Project Report_Praveen Kumar

8/17/2019 Project Report_Praveen Kumar

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Analysis & Design of Pipe Carrying

Bridge and TrestlesCentral Project and Engineering Division, Tata Steel Ltd

Trainee Dusi V Praveen Kumar

Civil Engineering Dept.NIT Silchar

Project Guide: Mr. Manoj Kumar

Head Project: Civil & StructuralTata Steel Ltd

2013


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CERTIFICATE OF APPROVAL

This is to certify that the project report on “ANALYSIS AND DESIGN OF PIPECARRYING BRIDGE AND TRESTLES”, submitted by DUSI V PRAVEEN KUMAR ofNIT Silchar is a record of efficient work done by him under my guidance. He hascompleted the job successfully. His duration of project was from 7 th May to 18 th June 2013.

The report contains sufficient calculations & relevant drawingsrelated to the subject matter under study & do fulfil all the requirements of thework assigned to him for the vacation training project.

We wish him every success in life.

Mr. Manoj Kumar

Head Project,Civil & StructuralEngineering & Projects

TATA STEEL LTD, JAMSHEDPUR


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TATA STEEL : a brief introductionEstablished in 1907, Tata Steel is among the top ten global steel companies withan annual crude steel capacity of over 28 million tonnes per annum (mtpa). It isnow one of the world's most geographically-diversified steel producers, withoperations in 26 countries and a commercial presence in over 50 countries.

The Tata Steel Group, with a turnover of US$ 22.8 billion in FY '10, has over80,000 employees across five continents and is a Fortune 500 company. Tatasteel completed 100 glorious years of existence on august 26, 2007 following theideals and philosophy laid down by its founder, Jamshedji Nusserwanji Tata. The

first private sector steel plant which started with a production capacity of1,00,000 tonnes has transformed into global giant.Tata Steel is a global player with a balanced presence in developed Europeanand fast growing Asia’s market and with a strong position in the construction,automotive, and packaging markets.Tata Steel is to be the global steel industry benchmark for value creation andcorporate citizenship.


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Tata Steel’s vision is to be the world’s steel industry benchmark through theexcellence of its people, its innovative approach and overall conduct.Underpinning this vision is a performance culture committed to aspirationtargets, safety and social responsibility, continuous improvement, openness andtransparency.

Tata Steel’s larger production facilities include those in India, the UK, theNetherlands, Thailand, Singapore, China and Australia. Operating companieswithin the Group include Tata Steel Limited (India), Tata Steel Europe Limited(formerly Corus), NatSteel, and Tata Steel Thailand (formerly Millennium Steel).


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STRUCTURAL ENGINEERING: Introduction

Structural engineering deals with planning, design and construction. These are

the most important steps for dealing any structural project. Design concernedwith either concrete or steel, first prepare the plans showing the arrangementfulfilling the purpose to be accomplished by the structure. For instance the plansprepared by structural engineer should be such that not only acceptablearrangement from functional, strength & adequacy requirements and also itshould be economical.

It is very important to calculate and analyse different types of load acting on astructure. Once this is done, structure analysis is made to determine differentstresses induced in various member of structure are ascertained. The size andshape of component member should be chosen such that they will be subjectedto stress not exceeding permissible stress value and then finally fieldconstruction is done.

Structural engineers are most commonly involved in the design of buildings and

large non building structures but they can also be involved in the design ofmachinery, medical equipment, vehicle or any item where structural integrityaffects the item’s function or safety. Structural engineers must ensure theirdesign satisfies given design criteria predicated on safety (e.g. structure mustnot collapse without due warning) or serviceability and performance (e.g.building sway must not cause discomfort to the occupants).Structuralengineering theory is based upon physical laws and empirical knowledge of thestructural performance of different materials and geometries.

Structural engineering design utilizes a number of simple structural elements tobuild up structural systems that can be very complex. Structural engineers areresponsible for making creative and efficient use of funds, structural elementsand materials to achieve these goals.


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TYPE OF LOADS AND THEIR DEFINITION

Estimation of total load acting on a structure is very difficult work, but sometime it is very important from safety point of view.

It is very necessary to consider the combination of load for which structure is tobe designed.

Types of loads

Dead loads

Live loadLongitudinal loadImpact loadWind loadHorizontal loadUpthrust loadEarth loadTemp effectDeformation stressSecondary stressErection stressSeismic force

dead load of structure:

Dead load of a structure means weight of a structure itself. The dead load of astructure is not known before it is designed. It is assumed. Dead load iscompared with actual dead load if the difference is significant, the assumeddead load is revised and the structure is redesigned. Dead load should includethe superimposed loads that are permanently attached to the structure.Different materials should have different dead loads.


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Live load on a structure:Live load is the movable or changeable load acting on a structure. It may changeits position. In case of roof the live load due to human load ,access load(ifpresent).

In case of tank, live load is human load and liquid load.

Wind load on a structure:

The wind load intensity at any height of a structure depends upon the velocityand density of air, shape and height of structure , topography of surrounding

ground surface and angle of wind attack.

Design wind sped at any height ‘z’ is given by,

VZ = Vz*K1*K2*K3

Where,

K1 = probability factor of risk coeffecient

K2 = height and structure factor based on terrain category

K3 = topography factor

Wind pressure = 0.6 (V z)2 N/mm 2.


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Project Details

Project Title: Analysis and Design of Pipe carrying steel structure Bridge andTrestle.

Given:

1. 2 numbers of CO pipe of diameter = 500mm.2. 1 number of water pipe of diameter = 300mm.

Other Data:Density of steel = 7850 kg/m 3

Muck (live load) = 1200 kg/m 3 [Note: Muck is sediment formation inside CO gas pipe which is assumed to beoccupying 1/3 rd area of the pipe (worst case) and is periodically removed.]

The pipes are continuous and the next trestle support is at 6m, hence theprojected pipe influence is 3m in length.

3 m20 m

1 . 6 5 m

1

. 7 5 m

1 . 6

5 m

1 0 m

3 m1.65 m


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Preliminary Planning:In order to proceed further, exterior has to be planned judiciously with someassumptions, which get refined with experience.

Depth of the Bridge:This depends on the type of loading, ie. Light, medium, heavy. As the loadingincreases light to heavy the depth should increase and this is governed by spanto depth ratio which can be enumerated as follows:

Basing on experience our project guide has guided us that this is mediumloading and span to depth ratio is taken as 12 and after fine tuning to a rationalno. depth is arrived at 1.65 m.

No. of panels in front view truss:Firstly to simplify the design some degree of symmetry is incorporated. Hence

the diagonal members are arranged in back slash up until middle and forwardslash for the rest. Usually to satisfy the above requirement and to avoidindeterminacy even no. of panels is preferred. Also along with this to ensurepractical feasibility and aesthetics the diagonal member is inclined at around

range.Keeping the above points in view, the end panels are of length 1.75m and theintermediate panels are of length 1.65m.

Accordingly the horizontal bracings in the top and bottom truss systems, (thatresist wind loads) are laced continuously.


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Estimation of Loads:

1. Dead Load (DL)a. DL of CO pipe (2 nos.)

Diameter ( Φ) = 500mm

Thickness (t) = 10mmDensity of steel (γ) = 78.5 kN/m DL = 2*π*(0.51 2 -0 .49 2)/4 kN/m

= 2.47kN/m.

b. DL of water pipe (1 no.) Diameter (Φ) = 300mm

Thickness (t) = 8mmDensity of steel (γ) = 78.5 kN/m DL = π*(..308 2 - .292 2)/4 kN/m

= .60kN/m.

c. DL of truss = 2kN/m (assumption)

Total DL acting on Top chords of Vertical Truss = 5.07 kN/m

DL acting on Each Top Chord of Vertical Truss = 2.535 kN/m

Also, DL of projected pipe on ends of truss = 4.605 kN

2. Live Loada. LL due to running water = 10* π*0.292 2/4 kN/m

= .669 kN/m b. LL due to Muck in CO pipes (2 Nos) :

Density of Muck = 12 kN/m 3 Area of CO pipe, A CO= π*.49

2/4 m 2 Assumption : 1/3 rd of cross sectional area of CO pipes is filled with

muck (worst case)So, Maximum LL due to Muck = 2*A CO*12 kN/m

= 1.5 kN/m

Total LL acting on Top chords of Vertical Truss = 2.169 kN/mLL acting on Each Top Chord of Vertical Truss = 1.085 kN/m


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3. Wind Load:Wind Loads are calculated according to IS 875 : 1984 Part 3

Basic Wind Speed, V b = 47 m/s

Design Wind Speed at any height z , Vz = k1.k2.k3.Vb

Where, k 1 = Probability Factor (Risk Coefficient),k2 = Terrain, height and structure factork3 = Topography factor

As per Code Clause,

k1 =1k2 =0.87 (Category III, Class B)k3 = 1

So, Design Wind Speed at any height z, V z = 1*0.87*1*47 m/s= 40.89 m/s

As per Code Clause,

Design Wind Pressure P z = 0.6*Vz2 N/m 2 = 1003.2 N/m 2

Design Wind Force, F z = Cf *Ae*Pz N

Where, C f = Force coefficient, depends on solidity Ratio ( φ )Ae = Effective area

Pz = Design Wind Speed

Wind Force on Pipe :

Wind Udl on Pipe, Wp = C f *Ae*Pz/L

Cf = 0.5 [As per IS 875 part 3, Fig. 6]L = Length of Pipe

So, Wind Udl on pipe, Wp = 0.6*0.7*D*P z= 0.6*0.7*0.5*1003.3


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= 210.693 N/m

Total Udl on all (three) pipes due to Wind, Wpt = 1.5*210.693= 316.0395 N/m

Wind Load Udl on Lateral Truss :

Total Wind load udl on Windward Side Truss Wt = C f *Ae*Pz/L

Cf on windward side depends on Solidity ratio ()

Solidity Ratio ( φ ) =

Assuming ISLC 125 for Top Chord and Bottom Chord, ISA 65X65 forVertical and Diagonal members,

Windward Side truss,Ae = 0.125 * 20 * 2 + (10 * 0.065 * (1.414 * 1.65)+2*0.065*2.405) + (13 * 1.65 * 0.065)

= 8.25 m 2

So, Solidity Ratio ( φ ) =

= 0.25Cf = 1.35 [as per IS 875 Part 3, Table 28]So, Wt1 = 1.35*8.25*1003.2/20

= 558.657 N/mSo, WL on Windward side Top Chord = 558.657/2 = 279.32 N/mand WL on Windward side Bottom Chord = 558.657/2 = 279.32 N/m

Leeward Side Truss,

Total Wind load udl on Windward Side Truss Wt = C f *Ae*Pz/LCf depends on screening factor η Since frame spacing ratio = 1 [as per IS 875 part 3]η = 0.85

Cf 0.85*1.35 =1.14So, Wt2 = 1.14*8.25*1003.2/20


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= 471.754 N/m

So, WL on Leeward side Top Chord = 471.754/2 = 235.877 N/mand WL on Leeward side Bottom Chord = 471.754/2 = 235.877 N/m

Transferring WL on Pipe to Top chords of Windward Side and Leewardside we have,

WL on Windward side Top Chord = 158.02+279.32 = 437.94 N/mWL on Leeward side Top Chord = 158.02+235.87 = 393.89 N/m

WL on Windward side Bottom Chord = 279.32 N/m

WL on Leeward side Bottom Chord = 235.877 N/m

Analysis of horizontal truss system can be done by loading totally onWindward side top chord and similarly Windward side bottom chord.

Wind Loading for Top Horizontal Truss,Windward side Chord = 437.94 + 393.89 = 831.83 N/m = 0.83183 kN/mEnd Joints (due to projected Pipe) = 0.316 *3 = 0.948 kN

Wind Loading for Bottom Horizontal Truss,Windward side Chord = 393.89 + 235.88 = 629.77 N/m


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Load Combinations :As per IS 800 : 2007, for the design of steel structures as per limit statedesign, following load combination have to be used

1. 1.5(DL + LL)2. 1.2(DL + LL + WL)

Fig 1: 1.5(DL + LL) case, Front View (kN,m units)

Fig2: 1.2(DL + LL + WL) case, Front View (kN,m units)

Fig3: 1.2(DL+LL+WL) case, Plan view, Top Horizontal truss (kN,m units)

Fig3:1.2(DL+LL+WL) case,Plan view, Bottom Horizontal truss (kN,m units)


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Structural Analysis :

Structural Analysis for above loads has been done manually using method of joints, and is verified using the software modelling.

The Udls shown above are transformed into joint loads according to their lengthof influence.

Method of Joints:After transforming udls to joint loads the internal Axial forces are calculated bydrawing FBD of each joint and finding out the unknown forces after resolvinginto mutually perpendicular axes.

Analysis for Load Combination 1.5(DL + LL):

Reactions:

R0 = R12 = = 66.09 kN

Joint 0:

F25 F25 = -R0 = - 66.09 kN

F0 = 0F0

R0

Joint 25:16.541 kN

F24

F25 F26


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Joint 1:F26 F27 F0 F1

Joint 24:F24 9.231kN

F23 F28

F27



F22 F30

F29



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F21 F32

F31



F20 F34

F33



F19 F36

F35


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F18 F37Joint 6:F36 F37 F38 F5 F6

Since the truss is symmetric, the forces in other members will be mirror images of this halfsection.


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Analysis for Load Combination 1.2(DL + LL+WL), front view:

Reactions:

R0 = R12 = = 52.87 kN

Joint 0:

F25 F25 = -R0 = - 52.83 kN

F0 F0 = 0

R0

Joint 25:13.227 kN

F24 F25 F26 Joint 1:F26 F27

F0 F1 Joint 24:F24 7.385kN

F23

F28 F27


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Joint 2:F28 F29

F1 F2


F22 F30 F29



F21 F32

F31



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F20

F34F33



F19 F36

F35 Joint 19:F19 7.167kN

F18 F37Joint 6:F36 F37 F38 F5 F6


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Since the truss is symmetric, the forces in other members will be mirror imagesof this half section. Axial forces are verified using SAP 2000 model.


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Analysis for Load Combination 1.2(DL + LL+WL), Plan View, Top Horizontal

truss:

Reactions:

R0,Top = R12 = = 11.12 kN

Joint 25:2.013 kN

F23 F24

Joint 0:

F24 F25 F0

R0 Joint 1:

F38

F0 F1 Joint 24:

F23 1.697kN F22

F25 F26 F38


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F21F39

Joint 2:F26 F39 F27 F1 F2

Joint 3:

F40 F2 F3


F20 F27 F28 F40 Joint 21:F20 1.697kN

F21

F41 Joint 4:F28 F41 F29

F3 F4


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Joint 5:F42 F4 F5


F18 F29 F30

F42 Joint 19:F18 1.697kN

F17 F43




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Analysis for Load Combination 1.2(DL + LL+WL), Plan View, Bottom Horizontal

truss:

Reactions:

R0 = R12 = = 7.55 kN

Joint 25:0.66 kN

F23 F24

Joint 0:

F24 F25 F0

R0 Joint 1:

F38

F0 F1 Joint 24:

F23 1.283kN F22

F25 F26 F38


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F21F39


Joint 3:F40

F2 F3

Joint 22:

F21 1.245kN F20 F27 F28

F40 Joint 21:F20 1.245kN

F21 F41


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Joint 5:

F42 F4 F5


F18 F29 F30

F42


F17 F43Joint 6:F30 F43 F31 F5 F6


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Summary of Above Structural Analysis:As we can see in the Axial force diagrams, the compressive forces are induced inthe top chords of each truss for vertical loading (DL+LL) and in case of WL thecompressive forces are induced in the top and bottom chords of windward side

truss and tensile forces in leeward side top and bottom chords.

This can be diagrammatically visualized as shown below:

Hence we can see the maximum compression occurs in Windward side topchord and maximum tensile forces occur in Leeward side bottom chord. Theresults of above analysis also prove the same.

So we design the top chords for maximum compressive forces and bottom chordmembers for maximum tensile forces, since the direction of wind loads arereversible in nature.

Comments on Structural Analysis:

As we have seen in the above analysis, the maximum forces were induced incase of 1.5(DL + LL) as compared to 1.2(DL+LL+WL) combination. This might beas a result of less wind forces acting because of choice of Terrain, height andstructure factor (k 2) while calculating design Wind Speeds.

(DL + LL)

(WL)


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Structural Design for Induced Forces:As shown in summary the maximum compressive forces are induced in topchord of windward side truss and similarly the maximum tensile forces areinduced in bottom chord of leeward side truss.

The wind can change its direction hence the top chord (both windward, leewardside) are designed for maximum compressive forces and the bottom chord (bothwindward, leeward side) are designed for maximum tensile forces.

Design is done using IS 800 : 2007 :

Maximum compressive forces are produced in case of 1.5(DL+LL), P = -164.54 kNMaximum tensile forces are produced in case of 1.5(DL+LL), P = 160.05 kN

Design of Compression Members [section 7 : IS 800:2007]

Where, Pd = design compressive force of sectionAe = effective area of sectionf cd = design compressive Stress of section

f cd depends on slenderness ratio and the type of steel

type of steel is Mild Steel (Fe250, f y =250Mpa)

Slenderness ratio

Where, K = 0.85 [for weld connections]L = length of memberr = minimum radius of gyration of the section.

Assuming ISMC 125 for compressive members,

Slenderness ratio

f cd for slenderness ratio 73.04 and f y 250 is

so,


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Since, the section is O.K for compression.Design for Tension Members [Section 6: IS 800:2007]

Where,

Td = design Tensile force of sectionAg = gross area of section

m0 = partial safety factor for failure in tension

Since, the section is O.K for tension.

Vertical Member maximum compressive force, P = -66.09 kNDiagonal Member maximum Tensile force, T = -72.22 kN





Slenderness ratio


Assuming ISA 6565, t=6mm for vertical compressive members,

Slenderness ratio



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so, Since, the section is O.K for compression.

Design for Tension Members [Section 6: IS 800:2007]

Where, Td = design Tensile force of sectionAg = gross area of section


Assuming ISA 6565, t=6mm for diagonal tensile members, Since, the section is O.K for tension.

Top truss Vertical member maximum compressive forces P = -2.01 kNTop truss Diagonal member maximum Tensile force T = 10.48 kN





Slenderness ratio

Where, K = 0.85 [for weld connections]

L = length of member


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r = minimum radius of gyration of the section.


Slenderness ratio



Design for Tension Members [Section 6: IS 800:2007]

Where, Td = design Tensile force of sectionAg = gross area of section


Assuming ISA 6060, t=6mm for diagonal tensile members, Since, the section is O.K for tension.

Bottom truss Vertical member maximum compressive forces P = -0.66 kN

Bottom truss Diagonal member maximum Tensile force T = 7.98 kN

Since the forces are less compared to Top truss Vertical member and Diagonalmember maximum forces, we can use the sections decided for top trussmembers.

So, Bottom truss Vertical member is provided with ISA 6060 t=6mAnd Bottom truss Diagonal member also is provided with ISA 6060 t=6m


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TRESTLES: They are vertical structures that transfer the loads from Pipes and bridge to thefoundations.As per the problem statement we have a 4 legged trestle and a 2 legged trestle.

Shown below are the frontal and cross sectional views of 4 legged trestle and 2legged trestle.

Load Combinations:Since we have DL, LL and WL acting on the trestles, we have following loadcombinations:

1. Pipe full condition : 1.2(DL+LL+WL)2. Pipe empty condition : 1.5(DL+WL)

Front view of 4 legged trestle Cross sectional viewof 4 legged trestle

Front viewof 2 legged trestle

Cross Sectional viewof 2 legged trestle

1.65 m1.65 m1.65 m


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Structural Analysis :The forces due to pipes and bridges transferred to trestles as reactions ontrestles. In addition to these loads the wind loads on trestles and Self weight(DL) of trestle also have to be assumed or calculated.

Analysis for Pipes Full condition :: 1.2(DL+LL+WL) case :

This analysis is split into 1.2DL, 1.2LL and 1.2WL and then superimposed to getfull picture.

Analysis for 1.2 DL :

Reactions from top due to pipe and bridge self-weight: Reaction on each column Self-Weight of trestle (assumed)Self-Weight of trestle acts on

Members just above supports

Axial forces induced in each columndue to DL of pipes and bridge = 17.98 kN

Axial forces induced in lower columnMembers = 22.98 kN

Foundation Reactions


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Analysis for 1.2 LL

Reactions from top due to pipe’s liv e load (LL):

Reaction on each column

Foundation Reactions Axial forces induced in each columndue to LL of pipes Analysis for 1.2WL :Wind forces act in two mutually perpendicular directions on the trestle. One is

Lateral and other one is longitudinal direction.

Longitudinal direction Lateral direction


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Analysis for Lateral direction (1.2WL) :

Reaction for 1.2WL on Top Horizontal Truss Reaction for 1.2WL on Bottom Horizontal Truss Total Transferable WL on trestle Wind Load on each column

Wind Load Trestle:

We have already calculated, Design Wind Speeds, Design WindPressure for this location.

We now need to calculate the Wind Loads on Windward side oftrestle and Leeward Side of trestle.

To initiate the calculation, we are assuming ISMC 200x75 forcolumns and ISA6060 section for horizontal and diagonal members

Solidity ratio

Windward side

[as per IS 875 part 3]

Wind

Plan view

Section view


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Wind Load on windward side (udl)

Frame spacing ratio ⁄ Solidity ratio

Shielding Factor, [as per IS 875 part 3, Table 29 ]Leeward side Wind Load on Leeward side (udl)

Total wind load on trestle Wind Load (WL) on each column Reactions due to Wind Loads on trestle :

Horizontal Reactions, Joint 1:

F14 4.67kN F

1

Joint 14:5.31 kN

F14 F25

F13


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Joint 13:F13

0.64 kN F15 F12

Joint 4:

F25 F1 F15

F24 F2

Joint 3:F2

F16 F3

Joint 12:F12 F24

0.64kN F16 F11 F23

Joint 11:F

11

0.64kN F17 F10

Joint 4:F23 F3

F17 F22 F4


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Joint 5:F18 F4

F5 Joint 10:F10 F22

F18 F9 F21

Joint 9:F9 0.66 kN F19

F8

Joint 6:F21 F5

F19 F20 F6

Joint 7:F6

F7

RH RWL, pushJoint 8:F8 F20

0.341 kN RH F7

RWL, uplift


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Superposition of above three cases gives the Load Combination1.2(DL+LL+WLLateral ).Hence, on superposition we get the following result:

Foundation Reaction

Foundation Reaction Horizontal Reactions R H = 6.62 kN

Analysis for Longitudinal direction (1.2WL longitudinal ) :Since the pipe in the cross sectional view is continuous, wind load acts only on

the cross braced frame of the truss at the support.

Assuming ISA 65X65 cross bracing at the support,

Solidity ratio

Windward side [as per IS 875 part 3]

RHRH

R1 R2


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Wind Load in sectional view of bridge

Wind Load on each column

Wind Load Trestle:We have already calculated, Design Wind Speeds, Design WindPressure for this location.


To initiate the calculation, we are assuming ISMC 200x75 forcolumns and ISA6060 section for horizontal and diagonalmembers

Solidity ratio

Windward side [as per IS 875 part 3]

Plan view

Wind


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Total wind load on trestle

Wind Load (WL) on each column Reactions due to Wind Loads on trestle :

Horizontal Reactions, Joint 1: F14 0.33kN

F1

Joint 14:

0.651 kN F14 F25

F13 Joint 13:

F13 0.64 kN F15

F12


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Joint 2:F25 F1

F15 F24 F2

Joint 3:

F2 F16

F3 Joint 12:

F12 F24 0.64kN F16

F11 F23 Joint 11:

F11 0.64kN F17 F10 Joint 4:

F23 F3 F17 F

22 F

4

Joint 5:F18 F4

F5


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Joint 10:F10 F22

F18 F9 F21

Joint 9:

F9 0.66 kN F19

F8

Joint 6:F21 F5

F19 F20

F6 Joint 7:

F6 F7 RH

RWL, push

Joint 8:

F8 F20 0.341 kN RH F7 RWL, uplift


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Superposition of above three cases gives the Load Combination1.2(DL+LL+WLlongitudinal ).Hence, on superposition we get the following result:

Foundation Reaction



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Analysis for Pipes Empty condition :: 1.5(DL+WL) case

This analysis is split into 1.5DL, and 1.5WL and then superimposed to get fullpicture.

Analysis for 1.5 DL

Reactions from top due to pipe and bridge self-weight:

Reaction on each column Self-Weight of trestle

(assumed)

Self-Weight of trestle acts on Members just above supports


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Analysis for 1.5WL :Wind forces act in two mutually perpendicular directions on the trestle. One isLateral and other one is longitudinal direction.




Wind

Plan view


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Wind Load Trestle:We have already calculated, Design Wind Speeds, DesignWind Pressure for this location.

We now need to calculate the Wind Loads on Windward sideof trestle and Leeward Side of trestle.


Solidity ratio

Windward side [as per IS 875 part 3 ]


Frame spacing ratio

⁄

Solidity ratio



Wind Load (WL) on each column


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Reactions due to Wind Loads on trestle :

Horizontal Reactions, Joint 1: F14 5.83kN

F1

Joint 14:6.63 kN

F14 F25 F13 Joint 13:

F13 0.80 kN F15

F12

Joint 4:

F25 F1 F15

F24 F2 Joint 3:

F2 F16

F3


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Joint 12:F12 F24

0.80kN F16 F11 F23

Joint 11:

F11 0.80kN F17

F10 Joint 4:

F23 F3 F17 F22 F4

Joint 5:

F18 F4 F5

Joint 10:F10 F22

F18 F9 F21

Joint 9:F9

0.83 kN F19 F8


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Joint 6:F21 F5

F19 F20

F6

Joint 7:

F6 F7

RH RWL, push

Joint 8:F8 F20

0.426 kN RH F7 RWL, uplift Superposition of above two cases gives the load combination1.5(DL + WLLateral ).

Foundation Reaction Foundation Reaction

Horizontal Reactions R H = 8.27 kN


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Analysis for Longitudinal direction (1.2WL longitudinal ) :Since the pipe in the cross sectional view is continuous, wind load acts only onthe cross braced frame of the truss at the support.

Assuming ISA 65X65 cross bracing at the support,

Solidity ratio

Windward side

[as per IS 875 part 3]

Wind Load in sectional view of bridge

Wind Load on each column

Plan view

Wind


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Solidity ratio Windward side [as per IS 875 part 3]


Frame spacing ratio ⁄

Solidity ratio

Shielding Factor, [as per IS 875 part 3, Table 29]Leeward side

Wind Load on Leeward side (udl) Total wind load on trestle

Wind Load (WL) on each column


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Horizontal Reactions, Joint 1:

F14 0.407kN F1

Joint 14:0.811 kN

F14 F25

F13 Joint 13:

F13 0.8 kN F15

F12 Joint 2:

F25 F1 F15

F24 F2 Joint 3:

F2 F16

F3


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Joint 12:F12 F24

0.8kN F16 F11 F23

Joint 11:

F11 0.8kN F17

F10 Joint 4:

F23 F3 F17 F22 F4

Joint 5:F18 F4

F5 Joint 10:F10 F22

0.8kN F18 F9 F21

Joint 9:

F9 0.83 kN F19

F8


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Joint 6:F21 F5

F19 F20

F6

Joint 7:

F6 F7

RH RWL, push

Joint 8:F8 F20

0.426 kN RH F7 RWL, uplift Superposition of above three cases gives the Load Combination1.2(DL+LL+WLlongitudinal ).Hence, on superposition we get the following result:

Foundation Reaction Foundation Reaction Horizontal Reactions R H = 2.84 kN


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Analysis for Pipes Full condition :: 1.2(DL+LL+WL) case :

This analysis is split into 1.2DL, 1.2LL and 1.2WL and then superimposed to getfull picture.

Analysis for 1.2 DL:


Self-Weight of trestle (assumed)Self-Weight of trestle acts on Members just above supports




R0


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Analysis for 1.2 LL

Reactions from top due to pipe’s live load (LL):


Axial forces induced in each column due to LL of pipes Analysis for 1.2WL :Wind forces act in two mutually perpendicular directions on the trestle. One isLateral and other one is longitudinal direction.


R0


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To initiate the calculation, we are assuming ISMB 200 for columnsand ISA6565 section for horizontal and diagonal members

Solidity ratio Windward side, [as per IS 875 part 3]


Wind

Plan view


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Horizontal Reactions, On analysing this 2 legged trestle as done in 4legged trestle using method of joints yieldsfollowing results:

Superposition of above three cases gives theLoad Combination 1.2(DL+LL+WLLateral ).Hence, on superposition we get the followingresult:Foundation Reaction


Axial force diagram for1.2WL

Axial force diagram for1.2(DL+LL+WL)


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Analysis for Pipes Empty condition :: 1.5(DL+WL) case

This analysis is split into 1.5DL, and 1.5WL and then superimposed to get fullpicture.

Analysis for 1.5 DL:


Self-Weight of trestle (assumed)Self-Weight of trestle acts on Members just above supports




R0


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Analysis for 1.5WL :Wind forces act in two mutually perpendicular directions on the trestle. One isLateral and other one is longitudinal direction.




Wind

Plan view


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To initiate the calculation, we are assuming ISMB 200 for columnsand ISA6565 section for horizontal and diagonal members

Solidity ratio

Windward side, [as per IS 875 part 3]






Horizontal Reactions, On analysing this 2 legged trestle as done in 4 legged trestle using method of joints yields following results:


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Superposition of above three cases gives the LoadCombination 1.2(DL+LL+WLLateral ).Hence, on superposition we get the following result:Foundation Reaction


Axial force diagram for1.5WL

Axial force diagram for1.5(DL+WL)


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TRESTLES:4-legged Trestle:Summary of Above Structural Analysis:As we can see in the Axial force diagrams, the Maximum Compressive force andMaximum tensile force occurs in pipe empty load combination,1.5(DL + WLlateral ).Accordingly,Maximum Compressive force, column = -114.13 kNMaximum tensile force, column = -39.63 kN

Maximum Compressive force diagonal members = -21.6 kNMaximum tensile force diagonal members = 23.47 kN

Again, since the Wind forces are reversible in nature, we design the Columnsection for Max compressive forces and diagonal members also for maximumcompressive forces.

Comments on Structural Analysis:As we have seen in the above analysis, the maximum forces were induced incase of 1.5(DL + WLLateral ), Pipe empty condition as compared to the case of1.5(DL+WLlongitudinal ), Pipe empty condition. This might be as a result of greaterWind loads in Lateral direction as compared to longitudinal direction as theeffective area in contact with wind forces is greater in lateral direction

Structural Design for Induced Forces:As shown in summary the column section are to be designed for maximumcompressive forces and the diagonal members also should be designed for their

maximum compressive forces.


Maximum compressive forces are produced in columns for the case of1.5(DL+WLlateral ), P = -114.13 kN


Where,


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Pd = design compressive force of sectionAe = effective area of sectionf cd = design compressive Stress of section



Slenderness ratio


Assuming ISMC 200 for compressive members,Slenderness ratio

f cd for slenderness ratio 73.04 and f y 250 is so, Since, the section is O.K for compression.

OR

Assuming ISA9090 t=10mm for compressive members,

Slenderness ratio

f cd for slenderness ratio 85.4 and f y 250 is so, Since, this section is also O.K for compression.


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Maximum compressive forces are produced in case of diagonal member for thecase of 1.5(DL+WLlateral ), P = -21.6 kN





Slenderness ratio



Slenderness ratio



The same section can be provided for horizontal member or ISA 6060 t=6mmcan be provided.


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2-legged Trestle:Summary of Above Structural Analysis:As we can see in the Axial force diagrams, the Maximum Compressive force andMaximum tensile force occurs in pipe empty load combination,1.5(DL + WLlateral ).Accordingly,Maximum Compressive force, column = -337.10 kNMaximum tensile force, column = -234.288 kN

Maximum Compressive force diagonal members = -67.73 kNMaximum tensile force diagonal members = 71.60 kN

Again, since the Wind forces are reversible in nature, we design the Columnsection for Max compressive forces and diagonal members also for maximumcompressive forces.

Comments on Structural Analysis:As we have seen in the above analysis, the maximum forces were induced incase of 1.5(DL + WLLateral ), Pipe empty condition as compared to the case of1.5(DL+WLlongitudinal ), Pipe empty condition. This might be as a result of greaterWind loads in Lateral direction as compared to longitudinal direction as theeffective area in contact with wind forces is greater in lateral direction

Structural Design for Induced Forces:As shown in summary the column section are to be designed for maximumcompressive forces and the diagonal members also should be designed for theirmaximum compressive forces.


Maximum compressive forces are produced in columns for the case of1.5(DL+WLlateral ), P = -337.10 kN


Where,


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Pd = design compressive force of sectionAe = effective area of sectionf cd = design compressive Stress of section



Slenderness ratio


Assuming ISMB 200 for compressive members,Slenderness ratio

f cd for slenderness ratio 73.04 and f y 250 is so, Since, the section is O.K for compression.

Maximum compressive forces are produced in case of diagonal member for thecase of 1.5(DL+WLlateral ), P = - 67.73 kN





Slenderness ratio Where, K = 0.85 [for weld connections]

L = length of member


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r = minimum radius of gyration of the section.


Slenderness ratio



The same section can be provided for horizontal member or ISA 6565 t=10mmcan be provided.


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37 3838 200 37

350

Design of Base plate:

4-legged trestle:

Maximum uplift = 56.71 kN25 % increase for future consideration = 1.25*56.71 = 70.8875 kN

Assume 25mm dia bolt

A = 392.69 mm2

Where, f ub = 400 MPa [IS 1367, 4.6 grade bolt]

Tdb = 141368.4/1.25 = 113.094 kN

No. of Bolts = 70.8875/113.094 = 0.622

Hence provide 2 bolts.

The dimensions of base plate are determined by providing minimum clearances

to the bolt and column and then summing up the resulting distance.

Edge clearance from bolt =1.5D


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Hole clearance from the section = 25+0.5DThickness of base plate (for ISMC 200 Column):

Permissible stresses in the base plate, f b = 189 MPaContact pressure between concrete of M25 grade and steel = 7 MPaConsidering bending about major column length:

√

Considering bending about minor column length:

√ So, choose 25 mm thickness base plate for 4-legged trestles.

Thickness of base plate (for ISA9090 Column):

Permissible stresses in the base plate, f b = 189 MPa

Contact pressure between concrete of M25 grade andsteel = 7 MPaConsidering bending about any column length(symmetric):

√ So, choose 25 mm thickness base plate for 4-legged trestles (ISA 9090 t=10mm).

220


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2-legged trestle

Maximum uplift = 234.20 kN25 % increase for future consideration = 1.25*234.20 = 292.75 kN

Assume 25mm dia bolt

A = 392.69 mm 2

Where, f ub = 400 MPa [IS 1367, 4.6 grade bolt] Tdb = 141368.4/1.25 = 113.094 kNNo. of Bolts = 292.75/113.094 = 2.588

Hence provide 4 bolts.

The dimensions of base plate are determined by providing minimum clearancesto the bolt and column and then summing up the resulting distance.

Edge clearance from bolt =1.5D

Hole clearance from the section = 25+0.5DThickness of base plate (for ISMB 200 Column):

Permissible stresses in the base plate, f b = 189 MPaContact pressure between concrete of M25 grade and steel = 7 MPaConsidering bending about major column length:

75

200

350

75


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√ Considering bending about minor column length:

√

So, choose 25 mm thickness base plate for 2-legged trestles.


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REFERENCES

www.google.comwww.Tatasteel.com IS : 875 (Part 3)-1897IS : 800 : 2007SP : 6 steel tables.

Books :

Design of Steel structures – SK Duggal.Design of steel structures – Ramamrutham
http://www.tatasteel.com/http://www.tatasteel.com/http://www.tatasteel.com/

Project Report_Praveen Kumar

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