CHAPTER J DESIGN OF CONNECTIONS

Transcription

1 J-1 CHAPTER J DESIGN OF CONNECTIONS INTRODUCTION Chapter J of the addresses the design and heking of onnetions. The hapter s primary fous is the design of welded and bolted onnetions. Design requirements for fillers, splies, olumn bases, onentrated fores, anhors rods, and other threaded parts are also overed. Speial requirements for onnetions subjet to fatigue are not overed in this hapter.

2 J- Example J.1 Fillet Weld in Longitudinal Shear Given: An ¼ in. 18-in. wide plate is fillet welded to a a-in. plate. Assume that the plates are ASTM A57 grade 50 and have been properly sized. Assume F EXX 70 ksi. Note that plates would normally be speified as ASTM A36, but F y 50 ksi plate has been used here to demonstrate requirements for long welds. Size the welds for the loads shown. Solution: Determine the maximum weld size Beause the overlapping plate is ¼ in., the maximum fillet weld size that an be used without speial notation (built out to obtain full-throat thikness as required in AISC Setion J.b) is a x-in. fillet weld. A x-in. fillet weld an be deposited in the flat or horizontal position in a single pass (true up to -in). Determine the required strength P u 1.(33 kips) + 1.6(100 kips) 00 kips P a 33 kips kips 133 kips Determine the length of weld required The design strength per inh of a x-in. fillet weld is φr n 1.39 (3) 4.17 kips/ in. Pu 00 kips 48 in. φrn 4.17 kips/in. or 4 in. of weld on eah side The allowable strength per inh of a x-in. fillet weld is R n /Ω 0.98 (3).78 kips/ in. Pa Ω 133 kips 48 in. Rn.78 kips/in. or 4 in. of weld on eah side. Manual Part 8

3 J-3 Chek the weld for length to weld size ratio l w 4 in. 18 > 100, in. Therefore Equation J-1 must be applied, and the length of weld inreased, sine the resulting β will redue the available strength below the required strength. Try a weld length of 7 in. The new length to weld size ratio is 7 in. /0.188 in. 144 Eqn. J-1 For this ratio β (l/w) M 1.0; (144) 0.91 Rehek the weld at its redued strength φr n ( )( ) (0.91) 4.17 kips/in. 54 in. 05 kips > P 00 kips u Therefore, use 7 in. of weld on eah side Rn Ω 137 kips > P 133 kips (0.91) (.78 kips/in. )( 54 in. ) a Therefore, use 7 in. of weld on eah side

4 J-4 Example J. Fillet Weld Loaded at an Angle Given: Design a fillet weld at the edge of a gusset plate to arry a fore of 50 kips due to dead load and a fore of 150 kips due to live load, at an angle of 60 degrees relative to the weld. Assume the beam and the gusset plate thikness and length have been properly sized. Solution: Calulate the required strength P u 1.(50 kips) + 1.6(150 kips) 300 kips P a 50 kips kips 00 kips Assume a -in. fillet weld is used on eah side. The shear strength of a -in. fillet weld is 5(1.39) 6.96 kip/in. And for two sides (6.96 kip/in.) 13.9 kip/in. 5(0.98) 4.64 kip/in. And for two sides (4.64 kip/in.) 9.8 kip/in. Beause the angle of the fore relative to the axis of the weld is 60 degrees, the strength of the weld an be inreased as follows: 1.5 k w 0.60FEXX ( sin θ ) (70) ( (0.866) ) 1.40

5 J-5 Find the inreased strength and the required length of weld Calulate the inreased strength kip/in.(1.40) 19.5 kip/in. Calulate the inreased apaity kip/in.(1.40) 13.0 kip/in. Determine the required length of weld. 300 kips/19.5 kip/in in. Determine the required length of weld. 00 kips/13.0 kip/in in. Use 16 in. Use 16 in.

6 J-6 Example J.3 Combined Tension and Shear in Bearing Type Connetions Given: A ¾-in. diameter, ASTM A35-N bolt is subjeted to a tension fore of 3.5 kips due to dead load and 1 kips due to live load, and a shear fore of 1.33 kips due to dead load and 4 kips due to live load. Chek the ombined stresses aording to the Equations J3-3a and J3-3b. Solution: Calulate the required tensile and shear strength Tension 1.(3.5) +1.6(1.0) 3.4 kips Shear 1.(1.33) + 1.6(4.00) 8.00 kips Tension kips Shear kips Calulate f v 8.00/ ksi φf nv Chek ombined tension and shear. Fnt F nt 1.3Fnt fv Fnt φfnv F nt 90 ksi, F nv 48 ksi 90 F nt 1.3(90) (18.1) 0.75(48) 71.7 < 90 R n F nt A b 71.7(0.44) 31.7 kips For ombined tension and shear φ 0.75 Design tensile strength φr n 0.75(31.7) 3.8 kips > 3.4 kips Calulate f v 5.33/ ksi F nv /Ω Chek ombined tension and shear. ΩFnt F nt 1.3Fnt fv Fnt Fnv F nt 90 ksi, F nv 48 ksi F nt 1.3(90).00(90) (1.1) < 90 R n F nt A b 71.6(0.44) 31.6 kips For ombined tension and shear Ω.00 Allowable tensile strength R n /Ω31.6/ kips > 15.5 kips Eq. J3-3a and J3-3b Table J3. Setion J3.6

7 J-7 Example J.4 Slip-Critial Connetion with Short Slotted Holes High-strength bolts in slip-ritial onnetions are permitted to be designed to prevent slip either as a servieability limit state or as a strength limit state. The most ommon design ase is design for slip as a servieability limit state. The design of slip as a strength limit state should only be applied when bolt slip an result in a onnetion geometry that will inrease the required strength beyond that of a strength limit state, suh as bearing or bolt shear. Suh onsiderations our only when oversized holes or slots parallel to the load are used, and when the slipped geometry inreases the demand on the onnetion. Examples inlude the ase of ponding in flat-roofed long span trusses, or the ase of shallow, short lateral braing. Given: Selet the number of ¾-in. ASTM A35 slip-ritial bolts with a Class A faying surfae that are required to support the loads shown when the onnetion plates have short slots transverse to the load. Selet the number of bolts required for slip resistane only. Assume that the onneted piees have short slots transverse to the load. Use a mean slip oeffiient of 0.35, whih orresponds to a Class A surfae. Solution: Calulate the required strength P u 1.(17 kips) + 1.6(51 kips) 10 kips P a 17 kips + 51 kips 68 kips For standard holes or slots transverse to the diretion of the load, a onnetion an be designed on the basis of the servieability limit state. For the servieability limit state: Find R n, where: µ 0.35 for Class A surfae D u 1.13 h s 0.85 (short slotted holes) T b 8 kips N s, number of slip planes φ 1.00 Ω 1.50 Setion J3.8 Table J3.1 R n µd u h s T b N R n 0.35(1.13)(0.85)(8)() 18.8 kips/bolt Eqn.J3-4

8 J-8 Determine the required number of bolts. 10 kips/1.00(18.8 kips/bolt) 5.4 bolts 18.8 kips/bolt 68 kips / 5.4 bolts 1.50 Manual Table 7-3 Use 6 bolts Use 6 bolts Given: Repeat the problem with the same loads, but assuming that the onneted piees have long slotted holes in the diretion of the load and that the deformed geometry of the onnetion would result in a ritial load inrease. Solution: P u 10 kips and P a 68 kips per the first solution For this onnetion, the designer has determined that oversized holes or slots parallel to the diretion of the load will result in a deformed geometry of the onnetion that reates a ritial load ase. Therefore, the onnetion is designed to prevent slip at the required strength level. φ 0.85 Ω 1.76 In addition, h s will hange beause we now have long slotted holes. Setion J3.8 Find R n µ 0.35 for Class A surfae D u 1.13 h s 0.70 (long slotted holes) T b 8 kips N s, number of slip planes Table J3.1 R n µd u h s T b N s R n 0.35(1.13)(0.70)(8)() 15.5 kips/bolt Eqn. J3-4

9 J-9 Determine the required number of bolts 10 kips 7.73 bolts 0.85(15.5 kips/bolt) Use 8 bolts 68 kips (1.76) 15.5 kips/bolt Use 8 bolts 7.63 bolts Manual Table 7-4

10 J-10 Example J.5 Combined Tension and Shear in a Slip-Critial Connetion. Beause the pretension of a bolt in a slip-ritial onnetion is used to reate the lamping fore that produes the shear strength of the onnetion, the available shear strength must be redued for any load that produes tension in the onnetion. Given: The slip-ritial bolt group shown below is subjeted to tension and shear. This onnetion is designed for slip as a servieability limit state. Use ¾-in. diameter ASTM A35 slip-ritial lass A bolts in standard holes. This example shows the design for bolt slip resistane only, and assumes that the beams and plates are adequate to transmit loads in a rigid fashion. Solution: The fastener pretension for a ¾-in. diameter ASTM A35 bolt is 8 kips D u 1.13 per Setion J3.8. N b number of bolts arrying the applied tension. Table J3.1 Determine the tension on bolts, hek tension on the bolts and find k s P u 1.(10 kips)+1.6(30kips) 60 kips P a 10 kips + 30kips 40 kips By geometry, By geometry, T u 4 5 V u kips 8 bolts 6 kips/bolt 60 kips 8 bolts 4.5 kips/bolt T a 4 5 V a kips 8 bolts 4 kips/bolt 40 kips 8 bolts 3 kips/bolt Chek bolt tension Chek bolt tension φr n 9.8 kips/bolt > 6 kips/bolt R n /Ω 19.9 kips/bolt > 4 kips/bolt Manual Table 7-

11 J-11 Combined tension and shear fator Tu 6 kips k s 1 1 DTN u b b 1.13(8 kips)(1) Combined tension and shear fator k s Ta DTN kips 1 u b b 1.13(8 kips)(1) Eqn. J3-5a and J3-5b Multiply the available shear strength of the bolts by the redution fator k s φr n 11.1 kips/bolt R n /Ω 7.38 kips/bolt Manual Table 7-3 Modify the slip resistane by k s and hek bolt shear k s φr n (0.810)(11.1 kips/bolt) kr s n (0.810)(7.38 kips/bolt) Ω 8.99 kips/bolt > 4.50 kips/bolt 5.98 kips/bolt > 3.00 kips/bolt

12 J-1 Example J.6 Bearing Strength of a Pin in a Drilled Hole Given: A 1-in. diameter pin is plaed in a drilled hole in a 1-in. thik steel plate. Determine the available bearing strength of the pinned onnetion. Material Properties: Plate ASTM A36 F y 36 ksi F u 58 ksi Solution: Calulate the projeted bearing area A pb p ( 1 in. )( 1 1 in. ) 1 1 in. dt Calulate nominal bearing strength 1 ( )( ) Rn 1.8FyApb ksi 1 in. 97. kips Eqn. J7-1 Calulate the available bearing strength φ 0.75 Ω.00 ( ) φ kips 7.9 kips R n 97. kips Rn / Ω 48.6 kips.00 Setion J7

13 J-13 Example J.7 Base Plate Bearing on Conrete Given: A W1 96 olumn bears on a 4 in. 4 in. onrete pedestal. The spae between the base plate and the onrete pedestal is grouted. Design the base plate to support the following loads in axial ompression: P D 115 kips P L 345 kips Material Properties: Column W1 96 ASTM A99 F y 50 ksi F u 65 ksi Base Plate ASTM A36 F y 36 ksi F u 58 ksi Conrete Pedestal f 3 ksi Grout f 4 ksi Geometri Properties: Column W1 96 d 1.7 in. b f 1. in. t f in. t w in. Manual Table 1-1 Solution: Calulate the required strength P kips kips 690 kips P 115 kips kips 460 kips u ( ) ( ) a A ped (4 in.)(4 in.) 576 in. A ol (1.7 in.)(1. in.) 155 in. Sine the pedestal area is larger than the olumn footprint area, but less than less than 4 times olumn footprint area, the onrete bearing area will be the geometrially similar area of the pedestal to the base plate.

14 J-14 Calulate the base plate area φ 0.60 Pu A1( req) φ 0.85 f ' Ω.50 PaΩ A1( req) 0.85 f ' Setion J8 690 kips 451 in ksi ( )( ) ( 460 kips) (.50) ( 0.85)( 3 ksi) 451 in. Note: The strength of the grout has onservatively been negleted, as its strength is greater than that of the onrete pedestal. Try a in. in. base plate Chek base plate dimensions Verify N d + 3 ( in. ) and + 3 ( in. ) B b f Table 14- d + ( 3 in. ) 1.7 in. + ( 3 in. ) 18.7 in. < in. ( ) ( ) b + 3 in. 1. in. + 3 in. 18. in. < in. f Base plate area, ( )( ) A1 NB in. in. 484 in. > 451 in. Note: A square base plate with a square anhor rod pattern will be used to minimize the hane for field and shop problems. Calulate the geometrially similar onrete bearing area Sine the pedestal is square and the base plate is a onentrially loated square, the full pedestal area is also the geometrially similar area. Therefore, A ( 4 in. )( 4 in. ) 576 in. Verify the onrete bearing strength φ 0.60 A Pp 0.85 f ' A1 A 1 φ φ Ω f ' A1 A Pp / Ω Ω A 1 Setion J8 Eqn. J8- ( )( )( ) ksi 484 in. 576 in. 484 in. ( ) ( ) ( ) ksi 484 in. 576 in in. 808 kips > 690 kips 539 kips > 460 kips

15 J-15 Note: It is permitted to take φ 0.65 per ACI Also note that as the area of the base plate approahes the area of onrete, the modifying A ratio,, approahes unity and Eqn. J8- onverges to Eqn. J8-1. A 1 Calulate required base plate thikness ( ) N 0.95d in in. m 4.97 in. Manual Part 14 ( ) B 0.8b f in in. n 6.1 in. ( 1.7 in. )( 1. in. ) db f n ' 3.11 in. 4 4 Calulate required base plate thikness Calulate required base plate thikness X 4dbfPu ( ) d + b φ P f p X 4dbfPaΩ ( + ) f d b P p 4( 1.7 in. )( 1. in. )( 690 kips) ( 1.7 in in. ) ( 808 kips) 4( 1.7 in. ) ( 1. in. )( 460 kips) ( 1.7 in in. ) ( 539 kips) X λ X X λ 1+ 1 X > 1, use λ > 1, use λ 1. Note: λ an always be onservatively taken as being equal to 1 λ n' ()( in. ) 3.11 in. ( ) ( ) l max m, n, λ n' max 4.97 in.,6.1 in.,3.11 in. 6.1 in.

16 J-16 f pu Pu 690 kips 1.43 ksi BN ( in. )( in. ) f pa Pa 460 kips 0.95 ksi BN ( in. )( in. ) t ( ) preq l f pu 0.9F y t ( ) preq l 3.33 f F y pa ( 6.1 in. ) ( ) ( ) 1.43 ksi ksi ( 6.1 in. ) ( ) ( 36 ksi) ksi 1.8 in. 1.8 in. Use a in. thik base plate.