CV3012 EC3 Design of Bolted Welded Connections Xx1

March 28, 2018 | Author: Peiling Teoh | Category: Welding, Screw, Structural Steel, Beam (Structure), Joining


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CV3012: Steel Design – Eurocode 3 Loadings – one way slabs for steel structures  Members subjected to axial loads only  Fully restrained beams  Completely unrestrained beams  Columns subjected to axial load and nominal moments  Columns subjected to axial load and moments  Angle, channel and tee sections under tension and compression  Typical structural connections CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 1 Types of Joints or Connections  Pinned joints  A pinned joint should be capable of transmitting the internal forces, without developing significant moments which might adversely affect the members or the structure as a whole.  A nominally pinned joint should be capable of accepting the resulting rotations under the design loads.  Rigid joints  Joints classified as rigid may be assumed to have sufficient rotational stiffness to justify analysis based on full continuity. CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 2 Typical Simple (Pinned) Connections NEd (a) Web Cleats NEd (b) End Plate NEd (c) Fin Plates CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 3 Typical Rigid (Moment) Connections MEd (a) Haunch Connection NEd MEd (b) Extended End Plate NEd CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 4 . A Typical Pinned Connection NEd CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 5 . Ed CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 6 .Bolt Subjected to Shear and Tension VEd Nt. A Typical Rigid Connection NEd F1 MEd Nt.Ed F1 CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 7 . Failure of Plate by Block Tearing and Plain Shear NEd NEd CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 8 A Typical Connection – Bracket Plate  Components check – bolt group under resultant shear, plain shear and block tearing of bracket plate. NEd = 75 kN 40 50 50 40 50 50 40 12 mm M20 class 8.8 bolt 203  203 UC 60 (Tf = 14.2) CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 9 A Typical Connection – Welded End Plate  Components check – bolt group under direct shear, end plate under plain shear & block tearing, strength of fillet welds, beam web under local shear across partial depth D. NEd = 65 kN D CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 10 bolt group B under direct shear. and reduced moment capacity at notched cross beam end NEd = 165 kN Main beam Cross beam Web cleat Bolt group B through main beam Bolt group A through cross beam CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 11 . block tearing of cross beam.A Typical Connection – Double Web Cleats  Components check – bolt group A under resultant shear. web cleat under plain shear and block tearing. Ed VEd MEd NEd Compression zone checks: column unstiffened flange. column web tension. tension zone. end plate in bending. column web crushing. beam web to end plate weld. CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 12 . Horizontal shear zone check: column web panel shear. NEd Nt. horizontal shear zone and vertical shear zone Tension zone checks: bolts in tension. column unstiffened flange. Vertical shear zone checks: Bolts in direct shear. column web bearing.A Typical Connection – Extended End Plate  Components check – compression zone. flange to end plate weld. A Typical Connection – Base Plate  Components check – base plate thickness and holding down bolts Nc.Ed fck= 30 N/mm2 t w1 w2 CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 13 . 10.9 bolts  Clause 3. end and edge distances  Clause 3.9 – Slip-resistant connections using 8. end and edge distances  Clause 3. nuts and washers  Clause 3.4 – Categories of bolted connections  Clause 3.2 – Design for block tearing  Clause 3.5(1) – Minimum spacing.Eurocode 3: Part 1-8: Design of Joints  Clause 3.5 – Positioning of holes for bolts and rivets  Clause 3.11(1) – Prying forces CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 14 .1 – Bolts.8 or 10.10 – Deductions for fastener holes  Clause 3.5(2) – Maximum spacing.6 – Design resistance of individual fasteners  Clause 3.8 – Long joints  Clause 3. 2.3 – Design resistance of fillet welds  Clause 4.8.5.2.2.5.3.2.12 – Anchor bolt in tension  Clause 6.3 – Column base subjected to axial forces and bending moments CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 15 .Eurocode 3: Part 1-8: Design of Joints  Clause 4.3.7 – Design resistance of butt welds  Clause 4.5.2 – Directional method  Clause 4.1 – General method  Clause 4.2.7.3 – Simplified method for design resistance of fillet weld  Clause 4.7 – Design moment resistance of beam-to-column joints and spices  Clause 6.8.8 – Design resistance of column bases with base plates  Clause 6.2 – Column base subjected only to axial forces  Clause 6.6.1 – Full penetration butt welds  Clause 6.3.5. Critical parts to be checked in connections  Base plate thickness  Plain shear  Block tearing  Bolts in shear and bearing  Bolts in tension  Design moment resistance of an unstiffened flange CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 16 . Effective Edge Distance EC3:1-8 – Clause 6.2.Base Plates .5(4) and Clause 6.8 2c+tf 2c+tw 2c+tw Effective edges offset by a distance c CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 17 .2. 2. A eff  (h  2c)(b  2c)  (h  2c  2t f )(b  t w )  (hb  2bc  2hc  4c 2 )  (hb  2bc  2bt f  ht w  2t w c  2t f t w )  4c 2  (2h  4b  2t w )c  (2bt f  ht w  2t f t w )  4c 2  (section perimeter) c  (section area) (h-2c-2tf) (tf+2c) a tw c (b + 2c) – (tw + 2c) tf b (tw+2c) = (b – tw) (b+2c) = 2a a h (h+2c) CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 18 . EC3:1-8 – Clause 6.5(4) Effective Area.Effective area. 5(4) where M0 = 1. EC3:1-8 – Clause 6. the compressive strength of concrete fcd is defined as fcd=(ccfck)/c where c=partial safety factor of concrete (1.5 EC3:1-8 – Clause 6.2. equate effective area Aeff with required area Areq to obtain the value of c  Substitute c into the following equation to get the base plates thickness t  3f   t  c  jd M 0  fy    0.85 for compression.2. fck=characteristic cylinder strength  It is shown that fcd = fjd  Eurocode 3:1-8 Clause 6.5(7)  Using the effective area method. cc=0.1 – Clause 3.5(4)  The required area Areq = NEd / fjd where Ned = applied force and fjd = design bearing strength  With reference to EC2:1.6(1).Base plates thickness.5 for persistent and transient design).0 (UK National Annex) and fy = yield strength of the base plate CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 19 .2.1. t M y .5   3f   c  jd M 0   fy    0.0 t=d 1. Wel. My.Base plate thickness t NEd Assume unit width cantilever strip.Ed  f y Wel.5 c x CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 20 . bending moment at column face assuming single action under the baseplate.   t  c  3w   f y  0.Ed = wl2/2 where l = c. For a cantilever subjected under UDL.y 2 c w 2  wc  f y t 2 6 2 2  t 2  6  wc  3wc 2fy fy 1.y  I y max x 2 1.0  d t     12 d 6 6 bd 3 2 2 and.0 c Therefore. cube = 30 N/mm2 t 600 600 CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 21 .Example 1 4300 kN 305 x 305 UC 137 fck. 1 Clause 3.4    255.5  4c 2  (2h  4b  2t w )c  (2bt f  ht w  2t f t w )  4c 2  1850.1  c  1.5  50 mm CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 22 .1 0.4 mm  3(14.17  3f   t  c  jd M 0  fy    0.5  303.0   0. cube  30 N/mm 2 f ck  25 N/mm 2 EC2 :1.1 Clause 2.2(3) Table 3. A req  Therefore.17)(1.1N UK National Annex Table NA.1 Clause 3.2c  17243.2.5 10 3 mm 2  Solving A req  A eff gives c  122.26 Now.0)   122.85 (compression) f cd  A eff EC2 :1.4.17 N/mm 2 1.Example 1 f cd   f EC2 :1.4(1) Table 2.5 (persistent and transient design)  cc  0.17 N/mm 2  f jd  f cd  14. N f Ed jd  4300 10 3 14.16(1) cc ck c f ck.85(25)  14.1. Aeff CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 23 .Extension to eccentric base plate Ned=1380 kN e  70 mm Med=185 kNm e Taking moment about point A : N Ed  (h 2  e)  M Ed  FEd  (h  e) N Ed  TEd  FEd h/2 h/2 z A w TEd FEd Assume uniform pressure and symmetrical around the compression flange b + 2c tf + 2c Effective portion of the compression zone. 2(3) Table 3.3(1) 320.3 2  Fc. Ed = 1324.1N UK National Annex Table NA. Ed  379. cube  30 N/mm 2 f ck  25 N/mm 2 EC2 :1.65  10. 5 Lever arm.85(25)  14.3 + 185 = Fc.5  70  Taking moment about A.16(1) cc ck c f ck.2.5 (persistent and transient design)  cc  0.8. z  h  e  tf 2  320.85 (compression) f cd  EC2 :1.17 N/mm 2 1.5  70)  10.1 Clause 3.1 Clause 3.7  379.65 mm 2 EC3 :1.1 Clause 2. r.1.Example 2 f cd   f EC2 :1.1  c  1.233 kN CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 24 .8  Clause 6.1 0.2.17 N/mm 2  f jd  f cd  14. r. 1380  ( 21.4.4(1) Table 2. 233 10 3  93.2.5  34.2(1) = 4c 2 + 661.8  Clause 6. A req  N f Ed jd  1324.17)(1 .8.) Therefore.0)  265.453 103 mm 2  Solving A req  A eff gives c  86.) CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 25 .64 Now.517 mm  40 mm (O.8c + 6709.5    86.  3f γ  t  c  jd M 0  fy   0.0   0.181 mm 14.181  3(14.Example 2 A eff = (t f + 2c)(b + 2c) = 4c 2 + (2t f + 2b)c + t f b Eurocode 3 :1.17 h + 2c and b + 2c < 600 mm (O.K.K. Rd is design tension resistance of the bolt Holding down bolts should be designed for the effects of factored loading.Anchor bolts in tension – EC3: Part 1-8 – Clause 6.6.Ed  Ft.12 Shear key Base plate Grout Concrete foundation Ft.Rd where Ft. CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 26 .2. Where they are required to resist tension they should be properly anchored into the foundation by a washer plate or other load distributing member embedded in the concrete. 7  The applied design moment Mj.Rigid beam-to-column connections – EC3: Part 1-8 – Clause 6. Mj.Rd FRd is total resistance of the fillet welds. CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 27 .Rd is the design moment resistance of the joint.0 M j.2.Ed  1.Ed should satisfy: M j.Rd where Mj. 7 FRd is the total tension resistance of the bolts. FRd is the total resistance of the fillet welds or total tension resistance of the bolts. FRd is the total tension resistance of the bolts.2. whichever is lower.Rigid beam-to-column connections – EC3: Part 1-8 – Clause 6. CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 28 . e.Plain Shear – EC3: Clause 6. Vpl.e.Rd in practice.  The plastic shear resistance is essentially defined as the yield strength in shear multiplied by a shear area Av.3  The usual approach is to use the plastic shear resistance Vpl. i. Av = A – Ah where A = gross area and Ah = total area of holes CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 29 . i. Rd    AV       3  fy  M0 where Av = effective shear area.2. 10.Block Tearing – EC3: Clause 3.2  Block tearing consists of failure in shear at the row of bolts along the shear face of the hole group accompanied by tensile rupture along the line of bolt holes on the tension face of the bolt group Ned = design tension force 1 small tension force 2 large shear force 3 small shear force 4 large tension force Block Tearing CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 30 . Anv = net area subjected to shear CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 31 .2. (3.Rd is given by Veff.Rd   f u A nt  M2   1  f y A nv   3   M0   Eqn.10.Block Shear – EC3: Clause 3.10)  where Ant = net area subjected to tension.2  For a symmetric bolt group subject to concentric loading the design block tearing resistance. Veff.5 f u A nt  M2   1  f y A nv   3   M0   Eqn.9)  For a bolt group subject to eccentric loading the design block shear tearing resistance Veff.2.1.Rd is given by Veff.Rd  0.1. (3. 1.1.2(2) Veff.10.Rd Lv Lv End Plate Lt Lt Block shear capacity : Veff.1.Rd Veff.Block Shear – EC3: Clause 3.Rd  f u Ant / M2  (1/ 3)f y Anv /  M0 CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 32 . 2(3) Veff. Rd  0.5fuAnt / M2  (1/ 3)f yAnv /  M0 CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 33 .Rd Lv Cut-off top flange of beam Lt Block shear capacity : Veff.10.2.2.Block Shear – EC3: Clause 3. bolt bearing resistance. plain shear resistance and block tear resistance of the connection Veff.8 6 mm thick S275 fin plate 120 70 Lv 70 40 Lt CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 34 .Example 3  Determine bolt shear resistance.1.Rd 70 40 M20 Grade 8. 0  354.0.Example 3 Plain shear resistance: f y  275 N/mm 2 . A v  300(6)  3(20  2)(6)  1404 mm2 A (f / 3) 1404(275/ 3) Vpl. Rd  430(462)/1.25.5(20  2)](6)  462 mm 2 A nv  [(120  70  70)  2.25  (1/ 3)(275)(1230)/1.0  M0 Block tear resistance : f u  430 N/mm 2 .1.22 kN CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 35 . f y  275 N/mm 2 .  M 0  1.91 kN 1.  M 2  1.5(20  2)](6)  1230 mm 2 Veff.  M 0  1. Rd  v y   222.0 A nt  [(70  40)  1. Failure Modes of Bolt and Plate Top Plate Bottom Plate NEd Shear NEd Bearing Bearing on Plate and Bolt Bolt under Shear CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 36 . 6 and 8.6.Rd is given by Fv.5 for classes 4.6 for classes 4.8 and v = 0.  Tension resistance per tensile stress area. Ft.1  Shear resistance per shear plane. v = 0. A = A if the shear plane passes through the unthreaded portion of the bolt. 6.Rd is given by Ft. As is the tensile stress area of the bolt.63 for countersunk bolt. 5.8.6. Rd   f A v ub  M2 where fub = ultimate tensile strength (Table 3.9 for hexagonal bolt.1). Fv. Rd  k f A 2 ub  s M2 where fub = ultimate tensile strength (Table 3. 5.Shear and tension resistance of fasterners – EC3:1-8 – Clause 3. CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 37 . k2 = 0.9. and k2 = 0. A = As if the shear plane passes through the threaded portion of the bolt.1).8 and 10.8. Nominal values of ultimate tensile strength fub for bolts CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 38 . Shear resistance of bolts CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 39 . Bearing and tension resistances of bolts CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 40 . Bolt resistance – Non Preloaded, Class 4.6 CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 41 Bolt resistance – Non Preloaded, Class 8.8 CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 42 Bolt resistance – Non Preloaded, Class 10.9 CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 43 8  When the distance Lj between centers of two end bolts in a joint exceeds more than 15d (d=nominal bolt diameter).75 200d Lj CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 44 .0 and Lf  0.Rd of all fasteners should be reduced by multiplying it by a reduction factor Lf given by: Lf 1 L j 15d but Lf 1.Long joints – EC3: Part 1-8 – Clause 3. the design shear resistance Fv. Packing plates – EC3: Part 1-8 – Clause 3. the design shear resistance Fv.1  Where the fasteners transmitting load in shear and bearing pass through packing of total thickness tp greater than one-third of the nominal diameter d.6.0 CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 45 .Rd of all fasteners should be multiplied by a reduction factor p given by: p  9d 8 d  3 tp but p 1. Ed p2 d0 e3 0.5 e1 p1 e4 e2 Ft.5d0 End. edge distances and spacing End and edge distances for slotted holes Symbols for end and edge distances and spacing of fasteners CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 46 .Positioning of holes for bolts – EC3:1-8 – Clause 3. for inner bolts: 0 p 1 3d 0  1 4 perpendicular to the direction of load transfer: e .for end bolts: d  e 1 3d d  . fub/fu or 1. Fb.5 0 CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 47 .Rd is given by Fb.0.6.1  Bearing resistance per shear plane.7 or 2.for inner bolts: k1 is the smallest of 1.for edge bolts: k1 is the smallest of 2.8 2  1.4 p d 2  1.7 or 2.Bearing resistance of bolts – EC3:1-8 – Clause 3.5 d 0 . parallel to the direction of load transfer: . Rd  k  f dt 1 b u  M2 where b is the smallest of d. 8 .6(800)(245) 10 3  94.8  M2  1.6.Clause 2.08kN 1.6 for Class 8.08  564.8 . Fv.Example 3 Shear resistance of the bolts :  v f ub A Fv.25 For six bolts.Clause 3.48 kN CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 48 . Rd   M2 EC3 :1. Rd  6  94.1(3) Table 3.4 v  0.2(2) EC 3 :1.1(1) Table 3.1 EC Bolt Table 0. Fv.8 .25 f ub  800 N/mm2 A  As  245 mm2 For one bolt. Rd  EC 3 :1.1.Clause 3. 8 .811 3d0 4 3(22) 4  d smallest of 1.0  d smallest of 0.811 for end bolts for inner bolts for end bolts for inner bolts CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 49 .86 or 1.1(1) Table 3.86 or 1.818 3d0 3(22) p  d  1  1  70  1  0.811or 1.1(1) Table 3.1 Parallel to the direction of load transfer : e  d  1  120  1.818 or 1. Rd  k1 bf u dt  M2 f u  430 N/mm 2 f ub 800   1.0  0.86 f u 430 EC3 :1.0  1.4 EC 3 :1-1 .Clause 3.6.Clause 3.2.Example 3 Bearing resistance of bolts : Fb. 7  2.695 kN  M2 1.7  1. Rd  for end bolts for inner bolts Bearing resistance of 6 bolts  2(103.18 kN CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 50 .39 d 22 0 1.5(0.25 Fb.8 2  1.695)  541.4 p d 2  1.75 or 2.Example 3 Perpendicular to the direction of load transfer : e 40 2.5 k1 b f u dt 2. Rd  1 b u   83.75 22 for edge bolts for inner bolts k1  smallest of 3.5  2.7  2.39 or 2.8  1.811)(430)(20)(6) Fb.25 k  f dt 2.7  3.2) 4(83.5(1.4 0 70  1.2 kN  M2 1.0)(430)(20)(6)   103. Minimum and maximum spacing. end and edge distances CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 51 . Bolts Components Bolt 22 mm < 22 mm Washer Nut CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 52 . 1. Slip should not occur at the serviceability limit state. c) Category C: Slip-resistant at ultimate limit state In this category preloaded bolts in accordance with 3. No preloading is required.4.4 Shear connections – EC3:1-8 – Clause 3. The design serviceability shear load should not exceed the design slip resistance. The design ultimate shear load should not exceed the design shear resistance.1.2(1) should be used.2(1) should be used. b) Category B: Slip-resistant at serviceability limit state In this category preloaded bolts in accordance with 3.9 should be used.6 up to class 10.1 Bolted connections loaded in shear should be designed as one of the following: a) Category A: Bearing type In this category bolts from class 4.Categories of Bolt Connections – EC3:1-8 – Clause 3. CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 53 . Slip should not occur at the ultimate limit state. The design ultimate shear load should not exceed the design slip resistance. The design checks for these connections are summarized in Table 3.2 Bolted connection loaded in tension should be designed as one of the following: a) Category D: non-preloaded In this category bolts from class 4.9 bolts with controlled tightening in conformity with 1.6 up to class 10.4.4 Tension connections – EC3:1-8 – Clause 3.2. No preloading is required.9 should be used. CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 54 .2. b) Category E: preloaded In this category preloaded 8.Categories of Bolt Connections – EC3:1-8 – Clause 3.7 Reference Standards: Group 7 should be used.8 and 10. 4 CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 55 .Categories of Bolt Connections – EC3:1-8 – Clause 3. Bolts under Single and Double Shear Single Shear Shear NEd NEd Double Shear Shear NEd/2 NEd NEd/2 CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 56 . 8.Bolted Connections  Non-preloaded bolts of Class 4. 20 mm.8 and 10. 16 mm.9 in S275  Diameters are 12 mm. 24 mm and 30 mm  Direct shear or direct tension connections Bolts in single and double shear joints CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 57 .6. Single Shear and Double Shear The shear stress in the bolts is given by Single shear: Shear stress. of bolts (n)  Root Area (A s ) NEd  n  f b  As Double shear: Shear stress. f b  or Load (NEd ) 2  No. f b  or Load (NEd ) No. of bolts (n)  Root Area (A s ) NEd  2  n  f b  As CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 58 . Eccentric Bolted Connections There are two types of eccentrically loaded connections:   bolt group in direct shear and torsion bolt group in direct shear and tension P P e e G T = P×e M = P×e P P (a) Bolt group in direct shear and torsion (b) Bolt group in direct shear and tension CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 59 .  G  cos  = r2/r1 Therefore. FT cos  = FT  r2/r1 CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 60 .Bolt Group in Direct Shear and Torsion  The moment applied in the plane of the connection tends to rotate the side plate about the bolt group centre of gravity.  A linear variation of loading due to moment is assumed with the both farthest from the centre of gravity of the group carrying the greatest load.  The direct shear is divided equally between the bolts. . Let the force due to the moment on bolt “A” be FT.  FT r1 (r12  r22  .) ( x 2   y 2 )  Pe The load FT due to the moment on the maximum loaded bolt “A” is given by FT   FT r1 Per1 2 2 x  y The load FS due to direct shear is given by FS  P No. the force on any bolt r2 from the centre of rotation is FTr2/r1. . of Bolts CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 61 . Then. .. and so on for all the other bolts in the group.  The moment of resistance of the bolt group is given by M R  FT   r1 r1 r1  FT 2 r  FT r1 r2 r1 r2  . 1.1.Ed  [ ( FT sin )2  ( FS  FTcos )2 ]1/2  ( FT2sin 2  FS2  2FSFTcosFT2cos2 )1/2  ( FT2  FS2  2FSFTcos )1/2 The total shear stress fb = FR/As should not exceed the permissible shear stress of the bolt as defined in EC3: Part 1-8 – Clause 3.Resolving the load FT vertically and horizontally produces Vertical load acting on bolt A = FS + FT cos  Horizontal load acting on bolt A = FT sin  Resultant load on bolt A is equal to FR  Fv. CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 62 . .Bolt Group in Direct Shear and Tension The centre of rotation is assumed to be at the bottom bolt of the group. The moment of resistance MR of the bolt group is given by M R  2 (FT y12 y1  FT y 22 y1  FT y32 y1  . . .)]  2FT  y 2 y1  Pe CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 63 . A Centre of rotation A bracket subjected to a load P at an eccentricity e. and the loads vary linearly.)  2FT y1 [(y12  y 22  y32  . . of Bolts  The load FS due to direct shear is given by FS   Tensile stress is given by f t   Shear stress is given by fs   The combined tensile and shear stresses should also satisfy the conditions given in EC3: Part 1-8 – Clause 3.4Ft.Ed   1.Ed  fs As Fv.6.1. The load FT due to the moment on the maximum loaded bolt “A” is given by FT  Pey1 2 y 2 P No. FT As FS As  FT  Ft.Ed  f t As  FS  Fv.0 Fv.Rd 1.Ed Ft.Rd CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 64 .  Bolt failure.Tension – No Prying Action 2Ft.Rd CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 65 .  There is no prying force. Ft.Rd Ft.Rd Mode – 1  Flange is fully rigid. Prying action due to flexible end plate CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 66 .Rd + Q  Tension in bolts is 90% of resistance value.Rd + Q Ft.  Total bolt force = Ft.Tension – Prying Action 2Ft.  There is prying force.  Bolt failure with flange Q Ft.2  Flange is flexible.Rd Mode .Rd + Q Q yielding.  Total bolt force Q = Ft.Rd + Q  Column flange fails by yielding.3  Flange is completely flexible.Tension – Prying Action 2Ft.Rd Mode .Rd + Q CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 Q 67 .Rd + Q Ft. Ft.  There is prying force but smaller in magnitude. Preloaded or High Strength Friction Grip (HSFG) Bolted Connections Preloaded Bolts Connections CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 68 . Difference between Non-preloaded and Preloaded Bolts NEd Friction Bearing Shear NEd Non-preloaded Bolted Connection in Shear Preloaded Bolted Connection in Shear CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 69 . Preloaded Bolts Tension and Tightening CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 70 . CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 71 . nuts and hardened steel washers.  The bolts are tightened to a predetermined shank tension so that the clamping force will transmit the force in the connected members by friction. hence this type of joint is useful where rigid connections are required.  The bolts do not act in shear or bearing as in non-preloaded bolted connections. and the conditions given in EC3: Part 1-8 – Clause 3.  There is no slip or movement between the connected parts.Preloaded or High Strength Friction Grip (HSFG) Bolts  HSFG bolts consist of high-strength steel bolts.6.1 has to be checked accordingly.  If there is a slip the bolts are then in tension and shear. 1 and As is the bolts root area. Rd  k n s  Fp.9 bolts EC3: Part 1-8 – Clause 3. n = number of friction surfaces.  = 0. Rd  k n s  M 3.C or Fs. CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 72 .25 at ultimate limit state. fub is obtained from Table 3.9 bolt should be taken as Fs.8 or 10.9.8 or 10.0 for bolts in normal holes.5 slip factor (coefficient of friction) and Fp.1 at serviceability limit state and M3 is 1. M3.Slip-resistant connections using class 8.C M3 where ks = 1. service is 1.7fubAs preloading force to be used in the above equation.service Fp.C = 0.1  The design slip resistance at serviceability limit state (SLS) and ultimate limit state (ULS) of a preloaded class 8. C   0.8 or 10.9. CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 73 .8F t. Fv. Rd   k n(F s p.9 bolts EC3: Part 1-8 – Clause 3.Ed.Ed or Ft. in addition to the shear force.2  Combined tension and shear If a slip-resistant connection is subjected to an applied tensile force. Ed ) M3 It applies for both slip-resistant at serviceability and slip-resistant at ultimate limit state.Slip-resistant connections using class 8.Ed or Fv.ser.Ed. the design slip resistance per bolt should be taken as follows: Fs.ser. tending to produce slip. Ft. 8 preloaded higher grade high strength friction grip (HSFG) bolts in S275 and designed to be non-slip in service.Example 4 A bracket shown below is constructed from a cut 300 x 200 x 77. and assuming the slip factor  = 0. the bottom of the bolts group. The bracket is subjected with a factored vertical design load of 380 kN acting at an eccentricity of 160 mm from the face of the column. A total of 8 numbers of M20 class 8. show that the proposed 8-bolt group is adequate under combined shear and tension action.3 kg/m UB with a 15 mm thick plate welded on top if it. Assuming the centre of rotation is at point A.5 and there is no prying force. CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 74 . and the loads vary linearly. 8 bolts Ft × y1/y1 70 Ft × y2/y1 y1 y2 70 Ft × y3/y1 y3 60 A CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 75 . M20 class 8.Example 4 380 kN Two bolts one in each side of flange of column 50 70 160 Maximum load due to moment Ft 8 No. Example 4 The moment of resistance MR of the bolt group is given by Applied moment is equal to the moment of resistance MR of the bolt group. Therefore. the tensile load per bolt Ft due to the moment on the maximum loaded bolt is given by CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 76 . C   0. i.8F M3 t.Ed or Ft.0  1  0.061  103 )  1.1 kN Therefore.Rd.5 ( 0.ser. Fv.Ed  Fs.25 Since Fv. tending to produce slip.8  93. it is not O.Example 4 The shear load per bolt Fv. Ft. the design slip resistance per bolt should be taken as follows: Fs.Ed. in addition to the shear force.K.Ed due to direct shear is given by If a slip-resistant connection is subjected to an applied tensile force. Ed ) 1. 47. Rd  k n(F s  25.1 kN p.ser.7  800  245  0.5 kN  25.Ed.e. CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 77 .Ed or Fv. Electrode Electrode. Flux. where heat is applied to the work pieces to melt and fuse to form a permanent bond. welding is used to construct and repair parts of many on-shore and offshore steel structures. Completed Weld CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 78 . Slag. Core Wire. Heat-affected Zone.Schematic Representation of Welding Process Welding is a process of permanently joining metal parts. Because of its strength. Typical electrode diameters vary from 2.0 mm and length from 350 to 450 mm. and as soon as the arc is struck. Open-circuit voltages are from 50 to 90 V. The wire is covered with an extruded flux coating. It is flexible as it can be manipulated in many situations. It requires a skilled welder. The MMAW Welding Process Different Type of Electrodes CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 79 .5 to 6.Manual Metal Arc Welding (MMAW) – This is the simplest form of welding process. Typical currents range from 50 to 400 amps and the deposition rates is from 20 to 100 g/min. The electrode is hand held and fed into the weld pool. Quality of welding is closely related to operator skill. the voltage falls from 20 to 35 V. FCAW Machines Mechanism and Spool of Electrode Wire of FCAW Machine CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 80 .Flux Core Arc Welding (FCAW) – This is a more automated method of welding as compared to MMAW. argon or helium). This method consists of the emission of wire and gas (CO2. The wires are supplied on a spool which is placed in a FCAW welding compartment. Thus this method is more convenient. welds can be continuous and lengthy. Compared to the MMAW method of welding. FCAW Welding Machine Welder Operating the FCAW Welding Machine CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 81 . without the need to change electrodes. The FCAW machines can be connected to a special traveling machine which facilitate the welding of lengthy welds such as a T-beam as shown below.The wire is emitted from a nozzle with a trigger to eject it. minimum cleaning required unlike normal welding processes and there is no visible arc light since the welding action is submerged under the flux that is used. SAW Machine Flux Used in the SAW Process CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 82 . the height of welding nozzle and voltage of the SAW machine. a compartment to store the spool of electrode wire and a control panel which allows the operator to control the speed of travel. which holds the flux. The machine is made up of a flux hopper.Submerged Arc Welding (SMAW) – SMAW is a form of welding that utilizes the use of flux. It is mainly used for long straight welds usually on flat surfaces. Advantages of the SMAW method is that its easily automated. Spool of Electrode Wire for SMAW SMAW Control Panel Operating the SMAW with Ease Hardened Slag Close up of Hardened Slag SMAW Producing an Excellent Quality Welds CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 83 . Weld Types – Fillet and Butt Welds Types of fillet and butt welds CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 84 . 6 kN = 138.155 kN/mm = 138.8 kN Total weld strength = 2 x 60mm x 1.Classification of Fillet Welds F/2 F F/2 F F/2 50 mm F/2 F F F F 60 mm Weld leg length = 6 mm (a) Side Shear (b) End Tension Total weld strength = 2 x 2 x 50mm x 0.6 kN CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 85 .942 kN/mm = 184. Design resistances of fillet welds CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 86 . the resultant of all the forces.Rd  f vw. Rd where Fw. welds. at every point along its length. Fw.Ed is the design value of the weld force per unit length.Rd is the design weld resistance per unit length.5.  The design resistance per unit length Fw. per unit length transmitted by the weld should satisfy the following criteria: Fw.d is the design strength of the weld.Design resistance of fillet welds EC3: Part 1-8 – Clause 4.3 Simplified method  The design resistance of a fillet weld may be assumed to be adequate if.d  a where fvw. a = throat thickness of the fillet CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 87 .3.Rd should be determined from: Fw. Ed  Fw. not necessarily along the weld axis. M2 = 1.Design resistance of fillet welds EC3: Part 1-8 – Clause 4.1.3.25 CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 88 . w = 0.3 Simplified method  The design shear strength fvw.d  fu 3 w  M2 where fu = 410 N/mm2 for S275.d of the weld should be determined from: f vw.85 from Table 4.5. 2 Directional method  In this method. Transverse welds are stronger than longitudinal welds up to 25% more CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 89 . the forces transmitted by a unit length of weld are resolved into components parallel and transverse to the longitudinal axis of the weld and stresses normal and transverse to the plane of its throat.Design resistance of fillet welds EC3: Part 1-8 – Clause 4.3.5. 083 when   0 2 1.0  cos  CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 90 .5 . K  1.25  Longitudinal resistance.T.5.L.25 when   45 and K  1. Fw.3.Rd = K  fvw.d  a × 1. Fw.0  Transverse resistance.2 FT K  1.d  a × 1.Rd = fvw.Design resistance of fillet welds EC3: Part 1-8 – Clause 4.0 1. 3.945 kN/mm run CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 91 .2 6mm wt = 6 x cos 45° 45° 6mm  Longitudinal capacity.Rd = fvw.L. a 6-mm fillet weld strength is = 222. Fw.Design resistance of fillet welds EC3: Part 1-8 – Clause 4.d  a As an example.707 = 0.80 x 6 x cos 45 = 222.80 x 6 x 0.5. 25 x 222. a 6-mm fillet weld strength is = 1.  = 45 and K = 1.80 x 6 x cos 45 = 1.707 = 1.5.Design resistance of fillet welds EC3: Part 1-8 – Clause 4.3. the weld should satisfy the following relationship: CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 92 .80 x 6 x 0. Fw.25 FT 450 450  Transverse capacity.d  a As an example.T.153 kN/mm run Throughout its length.Rd = K  fvw.25 x 222.2 For a transverse force parallel to one leg of an equal leg fillet weld that connects two elements that are at right angles to each other. Matching electrodes should have specified minimum tensile.7 Bevel angle h Throat thickness Root opening Backing plate The strength of butt welds should be taken as equal to that of the parent metal provided matching electrodes are used.Design resistance of butt welds EC3: Part 1-8 – Clause 4. yield strengths. elongation and impact values each equivalent or better than those specified for the parent materials. CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 93 .
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