Suspension Cable Bridges

UNIT 14 SUSPENSION BRIDGES

Structure 14.1 Introduction

Objectives

14.2 Analysis of a Light Unstiffened Suspension Bridge 14.2.1 Cable subjected to Concentrated Forces 14.2.2, Cable subjected to Uniformly Distributed Load along Horizontal Span 14.2.3 Analysis of Anchor Cables 14.2.4 The Catenary

14.3 Analysis of Light Stiffened Suspension Bridge 14.3.1 Introduction 14.3.2 Analysis of Cable Stifl~.led with Three-Bnged (iirder 14.33 Influence Line Diagrams (ILDs) for Three-Hinged Girder Suspension Bridge 14.3.4 Analysis of Suspended Cable Stiffened with Two-Hinged Girder

14.4 Supporting Towers I

14.5 Summary

14.6 Key Words

14.7 Answers to SAQs

14.1 INTRODUCTION

In many engineering structures, such as suspension bridges, transmission lines, aerial tramway, guy-wires for high towers etc, cables are suspsded between supports and subjected to vertical loads. The load of the bridge floor is transferred to a cable which is stretched over a span to be bridged. The cable is flexible and can adopt any curvature as required by the load. Bridges are of two types (a) unstiffened, and (b) stiffened. In unstiffened bridges the curve of the cable undergoes changes with passage of loads over the bridge and decking also gets disturbed. The decking of stiffened bridge is stiffened by

.girders or trusses which distribute the loads evenly over the entire cable. Various components of the suspension bridge are as follows :

(1) the cable, (6) vertical suspenders,

(2) decking, (7) supporting towers,

(3) main span, (8) side span,

(4) back stay, (9) anchorage, and

(5) saddle or pulley, (10) Stiffening girdet.

The unstiffened bridges subjected to concentrated and uniformly distributed load, shape of cable, tension, length of cable are discussed in Section 14.2.

General

Since flexible cable offers no resistance to bending, the resultant internal force on any cross-section of cable must act along the tangent to the cable at that section. The resist'ulce to bending offered by actual cables is usually relatively sml l and can be neglected without serious error. Thus, cable structures carry their load through tension which is most efficient way of re9isting loads.

A cable suspended between two supports at its ends may be subjected to different types of loading and the shape assumed by the cable, therefore, will depend upon the type of loading. Various types of loads are as follows :

(a) Vertically downward concentrated load(s)

(b) Distributed load

(i) The weight of a suspension bridge roadway is an example of this type of load, which is uniformly 'distributed along the horizontal span. Cables loaded uniformly along the span remains in the shape of a parabola.

Miscellaneous Topics

(ii) The weight of homogeneous cable of constant cross-secti'on is an example of load distributed uniformly dong the cable. A cable loaded uniformly along the cable assumes the shape of a catenary.

Assumptions

(a) The cable is perfectly flexible. The wire ropes and parallel wire cables are obviously quite flexible and have little flexural stiffness. Eye bar chains cannot carry any momentsbecause of the freedom of rotation at the hinges.

(b) The stiffening girder is straight, its moment of inertia is constant and it is tied to the cable throughout its length. This assumption takes into account the deformation of only the chord members of the truss and neglects the deformation of the web members. Also since suspenders are closely spaced, their loads may be assumed to be uniformly distributed without the introduction of serious error.

(c) The dead load of the truss (stiffening girder), cable and suspenders is uniform per unit horizontal length; therefore the initial curve of the cable becomes parabolah This is achieved in practice by attaching the girder, after the dead loads have been transferred to the cable, by adjusting the suspender forces during erection by meals of turn-buckles provided in them for this purpose.

(d) The form and magnitude of ordinates of the cable curve remain unchanged even after the application of the live loads. By this assumption, the cable always remains parabolic with the same central sag. Obviously, the stiffening girder must deflect under live loads, pulling the cable downwards along with it. These deflections and resulting errors will increase with larger span and shallower trusses. It is found that with small spans and deep trusses, the errors are small. The errors, however, are found to be always on the safe side.

Concept

To understand the structural behavior of the cable, consider a simply supported beam as shown in Figure 14.1 (a). Beam is subjected to a point load P at mid point. As the cable has zero moment carrying capacity, insert an internal hinge under load so that moment carrying capacity of beam at that point reduces to zero and beam assumes shape as shown in Figure 14.1 (b).

Considering the equilibrium 6f point B as shown in Figure 14.1 (c) tension in beam . P . T = - stn8, 2

and both horizontal components are T cos 8 which are equal and opposite at B.

P

a 1 c B

P

e ( c I

8

Figure 14.1

Objectives After studying this untt, you should be able to

evaluate the tension in the cable under various loading and calculate final length of the cable,

find the horizontal and vertical reactions at the point of support,

evaluate the moment at the base of suspension pier,

compute the bending moment and shear force in the stiffermg girder,

find the forces in the suspenders under various loadings, and

construct influence line diagram (ILD) for BM and SF in girders, reactions at cable supports etc.

I 14.2 ANALYSIS OF A LIGHT UNSTIFFENED SUSPENSION BRID,GE

Suspension Bridges

-

14.2.1 Cable subjected to Concentrated Forces Solution by General Theorem

Consider cable supported on horizontal span of 8 m and length of cable on application of load is 10 m as shown in Figure 14.2 (a). The freebody at B is shown in Figure 14.2 (c).

Figure 14.2

Tension in cable at point B

P 5P 2 sin 0 6

Horizontal component of T

Now the moment which is product of horizontal component at B and sag at point B, is

Assuming rigid beam as shown in Figure 14.2 (b). Bending moment at B is calculated as

which is same as calculated in Eq. (14.1).

From above, the following general cable theorem may be stated.

At any point on a cable acted upon by vertical loads, the product ofthe horizontal component of cable tension and the vertical distance from that point to the cable chord equals the bending moment which would occur at that section if the loads carried by the cable were acting on an end supported beam of the same span as that ofthe cable.

Example 14.1

Application of general cable theorem may be understood by following illustrative example :

Suppose that the loading on a cable is defined as shown in Figure 14.3 (a) and that the vertical distance from the cable to the cable chord is known at one point. Neglecting the weight of the cable itself, the bending moment at point C on an imaginary beain Figure 14.3 (b) may be found as follows.

Support reactions :


Giving, REV = 1.17 kN,

and RAv = 3.5 - 1.17 = 2.33 kN

Moment at point C, considering equilibrium of portion ABC, and applying general cable theorem

M c = 2 . 3 3 ~ 2 - 1 ~ 1 - H ~ 1 = 0

Figure 143

Segment AB

To detennine the distance of any other point such as B from the cable chord, the general cable theorem is applied for portion AB at section B, leading to

from which

YB = 0.64 m

The segment of the cable between A and B along a straight line, has length equal to

Since the horizontal component of cable tension is equal to 3.67.kN, the actual cable tension between A and B equals

Segment DE k 1

Generil cable theorem is applied for portion DE at section D, leading to

Length of segment DE = J[m = 2.10 m

Tension in segment DE, TDE = 3.67 x - = 3.85 kN [;b" ) Segment BC

\ /

Length of BC = ( I ) ~ + (1 - 0.64)' ] = 1.06 m

\ I Segment CD

Suspension Bridges

-

Length of CD = (2)2 + (1 - 0.64)~ ] = 2.03 nl

Solution by Graphical Methad

Figure 14.4 (a) shows a cord ABCD of negligible weight supported at A and D and carrying the load system W, and W2. Let the tensions in the parts AB, BC, CD of the cord be TAB, TBC and TCD rey9ectively.

Figure 14.4

Since the forces acting on the cord form a system in equilibrium, the following conditions must be satisfied.

Let the applied loads Wl and Wz be represented by nb and bc to a suitable scale. Through 'a' draw line ao parallel to AB and through 'c' draw line co parallel to CD and thus obtain point '0'.

Now abcoa is the force polygon corresponding to the system of forces keeping the cord in equilibrium. Draw od perpendicular to the load line alx, so that (la represents the vertical reaction RAV and cd represents the vertical reaction Rr>v.

Now consider the equilibrium of point B. The force acting at this point are T A ~ , TBc and Wl. TAB is represented by ao, WI is represented by ab, therefore, by principle of triangle of forces. Tsc is represented by ho. Similqly, other forces can be found out.

Example 14.2

A cable supported at A and E is loaded as shown in Figure 14.5 (a). Compute the reaction components and the tension in different portions of the cable.

Solution

Considering the cable as whole in equilibrium [Figure 14.5 (b)] and applying condition of equilibrium Fx r 0.


Thus,

We get,

8REv - 2H = 56

since the sag of the point C is known, we consider the equilibrium of the portion CDE of thc cable as shown in Figure 14.5 (c).

Similarly,

Giving, 4 R ~ v - 2.25 H = 8

On solving (i) and (ii) simult;meously, we get

R , q v = l l k N and H =16kN

Applying condition of equilibrium for the whole cable Fr = 0,

Giving RAv = 3kN

4kN 6kN CkN

Segment AB

Consider equlibriurn at left of point B

(ii)

Giving, YB = 0.875 m

Length of AB, L~ = d[ (2)' + (1.5 + 0.875 - 21'1 = 2.035

Segment BC Suspension Bridges

Segment CD

Applying general cable theorem at point D,

Segment DE

14.2.2 Cable subjected to Uniformly Distributed Load along Horizontal Span

Shape of Cable

Let ABC in Figure 14.6 (a) be a flexible cable supported and loaded as shown. The lowest point of the cable B, is selected as the origin for the co-ordinate axes. A free

-body diagram of portion of the cable immediately to the right of point B is shown in Figure 14.6 (b). The horizontal tension in the cable at B is denoted by TB and the tension at any other pointy (x, y) by Tp .

Here,

It gives Tp COS 8 = TB ; Tp sin 8 = qx (14.3) where q is the load per unit length (measured horizontally). Elimination of T from these equations gives

9x tan 8 = - TB

since the slope of the curve is at point P,

Miscellaneods Topics

Equating Eq. (14.4) and Eq. (14.5) and integrating,

when x = 0, y = 0, therefore, C = 0 and

Eq. (14.6) shows that shape of cable is a parabola with its vertex at the lowest point of the cable.

Tension in Cable

The elimination of 8 from Eq. (14.3) gives

Eq. (14.7) shows that the tension (T) in the cable varies from the minimum value, TB at the lowest point in the cable when x equal to zero, to maximum values at the supports of towers. By substituting the value of x and y at support in Eq. (14.6) and using Eq. (14.7) the tension at supports

Where fA and fc are vertical cmrdinates of A and C, while a and b are horizontal intercepts of A and C with respect to B as shown in Figure 14.6 (a). Eq. (14.8) can be used to determine the sag for a given tension and span in case of power transmission lines.

Length of Cable

Sometimes it is desirable to evaluate the length of cable, L, between the origin and one of supports. The length of the curve from B to A, LA is given by the following expression,

L J

From Eq. (14.61,

and thus,

when Eq. (14.9) is integrated directly, the result is a cumbersome hyperbolic ar logarithmic function. A mare useful expression for LA can be obtained by expanding the integrand by the binomial expansion and integrating term as follows :

This series converges to the correct length of LA for srnall value of fA la less than 112. Similarly, the length of the cable LC from the origin to the left support is

The series converges to the correct length of LC for all values of fc lb less than 112. Usually the ratio fA la or fc Ib is quite small, and the first two terms of the series give a sufficiently close approximation to the actual length. i Length of cable ABC, I

Elastic Stretch of Cable When a cable supports a load, it undergoes an elastic stretch, which is often of importance in determining cable sags and also for other purpose. By the definition of the modulos of elasticity.

An element of a cable of length day is subjected to tension Tas shown in Figure 14.6 (c). A convenient method of determining the elastic stretch of a cable consists in first determining T,, which by definition will be taken as that average tension which if . applied throughout the length of the cable will cause the same total elastic change of length as actually occurs. It can be expressed as

where A and E are assumed constant and is length of cable from B to P as shown in Figure 14.6 (c). Hence,

T = - Tds (4: Substituting values, for portion BA and BC for Figure 14.6 (a),

but from Eqs. (14.4) and (14.5),

Suspension Bridges


Example 14.3

A chain weighing 2.0 N/m is suspended between two poles with the same elevation, spaced 6.0 m apart. The pull of the chain on the poles is 35 N. Deternline the sag and the length of the chain, assuming the weight to be uniformly distributed along a horizontal axis.

Solution

Substituting the values,

6 q = 2 N/m, = - = 3 m and TA = 35 N in m. (14.8) 2

L J

Therefore, f~ = 1.7 m

f~ 1.7 The value of - = - = 0.6. using this value, 0 3

= 3.645 m for half the chain = Lc

Therefore, the total length of the chain is 2 LC = 7.29 m. SAQ a

A SO kN load is uniformly distributed over a span of 61 m. It is to be supporled by ;I

single flexible cable hung from two towers at the same level. The design staength of Ihe ci~h!es 1s 100 kN. neberrnine the miraimurn sap ;md lel~gtlm of cahle which can he used.

Exampk 14.4

A cable carries a load of 2 kN per horizontal meter as shown in Figure 14.7. The left support is lower than the right support by 3 m, and the sag measured from the left support is 3 m. The supports are 46 m apart. Determine the minimum and maximum tension in the cable and the length of the cable.

Solution

In this problem, assume A is right sup jx~ t and C is left support.

f A = 6 m , f c = 3 m

q = 2kN/m, a + h = 4 6 m

h = (46 - a)

Eq. (14.6) of shape of cable,

For segment AB,

and for segment BC, Suspension Bridges

2 x (46 - a)' TBC = (2 x 3)

Equating, TEA = TBC

d = 2 (46 - a12 = 2 [(4612 - 2 (46) (a) + $1

Figure 14.7

On solving equation, we get

= 27m

h = 19m. as a + / ) = 46m,

Tension at support A , using Eq. (14.8),

Tension at support C

f~ 6 NOW.- = - = 0.22. fC - - 2 a 27 b 19

= 0.16. applying equation of length


The I l cx~h i :~ cahl:. ne~!!\iu~y 15 K N , m ili~!\o\ hc't'.v~', !x !t it; \rriy~i>r!\ I I H I I r r ,q'.iu J , I I I I

left support 3 In L ~ O V C tile rr~lll supporr ?he ,ago[ t l r = cahi~, talc;r\urcill I r o r i ; h-rghcr el\d ms 5 m Find the li(ir~~ont:~l tlliunt. Inaxinlurn L~IISI:III 111 t l~c cable ant1 the 1cn:di of the cable

14.2.3 Analysis of Anchor Cables The suspension cables are supported on either sides on the supporting tower. The anchor cables transfers Qe tension of the suspension cable to the anchorage. There are generally two types of arrangements,

(a) cable passes over guide pulley at the support, and

(b) cable clamped to saddle on smooth rollers on the top of pier.

In the former cast, when the suspension cable passes over the guide pulley as shown in Figure 14.8 (a), it acts as anchor cable on the other side, the tension Tin the cable is the same on both the side.

Tower

Suspension Anchor cable cable

\

(b) Figure 14.8

Let 81 = inclination of the suspension cable with vertical, and

O2 = inclination of the anchor cable with vertical.

Pressure on the top of pier

Vp = TcosOI + T cos O2 = T(COS el + cos 02)

Horizontal force on, the top of the pier I

= TsinOl - Tsin82 = T(sin8, - sine2)

This horizontal force will cause bending moment in the tower.

If the cable is supported on a saddle mounted on rollers, as shown in Figure 14.8 (b) the saddle remains in equlibirium, horizontal component of suspension cable and the anchor cable will be equal.

TI sill = T2 sin O2 = H

The vertical pressure'on the pier is given by

Vp = TI cos O1 + T2cos02

Example 14.5

A suspension cable 180 m span and 18 m central dip carries a load of 1 ldrl per linear horizontal meter. Calculate the maximum and minimum tensions in the cable. Find horizontal and vertical forces in each pier under the following alternative conditions

(a) If the cable is firmly clamped to saddles carried on frictionless roller on the top of the piers.

(b) If the cable passes over frictiorlless pulley on me top of the piers. In each case the back stay is inclined at 30' to the horizontal.

Suspension Bridges

The maximum tension (T) in the cable is always at its ends while the minimum tension in the cable is at its lowest point 'B" and is equal to H.

Solution

Horizontal thrust in cable is also the minimum tension occuring at lowerst point of cable;

The vertical reaction,

which is @= (a) When cable is clamped to saddle the horizontal components of tension

balances. Let T, = tension in the back stay or anchore cable

T --- s - 225 - 260 kN

cos 30'

Total compression on pier,

= Tv sin 30' + TA sin 0

260 = T, sin 30' + R = - + 90 = 220 kN (Since, TA sin 0 = R)

2

(b) When the cable passes over the frictionless pulley, the tension in the back stay T, is equal to the tension in the cable TA.

Let the inclination of the cable be 0 with horizontal

Loadonpier= TAsin0+Tsin300= ~ + T s i n 3 0 ' (when TA=Ts=T)

Horizontal shear = T cos 0 - T cos 30' = 225 - 242.33 cos 30' = 15.14 kN SAQ 3

4 - ~ ~ \ p r n \ h j n ~ d l ~ l e of 120 1x1 spnn and 15 111 ce~llral dip carrles a load 01 5 kN ptbr Inlcar mc:re, calculate tlic nllnlmum and rrlax~rnuln teiisri)n Find thc Ilori7ontal and ver llc31 11)tces In p~cr. When

I 1h2 i:ible pasjcs over trictionle~s pulley, and

I h) the cable is clanlpcd to s;idcjles carrlcs 19y rollers. In each back stay is i!~clint-s: ;it 45" with Iir3rrizont:ll.

14.2.4 The Catenary Let ABC in Figure 14.9 (a) represent a flexible cable having a weight of q N per m length of cable. The origin is selected at a distance D below pomt B to simplify the final results. The distance D will be determined later.

I

I (a1

I

( b )

Figure 14.9

A free body diagram of the portion of the cable of length L from B to any point P (x, y) is as shown in Figure 14.9 (b). When the equations of equilibrium are applied to the free body diagram as shown in Figure 14.9 (b), the results are as follows :

Tcos 0 = TB, Tsin 0 = qs (14.13)

when T is eliminated from Eq. (14.13)

(14.14)

and TB. The equation may be written in terms of cartesian co-ordinates by using the following relationship, which applies to any plane curve.

4 ds 2 = [&I[.,)= s ine (%)

since is equal to s h 0 For the curve using Eq. (14.14) and differentiating' ds

and thus,

Integrating,

v = ( p ) s e c O t i m o d = [ $ ) s e c t ) + c ,

At the point x = 0, v = D the angle 0 is zero, therefore,

TB By selecting D = -, the constant of integration C1 becomes zero and

4

The slope of the curve is

4 2 - & = tan0 = n and dx = 1) dx

tiom which

Slpspension Bridges

when x is equal to zero, v is equal to D and cosh -' (1) is equal to zero, therefore C2 is equal to zero, and the preceding equation can be written

Eq. (14.16), the equation of the curve in cartesian co-ordinates, is called a catenary. When D TB .

is substituted for - in Eq. (14.14), the resulting equation can be combined with 4

Eq. (44.15) to obtain the relationship between y and s as follows :

Length of the cable thus s = D [y) = qyv

An equation relating x and s can be obtained by combining Eq. (14.16) and Eq. (14.17) to eliminate y as follows

y = ~ ~ c o s h [ ~ ~ =

from which s' - D2 = L? cosh2

or

and

The sag f in the cable is the difference between the value of y at the support (yA) and D. For I the right-hand af the cable in Figure 14.9 the sag is I

- ;i

1 also

1 If 0 is eliminated from Eq. (14.41, the result is as follows :

MisceHaneous Eq. (14.20) indicates that the tension in the cable at any point is equal to the product of the Topics load per meter of cable and the vertical distance of the point from the x axis.

The use of Eq. (14.15) to Eq. (14.19) requires the determillation of the length I), which must usually be done by a trial and error process or graphical solution. Various tabular and semigraphical procedures have been devised to facilitate the solution of these equations.

Example 14.6

A chain weighing 7 N per m is suspended between two poles with the same elevation, spaced 6 nl apart. The pull of the chain 011 the poles is 36 N. Determine sag and length of the chain, assuming the weight to be uniformly distributed along cable.

Solution

Here,

is the distance from the x axis to the support and T = TA is the tension in the chain at . the support.

T = qy where T = 36 N, q = 7 N h , y = y~

36 = 7 x yA , giving Y A = 5.14 1n

and

In this equation x is equal to 3 m when v is equal to y ~ . *refore,

5.14 D cosh.[5] = 5.14 3 cosh ($1 = '- D

The value of D can be determined graphically by plotting value of 5.1410 against I1 and also by plotting values of cosh (3lD) against D on the same graph. The ihtersection of the curves will determine the value of D. The value of D can also be determined by successive trials as shown in the following table.

The value of D = 4.00

fA = vA - D = 5.14 - 4.00 = 1.14 m

In equation,

cosh (3lD)

1.54 1.51 1.47 1.44 1.42 1.39 1.36 1.35 1.33 1.31 1.29

D

3 .O 3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8 3.9 4.0

The corresponding value of s, which equals SA, is one-half the length of the chain. The value of SA is as follows :

5.1410

1.7 1 1.66 1.61 1.56 1.5 1 1.47 1.43 1.39 1.35 1.32 1.29 .

Therefore, he \ength of thc chain in 6.58 m.

SAQ 4 (a) A uniform chain 37 nl long and weight 7 N/NI has a span 30 1x1. The supports

are at same elevation. Determiire Ule sag and the maxlmum ar~d minimunl tension in the chain.

(b) The towers lhat support a steel cahle with wr;~pped telephone llncs arc 750 111 apart. The combination of cable aid wires has a mass 0.75 kgtin. The maximum allowable tension in the cable is 4.5 kN. Calculate S, the length of the wire and~f, sag necessary to produce this tension.

14.3 ANALYSIS OF LIGHT STIFFENED SUSPENSION BRIDGES

14.3.1 Introduction Since the cable of the suspension bridge is the main load bearing member, the curvature of the cable of an unstiffened bridge changes as the load moves on the decking. To avoid this, the decking is stiffened by provision of either a three-hinged stiffelling girder or a two-hinged stiffening girder. It is assumed that the stiffening girders transfer a uniform or equal load to each suspender, irrespective of the position of the load on the decking.

14.3.2 Analysis of Cable Stiffened with Three-Hinged Girder A three-hinged stiffening girder supported at ends and connected to the cable (Figure 14.10) is subjected to any external load system Wl, and W2. Such a system is statically determinate. Let q per unit run be the uniformly distributed load transferred to the cable by the suspenders. Obviously, the girder will be subjected to an upward uniforr~lly distributed load q per unit run.

Figure 14.10

Let us consider the equilibrium of the cable and the stiffening girder separately. The cable is subjected to a uniformly distributed load q per unit run transmitted by suspenders as shown in Figure 14.1 1 . The equation of the cable with the center B as origin is as follows :

r )

v~rtical reaction at each support V = & 2 '

Horizontal reaction at each support,

, H = = @? , where f = dip of cable 2f Sf

Maximum tension, T =

Suspemion Bridges

Miscenanmus Topics

Figure 14.11

Now, consider the girder which is subjected to two load systems, l~arnely (a) the applied external load system Wl, and W2, (b) upward uniformly distributed load of q per unit run from the suspenders (Figure 14.12). Consider my section x of the girder at distance x from the point C. The bending moment at any section is due to the above two load systems. Let M be the actual bending moment at the section.

q L / 2 q L I Z I q per unit run I

Figure I* 12

@ li- ,, x 1 2 ] M = M x - [ 2 [ 2 x j - 4 [ (

where M, = bending moment at the section due to girder load considering the span as that of a simply supported girder. This bending.moment is called the beam moment at the section.

Now, consider the cable section at the corresponding point of the cable, we have

Applying this condition to the hinge B of the girder where the bending moment is zero we get

MB = 0 + Mx - H(f-y), = MB - H ( f - 0 ) = O

Giving, since at B , Mx = Mg, a d y 'F 0

Now it is easy to determine the u. d. 1. transmitted to the cable from the relation

After finding q, the vertical reaction Vat each support for the cable be determined from the relation V= qU2 and shear force (V) at any section of the girder is given by

where V, is the shear force at the section due to the applied external load system considering the span as that of a simply supported girder. This shear force is called the "beam shear" at the section. Now for the cable with the central point as origin the equation to the cable is given by

On differentiating,

or H tan 8 = qx

From the Eqs. (14.25) and (14.26), we get

Example 14.7

A suspension bridge of 200 m span has two three-hinged stiffening girders supported by two cables having a central dip of 25 m. The width of road way is 8 m. The roadway carries a dead load of 1/2 kN per sq. meter extending over the whole span and a live load of 1 kN per sq. meter extending over the left hand half of the bridge. a

Find the B.M. and S.F. at point 60 m and 150 m from left hinge. Also calculate maximum tension in the cable.

M DIP

'Z K N / &

$50 M

100 M - - IOOM IC

2 0 0 M SPAN a D

figure 14.13

Solution

There will be no B.M. and S.F. any where in the girder due to uniformly distributed dead load. As there are two cables half the live load will be transfered.

Live load per meter on each girder

Suspension Bridges


and VD = (4 x 100) - 100 = 300m

Considering equilibrium of portion DE, bending moment at hinge B is zero.

ME = (-100 x 100) + 25H = 0

H = 400kN

The equation of parabola is

M60 = Bending moment at section x due to extenla1 load - H a y

= + 2400 kN m (Sagging)

Crmider equilibrium of right portion of section at x = 150 m from left

Mlso = (+ 100 x 50) - (400 x 18.75)

= - 2500 kN m (Hogging)

S66 = (Beam shear)6o + H tan 6

= (- 700'+ (60 x 4)) + (400 x 0.2)

= 20 kN

SlsO = 100 + (400) (- 0.25) = 0

Again the equivalerit u.d.1. transferred to the cable due to live load per meter.

Equivalent u.d.1. transferred to the cable due to dead load per meter

Total q.= 2 + 2 = 4 kN/rn.

Maximum tension in the cable is given by using Eq. (14.8)

SAQ 5 A suspension bridge ciible of span 80 m and central d ~ p 8 m 1s suspended irom thc same level at two towers as shown in Figure 14.14. B e br~dge cable 1s stiffened hy a three hinged stiffening g~rder which carries a single concentrated load of 20 kN at a point 20 m from one end. Sketch the S.F. diagram for girder.

Figure > 4.14

14.3.3 Influence Line Diagrams (ILDs) for Three-Hinged Girder Suspension Bridge .

(a) Influence Line for Horizontal Reaction [Figure 14.15 (b)]

Let a unit rolling load be at a distance x from B (between B and A). The simply-supported reactions at the ends of the girder are as follows :

* H M DIP

MB L - 2 x Hence, horizontal reaction, H = - = - f 4f

L Thus, as the unit load moves from B to A the horizontal reaction H changes from - 4f

to Zero. Similar linear change occurs for H as load rolls from B to C by symmetry. This is shown in Figure 14.15 (b).

(b) Influence Line for Load Intensity (q) in Suspenders, Transmitted to she Cable [Figure 14.1 5 (c)]

L-2x For unit load placed x from B , we h o w that horizontal reaction H = - 4f

But we know that H = & S f

Suspendon Bridges

Equating the two

I 1, gives,

Thus, as Ule unit load inoves from B to A, the Ioad intensity, q in the suspenders 2

changes from - to zero as shown in Figure 14.15 (c). By symmetry, a similar straight L

line gives the influence line between B to C.

- I d . )

+ve B.M.

-ve B.M.

Figure 14.15

(c) Influence Line for Bending Montent at a Sectiun [Figure 14.15 (d)]

Let the section D at which the bending moment is sought be at a distance a from B; also let the corresponding point on the cable be at a height h above the lowest point B. Since the cable is parabolic,

Now, the bending moment at D,

MD = (Free girder bending moment) - (moment due horizontal reaction H)

= p D - ( f - h ) H

Hence, infuence line for MD

= (mfluence line ordinate of free B M) - (influence line ordinate of H) x (f- 0)

We h o w from our study of Block 1 that ILD for free BM at D is given by a triangle whose apex is below D, the altitude of the triangle being

This is shown as D3CA in Figure 14.15 (d)

Also, f - h = f--- 4 f a2 - f ( ~ ~ - 4 a2) is a oIIStant. r 2 12

Hence, the second term will be influence ordinates of H multiplied by this constant.

L But as discussed in section (a) above, the ILD for H is a triangle of altitude -

4f ' Hence, ILD for H (f - 6 ) will be a triangle with altitude,

Thus, we see that the ILD for the second terms has also the same height as the ILD of the first term but apex placed at point B. This is shown as B3CA in Figure 14.15 (d) where the two diagrams are placed on the same base AC. The influence line ordinates for MD is the difference of the two ordinates of AB3CA and AD3CA shown hatched in Figure 14.15 (d).

It may be observed that the influence line ordinates for MI) is negative (hogging) between the points C and E, and is positive (sagging) between the points E and A. Also it can be easily proved that the area$ of positive and negative portions are equal. Hence, it is concluded that when the girder carries a uniformly distributed load on the whole span, the bending moment at every section is zero.

(d) Maximum B.M. Diagram due to Unit Rolling Point Load

Suspension Bridges

From the above discussion the maximum positive (sagging) bending moment occurs at a section when the unit rolling load is at the section; while the maximum negative (hogging) bending moment occurs when the unit rolling load is at the centre of the span. Expressions for these are given below which you must try to verify :

C)n simplifying, we get

To find the greatest value of &,, let us differentiate it with respect to 'a'

Then -- d&m - L 2 - 12a2 = 0 da

It gives,

Substituting this value of a in the expressioil fior M,,, (positive), we get the absolute maximum positive bending moment as 0.096 L which you may verify. The ILD for M,, (positive) is shown at Figure 14.15 (e).

For the maximum negative moment which occurs when the load is placed at B we have,

For this value to be the greatest, we differentiate it with respect to a, aqd get

6 ( L - 4a) - o, Then -- -

da 2L

It gives L a = - 4

Substituting this value of a'in the expression for M,, (negative), we get the L

absolute maximum negative bending moment as - - The ILD for M,, 16'

(negative) will be parabolic as shown in Figure 14.15 ( f ) .

Maximum B.M. Diagram due to Uniformly Distributed Load

For a uniformly distributed load q per unit length the maximum positive moment will occur when it is on the span from E to A. Similarly, maximum negative moment will occur when it covers the span from E to C. And both these values are equal in magnitude as the two areas are equal. Hence, we must first find the position of the point E.

Let AE = (7.

From geometry,

Since B3B = D3D

We have

\ 1 On substituting the value of z, we get


Now, the area of either of thk. shaded triangles will be as follows :

= (Area of ADsCA) - (Area of AE3CA)

For this area to be maximum, we have to differentiate the expression with respect to a and equate to zero.

The expression obtained is 8a" 1 2 a 2 ~ - L~ = 0

The cubic equation can be solved by trial and error to give a = 0.266 L.

The absolute maximum bend~ng moment will have the value (verify by substituting) as + 0.01 883 q ~ 2 for this position,

Influence Line for Shear Force at a given Section of Girder (Figure 14.1 6)

Shear force at the section D,

Vn = (free girder S.F) - (H tan 8);

where tan 8 is the slope of the cable at the point D.

The equation of the parabolic cable is y = & 2H

We have, tan 8 = @ = dx H'

It gives, H tan 8 = qx

Thus, VD = (free girder S.F) - qx

For unit load, (q = I), we get

Vrl = (free girder S.F) -x

Here, following three cases may arise :

L Case (a) a < -

4 [see Figure 14.1 6 (b)]

L Case@) a = -

4 [see Figure 14:16 (c)]

L Case (c) a > -

4 [see Figure 14.1 6 (d)]

The maximum negative shear force at a section occurs when the rolling load just reaches the section (on the right-hand-side of it). The maximum positive shear occurs when the load has just corossed the section.

L Case (a) : At a section for which a i - there is a possibility that the maximun

4 shear force (negative) may occur when the load is at the centre of the span. From Figure 14.16, maximum negative shear force is given by the ordinate

For this maximum negative SF to be the largest, we differentiate it with L

respect to a and equate to zero, getting a = -. 8

On substituting this value of a in above equation, the absolute maximum negative SF is as follows :

Suspension Bddges

L This occurs at r r = - from the centre B of the girder. Similarly, the maximum

8 7

positive SF will occur at the same point having value - [Figure 14.16 (e)]. 16

L Case (b) : At a section for which a = - , both maximum positive and negative

4 1

SF will be - as can be easily seen. 2

L Case (c) : For a section a > q, there is no absolute maximum negative SF as it

goes on decreasing to 0 and maximum positive SF will go on increasing to +l.

Example 14.8

A suspension cable stiffened with a three-hinged girder has 100 m span and 10 m central dip. The girder carries a load of 0.2 kN/m. A live load of 10 kN rolls from left to right. Determine (a) maximum bending moment anywhere in the girder; and (b) maximum tension in the cable.

Solution

(a) Absolute maximum negative BM due to the point load occurs at

x = 0.211 L = 0.211 x 100 = 21.1 m.

Mm,, (negative) = - 0.096 WL = - 0.096 x 10 x 100 = - 96 kN m.

M,,, (positive) = 0.0625 WL = 0.0625 x 10 x 100 = 62.5 IcN m.

at x = - 0.25 L = 25 m from either end.

(b) Maximum cable tension

W L 1 0 ~ 1 0 0 = ~ ~ ~ Hnlax due to concentrated load = - = 4f 4x 10

Hnl, due to uniformly distributed load = !& 8f

Total H = 25+ 25 = 50 kN b

14.3.4 Analysis of Suspended Cable Stiffened with Two-Hinged Girder In this case, the girders supporting the roadway are hinged only at the ends. The structure in this case is statically indeterminate. The solution is possible by strain energy method, which lies beyond the scope of this unit. But by making the assumption that the girder is so rigid that the load on the girder is transmitted to the cable as uniformly distributed load, the analysis can be made simple. For instance consider the point loads Ql and Q2 on the stiffening girder as shown in Figure 14.17.

Uniformly distributed load transmitted to the cable,


In general, whatever may be the loading on the girder,

(Total load on the span of the girder) 4 =

span Horizontal reaction at each end of the cable,

Vertical reaction at each end of cable,

Maximum tension in the cable,

For any point PI on the girder,

Bending moment at PI = (Beam moment) - H Cf - y)

Shear force at P I = (Beam shear) - H tan 0 ; where tan 0 = & dx

as in the case of the three hinged stiffening girder.

The cables have considerablt sag at the centre of the span. In addition to that there are suspenders connecting the cable to the girder. Thus, the cable ends are to be raised sufficiently high above the girder level for which towers have to be provided at either end. The towers may be masonry piers for smaller spans. Whereas for larger spans steel trestles are used. The cables passing over the tower are supported on movable saddles or fixed pulleys (Figure 14.8). The portion of the cable passing over the tower is anchored to the abutments and is known as the back stay. The design of the tower depends upon (a) the horizontal thrust due to the cable which causes overturning moments and horizontal shear, and (b) the vertical component of the cable tension causing downward pressure on the tower.

14.5 SUMMARY

The cable is perfectly flexible. The wire ropes and parallel wire cables are flexible and have little tlexural stiffness.

The dead load of the truss, cable and suspenders is uniform per horizontal length.

At any point on a cable acted upon by vertical loads, the product of the horizontal component of cable tension and the vertical distance from that point

to the cable chord equals the bending moment which would occvr at that Suspension Bridges

section if the loads carried by the cable were acting on an end supported beam of the same span as that of the cable.

The length of the cable

When cable simply supported over guide pulley at the support the vertical pressure on the top of pier

Vp = T cos O1 + T cos 02 = T (cos O1 + cos 02) (see Figure 14.8)

and horizontal force force on the top of pier

H = 2 sin el - T sin 82 = T (sin O1 - sin 82) (see Figure 14.8)

When cable is clamped to saddle on smooth rollers on the top of pipe the vertical pressure

Vl, = T1 cos 0, + T2 cos O2

and horizontal force on the top of pier

H = T1 sin01 = T2 sin O2

14.6 KEY WORDS

Decking : These are the cross girders supporting longitudinal stringers and connected to suspenders at their ends.

Supporting Towers : These are built at either end of the span to provide the necessary sag to the cable. These may consist of masonry piers or steel trestles, the former being used for small spans only.

Back Stay : It is the portion,of the cable between the tower and the ancho;age.

Saddle or Pulley : These are provided at the top of the tower to facilitate the cable to undergo a change of slope.

14.7 ANSWERS TO SAQs

SAQ 1

Total length of cable = 61.68.

SAQ 4

(a) TA = 158.7 N, TB = 91.7 N, and fA = 9.56 m.

(b) Length of cable = 1133 m and fA = 382 ni.

SAQ 5 SF = -5kNatx=O

= -1Oor+lOkNatx=20rn.

= -5kNat x=80m.

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Suspension Cable Bridges