Unit 12 Bridges

Structure 12.1 Introduction Objectives 12.3 Types of Bridges 12.3 Plate Girder Bridges 12.3.1 Deck Type Plate Girder Railway Bridges 12.3.2 Half Through Type Plate Girder Bridges 12.4 Design of Deck Type Plate Girder Bridges 12.4.1 Design Problem 12.4.2 Design of Welded Plate Girder Bridge 12.4.3 Design Problem 12.4.4 Lateral Bracing System for Deck Type Plate Girder Bridges 12.4.5 Design Problem 12.4.6 Crossframes 12.4.7 Design Problem Lattice Type Railway Bridges 12.5.1 Component of Through Type Bridges 12.5.2 Types of Truss Girders 12.5.3 lnfluence Lines for Pratt Truss Members 12.5.4 Influence Lines for Warren Truss Members 12.5.5 Stringer Beams 12.5.6 Cross Girders 12.5.7 Calculation of Forces in the Members of Truss Girder 12.5.8 Problem on Calculation of Forces in Members of Truss Girder 12.5.9 Design of Top Chord Members 12.5.10 Design the Top Chord Members of the Truss Bridge 12.5.11 Bottom Chord Members 12.5.12 Design the Bottom Chord Members of the Truss Bridge 12.5.13 Diagonal and Vertical Members 12.5.14 Design an End Post for the Truss Bridge 12.5.15 Design the Diagonal Members for the Truss Bridge 12.5.16 Design the Vertical Members for the Truss Bridge 12.6 End Bearings . 12.6.1 IS Code Recommendations 12.6.2 Forces Acting on Bearings 12.6.3 Types of Bearings 12.6.4 Plate Bearings 12.6.5 Roller Bearing 12.6.6 Rocker Bearings 12.6.7 Knuckle Bearings 12.6.8 Spherical Bearing 12.6.9 Railway Board Roller Bearing 12.6.10 Rocker & Roller Bearing 12.6.1 1 Design Considerations for Elastomeric Bearings 12.6.12 Design of Rocker Bearing 12.6.13 Design a Rocker Bearing 12.6.14 Design of Rocker and Roller Bearing 12.6.15 Design a Rocker Roller Bearing 12.7 Summary 12.8 Answers to SAQs

2 INTRODUCTION

Bridges are mainly constructed over a gap in highways or railways. The railway bridges are constructed with steel. The bridge may plate girder type or lattice girder type. These may be deck type or through type. The details about their components. structural action and design procedures are discussed in this unit. The bridges are placed over bearings. The bearings may be rocker type, roller type or elastomeric bearing. Steel Structures Objectives

After studying this unit, you should be able to a know the different type of bridges, a design the deck type railway bridges using riveted, welded or welded with flange angles, a design the lattice girder type bridge, and a design the suitable bearings for railway bridges.

12.2 TYPES OF BRIDGES

Bridges are the structures which are used to provide means of communication over a gap. i) According to provided support to pedestrian loads of vehicular traffic loads bridges are classified as 1) Highway bridges: To carry highway traffic loads 2) Railway bridges: To carry railway traffic loads 3) Combine highway and railway bridges: To carry both highway and railway loads

4) Foot bridges: These are constructed to serve the purpose of providing means to pedestrians only. 5) Aqueduct bridges: These are used to carry canals and pipelines. ii) According to the location of floor, the bridges are classified as 1) Deck type bridge: These are two types: a) Deck type plate girder bridge: In this the floor is placed on the top flanges. b) Deck type truss girder bridge: In this type the floor is placed on the top chords

c) In these deck type bridges over the top of the traffic there will not be any bracing.

2) Through type bridge: In this type the floor is supported on the bottom of the maill load carrying members.

a) Through type plate girder bridge: The floor is supported on buttom flanges.

b) Through type truss girder bridge: The floor is supported on the bottom chords. In these through type bridges over the top of the traffic bracing is also done.

3) Half through type bridge: In between the top and the bottom of the main load carrying members the floor is laid. In this type of bridge load carrying members project above the floor level and over bracing is not provided. These are also double deck type bridges in which decks are provided at two different levels. In these double deck type bridges both the decks may be through type or one can be open deck type and the other can be through type. iii) According to the make up of main load carrying element steel bridges are Bridges classified under 1) I-beam bridges: Rolled steel wide flange I-beams are used as load carrying elements. In Railway bridges up to 15 m span - I-beams are used. In Highway bridges up to 25 m span - I-beams are used. 2) Plate girder bridges: Load carrying members are the plate girders. For span up to 30 m plate girder bridges may be used. But depth is limited up to 4 m. 3) Truss girder bridges: For more depths truss girder bridges are used as load carrying members from span 30 m to 250 m.

4) Suspension bridges or cable bridges: In this type load carrying members are the high tensile steel wire cables. These are used for span of 250 m for highway bridges. iv) According to the structural layout the main load carrying members are classified as

1) Simply supported span bridges: The total width of the bridge is divided into a number of individual spans and each span the load carrying member is simply supported at both the ends. Suitability: Where uneven settlement of foundation may take place. 2) Continuous span bridges: In this type load carrying member of the bridge is continuous over more than two supports. Suitability: When uneven settlement of foundation does not occur. Advantages: Saving of material running is as high as 10 to 20 per cent. They can take higher loads.

3) Cantilever bridges: It is defined as a bridge in which one or more of its trusses are extended beyond their supports forming the cantilever arms. 4) Arch bridges: To upgrade the beauty of its surroundings a) Solid ribbed steel arch bridges-used for highway Span: 150 m to 200 m bridges b) Braced rib a~ches-used for highway bridges Span: 500 m to 600 m c) Spandrel braced arches: used for Railway bridges Spans: up to b50 m d) Tied Arch: In this horizontal ties resist the horizontal thrust of the arch 5) Rigid frame bridges: Here rigid frames are load carrying members Spans: 10 m to 25 m Suitability: For rigid foundations v) According to type of fasteners used 1) Riveted bridges: They are rigid and have more secondary stresses Steel Structures 2) Welded bridges 3) Pinconnected and bolted bridges vi) According to level of crossing railway track and highway 1) Overbridges: When the highway bridge is carried over the railway track by means of a bridge. 2) Underbridges: When the highway bridge is carried under the railway track by means of a bridge. vii) According to the nature of movement of the bridge girders 1) Fixed (permanent bridges): They remain in one portion without any movement 2) Movable bridges: They can be opened in horizontal or in vertical directions to allow over or channel traffic to pass. Loads The loads, which should be taken into account in the design of railway bridge are as follows: a) Dead loads: It includes weight of rails, sleepers floor system and the supporting structure and the dead weight of the structure. b) Live loads: Live loads are due to the train loadings like main lines axle loads, engineer, pedestrian traffic uniformly distributed over the foot way.

C) Impact load: The dynamic effect of the moving load is taken care of by increasing the live load by a certain factor called impact factor. This impact factor should depend on many variables like the type of loading, speed, type of the structure, material of the structure loaded length. d) Centrifugal load: (Loads due to curvature of track) When a curve is there where railway bridge is situated, the structure will be affected by the centrifugalaction of moving vehicles Centrifugal force can be found by formula

where, C = centrifugal force in tonnestmetre of span W = equivalent distributed live load in tonnestmetre run V = maximum speed in kmph R = radius of curvature in metres e) Wind load: Wind loads are the lateral loads which are caused due to the interference in the flow of wind by the moving load on the structure and the bridge structure itself. The intensity of wind pressure depends on the wind velocity which in turn depends upon the height of the structure above the mean retarding surface.

f) Racking forces: Lateral forces are applied by the train due to its small lateral movement while moving on straight track.

g) Longitudinal forces: The longitudinal loads are caused due to i) the tractive effort of the driving wheels of the locomotives. ii) the braking effect resulting from the application of the brakes to all braked wheels. iii) the resistance offered by bearings to the movement at the roller end. Bridges The live loads and longitudinal loads on railway bridges are given in the Tables 12.1 to 12.5. Live loads on foot bridges and foot paths attached to railway bridges 1) The live load due to pedestrian traffic shall be treated as uniformly distributed over the foot way. For the design of foot bridges or foot paths on railway bridges, the live load including impact shall be taken as 490 kg/m2 of the foot path area. 2) Live load on foot path for the purpose of designing the main girders shall be taken as follows. i) For effective span of 7.5 m or less = 415 kg/m2 ii) For effective span over 7.5 m but not exceeding 30 m. An intensity of load reducing uniformly from 415 kg/m2 for a span of 7.5 m to 295 kg/m2 for a span of 30 m. iii) For effective span of over 30 m, live load is taken according to the formula:

where, P = live load in kg/m2 L = effective span of bridge in m. W = width of foot way in m. 3) Kerbs 600 mm or more in width shall be designed for the load of 490 kg/m2 in addition to the lateral loading of 750 kg/m run of the kerb, applied horizontally at the top. If the width of kerb is less than 600 mm, no live load shall be applied. SAQ 1 1) What is a bridge? 2) What are the types of bridges based on material of construction? 3) What are deck type bridges? 4) Differentiate between deck type and through type bridges. 5) What are the uses of aqueduct bridges? 6) What is the difference between plate girder bridges and truss-girder bridges?

7) Enumerate the types of arch bridges? 8) Differentiate between dead loads and live loads. 9) What is impact load? How can you consider it in the design of railway bridges? 10) Enumerate the different loads acting on the railway bridges. 11) What are the factors contributing to the longitudinal forces? 12) What is the live load acting on foot paths? 13) What is the,live load acting on Kerb? Steel Structures 12.3 PLATE GIRDER BRIDGES

These are solid web girders bridges. When the span of the bridges is less, simple I-girders may be used. fir large spans and heavy loads, plate girders built up of plates are used. These are very popular in railways. These can be easily transported. These are effectively used for spans up to 30 m for railway bridges. For more details about plate girders, you may refer to Unit 10. These are two types: 1) Deck type plate girder bridges. 2) Half through type plate girder bridges In railways generally deck type bridge is generally preferred. The half through type plate girder bridge is preferred, where a sufficient head room cannot be provided or the additional cost to raise the embankment is higher. 12.3.1 Deck Type Plate Girder Railway Bridges The components are: I) Rails, 2) Sleepers, 3) Plate girders, 4) Horizontal bracing, 5) Cross-frames, and 6) Bearings.

(a) Elevation

(b) Top Plan

CROSS FRAME

(c) 1'1:1n 81 Ilori/onl.~l id) Cross Frames .Sru\\ 15raci11g Figure 12.1 In this type bridge, the plate girders are the main load carrying members. These are placed at a spacing of 2 to 3 m clc, to resist the wind and other transverse 1 loads. The minimum spacing is about -x span, or depth of girder. Sleepers are 20 placed directly over the top surface of the plate girder. These reduce the impact, though spring action. The reduction depends on the length of the sleepers. Bracing is provided to resist lateral and longitudinal loads. Main horizontal bracing (a horizontal) is provided in between the top flanges of the girder and at the level of bottom flanges also. These bracing are designed to resist a transverse shear equal to 2.5% of the total compressive force by the flanges. Cross-frames are provided in parallel-vertical girders, to transfer the load at top flanges to the bottom flanges. 12.3.2 HaIf Through Type Plate Girder Bridges The load is carried at the lower flanges with the provision of cross beams. These consist of: 1) Rails 2) Sleepers 3) Stringers 4) Cross beams 5) Plate girders 6) Bearings 7) Laterals 8) Triangular gusset plates for transverse bracing.

(a) Elevation

(b) Top Plan

(c) Plan at Cross-girder (d) Cross View(Ha1n Level Figure 12.2 Steel Structures This type of bridge consists of two plate girders. The plate girders receive the load through the cross-beams. The beams span right angles to the main plate girders. The beams are provided at some suitable spacing and are connected to the girders with framed connection. The stringers are provided parallel to the main girders and are supported and framed into cross-beams. These carry sleepers. Horizontal bracing system is provided to transmit the lateral forces to the bearings. This is provided at the level of lower flange. The triangular gusset plates are provided at the top flange. These are connected with inner sides of cross-beams.

12.4 DESIGN OF DECK TYPE REVETED PLATE GIRDER BRIDGES

The Given Data Span of the bridge, spacing of girders. Step I: Computation of loads BM and SF Assume the self weight of girder, main rails, guard rails, sleepers, fastenings as 25 kN/m. Take EUDL for BM and EUDL for SF from' steel bridge code tables. Calculate the value of CDA. w L~ Calculate the max. BM, M = - 8 WL Calculate the max. SF, V = - 2 Step 2: Design of web plate Calculate the economical depth of web plate,

dz4.5

Here M = maximum bending moment, N-mm %, = allowable bending stream in compression, ~/mm'

= 0.66 x 250 = 165 ~/mm' Thickness of web plate Allowable shear stress = 0.4 f, = 0.4~250

zva= 100 ~/rnm~ v thickness, t = - rva d where, V = shear force (N) d = depth of web plate (mm) .r, = allowable shear stress (~/rnm~)

The minimum thickness should be 6 mm. Generally, a thickness of 10 mm, 12 mm or 16 mm is preferred for bridges. Step 3: Design of flange elements M Net flange area required =- abc . de where, M = maximum moment (N-mm)

obc= 165 ~/rnrn~

de = effective depth of plate plate girder (mm) = C to c distance of flange elements (may be taken as equal to 6) Gross area of flange required = 1.25 x Net area of flange required Gross area of flange angles required 1 = - x Gross area of flange required. Try for 2 angles of equal sides. 3 Gross area of flange plate 2 = - x Gross area of flange required. 3 Minimum number of two plates is required. Step 4: Check for flexural stresses Calculate the Gross area of flange Calculate the deductions for rivet holes Calculate the Net area of flange :. Net area of flange = (Gross area of flange - deductions for rivet holes) Calculate the Gross Moment of Inertia of section abbut N.A. Calculate the actual bending stress in M Compression, obc,cal = -x ymax I where, M = maximum bending moment (N-mm) I = moment of inertia (mm4) y,,= y,,= distance of the extreme Compression fibre from N.A (mm) Calculate the actual bending stress in tension. - Gross area of flange Obt9b'." - ObC, ~al Net area of flange t Step 5: Rivet connections I I a) Flange angle to web Calculate the strength of rivets in

2 double shear = 2 x - d i) 4 ii) bearing = d t opf where, d = gross diameter (mm) t = min thick of element to be connected (mm) = allowable shear stress in fastenings (100 ~/mm~) opf= allowable bearing stress in fastenings (300 ~/mm~) Calculate the Rivet value (R), which is the least value of the above two. Calculate the pitch of rivets, p Gross area of flange - - Y (Gross area of flange - web component area1 where, V = max. SF (N) d = depth (mm) Max. allowable pitch is 12t or 200 mm, whichever is less b) Flange angles to flange plate Calculate the strength of rivets in

2 i) Single shear = - d 4 ii) Bearing = dt opf Calculate the rivet value (R) Calculate the pitch of rivets, p (Gross area of flange) --- Rd x V (Gross area of flange plates) Maximum allowable pitch is 12t or 200 mm, whichever is less. Step 6: Curtailment of flange plates Calculate the theoretical length of the plate to be curtailed.

,/ ,/ ~reaof the plates be curtailed x=lx Gross area of flange Calculate the bending stress in tension in the plate to be curtailed

where, MI = BM at TPC (N-mm)

I, = MI of the section at TPC) (mm4)

y, ,,, = distance of centroidal axis of the curtailed plate from N.A. (mm). Calciilate the force in the plate. F = (bending stress x area of plate). N Calcylate the strength of rivets in single shear and bearing. Calculate the rivet value (R) F Calculate the number of rivets, n = -.R Calculate the minimum pitch, p = 2.5 x nominal dia. providing rivets in two rows. n Calculate the extra length of plate required for connecting rivets = - 2 'P Calculate the actual length of the plate to be curtailed = (theoretical length + n. PI. Step 7: Design of bearing stiffeners These are provided at the supports and under concentrated loads. Calculate the allowable bearing stress, o,,

Calculate the bearing area required - Reaction or concentrated load (mm2)

Try 4 equal angle sections such that the area provided is more than the bearing area required. Calculate the length of the bearing stiffener = [ d - 2 x thickness of web plates ] (mm) Calculate the effective length of the bearing stiffener = 0.7 x Actual length of the bearing stiffener. In calculating, Gross area and moment of inertia, the length 20 times the thickness of web is effective in bearing. Calculate the Gross area of the stiffener Calculate the Gross M.I. about the centre line of web. Calculate minimum radius of gyration Moment of Inertia = area Effective length Calculate the slenderness ratio = Radius of gyrating Calculate the value of Allowable Compressive stream (o,,)from Table 5.1 of IS: 800 - 1984. Calculate the load carrying capacity = a, x Gross area.

It should be greater then the reaction or the concentrated load. Design of riveted connection Calculate the strength of rivets in double shear and bearing. Calculate the Rivet value (R) Calculate the number of rivets required, Reaction or Concentrated load n = R Provide the rivets in two rows. Step 8: Necessity of intermediate stiffener Calculate d, d,, d7 Steel Structures

dl dl 6 dl i) Calculate - G , 85' -1344 and 816 If the thickness of web is less than any one of the above values, provide vertical stiffeners. If the thickness of web is more, no intermediate stiffeners are required. clear panel (smallest) -d2 d2 * ii) Calculate 9 200 and 180 -3200 If the thickness of web is less than any one of the above values a horizontal stiffener at a distance of 2/5. The distance of compression flange to NA. Smallest clean panel d2 d2 iii) Calculate - and2 @- 180 ' 250 4000 If the thickness of web is less than any one of the above values, a horizontal stiffener is provided at N.A. Step 9: ~esi~nof vertical stifSener Panel dimension (C) Minimum = 180 t or 0.33d Maximum = 270 t or 1.50d. d Using Table 6.6 (IS: 800), for ;and T,,,,,~, find the value of C

:. Choose the values of C. Calculate the moment of inertia required

where, C = panel dimension t = minimum thickness of web required, can be computed as follows: c d Calculate - - d't From Table 6.6 (IS 800) calculate z,, v Calculate t = - dzva

Generally 2 angle seCtions are used. Calculate the MoI about the centre line of the web. It should be greater than the MoI required Design of rivets Calculate the rivet value (R)

Calculate the shear force, S = -25t2 N/mm h t = web thickness h = outstand R Calculate the pitch = - S Maximum allowed pitch of 32 t or 300 mm, whichever is less. 2 Step 10: Design of horizontal stiffener at - th of the distance to NA from comp. 5 flange Calculate the value of Imq = 4 ct3. Generally one pair 2 of equal angle section is preferred. Calculate the MoI of the stiffener about the centre line of the web. Connections Calculate the strength of rivets in double shearing and bearing Calculate the rivet value (R)

Calculate the shear flow (S) = -125 N/mm h ' R Calculate the pitch = - S

Maximum allowable pitch is 32 t or 300 mm whichever in less. Step 11: Design of horizontal stiffener at NA Calculate the value of I req = 4 t3 Generally one pair of equal angle section is preferred. Calculate the MoI of the stiffener about the centre line of the web. R Connection: p = - < 32 t S c 200 mm. 12.4.1 Design Problem Design a riveted plate girder bridges for BG Main line: Span of bridge = 24 m. Step 1: Computation of loads, BM and SF Assume self wt = 25 kN/m, Total live load for BM, EUDL = 2034 kN. Total live load for SF, EUDl = 2034 kN. CDA = 0.417 Dead load = 25 x 24 = 600 kN, Total load for BM = 600 + 1.417 x 2034 = 3482.2 kN Total load for SF = 600 + 1.417 x 223 1 = 3761.3 kN. I Steel Structures Total load per track, for BM = 3482'2 - 1741.1 kN. -2 3761.3 Total load per track, for SF =yj-- = 1880.65 kN.

W1- 1741'1 Max. BM, M = - - 24 = 5223.3 kN-m. 8 8

Max. SF, V=- = 1880.65 = 940.32 kN. 2 2 Step 2: Design of web plate M = 5223.3 kN-m

o,, = 165 ~/mm~

Economical depth,

Adopt d = 1600 mm.

v Thickness of web plate, t = - d. %a 940.32~lo3 - = 5.9 mm. 1600x 100 Adopt 12 mm thick web plate of 1600 mm depth Step 3: Design of flange M Net flange area required =- 4c.d

Gross area of flange reqd. = 1.25 x 19785.2 = 24731.5 mmL

Gross area of flange angles reqd. = 24731'5 = 8243.8 rnm2 3

Gross area of one flange angle = -8243m8- 4121.9 mm2 2 Let us adopt 21SA 200 200, 15 mm.

(A = 5780 mrn2, I,, = 2197.7 x lo4mm4

C,, = 54.9 mm)

2 Gross area of flange plate = 7 x 2473 1.5 = 16487.6 mm2

Gross area of one flange plate = 16487m6= 8243.8 rnm2 2 I Adopt 2 plates of 500 x 20 mm. as flange plates Bridges' I Step 4: Check for flexure stresses Using 22 mm diameter power driven shop rivets.

Gross area of flange = 2 x 5780 + 2 x (500 x 20) + 200 x 12 = 33960 mm2 Deductions for rivet holes = 2 x (23.5) x (40 + 15) + 23.5 (2 x 15 + 12) = 3572 mm2 Net area of flange = 33960 - 3572

l2 16003 Moment of Inertia I = + 4 2197.7~ 10' x 5780 (800 - 54.912] 12 [

- 1600 + 2 x 40 - 1680 - 840 mm, Ymax - 2 2 M obc, ca~= -j- Ymax

Gross area of flange obr, COI= obc, cal Net area of flange

< 165 ~/mm~(Safe) Figure 123

Stap 5: Curtailment of flange plates Theoretical length of curtailed plate, d~reaof plate to be curtailed x=2 Gross area of flange

1741.1. Reaction = - 2

Figure 12.4 Bending stress = -Ml x y 11

Force in ,the plate = Bending stress x area

Strength of rivets in n Single shear = - (23.5)' x 100 = 43.37 kN a) 4 1000 300 Bearing = 23.5 x 20 x -= 141 kN. b) 1000 :. Rivet value, R = 43.37 kN. 1016.4 No. of rivets required = -- 23.4 2: 24. 43.37 Adopting a pitch of 60 mm and two rows of rivets 24 Extra length required for rivets = - x 60 = 720 mm = 0.72 m. 2 :. Actual length of the plate to be curtailed = 13.02 + 2 x 0.72 = 14.46 m. Step 6: Riveted connections a) Flange angles to web Strength of rivets in i) double shear = 2 x 43.37 = 86.74 kN. 300 bearing = 23.5 x 12 x -= 84.6 kN. ii) 1000 . Rivet value, R = 84.6 kN. V = 940.32 kN. Gross area of flange Gross area of flangeweb component 1

= 154.9 mm. Maximum allowable spacing = 12 t or 200 mm. mm. Adopt 22 mm diameter power driven shop rivets at 140 mm c/c b) Flange angles with flange plate Strength of rivets in i) single shear = 43.37 kN.

ii) bearing = 23.5 x 15 x -300 - 105.75 kN. 1000 :. Rivet value, R = 43.37 kN. V = 940.32 kN. ~d Gross area of flange Pitch, p=x V Grossarea of flange plates

= 250.6 mm. Maximum allowable spacing = 12 t or 200 mm = 12x 15 or 200 mm = 180 mm. Adopt 22 mm dia. P.D.S rivets @ 180 mm c/c in a staggered manner. Step 7: Design of bearing stiffener V = 940.32 kN o, = 187.5 ~/mm~

Bearing area required = -= 940'32x lo3 = 5015.04 mm2 OP 187.5 Adopt 4ISA 9090, 10 mm. (A = 1703 mm2, I = 126.7 x lo4 mm4, Cxx = 25.9 mm) Length of the bearing stiffener = 1600 - 2 x 1.5 = 1570 mm. Effective length = 0.7 x 1570 = 1099 mm.

Figure 12.5: Bearing blil'l'e~wr Gross area

A =4x 1703+ 330x 12= 10772mm2 M.o.1. about its centre line Steel Structures Radius of gyration,

r= = 43.2 mm. 10772 Slenderness ratio,

From Table 5.1 of IS: 800-1984,

o,, = 146.5 ~/mm~

P = oa, .A = 146.5 x 10772 N

= 1578 N > V (Safe) Connections Rivet value, R = 70.5 kN. V 940.32 No. of rivets, n = - = -= 13.33 R 70.50 Provide 14 rivets in two rows @ 220 mm c/c Step 8: Necessity of intermediate stiffeners t = 12 mm. (thickness of webs) d, = 1600- 2 x 15 = 1570 mm.

dl 6 1570m I= = 18.47 mm. 1344 1344

dl - 15706 6-= 13.47 mm. 816 816 :. Vertical intermediate stiffeners are required. 4 1570 ii) %==- - 7.85 mm.

:. No horizontal stiffeners are required. Stap 9: Design of vertical stiffeners Panel dimension (c)

Minimum value 0.33 d = 0.33 X 1600 = 528 mm.

Maximum value: 1.5d = 1.50 x 1600 From Table 6.6, C = 1.5d = 2400 mm. Bridges Adopt 1600 mm panels.

t is the greater of

d2 1570 " ~=~-- 3.9 mm.

2 1570a ii) -- = 3.88 mm. 6400 - 6400

V 940.32~lo3 iii) -- = 5.88 mm. d~,,1600x100

:. t = 5.88 mm.

= 48.8 x lo4 mm4 Let us try 21SA 9Q90, 10 mm. Moment of Inertia about the centre line

126.7~l@+ 1703

Figure 12.6 Design of rivets Using 22 mm dia P.D.S rivets. Rivet values R = 70.5 kN.

125 t' Shear flow, S = - h

70'5 Pitch p =-= lo3 = 352. J mm. S 200 Steel Structures Max. allowable pitch, p = 32 t or 300 mm

= 32 x 12 or 300= 384 or 300 = 300 mm. Adopt 22 mm dia. P.D.S. rivets @ 300 mm clc 12.4.2 Design of Welded Plate Girder Bridge For more details refer Unit 10. Step I: Computation of loads, BM and SF Assume the self wt. of girder and track as 25 kN1m Take EUDL for BM and EUDL for SF from tables of steel bridges. Calculate the value of CDA. WL Calcutta the max. BM, M = - 8 W Calculate the max. SF, V = - 2 Step 2: Design of web plate Calculate the economical depth

where, M = max. BM (N-mm)

Calculate r,,, = 0.44 =I00 ~/rnm* v Calculate the thickness, t = - d. ="a Adopt 12 mm, 16 mrnt20 mm thickness. Step 3: Design of flange plate M Flange area required, Af= - 9-d LL Flange width, B = - to - 40 45 A Thickness = B Step 4: CIleck for flexure Calculate Gross area (A), M.o.1 (Ixx and lyy)

Calculate ryy= A

Calculate elastic critical stress Kb) From Table 6.2 of IS:800, find permissible bending stress (obc) Calculate bending stress, It should be less than oh,.

Otherwise, revise the section. Stel-' 5: Welded corlnection between the flange and web VA 7 Calculate shear force per mm length, F = I where, V = ~naxirnumshear force (N) A y-= moment of the area of flange plate about centroidal horizontal axis = (Area of flange plate) x (distance between centroid and c.g. of flange plate) 1 = lxx,.

Adopting continuous weld, on both sides, strength of weld = 2 x 0.7 S x 110 Here shear strength of weld is 110 ~/mm~

From the above equation find the size of the weld. Szey 6: Design of bearing stiffener

a, = 187.5 N/mm 2 V Calculate the bearing area required = - DP Adopt 2 flats such that the outstand should not be greater than 12t. The web length about 20 t in effective in bearing.

Calculate A. I about centre line of web, r = di,!, o, (Tables 5.1 of IS: 800). Calculate the load carrying capacity P = oUcA.

It should be greater than V. Design of connections Adopt size of weld as 6 mm (S) Length of intermittent weld, L = lot. Strength of weld = 0.7 S L x 110 N. = 77 SL N Required strength of weldlmm length --- v (Since the weld is provided at four corners of bearing stiffener) 4d :. Spacing of intermittent weld Steel Structures < 16t < 300 mm. Step 7: Necessity of intermediate stiffeners v Calculate z,, = - (d = d - ,,, dt 1-4)

dl d, dfy dl J~IIU,c',, i) Calculate - and 85 ' 1344 816 If the thickness of web is less than any one of the above values, intermediate vertical stiffeners are required.

d2 d2 dfy ii) Calculate - 200 ' -3200 If the thickness of web is less than any one of the above values, a 2 horizontal stiffener is provided at a distance of - of the distance from 5 the compression flange to NA. d2 d2 iii) Calculate - -a 250 ' 4000 If the thickness of web in less than any one of the above values, a horizontal stiffener is required at N.A. Step 8: Design of vertical stiffener Panel dimension (c) Minimum: 180t or 0.33d Maximum: 270t or 1.50d d Using Table 6.6 (I S: 800), for - and t z ,,, ,.,, , find the value of C. :. Choose the value of C d3 t3 Calculate required 1 = 1.5 -2 Choose a single flat. Find the moment of inertia about centre line of the web. It should be greater than required I. Design of weld 125 t' Calculate shear forcelmm = -N/mm. h Assume the size of weld S as 3 mm Length of intermittent weld, L = 10t Strength of weld = (0.7 S) L x 110 = 77 SL N

< 300 mm. 2 Step 9: Horizontal stiffener at - distance comp. flange L 5 , Refer step 10 of Section 12.4 to NA For Design of weld refer step 8 of this procedure Step 10: Horizontal stifener at NA Refer step 11 of section 12.4 For design of weld refer step 9. 12.4.3 Design Problem Example 12.1 ' Design a welded plate girder bridge for Broad Gauge Main line of span 24 m. Solution Step I: Computation of loads, BM an SF For details refer, step 1 of section 12.4.1 M = 5223.3 kN-m V = 940.32 kN Step 2: Design of web plate

Economical depth, d = 5 Gb

Adopt d = 1600 mm.

T,, = 100 ~/mm~

Thickness of web, t =A d - =,a 940.32 x lo3 - = 5.9 mm. 1600x 100 Adopt 12 mm thick web plate. Step 3: Design of flange plate M Area of flange required = - 0b.d

LL Fla, ;e width = - or - 40 45

Adopt 600, width 19785'22 Thickness of flange = = 32-98 600 Steel Structures Adopt 600 mm x 50 mrn flange plate. Step 4: Check for flexure

Figure 12.7 Gross area, A = 2 x 600 x 50 + 1600 x 12 = 79200 mm2

Radius of gyration, ryy- A

Let us provide the cross beam at 4 mclc.

From Table 6.5, (IS: 800-1984), X = 1713 Y = 1656

C2 Elastic critical stress, Lb = kl (X+ k2 Y) - c 1 c,= c* Y = 1, k, = 1.0 (Table 6.3)

w = 0.5, k2 = 0 (Table 6.4)

f,,= 1.0 (1713+0) x 1 = 1713~/rnm~ From Table 6.2 of IS: 800, Actual bending stress,

< 157 ~/rnrn~(:. Safe) Step 5: Connection between the flange and web

Shear forcelmm length -_VAy I

= 517.2 Nlmm Let the weld is provided are both faces of the web continuously. Strength of weld = 2 x (0.7 S) x 110 = 154 S N/mm :. 154 S = 517.2 :. S = 3.36 mm Adopt 5 mm size weld.

; Step 6: Design of bearing stiffener

Allowable bearing ob= 187.5 ~/mm~ V 940.23~10' Bearing Area required = - = OP 187.5

Figure 12.8: Rearing Stiffener Adopting two plates, 5014.6 Area of one plate required = --- 2

= 2507.3 mm2 Adopt 2 plates of 240 x 20 mm plate A =2~240~20+340x12

= 13680 rnm2 Moment of inertia about centre line.

Radius of gyration, r = 13680

= 120.47 mm. Length, = d = 1600 mm.

Effective length, 1 = 0.7 d =0.7x 1600 = 1120mm 1 1120 Slenderness ratio, h = - = -= 9.3. r 120.47

From Table 5.1 (IS: 800), o,, = 150~/mm~ :. Load carrying capacity = a,, A

= 150 x 13680 N = 2052 kN >V (OK) Design of Connection Let us adopt size of weld, S = 6 mm.

Length of weld = 10 t = 10 x 20 = 200 mm

Strength of weld = 0.7 S L x 110 =77x6x200N = 92.4 kN. Required weld strength per mm. length

92.4 x lo3 Spacing of Weld = = 629 mm. 146.9 Max. allowable spacing = 16t or 300 mm = 16 x 20 or 300 mm = 300 mm. Bridges Adopt 6 mm size weld of length 200 mm @ 300 mm c/c Step 7: Necessity of intermediate stiffeners

160066- y= = 18.82 mm. 1344 1344

Thickness of web is less than the above values, vertical intermediate

stiffeness are required. I

No horizontal stiffener is required. Step 8: Design of intermediate vertical stiffeeneer Panel dimension min = 0.33 d = 0.33 x 1600 = 533 mm. Maximum 1.5d = 1.5 x 1600 = 2400 mm d For z,, = 49 ~/mrn~and - = 133.33, t C = 1.5d. Adopt a panel dimension of 1600 mm

d3 t3 I required = 1.5 -Z- C

4 1600 ii) t=-=-- - 4 mm. 400 400

d2q 1600a iii) t=L== 3.95 mm. 6400 6400 Adopt t = 5.88 mm

1'1 Steel Structures Let us try .a flat section 110 x 10 mm

Moment of Inertia about the face of web = lox 'lo3 =443.7x 104mm4 3

' 'required (OK)

FLANGE

Figure 12.9: Vertical Stiffener

Size of weld, S = 3 mm Length of weld, L= lot= lox 10= 100mm. Max. allowable spacing = 16 t or 300 mm = 16x 10= 160mm. Strength of weld = (0.7 S ) L (110) =77SLN. =77x3x 100=23.lkN.

t2 Shear force per mm length = 125 - N/mm h

23.1 x ld Spacing - = 203 mm 113.64 Adopt 3 mm size weld of 100 mm length @ 160 mm c/c Example 12 Design a welded plate girder bridge for B.G. main line for a span of 24 m. Solution For mole details refer Section 12.4.1 Step I: Computation of loads, BM and SF Refer Section 12.4.1 Step 2: Design of web plate Bridges For more details refer section 12.4.1 Adopt 1600 mm depth x 12 mm thick web plate. Step 3: Design of flange Adopt 21SA 200 200, 15 rnm flange angle. Adopt 2 plates of 500 mm x 200 mm. Step 4: Check for flexural stresses

Figure 12.10

Gross area, A = 2 x 500 x 20 + 2 x 450 x 20 + 1600 x 12

+ 4 x 5780 = 80320 mm2 Moment of inertia ;bout XX-axis, Ixx

Moment of inertia about YY-axis, I,, Steel Structures = 8.94 x lo4 mm4

Radius of gyration, ry =

Let us provide cross frames at 4 m clc L = 4 m = 4000 mm.

D 1680 ---- - 42 T- 40 From Table 6.5 of IS: 800 - 1984, X = 1702

c2 Elastic critical stress, fcb = K1 (X + K2 Y)- Cl Here, Cl = C2

For Y=l,Kl=l.O

:. fcb = 1.0 (1702 + 0) x 1 = 1702N/mm 2

From Table 6.2 of IS: 800, obc= 157N/mm2

Allowable bending stress, obc= 157 N/mm2 M Actual bending stress, obc,cal = - x Ymax I

= 101.1 N/rnrn2 < obc ;. Safe. Step 5: Design of welded connections

a) Fillet weld connecting the flange plates Adopt 6 mm size weld Min. length = 4 x 6 = 24 mm. Adopt 50 rnm length welds. Strength of weld = (0.7 S)(L) (1 10) N =77x6x50N=23.1 kN VAy Shear forcelmm length = Figure 12.11 I

Steel Structures Spacing of weld, lesser of the following

ii) 300mm 69.3 x lo3 iii) = 125.4 mm. F 552.7 :. Adopt 6 mm size weld of 150 mm length at 125 mm c/c Step 6: Curtailment of top plate Theoretical length of the plate to be curtailed,

Area of the plate to be curtailed x=l 4 ~rossarea of flange

Referring section: 12.4.1, M, = 3685.99 kN-m Figure 12.13

MI Bending stress is the plate =- x y 11

= 103.28 ~/rnm' Tensile force in the plate = 103.28 x (500 x 20) = 1032.8 kN Adopting 6 mm size weld. Strength of weld provided in transverse direction

The strength to be resisted by longitudinal filled welded = 1032.8 - 207.9

Strength of weld = 2 x (0.7 x 6) x L x 110

:. 0.924 L = 824.9 L = 892.7 mm. Adopt 900 mm length weld. :. Actual length of plate to be curtailed I2

= 13.02 + 2 x 0.9 = 14.82 m. Figure 12.14 Step 7: Design of bearing stiffener Bridges 1 Ado~t4 ISA 9090. 10 mm. as bearing stiffener

Figure 12.15: Bearing Stiffener I Design of weld Strength of weld required for mm length

Refer Section 12.4.3

Adopt 6 mm size weld of 50 mm length @ 150 ~JII c/c Step 8: Necessity of intermediate stiffeners Vertical intermediate stiffeners lare required. Step 9: Design of vertical stiffener Adopt 2 ISA 9090, 10 mm @ 1600 mm c/c Design of weld Shear flow = 200 N1 mm. Adopting 6 mm size weld of 100 mm length,

Figure 12.16: Vertical Stiffener

Strength of weld = 2 X 23.1 = 46.2 kN. Spacing of weld, lesser of the following, i) 16t=16~15=240mm ii) 300 mm

iii) 2~23.1~l~~=~~~~~ 200 Adopt 6 mm size weld of 100 mm length @ 200 mm c/c. 12.4.4 Lateral Bracing System for Deck Type Plate Girder Bridges

The lateral bracing system is generally provided at the top level of the flanges. It consists of diagonals and cross members. The system may consist of single diagonals, double diagonals, or K-type diagonals. The diagonals are provided in 1 square type panels. The lateral bracing system are shown is Figure 12.17. I Steel Structures J Wind load intensity

AWind load intensity

/Wind load intrnsi ty

(e) Figure 12.17: Lateral Bracing Systems Let p = intensity of wind load per unit length (maximum value of loaded and unloaded cases) The bracing system can be assumed to be a simply supported girder. P . L= Maximum bending moment, M = 7

where, P = PL = Total wind load. Due to this bending moment, equal and opposite axial force will be developed the loaded flange of the girder.

PL F = - where, S = Spacing of girders. 8s F PL Stress, a=-=- A 8SA where, A = area of flange. At the bottom flange, another lateral bracing system is provided to take 114 of the load taken by the top lateral bracing sqstem. 12.4.5 Design Problem Bridges Design the lateral bracing system for a deck type plate girder bridge over a B.G. Main line for a span of 24 m. The spacing of girder is 2 m. Take intensity of wind pressure as 1.5 kNlm2. The depth of girder is 1600 mm. Design of top lateral bracing Step 1: Computation of wind load Wind load on both girder = 1.5 (1 + 0.25) x 1.6 = 3.00 kN/m Wind load on train = 1.5 x 3.5 = 5.25 kN/m Racking force (assume) = 5.75 kN/m Total wind loads p = 14 kN1m Step 2: Forces in the members Let us divide the bracing system of 2 m size No. of panels is 12. The force acting at intermediate node = 14 x 2 = 28 kN The force acting at end node = 14 x 1 = 14 kN

I2 Panels @ 2 m R =I68 kN I68kN

Figure 12.18

14 x 24 Reaction, R = -= 168kN. 2

8 = 45" Compressive force in the end strut = 168 kN

Tensile Force is the end diagonal = sin 45" = 154 6kN= 217.79 kN Step 3: Design of end-strut Let us provide to angles on the same side of angles plate. Force, P = 168 kN.

Assume a,, = 70 N/mm 2

P 168x103 Grosss area required = -= aac 7 0 Provide 2 ISA 9090, 10 mm Area, A = 3406 mm2 rmin= 27.3 mm Effective length, 1 = 2000 mm 1 2000 Slenderness ratio h = -= -- 73.3. rmin 27.3

From Table 5.1 (IS : 800) o,, = 109 ~/mm~ Load carrying capacity = o,, . A = 109 x 3406 N = 37 1.3 kN > 168 kN (Safe) 4: a)Design of rivets for end strut Using 20 mm diameter power driven shop rivets. Gross diameter = 20 + 1.5 = 21.5 mm. Strength of rivets in n Single shear = - (21 .n2x = 36.3 kN i) 4 1000

lOm m thick gusset plate

Figure 12.19: End Strut

ii) Bearing =21.5xlOx-- 300 - 64.5 kN 1000 :. Rivet value, R = 36.3 kN. P 168 No. of rivets required = - = -= 6 R 36.3 Pitch = 2.5 x 20 = 50 mm. Edge distance = 30 mm b) Design of welding for end strut

lOmm gusset plate ISA 90x90~ /

IFigure 12.20: End Strut Let us adopt 3 mm size weld Bridges Strength of end weld = (0.7 x 3) (2 x 90) (1 10) = 41.58 kN Strength to be resisted by longitudinal fillet Welds = 168 - 41.58 = 126.42 kN Strength of longitudinal weld

= 2 (0.7 x 3) (L) (1 10) = 462 L N.

:. 462 L = 126.42 X lo3 - L = 273.6 mm Adopt 300 mm length weld Step 5: Design of diagonal member Tensile force, P = 217.79 kN

a,, = 0.64 = 150 ~/md

217.79 x lo3 Net area required = = 1452 mm2 150

Gross area required = 1.4 x 1452 = 2032 mm2 Adopting 2 ISA. 90mm x 90mm x lOmm Gross area = 3406 mm2 If 20 mm diameter power driven shop rivets are used, deduction for rivet holes =2x21.5x10=430mm2 Net area provided = 3406 - 430 = 2976 mm2 > Net area required (OK) Step 6: a)Design of rivets for diagonal number 217.79 No. of rivets = ---5.99~ 6 36.3 Adopt 3 rivets for each angle at a pitch of 50 mm and end distance of '30 mm. (Ref. Figure 12.19) b) Design of welding jor diagonal

Adopting 3 mm size weld, the strength of end weld = 41.58 kN. Strength of fillets welds to be provided by longitudinal welding = 217.79-41.58 = 176.21 kN Strength of longitudinal weld Steel Structures

:. :. L = 381 mm. Adopt 400 mm. longitudinal weld. 12.4.6 Crossframes These are provided between two girders in the vertical plane. These are provided at all plane points of horizontal bracing. Generally, they consist of two diagonal members, top and bottom member. The diagonal member is subjected to maximum load. The top and bottom member are in lateral bracing system.

Figure 12.21: Cross Frame Design procedure Calculate the force in the diagonal member

o,,= 150 ~/mm~= P sec 0. Calculate the net area required for the member. Calculate the gross area required = 1.4 x Net area required Choose 2 IS equal angles on one side of the gusset plate. Calculate the net effective area provided. It should be greater than the net area required. Connection design

1) Rivet: Calculate the rivet value. Calculate the no. of rivets. 2) Welding: Assume size of weld. Calculate the strength of end weld. Find the length of filled weld required. 12.4.8 Design Problem Example 12.2 Design cr6ssframes for the problem'in sub-section 12.4.1 & 12.4.6 From sub-section 12.4.1, the section is as shown in Figure 12.22. Bridges Solution From sub-section 12.4.6,

Force at the end panel point = 168 kN. Force at the intermediate panel point = 28 kN. Design of end crossframe 16 tan 8='=0.8 2 sec 8 = 1.28 Force in the diagonal

= 168 x 1.28 = 215.14kN o,, = 150 ~/mm~

Net area required = 215'14 lo3 = 1434.3 mm2 150

Gross area required = 1.4 x 1434.3 = 2008 mm2 Let us try 21SA 80 x 80 x 8 mm

Gross area = 2 x 1221 = 2442 mm2 Using 20 mm dia, P.D.S rivets, reduction for rivet holes

= 2 x 21.5 x 8 = 344 mm2

Net afea provided = 2442 - 344 = 2098 mm2 - +-----2m-- > Net area required :. Safe

Design of connections Figure 1223 a) Rivet Rivet value = 36.3 kN. 215 14 :. No of rivets = --- = 5.9 -- 6 36.3 Provide 3 rivets for each angle.

Figure 12.24: Diagonal of Crossfmme Steel Structures b) Welding Adopting 3 mm size weld

Figure 12.25 Strength of weld = (0.7 x 3) (2 L + 160) x 1 10 = 462 (L+ 80) N. Force = 215.14 kN L = 386 mm

Adopt L = 400 mm Design of intermediate crossframe Force in the diagonal = P sec 8 = 28 x I .28 = 35.84 kN.

35.84 x lo3 Net area required = = 238.9 mm2 150 Gross area required = 1.4 x 238.9 = 334.5 1 mm2 Let us try for 2ISA 50 50, 6 mm. Gross area = 2 x 568 = 1136 mm2 Using 20 mm dia. P.D.S. rivets,

Reduction for rivet holes = 2 x 21.5 x 6 = 258 mm2 Net area provided = 1136 - 258

= 878 mm2 Net area> required (OK) Connection 35.84 a) Rivets: No. of rivets = - 1. 36.3 - Use one rivet for each angle. b) Welding: Using 3 mm size weld, Strength of weld = (0.7 x 3) (2 L + 100) x 110 = 462 (L+ 50)N Force = 35.84 kN.

:. L = 27.6 mm

Adopt 5 mm, L = 400 mm Design of intermediate crossframe Bridges

Force in the diagonal = P sec 8 = 28 x 1.28 = 35.84 kN.

35.84 x lo3 2 Net area required = -- = 238.9 mm 150

Gross area required = 1.4 x 238.9 = 334.51 mm2 Let us try for 2ISA 50 50, 6 mm.

Gross area = 2 x 568 = 1136 mm2 Using 20 mm dia. P.D.S. rivets,

Reduction for rivet holes = 2 x 21.5 x 6 = 258 mm2

Net area provided = 1136 - 258

Net area> required (OK) Connection 35.84 Rivets: No. of rivets = -- 1. 36.3 - Use one rivet for each angle. b) Welding: Using 3 mm size weld, Strength of weld = (0.7 x 3j (2 L + 100) x 1 10 =462(L+50)N Force = 35.84 kN.

Adopt 50 mm length weld.O mm length weld. SAQ 2

1) Design a deck type riveted plate girder bridge for BG Main line for a span of 20 m. 2) Design a deck type welded plate girder bridge for BG branch line for a span of 16 m. 3) Design a deck type welded plate girder bridge for BG Main line for a span of 22 m. Use flange angles. 4) Design a lateral bracing system for pr. (I), (2), & (3) 5) Design cross-frames for pr. (I), (2) & (3). 6) Enumerate the types of lateral bracing system. Steel Structures Table 12.1: Equivalent Uniformly Distributed Live Loads (EUDLL) on each Track and Impact Factors for Broad Gauge Bridges

BM (kN) (kN) L Total Load for Total Load for SF Impact Factor (Metres) 20 M.L. I B.L. B.L. B.L. 14+L L Total Load for BM (kN) Total Load for SF (kN) Impact Factor dges (Metres) 20 M.L. B.L. M.L. B.L. 14 + L

44 3910 3085 4218 3316 0.345 46 4080 3222 4384 3448 0.333 48 4240 3345 4549 3576 0.323

50 4380 3470 4713 3702 0.313 55 4775 3765 5 122 4010 0.290 69 5148 4052 5528 4310 0.270

65 5440 4335 5916 4600 0.253 70 5918 4610 6322 4882 0.238 75 6280 4865 6700 5154 0.225

80 6670 5130 7109 5438 0.213 85 7035 5390 7504 5708 0.202 90 7420 5680 7898 5978 0.192

95 7800 5913 8278 6246 0.183 100 8200 6160 8686 6512 0.175 105 8580 6425 9062 6772 0.168

110 8970 6685 9456 7042 0.161 115 9350 6945 9848 7302 0.155 120 9730 7195 10253 7564 0.149

125 10100 7450 10724 7824 0.144 130 10485 7700 11 133 8804 0.139

Note : The intermediate values may be found by linear interpolation. The values of loads have been converted from metric tonnes to kilo-Newtons.

Table 12.2: Equivalent Uniformly Distributed Live Load (EUDLL) in on each Track and Impact Factors for Metre Gauge Bridges

Total Load for BM (kN) Total Load for SF (kN) Impact L Factor (Metres) 20 M.L. B.L. C M.L. B.L. C 14 + L

1.O 264 214 162 264 214 162 1.000 1.5 364 214 162 29 1 236 192 1,000 2.0 264 214 174 350 284 138 1.OOO

2.5 282 229 20 1 386 313 255 1.OOO 3.0 320 257 223 437 354 289 1.OOO 3.5 386 313 260 488 394 317 1.000

4.0 437 354 289 525 426 350 1.000 4.5 477 386 3 10 579 470 383 1.OW 5.0 508 41 1 338 627 508 409 1.000

5.5 550 446 365 666 540 43 1 1.000 6.0 593 480 389 699 566 454 1.000 6.5 623 508 409 73 1 60 1 482 0.976

7.0 657 542 437 770 633 503 0.952 7.5 690 576 459 806 660 523 0.931 8.0 728 605 482 835 684 541 0.909

8.5 76 1 630 499 862 706 560 0.889 9.0 789 653 516 888 7 30 582 0.870 9.5 816 674 537 922 756 602 0.85 1 Steel Structures Total Load for BM (kN) Total Load for SF (kN) Impact L I I Factor 20 (Metres) M.L. B.L. M.L. 1 B.L. I C I

Note : The intermediate values may be found by linear interpolation. The values of loads have been converted from metric tonnes to kilo Newtons. Table 12.3: Equivalent Uniformly Distributed Live Load (EUDLL) on each Track and Impact Bridges Factor for 762 mm Narrow Gauge Bridges

Total Load for BM (kN) Total Load for SF (kN) Impact L Factor (Metres) H Class A Class B Class H Class , A Class B Class 90 Loading Loading Loading Loading Loading Loading 90 + L

120 242 ' 193 122 270 220 160 315 255

190 370 300 220 405 330 240 430 370

225 460 400 270 495 430 295 5 15 460

340 535 485 380 555 510 405 580 530

425 610 550 445 640 570 455 670 590

465 695 615 475 725 630 480 755 650

485 785 670 500 850 720 520 915 765

550 970 8 10 600 1030 855 700 1090 900

770 1140 960 800 1200 1010

840 1270 1060 870 1340 1100

900 1410 1150 940 1470 1190

970 1520 1230 1000 1560 1280

1040 1600 1320 1070 1640 1370

11 10 1690 1410 1140 1730 1450

1180 1770 1500 1210 1820 1540

1240 1860 1590 1310 1940 1670

1380 2010 1760 1450 2090 1840 Steel Structures Total Load for BM (kN) Total Load for SF (kN) Impact L Factor (Metres) H Class A Class B Class H Class A Class B Class 90 Loading Loading Loading Loading Loading Loading 90+L

38 2010 1810 1520 2170 1920 1560 0.703 40 2080 1890 1580 2240 1990 1620 0.692

42 2140 1960 1650 23 10 2070 1690 0.662 44 2210 2020 1720 2380 2140 1750 0.672

46 2280 2090 1780 2450 2210 1800 0.662 48 2340 2150 1850 2510 2280 1860 0.652

50 2400 2220 1910 2580 2350 1920 0.643 55 2560 2390 2060 2740 2510 2070 0.621

60 2720 2540 2190 2910 2680 2220 0.600 65 2830 2690 2320 3080 2840 2360 0.581

70 2880 2840 2420 3240 3000 2500 0.563 75 2910 2990 2490 3400 3150 2630 0.545

Note : The intermediate values may be found by linear interpolation. (The values of loads have been converted from metric-tonnes to kilo-Newtons.

Table 12.4: Longitudinal Loads (Without Deduction for Dispersion) for broad gauge 1676 mm Bridges L Tractive Efforts (kN) Braking Force (kN) (Metres) M.L. B.L. M.L. B.L. 19 405 315 360 280 20 416 324 374 29 1 2 1 424 33 1 384 300 22 43 1 335 393 304 23 435 339 399 311 24 440 344 408 318 25 448 350 416 325 26 455 352 427 33 1 27 458 356 432 336 28 465 36 1 442 343 29 469 366 447 349 30 476 368 457 353 32 476 368 34 476 368 36 476 368 38 476 368 40 476 368 42 476 368 44 476 368 46 476 368 48 476 368 52 476 368 55 476 368 60 476 358 65 476 368 70 476 368 75 476 368 80 476 368 85 476 368 90 476 368 95 476 368 100 476 368 105 476 1 368 110 476 368 115 476 368 120 476 368 I 125 476 1 368 130 476 368

Note: Intermediate values may be found by linear interpolation. The values of loads have been converted from metric-tonnes to kilo-Newton. Table 125: Longitudinal Loads (Without Deduction for Dispersion) for Metre Gauge 1 m

L Tractive Efforts (kN) Braking Force (kN) (Metres) M.L. B.L. C M.L. B.L. C

1.0 91 74 56 5 8 47 35 1.5 85 72 54 57 46 35 2.0 86 70 57 56 46 37

2.5 89 73 64 60 49 43 3.0 99 79 69 67 54 44 3.5 116 94 78 80 65 54 Steel Structures Braking Force (kN) -I. Tractive Efforts (I&) (Metres) M.L. B.L. 1 M.L. B.L. 128 104 73 137 111 79 142 115 83

151 159 165

169 174 --- - Bridges L Tractove Efforts (kN) Braking Force (kN) 1 I (Metres) M.L. I B.L. I C

Note : Intermediate valu-s may be found by linear interpolation. The values of loads have been converted from i~~erric-ronnesto kilo-Newton.

12.5 LATTICE TYPE RAILWAY BRIDGES

When the span of the bridge is more, the plate girder type is more uneconomical. In such cases, truss type bridges are preferred. These are also called as lattice girders or open web girders. These may be deck type, through type or half through type- 12.5.1 Components of Through Type Bridge 1) Main truss girders 2) Floor 3) Lateral bracing system-top and bottom 4) Portal bracing 5) Sway bracing The Figure 12.26 shows a through type plate girder

(a) Top Lateral Bracing System

/ \11 _Topc.hod member

(b) Elevation of Truss Bridge Figure l2.26(c): Bottom Lateral Bracing System Generally, the bridge consists of two trusses arranged at some spacing. The cross-girders or floor beams connect the correspondent lower points. The stringer beams run parallel to the span. The stringer beams distribute the loads to the floor beams. These support sleepers on which the rails can be laid longitudinally. Hence, the loads from the moving trains distribute from rails to sleepers, from sleepers to stringer beams, from stringer beams to floor beams, from floor beams to lower panel points of truss girder. Lateral bracing system is provided at top and bottom to receive lateral loads, wind loads, seismic loads and racking forces. The loads acting on bottom lateral bracing system is transferred to the bearings. The loads acting on the top lateral bracing system is transferred to the end posts. Sway bracing are provided, to keep the rectangular shape, at each panel point. 12.5.2 Types of Truss Girders

(a) Pratt Truss

(b) Warren Truss

---.. (c) Warren Truss With Verticals

(d) N-Type Truss

Steel Structures

Figure 12.28(c): ILD for RR

ILD for LOUl : In panel L&, the SF is taken by LOUl.At joint Lo Ey=O P,, U, . sin 0 + RL = 0

p u=-y=RL sin0 -[ I -$)cosecO The above equation is valid from 8- 9,L, I 5 I At x = a, PL, U,= - - cosec 0 Figure 12.28(d): Joint rLo 6

x = 6a7PL, U1= 0 I ILD for top chord members \&yN Member U, U2 Figure 12.28(e): ILD for L~U~ Pass section 1-1 and consider the section as shown Figure 12.28(f).

Figure 12.28(f) To get ILD for Ul U2, moment is to be taken about the point L2 where Ul L2 and L, L2 are meeting Unit load rolling from Lo to L2 (x + 0 to 2 a) Taking moments about L2

(1 -$).2a-1 (2a-x)+PIII[12xh=O Bridges

Unit load is moving from & to L6 (x + 2a to 6a)

kigure 12.28 (g) Taking moments about &,

Figure 12.28(h): ILD for UJUz Member U2 U3 Pass section 2 - 2 and consider the section as shown in Figure 12.28(i).

- &C;L Figure 12.28(i) To get ILD for member U2 U3, take moments about the Joints L3 where U2 L3 and & L3 are meeting. Unit load moving from to L3 (x -+ 0 to 3a) Taking moments about &-, Steel Structures

Unit load moving from L, ro L6 : (X 4 3a to 6a) Taking moments about 5

Figure 12.28(k): ILD for U2 U3

ILD for bottom chord members Member Lo L,

Coltsider Joint (from Figure? 12.28(d)).

x=6a,PL =O Figure 12.28(1): ILD for LOLl 0 1 Member L1 &

Pass section 1 - 1, and consider the section as shown in Figure 12.28(e), To get ILD for L1L2, moment is to be taken about U, where the member U, and U1 U2are meeting. Unit load moving from Lo to L, (x + 0 to a) Taking moments about joint U1

5a-5 + cat- s x=a,PL .cote lLz-6h 6

Unit load moving from Ll to Lg : (X a to 6 a) + L 0 L, Lb Taking moments about joint U, (from Figure 12.28(g)) Figure 12.28(m): LLD for L~ L~ 1 ..-

Bridges

Member & 5 Pass section 2 - 2 and consider the section as shown in Figure 12.28(i). To get ILD for & L3 , Moment is to be taken about U2 where the members U2L3 and U2 U3 are meeting. Unit load moving from to L1 (x + 2 a) Taking moments about Joint U2 ,

Unit load moving from & ?0-L6 : (x + 2 a to 6 a) Taking moments about joint U2, and from Figure 12.28Cj),

x=6a, PL2%=0 Figure 12.28(n): ILD for L2L~

ILD for diagonal members The diagonal members take the shear in the panel Member U1&

Pass section 1 - 1, and consider the section as shown in Figure 12.28(f). Unit load moving from Lo to L, (x + 0 to a) zy=o Steel Structures

1 x = a, PU ,= 3.cosec 0. I2

Unit load moving from 4 to L6 : (x + 2a to 6a) Cy = 0.

3 [ 1 - 2)- P,, sin 0 = 0

pLIi=(1 -~)CoseCO

2 At x=2a,Pu12 ,=jcosecO x=6a,PU, =O 12 When unit load is moving from L, to L, i.e., in the panel, the force in member U, 4 changes from compression to tension.

Figure 12.28(0): ILD for UILz Member U2 L,

Pass section 2 - 2, and consider the section as shown in Figure 12.28(i) Unit Load moving from to L, (x + to 2 a)

Cy = 0

Unit load moving from 5 to L6 (X + 3 a to 6 a) Cy=O

3 -)-Pu2isin0=o

Pu2L3 = ( 1- 4cosec 0 1 At x = 3 a, Pu ,= 5cosec 0 2 2 Bridges

Figure 12.28(p): ILD for UZL3 ILD for Vertical Members Mernber U1L,

Consider to Joint L1

Zy= 0 P, ,= Load at Joint L, I1 When unit load is at Lo, P, = 0 I I

at L,, P,., = + 1 I I

at L,, P, = 0 I I

L o L, '- L L6 Figure 12.28(r): U2 Lz Mernber U2L2

Pass section 3 - 3 and consider the section as shown in Figure 12.28(s):

Figure 12.28(s)

Unit load moving from Lo to L2 : (x -+ 0 2 a) - - Steel Structures Unit load mov ling from L3 to L6 : (x + 3a to 6a)

Figure 12.28(t)

3 [1-&)+Pu2i=o

pUlL2=-[l-&I 1 Pu2 L2 = - 7 PUL=o 2 2

Tens~on

Lo ampress

Figure 12.28(u): ILD for V2 L2 Member U3L3 Since the unit load is moving on bottom chord, the force in U35 is always zero. The ILD is as shown. i + Li L6

Figure 12.28(v): 1LL) for Uj LJ 12.5.4 Influence Lines for Warren Truss Members

Figure 12.29(a): Warren Truss ILD for top chord member Bridges Member U1 U2

Pass section (1) - (I), consider the section as shown below.

higure IZ.LY(b) To got ILD for member U1 U2 take moments about the Joint L2 where the members L, L, and L, L2 are meeting.

Unit load moving front L1 to 4 : (x + 0 to a)

C.Mk=O* 1-- a-l(a-x)+Pu .h=O i 4 12 a [a-~-u+x)=-Pu 12h

-2x -2xx2 -4x =3 Pulu2=3h=3=

At x = 0, Pul u2= 0

-4 x = a, PO, u2=

Unit load moving from 4 to L4 (X + a to 3a)

1 Figure 12.29(c)

Figure 12.2Y(d): ILD for U,U2 Steel Structures Member U2 U3

Pass section (2) - (2) consider the section as below:

Figure 12.2Y(e) To get ILD for member L12 U3, take moments about the joints L3 where the member U2 4 and L24.

Unit load moving from Lo to L1 : (x -+ 0 to 2a)

Unit load moving from 4 to L4 : (x -+ 2a to 3a)

Figure 12.29(f) ZML3=0

- 4 At ~=2a,P,~~~=3-[l-

3E Figure 12.29(g): ILD for UZU3 ZLD for bottbm chord members Bridges Members L1 L2 Consider the Figure 12.31(b) taking moments about U1 ! Unit load moving from L1 to U1 :

Unit load in between U, and 4 : (Figure 12.29(c)) x dgto3 a (2 1

Since the load is ioving on the bottom chord member, linear variation starts from L, to &.

Figure 12.29(11): ILL) fur L1 Lz

Member &L3 : Unit load moving from L1 to U2

Consider Figure 12.29(e) Taking Moments about U2, Steel Slructures

Unit load moving from U2 to L4 x + -to 3 a r\ 3i 1 Taking moments about U2, (Ref. Figure 12.29(f))

Since the load is moving on the bottom chord members, linear -. .. L2 to 5.

Figure 12.2Y(i): ILD for L2 Lj

Similar to ILD for L, &

Figure 12.2YG): ILD for L3 L4 Similarly, using this procedure you are-also to draw the influence line diagram for Bridges any member of any type of truss. 12.5.5 Stringer Beams These are the longitudinal beams provided parallel to the track. The span of the stringer beam equal to the spacing of cross-beam. The loads acting are 1) Weight of stock rails, guard~rails,fastenings, sleepers, self wt. 2) Live load from rolling stock. 3) Impact load The stringer beams are braced to resist the effect of wind load. The stringer beam is checked for stress due to dead load, live load, wind effect, racking force and longitudinal force. Example 12.4 Design a stringer beam for a through type truss bridge of span 60 m. The cross beams are provided at 6 m clc. The stringer beams are at 2 m clc. Solution Step 1: Calculation of loads Assume weight of stock rails, guard rails, fastenings, sleepers and self wt. as 8 kN/m. of track. From table, EUDL for BM = 763.9 kN EUDL for SF = 918.9 kN To calculate CDA, L = 1.5 x c to c spacing of cross-beams

8 8 CDA =0.15+- =0.15+--(6 + 9) - 0.683. (6 + L) Impact EUDL for B M = 0.683 x 763.9 = 522 kN. Impact EUDL for SF = 0.683 x 918.9 = 627.6 kN. Total load for BM per track = 48 + 763.9 + 522 = 1333.9 kN Total load for SF per track = 48 + 918.9 + 627.6 = 1594.5 kN Total load/stringer beam for BM = 666.95 kN. for SF = 797.25 kN. Stop 2: Calculation of BM and SF

Step 3: Design of section

Permissible bending stress, obc= 165 ~/mrn~. M Z required = - Ohr Steel Structures

= 3031.58 x lo3 mm3. Let us select IS MB 600 with one plate 300 mm x 10 mm one each flange.

figure 1Z.JU Gross area = 15621 + 2 x 300 x 10 = 21621 mm2

I,, = 91813 x lo4 + 2 1 300n103 +loox lOP+fj 1 = 147633 x lo4 mm4

z,, = 147633 lo' = 4762.4 x lo3 rnm4 3 10

Gross area of flange = 300 x 10 + 210x 20.8 = 7368 mm" Using 20 mm dia. P.D.S rivets,

Reduction for rivets = 2 1.5 x (10 + 20.8) = 662.2 mm2 :. Net area of flange = 7368 - 662.2 = 6705.8 mm2 M Stress in top flange = -. Y,,,,, I

Gross area Stress in bottom flange = 105 x Net area

< 165 ~/mm~(Safe) v Shear stress = - dtw Bridges

< 100 ~/mm~ (:. Safe) Step 4: Design oj' conrzection

Strength ol I i\,ctq in

7t a) Single shear = - (21.5) x -loo- 36.3 kN 4 1000

Bearing =21.5 x lox--300 - 64.5 kN b) 1000 . . Rivet value, R = 36.3 kN. VAY Shear stresslmm length = - I

= 246 N/mm. Pitch of rivets, lesser of the following i) 12 t = 12 x 10 = 120 mm ii) 200 mm

iii) 36'3 lo3> 147.6 mm. 246 Provide 20 mm dia. P.D.S rivets at a staggered pitch of 120 mm clc Step 5: Bracing system for strirzger beams Let us sub-divide the panel length 6 m into three sub-panels, 6 Length of sub-panel = - = 2 m 3

Length of the diagonal = 2 fim Calculation of lateral loads a) Wind load on exposed Area of stringer above chord (let the height of stringer above the top of bottom chord is 0.4 m) = 0.4 x 1.5 = 0.6 kN/m Intensity of wind load = 1.5 k~/m~ b) Wind load on moving train = 3.5 x 1.5 = 5.25 kN/m (height of bogie as 3.5 m) c) Racking force Total lateral load

= 12 kN/m (say)

Load at end panel = 12 x 1 = 12 kN.

Load at intermediate panel = 12 x 2 = 24 kN. Steel Structures

Figure 12.31 Force in the diagonal = (36 - 12) fi = 24 fikN= 33.94 kN. (Comp.) Let us assume o,, = 60 ~/mm~

= 565.67 rnm2 Let us try ISA 100 x 100 x 6 mm A = 1167 mm2 rmi, = 19.5 mm. A=-_-=1 26x1000 r min 19.5 From Table 5.1 = 145

(IS: 800), o,, = 48 N/rnrn2

Load carrying capacity = o,, . A = 48 x 1167 N = 56 kN

33.94 No. of 20 mm dia, rivets = -- 1 36.3 - Adopt 2 rivets. 12.5.6 Cross Girders These are the floor beams or girders provided at each panel point. These are provided to join the lower panel points of truss girders. The span of the girder generally equals the spacing of the truss girder. The loads acting on cross girders are self weight, load (reaction) from stringer beams. To calculate, the maximum BM and SF, two adjoining panel lengths should be loaded. For this loaded length, the live load reaction from each stringer beam will be 1/4 of total EUDL per track. Generally, the cross girder is a plate girder since the stringer beam is to be housed into this.

-- Design Procedure for Cross Girder Bridges Step I: Calculation of loads

1) Assume self weight as 5 kN/m. 2) To get max. BM and SF, two adjacent lengths should be loaded. 1 For this loaded length (= 2 x span of stringer beam), take total EUDL for 1 BM and SF from railway bridge codes. 3) The impact factor (CDA) is to be calculated for a span of 2.5 x span of the stringer beam. I 4) Calculate total load for BM and SF. ) Step 2: Calculation of Max. BM and SF I Calculate maximum BM and SF from loading diagram. Step 3: Section of the girder It is designed as a plate girder. For more details refer Unit 10. Step 4: Connection of flange angles with web

a) Calculate strength of rivets in double shear and bearing. b) Calculate the rivet value, R VAY Calculate the shear flow, F = ---- c) 1 - R d) Calculate the pitch of rivets, p = - . F The maximum allowable pitch is 12t or 200 mm, whichever is less. Step 5: Connection between stringer beam and cross girder Stringer beam is connected to the web of the cross girder. Generally framed connection is used. For more details refer Unit 7.

a) Calculate strength of rivets in double shear and bearing b) Calculate the rivet value, R. Reaction Calculate the number of rivets, n = c) R . d) Choose the size and length of the framing angle to accommodate the rivets. Example 12.5 Design a cross-girder for the following data: Stringer beam = ISMB 600 with 300 x 10 mm flange plates. Span of the bridge = 60 mm. Type of bridge - through type bridge with truss girders. Spacing of the truss girders = 5.2 m. Type of truss - 10 panels of equal length pratt type truss.

Type of track - B.G Main line. , Assume any suitable data, if necessary. Steel Structures Solution Step I: Calculation of loads Assume self weight = 5 kNIm. 60 Span of stringer beam = panel length = - = 6 m 10

For maximum BM and SF, loaded length = 2 x 6 = 12 m. From railway bridge code Table 2.1 Total EUDL for BM = 1228 kN. Total EUDL for SF= 1359 kN. For Impact (CDA), L = 2.5 x 6 = 15 m. CDA = 0.531 Assume dead load of stock rails, guard rails, fastenings, stringer beam as 8 kN/m.

Table 12.6: Various Loads on Cross Girders

Loading For BM (kN) For SF (kN) 1 1 1) Dead load -~(8~6)=24 -~(8~6)=24 2 2 . 1 1 2) Live load from stringer bean1 - x 1228 = 307 - x 1359 = 339.75 4 4 3) Impact load 163 180.40 - Total load 494 544.15

Step 2: Calculation of Max. BM and SF Assume the C to C distance of stringer beam be 2 m. t 111r.3~

5 z2 Max. BM, M = 5 x -+ 494 x 1.6 = 807.3 kN m. 8

Figure 12.33

5 2 Max. SF. V = 5 x A 544.15 = 557.15 kN 2 Stap 3: Section of the girder Bridges a) Design of Web

o,, = 165 ~/mm~.

3M Economical depth of web, d = 5& Obc

= 5q807.3 x lo6 = 764 mm. 165 Adopt 1000 mm depth

v 557.15 t=-- - = 5.57 mm. '"a. d 100 x 1000 Adopt 1000 mm x 10 mm web plate. b) Design of flange M Net area of flange required = - O6c .d

- 807'3 lo6= 4892.72 mm. 165 x 1000 Adopt 2 ISA 154150, 12 mm.

(Area provided = 2 x 3459 = 69 18 mm2) c) Check for stresses Gross area of flange = 2 x 3459 + 150 x 10 = 8418 mm2 Reduction for rivet holes = 21.5 (12 x 2 + 10)

(Using 20 mm dia. P.D.S rivets) = 731 mm2 Net area of flange =8418-731

Figure 12.34: Section of Cross Girder M obc, cai = 7 Ymax

Gross area of flange (Jbr. caI= 4c.car ~~t of flange

= 117.16 ~/mrn~

< 165 N/mm2 (:. Safe) Step 4: Connection of flange angles with web Using 20 mm diameter power driven shop rivets, Gross diameter = 20 + 1.5 = 21.5 mm. Strength of rivets in n double shear = 2 x - (2 1.5)' x -!@ = 72.6 kN. i) 4 1000

ii) bearing = (21.5 x 10) x -300 - 64.5kN. 1000 . . Rivet value, R = 64.5 kN VAY Shear flow, F = - I

= 465.73 N/mm. Pitch of the rivets, p, will be lesser of: i) 12t = 12 x 10 = 120 mm ii) 200 mm

64.5 x ld iii) = 138.49 mm. 465.73 Adopt 20 mm diameter P.D.S rivets @ 120 mm clc. Step 5: Connection between stringer bean? and cross girder Load, Reaction from stringer beam = 544.15 kN. Rivet value, R = 64.5 kN. 544.15 No. of rivets, n =- 10 rivets. 64.5 - Use 5 rivet for connecting stringer beam with cleat angles. Use 10 rivets for connecting cross girder with cleat angles. The size of cleat angle: ISA 200 100, 10 mm. Length of the angle Pitch = 50 mm Edge distance = 300 mm. Length of the angle required = 4 x 50 + 30 = 230 mm Adopt 300 mm length cleat angle.

Cross Girder

+ 300mm --- +t . L A b - -3trimBer Bebm P -

4 C I 1

(a) Elevation

Web OF Cross Girder -- --u1511 ZOOLOO, lomm

I I L 1 t~ebof Stringer Beam - (b) Wan Figure 12.35 12.5.7 Calculation of Forces in the Members of*~russGirder The following loads induce forces in the members

I a) Dead weight of stock rails, guard rails, fastenings, sleepers, stringer beams and cross-beams, truss. (In the absence of data, assume a DL of I ! 10 kN/m) I b) Live load from railway bridge codes as per loaded length. c) Impact load from railway bridge codes as per loaded length. d) Longitudinal load for bottom chord members. The longitudinal load is the maximum of tractive effort or bracking force. It may be compressive or tensile. The nature of the force depends upon the direction of train. Steel Structures e) Wind load Wind pressure may be taken as 1.5 ~/m~.To take the effect of wind on' leeward surface, the area may be increased by 50%. Bottom lateral truss system takes the wind load acting on train, truss projections below the top of the train, stringer beams, track, etc. Top lateral truss system takes the wind load acting on truss projection above the top of the train. i) Bottom lateral truss effect Bottom lateral truss will consist of Chord members - bottom chord members of the truss girders.

Figure 12.36: Bottom Lateral Truss System Struts (horizontal)-cross-members. Diagonals It is a statically indeterminate Assumptions

1) The forces in the chord members of a panel are equal and opposite. 2) The diagonals equally share the shear in the panel. Calculate the total load on the bottom chord W = (wind load intensity) x (projected area for bottom chord) W Calculate the force acting at intermediate panel point, P = No. of panels

Calculate the force at end panel points = E 2'

W Calculate the Reaction at the end, R = -. - 2

Calculate the Moment at middle of panel (M) M Calculate the force in the member of the panel = - . S

Where S = Spacing of Girder. For example, to find the force in the member Lo L,. Force in 4 L3

ii) Top lateral truss effect

Top lateral struss will consist of : Chord member-Top chord members of the truss girders.

Figure 12.37: Top Lateral Truss System Struts. Diagonals. It is a statically indeterminate system. Calculate the total load on the top chord W1= (wind toad intensity) x (projected area for top chord) Calculate the force acting at intermediate panel

WI Point = = P, no. of panels Calculate the force acting at end panel

PI Point = - 2

\\', Calculate the reaction, R1 = ,-. &

Calculate the moment M1 at the niid point of panel.

MI Calculate the force in the member = - S Steel Structures

iii) Overturning eflecr

Figure 12.38: Overturning Effect Leeward truss is subjected to additional vertical; stresses due to overturning effect of wind. Calculate the value of 2 R2

This reaction 2 R2 is UDL

2 R2 Intensity of load = -. kN/m Span It is expressed as % of DL :. :. The forces in the members can be easily computed. iv) Portal efect

Figure 12.39: Portal Frame The lateral load will consist of i) wind load reaction from top chord. ii) 1.25% of the force is two end top chord. Calculate the lcngth of the post : (L) Bridges

H (c + h,) Calculate the force in end post, P3 = S a Calculate the force in bottom chord member = P3 x -.L v) Sway effect

Figure 12.40: Sway Frame

Calculate BM in verticals = I The final forces in the members of the truss girder are calculated by algebraic sum. 12.5.8 Problem on Calculation of Forces in Members of Truss Girder

; Calculate the forces in the members of the truss girder for the following date. - Effective span = 36 m. i i Spacing of stringer beam = 2 m. I Type of track- B.G Main line i / Spacing of the girders: 5.4 m. Height of the truss girder: 7.0 m. ! Assume any suitable data i Step I: Configuration of the truss

Figure 12.41: Truss Girder Steel Structures Let us adopt pratt type truss The panel dimension is 6.0 m. Step 2: Calculation of dead load Assume weight of track, floor beam, etc. = 5.0 kN/m Self Wt. = 5.0 kN/m

Step 3: Influence lines for member forces Top chord members Member UlU2 (Compressive)

Member U2U, : (Compressive)

Figure 12.42: ILD for I11 Uz and Uz UJ 18 -144 Area of ILD -X-x36=- 2 7 7 Bottom chord members Member L,, Ll& L, L2 : (Tension)

Figure 12.43 : ILD for 4LI & LILz

15 +90 , Area of ILD =-x-x36=- 2 7 7 Member & L3 :(Tension) , Figure 12.44 : ILD for LZLJ 19 + 162 Area of ILD = - x - x 36 = - 2 7 7 Diagonal Members 7 Member Lo U1(compressive) tan 0 = - 6 5 PL = 6 cosec 8 cosec 8 = 1.317 0 1

bipure 12.45 : ILD for Lo U1 1 Area df ILD=-x l.i2~36=-20.16 2 Member U1

Figure 12.46 : ILD for UILz 6-x - 4x=6-x x=1.2m 1 4 - cosec0 - cosec 8 6 6 Area of ILD 11 -ve portion = - x - cosec 8 x 7.2 = 0.79 2 6

+ve portion = x cosec 0 x 28.8 = 12.64. 2 6

-L cosec 8 -3 cosec 0 6 h Steel Structures

Member Lz U3

Figure 12.47 : ILD for Lz UJ Area of ILD 12 + ve portion = - x - cosec 0 x 14.4 = 3.16 2 6 13 -ve portion =-X-cosec0~21.6=7.12 2 6 Vertical Members Member UIL,

Figure 12.48 : ILD from UI Li 1 Area of ILD =-x 1 x 12=6. 2 Member U2

I 1 '-1 '-6 Figure 12.49 : ILD for U2 L2 Area of ILD = 0 Member U35

,&Tension L, LZ L3 '-L, 6

Figure 1250 : ILD for UJLJ Step 4: Calculation of forces due to DL, LL & IL D.L. = 10 kN/m. Top chord members Bridges I (U, UZ u2 U3) L I Area of ILD =-- 144 7

Loaded length = 36 m. Total EUDL for BM = 2539 kN.

8 8 CDA =0.15+-- -0.15+-- 6+L 6 + 36 - 0'34 ZL = 0.34~35.26= 11.99 kN/m Total load = DL + LL + IL = 10 + (35.26 + 1 1.99 = 57.25 kN/m. Force in the member U1U2 and U, Uj = (Total load) x (Area of ILD)

Bottom chord members Member LI & L, 4

Loaded length = 36 m, Total load = DL + LL + IL = 57.25 kN/m. r 90 Force in Members Lo L, &L, 4 = 57.25 x 7 = + 736.07 kN.

Member 4 & +I62 Area of ILD = - 7 Loaded length = 36 m. Total load = DL + LL + ZL = 57.25 kN/m. 162 Force in Member L2 & = 57.25 x -= + 1324.93 kN. 7 Diagonal Members: Member LOUl Arca of ILD = -20.16 Loaded length = 36 m. Dl, = 10 kN/m. CDA = 0.34

Total EUDL for SF = -2757 - 1378.5 kN. 2 Steel Structures Total load = DL + LL + IL

= I0+ 38.29+ 13.02 = 61.3 1 kN/m. Force in L, U, =(-20.16) (61.31)

= -1235.98 kN Men~herUI Kb

-1.e portion: Area of ILD = -0.79

Loaded length = 7.2 m.

Total UDL for Sf:= 1004'8X= 502.44 kN. 2

IL = 0.756 x 69.78 = 52.76 kN/m. LL + IC = 69.78 + 52.76 = 122.53 kN/ni Comp. force in Member due to (LL+IL) = -0.79 x 122.53 = -96.8 1 kN Net area = 12.64 - 0.79 = + 11.85

Tensile force in Member due to DL = + 11.85 x 10 = + 118.5 kN. :. Net force in Member due to (DL + LL + IL) = + 118.5 - 96.81 = + 21.69 kN +ve portion : Area of ILD = + 12.63 Loaded length = 28.8 rn.

Total udl for SF = -2587 - 1293.5 kN 2

CDA =0.15+ 8 = 0.38 6 + 28.8 IL = 0.38 x 44.91 = 17.07 kN/m. LL + IL = 44.91 + 17.07 = 61.98 kN/m. Force in member due to (LL + IL) = +(12.64) x 61.98 = +783.37 kN Force in member due to (DL + LL + IL) = +118.5 + 783.37 = +901.87 kN. Mernber L2 U3

Net area = +3.16 - 7.12 = -3.96 Force in the member due to DL = (-3.96) x 10 = -39.6 kN. +v.e porriarz : area of ILD = + 3.16 Loaded length = 14.4 rn.

Total EUDL for S F = -1508'2 - 754.2 kN. 2

Force in the menlber due to (LL + IL) = (52.38 + 28.40) x (+3.16) = +255.26 kN. Net force in the ~nernberdue to (DL + LL + lL) = +255.26 - 39.6 = +215.66 kN.

-1~portion : Area of ILD = -7.12 Loaded length = 21.6 m. 2072.2 Total EUDL for S F = -- 1036.1 kN 2

LL=Z-- 47.97 kN/m 21.6 8 CDA = 0.15 + = 0.44. 6 + 21.6 IL = 0.44 x 47.97 = 21.10 kN/m (LL + IL) = 47.97 + 2 1.10 = 69.07 kN1m. Force in the member due to (LL + lL) = (-7.12) (69.07) = - Force in the member due to (DL+ LL+ IL)

= -39.6 - 491.78 = -531.38 kN. Vertical Metnber Menrbcr U, L, & U, L,

Area of ILD = + 6

Loaded length = 12 111. 1359 kN. Total EUDL for S F = - 2

CDA = 0.504

IL = 0.594 x 56.63 = 33.63 kN/m DL = 10 kN /m (DL+ LL+ lL) = 10 + 56.63 + 33.63 = 100.26 kN 1111. Force in the ~ncn~berdue to (DL+ LL+ lL) Steel Structures = (+6) (100.26) = + 601.56 kN. Member U2L2 No force. Step 5: Force in the member due to longitudinal load ' The truss is hinged at Lo and on robbers at Lg.

For Member Lo L,

Loaded length = from Lo to L6 = 36 m. 735.5 (From table) Tractive effort = = 367.75 kN +- 2 + For Member L, L2

Loaded length = L, to L6 = 30 m.

1 Tractive effort = - x 637.4 = f 3 18.7 kN 2 For Member L2 L3

Loaded length = to L6 = 24 m.

1 Tractive effort = - x 588.4= f 294.2 kN. 2 Step 6: Bottom lateral truss effect Wind pressure = 1.5 kN/m2 Projected area : i) Projected area of stringer beam, sleepers, up to the top of rail = 1.20 x 36 = 43.2 m2

ii) Exposed area of moving train = 3.5 x 36 = 126 m2. iii) Area of truss components = 1.8 m2 To take the effect on leeward side, increase the area of truss components and stock by 50%. Total area = 1.50 (43.2 + 1.8) + 126 = 193.5 m. Total load due to wind = 193.5 x 1.5 = 290.25 kN

Reaction = 290.25 = 145.13 kN. 2

Load at intermediate panel point = -290'25 - 48.4 kN. 6 48.4 Load at end panel point = -= 24.2 kN. 2 1 Force in the Member = k - (Moment at the mid-point of the member) 5.4 Bridges

7: Top lateral truss effect Intensity of wind pressure = 1.5 k~/m~ Projected area: i) Top chord = 0.4 x 24 = 9.6 rn2 (Depth = 400 rnm) 20 Gussets (20% Top chord) = 9.6 x -= 1.9 rn2 ii) 100 iii) Truss Members = 8.5 rn2 Total projected area = 1.5 (9.6 + 1.9 + 8.5) = 30 m2 (Taking 50% for leeward area) Total load = 1.5 x 30 = 45 kN

Reaction = 45 = 22.5 kN. 2

Load at intermediate panel point = = 11.25 kN. 4

Load at end point = 11.25 = 5.63 kN 2

0S.E.WU u.soL(L Figure 12.52: Bottom ~a(erajTruss Steel Structures Step 8: Overturning Effect

Figure 12.53: Overturning Effect Height of the train = 3.5 m. Height of the support = 0.6 m. Standard value Height of the axle from sleeper 0.4 m. The bearings are provided at 0.6 m before the centre line of the bottom chord. Taking moment about bearing, 2R x 5.4 = 101.25 x 0.6 + 45 (7 + 0.6)

Vertical load on leeward truss

36 --- Forces in the member Step 9: Portal Bridges

+!.lo- 6I Ion+ 1 -

W6m

, POC

m - 1- 1- 5.Mm

Figure 1254 : Portal Effect

Length of the end post - 4- 6 + 7 = 9.22 m. H = wind load reaction from top chord + 1.25% of total axial force in the end top chord

= 22.5 + 29.4 = 51.9 kN Axial load due to: 1 Wind load reaction = -(22.5) (1.7 + 3.76) = f 22.75 kN. i) 5.4 ii) 1.25% of total axial force in two end top chord

6 Force in , L, and & &) = 22.75 x - (u, 9.22 = + 14.80 kN BM in U,,in absence of wind 55.27 kN/m

51 9 BM in Lo U,,with wind = -X 3.76 = 97.57 kNm 2 Step 10: Sway egec.1

POC

m 1- 1- bWn4

Figure 12.55: : Sway Effect Steel Structures 11.25 (1.4 + 2.8) Force in verticals = - 2 5.4

BM in verticals = -x 2.8 = 7.9 kN-rn 2x2 Step 11: Total wind load effect Top chord member PuI u2= + 9.37 7: 125.03 = 1 1 6.25 kN

Bottom chord member P, ,=+ 67.18+-78.43+14.8 =f160.4kN 0 1

Diagonal member - P, ,= + 122.98T 22.75 =T 145.73 kN 0 I

Vertical members =& 36.6&4.4=+41kN. Pul L,

Step 12: Total forces in member (DL+ LL + IL) + WL + longitudinal effect Top chord members (Maximum values) P, ,=- 1177.71T 116.25~-1293.96kN. 0 2

Bottom chord members = 736.07f 367.75 f 160.41 = 1264.23kN. P~o~,+ +

Diagonal Member PLou, = (- 1235.98- 29.73) T 145.73=- 1411.44kN.

P,, L2 = + 901.87 + 72.29 = + 974.16 kN (ii) -491.78T24.16=-515.94kN. Vertical mentber

12.5.9 Design of Top Chord Members In through type truss bridges, the top chord members are subjected to compression. Generally the following sections are adopted (Figure 12.56).

r~op cover Plate

Channel

------'--I------

(d) Figure 12.56 The force in the chord members is high. Hence, built up members with less r number components is preferable. Top cover plate is necessary to avoid entering of rain water. The bottom of the section is to be provided with lacing. The number of cover plate should be only one. The thickness of cover plate should be as small as possible. The webs should be as thick as possible. The section should be such that 1 r,, = r,,. The depth of top chord members is about - the height of the truss (c to 10 c distance between top and bottom chords). The clear width = I [ii]height of the truss +2 x thickness of gusset plate. Uniform section is to be provlded throughout the top chord length. The section for end posts is also same as that for top chord. i If the members are subjected to axial load and BM, then

(Jar.cn~ 'bc, cal a) -+- 1 1.0 for normal loads. oat Ob,

oar, cal Obc cal b) -+ -- 1 1.167 for occasional loads. oac Obc 1 Normal loads are DL, LL and IL. Occasional loads are DL,%LLand IL and WL. Steel Structures -, Under occassional loads, the permissible \trcs\c.s arc to be incrcasetl by 16" %. 3 Design .src>p.s Data: Forcc in the member (i) under normal loads (ii) under occasional loads

a) Desigrr qf .s.vectio~l

1) Assunie o,,,.= 100 ~/mtn' 2) Calculate Gross area reouil-ed: Force under norlnal loads A required = OfK Force under occasional loads A required = -- 1 . 167 o~,~.

3) Choose the suitable sections (Ref. Figure 12.58) - 4) Calculate ~rossarea (A), centroidnl di.\t~uice( Y ), M.o.1 about XX-axis (Ixx) and MoI about YY-axis (I,,)

5) Calculate r,,, ryy and r ,,,,,, 6) Calculate Effective length =0.85 x length of the ~tieniber.

7) Calculate slenderness ratio

8) Find o,,. from Table 5.1 (IS : 800) 9) Calculate the load carrying capacity:

Y under Norrnal load) = o,,,.X A

PI ( under occasion and load) = 1.167 P

There should be less than the meniher forces under corresponding load conclitions. b) Desigrz of lacirzg

1) Assume the angle of lacing angle, 0 =48 to 70°, generally 60" 2) Calculate the shear force = 1.25% of axi;rl force due to normal loads. 3) Calculate the force in the lacing ~nelnber Shear force F= sin 0 4) Calculate the distance hetwecn two clid rivct lincs ((1) 5) Calculate the effective Icngrh of lacing mcnihcr

6) Colc~llatethe slentlcl.lic\s of lacing member 7) Find o,, fro111 Tablc 5.1 (IS : 800)

8) Calculate the load carning capacity of lacing mcmbcr = u,, x arca of lacing mcmbcr It should bc lnorc than thc forcc in thc lacing mcrnbcr

9) Calculation of dim~ctcrof rivet Adopting only onc rivct.

From thc abovc cquation \vc can calculate thc diamctcr. 12.5.10 Design the Top Chord Members of the Truss Bridge (Given in Sec: 12.5.8) Data: From Scc: 12.5.8 Met?7ho. 11, 1J2 &[I4 1J5

Forcc undcr Normal loads = 1 177 7 1 kN (Comprcssivc)

Forcc undcr occasional loads = 1293.96 kN (Cornprcssive) Met?7hcr 1J2 lJj & 1J3 1J4

Forcc undcr Nornial loads = 1 177.71 kN (Compressive)

Forcc undcr occasional loads = 128 1.46 kN (Cornprcssive) Sincc thc samc scction is providcd throughout thc lcngth of top chord, takc the maximum valucs of forcc.

Forcc undcr Normal loads = 1177.71 kN

Forcc undcr occasional loads = 1293.96 kN.

1) Assumc o,, = 100 ~/mm~ 2) A rcquircd

1177 71 x lo3 a) Normal loads = - = 1 1777.1 mm2 100 1293.96 x 1o3 b) Occasional loads = = 11087.9 mm2 1.167~700 I 3) Mas dcpth = x 7000 = 700 mm. 10 Lct us assumc 400 nim dcpth For ISA 75 75, 8 mm anglc.

2 A = 1138 mm

I = 59 x 10%m4

C7.yx= 2 1 .4mm Steel Structures

- .---- . Calculation of Y,!, and !, 10 ~,=500x10=5000mm2, y,=-=5mmfrom top 2

A2 = 2 x 1138 = 2276 mm2; y2 = 10 + 21.4 = 31.4 mm.

A, = 2 x 1138 = 2276 rnm2; y, 10 + 400 - 21.4 = 388.6 mm.

-y=- CAY CA 6) Effective length, 1 = 0.85 x 600 = 5 100mm. 2 5100 7) Slenderness ratio, A = -- - rmi, 146.5

8) Form Table 5.1 (IS : 800), o,, = 142N/mm2 9) Load carrying capacity a) P=o,,xA

= 142 x 2353 = 3173.98 kN > ~brcein member under normal load,

b) P, = 1.167 x 3173.98 = 3704.0kN > Member force under occasional load Design of lacing system 1) Let angle of lacing member = 60" 1.25 S.F. in lacing system = -x 1177.71 2) 100

Force in lacing member = -14'72 - 17.00 kN. 3) sin 60"

4) Distance between two end rivets = 300 + 2 x 16 + 2 x 35 = 402 mm. 402 Effective length of lacing member 1 = -= 464.2 mm. 5) sln 60° 6) Let us adopt ISA 7550, 8 mm. A = 938 m2 , rmi,= 10.6mm

A=-=--1 464.2 - 44. rmin 10.6 7) From Table 5.1 (IS: 800), ol, 136N/mm2

8) Load carrying capacity = 130 938

I > Force in lacing member (ob) I 9) Diameter of rivets Using one rivet for connection.

Adotp 14 mm dia rivets. Sled Structures

Figure 12.58: Lacing for Top Chord Members 12.5.11 Bottom Chord Members In through type truss bridges, the bottom chord members are subjected to tension. The following sections are generally adopted (Figure 12.59).

,----ILocan9------_

II;'l------Lo- - c ;n-* ------

(a)

Ib) (c) Figure 12.59 No cover plates are required, either at top or at bottom of the section to avoid cdlw%ionof rain later. If the member are subjected to moments in addition to axial tension, it can be checked by:

Design Steps Data: Maximum force under normal loads Maximum force under occassional loads Design of Mernber Bridges I

1) Assume o,, = 0.9 X 150 = 135 N/mm2 (for normal loads)

oat= 1.167 x 150 ~/mm~(for occasional loads)

2) Calculate the Net area required Maximum force under Normal load a) For normal loads = 135 Max. force under occasional load b) For occasional loads = 1.167~150 3) Choose the suitable section 4) Calculate Gross area 5) Calculate reduction for rivet holes 6) Calculate Net area. If should be greater than net area required. 12.5.12 Design the Bottom Chord Members of the Truss Bi-idge (Given in Sec: 12.5.8.) Data : From Section : 12.5.8 Member lo L,

Maximum force under normal loads = 736.07 kN (tension) Maximum force under occasional loads = 1264.23 kN (tension) I Member L, &

Maximum force under normal loads = 736.0 kN (tension) Maximum force under occasional loads = 1322.66 kN (tension)

Member Lz L., I 1 Maximum force under normal loads = 1324.93 kN (tension) i Maximum force under occasional loads = + 2003.46 kN (tension) In member Lo L, and L, L, the maximum force are almost equal. Hence, one section is designed for both Lo L, and L, I?. ) Members Lo LI and LI & Maximum force under normal loads = 736.07 kN Maximum force under occasional loads = 1322;66 kN.

1) For normal loads, oat= 0.9 X 150 = 135 N/mm2.

For occasional loads, o,, = 1.67 x 150 ~/mm~. 2) Net area required for

Normal loads = 736.07 lo3= 4907.13 mm2 a) 135 1322.66 x lo3 b) Occasional loads = = 7555.9 mm2 1.167~150 Steel Structures 3) Choose the section as shown

- - -- -eao mm-d

Figure 12.60 Gross area =4x 1138+2~300x10 = 10552 mm2

Deductions for rivet holes = 4 x 21.5 x 18 + 2 x 21.5 x 8 = 1892 mrn2 (Using 20 mm diameter rivets) Net are provided = 10552 - 1892 = 8660 mm2 Net area required (OK) Member 5

Maximum force under normal loads = 1324.93 kN (tension) Maximum force under occasional loads = 2003.46 kN (tension) 1) For normal loads, o ,, = 0.9 x 150 = 135 ~/mm~

For occasional loads, a,,= 1.167 x 150 ~/mm~. 2) Net area required for 1324.93 x 10' a) Normal loads = =9814 mm2 135

2003.46 x 10' b) Occasional loads = = 1445 mmz 1.167~150 3) Choose the section:

Figure 12.61 Gross area = 4 x 1138 + 2 x 300 x 20 = 16552 mrn2. Using 20 mm dia. rivets,

Reduction for rivet holes = 4 x 21.5 x (20 + 8) + 2 x 21.5 x 8 = 2752 rnrn2.

Net area provided = 16552 - 27$2 = 13800 mm2 > Net area required (safe) 12.5.13 Diagonal and Vertical Members The vertical members in pratt truss are subjected to compressive force. The alternate diagonals in a warren truss are subjected to compressive force. In some cases, the members which are subjected to tension are also designed as compression member in case of stress reversal. Generally, there are designed as built up I-section. Some commonly used compression member section are shown in Figure 12.62.

Figure 12.62: Common Sections for Webs in Compression Some members are subjected to tension. These are designed as built up I-section. Some commonly used sections are shown in Figure 12.63. (c) (d) Figure 12.63. Canman SPctiolro for Webs in Tension 12.5.14 Design an End Past Par the Truss Bridge (Given in Section 123.8) Data : From section: 12.5.8 Member & U1

Maximum force under normal loads = 1265.71 (Compression)

Maximum f~li~~under occasional loads = 14 11.44 kN, (Compression) BM under normal loads = 55.27 IrN-m. BM under occasional loads = 97.57 kN-m.

1) ow for normal loads = 100 N/mm2 (let)

o,, for occasional loads = 1.167 x 100 N/mm2 (let)

Gmma required for normal loads = 1265.71 lo3 = 12657.1 mm2 2) 100 Gross area required for occasional loads

3); Choose the section

aoexibmm Plate r A, I I I I I I I I I I I CaOGxI6rnrn I I I I PLate I I - - I I I I I I I I I I ------pG1;5~@t , r------Hate ~300mm.! ~300mm.! 1 Figure 12.64 Effective length, 1 = 0.70 x 9220 = 6454 mm 1 6454 6) Slenderness ratio, h = -= -- 'min 109.68 - 58'84' 7) oocfrom Table 5.1 (IS: 800) = 123 N/mm2

9) Check for stresses under normal loads

10) Check for stresses under occasional loads

M Obc, cal = 7 y = 97'57 lo6 = = 52-21N/rnm2. 2.8 x 10' (y)

c 1.167 (Safe) Steel Structures 12.5.15 Design the Diagonal Members for the Truss Bridge (Given in Section 12.5.8) Data: From section: 12.5.8 Member U1L2 Force under normal loads = + 901.87 kN Force under occasional loads = + 974.16 kN. Member Lz U3 Force under normal loads = + 2 15.66 kN and - 49 1.78 kN

Force under occasional loads = + 239.82 kN and - 515.94 kN. Member UI L2 : (tension)

1) oat under normal loads = 0.9 x 150 = 135 ~/rnm~.

oatunder occasional loads = 1.167 x 135 ~/mrn~

2) Net area required

901.87 x lo3 a) Under normal loads = = 6680.5 mrn2 135 974.16~lo3 b) Under occasional loads = = 6183.4 mm2 1.167 x 135 3) Section

Figure 12.65

4) Gross area = 2 x 4564 = 9 128 mm2. Using 20 mm dia. rivets.

Reduction for rivet holes = 2 x 21.5 x 13.6 = 584.8mm2.

Net area = 9128 - 584.8 = 8543.2 mm2. > Net area required. Hence, Safe. Member Lz U, : It is primarily Compression Member

Compressive force under normal loads = 491.78 kN Compressive force under occasional loads = 515.94 kN.

1) Assume q, = 100~/mm'for normal loads o,, = 1.167 x 100 ~/mm*for occasional loads. 2) Gross area required for 491.78 x lo3 Normal loads = =4917.8 mm2 100 515'94x ld Occasional loads = =4421 mm2 1.167~100 3) Section

Figure 12.66 For ISMC 250, A = 3867 mm2 Cyy= 23 m'm.

6) Effective length, 1 = 0.70 x 9220 = 6454 mm 1 6454 7) Slenderness ratio, ?LC- = -= 65 'min 99.35

8) o,, ,for Table 5.1 (IS : 800) = 1 17 ~/mm~

9) Load carrying capacity = 117 x 7734 = 904.9 kN. > Maximum load, (Safe) Check for tension Using 20 mm dia. rivets,

Reduction for rivets = 2 x 2 1.5 x 14.1 = 606.3 mm2

Net area provided = 7734 - 606.3 = 7127.7 mm2 Tensile force under normal loads = 215.66 kN Steel Structures Net area required = 215.66 lo3= 1597.5 rnrn2. 135 < Net area provided. (OK) Tensile force under occasional loads = 239.82 kN. oat= 1.167 x 135 ~/m& 239.82 x lo3 Net area required = 1.167 x 135

< Net area provided. (OK) 12.5.16 Design the Vertical Members for the Truss Bridge (Given in Section 12.5.8) Date : From section: 12.5.8 Members U1 L1 and U3 5 Max. Force under normal loads = + 601.56 kN (Tension) Max. Force under occasional loads = + 642.56 kN (Tension) BM under occasional load = 7.9 kN - m.

1) o,, for normal loads = 0.9 x 150 = 135 ~/mm'.

o,, for occasional loads = 1.167 x 135 ~lrnrn' 2) Net area required for:

601.56 x lo3 = ,56 mm2 Normal loads = 135 642.56 x ld Occasional loads = = 4078.6 mm2 1.167 x 135 3) Section

Figure 12.67

4) Gross area = 300 x 10 + 4 x 1138 = 7552 mm2. Using 20 mm dia. rivets,

Reduction for rivet holes = 2 x 2 1.5 x (10 + 2 x 8) = 1 118 mm2.

Net area provided = 7552 - 1118 = 6434 rnm2 > Since BM is very less No check is required for combined stresses

12.6 END BEARINGS

Bearing is a part of a bridge which receives all the forces from the structure above and transmits the same to the supports. These are provided at the ends of the span of the main girders. The functions of the bearings are a) Transmission of the end reaction to the pier. b) Provision of free movement of the lower chord in horizontal direction. c) The change in the angle of the girder is allowed. d) Reduction in impact due to live load. e) Damping of structural vibration. l f) Limitation on the transmission of sound waves. 12.6.1 IS Code Recommendations IS: 1995 - 1961 stipulates the following specification:

1) For all spans in excess of 9 m provision shall be made for changes in length due to temperature and stress variation. The provisions for expansion and contraction should be such as to permit movement of the free bearings to the extent of not less than 25 mm for every 30 mm of length. 2) For spans greater than 15 m on rigid piers or abutments, bearings which will permit angular deflection of the girder ends shall be provided and at one end there shall be a roller, rocker or other effective type of I expansion bearing. I 3) For wide bridge and skew spans, consideration shall be given to lateral I expansion and contraction. 4) In the design of bearings, provision shall be made for the transmission of longitudinal and lateral forces to the bearings and the supporting structures. 5) In seismic zones excess movement of the spans may be prevented by connecting the top saddle of the bearing to the bottom saddle through hinged connections. I 6) Roller bearings for spans above 35 m should preferably be protected from dirt by oil or grease boxes.

7) Provision shall be made against any uplift to which the bearings may be i subjected. i 8) Rollers and Rockers shall be located by means of dowels, lugs or keys. 1 9) All bearings shall be designed to permit inspection and maintenance. I 12.6.2 Forces Acting on Bearings The following forces are considered in the design of bearings: 1) Reaction at the support 2) Longitudinal loads Uplift forces Transverse forces. Types of Bearings Based on functional behaviours 1) Fired bearings: These allow only rotation. 2) Free bearings (Expansion bearings) These allow both rotation and translation. The movement may be due to creep in concrete, shrinkage, settlement, thermal stresses, braking force, uplift force. Generally one end of the bridge is provided with fixed bearing and the other with expansion bearing. If both the ends have fixed type bearings, internal stresses will develop in bridge structure. For a simply supported bridge girder, fired bearing is at one end and free bearing at the other. But for continuous bridge girder, fixed bearing is at one end and free bearings at the other ends. Based on material used for bearings 1) Mechanical bearings a) Ferrous bearings b) Non-ferrous bearings 2) Elastomeric bearings 3) Combined mechanical and elastomeric bearings. Mechanical Bearings: These allow longitudinal movements and end rotations by sliding, rocking or rolling. These are most commonly used. There are made of metals. a) Ferrous Bearings: These are the most commonly used mechanical bearing. These allow longitudinal movements by sliding of steel on steel and cast-iron. These allow rotations on pins, rollers and rockers. The types of ferrous bearings are i) Plate bearings ii) Roller bearings iii) Rocker bearings iv) Spherical bearings v) Knuckle-pin bearings vi) Railway board bearings. b) Non-ferrous Bearings: These are made from aluminium alloys. These should be used with copper. "Ball and Socket type bearing" is the most commonly used bearing. These permit rotation up to 0.09 radians. Elastomeric bearings: Elastomer is a polymeric substance obtained after vulcanisation. It possesses rubber like properties. It regains original shape completely upon unloading. These bearings allow longitudinal movement and rotation. These consist of one or more internal layers of elastomer bond to internal steel laminates by the process of Vulcanisation. These allow a horizontal Bridges movement of 70 mm and a rotation of 0.02 radians.

3) Combined mechanical and elastomeric bearing: The mechanical bearings provide the longitudinal movement and the elastomeric bearings provide the rotation. 12.6.4 Plate Bearings This is also called as sliding bearings. This does not allow rotation. This is recommended for use for spans less than 15 m. These can be fabricated out of steel, Teflon and Bronze. These are only used for straight bridges. These are unstable under lateral forces. This bearing contains sole plate and bed plate. The sole plate is attached to the bottom of the Bridge Supporting Structure. The bed plate is fixed to the supporting structure. The size of the plates depends upon end reaction and allowable compressive strength of the material used for supporting structure. For sliding, Pintles are fixed in the bed plate. Different types of plate bearings are shown in Figures 12.68 to 12.71.

3ole PLde Bed Pulte

Figure 12.68: Sliding Bearing bGirder

Deep cast Iron bed Plate

Figure 12.69: Cast-Iron Bed Plate Bearing

Figure 12.70: Curved Sole Plate Bearing Steel Structures

(a) (b) Figure 12.71: Provision of Pintles 12.6.5 Ro3er Bearings This type of bt?ring allows both translation and rotation. Figures 12.72 and 12.73 show a typical riller bearing.

Figure 12.72 : Single Roller Bearing

wPier Figure 12.73 : Multiple Rollers Bearing This is generally used for long span bridges. The roller is kept in position by using keys, lugs. A complete circular roller is provided for small sizes. For large size the segmental rollers are used. To take heavy loads number of rollers are used. Multiple Rollers permit horizontal moment very easily. For proper placing of roller spacer plates used. In the bottom stoppers are provided to avoid the movement of Rollers beyond the bottom plate. 12.6.6 Rocker Bearings These are suitable for straight steel bridges. This type of Rocker bearing consist of a pin to accommodate very high deflections. These are used for spans more than 15 m. These are made of steel. These are connected with the substructure through a dead plate. These are bolted or welded to the bridge structure component. These do not permit horizontal movement. These allow only Rotation. Different types of Rocker bearings were shown in Figures 13 74 to 12.76. IRC specifications for Rocker pins: The diameter of the Rocker pin shall not be less than 60 mm. The pin shall be Bridges fitted to a depth 0.5d in the groove. The minimum clean ends above the top surface of the Rocker pin shall be 2.5 mm.

Figure 12.74 : Rocker Bearing with Curved Sole Plate

Figure 12.75 : Rocker Bearing with Curved Bed Plate

Figure 12.76 : Kocker Bearing with Pin 12.6.7 Knuckle Bearings This is a special form of Roller bearing. A knuckle pin is inserted between top and bottom castings. The bottom casting is attacked to the top of the Rollers. The top casting is attached bottom of the Bridge girder. Figure 12.77 shows a typical knuckle pin bearing. , 12.6.8 Spherical Bearing In this type of bearing a spherical convex Rocker rotater in a spherical concave seating. The seating is attached to base plate. The Rocker is attached with the distributed plate which is with the bottom of bridge girder. 12.6.9 Railway Board Roller Bearing Figure 12.78 shows railway board roller bearing. Top castings is attached to the bottom of the bridge girder. The bottom casting is attached on the top of the rollers. The total height of the bearing is ahout lm. The pin is subjected to only Steel Structures bearing pressure. The whole bearing is enclosed ir? n sheet metal box, filled with lubricating oil.

r------Girder

p\n over cont,neou%

Bottom cust~n4

se*mentd totiers

Figure 12.78 12.6.10 Rocker & Roller Bearing A typical rocker and roller bearing is shown in Figure 12.79

Figure 1279: Rocker and Roller Bearing These bearings allow translation and rotation. The nest of rollers facilitates translation. The r ~ckerpin facilitates rotation. The main drawback is collection of dust. IS code provisions Bridges 1) The minimum diameter of the roller shall be 75 mm. 2) The ratio of the length of the roller to the diameter shall be not more than 6

3) In the case of multiple rollers the gap between the rollers shall be not less than 50 mm. The allowable working loads per unit length of cylindrical rollers [on flat surface] shall be taken as: For mild steel

1) Single and double rollers 80 N/mm of length 2) Three or more rollers: 5D N/mm of length. For high tensile steel

1) Single and double roller: 10 D, N/mm of length 2) Three or more roller: 70 N/mm of length. where D is the diameter of roller in mm. IS code provision: [IS 1915-19611

1) Cylindrical roller on curved surfaces: The permissible load per length shall be follows:

a) Single or double rollers: - (l/Dsl) - (l/Dsz) I,,/mrn

k4 Three or more rollers: - kN/mm b) 101.9

where, k3 and k4 are as follows:

k3 k4 Steel conforming to IS : 226 - 1958 0.80 0.50 Steel conforming to IS : 961 - 1957 1.10 0.70 and Ds, and s4 are the diameters in mm, of the convex and concave surfaces respectively.

2) Pins: The permissible stresses in pins shall be as follows: Steel conforming to Steel conforming to IS : 226 - 1958 (~/rnrn') IS : 961-1957 (~/mm~) 1) In Shear 9 8 137 2) In Bearing 206 294 3) In Bending 206 294 For turned and fitted knuckle pins and spheres in bearing. The permissible stress on projected area shall not exceed 118 N/mm2 for steel conforming to IS : 226 - 1958. No pins shall be of diameter less than 100 mm. For railway bridges and road bridges the permissible stresses for pin (including knuckle pins) are given in above Table 28.13. However, as per IS : 800 - 1984 permissible stresses in pin of mild stefi conforming to IS : 226 shall be : In shear : 100 N/mm2 , in bearing: 300 ~/mrn~and in bending : 0.66 0,. Steel Structures 3) Sliding Rearing: The permissible pressure for steel sliding on steel, hard copper alloys, or on cast iron shall not exceed 31 ~lmm*. 12.6.11 Design Considerations for Elastoweric Bearings IRC Specifications I) An elastic bearing should satisfy the following conditions as per IRC 83 (part 11) code. a) Hardness should be 60 & 5 degrees on IRHD scale (international) rubber hardness scale)

b) Minimum tensile strength should be.17 MPa c) Minimum elongation at break shall be 400 d) Shear modulus of elastometer shall be in range from 0.8 MPa to 1.20 MPa e) Adhesion strength of elastometer to steel plates shall not be less than 7 kN1m. Design of unreinforced electrometric bearings I) Plan dimensions: The preferred dimensions of elastomeric bearings are given in Table 12.1 below. Hower interpolation of plan dimensions can be made if the situation warrants. 11) The vertical (axial) stiffness of the elastometer is represented by its shape factor. The shape factor S of the elastometer is given by ratio ab (Loaded surface area)/(Surface area free to bulge) 2 t (a + b) where, a and b are plan dimensions of the pad, and t is the thickness of pad. Table 12.1: Standard Plan Dimensions of Elastometric Bearings (IRC: 83-1983, Part 11)

Size Width (a) Length @) Index No. mm mm

111) Thickness: The thickness of a bearing is governed by its shear movement. If x is the translational shear deformation (Figure 12.80 then

Figure 12.80 : Iiearillg I)i111er1siul1snlld I)el'urn~atiun x = t tan @

Hc H, tan @=- GA where, G = modulus of rigidity in ~/rnrn~ Hc = sustained horizontal load in newton.

Hs = sustained dynamic horizontal load in newton

The value of x should be less than 0.7t, such that t > 1.43 x IV) Average compressive stress : This is given by

where, P = total vertical load in newton a-x Ae = effective plan area excluding shear deformation in mm2 =- b The average stress so calculated should be less than 245. V) To prevent slip : The slip of a bearing is due to high horizontal force and low vertical force. To avoid slip, the following conditions need to be met with

where, PC and P., = sustained and dynamic vertical load, respectively in newton.

H, and Hs = sustained and dynamic horizontal load, respectively in newton p = coefficient of friction (average value = 0.3 ) VI) In order that a bearing does not overturn or topple, the thickness of bearing is restricted to less than a/5. 12.6.12 Design of Rocker Bearing Data: Vertical load due to (DL+ LL+ IL) (W) Vertical load due to wind (W) t Lateral load (F). Design procedure i Step 1: Size of base plate I W I Area of base plate requires = I Safe bearing pressure I Generally square plate is adopted. Step 2: Din. of pin 2 100 mnz. (d) Steel Structures Step 3: Thickness of vertical plates of bearing plate: - W 'd x allowable stress in bearing Step 4: Check for stresses in pin Calculate max. moment in pin, M = (Load on each plate) x (distance between the centre of outer plates of castings) M Calculate max. bending stress - ? d3 32 It should be less than allowable bending stress. 4 Load on each plate Calculate the max. shear stress = -

It should be less than the allowable shear stress. Step 5: Check for bearing pressure Calculate total vertical load (W + W1) w+ w, Calculate the direction stress (ol)= - a2

Calculate the moment due to wind, MI = (lateral load due to wind x 0.3 m) Assuming that the pin is at 0.3 m above the bottom of base plate. 6 MI Calculate stress due to MI = o2= --j-. a Calculate the moment due to wind, M2 = (longitudinal load) x 0.3. 6 M2 Calculate the stress due to M2 = o3- 3- a

Calculate the total stress, a = 0,+ o2+ 03. It should be less than 1.33 x allowable stress. Step 6: Design of base plate

0 Calculate upward pressure, p = - 1.167 Calculate max. moment (M3) is the base plate. Calculate the thickness (t). t2 -X 185=M3. 6 Step 7: Check for bending stress in the plate - Calculate Y, I,, Z Calculate bending stress b

I It should be less than 185 N/mm2. Bridges 12.6.13 Design a Rocker Bearing Data Vertical load due to (DL+ LL+ 1L) = 900 kN. Vertical load due to wind = 200 kN. Lateral load due to wind = 250 kN. Longitudinal load = 300 kN. Solution Step 1: Size of base plate Assume the allowable bearing stress on concrete as 4 N/mm2 900 lo3 Area of base plate required = = 22.5 x lo4 rn2 4 Let us adopt 700 mm x 700 mm base plate Step 2: Diameter of pin Minimum diameter = 100 mm Adopt 125 mm diameter. Step 3: Thickness of vertical bearing plates I Allowable bearing stress (as per code) = 208 N/mm2 900 x lo3 Total thickness = = 34.6 mm. 125 x 208 Adopt 3 plates of 25 mm thick each. Step 4: Check for stresses in pin 900 Downward load on each plate = -= 300 kN. 3 Allowable shear stress = 100 (as per code) 300x lo3 Maximum shear stress - -4 (125)~ 4 = 32.6 N/mm2 < 100 ~/mm~(:. Safe) Provide a clear gap of 5 mm between the two outer plates of top casting and bottom casting.

Figure 12.81

Moment in pin, M = 300x 30 x lo3 = 9 x lo6 N - rnm. Steel Structures Allowable bending stress = 208 N/mm2

< 208 N/mm2 :. Safe Step 5: Check for bearing pressure Total vertical load = 900 + 200 = 1100 kN.

Direct stress, ol = loo lo' = 2.24 N/~~~. '700 x 700 Allowable bearing pressure = 1.33 x 4 = 5.32 N/'rnm2 Let us assume the centre line of the pin is at 300 mm above the bottom of the base plate. 1 Lateral load due to wind = - x 250 = 62.5 kN. 4

Moment due to lateral load, MI = (62.5 x lo3)(300) = 18.75 x lo6 N-mm.

6 Ml Stress due to moment MI, o2=- a3

Moment due to longitudinal load, M2 = (300 x lo3) (300) =90x 106N-mm.

6 M2 Stress due to moment M2, o3= a

. Maximum Stress, o = ol + o2+ o3

= 2.24 + 0.33 + 1.57 = 4.14 N/mm2 < 5.32N/mm2 :. Safe Step 7: Design of base plate

4.14 2 Upward pressure, p = -= 3.55 N/mm . 1.167 Projection = 100 mm.

t2 - x 185 = 1.78 x lo4 :. t = 24.0 mm. 6

Adopt 25 --.-.tl1irl.2p.s. Bridges

Figure 12.82 Step 7: Check jor bending stress in the plate - 700x25~12.5+3x212.5x25x 131.25 Y = = 69.1 mm. 700~25+3x212.5~25

(350)~ Max. moment, MmaX= (3.55) (700) = 152.2 x lo6N - rnm. 2

Mmax :. Bending stress = - 1 =',ax

= 143.7 ~/rnrn~ < 185 N mm2 :. Safe

r--7Girder

ease Plot?

Figure 12.83 12.6.15 Design of Rocker and Roller Bearing 1) Design of Rocker bearing and Base plate Follow the design steps given in Sec 12.6.13. Steel Structures 2) Design of Rollers Step 1: Assume the diameter of roller Minimum diameter should be 75 mm. Step 2: Calculate the load on rollers (P) for a) Multiple roller = 5D Nlmm. length b) Single & double roller = 8 DNImm length Step 3: Calculate the total length of rollers Total vertical load L= 1.167 (P) By choosing suitable length, calculate the number of rollers. Step 4: Design of base plate below rollers Size of base plate is same as in part 1. Assume the thickness of base plate as 30 mm or 50 mm. Calculate the height of rocker pin from the bottom of the base plate. Calculate the moment due to lateral load (MI) W+Wl 6M1 Calculate the max. pressure o = + - a a 3 It should be less than bearing pressure. Calculate maximum moment due to upward pressure

where, b = projection of plate beyond C.L. of last roller.

It should be greater than Mma, . 12.6.16 Design a Rocker Roller Bearing for the Data (Given is Section 12.6.14) Design of Rocker bearing Same as is Sec: 12.6.14 Design of Roller bearing 1) Diameter of Roller = 100 mm 2) Load allowed on roller = 5 D Nlmm = 5 x 100 = 500 N/mm 3) Total load = 900 + 200 = 1100 kN.

Total length of roller = 1100x'103= 2200 mm 500 Adopt 4 rollers of 600 mm length. 4) Size of base plate = 700 mm x 700 mm. Bridges 1 Assume the thickness of base plate = 30 mm. Height of centre of Rocker pin from the bottom of base plate = 30 + 100 + 300 = 430 mm. Moment due to lateral load, M, = 62.5 x 0.43 = 26.875 Id\l-m

1lOOx lo3 + 6 x 26.875 x lo6 Maximum pressure = c = 700 x 700 (700)~

= 2.72 N/mrn2 < 1.33 x 4 ~/mm~. (O.K.)

Upward pressure, p = 2'72 - 2.33 N/rnm2. -1.167 Providing a clear spacing between rollers as 50 mm. Projection from centre of outer roller 700-4x100-3x50+~=125mm b = 2 2

P b2 1252 Max. moment = = 2\33 x N-mdmm -2 -2

= 1.82 x lo4N-dmm length.

I (7 f \ r Ioernm Dia I, Rollers

I Pier I

Figure 12.84 : Rocker and Roller Bearing SAQ 3 1) A pratt truss consisting equilateral triangles of 12 panels of 3 m. each is used for B.G. Main line loading. Find the maximum forces in the members meeting at the fourth joint in the bottom chord. Steel Structures 2) Design a compression chord member of a truss bridge having a length of 8.0 m. for the forces detailed below. Force in the member due to (DL+ LL+ IL) = 1600 kN. Force in the member due to (DL+ LL + lL+ WL) = 1900 kN. Use only channel and plate sections. 3) Design a compression chord member a truss bridge using plates and angles. Panel length is 6.0 m. Force in the member due to DL+ LLt ZL is 3000 kN 4) Design the members of a truss bridge at a Joint are given below: Vertical: +I00 kN and -800 kN under normal loading + 200 kN and -1 100 kN under occasional loading.

Diagonal: 150 kN and +SO0 kN under normal loading - 250 kN and +I200 kN under occasional loading. 5) Design a rocker bearing at one end and the roller bearing at the other end for a plate girder bridge of 30 m span given that the total reaction due to DL, LL and IL is 1500 kN. The vertical reaction due to overturning of wind at each end is 100 kN. Lateral wind load at each bearing is 50 kN. 6) Design the rocker and roller bearing for a railway truss bridge for a railway truss bridge of BG Main line loading for a span of 30 m. The total line load for shear = 2680 kN. Assume self weight and superimposed dead load together as 4 kN/m track. Tractive effort = 640 kN. Breaking force = 500 kN.

12.7 SUMMARY

In this unit, you have studied about steel bridges used for railways. The plate girder bridges and truss bridges are generally used. Now, you are able to do the design the above said bridge. You are aquainted with the knowledge of bridge design (Steel). In the last section, you studied about end bearings which are the most importent component of bridge structure. By going these section, you can design the bearings also.

12.8 ANSWERS TO SAQs

SAQ 1 Refer Preceding text given in the block SAQ 2 1) Sub-section 12.5.8 2) Sub-section 12.5.10 3) Sub-section 12.5.10 4) Sub-section 12.5.15 5) Sub-section 12.6.15 & 12.6.16 6) Sub-section 12.6.16