___ Fasteners © CONTENTS § Location Details of Fasteners * Bearing Type Bots JOIN TELEGRAM & Friction Grip Type Bolts & Welds and Welding @ CSVTU EXAM % Advantages and Disadvantages of Welded Connections & Lap and Butt Joints % Truss Joint Connections by Bolts and Welds SED2 UNIT2 CSVTU GATE SEMESTER BOOKA fastener is a hard WATE ACAD FUBLIe4 ware devie Ro, Fastener may by vice that mechanically joins e a vet bolt pin and we Rivet : Rivet capacity is problemage, °° Sl inconsistent The capacities of river Bolt : Bolts are used contrast to all other {0 or more object 147 BUSS Plates, brackets, connecting mas and Connecee 4 ‘amping force, so determining ‘ ‘oni connection 's best done considering only the bea ‘ re ar " han ay other type of fastener, They ee a Be of fastener, ey tu strength steel bolts tightened ton ten, ecial equipment. Generaty | Some a high tension ai Nghtly larger than the nominal bolt size, calraieenn x Pin:: Pins are generally smooth large di These fasteners ate not very load direction and are ‘in shear. Weld : Welding is a highly specialized skin, | Welding consists of joining two steel sections by establishing easel by between them through the application of pressure or through fusion. [EBA Location Details of Fasteners Question 4 What are the location details of fasteners as per code provision? [1S 800-2007 Page 73 Clause 10.2) 1. Clearances for hole for fasteners 3. Maximum spacing 4. Clearance for hole for fasten short slotted or long slotted hole. peat aed bolts, bots wl a oe Cage “specified, the diameter of standard xe holes for fasteners shall be as given in table. Ans. 2. Minimum spacing 4, Bdge and end distance ers : Bolts may be located in standard size, ore) for hole Je 1: Standard clearance ci a eainal sie Zize of the hole= nominal diameter! ty S.N. lor of fasteners, 7 Jearess! mm Ee" syotsCATIONS* 2-8 gars CADE = @) o 6) To 30 [aa 20 a [eo 20 eo [80 Tage than 30 [a0 2 3 4a ‘Oversize hole : Holes of size anger than the standard clearance holes, as given in table may be used in slip resistant connections and hold down bolted ‘hole not larger than the resistant connections) (©) Short and long slots : Slotted holes of size larger than the standard clearance hole, as given in table may be used in slip resistant connections and hole down bolted connections, only where specified, provided the oversize holes in is covered by a cover plate of sufficiently large size and thickness and having a hole of size not larger than the standard clearance hole (and hardness washer in slip resistant connection). Minimum spacing : The distance between centre of fasteners shall not be less than 25 times the nominal diameter ofthe fastener. ‘Maximum Spacing : In general the distance between the centres of any two adjacent fasteners shall not exceed 35 ¢ or 300 mum, whichever isles. In tension members : 16 t or 200 mm, whichever is ess. In compression member :12 f or 200 mm, whichever is less. ‘where 1 isthe thickness of the thinner plate. Edge and End distance: (@) The edge distance is the distance at right angles to the direction of stress from the centre ofa hole to the adjacent edge. The end distance is the distance in the direction of stress from the centre of a hole tothe end of the element (©) The minimum edge and end distance from the centre of any hole to the nearest edge of a plate shall not be less than 117 times the hole diameter in case of sheared or hand-flame cut edges sand 15 times the hole di se of rolled, machine-lame cut, sawn and planed edges. (©) The maximum edge distance to the nearest line of fasteners from an edge of snyunetifned pats at ccd 121 whore (32) ar the thickness of the thinner outer plate.Structural Enginceting Design “UL 2-4 [EIA Bolded Connection | ‘Guestion 2 Whats abot? ‘Ans. ‘A bolt may be defined as a material wit threaded at the other end to receive a mut. “em ih a head at one of the end and a shan, Grip Steel 4 washers ; a | a Nut | Washers are usually provided under the bolt as well as under the nut fo serve ts purposes : 1. Todistribute the clamping pressu 2. Toprevent the threaded portion of the ‘Question 3 ‘What s the purpose of bolted connection? of the connections, the parts to be connect! and nut. Ifthe connection is subje! ire on the bolted member, ‘bolt from bearing on the connecting piecs ‘Ans. In order to assure proper functioning ‘must be tightly clamped between the bolt head to vibrations, the nuts must be locked in position. ‘Question 4 ‘What are the merits and demerits of bolted connection over riveted connection? “Ans, Merits of bolted connection over riveted connection : 1. The erection of the structure can be speeded up. 2. Less skilled persons are required. 3, The overall cost of bolted construction is cheaper than that of rive because of reduced labour and equipment costs and the smaller number required resisting the same load. Demerits of bolted correction are as follows: 1. Costof material is high, about double than that of rivets. rut te | ave ACADEMY PUBLICATIONS® 2-5 2, The tensile strength of the bolt is reduced because of area the thread and also due to stress concentration. ‘edeton atthe 0k of 43, Normally, these are of a loose fit excepting turned bolt i en ts and hence their strength is 44, When subjected to vibrations or shocks, bolts may get loose. Question 5 ‘What are the different type of bolts? “The classification of bolts are as follows: 1 basis of type of shank : (@) Unfinished bolts (ordinary, common, rough or black bolts) () Tumed bolts 2. On basis of strength and material : (@) Ordinary structural bolt 3. Onbasis of shape of head and nut : (@) Square bolt 4. On basis of pitch and fit of thread : {@) Standard pitch bolt (© Fine pitch bolt ‘The most commonly used bolts are: © Unfinished bott 2.3:4/Unfinished Bolt () High strength steel bolt (&) Hexagonal bolt (©) Coarse pitch bolt (ii) High strength steel bolt ‘Question 6 Write short notes on unfinished bolt, An i a en ein el em ld le usualy of grade 46. Buck bot ave adequate eength ac ductity when propety we but wile ening thet aug ight (Snag eh id tightness that exsts when all plies in ajoint ae infirm contact) will twist off easily if tightened too much. ‘These are used in light structures under static Toads such as small trusses, puslin, bracings and platforms. ‘They are also used as temporary fasteners during erection where SRG Dols oF ‘Welding are used as permanent fasteners. They are not recommended for connections subjected to impact, fatigue or dynamic loads. Commonly used bolt diameters are 16, 20, 24,30 and 36 mm. ‘Thus bolts are designated as M16, M20, M24 etc.Structural Engineering re ee Sea eee ee 2.3.2 High Strength Friction Grip Bolt ‘Question 7 ts may be tightened until they have very high tensile stresses, two or more times that of ordinary bolts, so that the connected par, load earying elements jointed as shown in (Figure (b). Due to this fiction, the slip the joint. which is there in joints with ordinary bolts, is eliminated. This fiction | developed by applying a load normal to the joint by tightening these bolts to prod load. That is why these bolts are also known as frition-type bolts. 6. ‘uestior re ACADEMY PUBLICATIONS * # Due to high frictional resistance available due to high clamping force shearing an 2-7 Fasteners stresses do not occur in the bolts, J. Bven outside the bolt hole frictional resistance is available. Hence the load through the net section is reduced. . Stress concentration in the holes and close to them are eliminated. The fatigue increased. The joint can withstand pulsating or alternating loads, The bolts are in high tension close to proof load and hence the muts are unlikely loosened. Point out the differences between bearing type bolt and HSFG bolt. ‘Ans, Differences between bearing type bolt and HSFG bolt: S.No. Bearing type bolt SFG bolt ‘Waster weer 1. | Bolts bear against the holes, to | Force transfer occurs between the plaice Paral Waster shank | transfer the force. due to clamping force generated by the LZ Z pre-tensioning of the bolts efeefeeeeeeseeeeeeellfefdb. LALIT | % | Bearing type bolts are used in| Friction type bolts are used as suuctaral ,, flange type joints to provide joints for load transfer across structural y jointintegrity. members by shear mode. (Types of high strength bots ERA types of Bolted Joints Cn 0 Tea net ee it Ler | Wit are the types of bolted joints? thetastt Foor Ans. 1. Lay ‘The Plates to be connected overlap each other. The lap may have CP Staggered or chain riveting. 2 Butt (©) Slip resistance Fig High strength bolted connection 2.3.3 Advantages of HSFG Bolts Question 8 What are the advantages of HSFG bolts? Ans. Advantages of HSFG bolts are given below ; 1. The joints absolutely rigid since the plates do not slip. | 2. The bolts which are in high tension provide considerable clamping force on + Plates due to which large frictional resistance against slipping, of the plate? available strength of the joint becomes very high. {CSVTU May 2016 Oar cer ce mm : +The plates to be connected butt against each other and the connection 's made by providing a cover Pate on one or both sides of the joint. The butt joint ‘ay havea single row of staggered or chain riveting, =. a Fig, Single bolted lap joint = = 5— Hig. Single bolted double cover butt jointESA, Terminologies Question 11 Define the following terms in relation to bolted joints. Nominal diameter 2. Effective or gross diameter 3. Grossarea 4, Netarea 5. Pitch 6. Gauge 7. Edge distance 8. Proofload 9. Slip factor 10, Property class ‘Ans. 1, Nominal diameter: The diameter of the shank of a rivet before riveting, is called the nominal diameter. For a bolt, the diameter of the unthreaded portion of the shank is called its nominal diameter. 2. Effective diameter or gross diameter : The effective or gross diameter ofa rivets ‘equal to the diameter of the hole it fills after riveting, For a bolt the nominal | diameter is same as the gross diameter. | Gross area : The gross area of a rivet or bolt i given by its gross diameter. Net area: The net area of a boltis the area at the root of the thread, Pitch : The distance between centres of any two adjacent rivet Diagonal pitch : The distance between centres of any two adjacent rivets in the diagonal direction is called diagonal pitch. Staggered pitch : The distance between any two consecutive rivets in a zig-zag riveting, measured parallel to the direction of stress in the member is called staggered pitch, 5. Gauge: A row of rivets parallel to the direction of force is called a gauge line. The normal distance between two adjacent gauge lines is called the gauge distance. ae 7. Proof load : Initial tension in HSFG bolts is known as proof load of the bolt. 8. Slip factor : Coefficient of friction in friction type joint is known as slip factor. 9. Property Class : Bolts are grounded under different grades or property classes depending upon their strength. For example, property class 4.6 indicates that tt ‘nominal yield stress is 0.6%400 = 240 MPa . FXG Design of Bolted Connection ‘Question 42 What are the design consideration for black bolts/ unfinished bolts in bolt! conection? Or Wat are the types of failure unfinished associated with bolted connection? we ne ope ot! failures in a shear connection are: 1. shearing of bolts 2. Beating failure of bott 3. Tensile failure of bolt 4. Bending of bolts Trerefore, the bolted connections should be checked for shear capacty bearing capacity, tension capacity and bending. shep 6): Shear capacity of bolt: 18800-2007, Page 75 clause 103. ‘The design strength of the bolt, 7a.) as governed shear strength this given by os, aa where, Vz4 Nominal shear capacity of bolt “tu ~Pattial safety factor forthe material ofthe bolt tna 7125 (as per code) ty and Yun FO ha Aa By Ba Bo ‘fa “Ultimate tensile strength of the bolt material n, = Number of shear planes with thread intercepting the shear plane n, = Number of shear planes without threads intercepting the shear plane Yao = Partial safety factor for the bolt material = 1.25, 44 =Nominal plain shank area ofthe bolt ‘Ay =Net shear area of the bolt at threads = Area corresponding to root diameter at the thread Ay =Fld-0.9382 97 [where p= Pitch of threads in mm] ‘= 0.78 to 0.80 of gross area of shank. Ag 078d Or The table below gives the net area at threads for Grade 4.6 bolts Table : Net area at threads section of bolts Bolt diameter (mm) 20 [2 [24 [27 [30 [36 ‘Area at root of threads (am') | 843 | 157 | 245 | 903 | 353 | 459 | 561 | 817 ‘The nominal shear capacity of the bolt given above has to be reduced inthe ease of long joints, in the case of large grip length and when thick packing plates are Provided.‘Structural Enginooring Design ~ 11 Cat GATE ACADEMY PUBLICATION, (2 Redaction fat frlong joins iy When helen of et of apes end connection in a compression o ton menor containing more than yo bots (that the stane between he tan he ast ows of bt in he oy measured in the det ofthe lad tanse) exceeds 15d inthe dean of load the nominal shear capacity shall be reduced by the factor given by, By =LaTs Sis for 0.7558, st where, d= Nominal diameter of the bolt (6) Reduction factor for large grip lengths ,: When the grip length J, (equal to th total thickness of the connected plates) exceeds 5 times the diameter d of te bolts, the shear capacity shall be reduced by the factor fi, given by, ad Tae, A, stall not be more than, given above the grip length J, shall inno case be ‘greater than 8d. | (2 Reduction factor for packing plates i, : The design shear capacity of 63s carrying shear through a packing plate in excess of 6 mm shall be decreased by 0/ factor givenby, By =(1-001254,), where, ty =Thickness ofthe thicker packing plate in mm. Considering all the above provisions, the nominal shear capacity of the bolts given by, Kar Pettcdy +m AID PaDre Step (i): Bearing capacity of bolt: Be yg nte te Where, Fy Nominal beating strength of bolt =2.5K tf K, issmalleror £2 9 25,fa, sateror 353-025-4410 «p= End and pitch distance of the bolt along bearing direction 4,~Diameter of the hole ‘Sar J,=Ultimate tensile stress of the bolt and ultimate tensile stress of th Plate respectively. yenfy PUBLICATIONS® 2-44 ‘d=Nominal diameter of the bolt -=Summation of the thickness of the connected plates experiencing bearing stress in the same direction, or if the bolts are counter sunik, the thickness of the plate minus one half of the depth of counter sinking, ‘Ya: =Partial safety factor forthe material of the bolt=1.25 step Gil): Tension Capacity of bolts: A bolt subjected to a factored tensile force T, shall satisfy TsTo role hese Tae Ty 2090 fad < Saha were, fa Ulinate tele sess fhe bolt fa ield sees ofthe bolt ‘A _Nettenae sess ares Arua the atom ofthe ead and 4. eStankarea ofthe bolt 1 =Porlal sft fctor for esistanegovered by yiding=110 ‘aaa sft fco fr material of bole | ‘Step (iv) ; Bolts subjected to shear and tension : A bolt required to resist both shear ‘ tition athe sme tine slot the conden Me) fh fel ay« SHOR, = Faced ser fo actg cn the Yt Y= Delg shea cpacy of he bol 1 = Facore tense foree ating othe bl 1 «Design tele capt ofthe balk Question 13 Daw iteracon curve orbEe F ring Design-II___2-42 GATE ACADEMY PUBLICATION, seo PUBLICATIONS* 2-43 Fasteners ‘The interaction curve illustrates the use of the equation ons. 5 + is the width of the plate in mm, y ny mber of holes along the width () 2) s10 isthe 6 Fa) “Ts +, perpendicular to the direction of load, ‘Question 14 What are the assumptions in the analysis of bolt joints? ‘Ans. In the generally adopted design of bolted joints the following assumptions made: We generally assume that all bolted connections are of the bearing type and 4, effect of frictional resistance against slip between the contracting. plates | ‘gnorable. The plates are taken to directly bear on the bolts while transmitting ¢ Toads. 2. Any deformation on the plates due to loads is ignored. 3. Shear load on a bolt is assumed to be uniformly resisted by the bolt section. | 4 5. 4, isthe diameter ofthe hole in mm, and 1 isthe thickness ofthe plate in mm. 4 fect of stress concentration in plates around bolt holes is ignored. | AL A= (b-nayt The bearing stress between the bolt and the plate is taken to be uniformton Fig(a) Chain of holes in rows bearing area 6 Bending ofthe bol is ignored. © _ raitue EZA\ entctency of Joint Las Question 75 Whats eticieney of joint? ‘Ans. “eis th oof strength of ont per pitch eng to the srengih of ald plate por PE 8 Direction of , iret sexe A | 8 - ‘Strength of joint perpitch length e | Efficiency = ——“ngtof joint perpitchlength _ i i send ‘Strength of solid plate per pitch length 00 i L i The tension capacity 7, ofthe plate as expressed as a fn OAL Fig) Staggered holes Ya ane Fig. Tension capacity of plates ‘where J, isthe ultimate stress of material in MPa, ‘Online A: 4 ae nae A; is the net effective area of the plate in mm?, and 3 ‘a Hs the partial safety factor taken as 1.25 in the code. CntiwB: Aaa aay + ps load caring capaty ofthe plate depends on the nt eectve ares of Fi 2 cane ild n tur depends on the arrangement ofthe holes f the holes are 1 Online: 4, =Afb—4d+ Zi Py staggered, the net area 4, can be easily computed as, A=(o-nd,y ‘Where d, is the hole diameter.ae Serctral Ropnering Dos 34 I ‘Question 16 Draw cross-section of a double bolted double cover butt joint. zoTee +} Jeoieo| (> eoleo . es HEAP — Double-cover double beled butt joint But joint, boltin double shear EEA, besign Strength of HSFG Bolts ‘Guestion 47 What are the design considerations for HSEG bol? ‘Ans. 1. Design slip resistance: [1S 600-2007, Page 76 cause 103) } Design for friction type bolting in which slip is required is to be limita sx nl fated design sheaf, Fn the ners of conn at which slip cent be tolerate, shal satisty the following: Vy ay Vig =Nominal shear capacity of a bolt as governed by slip for friction §p connection, calculated as follows Bag = Wyre Fe H, = Coefficient of friction factor) (1, =055) n= Number of effective interface offering frictional resistance to slip 4, =1.0{or fasteners in clearance holes =0.85 for fasteners in oversized and short slotted holes and for fastenes? long slotted holes loaded perpendicular to the slot = 0.7 for fasteners in long slotted holes loaded parallel tothe slot sistance is designed at service load) lip resistance is designed at ultimate load) .=Minimum bolt tension (proof load) at installation and may be taken as Asf ‘4g =Net area ofthe bolt at threads, S=Proot stress (20170 f,) GATE ACADEMY PUBLICATIO9, acADEMY PUBLICATIONS * 2-16 oat: Fasteners 7 Long joint Similar to black bolts, the desig sip resistance Vy for parallel shank first and last rows of bolts in the direction of the load transfer exceeds 15d in the direction of the load) by a factor fy, given by | 4 put 0.75 250 kN Hence safe ‘Question 19 Design a lap joint between two plates as shown in figure so as to transmit a factored Joad of 70 kN using M16 bolts of grade 4.6 and grade 410 plates. Given data: /,, =Partial safety factor for the material of bolt= 1.25, Factored load = 70 kN, Grade of bol “Ultimate tensile strength of bolt J, For example if grade of is46 4, =400g - ACADEMY PUBLICATION, actural Engineering DAD ___—2-18 GATE 0r8, “Nominat diameter of bolt (4) =16 mam(MI6 bol) ~ “Unimate tensile strength of plate (/,) = 410N/mm' | srength calculation : a Hole diameter = Nominal diameter +2.mm 216+2 mm=18 mm (Shear capacity of boltis given by, 1 800-2007 page 75 clause 1033 vy, =fe( moder de) @ | Ble) de Bam HE = . JT row =F oo o 40 mm ep _p ef et 30 00 4030 Fe From figure, since the top plate is only 12 mm, it is assumed that the shear plan through the threaded portion and hence 1, = 1, =Number of shear planes with threads intercepting the shear plane m,=1 | Soo =400 | nl From table2, fr bolt diameter 16 mm, Ag #157 mm? Or Ao =078E a 0.762067 =15Tmm? Fag} =157 n=0 (since n, is valid :. n,=0) te =125 Substituting values in equation (i), S(R2) 2900 | 125 | | Question 20 AGADEMY PUBLICATIONS ® 2-10 {G) Bearing capacity of thinner plate is given by, 25 Kyat fy To &, Is smaller of 30 ® 3a, 3x18 0585, (e=30 mm from figure, d,=hole diameter =18:um) P 40 F-025-9 925. © fr -025= 8-025 00.49 Minimum pitch = 25d =25%16= 40 mm fo 100 |, © pray @ 1 Now bolt value is the smaller of shear capacity and bearing capacity Bolt value = 29 XN Now required number of bolt = ftoredload _ 70 _ 5 4) requi o — Jo7241=3 bolts Bolt value = 29 kN Provided 3 nos. of bolt Minimum pitch =2.5xd=40 mm Minimum edge distance =1.7xholediameter = 1.7x18=30 ram Ans. ny Plates of a6 mm thick tank are connected by a single bolted lap joint with 20 mm meter bolts at 60 mm pitch. Calculate the efficiency of the join take f, of plate as rade bolts. “ = Partial safety factor for material of bolt =1.25 ‘Thickness of plate ¢=6 mm, Diameter of bolt =20 mm Hole diameter = Diameterofbolt+2 mm =20+2-=22 mm Pitch 410 MPa and assume 46 Step (i) Given data: V,, p= 60 mm Sg =410 Nim?Structural Baginoving Design“ Il 2+20___GATEACADEMY PUBLICATION, Grade of bolt = 4.6 —™~ e f= 400 Nios? / Calculation of strength of joint pr pitch ngth willbe minimum of 1. Strength of boltin single shear 2. Strength ofboltin bearing 2. Strength of ne eton pe Step Gi): Strength of bolt in =. to [1S 800-2007 Page 75 Clause 10.3.3] Vos Vag = Fetty hy ty dg) oe Yo Since», 0 becruse thickness of plate it only 6 mm “2. Ttisassumed that the shear plane s through threaded position : mn0 From table, for bolt diameter 20 mm, 4,, = 245 mm” or Step (ii) : Strength of bolt in beating: 25K to Kew ‘Step Gv): Stength of the net section per pitch length, hg 4, =Net area= (pitch—hole diameter)xthicknessof plate 2910 (60-226 675, “+ Strength of joint per pitch length = 45.26 kN (Minimum of 1,2, 3) ofl) r step 0): Strength of plate per pitch length = 22LaA Te o9xatonsox6 eS = 1063 as ; Strengthof jointperpiteh length 45.3 it efficiency = Senet oF joint perpiteh length _ 453 yy Joint Strength of solid plate pe pitch length 1963100" 426% Ans, Question 24 ; Calculate the efficiency ofa zigzag double bolted lap joint as shown in igure. Assume Fe i0 grade plate and grade 6 bolts of diameter 20mm and 8 mm thick plater, 1 2 ‘Sol. Step () Given data: 1, = Partial safety factor for material of bolt= 125 Grade of bolt = 4.6 So =400 Nine? Grade of plate = Fe410 Diameter of bolt (d) = 20 mm ‘Thickness of plate = 8 mm Step (ii): Shearing capacity of bolt :[ TS 800-2007 page 75 clause 10.3.3] = Sa {trda tm Ay svell) Paya GES) Since plate thickness is only 8 mun, “+ Assuming thatthe shear plane is through the threaded portion n=0,,=12-22 GATE ACADEMY PUBLICATION, ‘ sign - Siructua Bagperas Pe 2 $e For20mm bolt diameter Ay =0.78%(20)' = 245mm? Ang, ‘Step (ii) : Bearing capacity of 8 mm plate: 2SKydtofy Yo 4, isminimum of <= 35 3d, "3x22 B-025- 2 025-051 f-02s- 52-025 Vay = iy 2053 (assuming ¢=35 mm) Minimum pitch =2.5xd =2.5x20 =50 + Choosing X,=051,d=20mm, 1=8 mm, f, =410 Nimm*, y,,=1.25 ‘Substituting values in equation (i pq x BEOSVDDNB AIO 669 aa 125 Consider the 100 mm of length of the joint shaded as shown in fig, which represents {yplcal condition. Line 1-1 has one bolt hole 22mm in diameter. Line 2-2 also has one bolt hole, but if plate-A in tearing along 2-2 ~ Strength of the net section of plate 22 one bolt strength. Hence for plate 4, net section 1-1 is critical here, Step (iv): Net tensile strength of plate: 0S A, Yo where, 4, = (b-d,)xt = Width of plate =100 mm (rom figure) 4, =Hole diameter =22 mm 4 2Thickness of plate = mm “Substituting values in equation (iy 8954105{ 000-22). oa ne ACADEMY PUBLICATIONS ® 2.23 eee ot values will be uiniman of shearing capacity of bolt = 45.3 kN Bearing capacity of plate = 669 kN [Net tensile strength of plate =184:2kN Bolt value = 453 kN sirength of two bolts (double bolted lap join) =2xBoltwlue «2483905 kN ‘Therefore, strength ofthejoint = 506 k (smaller of 184.2 and 906 kN) 09.410 000%8) 09f, Grossetrength of pate = 2% 4 2288100078) ayy joint = Stent joi, 5p _ 90.6 Ana ‘ = 1100 = 251000384 . Efficiency of joint = srengthot plate”! “236 ms ‘Question 22 A single-bolted-cover butt joint is used to connect two plates 6 mm thick (ee figure). Assuming the bolts of 20 mm diameter at 60 mm pitch, calculate the efficiency of the, [Use 410 MPa plates and 4.6 grade bolts. ae eT Bolt diameter, d= 20mm, Pitch, P = 60mm “Yoa = Partial safety factor fall material of bolt =1.25 7 : onl pas 3 ‘The bolts will cu at section (1) and section (2) due to double cover buttjoint ~ The bolts are in double shear Step Gi) : Shearing capacity of bolt (in double sheat) [IS 800-2007 page 75 clause 1033) For double shear >| Geer 2h, =o =o VB Ye A= 0.7800" =245 mm? af BCE) -m00 kvee Seeuctural Engines * aces Easton. PUBLICATIONS cADEMY PUBLICATIONS js : 2-26 GATE ACADEMY. er 7 _ se aie So esha teal ofonebols 1S PRRT gs oS ‘Taicknes of cover plate $= 56.03.75 mm oa “han sal i i ate a “Therefore provide 4 mm thick cover pl Se cleats in the calculations are ‘The thicknesses to be considered i th of the cover plates, whichever is less. Hence, un . =a 4 bolts of 16 mm diameter Step Gli): Strength ofthe belt in bearing ==" (essuming ¢=35 mm) T 319009) L Jestoke dq shale she ane} 2am SoS J 2mm i ) | ‘Step Gi) : Design bearing strength of one bolt : [1S 800-2007 page 75 clause 10.3.3) Mae = L125 Ky dt fl) K, is the least of the following : ‘The strength of the plate in bearing is the minimum. Hence, «30 Pp 40 i at =056 2 £-025=40 -025-049 ‘3, Fatt 3a, Bae Strength of the joint = 48 kN 09 SupA Lo 400 Step (): Strength ofthe slid plate per pitch length = 7 a Ba Ty n 098 41 4 Ky=049 (east of 1, 2,3, 4) #12 mm (lesser of 12mm and 60 min) Joint etikency }x100= 45.17%. Vaa= Lo(25n0 49,1612<400)= 75964 Question 23 = ‘Two pats 12mm 60 mm are connected in a lp joint with 4 bolts of 16 ma diamet| ‘Lesser design strength of one bolt =28.9 kN 45 shown in igure. Determine te strength o the join. Total design strength of four bolts = 4x28,9=115.9 kN Ultimate strength of bolt material = 400 N/mm? Step Gv) : Desi : 103. timate strength for plate = 410 Ninm? "ep Gv) : Design tensile strength of plate: [1S 800-2007 page 76 clause 10.51 I. Step (i) : Given data: Szsofmain plate =I2mm«60mm, f, =400NImm?, f, Baa 11094, /1 = =410Nmm?, “ =16+2=18 mm Bolt diameter 16mm , Diameter ofbolt hole dya oN savestigate the safety ofa lap joint connecting two plates tothe following party, _/, Te platesare 12 mun and 15m thick Each plates 75 mm wide | Factored load = 75 KN Use 16 mm diameter bolts of grade 4 as shown in figure and plates of, ol. Step @): Given data 4.6 grade bolls are used. Ultimate strength of bolt material F fog = 4100 = 400 Nim? Bolt value = 28975 N Yield stress of bolt material = f, =0.6x400 = 240 Nim? : For plates, f, =410 Ninm*, f, =250 Nimum* Number of bolts required = 2~ Provided 3 bolts of 16 mm diameter: Figure shows the arrangement of the bolts QO Oo @ iE ‘Check for the strength of the plate i i i i Step (iv) : Design strength of the plate (rupture consideration) H fl i i {1S 800-2007 page 76 clause 10.35] He3sele— 40—o}e— 40 —9}e—35 >] 1 1 Tu = E094, f= 7 ygl09C0S-10)12%410]=201917 N ‘sum F f2mm ‘ T, >75000N Hence, design is safe. ‘Ans. ; 1 “| [question 25 Diameter of bolt =16 mm | 2, Design a lap joint connecting two plates 120 mmx8 mm to transmit a factored load of Diameter of bolt hole =16+2=18 mm °© 120 KN. Use 12 mm diameter bolts of grade 4.6 and plates of grade 410. Step (ii): Design strength of a bolt : [15 800-2007 clause 10.3.3] Sol, Step (i) : Given data : For bolts of 4.6 grade, Design shear strength of a bolt =. -1 [Finds +n, 4] Uttinigte strength of bolt material, fi, = 4%100 = 400 Nim? Yield stress of bolt material = f, = 0.6%400 = 240 Nimm* ‘The boltis in single shear, n, =1, n, =0, pa emns For plates, f, = 410 Ni 50 Nim LL ood a Minimum edge distance olts)=20 mm rah a hilt aos Ea St i Dagnbengwrenghfselt 5 00.007 page cause. aim =25xboltdiameter= 2.512=30 mm emit p=30mm ki any PSh aL step Gi): Design shearing strength of the bolt [1S 800-2007 page 75 clause 10.33] isthe least ofthe following : Ss rele] n=l, n,=0 * Fara 075=049 [Providea pitch p=40 mm) | : v= ea a grt na rE ats n1 step Gi: Design bearing strength of bolt = Vag = 125K tf) TIS 800-2007 page 75 clause 10.34) £; Is the least ofthe following : ii) 2025-2 025-0519 (0) Forse sry -025 0519 (iv) 3400] = 39398 N + Boltvalue=16298 N- (Minimum of Vig and Vy ) Number of bolts required = “2 Provide bolts of 2 mm diameter "igure shows the arrangement ofthe bolts ‘Checkfor the strength ofthe pate Step (iv): Design strength of the plate from rupture consideration | 1S 800-2007 page 76 clause 10351 14 T= 224L A Ym 128 ‘Factored load on the connection =1 ‘+ The designis safe, 120-2%13)8x410] = 221990 N 120 kN =120000 N i , scatsne PUBLICATIONS © 8 ppp —_— Ba 00H 7 ng the design strength of. 22m dlametr bolt for the cases given below Det int ie ae cover butt joint with 12 mm cover plate (0 Double enimum edge distance = ¢ =1.5dy =1.5%24 =36 mm, say 40 mm atthe pitch ofboltsbe p=2.5d =2.5%22.=58 mm, say 60 mm step i : Tap joi oer ols are in single shear {1S 600-2007 page 75 formalas] {@ Strength of abottin single shear Pg wat Eee 2 eT 4 (i) Bearing strength of abott Yoo = LOSK AL) 1400 a 078 2x2 |= 4781 Ae =755 m N K, =Least of the following K, =0556 e 40 arpa 16x 400] =156569.5 NV + Design strength ofthe bol = Lesser of () and (i) = $4781 N= 54.781 KN Step (ili) : Single co joint with 10 mm cover plate : The bolts are in single shear, (© Strength ofa bolt in single shear LL nt 1400] Vg = Le hy = 2S ore ex |= 54781N Yo v3 125 + 4 |Structural Engineering Design * I 2-30 (6) Strength of a bolt in bearing =Voe = LosKat) In this case 1=lesser ofthe thickness of the plates =12 m= Design strength ofthe bl 4781 N = $4.781 kN Double cover butt joint with 10 mm cover plates: are in double shear ‘Strength of a bolt in double shear Lk vw tf Sea, | =2nsarer «109560 N oe serengthta boltinbesrng 1 Von ZrRsKia hd ‘Thickness ofthe main plate = 16: samofthethiknes of th cove pltes=10 10=20 am ‘telesser of the above =16 mm 2216x400] = 1565696 N ‘Question 27 ‘Two plates 250 mmx12 mmare to be connected in a double cover butt joint = 20 mm diameter bolts. The factored tensile force on the plates is $00 KN.Desin® connection. The cover plates are 8 mm thick. The bolts may be arranged in dian pattern. ‘Step (Given data: Size of plate =250x12mm, Diameter of bolt=d = 20 am Diameter ofolthole =4, =20¢2=22 mm ‘Trebolts aein double sear. ‘Step (i): Design shearing strength of 1 bolt in double shear US 800-2007 page 75 clause 1033.3] raat fe |: ao) ela" | Tas yO 220 ey = 905478) 4 GATE ACADEMY PUBLICA yas capenty PUBLICATIONS” 2-34 red = 5. rome f DORSEY =p. S52 provide 6 bolts. Progr shows the proposed arrangement of ols OV .O4 250mmx12mm OF 40, 60 60, 40 40 60 60 40, = mort oa cai Sie c Step (ii): Design bearing strength of 1 bolt : 15 800-2007 page 75 clause 103.3] = besKa sl K, = Least of the following : --%. 3a, “32g 08 (iy £2400 © ooo “Design bearing strength of bolt + Design strength of 6 bolts =6: Step (iv) : Design strength of the plate from rupture consideration [1S 800-2007 page 76 clause 10.3.5] mail 0.9{250~(3x22)x12}x410x107 aya 5 651.80 KN Factored load on the connection =$00 kN ‘The design is safe. Fasteners aon the shearing strength ofthe boltStruc \TE ACADEMY PUBLICA, i -m_2:82 oat | tural Engineering Desie Determine the adequacy figure when 20-mm-diameter g. ine the adequacy of the fasteners in figure * bolts are used, Assume that the strength of the column flange and the struct ts are - sections do not govern the design. Neglect prying action. % _acaDEMY PUBLICATIONS © 2-33 ws From table, for bolt diameter 20 mam, step Gv) : Strength in ten 09. -! sah {1S 800-2007 page 76 clause 10:36] A, = Pitch —hole diameter) Thickness of pate = 2.9410(60~22)x6 12s =6731N ny (mY soprtqton (22) (cro (25/453) + 13 /67.3)' = 0.55<1.0 Hence, the six grade 4.6 bolts of 20-mm-diameter are sufficient to carry the load of 250 KN applied atthe joint. Ans. ‘Sol. Step G) : Given data : Diameter of bolt = 20 mm, Grade of bolt = 4.6 vr Hole diameter = 20 +2 mm=22 mm {EBD Truss Joint Connection by Bolts Figure ‘Typical truss joints are shown in figure which use gusset plates in order to accommodate the bolts. These gusset plates are connected to the members ofthe truss by means of bolts ‘oF welds using lap or butt joints depending on whether one or two angles are used for the ‘member. Note that in Fig. (c), end plates are used to facilitate the connection of shop Step (i : The factored loads 7, and¥, per bolt ‘welded parts of a truss at the site using bolts. Teen component 2, =280+4=200 1 Siarcmponent f, =250< 130 Tension 7, Shear Vy = 7 : i — For a 20-mm-diameter bolt. Tees : Step (i): Strength in single shear oS Vous Yat ay nt25 Sige l | 7 ls. (a) Groove welded joint between Tees Fig.(b) Bolted joint with double angles Yas Re Aa, Ay x Based on the experimental study on gusset plates commonly found in Warren type Since n,=0 becuse thi ts | Prdass Whitmore (195) proposed thatthe maximum due srs in a gust plate rom : anes of pte it only 6 mm 'n individual member could be estimated adequately by ensuring that the member force is 7 Msessumed thatthe shear plane is through threaded position =o Aistebuted uniformly over an effective area given by 30° dispersion from the outer row of | fasteners as shown in fig. (a). “yA Mepn nner tenner ere.Structural Enginooring Design - II 2-34 GATE ACADEMY PUBLICATION,, “Susur Rasinorne Design I_2-84_ GATE ACADEMY UMLACATIOn, Fig. (SDA Block Shear Failure Failure ocursin shear at row of bolt holes parallel othe applied Toads, accompanied tensile rupture along a perpendicular face. This type of failure results in a block of mater) being tom out by the applied shear force. The block shear strength T, of a connection taken the smaller of, Sy 9f.he tm -| bes © Rooleupiovnte en re age O he Soh Bin toe 4 AM, wer Be cee en wl 4 me area, respectively, in shear along @ | 4s AoADEM PUBLICATIONS 2-35 mgt and A, are the minimum gross and net area, respectively, in enson fom the hes tothe tow ofthe angle oF next ast row of bolt in gusset plates, pane Sy tos and Yaa the partial sa ultimate and yield stress of the material, respectively. for material resistance governed by yielding (aio) and governed by ultimate stress (1.25), respectively. ‘Note that if the connection is only required to function as: ‘Ron-slip under service loads, the interaction equation has to be checked at service stage. [EAA Numerical on Truss Joint Connection by Bolts & Block Shear Failure Guestion 29 A member of a truss consists of two angles ISA 754756 placed back to back. It carties an ultimate tensile load of 150 KN and is connected to a gusset plate mm thick placed in between the two connected legs. Determine the number of 16-miniameter 4.6 grade ordinary bolts required forthe joint, Assume /, of plate as 40 MPa Given data : ISA 75x75%6mm Ultimate tensile load ~ 150 KN, Thickness of gusset Bolt diam =16 mm, Hole diameter (d,) = + Sy = S10MPa ‘The arrangement of joint is as shown in figure. The bolts are in double shear. They ‘bear against 8mm gusset and two G-mm angles, the former controlling the value in beati (9 Strength in double shear for 16-mn-iameter 4.6 grade bolts 4,2 078x24" = 0.78% (6) =15683mm" 1" 1S 800-2007 page 75 clause 103.3] => Bt Atm ds M3 te m=1,n=0 57.95 = S8KN = 2x29=58 kNes ee GATE ACADEMY PUBLICATION, | MY PUBLICATIONS® ater 8, ACADE ce oo SS 25K 058x16%8X410 {a Strength in bearing on $m plate 2.5hdh/ = 125 = 60.7eN Fo Assume 8mm thick gusset, dhe gusset plate is sandwiched between the angles and hence the bolts will bein double shear. (9 fn S2 e097 For 16-mm diameter property class 4.6 bolt, & | Strength in double shear {1S 800-2007 page 75 clause 10.3.3) da wr cette Nt n=l nn0 | ? = 156.83 mm? f x (essuming e=35mm) P9252 0 9 25= i (F-02505 025-049 Lo 400 Similarly, twvo 6 mm thick angles © 4-097 16410) <9) 51 “4 On ‘Thus, Strength of the bol imum of strength in shear and bearing) A= 049 (minimum of a, b,c é) 25x0.49x16x8%410 Required number of bolts = A=258~3 ass) SLA KN | Hence, ‘Therefore, provide three bolts as shown in figure. ee EY | Strength of bolt =51.4kN (Minimum of strength in shear and bearing) Question 30 i Nts for 200 kN=200/51.4=3.9 (Hence provide four Design a connection of a truss joint as shown in figure using M16 back bell = 200 kN =200/51.4 =3.9 (Hence provide four bolts) Property class 46 and grade 410 steel Assume thatthe members shown are capa) ols For 150 kN = 150/51.4=2.9 (Hence provide three balls) Bolts for 350 kN =350/51.4 ~ 6. (Hence provide seven bolts) L—_ GATE ACADEMY PUBLICATION, Stractural Engineering Design “1 Provide edge distance =2x16=32, say 35 mm =40 mm (Check for gusset plate Distance from first bolt toast bolt in member earrying 200 KN =37=3%40 =120 mq Whitmore effective width = 2tan30? x120 =138.56 mm Capacity of pat = 224 = SRST 30 Hence the connection is safe The connection should be checked for block shear fakin, “The required calculations are shown a follows: Net length of shear face = (340+35)—3.5x18=92 mm [Net length of tension face =35~0.5x18 =26 mm Ag =6«(9%404+35) =930 man? Ay = 6x92 = 552 mn? 4 =6x35= 210 mn? 4, =6x26=156 mm? Aghy Hh] 30x250 , 0.9%410%156 _ 168 08 gy Ta - = *\Bty tm | Vand 135 09f,Aa , SrAy | _0:9%410%552 , 250%210 Ani - +54). + 250210 14.8 Thus, Ta =141.8 2N (inimum of Ty & Ty)? 200/2= 100 EN Hence, the connection is safe. Adopt the structural details as shown in fig, ANS 2001 (92001 350«—- O15" be rs1se5x6 Exact gusset shape ‘depends upon scale Gaestion 31 ‘Design joint Q of a truss as shown in fig.1 Use Ma oe attr he hes (2270x70x10) 22 80x80x10) ae ‘Schedule of load and section atjoint Q. Member Desiga Toad Seaion Used PO TORY 2arOxToxi9 wR 225KN 2280x8010 es 75kN ALTSK7Sx8 or THON Zax txe ICSVTU May 2015), |. Assume 8mm thick gusset, the gusset plate is sandwiched between the angles and hence the bolts will bein double shear. For 6mm diameter property cass 46 bolt, Strength in double shear [1S 800-2007 page 75 clause 103.3] =f token NS nan, 90 = pyrex’ 4,-0.7 = 0.784% (6) =156830* 400/1%15683)]_ af 409 =57.95 = 58kN {s0( S85) Sass Strength in bearing =2.5k, -dt-f, 4, willbe smaller of \ @ £-35 2 e=35om) 37a (assuming » P0252 025-049 (f-005 5 ) ) 2-025-025-0498 34,075" 549705" La 400 _ Jo 810°" @1Sc cieeemnee 2 enim | So ibe) ssa PUBLICATIONS tal 5-049 (eninimum of a,b, ¢ d) | gyrp can a = 2SHOAD BAO 5 yay and to Be al 123 | 194105582 250% Strengthof bolt =S14kN (Minimum of strength in shear and bearing) ; no ol Tq “141. KN (nininum off, & )>200/2-100 : ots formember PQ. 1D AN = #943. gene provide four bot tence the connections safe, Adopt the structural deals asshownin fig.) Ans 2 238 " | = 2.0437 (hence provide five bolts) Bots for member QR, 225 KN = 225.0437 (hence p sf - Bolts for member QS, 75 KN = 75. =1.46 (hence provide 2 bolts) Bolts for member QT, 110 kN 14 (hence provide 3 bolts) ‘Question 32 An ISMB 600 is connected to a column by web cleats with a single row of bolts. If the reaction is 350 KN and there are four 20 mm diameter bolts through the web, asin. figure. Check ifthe section is adequate for block shear failure. ann Capacity of plate = 224A. = 0:9%138.56%8%410 _ 597.99 gy > 200 kN Hence the connection is safe. The connection should be checked for block shear failure ‘The required calculations are shown as follows : Netlength of shear face = (3x40-+35)-3.5x18=92 mm. en iain Net length of tension face =35~0.5x18=26 mm | $0L Given data: Bolt diameter (d) = 20mm 4, =6x(3%40435) =930 mm? From steel table page 4, for ISMB 600 properties are: =6x92=552 mm? Web thickness = 12 mm ees | 5 e epee Block shear capacity will be minimum of A, = 6226 =156 mm? | rw Aah, 0h n [#4 995, 4, eB” 135 fae aie Design strength ofthe web =250 Nim? $30%250 | 0.9410 | Tom a 156 xt Tas 716808 kNYr YUBLICATIONS* 2. ess sen tate ferent welding process are as follows ‘Welding : In Forge welding the edges tobe joined are heated tothe plastic state and then joined by applying an extremely high extemal medarie, ding, a mixture of iron oxide and aluminium ited. Iron oxide reduced to molten metal is deposited at the jeint by a mould constructed around the joint. 3, Gas Welding : In gas welding, the edges to be joined are melted by an oxyacetylene gas flame. Additional metal is filled by melting a welding rd inthe flame. “Therefore, The vale of Ty =466324N esc eng ea eae wey ef w pa “ "hie anes pressed together and current i p ‘one part oofter 5 tee A oe Ae the flow of current at the joint increases, the temperature. When welding {EXEA Welded Connections 1ed, mechanical pressures applied of forge weld i i a method of connecting Welding : In elecricare welding, het is applied by means of an that fusion occu tlectric are struck between the parts to be welded, and an electrode held by 2 “The different processes of arc welding that are used in structural steel application x two pieces of metal by heating to a plastic or fit 5. Electric trelder or automatic machine. During welding the electrode melts and fills the gap astollows: , at the joint. Shielded metal arc welding (SMAW) For structural works, only clecticare welding is used. Submerged arc welding GAW) Gas shielded metal arc welding (GMAW) Qeestien 28, Flux core are welding (FCAW) = Whatare the diferent proce ofr vein? Electro slag welding (ESW) 18. The different arc welding processes are: Stud welding (SW) 1. Shielded metal arc welding (SMAW) 1. Submerged arc welding (SAW) Question 33, 2 3. Gas-shielded mata are welding GMAW) 4 s Draw a flow chart of welding processes. |. Flux core are welding (FCAW) Electro slag welding (ESW) 6 Stud welding SW) Question 36 are chosen Balt the parameter on which diferent wing process enemy swaw | ‘abit shop, ot | Weal TTecaion of ie welding operation + 1 welding ls done in abrcation sop FCA SAW, GMAW, FAW and ESW canbe used. SW ae | For field applications SMAW is prefered.senimuing migwtt —_frH4 _—_CATEACADEEURCH gg | 5 “pecuraey of setting up SAW, spray transfer GMAW, and ESW require a setup. Penetration of weld : Penetration of FCAW and SAW is better than SMAW, Volume of weld to be deposited : FCAW, GMAW, and ESW aye yp dey ton rates. 5. Postion of welding : SAW and BSW are not suitable for overhead pox, | “A FCAW and GMAW can be use in all positions. (ERED advantages & Disadvantages of Welded Joints arate ein dene of wel oss rt CSVTU Dang ‘Ans. Merits of welded joins: 1. Asnotoles are required for welding, the structural members are more effective, taking load. 2. Theoverall weight of structural stee! required is reduced by the use of welds joints, if 3. Welded joints are often economical as less labour and material are required fora joint. | 4. The welded connections look better than the usually pleasant structure are formal, than bolted or riveted joints. 5. Thespeed of fabrication is higher with the welding process. 6. Any shape of joint can be made with ease. 7. The welding process requires less working space than the riveting process. 8. Complete rigid joints can be provided with the welding process. 9. Nonoise is produced in the welding process asin the riveting process. Disadvantages of welded joints : 1. Skilled labour and electricity ae required for welding. 2 Internal stresses and warping are produced due to uneven heating and cooling 5 Welded joints are more brittle due to uneven cooling and heating and therefor their ftique strength is less than the members joined, ‘ eee ‘ait pockets, slag inclusion and incomplete penetration a*° | 5. Complete supervsions required for finding defects in weld scaDEMY PUBLICATIONS ® A Types of Welds — cation 38 - & What The welds may be grouped into four types as follows: [ are the types of welds? Explain in brief about tele properticn {CSVTU May 2016), 1. Groove welds 3, Slot welds 4. Plug welds Li Le a2 =, 2. Fillet welds DAMA (@) Groove welds (Fillet welds Ends shall be semi circular iit BILie ees a Section AA Secon AA Slot weld Plug welds © © 1. Groove Welds : Groove welds are used to connect structural members that are aligned in the same plane and often used in butt joints. Groove welds may_ also be used in T-connections. The grooves have a slope of30° 60°. The Square groove weld is used to connect plates up to 8—mm thickness. 2 Fillet weld : Fillet welds are most widely used due to their economy, ease of fabrication, and adoptability at site. They are approximately triangular in ‘cross section and a few example of application of fillet weld are shown in Fig. Unlike groove welds, they require less precision in ‘fitting up’ two sections, due to the overlapping of pieces. Hence, they are adopted in field as well as shop welding.Structural Engineering Design -II___2-46 SATE ACADENT FURL qapenty PUBLICATIONS 2-47 i Slot and plug welds ae not used exclusively in stel construction yw SE*T Comer joint: Comer joints are used to form balltuprecunguavbomaren becomes impossible to use filet welds or when the length ofthe iy "Sy which may be used as columns or beams toressthigh onion force Tut, ot and pug welds are used t supplement the filet ye : welds are also occasionally used on fill up holes in connections, uch a"Y cee we mney co stn oa align members prior to welding. They are also assumed to fai jt Moreover, the inspection of these welds is difficult. Slot and plug yas" wey, useful in preventing overlapping parts from bucking. . Fig. Corner joint RTA types of Welded Joints 4. Edge joint : Edge joints are not used in structural engineering applications they | ‘are used to keep two or more plates in position. Question 39 What are the types of welded joints? ‘Ans, The five basic types of welded joints are : 1 Buttjoint: A butt joint is used to join the ends of flat plates of nearly equal thickness. The butt joint obtained from a full penetration groove weld has 1094, efficiency. Fig. Butt joint a Transverse efficient na fillet welds. Comment. 2. Lap joint: Lap joints are most commonly used because they offer eas offing RK Mee eT ee ne : and ease of jointing, Lap joints utilize fillet welds and hence are well suited 6 | ICSVTU Dex 20181 shop as well s field welding. ‘Ans. The transverse fillet welds will be subjected to more uniform stress distribution and amped therefore are more efficient, which is shown in figure. | - Fig, Lap joint Pp —P 3. Tjoint: T joints are often used to fabricate built -up sections such as T- saps! + shapes plate girders, hangers, brackets, and stiffeners, where two plates ae) at right angles. T joints can be made by either using fillet or groove weld 1. More load at the ends in longitudinal fillet welds leading to non-uniform stress distribution in the weld. tt _id Double bevel groove tress distribution. Ss " 2 Transverse fillet welds loaded uniformly lead to more uniform stress distribu Fig. joints +e rp 1uctural Enginoering Design -I1 EA Weta specifications iestion 44 ~ > y PUBLICATIONS? 2-49 capes GATE ACADEMY: PUBLICATIOg, wee grfective length of weld : The effective length fet that length which is ofthe specified si fe eld shall be taken as only ed Sarin eg Se Seach of weld ade he ete ee pa size, but rot less than four times the size ofthe weld. oe ‘What are the weld specification as per code? a | What are the design specification for welded joints as per 1S:800-2007? ‘The effective area of a plug weld shall be consicered athe ominal area of the hole in the plane ofthe faying surface. These weld shall not be dened carry stresses. 4, End Return: Us 900-2007 where the depth of penetration beyond the root age 78, clause 105.11) Rig, minimum of 24mm, the sie ofthe fillet should be taken a5 the minimum te hus 2.4mm nd return The size of weld shall not be less than 3 mm S23mm | Fillet welds terminating at the ends or sides of parts should be ‘Thickness of thicker panl plate | Minimum size @) continously around the comers fora distance of notes than twice thes ofthe Up toand including 10mm 3mm ‘weld, unless itis impractical to do so. Tiss particularly important on the tens i \ end of parts carrying bending loads. a Over 10mm and including 20 mm Sm (Over 20mm and including 32: 6: ¢ 5, Intermittent weld : [IS 800-2007, Page 79 clause 1055] 20mm and including 32mm mm Unless otherwise specified, the intermittent fillet welding shall have an Over 32mm and including S0mm fmm 2. Effective throat thickness : The effective | eticcive length of not les than four ines the weld size, with «minimum of thickness of a fillet shall not he 0 mm. than 3mm and shall generally not exceed under special circumstit, where ris the thickness of the thinner plate o ‘The clear spacing between the effective lengths of intermittent fillet weld ents being welded, shall not exceed 12 and 16 times the thickness of thinner plate joined, for For the purpose of stress calculation in fillet welds joining faces inctined Compression and tension joint respectively, and inno case be more han 200 mn. cach thes the effective throat thickness shall be taken as K times the fils, prereset mie anmmear teehee 2 - length of not less than four ties the weld size ee Ee constant depending upon the angle between fusion faces. As gi the effective length of welds shall not be more than 16 times the thickness of the inthe table below. thinner part joined. The intermittent welds shall not be used in positions subject to Value of K different Angle between fusion fi dynamic, repetitive and alternating stresses. ‘Angle between . Fedoatane” | 6-30 | o1-i00" | 101-106 | 107-115 | 10-2! BEA design stresses in Welds Sonsant x | ~o70 [as [om | ass] a) ievstovPtag weld: 1s 207 Page 79 late 7057 Throat thickness | ‘The design stress fora fillet weld fy shallbe based on the throat aren and shall Sit wea fiven by fun Where, f i Smaller of uitimate sess of weld of parent ictal | Yn =Partial safely factor = 1.25 fr shop fabrications, and 1.50 forPp = zl are ACADEMY PUBL sncu EagoneDeen-D__ 2-80 cen. | 7 Groove orbott weld: ; For single grooved welds the throat thickness taken equal 9 he thickness. | thinner plate. vor double grooved welds the throat thickness taken equal to the thickney the thinner plate. : Design stress forthe weld = Joy =~ Design strength of weld fora length L= Pa, = Lifus = —°* | 1, = Throat thickness Site weld: , at “The design strength in shear and tension for sit structural members shall be calculated same as tha safely factor of 15. Sua {15 800-2007 Page 79 Clause 10579 | fe during erection weld but using a part | wf tm oh inn af | Sa Burs | 4 Long joint: [1S 800-2007, Page 79 clause 1057.3) When the length of the welded joint, {,of a splice or end conection ina compressa | cor tension element is greater than 150 1, The design capacity o Sag Shall be reduced by the factor 024 n12-2Zhsio Be ol2-FaS Where, 1, = Length of the joint in the direetion of the force transfer, and 1, = Throat size of the weld. ‘Question 42 What are the assumptions made in analysis of welded joints? Ans. The following assumptions are made in analysis of welded joints: 1. The welds connecting various parts are homogenous, isotropic and elastic: | mpd connected by the welds are rigid and their deformation 36H") nly stresses due to extemal ores are considered. The effect of residual | tress concentrations, and the shape of the weld are neglected. | Lo 2 3. a _soapente PUBL | Da numerical Based on Welded Connections ‘ques tion 43 in groove weld is provided fo connect wo plates of thickness 18mm and 6 mm The ICATIONS® 261 factored tensile force on the joint is 450 KN. The length of the weld povided ‘o0 mm - Investigate the safety of the joint if “ ee {a) Single V-groove weld is provided (b) Double V-groove weld is provided. Assume Shop welding ‘Given data : Length of weld = 160 mm, Factored vnsile force on int = ASD IN (a) When a single V-groove weld is provided : For this case the throat thickness of the weld, iene Wom 1, =8 (Thickness of thinner plate) ‘This is less than the factored load 450 kN .. The connection is not safe (b) When a double V-groove weld is provided : Design strength of the weld =I. fos Find the design strength of 6 mm shop fillet weld permmlength of weld. Take J, = 410 Winn’, Given data Size of weld s = 6mm, F.=490Nimm?, 7,,=1.25 (Shop fabrication) ‘ In case of fillet weld tus so 207 page 79 esse 10571 Design stress for the weld = fos =‘Structural Engineoring Dosign - II Design strength of the weld per mm Lexi hae aking L, = 1mm) mhersesere | Ang, ‘Question 45 Two plates A and B are connected by a pair of fillet welds as shown in figure. Find yy Size ofthe weld required if the plate B is subjected a factored pul of 2500 N per pe width. Assume shop welding, capes PUBLICATIONS? engi of weld = 80m ofthe lengths of ams of eaage ion 47 pesgnstrenath ofthe weld = Zora of wg =100 1 > 80 aN jde 10 mim weld provide 10 A sommBinm plate i Wo be conned 63120 mmc mm ale nap to woven factored ten oe of 151%, provdings om se wl Coogi connection. Sol. Given data: Factored pall = 2500 Nimim width ri In case of filet weld 0s 800-2007 page 79, clause 103373, | 5 L Design stress forthe weld = = 7 12 mm phate Al SSB eke “yrs Nim? | Let the size of the weld be s mm ‘Design strength of two fillet welds per mm length = [Lf] = 2x 529. 9.4] = 2500N/mam, Question 46 A splice has become necessary in atong tension member carrying a factored tension | | S0KN. The member is a single angle 60 mmx 40 mm%8 mm, Design a single V-groove | welded joint Given data: Throat thickness ofthe V-groove weld =5( 3 Factored tension =80 kN, | ISA 60mm 40mmx8mm Given data : Factored tensile force = 125 1N, Size of site weld = 6 mim , Thickness ofthe plate =1=8.nm newest a ee IS 800-2007 page 79 clause 105.72) ‘Throat thickness of weld =, =0.7 5=0.7%6=42 mm Design stress forthe weld= j., =Z Bre 7.8 Nim Design strength per mm length of weld = Soa *LXt, where (1, = 0.75) =1578x1x0.7x6= 662.8 Nimm Factored load =125 kN Blfective length of weld required = 25210" 169 mm Peabiding only longitudinal weld, length of weld on ech side - 2 =94.5 mm say 95 mm Minimum length of each longitudinal weld = perpendicular distance between the longitudinal weld = 89 mm Each end return =25=2x6=12 mm oer)yy es et is PI Structural Enginooring Design-I_2- 54 GATE ACADEMY PUBLICATION, goyer cmos ats AC ‘Question 48 S| use: _— [Ate bar100 mm>x16 mm is welded to another plate as shows in figure I e = toa factored pull of 300 kN. Find the minimum overlap required if & 370) welds are use. 7 Given data: Factored pull=370 WN, Siz of ste wld hoop thickness ofthe Weld #,=0.75= 0.74642 mig incase of site weld 1S 800-2007 page 79 clause 1057.2 Design stress forthe weld _ fh. _at0 ; fous Fie FS 01578 Nin Design strength per mm length of weld 20,1, «157834226618 Ninn Factored pull =370 KN oh ‘Sol. Given data : Factored pull =300KN, Size of fillet weld = 8mm ‘Throat thickness of the weld =(, = 0.7 =0.7x8=5.6 mm oe weld required = 4x= 200210 655 9 sam In case of ste fillet weld [1S 800-2007 page 79 clause 10.5.7.2] | Length of weld req eong 582 Design stress for the weld whan 7.8 Nim? 4 #91396 mm ten | Provide an overlap of 140 mm a Design strength per mm length of the weld f., 4, x4, =157.8%1x5.6 = 883.7 Nima(/-| Let the overlap be x mm . fos 2 Length of weld required Figure shows a welded lap joint provided to connect two plates {0 mm:12 mam with 8 mm fillet weld. Investigate the safety of the design, Assume shop welding, 30x10" 883.7 x=60.74 mm Provide an overlap x=70 mm Ans ‘Question 49 2x+2(100)= =339.48mm For the welded joint shoven in figure Find the minimum overlap x fora factored pl of 370 KNif 6 mm site fillet welds are used. x __y J 70 —r}e— 100 —F}e— 70 4} po ee ree So, Gp ] Given data Size of fillet weld, s=8mm Throat thickness of weld =/, =0.7x8= 5.6 mmr Design stress forthe weld = Joy i Design strength permmlength of weld = fo. «2, ‘otal length of weld provided = For(5)and (6), [= 2x50 Since the design strength of the weld is greater than the design strength ofthe plate, the connection is safe. Aas ‘Question 54 ‘A 150 mmx10 mm plate and a180 mmx10 mm plate are to be connected in a lap joint by shop weld. Design the connection for the full strength of the 150 mm x10 zmm plate. |. Given data :Sizeof plate 1= 150mm x 10mm Size of plate 2~180 mm x 10 mm Design tengo 150mm mm plate= 48 Size of the weld, Minimum size = 3 mm Maximum size =10-1.5 =8.5 mm Provide 8 mm shop weld. [IS 800-2007 page 79 clause 10.5.7.1] i ee Design strength per mm length ofthe weld =189.4x0.7x8 = 1060.6 Nim =1894 Némm? cap SMY PUBLICATIONS 2-87 3 Fastenert insuct an arrangement the distance between the lngiudinal wale wal az he. This distance shall not exceed 16,10 =160 mm a ott tus therefore provide two longitudinal and one transverse weld Length of the transverse weld =150 mm 322-150 F786 mm, say 90 mm ‘Ans, Length of each longitudinal weld = 150%10 180500 10mm 10 mm ‘Question 52 Design a double cover butt joint to connect two plates 275 mm»12 mn to mobilize {ull tensile strength of the plate. Use shop weld. Sol. Given data: Two plates of 27Smmx12mm_ 2 Fall tensile strength of the plate a. Width of cover plates =275-2*20 =235 mm Provide 235 mm8 mm cover plates. ‘Area of the two cover plates = 2(235%}=3760 mn" of main plat) ‘Area of the cover plates should be not less than 1.05 x area of main “Area of cover plates provide i satisfactory:Structural Engineoring Design -I|__ 2-58. GATE ACADEMY PUD 235mmx8mm Ea Size of the weld Minimum size:5 mm (for plate of thickness 10 mm to20 mm ) Maximum size1.$ mm les than the thickness of thinner plate= 8—1.5= 65 mm Provide $ mm weld. Thwoat thickness of the weld =f, =0.7x5 #35 mm | {18800-2007 Page 78, Clause 10.75.11 Dasgs stress forthe weld = f= = Mme Designs strength per mm length of weld =I Length of weld required = 750000, ae 5609 tetotinronin flt=2e2) 4 233 CATIONS ictura! Engineoring Devign “Il __2-68 _eaTaeesnuucarionyy LA, 25000010) gn tension for the tie bar =——£ =. pau DR tm 110 <2 N ‘otal length of weld = 2(120)+2(70) +100 mm = 480 mm 115 600-2007, page 79, clause 10.75.1] t 410 i for the weld, f.. = fae = : Design stress for Soa a gg 1ST Nr Latthe size ofthe weld bes mm. Design strength of the weld permm length Design strength of the whole weld =(157, 52429 mm Provide 5 mm fillet weld, ee PZD Fitet Weld for Truss Member Truss member are composed of singleangle or doubleangle sections. The following Points should be borne in mind while designing, the weld. 1. The calculated weld length is placed as longitudinal fillet welds either on the two ‘es parallel to the ais ofthe load (ig) or an tre sides as sho in FO) 4. transverse welds along with longitudinal welds. longitudinal fillet weld should never be placed on one side only asthere willbe possi of at oe The centre of gravity of the weld should coincide with the ee will be placed ‘Red as truss member. I the member is aymmetio We WEN TT, ok 2 truss member, the lengths of longitial filet w= ‘hetwo sides, as shown in Fig. (a), toacheve the above cnATEACADEIT INL | Re Fig) Let iy; = Length of longitudinal fillet welds on two sides ‘3,2, =Factored design loads along lengths ,and J, respectively P= Factored load acting on the centroid of the section Taking moment about the line passing through length J, Fig.(a)), Ph-Ph, =0 Or fh afk Bay Similarly, well) ‘Once the factored design force FP, are known the fillet weld lengths can be designe. | Te eictvenen fle wold canbe crus hy ocetig the tw ie ma an end weld as shown in Fig. (b). Further, when the weld length to be provided ami be accommodated over the two parallel sides because of limited overlap, it may bet ‘only choice. This arrangement also reduces the size of gusset plate and results in ca Let 1, L, = Length of longitudinal fillet welds ‘= Length of end fillet weld P= Factored load acting on the centroid of the section ‘7, = Fectored design force in the end fillet weld =n, L fa Fre “ Total length of weld requited=2,+1,4h no aking moment about the ine passing thr ih 1, Ph, & h h Bh R A Ph ao sel apsp PUBLICATIONS? Eg moment abou elie pasting trough oa numerical Based on Fillet Weld for Truss Member 64 | Taaestton . | | ram>50 mm>x8 mm angle is to be connected to a gustet plate by 6 mmfllt Gol. 2-01 A Ph+BZ-Ph=0 eto) since fis knowns the equation () to) ean be solved for Rand Once the design fore Rand Fare known the filet weld enghs Land Ieanbe designed, AT welds at the extremities of the longer leg. Design the welded connection {erresponding to the full tensile strength ofthe angle. Assume shop welding. Given data :ISA 7Smm%50mm%8mm Lpdg _ 250x938 3182 N Fulltensle strength ofthe angle = ? = -"t-= This force acts at the level of the longitudinal centroidal axis of the angle (see figure). Let the Force transmitted to the welds atthe lower and upper edges ofthe longer leg of the angle be Rand P,. R+P, =P =213182N Taking moments about the bottom weld, 2x75 = 213182252 2 =71629N : 2 =213182-71629= 141553 N Design stress in the weld fy “ey =189.4 Nom? 110 21994x1x42 = 79548 Nim Design strength ofthe weld per mm length‘he tension member of «truss consists of 2 angles 80 mmx30-mmx8 mm ang welded on ether side ofa gusset pte, The member i subjected toa factored tag force of275 EN. Design a fillet weld connection. Assume shop welding, ‘Given data 2 ISA 80mm»S0mmx8mm, Factored tensile force = 275kN Let Sim welds be provided. [ts 800-2007 page 79 clause 10573 fa hr Design stress fr weld, fog =" =1894 Nimm? Bu 25 Factored tensile force on the member=275 EN acting along the longitudinal cen ‘axis of the member, See figure. Let the fore transmitted tothe welds a the lower and upper edges of the loge ofthe member by P and 2. FAR = P=275000N Taking moments about the bottom welds ae R= 93843. Gok, Given data :1SA 1 Fagg 71366 mm 3137 mm ich top weld =), = Length of ea* Ans. thick Use 5 0m field weld. The welding scheme is shown in igure, Factored ten: '=200 KN, Gusset plate thickness = 8 mm ‘Size of weld~s~5 mm, Throat thickness= 1, =0.7x5=3.5 mm {1S 800.2007 page 79 clause 105.7:1] Design stress for the weld, a fu A = Sos Design strength per mm length of weld Length of weld required = 2 Provide x = 85 mmand x, =203 mm‘Question 57 etemine the strength and efficiency of the lap joint shown in figure. The jy, bo mm diameter and of grade 46. The two plats © be joined are 10 mm ang thick steal is of grade Fe 410). Step Gb Given data ForFe 410 grade of stel: /,=410MPa Forbolts of grade 4.6: f, =400MPa Bolt diameter (d)= 20mm, Hole diameter (4,)=20+2mm = 22mm a Ni For20 mm diameter bolt: rete -anct eet | Pa Ayn 0rbxhd = 0785 | an 4d, and d, = 22mm | aaa “=Partial safety factor for material of bolt = 1.25 | pan 50mm’ 1, ~Partial ae facior for resistance governed by ultimate stress=125 | The bolts will be in single shear and bearing. The bearing strength ofthe bolt wil governed by the thickness ofthe thinner plate. Hence, = 10mm Step (i: Stength ofthe bolt in single shear: [1S 80-2007 page 75 clause 1033) / sirength of solid plate per pitch length =0:9x 101 00x10=295.24N Cae oze iciency ofthe joint (two bolts fall in one pitch) | Step (ii) : Strength of the bolt in bearing : [1S 800-2007 page 75 clause 103.4] Guestion 582 ‘Ailetmember consists of an ISA 8080mmx8mm (Fe-410 grade steel is welded to 2 Sf ett pine tes Doge wi rt nd ple Se strength ofthe member. Sketch the welded connection. [CSVTU May 2015, Dec 2015 Sol Given data ISA 80*80x8 mm, From steel table, 4, =1220 mm", C, =242 mm, For Fe-410, f, =410 MPa, f,=250 MPa Partial safety factor for steel = Ym, =11 (P=25d=2.5x20=500m Partial safety factor for site welding = Yq. "15 Design strength of member Here, fy isleastof 5 % (assuming e= 33mm) Hence, Strength ofthe joint per pitch length in bearing =2x80.0=160.0KN (robot finance) | ‘a4 GATE ACADEMY PUnyp, iy