Mba admission in india

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2. CABLES AND ARCHES2. CABLES AND ARCHES

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2.1 INTRODUCTION2.1 INTRODUCTION2.1 Introduction Cables carry applied loads & develop mostly tensile stresses -

Loads applied through hangers - Cables near the end supporting structures experience bending moments and shear forces

Arches carry applied loads and develop mainly in-plane compressive stresses; three-hinged, two-hinged and fixed arches - Loads applied through ribs - Arch sections near the rib supports and and arches, other than three-hinged arches, experience bending moments and shear forces


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2.1 INTRODUCTION(Cont’d)


2.1 INTRODUCTION 2.1 INTRODUCTION (Cont’d)(Cont’d)

In cables, the loads applied through hangers is considered to be a uniformly distributed load - in the same manner, the loads distributed to the arches through the ribs are considered to be uniformly distributed

Cable type structures - Suspension roof, suspension bridges, cable cars, guy-lines, transmission lines, etc.

Arch type structures - Arches, domes, shells, vaults


2.2 ANALYSIS OF CABLE2.2 ANALYSIS OF CABLE

2.2.1 Assumptions Cable is flexible and in-extensible; hence does not resist any

bending moment or shear force (this is not always true - e.g., fatigue of cables); self weight of cable neglected when external loads act on the cable

Since only axial tensile forces are carried by the cable, the force in the cable is tangential to the cable profile

Since it is in-extensible, the length is always constant; as a consequence of the cable profile not changing its length and form, it is assumed to be a rigid body during analysis

Even when a moving load is acting on the cable, the load is assumed to be uniformly distributed over the cable (since the cable profile is not assumed to change)


2.2 ANALYSIS OF CABLE 2.2 ANALYSIS OF CABLE (Cont’d)(Cont’d)

2.2.2 Cables subjected to concentrated loads When the weight of the cable is neglected in analysis and is

subjected to only concentrated loads, the cable takes the form of several straight line segments; the shape is called as funicular polygon. Consider for instance the cable shown in Figure 5.1

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L

L2L1

L3C

B

D

A

ycyD

P1

P2Figure 5.1



2.2.2 Cable under concentrated loads (Cont’d) In figure 5.1, the known parameters are L1, L2, L3, P1 & P2 - the

unknowns are the four support reactions at A and B, the three cable tensions (in the three segments) and the two sags (yC and yD) - 9 unknowns

Eight force equilibrium equations can be written at the four nodes and we need to have one more condition to solve the problem - This is met by assuming something about the cable, either its total length, or one of its sags (say yC or yD)



2.2.2 Cable under concentrated loads (Cont’d)Problem 5.1: Determine the tension in each

segment of the cable, shown below, and the total length of the cable


2.2 ANALYSIS OF CABLE - FOR 2.2 ANALYSIS OF CABLE - FOR CONCENTRATED LOADS CONCENTRATED LOADS (Cont’d)(Cont’d)

10

)5(;062.8)47( 2222 yBCf tAB

8683.0)062.8/ (7)s i n(;4962.0)062.8/ (4)cos ( 11

)5(

)s i n(;)5(/5)cos (222

222

y

yy

)s i n(;]3)3[ (

3)cos (;]3)3[ ( 322322

yyCD

3)3()t an(;]3)3[ (/)3( 3

22 yyy

C o n s i d e r i n g h o r i z o n t a l a n d v e r t i c a l e q u i l i b r i u m a t

B ,

0;0 VH FF

) ;cos (/)4962.0(0.0)cos ()cos ( 221 BABCBCBA

a n d 0)s i n(5)s i n( 21 BCBA ;). . . . . . . (. . . . . . . . . .) ] . . . . . . . .t an(4962.08683.0/ [5 2 IBA


2.2 ANALYSIS OF CABLE - FOR 2.2 ANALYSIS OF CABLE - FOR CONCENTRATED LOADS CONCENTRATED LOADS (Cont’d)(Cont’d)

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C o n s id e r in g e q u ilib r iu m a t C , 0&,0 VH FF ; ));/(cos()4962.0(;0)cos()cos( 332 BACDCDBC

010)sin()sin( 32 CDBC ;).........())........tan(4962.0)tan(4962.0/(10 32 IIBA

D iv id in g e q u a tio n (I ) b y (I I ) ,

2/1)]tan()(tan(4962.0/[)]tan(4962.08683.0[ 322 S u b s titu tin g fo r )tan( 2 a n d )tan( 3 in te rm s o f y a n d so lv in g , y = 2 .6 7 8 4 f t B A = 8 .2 9 8 8 k ip s ; B C = 4 .6 7 1 4 k ip s a n d C D = 8 .8 1 5 k ip s ; T o ta l le n g th o f c a b le = 8 .0 6 2 + 5 .6 7 2 + 6 .4 2 2 = 2 0 .5 1 6 f tadmission.edhole.com

2.3 CABLES SUBJECTED TO UNIFORMLY 2.3 CABLES SUBJECTED TO UNIFORMLY DISTRIBUTED LOADSDISTRIBUTED LOADS

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).......(0)cos()()cos(;0 ATTTF x

).......(0)sin()()sin(;0 0 BTTXwTF y

)......(0)sin()cos()2/)((;0 0 CxTyTxxwM o admission.edhole.com

2.3 CABLES SUBJECTED TO UNIFORMLY 2.3 CABLES SUBJECTED TO UNIFORMLY

DISTRIBUTED LOADSDISTRIBUTED LOADS (Cont’d)(Cont’d)

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E q u a t i o n ( A ) r e d u c e s t o : ;0)sin()cos( TT

;0)]cos([ Tdx

di n t e g r a t i n g )........(tan)cos( DFtConsT H

E q u a t i o n ( B ) r e d u c e s t o : ;0)cos()sin( 0 xwTT

t h i s e q u a t i o n c a n b e r e w r i t t e n a s )....()]sin([ 0 EwTdx

d

E q u a t i o n ( C ) r e d u c e s t o ;0)sin()cos( xTyT t h i s e q u a t i o n

r e d u c e s t o ).........().........tan( Fdx

dy

F r o m e q u a t i o n ( E ) , o n e g e t s ).......()sin( 0 GxwT , u s i n g t h e c o n d i t i o n t h a t

a t x = 0 , 0 F r o m e q u a t i o n ( D ) a n d ( G ) , d i v i d i n g o n e b y t h e o t h e r ( G / D ) ,

o n e o b t a i n s dx

dyFxw H /)tan( 0 f r o m E q n . ( F ) ; a n d i n t e g r a t i n g f u r t h e r ,

.)2/(20 BFxwy H A t x = 0 , y = 0 . T h i s l e a d s t o t h e f i n a l f o r m g i v e n b y

)2/(20 HFxwy


2.3 CABLES SUBJECTED TO UNIFORMLY 2.3 CABLES SUBJECTED TO UNIFORMLY DISTRIBUTED LOADS DISTRIBUTED LOADS (Cont’d)(Cont’d)

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)2/(20 HFxwy … . . T h i s i s t h e e q u a t i o n f o r a p a r a b o l a .

U s i n g t h e c o n d i t i o n , a t x = L , y = h , o n e o b t a i n s t h a t )2/(20 hLwF H ;

h e n c e 2)/( Lxhy C o n s i d e r i n g t h e p o i n t B , 2

02 )([ LwFT Hmas

]1)2/([()(])())2/([( 20

20

20 hLLwLwhLw


2.4 ADDITIONAL CONSIDERATIONS FOR CABLE 2.4 ADDITIONAL CONSIDERATIONS FOR CABLE SUPPORTED STRUCTURESSUPPORTED STRUCTURES

Forces on cable bridges: Wind drag and lift forces - Aero-elastic effects should be considered (vortex-induced oscillations, flutter, torsional divergence or lateral buckling, galloping and buffeting).

Wind tunnel tests: To examine the aerodynamic behavior Precaution to be taken against: Torsional divergence or

lateral buckling due to twist in bridge; Aero-elastic stability caused by geometry of deck, frequencies of vibration and mechanical damping present; Galloping due to self-excited oscillations; Buffeting due to unsteady loading caused by velocity fluctuations in the wind flow


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