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  • 1. BIRLA INSTITUTE OF TECHNOLOGY AND SCIENCE, PILANI INSTRUCTION DIVISION FIRST SEMESTER 2012-2013 Course Handout Part II Date: 03/08/2012In addition to part -I (General Handout for all courses appended to the time table) thisportion gives further specific details regarding the course.Course No. : CE C381Course Title : Design of Steel StructuresInstructor-in-charge : MANOJ KUMAR1. Scope and Objective of the CourseThe course intends to impart design skills to common type of Civil Engineering SteelStructures as found in practice as per revised code IS 800: 2007. An understanding ofbasic design concepts, loads and stresses to be used as per Indian standards forsteel design work will be developed. The course deals with designing of steelstructural elements subjected to axial tension, axial Compression, bending, combinedtwisting and bending. Moreover, emphasis will be also given to the special structuressuch as beam-column, trusses, and plate girders. In addition, analysis and design ofvarious types of connections such as bolted and welded will be discussed for use infabrication of tension, compression, and flexural members in the framed structures.All design approaches will be based on Limit State of strengths and serviceability.Furthermore, a special chapter on Plastic design of steel will also be introduced.
  • 2. Text Book S. K. Duggal, “Limit State Design of Steel Structures”, Tata McGraw Hill, New Delhi, 2010Reference Books(i) N. Subramanian, ‘Steel Structures: Design and Practice’, Oxford University Press, New Delhi, 2011.(ii) N. Subramanian, “Design of Steel Structures”, Oxford University Press, New Delhi, 2010.(iii) Teaching Material on Structural Steel Design, by Institute for Steel Development and growth (INSDAG), Calcutta, http://www.steel- insdag.org/new/contents.asp .(iv) IS 800:2007 ‘Code of practice for General construction in steel’ Bureau of Indian Standards, New Delhi(v) IS 875 : 1987 (parts I – IV) “Code of practice for design Loads”, Bureau of Indian Standards.
  • 3. Lectur Learning Topics to be covered Referencee Nos. Objectives Chap. No. of TB3 General Introduction, Advantages & Chapter 1 considerations Disadvantages of steel as structural material, properties of structural steel, rolled steel sections, Loads considered for structural design, basis for design, design philosophies3 Introduction to Bending of beams, Re-distribution of Chapter 2 Plastic Design moments and Reserve of Strength, Shape factor, Load factor, Mechanisms, Plastic Analysis and Design of simple beams and frames3 Introduction to Limit States (LS) design method for Chapter 3 Limit State Steel: LS of strength, LS of Design Serviceability, probabilistic basis for design, design criterion3 Design of Types of connections, Introduction to Chapter 4 bolted and Riveted joints, Design of bolted Pined connections, Design of pin connections connections
  • 4. Lecture Learning Topics to be covered Ref.ChapNos. Objectives No. of TB3 Welded Types of welds and their symbols, Design of Chapter 5 Connections Groove welds, Design of Fillet welds: Fillet weld specifications, Design strength, Design of welds3 Design of Types of tension members, Net sectional area, Chapter 6 Tension net effective area, design strength of tension Members members, slenderness area, design of tension members, lug angles, splices, gusset plate4 Design of Effective length, Slenderness ratio, types of Chapter 7 Compressio compression members, Design strength of n Members compression members, design of axially loaded compression members, Design of built-up columns, design of Lacings and Battens4 Design of Types of beam sections, behavior of beams in Chapter 8 Beams flexure, lateral stability of beams, bending strength of (i) laterally supported and (ii) laterally unsupported beams, shear strength of beams, web buckling, web crippling, deflection, design of rolled beams, design of built-up beams, beam bearing plates
  • 5. Lecture Learning Topics to be covered Ref.ChapNos. Objectives No. of TB3 Members Design of crane members, behavior of beam Chapter 9 subjected columns, design of beam columns to axial load and moment3 Column Types of column bases, design of slab bases, Chapter 10 bases and design of gusset bases, design of bases of caps columns subjected to axial load and moment4 Design of Elements of plate girders, general design Chapter 11 plate considerations, proportioning of web, girders proportioning of flange, flexural strength and shear strength, design of plate girder, design of stiffeners, flange curtailment, design steps2 Gantry Loads, fatigue effects, design of Crane girders Chapter 12 Girders4 Eccentric Beam column connections, Un-stiffened and Chapter 13 connection stiffened connections, Bolted bracket connections, Welded bracket connections
  • 6. S. No Evaluation Duration Weight Date & Time Remarks Component age 1 Mid-Sem Test 90 mts 35 04/10 2:00 – 3:30 PM CB 2 Tutorials 50 mts 20 Every M 8:00 – 8:50 AM OB 3 Comprehensive 3 Hrs 45 01/12 2:00 – 5:00 PM CB Examination5. Make-up Policy No Make-up will be given for Tutorials. Make-up for Mid-Sem Test will be given only for genuine cases if applied in advance.6. Chamber Consultation Hour: To be announced in the class. Students must adhere to the announced timing.7. Notice: Notice if any, concerning this course will be displayed on the Civil Engg. Dept. Notice Board.
  • 7. 5. Make-up Policy No Make-up will be given for Tutorials. Make-up for Mid-Sem Test will be given only for genuine cases if applied in advance.6. Chamber Consultation Hour: To be announced in the class. Students must adhere to the announced timing.7. Notice: Notice if any, concerning this course will be displayed on the Civil Engg. Dept. Notice Board. Instructor-in-charge CE C381
  • 8. • steel structure an assemblage of a group of members (elements)• Members- sustain their share of applied forces and transfer them• safely to the ground.• Depending on the orientation of the member in the structure and its structural use, the member is subjected to forces either• (i) axial, (ii) bending, or (iii) torsion, or a combination thereof.• Axial load  tensile—Tension Members (tie), Compressive – Compressive Members (Strut)• Flexural Force  Beams and girders• Torsion  Shafts (Not discussed here)• Steel members are connected using the rivets, welds, bolts, pins• The connection between steel members Joints• Joints Rigid — can transfer moments) Flexible —can transfer axial loads (shears); Semi-rigid —that fall in between rigid and flexible• Steel structures are used in:• Roof trusses for factories, railway station platforms, cinema halls, auditoriums• Bridges for railways• Crane girders in industry• Water tanks• Telephone Towers
  • 9. Steel Sections:Steel sections are rolled in industry in the standard shapes  called rolled sectionsThe shapes of rolled sections are:Steel I-Sections; Channel Sections; Angle Sections; Tee Sections;Steel Bars; Steel Tubes; Steel Plates;
  • 10. ADVANTAGES OF STEEL AS A STRUCTURAL MATERIALhigh strength per unit weight small section  little self-weight members resists heavy loads  smaller column sections lesser columns in buildings easy to transport  prefabricated members can be used Steel  ductile material No sudden failure Steel  may be bent, hammered, sheared or even the bolt holes may be punched without any visible damage. Steel Properly maintained steel structures have a long life. Steel properties mostly do not change with time Additions and alterations can be made easily Can be erected at a faster rate. Highest scrap value amongst all building materials and can be reused and recycled
  • 11. DISADVANTAGES OF STEEL AS A STRUCTURAL MATERIALWhen placed in exposed conditions corrosion  require frequent painting and maintenanceStrength reduces drastically in fire  Needs fire-proof treatment  Needs additional CostExcellent heat conductor  may transmit enough heat from a burning location to adjoining room.Fatigue  one of the major drawbacksAt stress concentration locations  steel may lose its ductility (tearing of steel)Fatigue at very low temperatures aggravate the situation
  • 12. Stress-Strain Curve for Steel
  • 13. Stress-Strain Curve for SteelPoint A Limit of Proportionality; B elastic limit;C’  upper yield point (strain increase with out increase in stress)  upto C’, stress is elastic  obtained by loading the specimen rapidly  not found in hot rolled steel due to residual stress  no practical significanceC  Lower yield point,  stress at lower yield point  called yield stress, fy.  obtained by loading the specimen slowlyCD  plastic yielding; strain at D = approx 10 times of strain at yield, provided ductilityDE  strain hardening, Presently this strength is not used in designE  Ultimate stress, after this point, section area reduced locally neckingEF Strain softening, stresses reduces in this zone and finally specimen breaks at point F If fractured section makes cup-and-cone arrangement  Ductile failure
  • 14. Notes:• Same curve for tension as well as in compression.• Actual behavior is different and indicates an apparently reduced yield stress in compression.• Divergence from the ideal path is called the Bauchinger effect.• The actual stress-strain curve may be idealized into bilinear or tri-linear form• At High Temperature, curve will be more rounded with no clear yield pointProperties of Structural Steel:Ultimate Strength or Minimum Guaranteed ultimate strength or Engineering ultimate strength Ultimate Tensile Load Ultimate Tensile strength (UTS ) Original area of cross sec tionActual Ultimate Strength Ultimate Tensile Load Actual Ultimate Tensile strength (UTS ) area of cross sec tion at breaking po intSince area of cross-section varies with load, it becomes difficult to measure area atdifferent load stage
  • 15. • Characteristic Ultimate Strength:• The strength below which not more than 5% of samples falls. fk f m ean 1.64 2 f m ean f where , s tan dard deviation n 1
  • 16. Design Strength: In order to incorporate the reduction in strength due to corrosionand accidental damage, the partial safety factor of 1.1 is used. ThusDesign strength of steel = Characteristic strength/partial safety factor (1.1)Ductility: capacity of steel to undergo large inelastic deformation without significantloss of strength or stiffness% Elongation elongated length gauge length % elongation 100; gauge length gauge length 5.65 A0 ; A0 area of cross sec tionToughness: Capacity to absorb energy, measure of fracture resistance underimpact. Area under the stress-strain curve is a measure of toughness
  • 17. Properties of Structural SteelTwo types of steel in India: (i) Standard Structural steel , (ii) Micro-alloyed medium /High strength steelStandard Structural Steel Designated as Fe 410 (IS 2062)Characteristic yield strength for thickness < 20 mm  250 MPa for thickness 20-40 mm  240 MPa for thickness > 40 mm  230 MPaAvailable in three grades Grade A  used for structures subjected to normal conditions Grade B  used for situations where severe fluctuations are there but temp > 00 C Grade C  may be used upto – 400 C and have high impact propertiesModulus of Elasticity (E) = 2 105 N/mm2Shear Modulus (G) = 0.769 105 N/mm2Poission’s Ratio ( ) in elastic range = 0.3 in plastic range = 0.5Coefficient of Thermal Expansion = 12 10-6 / 0CClassification Based on manufacturing process, Two types of Section: (i) Cold Formed Sections, (ii) Hot rolled SectionsCold formed sections: produced by steel strips (thickness < 8mm )  Light in weight  used for smaller loads where hot rolled becomes un-economicalHot Rolled Sections  Simply called as Rolled Sections  more commonly used as structural steel
  • 18. Rolled Sections• Sections produced by hot rolling process in rolling mill,• Due to hot-rolling  no loss of ductility• In India available standard Sections as per IS 808:1989 are:• Indian Standard Junior Beams (ISJB)• Indian Standard Light-weight Beams (ISLB)• Indian Standard Medium weight Beams (ISMB)• Indian Standard Wide-Flange Beams (ISWB)• Indian Standard Heavy weight Beams (ISHB)• Indian Standard Column-Sections (ISSC)• Indian Standard Junior Channels (ISJC)• Indian Standard Junior Channels (ISJC)• Indian Standard Light weight Channels (ISLC)• Indian Standard Medium-weight Channels (ISMC)• Indian Standard Angles (ISA)• Indian Standard Normal Tee-Sections (ISNT)• Indian Standard Deep-Legged Tee-Sections (ISDT)• Indian Standard Light weight Tee-Sections (ISLT)• Indian Standard Medium-weight Tee-Sections (ISMT)
  • 19. LOADS(1i Dead Load; (ii) Live loads; (iii) Environmental loadsDEAD LOADS (IS 875: Part I):(i) Due to gravity: acts in the direction of gravity (due to self weight) DL not known before design, so initially is assumed and later on is checked (ii) Superimposed loads: permanent loads (such as partition walls)LIVE LOADS (IS 875: Part IV):Loads which may change in position and magnitude (Furniture, equipments, occupants)Also some reduction in live loads are made for residential buildings.No. of floors carried by member % reduction of Live load on all floors aboveunder consideration the member under consideration 1 0 2 10 3 20 4 30 5 to 10 40 Over 10 50
  • 20. Impact Load:• When a load is applied suddenly or load is in motion,• Used for Lifts and Industrial buildings Frames supporting lifts and hoists 100 Foundations, footings, piers supporting lifts and hoisting apparatus 40 Light machinery shaft motor units 20 Reciprocating machinery or power units 50 Installed machinery 20Earth pressure: Used for Underground structures such as basement, retaining wallsWater current force: For piers and abutmentsThermal Loads: due to Temperature variations, may be up to 25% of LL in bridgesand TrussesEnvironmental Loads: Wind LoadsPressure, suction, upliftMore for tall structuresDesign wind pressure at height ‘z’ above mean ground level, P = 0.6Vz2 (N/m2)Vz = Vb k1 k2 k3k1  probability or risk factor, k2  terrine (height) factor, k3 topography factorVb = basic design speed, increases with height (constant up 10 m from MGL)Wind Force: F = Cf Ae pz where Cf = force coefficient for buildings depends on shape of structure
  • 21. BASIS FOR DESIGNSteel Structure are designed for• Integrity: Its constituent parts should constitute a stable and robust structure under normal loading and Columns must be anchored in two directions at right angles• Stability: Remain fit with adequate reliability and are able to sustain all actions• Durability: Not seriously damaged (collapse) under accidental events : Sway resistance is distributed throughout the buildingMETHODS OF ANALYSISWorking Stress Method:• An elastic method of design material behavior elastic• Based on concept that maximum probable stresses due to applied loads ≤ permissible stresses• Permissible stress = Material strength (yield strength) / Factor of safety• Factor of safety is used (only to strength) to make structure safe• Factor of Safety accounts: – Overloading under certain circumstances – Secondary stresses due to fabrication, erection and thermal – Stress concentrations – Unpredictable natural calamities• Disadvantage: (i) only limited material capacity is used, (ii) no check at overloading
  • 22. Plastic Method• Steel  ductile material major portion of curve lies beyond the elastic limit (from the stress–strain curve)• higher strength after elastic limit  called reserve strength  used in plastic design method• Plastic method  based on failure conditions rather than working load conditions• Structure designed for collapse load rather than elastic loads  excessive deformations at collapse• Design Load in plastic design (Collapse Load) = working loads load factor• At plastic stage Plastic Hinge formation  infinite rotation  More P Hinges  Collapse of structure• Since actual loads are less than design loads  Structure do not collapses• Disadvantage: Design at ultimate loads only , no check for serviceabilityLimit State Method• also known as load and resistance factor method• overcomes the drawbacks of (i) Working stress and (ii) Plastic Design• Limit State a state beyond which the structure is unable to function satisfactorily• Two major categories of limit states:• limit state of strength load carrying capacity (plastic strength, fracture, buckling, fatigue)• limit state of serviceability  perform. of str at service load (deflection., durability, vibrations, fire, etc)
  • 23. Limit states of strength:• associated with failures (or imminent failure), under the action of probable and most unfavorable combination of loads on the structure using the appropriate partial safety factors, which may endanger the safety of life and property.It include:• Loss of equilibrium of the structure as a whole or any of its parts or components• Loss of stability of the structure (including the effect of sway where appropriate and overturning) or any of its parts including supports and foundations.• Failure by excessive deformation, rupture of the structure or any of its parts or components,• Fracture due to fatigue,• Brittle fracture.Limit state of serviceability:• Deformation and deflections, which may adversely affect the appearance or effective use of the structure or may cause improper functioning of equipment or services or may cause damages to finishes and non-structural members.It includes:• Vibrations in the structure or any of its components causing discomfort to people, damages to the structure, its contents or which may limit its functional effectiveness. Special consideration shall be given to systems susceptible to vibration, such as large open floor areas free of partitions to ensure that such vibrations are acceptable for the intended use and occupancy (see Annex C).• Repairable damage or crack due to fatigue.• Corrosion and durability,• Fire.
  • 24. Objective of LSM design• structure not to become unfit for use with an acceptable target reliability OR• Very low probability to reach structure to LS during its lifetimeIn Limit state design method• Loads Characteristic loads (loads with small probability exceeding this value)• Strength  Characteristic strength (strength with small probability less than this value)• Checks are made for serviceability• However, in the LSDM, Structures are designed using the ‘Design Strength’ and ‘Design Load’.• Design Load = Partial safety factors for load ( f) Charact. load• Design Strength = Charact. strength / Partial safety factors for material ( m)Partial Safety Factor for Loads:• Partial safety factors are different for different load combinations:• DL + LL / CL case for DL 1.5 for LL/CL 1.5• DL + LL/CL + WL/EL case for DL 1.2 for LL/CL 1.2 for WL/EL 1.2• DL + WL/EL case for DL 1.5 (0.9) ------ for WL/EQ 1.5• DL + ER case for DL 1.2 (0.9) for ER 1.2
  • 25. Design Strength• Design Strength, Sd, is obtained as given below from ultimate strength, Su and partial safety factors for materials, m given in Table 5 of code. Sd=Su/ mwhere partial safety factor for materials, m account for:• Possibility of unfavorable deviation of material strength from the characteristic value,• Possibility of unfavorable variation of member sizes,• Possibility of unfavorable reduction in member strength due to fabrication and tolerances, and• Uncertainty in the calculation of strength of the members.
  • 26. Introduction to riveted connections:• Riveted connections are obsolete• Rivet made up of round ductile material round bar called ‘shank’• Classified based on head shape: ‘snap (common)’, ‘pan’, ‘flat countersunk’, ‘round countersunk’• Diameter of shank  nominal diameter• Two types Hot Driven and Cold Driven also shop rivets and Field RivetsHot Driven Rivet• Rivets are first heated  increase in diameter  heated diameter > shank diameter  gross diameter• On cooling length of rivet reduces  joint becomes tighter,  Diameter of rivet reduces  some space remains between rivet and holeCold Driven Rivet• Needs high pressure• Strength of cold driven rivets > hot driven rivetsRivet Pattern:• Chain Pattern; Staggered Pattern; Diamond Pattern; Staggered DiamondDesign of Riveted Sections are Same as Bolted Connections, with following differences• In Case of Rivets, Diameter of Rivet = Diameter of Hole,• in case of Bolt , Diameter of bolt = nominal diameter of bolt• Design Stress of rivet > Design Stress of bolts
  • 27. BOLTED CONNECTION• Consists of bolt (shank with a head at one end and threaded at other end), nuts and washers.• Washers are used to: – Distribute the clamping pressure on bolted member, – to prevent the threaded portion• If the section is subjected to vibrations  nuts are lockedAdvantages of Connections over Riveted connections:• Speedup erection of structure,• Needs less skilled persons• Overall cost of bolts less as compared to Rivet due to Reduction in labour and equipment costObjections on use of bolts:• Cost of bolts > cost of rivet material• Tensile strength of bolt < tensile strength of rivet (due reduction in area of cross-section at root of thread)• May loose due to vibrations and shocks
  • 28. Methods of making Holes for bolts:(1) Drilling ,(2) Punching simple: – saves time and cost but reduction in ductility and toughness – As per IS:800-2007permits punching, only when • material yield stress < 360 MPA, • thickness < 5600/fy mm • If punching is to be used, holes are punched 2 mm less than required and 2 mm is drilledTYPES OF BOLTED CONNECTIONS:Classification On the basis of Resultant Force Transferred• Concentric when load passes through CG of section (in case of axial loads)• Eccentric  load is away from CG of connection (such as in channels)• Moment resisting joint subjected to moments (beam-column connection)• Classification On the basis of Type of Force:• Shear connection  Load is transferred through shear (lap joint, butt joint)• Tension connections  load transfer by tension on bolts (hanger connections)• Combined shear and tension connection (inclined member connected to a bracket) (bracing connections)
  • 29. Types of Bolts:(1) Unfinished Bolts; (2)High Strength boltsUnfinished bolts• Also known as Ordinary, rough or black bolts• Used for connecting light structures for static and secondary members (purlins, bracings, trusses etc)• Not suitable for vibrations and fatigue• Made from mild steel rods forged from low carbon steel• Heads are made square or hexagonal (costly but better appearance)• Available in 5 mm to 36 mm diameter designated as M5 to M36• In structural steel generally bolts used are M16, M20, M24 and M30• Ratio of net tensile area / nominal plane shank = 0.78 (as per IS 1367)• In IS 800: above ratio is taken as 1.0• As per IS 800  net tensile area is considered at root called stress area or proof area• Bolts are designated by i.g. 4.6  Ultimate stress 400 MPa and yield stress 0.6x400 = 240 MPa• Not significant clamping stress is developed• Force is transferred through the interlocking and bearing  bearing type joint
  • 30. High Strength Bolts• Made from medium carbon heat-treated steel and from alloy steel• Tightened until they have very high tensile stress (twice ordinary bolts)• Load resisted at high stress  called proof load• Connected parts are clamped together• Loads are transferred primarily by friction not by shear  called friction bolts• to avoid slip surfaces to be connected must be free from rust, paint, grease• Better vibration and impact resistance• Available in 5 mm to 36 mm diameter designated as M5 to M36• Commonly used grades of bolts are 8.8S, 10.9S (written at cap of bolt) – 8.8S (diameter < 16 mm) ult. Stress 800 MPa yield stress 640 MPa – 8.8S (diameter > 16 mm) ult. Stress 830 MPa yield stress 660 MPa – 10.9S ult. Stress 1040 MPa yield stress 940 MPaAdvantages of High strength bolts:• No slip between elements connected rigid joint high strength of connection• No shearing and bearing failure• No stress concentration in the hole  more fatigue strength• Uniform tensile stress in bolt• No loosening in bolts• Less man power (compared to rivets)  cost saving• Less Noise nuisance• Less number of bolts are required as compared to rivets
  • 31. Types of bolted joints Two types: (i) Lap Joint, (ii) Butt JointLap Joints• Two members are overlapped and connected together• May be single bolted or two bolted joint• Loading axis of members do not match  Load is eccentric  uneven stress distribution• Bending of joint  couple is formed  bolt may fail in tension  at least two bolts must be usedButt Joint• Members are placed end to end• Cover plates are provided in two ways  (i) single cover plate butt jt. (ii) Double cover plate butt jt.• Double cover plate joints better than lap joint since: – SF in double cover plate = (1/2) of SF in lap joint – Shear strength of double cover plate = 2 x Shear strength of Lap joint – In double cover plate joint  no bending
  • 32. Failure of Bolted Joints:Six-types of failures:(i) Shear failure of bolt,• Occurs when shear stress in bolt > Nominal shear stressShear failure : Two types  (i) Single shear Failure  shear failure at one section of bolt  occurs in case of Lap joint, (ii) Double Shear Failure  shear failure at two sections of bolt  occurs in case of Butt joint(ii)) Bearing Failure of bolt and (iii) Bearing Failure of Plate• In general, transfer of force in connecting parts  through bearing action• half circumference of plate in contact with bolt get crushed (plate weaker than bolt)• half circumference of bolt in contact with plate get crushed (bolt weaker than plate)• or partially both plate and bolt are get crushed(iv) Tension Failure of bolt• If bolt in Tension and Tensile stress in bolt > Permissible stress Tension Failure at root of thread (weak)
  • 33. (v) Tension or Tearing Rupture failure of plates• In plates, holes in plate for connection  Reduction in net effective area of plate  Tension Failure of Plate• Tension Failure of Plate may be prevented by  (i) fewer holes, (ii) staggered holes• Plate breaks along the bolt line(vi) Shear Failure of Plate• Due to insufficient end distance (distance from end of plate from center of nearest hole measured along force direction)• portion of plate of width equal to diameter of hole is sheared• to prevent shear failure  provide enough end distance(vi) Block Failure • A combination of shear failure and Tension Failure• A portion of plate (block) shears along the force direction
  • 34. Pitch: C/C distance between individual fasteners (bolts) in a line/ rows (measured parallel to load/stress)- p• If bolts are in zig-zag pattern  distance measured parallel to direction of load/stress  staggered pitch- ps Line/row Edge distance p End distanceGauge: distance between adjacent gauge (bolt) lines (measured perpendicular to force)Limitations on Pitch:• Minimum Pitch:• Pitch 2.5 nominal diameter of bolt  called Minimum pitch• Pitch minimum pitch due to following reasons:• To prevent bearing failure between two bolts• for sufficient space to tight the bolts• to avoid overlapping of washers• to avoid tear-out of plate (between bolts in a rough
  • 35. Maximum Pitch:• In Tension Members: Pitch 16t or 200 mm, whichever is less (where t = thickness of thinner plate)• In compression members: Pitch 12t or 200 mm, whichever is less, (where t = thickness of thinner plate)• In compression members, where forces are transmitted through the butting facing:• Pitch 4.5d for a distance of 1.5 b from the butting faces, where b= width of member• Pitch for edge row of the outside plate (100 mm + 4t) or 200 mm, whichever is less• If bolts are staggered at equal interval and gauge 75 mm,  pitch for tension and compression members may be increased by 50% provided Pitch 32t or 300 mmPitch Maximum Pitch due to following reasons:• To reduce the length of connection and gusset plateto have compact joint• For Long joints (> 15 diameter of bolt)  end bolts are stressed more  progressive joint failure – called unbuttoning• Unbuttoning is controlled by limiting the maximum pitch• In case of built-up compression member joints  buckling of cover plates• In case of built-up Tension Member joint  connected plates apt (tends) to gap apart (from cover plate in transverse direction)
  • 36. • Edge Distance:• Distance from center of any (extreme) bolt hole to edge of plate (measured perp. to load direction)• End Distance:• C/C distance of bolt holes to the edge of an element (measured along load direction)• Minimum Edge Distance and Minimum end distance (book table is given for up to 32 mm dia) – 1.7 hole diameter  in case of sheared or hand-flame cut edges (uneven) – 1.5 hole diameter  in case of rolled, machine-flame cut, sawn(saw) & planed (plane) edgesIf Edge distance < mini. Edge distance and End distance < mini. end distance:• Plate may fail in tension• Steel of plate opposite the hole may bulge out (in direction perp to load)  crackHole Diameter:• Hole diameter = Nominal diameter of bolt + clearance• For nominal diameter from 12 mm to 14 mm  clearance = 1. 0 mm• For nominal diameter from 16 mm to 24 mm  clearance = 2. 0 mm• For nominal diameter > 24 mm  clearance = 3. 0 mm
  • 37. Maximum Edge Distance• Edge distance to the nearest line of fasteners from an edge of any un- stiffened part 12 t , where 250 / f y and t = thickness of the thinner outer plate.• Above is valid for fasteners interconnecting the components of back to back tension members.• Where members are exposed to corrosive influences, edge distance (40 mm+4t), (t=thick. of thinner plate)If Edge distance > maximum edge distance• Edges may separate  Moisture may reach between parts  Corrosion problem in joint
  • 38. Bearing Type Connections (in case of un-finished or ordinary bolts):• Load transferred > friction resistance  bearing action• Bolts in Bearing type connection are checked for (i) shear, and (ii) bearing Load to betransferred No. of bolts required Strength of one boltSince single bolt may fail and result in collapse  Minimum no. of = 2 or 3Strength BoltStrength of bolt = Minimum of (i) Strength of bolt in bearing, and (ii) Strength of bolt in shearingStrength of bolt connection = strength of one bolt no. of boltsStrength of joint = Minimum of (i) strength of bolt or bolt group, and (ii) net tensile Strength of plateNote: Bearing plane is considered in the threaded portion for safe design (since bolts may be put in both ways)
  • 39. Determination of Shearing Strength of Bolt:Shear strength of bolt depends on (i) Ult. tensile strength of bolt, fub(ii) No. of shear planes with threads, nn(iii) No. of shear planes without threads (shanks), ns(iv) Nominal area of shank, Asb,(v) Net stress area of bolt Anb fShear capacity of bolt, V ub n A n A nsb 3 n nb s sbReduction factor in shear for Long JointsIf the length of joint > 15d  Long SectionIf the section is long  Stress in outer bolts > inner bolts  Need to apply Reduction factor ( ij) ij accounts for overloading of the end bolts lj βlj 1.075 - but 0.75 β lj 1.0 200 dWhere, lj = length of joint = distance between first and last row of bolts measured in direction of loadFor uniform stress section (i.e. all bolts carry equal stress)  ij = 1
  • 40. Reduction factor in shear for Large grip length (i.e. more thickness of plates)more thickness of plates more grip length of bolt  More Bending Moment in SectionFor the safe design  Need to apply a reduction factor for large grip length ( lg)If total thickness of the connected plates > 5 nominal diameter of the bolt more grip length  Use lg 8d lg 8d and lg lj 3d lgReduction factor for packing PlatesIf packing plate thickness > 6 mm  bending is developed in shankNeed to apply a reduction factor in shear capacity ( pkg) pkg 1 0.0125 t pkg where tpkg = thickness of packing plateNominal Strength of JointThe nominal shear strength of the bolt taking in to account reduction factors f V ub n A n A lj lg pkg nsb 3 n nb s sb
  • 41. Factor of Safety of material ( mb)For safety of joint in shear, a factor of safety for material is used ( mb) Vnsb f V ub n A n A lj lg pkg sb mb 3 m b n nb s sb where mb = partial safety factor of material of bolt = 1.25 For 4.6 grade bolt, fsb = 400 MPa 400 V n A n A lj lg pkg sb 3 1.25 n nb s sb V 184.75 n A n A sb n nb s sb lj lg pkg Vsb 184.75 Ae lj lg pkgWhere , Ae n A n A for bolts in sin gle shear n nb s sb 2 n A n A for bolts in Double shear n nb s sbFor normal bolts and members  threads are excluded,if considered  very conservative design
  • 42. Bearing Strength (Capacity) of boltDue to Bearing  hole elongates Due to Excessive bearing  tearing of plateTo avoiding excessive elongation of hole  Bearing stress not greater than Nominal bearing strength of boltNominal bearing strength of bolt = projected bearing area ultimate Tensile stress Vnpb 2.5 k b d t f u e p f ub Where, k b smaller of , 0.25 , and 1.0 3d 0 3d 0 fud = nominal diameter of bolt;d0 = diameter of holep = pitch of the bolt (along bearing direction);e = end distance of the bolt (along bearing direction)fub = ult tensile stress of bolt;fu = ult. Tensile stress of platet = total thickness of the connected plates subjected to bearing stressing in the same directionin case of countersunk bolts  t = plate thickness – (1/2) depth of counter sunkFor safety of joint: need to use a Factor of safety, mb, mb = 1.25 Vnpb fu V pb V pb 2.5 kb d .t. mb mb
  • 43. Tensile Strength of Plate Net Area, An B n dh t ( for chain bolting) m 2 psi Net Area, An B nd h t ( for staggered bolting ) i 1 4 gi fu Tensile strength of plate, Tnd 0.9 An m1 Where, fu = ultimate stress in MPa; An = net effective area in mm2 m1 = partial safety factor = 1.25 Strength and Efficiency of Joint Strength of bolted joint = minimum of (strength of connection based on (i) shear, (ii) bearing , (iii) strength of main member) Strength of bolted jo int per pitch lengthEfficiency of connection , 100 Strength of solid plate without deductions for holes per pitch length
  • 44. Tension Capacity of bolt If the bolts are subjected to tension  Tensile stress in bolt Tensile strength of bolt Tensile of bolt is checked in (i) Shank zone as well as in (ii) threaded zone Tb 0.9 f ub A nb in threaded zone Tnb 0.9 f ub b A nbFor bolt to be safe in tension, Tension force in bolt, Tdb Tdb mb mb mbAlso the bearing strength is to be checked shank zone Tdb f yb A sb in shank zone m0 Where, fub = ult tensile stress of bolt; fyb = yield stress of bolt Anb = net tensile stress area of bolt; Asb = shank area of bolt mb = partial safety factor for bolt material = 1.25 m0 = partial safety factor for bolt material governed by yielding= 1.10
  • 45. Bolt subjected to combined shear and tension 2 2 Vb Tb 1.0 Vdb Tdb Vb = factored shear force on bolt; Tb = factored Tensile force on bolt; Vdb = design shear capacity Tdb = design Tension capacity
  • 46. Example: Two plates are connected by single bolted double cover butt joint using M20bolts at 60 mm pitch; steel grade 410 MPa and bolts 4.6 grade; Calculate bolt efficiency Assume bolt pitch = 60 mm End distance = 30 mm Strength of M20 bolt in shear For M20 bolts  Diameter of bolt = 20 mm  Diameter of bolt hole = 22 mm For Fe 410 Grade steel Net Tensile Stress area = 245 mm2 (from Table 4.3 of TB) For Fe 410 Grade steel  ult strength fu = 410 MPa For 4.6 grade bolt  ult strength of bolt, fub = 400 MPa Partial safety factor for bolt material = 1.25 fStrength of bolt in shear V ub n A n A lj lg pkg nsb n nb s sb mb 3Since, there is only one line of bolts  No question of overloading  lj = 1Since thickness of plates to be connected < 5 nominal dai of bolt i.e.5 20 mm lg = 1Since, size of connecting plates is equal  no packing plate   pkg = 1 f Strength of bolt in shear V ub n A n A nsb 3 n nb s sb mb
  • 47. f Strength of bolt in shear V ub n A n A sb n nb s sb mb 3 Due to double cover butt joint  The bolt will be in double shear, Assuming, both the shear planes in net area section  nn = 2, ns = 0, Anb = 157 mm2 f f f fV ub n A n A ub 2 A 0. A ub 2 A 0. A 2A ub sb 3 n nb s sb nb sb nb sb nb mb 3 mb 3 mb 3 400V 2 245 90.53 kN sb 1.25 3 2.5 k b d t f u Strength of M20 bolt in bearing Vnpb mb e p fub Where, k b smaller of , 0.25 , and 1.0 3d0 3d0 fu d = nominal diameter of bolt = 20 mm ; d0 = diameter of hole = 22 mm p = pitch of the bolt = 60 mm; e = end distance of the bolt = 30 mm fub = ult tensile stress of bolt = 400 MPa; fu = ult. Tensile stress of plate = 410 MPa t = minimum of (thickness of the connected plates, Sum of cover plates) 6 mm 30 60 400Where, k b smaller of , 0.25 , and 1.0 3 22 3 22 410 Smaller of 0.454, 0.659, 0.975, 1.0 0.454
  • 48. 2.5 k b d t f u 2.5 0.454 20 6 400 Vnpb 43.58 kN mb 1.25 Strength of bolt = Minimum of strength bolt in shear and in bearing = minimum of (90.53 kN and 43.58 kN) = 43.58 kN Determination of Strength joint per pitch length Strength of plate per pitch length in Tension fu fu Tensile strength of plate, Tnd 0.9 An 0.9 p d t m1 m1 410 0.9 60 22 6 67.3 kN 1.25 Strength of joint per pitch length = Minimum of (strength bolt, strength of plate) = minimum of (43.58kN, 67.3 kN) = 43.58 kN Strength of solid plate per pitch length in Tension fu 410Tensile strength of plate, Tnd 0.9 Ap 0.9 60 6 106.27 kN ( m1 part saf fact ) m1 1.25 Strength of joint per pitch length, 43.58Efficiency of join 41% Strength of solid plate per pitch length 106.27
  • 49. Example: In previous example, find the efficiency of joint if joint have two lines Shear strength of bolts = 2 90.53 kN = 181.06 kN Bearing strength of bolts = 2 43.58 kN = 87.16 kN Strength of plate = same as in one line case = 67.3 kN Strength of joint = Minimum of above three = 67.3 kN Strength of solid plate = 106.27 kN 67.3 Efficiency of join 63.3% 106.27
  • 50. Slip Critical ConnectionsAt service loads  load < Friction resistance  No significant slipLarge Member force or when connection length limited  slip  bearing actionTo avoid bearing action  High Strength Friction Grip (HSFG) bolts most suitable Joint with HSFG bolts  called slip resistant connectionAt service loads No slip  behave as Slip resistant connectionAt ultimate loads  Slip develops  Bearing  Bearing type connectionAs the load go on increasing  Slip occurs at particular load called critical load Connection is called slip critical ConnectionIn Bridges  Loading & unloading  fatigue Slip critical connections not good use slip resistant connectionTheoretically, at service loads  In slip critical connections  No Slip  Load transferby Friction only (i.e. no shear, no bearing)In practice,  Slip occurs  Slip Critical connections are designed for bearing alsoPrincipal of HSFG BoltsBolts dia > Bolt dia  bolt not fills complete holeNo bearing and shearing at loads < slip critical loadsSince, In the tightened to a very high load  90% of proof loadBolts  Initially load is transferred by  ) Friction action (load<friction resis)When Load > Friction resistance  Load is transferred by (i) shear and (ii) bearing actionsGap between shank and hole  At No Slip stage shearing load transfer by Friction onlyFriction force developed depends (i) Tension in bolt and (ii) Friction Coeff. between plate and nut/washer
  • 51. Design of HSFG Bolts Horizontal Friction Resistance , F fT Where,T Tensionin Bolt and f slip factor Types of HSFG bolt connections: HSFG Parallel Shank bolts  no slip at service loads but may slip at ultimate load  Slip critical connections HSFG Waisted Shank bolts  No slip at service as well as at ultimate loads  Slip resistant connections
  • 52. Shear Connections with HSFG Bolts Slip Resistance , Vnsf µf n e K h Fone = number of effective interfaces offering frictional resistance to slipKh = 1.0 for fasteners in clearance holes (gap is there) = 0.85 for fasteners in oversized and short slotted holes and for fasteners in the long slotted holes loaded perpendicular to slot = 0.7 for fasteners in long slotted holes loaded parallel to the slot.Fo = minimum bolt tension (proof load) at installation ( = Anb fo) Anb = net area of bolt at the threads; fo = proof stress (= 0.70 fub), where fub = ultimate stress of bolt. Vnsf µf n e K h FoDesign Slip Resistance , Vdf mf mf γmf = 1.10 (if slip resistance is designed at service load) γmf = 1.25 (if slip resistance is designed at ultimate load)Correction For Long JointThe above determined load is to be multiplied by long joint correction factor ljβ lj 1.075 - but 0.75 β lj 1.0 200 d d = nominal shank diameter of bolt
  • 53. Bearing Strength of HSFG (High-strength Shear-critical Friction Grip)At ultimate load  slip takes place  Need to check the strength of connectiondue to bearing at ult. Load In the same way as in case of black bolts.Tensile Strength of HSFG BoltsSame as for black bolts denoted as TnfCombined Shear and Tension For Slip Critical Connections 2 2 Vf Tf 1.0 Vdf Tdf Vf = applied factored shear force at design load; Tf = externally applied factored Tensile force at design load Vdf = design shear strength Tdf = design Tensile strength
  • 54. Prying Action When a channel section, with flexible flanges (in tension), is connected to roof  flange bends  the CG of the compressive stress (prestress) in the plate shifts towards the end  additional force in bolt called Prying force l f 0 be t 4 Te lvPrying force is given as, Q v Te ( neglecting Second term in braket ) 2le 27 le lv2 2 le
  • 55. l f 0 be t 4 Te lvPrying force is given as, Q v Te ( neglecting Second term in braket ) 2le 27 le lv2 2 leWhere, lv = distance from bolt center-line of bolt to the (i) toe of the fillet weld or (ii) half the root radius for the rolled section (Fig)Le = distance between prying force and bolt center-line which is taken minimum of f0 (i) end distance and (ii) 1.1 t fy = 1 for pre-tensioned bolts, = 2 for non-pre-tensioned bolts, = 1.5 for Limit State Design; be = effective width of flange; f0 = proof stress; t = thickness of end plate,To minimise the Prying Force:• Use fully tensioned bolts (i.e. applied tensile force in the plate pre-stress in the plate)• Use thick plate or stiffened the plate• Limit the distance between bolt and plate edgeDesign of Connection Subjected to Prying ForceBased on Trial and error  Since the prying force depends on section thicknessand no. of bolts (controls the edge distance)Based on plastic analysis, the thickness of the end plate may be determined as 4.40 M p t min f y be
  • 56. PIN CONNECTIONSProvided where rotations are allowedFor satisfactory function  minimum friction is requiredPinned connections make the structure determinateLarge force  since only one pine is provided rather than more as in case of boltsPins are available in diameter 9 m to 330 mmPins application:(i) Tie Rod (ii) Diagonal bracings in beams and columns (iii) TrussStrength of Pined ConnectionsShear Capacity:(a) If no rotation is required and the pin is not intended to be removed; Shear capacity = 0.6 fyp A(b) If rotation is required or the pin is intended to be removed; Shear capacity = 0.5 fyp AWhere, fyp = design strength of pin, A = Cross-sectional area of pin
  • 57. Bearing Capacity(a) If no rotation is required and the pin is not intended to be removed; Bearing capacity = 1.5 fy d t(b) If rotation is required or the pin is intended to be removed; Bearing capacity = 0.8 fy d tWhere, fy = lower of design strength of pin and connected part, d = diameter of pin; t = thickness of plateFlexural Capacity of PinThere is gap between the connected members due to following reasons: (i) To prevent friction (ii) To allow bolt heads if built-up connection (iii) To facilitate painting(a) If no rotation is required and the pin is not intended to be removed; Moment capacity = 1.5 fyp Z(b) If rotation is required or the pin is intended to be removed; Moment capacity = 0.8 fyp ZWhere Z = Section modulus of pinNote: Flexure is more critical  Pin diameter is generally governed by Flexure 13 3 d MuIf pin is cylindrica l, M u 1.5 f yp Z M u 1.5 f yp d 21.33 32 f yp
  • 58. WELDED CONNECTIONStwo pieces of metal connected by heating them to a plastic or fluid state called fusionWelding process  two types  Electric welding, (ii) Gas WeldingAssumptions in Welded JointsWelded are homogeneous, isotropic and elasticWelds are rigid and no deformation with-in weldsNo residual stresses in welds (due welding process
  • 59. Advantages of welded connections over bolted connections:• No deductions for hole  gross section is effective  More efficient use of material•small size of gusset plates  more compact joint• Economical due to saving in time in preparing drawings and fabrication• Fast speed of fabrication and erection• No connecting plates  Less weight of structure  economical• Better for fatigue, impact and vibrations (earlier it was assumed not good in fatigue)• Produces rigid connection  produce one piece Construction  less deformation• requires less depth of beam reduced overall ht of building• Less noise pollution• Watertight / airtight connections  good for liquid/gas storage tanks• Avoids problem of hole alignment• Needs power supply at siteDisadvantages of welded Joints:• Needs skilled labour• Needs costly equipments• Difficult to inspect the joints  needs NDT testing methods such as Magnetic particle method, dye penetration method, ultrasonic method, radiography• Welded joints if over rigid (than members) may fail in fatigue  cracking in members
  • 60. Types of Welded ConnectionsWelds classification based weld procedure (i) Fillet weld, (ii) butt or groove weld (iii) plug weld (iv) slot weldclassification based on its location (i) flat weld; (ii) horizontal weld (iii) Vertical weld (iv) overhead weldWelds classification based on type of joint (i) butt welded joint (ii) lap welded joint (iii) Tee Welded joint (iv) corner welded
  • 61. Butt Joints•Used when members to be joined are in a line or aligned in the same plane•Needs edge preparation costly and time consuming•The grooves have a slope of 30 -600•If two plates are of different thicknesses  thicker plate is made thin near joint Butt welding of parts of unequal thickness and/or unequal width Reinforcement: •makes the butt joint stronger under static load •Smoothen the flow of forces •concentration develops in case of fatigue loads  leading to cracking and early. •Not greater than 0.75 mm to 3 mm Extra reinforcement removed by machine
  • 62. Types of Butt (groove) Welds:(i) Single and double square (ii) Single and double V (iii & iv) Single anddouble Bevel (v & vi) Single and double U (vii & viii) single and double JIncomplete penetration welds :- single V, single bevel,Incomplete penetration  stress concentration  Complete penetration is better thanincomplete oneSquare welds: Easy but are used for plate up to 8 mm thickness only
  • 63. Fillet Joints:•used to joint two members in different planes  lap joints•Easy to make•Needs less material preparation•High stress concentration•For given amount of weld  flat welds are poor than butt welds•flat welds more common than butt (groove ) weldsTypes of Fillet WeldsTriangular in shape: when two perpendicular members are connected and in lap jointsMay be (i) Concave or (ii) Convex, (iii) Mitre Lap Joints Advantage of lap joints plates with different thicknesses can be joined Drawback of a lap joint  introduces some eccentricity of loads (May be avoided if double lap joint is used)
  • 64. Welded Lap Joints T-Joints
  • 65. Defects in Welds
  • 66. DESIGN OF BUTT JOINTSForces in butt joints:- (i) Axial:  Tension or Compression), and (ii) shear if anyDesign SpecificationsReinforcement:Extra weld metal (above the plate level)Reinforcement is required: to avoid error in thickness of weld to increase in static load capacity  increase in efficiency of jointReinforcement thickness  at least 10% of plate thickness but not more than 3 mm Reinforcement is made at the time of welding  later they are dressed flush Reinforcement is ignored in calculationReinforcement is not provided in case of vibrations and impact  to avoid failure due to stress concentrationSize of Groove weldSize of groove weld = size of throat  called effective throat thickness (te)In case of complete penetration of the groove weld  Throat dimensions = thickness of thinner member Effective Throat thickness = 7/8 of thickness of thinner member In calculation, Effective Throat thickness = 5/8 of thickness of thinner memberEffective area of weld: Effective area of weld = Throat thickness (te) effective length of weld (Lw) (measured along width)Effective length of weld = length over which required size of weld is done
  • 67. Design StrengthDesign Strength of groove weld in tension/compression f y Lw te Tdw mwfy = smaller of ULTIMATE stress of weld (fyw) and parent material (fy)Lw = effective length of weld in mmte = effective throat thickness of throat in mm mw = partial safety factor = 1.25 for shop weld and 1.5 for site weldDesign Strength of groove weld in shear f yw1 Lw te Vdw mwfyw1 = smaller of yield stress of weld (fyw/ 3) and parent material (fy/ 3)Design Procedure:In Case of Complete penetration  weld strength = member strength  nocalculation requiredIn case of incomplete penetration  determine throat thickness  calculate lengthrequired to develop strength of weld equal to member strength
  • 68. Design of Fillet WeldUsed when two members overlap each otherStresses developed  (i) Direct stress –Minor (ii) Shear stress  MajorTwo more widely used fillet weld shapes  concave and convex (mitre not common)Concave weld less penetration than convex smaller throat (than convex) on cooling  outer face in tension  cracks  Not GoodConvex welds  More penetration  Large throat  convex weld stronger  on cooling  Compression in outer face due to shrinkage  GoodNote: If concave welds are desired  in first pass they are made convex and in second concave
  • 69. Size of Fillet weldsWeld size = minimum leg length of weldLeg length = distance from the root to the toe of fillet weldLeg lengths are measured along the largest right angle triangle inscribed within weldThroat size  perpendicular distanceSize of weld  (i) equal and (ii) unequalEqual is preferredIn some circumstances  unequal is used to increase the throat size (hencestrength) (in butt angle connected to plate where thickness of weld on angle islimited)Weld Size SpecificationsThicker plates  Heat dissipation in horizontally as well as vertically  Lesser fusion depth due to high heat dissipation  Need to limit the maximum thickness of plateThin Plates: Heat dissipation mostly along horizontally  Lesser depth may be sufficient
  • 70. Maximum Size of weldTo avoid the melting of base materialIf plate thickness < 6 mm  Max Size of weld = Thickness o plateIf plate thickness > 6 mm  Max size of weld (Thick. of thinner member – 1.5 mm) (to avoid over stressing of weld at ends)Max size of weld for round toes (3/4) thickness of toe thicknessMinimum Size of weldIn a thick member, Small size of weld  no proper bond between weld and memberThickness of Thicker member Minimum Weld size (as per IS 800)0 – 10 mm 3mm10 – 20 mm 5 mm20 – 32 mm 6 mm32 – 50 mm 8 in first run finally 10 mmAlso Minimum weld size thickness of thinner memberGenerally minimum size of weld is preferred due to following reasons:Only one run is required (more the size  needs more than one runs such as in platesof thick > 32 mm)Less weld size – cheaper
  • 71. Effective Throat ThicknessShortest distance from root of fillet to the face line of weld i.e perpendicular distanceEffective throat thickness = K Size of weld = K SIn equal welds , K = 0.707  size of weld = 0.707 SFor other angles K is given asAngle between fusion faces 600 – 900 910 – 1000 1010 – 1060 1070 – 1130 1140 – 1200 K 0.70 0.65 0.6 0.55 0.5Effective thickness of throat 3 mm 0.7 thickness of thinner member thickness of thinner member in spl circumstances
  • 72. Effective Length of WeldEffective length of weld = Overall length of weld – 2 SActual length of weld 4SSlots in WeldsIn the Tension/compression members, if l < d  Stress concentration at endsTo avoid, this stress concentration, Code puts limit on length l asDistance of longitudinal fillet welds 16 thickness of thinner memberIf above condition is not satisfied, slots are made in plateThese slots are welded by welding of same strength as longitudinal weldSome times slots are made to increase the length of weldIf slots are made, they need to be cheeked just behind the weld for failure
  • 73. Correction for Long Welds If length of weld > 150 throat size of weld  reduction in weld strength 0.2 l j 1.2 1.0 where, lj = length of joint and tt throat size of weld 150 t1Effective Area = Effective length of weld Effective throat size of weldOverlap of plate in lap joints: 4 thickness of thinner plate or 40 mm whichever lessTransverse spacing of welds Length of weld on either side Transverse spacing of welds
  • 74. Design Strength of Weld Design Strength of Fillet Weld (in tension or shear): f wn fu Design stress of fillet weld, f wd Where, f wn nominal strength of weld mw 3 fu fu Design strength of weld, Pdw Lw tt f wn Lw tt Lw K S 3 mw 3 mw Lw = effective length, tt = throat thickness, S = Size of weld ; fu = smaller of ult strength of weld and parent material mw = partial safety factor = 1.25 for shop weld and 1.5 for site weldDesign StepsWeld may be subjected to axial load shear  Shear critical  shear controls the designAssume size of weld based on thickness of memberDetermine length of weld = force transmitted / strength of weldLength of weld is provided in sides and should not be less than ldAlso provide end returns of length equal to 2SIf length of weld > 150 tt  apply length correctionCommonly weld is provided on all three sides (in this case no need to check transverse spacing)
  • 75. Example on Butt or Groove Weld
  • 76. Example on Fillet Weld : Design a connection to joint two plates of size 250 x 12 mmof grade Fe 410, to mobilize full plate tensile strength using shop fillet welds, using (i) a double cover butt joint with 8mm cover plate (ii) lap joint(i) Connection using Double cover plate Butt jointAssume width of cover plate = 250 - 2 x 15 = 220 mmArea of cover plate = 220 x 8 = 1760 mm2Required area of cover plate = 1.05 x 250 x 12/2 = 1575 mm2 < 1760 mm2For the 8 mm thick plate well size between 3 and 8 mm  Let us use a 5 mm fillet weldStrength of the 5-mm weld = 410/( 3 1.25) x 0.7 x 5 = 661.5 N/mmRequired length of weld = 681.82 x 1000/661.5 = 1031 mm, say 1040 mmLength of the connection = [(1040 - 2 x 220)/4] x 2 = 300 mm < 150 tt i.e. 150 x 0.7 x 8  Joint is not a long jointHence, provide two cover plates of size 300 x 220 mm
  • 77. (ii) Connection using lap joint (large force) f y Ag 250 250 12 1 Plate strength 681.82 kN m0 1.1 1000Minimum = 5 mm ( from Table)Maximum = 12 - 1.5 = 10.5 (clause of IS 800)Assuming weld size s = 8 mm (in order to reduce the connection length not taken minimum) fy ks 410 0.7 8 1strength of weld . 1058 N / mm 3 m0 3 1.25 1000 681.82 1000Required length of weld 644 mm say 650 mm 1058Weld length available at end = 150 mmLength of weld on one side = (650 - 150)/2 = 250 mm < 150 x 0.7 x 8 = 840 mm  not a long joint OK
  • 78. Example on lap joint (limited force)Determine the size and length of the site fillet weld for the lap joint to transmit afactored load of 120 kN through a 8mm thick and 75 mm wide plate. Steel Fe 410SolutionMinimum size of weld for a 8-mm thick section = 3 mmMaximum size of weld = 8 - 1.5 = 6.5 mmChoose the size of weld as 6 mm (in order to reduce the connection length)Effective throat thickness = ks = 0.7 x 6 = 4.2 mm fy ks 410 0.7 6 1strength of weld . 662.7 N / mm 3 m0 3 1.5 1000 120 103 Required length of weld 181mm 662.7Assuming that there are only two longitudinal (side) welds,Length to be provided on each side = 181/2 = 90.5 mm > 75 mm OKHence, provide 95 mm weld on each side with an end return of 2s i.e. 2 6 = 12 mm.Therefore, the overall length of the weld provided = 2 x (90.5 + 2 x 6) = 205 mm
  • 79. Alternative solutions:Minimum size of weld for a 8 mm thick section = 3 mmMaximum size of weld = 8 - 1.5 = 6.5 mmLet us choose the size of weld as minimum specified i.e. 3 mm fy ks 410 0.7 3 1strength of weld . 331.4 N / mm 3 m0 3 1.5 1000 120 103Required length of weld 362.1 mm say 363 mm 331.4Two possible solutions are shown in Figure(i) If only longitudinal welds are provided, Length of each side weld = 363/2 = 181.5 mm Total length on each side including end return = 181.5 + 2 x 3 = 187.5 mm(ii) If welds are provided on three sides Length of each side weld = (363 - 75)/2 = 144 mm > 75 mm (width of the plate) OKThis solution is preferred since  connection is more compact  provides better stress distribution
  • 80. Fillet Weld for Truss Members: The weld may be done on (i) two sides (one side not recommended) or (ii) three sides i.e. two sides as well as at end For the angle section CG not at mid of width,  Weld is done such that CG of weld coincides with CG of angle section If welding is done on two sides only Ph2 PTaking moment about P1 P2 h Ph2 0 P2 h Ph1Taking moment about P2 P h Ph1 0 1 P1 h Ph2 P h h2 Ph1 alternativ ely, P P P2 1 P h h h
  • 81. If welding is done on three sides P2 P3 P P1 fu Strength of end weld, P3 h. tt 3 mw h Ph2 P3Taking moment about P1 Ph2 P2 h P3 . 0 P2 2 h 2 h Ph1 P3 Taking moment about P2 P h P3 1 Ph1 0 P1 2 h 2 Ph2 P3 Ph2 P3 Ph1 P3alternativ ely, P1 P P2 P3 P P3 P h 2 h 2 h 2
  • 82. Design of intermittent weldWhen design force for weld small  use smaller size of weld and provide on full lengthWhen loads are very small  Length required to weld (even with smallest size) < available length  Intermittent weld is selectedIntermittent weld  discontinuous welds  two types (i) chain, (ii) staggered – better than chain weld 8 (140) 70 (140) Shop weld 50 50 10 50 (100) 50 (100) 10 mm weld 100 8 mm weld 140 70 140 70Design Specifications for Intermittent Weld:Minimum effective length of intermittent weld = 4 weld size (except for plate girder)Clear spacing between the intermittent weld in compression 12t also 200 mmClear spacing between the intermittent weld in Tension 16t also 200 mmIf end weld not done: Length of longitudinal weld at the end width of the memberIf end weld done:Total Length of weld (end long. + Transverse at the end) 2 width of member
  • 83. Types of welds in tensionTwo types: (i) Single sided (ii) Double sidedSingle side weld  eccentricity between the line of action of the load and the throat centroid  creates a moment on the weld throat  should be avoided in practice
  • 84. Specifications for Plug or Slot WeldsIf t = thickness of plate in which slot is made thenWidth or diameter of slot 3t and also 25 mmCorners of slot made curved  Corner radius of slotted hole 1.5t and also 12 mmClear distance between two holes (slots) 2t and also 25 mmStresses due to individual forcesIn case of beam column, joint are subjected to (i) axial force (tensile or compressive due to bending) as well as (ii) shear force.If a joint is subjected to axial force P and shear load Q, then P Axial stress (compression or tension ), f a t t lw Q and Shear Stress, q t t lw
  • 85. Combination of normal and shear stressesFor Fillet WeldWhen shear stress are in addition to tension or compression fuEquivalent stress, f e f a2 3q 2 3 mw fuIf (normal stress shear stress) f wd No Need to check Equivalent stress 3 mwFor Butt WeldsNo Need to check for Equivalent stress Since:Butt welds joints are generally axially loadedIn single or double bevel butt joints  (Normal stress + Shear Stress) < fwdCombination of bending normal, shear and bearing stressesIf a joint is subjected to (i) bending stress; (ii) Shear stress; and (iii) bearing stress,then the equivalent stress are determined as fe f b2 2 f br f b f br 3q2 permissible stress for parent material
  • 86. TENSION MEMBERSMembers subjected to axial force only (i.e. not eccentric loads)Examples of Tension Members truss and tower members, tie member in bridgesThe fixidity at ends of truss  develops moment at ends  neglected in design orpermissible stresses are reducedSelf weight of members  bending moment in member  bending stresses  small NeglectedIn case of bolted members Holes are made for bolts  reduction in area  calledNet areaNet area = gross area – deduction of area due to holes
  • 87. Preferably put one bolt in a line (vertical line)If force is large  More number of bolts  bolts may be arranged in (i) Chain and (ii)staggeredTo increase the net area  Staggered better choiceTypes of Tension Members :(i) Single Sections  Single flat bars, Single angle sections, Single channel sections,Single I sections(ii) Built-up SectionsBuilt-up section are selected whenLarge cross-sectional area is requiredFor given area  more moment of inertia is requiredLoad reversal Can resist tensile as well as compressive loads
  • 88. In built-up sections:•Tie plates are provided at regular interval to: •Minimize the slenderness ration and •To transfer any unequal load from one member to other•These plate are not considered to increase area of section
  • 89. Net Sectional Area in Plates: Same as discussed in connection chapter Net area for chain bolting An B n dh t m 2 psiNet area for Staggered bolting An B nd h t i 1 4 giIf pitch and gauge lengths are same for all bolts, then p1= p2 = …… = p 2 and g1= g2 = …… = g p An B nd h m t where m = no. of staggered pitches 4g
  • 90. Net Sectional Area in Angle SectionsAngles very common sections, angles have two legsAngles may be connected by (i) both legs or by (ii) Single LegWhen angles are connected by both legs:- for the analysis angle is unfolded (developed)- now the section may be treated as flat (plate)- if ga and gb are the distance of first bolts from root and g1 (in book gf ) is the distance of second bolt from first bolt, then after unfolding the angle, the first gauge length will be g2 = ga + gb - t
  • 91. When angles are connected by Single leg only:The leg connected to gusset plate  called connecting legOther leg (not connected)  called outstanding legDue to single leg connection  Non-Uniform stress distribution in connecting leg Non-uniform stress distribution due to following reasons:(i) Load transfer from connecting leg to plate along the CG  eccentricity(ii) strain at junction of connecting leg and outstanding leg > strain at free end ofconnecting leg shear-lag
  • 92. Net Effective Area of SectionNet Effective Area of Section Depends on(i) Shear Lag effect(ii) Ductility of plate material(iii) Method used for hole formation(iv) Geometry factorNet effective Area of Plate = k1 k2 k3 k4 AnWhere, An = Net area of plate after making deduction for holes k1 = factor to consider the Shear Lag effect k2 = factor to consider influence of ductility of plate material k3 = factor to consider influence of method used for hole formation k4 = factor to consider the influence of geometry of connectionDuctility factor (k1):Ductile material of plate  better stress re-distribution at bolt holes at higher loadsNo stress concentration  stress are distributed better on whole width of platek1 0.82 0.0032 RFor Ductile material  k1 = 1.0 and for brittle material, R = 1  k1 = 0.8232For Common structural steels  assumed to have sufficient ductility  k1 = 1.0
  • 93. Factor for method of fabrication or hole forming factor (k2):Methods of bolt hole formation (i) Punching, and (ii) DrillingPunching  Shear deformation in material around the hole  10-15% reduction material strength than drillingIn some Design Specifications: For Drilled holes k2 = 1.0; For Punched holes k2 = 0.85As per IS 800:2007  k2 = 1.0 Provided diameter of hole is increased by 2 mm in calculation of net areaGeometry factor (k3)Small bolt diameter  Less gauge can be providedif (g/d) less  less material between holes at critical section  force transferred to adjacent bolt  less deformation in material  more uniform stress in plate more efficiency of joint k3 1.60 0.70 Ane Ag K3 varies between 0.9 to 1.14; As per IS 800: 2007  k3 = 1.0
  • 94. Shear lag factor (k4):In angle- and T- sections subjected to axial force  non-uniform stress distribution near joint  shear lagStress near the junction of connecting leg and outstanding leg  high as compared to at toesThese non-uniform stresses  becomes uniform after some distance  called Transition Length Failure of member occurs earlier than predicted  Due to shear lag Due to shear lag  more load is taken by connection leg and some by portion of outstanding leg near to junction  unequal angles are generally preferred (small outstanding leg) To incorporate shear lag factor, factor k4 is used k4 is calculated as k4 1 x Lx distance from the face of the gusset plate to the centroid of the connected areaL = Length of connection
  • 95. The length of the connection for bolted and welded connections is calculated as;; Small value of x and large value of ‘L’  large k4  large effective area  more strength of joint In the IS 800:2007: k1= k2 = k3 = 1.0 and k4 =  Ane An 0.6 for no. of bolts 2 0.7 for no. of bolts 3 0.8 for welds
  • 96. Design Strength of Tension Member:In Tension members, at the joint  holes in member, remaining length without holeTension Members without holes (beyond the joint):As load is increased  yielding of material  due to strain hardening, strength more than yield strengthBut more elongation before fracture  unserviceableTension Member with holes (within joint):Due to excessive load --: there may be (i) net area failure, and (ii) Block shear failureThus, design strength of a tension member will be minimum of the following:(i) Gross-section yielding: more yielding  more displacements before the fracture failure(ii) Net Section Failure  failure at net cross-section(iii) Block shear failure  a segment of block of material at end of member shears out in smaller jointsThus there are two Limit States are to be considered:Limit state of yielding in the gross-section  to avoid excessive deformation in member  to control excessive deformation  stress in gross-section < fyLimit State of fracture in the net section  to avoid fracture in net section, stress in net section < tensile strengthLimit State of block shear : combined shear and tension failure