STRUCTURAL ANALYSIS-1

CABLES AND ARCHES

BHUDIYA DINESH- 38

HIRANI SOHAN -10MAYANI KAILASH -37

USMAN KHATRI - 79

ARCHES

An arch looks like curved girder , either a solid rib or braced supported at its ends and carrying transverse loads which are frequently vertical .

The early Indian railway and highways bridges also use masonry arches. Arches are also used in buildings to carry loads over doorways, windows etc., as well as to add an aesthetic touch to building.

Linear Arch (Theoretical Arch)

Consider a system of joint link work inverted about AB, with load as show in fig.1 . Under a given system of loading ,every link will be in a state of compression . the magnitude of pushes T1,T2,T3….etc. can be known by the rays Od, Oe , Of, etc. in the force polygon. The actual lines of action of thrusts T1,T2,T3, etc is known as the linear arch or theoretical arch .

In practice the arch is made parabolic, circular or elliptic in shape for easy construction and aesthetic appearance . Such is called actual arch.

Types of Arches1. Three hinged arch2. Two hinged arch3. Single hinged arch4. Fixed arch.

Eddy’s theorem

Consider a section at P distance x from left hinge. Let the other co ordinate of P be y. For the given system of loads, the linear arch can be constructed, if H is known. Since funicular polygon represents the bending moment diagram to some scale, the vertical intercept P1P2 at the section P will Give the bending Moment Due to external load system.

Let the arch is drawn to a scale of 1 cm = P meters. Let the load diagram is plotted to a scale of 1 cm = q kN, and if the distance of pole O from the load line is r, the scale of bending moment diagram will be ,

1 cm = pqr kN.mNow, theoretically, B.M. at P is given by Mp = V1x – W1(x-a)-Hy = μx-Hy

Where μx = = V1x – W1(x-a)-Hy = usual bending moment at a section due to load system on a simply supported beam.

From fig , we have μx= (p1p2)x scale of B.M. diagram =p1p2 (pqr) Hy=(pp2)x scale of B.M. diagram =pp2 (pqr) hence, Mp= μx-Hy = p1p2 (pqr) - pp2 (pqr)

Edyy’s theorem:-○ “ The bending moment at any section of an arch is

equal to the vertical intercept between the linear arch and the center line of actual arch “

Three hinged arch A three hinged arch has two hinges at abutments and one hinge at the

crown. It is a statically determinate structure. No. of reaction components = 4 (2 each support) No. of static equilibrium equations available = 3+1 = 4 One extra equation is available from the fact that the B.M. at the hinge

at the crown is zero.

Let the arch is subjected to a number of loads w1,w2,w3,…etc. Science B.M. at c is zero. Mc= μc -Hy=0 H= μc /y …..horizontal thrust.

Resolving forces along the section at p, F=Vcos ø – H sin ø…….1…….Radial shear… Similarly,resolving forces normal to the section, N=Vsin ø + Hcos ø…….2…..normal thrust….

Three hinged parabolic arch:- The equ. Of parabola, with origin at the left hand hinge A is

given by Y=Kx(L-X)…..1.where k = constant When x=L/2 and y=r=central rise.. We get from equ (1) r =k.(L/2)(L-L/2)=k.L.L/4 L= span of

arch K=4r/LL r=central

rise y=4r/LL(x)(L-x)…this equ. Of parabolic arch x= distance

from supprot A B

Three hinged circular arch Consider the centre line of the arch to be segment of a circle of

radius R,subtending an angle of 2 ø at the centre. It is more convenient to have the origin at D. the middle of the span.

Let (x,y) be the co-ordinates of the point p.

From ∆ OC1p, x= horizontal distance mesured from crown C. Op2=oc12+c1p2 R2={(R-r)+y}2+x2…….1

By property of circle, (2R-r)r=L/2 L/2……2

UNIFORMLY LOADED CABLE

Horizontal Reaction : When The Chord IS Horizontal,

Due To Symmetry. Va=Vb=wL/2So the Equation We Get isH = wL^2/8d..... Horizontal

Reaction

Cable Tension At Ends : The Cable Tension T At Any End Is The Resultant Of Vertical &

Horizontal At The End. T= wL/2√1+(L2/16d^2)

Shape OF The Cable :Under The Uniformly Distributed Load

The Cable Takes The Form Of a Parabola,The Equation Of Cable, With Either Support as origin Will Be,

y=4d/L2 x(L-x). Length OF The Cable :

When Both Ends Of The Cable Are At The Same Level.The Length Of The Cable,S = L+ (8/3)(d^2/L)....

The Main Suspension Cable Is Supported On Either Sides On The Supporting Towers. The Anchor Cables Transfer The Tension Of The Suspension Cable To The Anchorage Consisting Of Huge Mass Of Concrete.

INCASE OF GUIDED PULLY ARRENGEMENT: β1 =inclination of the suspention cable with

vertical β2 =inclination of anchor cable with vertical Vertical preassure on top of the pier, Vp = T (cos β1 + cos β2 )

Horizontal forces on the top of the pier Hp = T (sin β1 + sin β2 ) the horizontal foece will cause

bendingmoment in the tower.

In case of saddle on roller arrangement , rollers do not have any horizontal reaction. Therefore, the horizontal components of the tensions in the suspension cable and the anchor cable will be equal.

T1 sin β1 = T2 sin β2 = H The vertical preassure on the top of

the pier is, Vp =T1 cos β1 + T2 cos β2

Examples