckling will not occur This is understood by considering the potential energy.

The earth's surface is the zero potential energy level. The potential energy of the ród is:

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The potential energy of the displaced liquid i

E=O.5gAl2

-The total potential energy is then:

=E+E0=o.5gA1Z(pPo)

For

p5> Po is E > O. In a possible buckled state the combined centre

of mass of both the rod and the liquid shall rise with respect

to the straight situation This means the potential energy of of

the system in the buckled state Eh > E But buckling is only possible when

the potential energy in the system decreases Conclusion no buckling will occur.

Buckling for the described system is possible only when P5 < Po' in words when the density of the rod material is

smaller than the density of the surrounding liquid.

(3.5)

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WEGEMT slendér structures jan'91 4 Derivati9fl of basic relations

To derive the formulas for a (steel) riser first the equations for a solid bar instead c'ff for a pipe are derived A circular cross section with constant diameter is assumed The coordinate system x,y is a member coordinate system In the undeforifled state the x-axis coincides with the bar axis and is parallel to the global Z-axis of section 2 The analysis will be restricted

to the two dimenSional case.

The forces acting on a small section with length ds of the bar

are shown in Fig 4 1 In the unloaded and undeformed situation the axis of the bar coincides with the x-axiS The independent variable is the coordinate s measured along the (deformed) bar axis.

e angle between member axis and and positive

x-axis

T = axial tensile force V - shear

M

bending moment

b hydrostatic force per unit length perpendicular to

the. mémber axis

-q all other forces perpendicular to the member axis per

unit length

w weight of the bar = Ap5g

A cross sectional area of the bär

= density of the mäterial of the bar g acceleration of gravity

radius of curvature f the deformed member axis,with

i

r

dB ds

(4.1)

The magnitude of the hydrostatic force b will be derived in section

5 For the time being it is sufficient to assume

that the direction

of b is as indicated as i Fig 4 1 The magnitude of b is not so obvious as will be shown later.

Constder in Fig 4.1 the equilibtitm o the forces ix. a direction parallel t the vectors b d and q ds:

_cos(4O)+(V+d) cos(-.dD)+

-Tsìn(-dB) -(T+d')

sin(--dB) + -w sinO

ds-qd$+bdSQ

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WEGEMT slender strUctures jan'91 'IR M- (M+dt'Í) + Vrsin(*dO)

+(V+dV)rsirk(--d6) +

Tr(ì-cqS(-dO))

-(T+dT) r (1-c6s(-c))=O

worked

Out: =0 now are:

dTr(1_cos(4-O)) and dT/rsin(4.c)

small of higher order änd are neglected, further

sin(c)--dO

then is:

(4,7)

(4,8)

(4,9)

(4,10)

worked out:

dVcos(-.a)

(4, 3)

with

cos (4c)

=1 _{sin(*c) =-.c}

(4,4)

we obtain:

dv- TaO-wdssinO-qds+bdS = D

(4 ¡5)

or:

f=2

+wslriO+q-b

(4,6)

equilibrium of moments with respect to the point Q (see Fig 4.2) yields:

_dM+2Vr-4d8=O or:

d4=VrdO

and because:zdO=ds

_d

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7 Equilibriurn of forces in the direction perpendicular to the direction of the vectors b ds and q ds (sée Fig 4.1) yields:

-Tcos(--dO)

+(T+dT)äoS(-0)

-+Vsin(-o) +(V+dV)

sin(4) -wcoaßds'O

worked out:

dTcos(-.a)

Now is

dVsin(4.c)

small of higher order and is neglected. Further is

and

this yields:

dT+VdO-wdscÔSO=O

or:

Thê x and y coordinates are related with 9 and s by:

.=cosO and

..=3inO

ds

The. deformation añd the bendig moment are related through:

1_M

r El

according to the Eui.erian beam theory.. Because off:

also

yields:

For small rotàticns s -' x.

The derivatives with respect to s are replaced by derivatives with

respect to x.

d!=_t'. +wcosO

ds

(4,12) (4,13) (4,14) (4,15) (4,16) (4,17) (4,18) (4,20)

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8

And further is:

for the bending moment:

dM

_y

dx

and for the

tension:

sinOO=-a11d

---(

d

dc

r

d. dic

(4,23)

The equation for the shear then yields:

or:

_(4,22)

d2y

(4,21)

¿k;

P4,7. f4 = - V -

-w or:

_(4,24)

+ w dic

EI

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dk

EI

dM

_{- y}

dx dx

EI

dT

-dx

EI

dx

dBM

d. EI

dM

dx

dV_

_-b

dx

q

wO + q - b

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Wé rewrite all derived formulas agath in such a way that all

differential quotients appear once in the left hand term For

comparison the "nçrmal" Euler beam equations are also presented for a transverse distributed load q - b.

Beam, transverse load q - b, axi tensiofl large rotations: = sinO da

c. M

da EI (4,25)

.._rTf1+wsin8+_b

ds EI

Beam, transverse load q - b,, axial tension, smal rotations:

(4,26)

Beam, transverse load q - b, no axial tension, "normal" e4uations

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and the assumed direction in Fig 4.1 appears to be correct.

For small values of the curvature yields:

=Ap0gsiflO

Ap0g(d-*' -4

dgcos8)+PA

(5,2)

i-;i =Ap2g(dX- -

dscosO}+P8A

p5 is the atmospheric pressure at the water surface. The vector b ds is completely arbitrary The components are b ds and b1

ds.

The vector VB is the buoyäncy force, working vertical upwards under the condition that also the cross sectional surfaces are also exposed to the hydrostatic pressure;

=Apgd5

The vector equation for the equilibrium of the forces is

Now solve the magnitude and the direction of b from these euations.

The dérivation is presented in the appendix. The result is:

=A

(p0g(siflO- (d-x)

(5,5)

(5,6)

(5,3)

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-Z=T ±1W

sinO+q

ds

0EI

Ñow consider the differential equation for T in (4.25) and

substitute (6.2) and (6.3). Then we obtain:

--

_{(T0-Ap-g(d-X)).}

= +(w0+Ap0g) cosO

With

or:

dT

dx

M

_1+Ap0g-.--

_V_Eii

+w9côsO4-Ap0gCO36

= cosO

ds

equation (6.5) becomes:

dT

M

.4

=-V-1

wcosO

(6,4) (6,5) (6,6) (6,7)

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6 Efféctive teúsion

and

effect:ive weight

Substitute (5 ..) in 4 25) and neglect the infiuence of the atmospheric pressure, this results in:

f=T.

48jo+q_Ap0gsino+Apog(d) --

_(6,1)

Or rearranged:

We now call:

T0=Tog(d

(6,2)

the effective tension and

w0 =

w-Ap0g

(6,3)

the effective weight.

The effective tension in a bar is. the real tensioh (that is the tension that causes axial stress and strain that can be measured by strain gauges) in the bar increased by the weight of a column liquid with density p0, with the same cross section as the bar and a height equal to the vertical distance of the point under consideration to the water surface The effective weight per unit length of the

bar is its mass per unit length multiplied by the acceleration of gravity

(this is the weight "in air") dtirtished by the weight of the

displäced water per unit length.

Using effective tension and effective weight equation (6 1)

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The set of equations (4.25) now

becoêS:

= sinO. ds

cfM

d.. EI dÑ . T0 .. +w,sinO + q - - V - + w .cos O

And in a simi),ar way the set (4.26) becomes:

dx

EI dM

_y

V=Tefi +w,0+q

!

+w dic

EI

(6,8)

(6,9)

obviously the efféctiv weight and the effettive tension control the bending behaviour and therefore the stability öf the submerged bar Using these quantities explicitly in the mathematical description

considerably simplifies the equations

Consider again the bar suspended in the deep hole of section 3

Calculating the effective tension in the bar with expression (6 2) it appears that the effective tension vanishes at the lower end and increases when x increases The effective tension is positive in the entire bar and therefore no buckling will occur.

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WECEMT slender structures jan'91

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7 Pipes

The behaviour of the solid bar submerged in water has little practical application (though the equations (4 25) and (6 8) also

apply to steél wire ançhor cables etc.). The subject of this course is however risers and pipelines The influence of the surroundig water on the outside of a pipe is

of

course completely the same as

the influence on a solid bar The influence of the liquid inside the pipe can be derived in a similar way as for outside and will not be repeated here The convenience of the use of effective weight and

effective tension now becomes more clear The equations (6 8) and (6.9) remain valid, only the expéssions for w0 and T0 have to be adj usted:

= T+A0p0g(d0-X)

-A1pg(d1-X)

(7,1)

We = w-A0 p0g+Apg

with:

A0 - outside cross sectional, area of the pipe

A inside cross sectional area of the pipe

Po density of the liquid outside the pipe

Pi density of the liquid inside the pipe

d0 - X = hydrostatic head of the liquid outside the pipe d - x hydrostatic head

of

the 1iquid insidé the pipe

The idea of the effective weight and effective tetision are explained in a somewhat. different but clear way in [2].

8 The effect of internal, pressure on the bending stiffness

Ïs it likely that the bending stiffness and thus the buckling

behaviour is affècted by the internai pressure in the pipe? This

question in various formulations has often been the subject of discussion between people working on risers or similar slender pipe structuìeS.

Consider equations (6 9) and (6 10) The internal pressure does not explicitly appear in these equations. One of the controling

parameters for the shape of the riser is T (7.1) in which the hydrostatic head inside the pipe appears Now hydrostatic head is nothing else than pressure therefore

one

might conclude that when

overpressure is applied at the top of the riser this pressure has to be treated as an increase of the hydrostatic head resulting in a

decrease of the effective tension and increasing the danger of

buckling (Although people are apt to think intuitively that internal pressure increases the bending stiffness) This however is not the

case Because to apply overpressure inside the riser, the riser has

to be provided with closed ends. The internal pressure' results in a force on the closed end and increases the real tension in the same amount as the increase of the hydrostatic head and the effective

tension remains unchanged For a riser, where for simplicity the outside hydrostatic head has been omitted, this has been illustrated in Fig 8.1. See [2] for an exhaustive discussion.

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WEGEMT slender structúres jan'91

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9 The eq,iatioflS

for pipelInes

There is

no principal

dffèrence

in the

equations

for risers and for pipelines But it is rather natural for risers to

select the pipe bound x-axis vertical in the uxideformed situation, while for pipeline problems it is more convenient to take the pipe bound x-axis

horizontal in the undeformed situation This means that the equations derived in the previous sections are riser equations rather than pipeline equations The pipeline equations are presented

below with reference to Fig 9.]..

.=sinO

ds

a'8M

ds

EI

_(9,1)

ds

w osO

!±=_v.!L+w.sinO

10 Riser dynamics

This course thainly deals with hard piped risers. The

defôrmati0,

displacement and rotation is usually small for this type of risers therefore equations (6.9) w.11 be used rather than (6.8).

The dynamic effect, and this is utLderstoó4 to be the

bending

vibration, is introduced as the inertia force

lateral to the to the pipé

axis.

This force is a compÌlént of the

load q in (6.9); d2

q-

m-

_dt2

where m includes the mass of

the

teei pipe and everything attached to it, the mass of the liquid in the pipe and the

added mass This results in:

¿ri =

A6)g

+ A11 f

Ç

with:

m mass pet unit length

A8 cross sectional area of steel

- cross sectional area defined by the inner diameter A0 cross sectional area defined by the outer diameter

p5 = density of pipe material (steel) p density of fluid

inside the riser

p0 density of sea water

coefficient of adde mass

The differential equation for

the

shear ï

(6.9) then becomes: M _o

-

+ We +

q

-

'7T

(10,1)

(10., 3)

(10,2)

op0

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15

11 Special effects due to fluid flow 11.1 IntroductiOn

Up till now no attentton has been paid to fluid motion inside the pipe or in other words it was assumed that the fluid flow

velocity

inside the pipe vas zero Obviously this is usually not true for risers in operation The effect of fluid flow on riser

behaviour

iS

-not very great and can often be neglected n practical cases. In

literature the effects to be discussed are treated with the use of the momentum theorem This is an excellent tool to obtain

overall effects no matter how complicated the flow pattern is

In our case we are interested in the detailed distribution of the forces and this will be obtained by considering the fluid particles.

11.. 2 Flow through a curved pipe

Fig 11.1 shows an element of a riser. A fluid particle that flows through this pipe element experiences a centripeta1 acceleration

v2/r. Where y is the velocity of the particle (that is assumed to be equal to the average fluid velocity with respect to the

pipe) and

r is the local bendig radius To obtain this centripeta], acceleration the pipe wall excerts a centripeta]' force on the fluid particle and consequently the pipe wall experiences an opposite and equal force, a centrifugal force When it is assumed that all particles in a cross section f011ow the same path thé centrifugal force q per unit length on the pipe wall is

qf

=

-A1pj--Compare the sign convention with Fig 4.1.. 11.3 Cor-jolis effect

Consider the straight vertical pipe in Fig 11 2

The pipe oscillates around a hinge at the lower end and has an ins tntaneouS

anEular

velocity dO/dt. The fluid velocity inside the pipe is directed

upwards and has the magnitude y. A body of fluid that covers the internal cross section of the pipe and has a length dx experiences a coriolis acceleration .a - 2 dO/dt y in the positive y direction. The inertia force or the force on the pipe wall is q per unit

length.

= -:2VAPj

(11,2)

The cor-jolis force is opposed to the direction of the velocity due to the oscillation around the hinge It has a positive damping effect

When the fluid velocity is reversed, from top to bottom,

the cortolis force has the same direction as the oscillating velocity and has a

negative damping effect As stated before this effect is

usually small.

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Literature

1. P. Tendrup Pedersen, Equilibrium of offshore

cables and pipelines diring laying

International Shipbuilding Progress, vol 22, Dec

1975, no 256

2 C P Sparks, The influence of tension, pressure

and weight on pipe and riser deformations and stresses.

Transactions of the ASME, Journal of Energy Resources

TechnolOgY, vol

106, March 1984:

3 B G Burke, Analysis of marine risers

for deep water OTC 1771, 1971.

WEGEMT slender structures jan'9l

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a. hydrostatic pressure

on a verticäl

sub-merged cylinder

Fig 2.1: Hydrostatic pressure on

an area dA.

a

*

dF = pidA

ifth p=-pgZ

N=pgaA

Fig. 2.2: Hydrostatic load on a

vertical partially submerged c4inder

dN:__

dZ

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b. resulting

c. resulting

axial internal

force N due

force. N due

to hydrostatic

to hydrostatics

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L=1000 m

H

guide

//

I

F=p*A p=pisg*L

E::

earth surface

hole, filled with

drilling mud

/j

steel bar

cross sectiOnal area A=1 O .m

=264m

Fig. 3i: Bar suspended ¡n. a

druid hole filled with

drilling mud.

TR18

distribution

of normal

force N

tensiofl pression

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Fig. 4.1: Forces on o

small, deformed section

òf a submerged bar.

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Lever arms of V with

respect to Q

Lever arms of T with respect

o Q

Fig. 4.2: Moment lever arms of Shear and tension.

(auxiliary diagram of Fig. 4.1)

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Fig. 5.1:

Hydrostatic forces on .0 small, deformed section

of a submerged. bar.

r

(23)

X

A

/1/

'ft

T

=

(2)

Te (x) =

cose 1: open riser

= F - w(d - x)

Te'°(X)

=

T°(x) Ap1g(d-

)

cose .2:

closed riser, overpressûre at the top

F + w(d x) + pIA.

T(2)()

_-'

_{A1p1g(d x) - pA1}

T

(x) +p A1pg(d1 x) - p1A

T°(x) A1pg(d x) = T1(x)

Fig. 8.1:

Internal overpressure and effective tension.

TR22

T.

does not change by application

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V

y

:z:

Fig. 9.1:

Suitable coordinates for pfpe laying.

TR23

d

T

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V

Fig. 11.1: Centrifugal load on O curved pipe.

V2/r

dF= dm

2

=ftp1 rde

dF

= dF

2

-= ---Ap rd9-!_

1R24

2 V

r

= A1p1

dF.

ra

qf=

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coriolis acceleration: a = 2.

y

dt

dF (on the flùid). = a dm

.= A1p dx o

dF (on the pipewaH) = - Ajpdx a

dF

dO

q=

= - Apa= 2 Apv

Fig. 11.2:

Coriolis. load on oscillating pipe.

Qdl.C.L4&

;-TR25

5trrcL

£

₍₄

1

s>çJ

cr

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APP- i

Appendix

From section

5:

+ + _{ds = V}

For the components in x-direction:

I cos (8 - 1/2dO) - cos (0 + 1/2d8) + bds =

For the components in y-direction:

sin (8 - 1/2do) - sin (0 + 1/2dO) + bds = O

Elaboration of thé equation in x-direction yields:

Ap0g {d - x + 1/2 ds cos 8} cos (0 - 1/2d9) + pA cos (8 - 1/2dO)

- Ap0g (d - x 1/2 ds cosO) cos (8 + 1/2do) + bds

- p5A cos (8 + 1/2dO) = Ap0g ds

Ap0g [(d -

x)

{cös (8 - 1/2dO) - cos (8 + 1/2d8)} +

1/2 ds cos 8{cos (8 - 1/2do) + cos (O + 1/2d8)}] + bds +

p5A sinOd8 = Ap0g ds

Now is:

cos (8 - 1/2do) - cos (8 + 1/2do) = -2 sinO sin(-1/2dO)

= 2 sinO sin 1/2do = sinO dO

and

cos (8 - 1/2dO) + cos (e + 1/2dO) = 2 cc:O cos (-1/2do)

= _{2 cosO cos 1/2do} = 2 cosO

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So:

Ap0g[(d-x)sin8 d8 + 1/2 ds cosO 2 cosO]

+ ds Ap,o .g ds

-Ap0 g[(d-x) sinO dB + ds cos2B] + Ap0g ds sinO dO b ds

bds

= Ap0g[ds-ds cos28 - (d-x) smB

_de]

-p5.A sinO dO

bds

= Ap0g

[S1fl20 ds

- (d-x)

sinO dO] sinO dO

dO dG

=

Ap0g sinG [sinO -

_{(d-x)ã---] -}

pA sin8

For the components 1n y-direction:

Ap0g {(d-x) + 1/2ds cosO}

sin

(8-1/2do) +

pA sin(8-dO)

-Ap0g{(d-x) 1/2ds cos0} sin (0+1/2do) +

bds

_-

_PA

sin(9d8)=O

Ap0g[(d-x) {sin(8-I/2d8) - sin (8+1/2d8} +

1/2ds cosO {sth(8-1/2d8) + sin(0+1/2d8)}] +

bds

= O

Now is:

sin(8-1/2d0) sin(8+i/2d0) .2 cosO sin(-1/2d8)

= -

cosO dO

and

siñ(8-1/2d8)

+

sin(8+1/2d0)

2 sinO cos(-1/2d8)

2 sinO

So:

Ap0g[-(d-x)cos8 d8+ ds cosO.sinel

bds -

PA cos0d8

= O ds = Ap0 g[(d-x)cose d8 - ds ccsO sinO]

-

d8

= Ap0 g[(d-x) cose - cose sinO] b y b y t

an

dO .

b = apòg cose[(d-x)--- - sine] = Ap0g cose[sine

-cose 1

bx

sin8tanO

APP- 2

d8

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bids = -Ap0g cose [ds sin8 - (d-x) d] + pA cbsOd8

= -Ap0 g cose [sinO - (d-x) - J + pA cose

The magnitude ofb is:

II=

b2 + b2)

The direction of is:

= cose

b - sinO

te

X

i.e perpendicular to the tangent of the bended beam

_{element, as}

alreay assumed in section 4.

In addition yields:

sinO

= -ICI

cose

-de . de

Ibi

= A0(p0g sine

- _a ) - A.(p.g sinO

Apg [sinO

-A[p0g sjnO dO d-x) dO px a APP- ₃

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ZEcT toJ

O

dO/)

Il)

_4_

2...