Beam F y = 50 ksi F u = 65 ksi Manual Table 2-3

Transcription

1 I-6 Example I-1 Composite Beam Design Given: A series of 45-ft. span omposite beams at 10 ft. o/ are arrying the loads shown below. The beams are ASTM A99 and are unshored. The onrete has f = 4 ksi. Design a typial floor beam with 3 in. 18 gage omposite dek, and 4½ in. normal weight onrete above the dek, for fire protetion and mass. Selet an appropriate beam and determine the required number of shear studs. Solution: Material Properties: Conrete f = 4 ksi Beam F y = 50 ksi F u = 65 ksi Table -3 Loads: Dead load: Slab = kip/ft Beam weight = kip/ft (assumed) Misellaneous = kip/ft (eiling et.) Live load: Non-redued = 0.10 kips/ft Sine eah beam is spaed at 10 ft. o.. Total dead load Total live load = kip/ft (10 ft.) = 0.93 kips/ft. = 0.10 kip/ft (10ft.) = 1.00 kips/ft. Constrution dead load (unshored) = kip/ft (10 ft) = 0.83 kips/ft Constrution live load (unshored) = 0.00 kip/ft (10 ft) = 0.0 kips/ft

2 I-7 Determine the required flexural strength w u = 1.(0.93 kip/ft) + 1.6(1.0 kip/ft) =.7 kip/ft.7 kip/ft(45 ft) M u = = 687 kip-ft. 8 w a = 0.93 kip/ft kip/ft = 1.93 kip/ft 1.93 kip/ft(45 ft) M a = = 489 kip-ft. 8 Use Tables 3-19, 3-0 and 3-1 from the to selet an appropriate member Determine b eff The effetive width of the onrete slab is the sum of the effetive widths for eah side of the beam enterline, whih shall not exeed: Setion I.3.1.1a (1) one-eighth of the beam span, enter to enter of supports 45 ft. ( ) = 11.3 ft. 8 () one-half the distane to enter-line of the adjaent beam 10 ft. ( ) = 10.0 ft. Controls (3) the distane to the edge of the slab Not appliable Calulate the moment arm for the onrete fore measured from the top of the steel shape, Y. Assume a = 1.0 in. (Some assumption must be made to start the design proess. An assumption of 1.0 in. has proven to be a reasonable starting point in many design problems.) Y = t slab a/ = 7.5 ½ = 7.0 in. Enter Table 3-19 with the required strength and Y=7.0 in. Selet a beam and neutral axis loation that indiates suffiient available strength. Table 3-19 Selet a W1 50 as a trial beam. When PNA loation 5 (BFL), this omposite shape has an available strength of: φ b M n = 770 kip-ft > 687 kip-ft o.k. M n /Ω b = 51 kip-ft > 489 kip-ft o.k. Table 3-19 Note that the required PNA loation for and differ. This is beause the live to dead load ratio in this example is not equal to 3. Thus, the PNA loation requiring the most shear transfer is seleted to be aeptable for. It will be onservative for.

3 I-8 Chek the beam defletions and available strength Chek the defletion of the beam under onstrution, onsidering only the weight of onrete as ontributing to the onstrution dead load. Limit defletion to a maximum of.5 in. to failitate onrete plaement. I req ( )( ) ( ) 4 5 w kip/ft 45 ft 178 in. /ft DLl = = = 1, 060 in. 384 E 384 9,000 ksi.5 in. ( )( ) From Table 3-0, a W1 50 has I x = 984 in. 4, therefore this member does not satisfy the defletion riteria under onstrution. Using Table 3-0, revise the trial member seletion to a W1 55, whih has I x = 1140 in. 4, as noted in parenthesis below the shape designation. Chek seleted member strength as an un-shored beam under onstrution loads assuming adequate lateral braing through the dek attahment to the beam flange. Calulate the required strength Calulate the required strength DL = 1.4 (0.83 kips/ft) = 1.16 kips/ft DL+LL = = 1.03 kips.ft 1.DL+1.6LL = 1. (0.83) + 1.6(0.0) = 1.3 klf 1.31 kip/ft(45 ft) M u (unshored) = 8 =331 kip-ft The design strength for a W1 55 is 473 kip-ft > 331 kip-ft o.k kips/ft(45 ft) M a (unshored) = 8 = 60 kip-ft The allowable strength for a W1 55 is 314 kip-ft > 60 kip-ft o.k. For a W1 55 with Y=7.0 in, the member has suffiient available strength when the PNA is at loation 6 and Q = 9 kips. n Table 3-19 φ b M n =767 kip-ft > 687 kip-ft o.k. M n /Ω b = 510 kip-ft > 489 kip-ft o.k. Table 3-19 Chek a Qn 9 kips a = = = in. ' 0.85 fb 0.85(4 ksi)(10 ft.)(1 in./ft.) in. < 1.0 in. assumed o.k. Chek live load defletion LL < l/360 = ((45 ft.)(1 in./ft))/360 = 1.5 in. A lower bound moment of inertia for omposite beams is tabulated in Table 3-0.

4 I-9 For a W1 55 with Y=7.0 and the PNA at loation 6, I LB = 440 in ( )( ) ( ) 4 ( )( ) 4 5 w kip/ft 45 ft 178 in. /ft LLl LL = = = 1.30 in. 384 EI 384 9,000 ksi 440 in. LB 1.30 in. < 1.5 in. o.k. Determine if the beam has suffiient available shear strength 45ft V u = (.7kip/ft) = 61. kips φv n = 34 kips > 61. kips o.k. 45ft V a = ( 1.93kip/ft) = 43.4 kips V n /Ω = 156 kips > 43.4 kips o.k. Table 3-0 Table 3-3 Determine the required number of shear stud onnetors Using perpendiular dek with one ¾-in. diameter weak stud per rib in normal weight 4 ksi onrete. Q n = 17. kips/stud Table 3-1 Q Q n n = 9 kips = 17, on eah side of the beam. 17. kips Setion 3.d(5) Total number of shear onnetors; use (17) = 34 shear onnetors. Chek the spaing of shear onnetors Sine eah flute is 1 in., use one stud every flute, starting at eah support, and proeed for 17 studs on eah end of the span. 6d stud < 1 in. < 8t slab, therefore, the shear stud spaing requirements are met. The studs are to be 5 in. long, so that they will extend a minimum of 1 in. into slab. Setion 3.d(6) Setion I3. (b)

5 I-10 Example I- Filled Composite Column in Axial Compression Given: Determine if a 14-ft long HSS10 6 a ASTM A500 grade B olumn filled with f = 5 ksi normal weight onrete an support a dead load of 56 kips and a live load of 168 kips in axial ompression. The olumn is pinned at both ends and the onrete at the base bears diretly on the base plate. At the top, the load is transferred to the onrete in diret bearing. Solution: Calulate the required ompressive strength P u = 1.(56 kips) + 1.6(168 kips) = 336 kips P a = 56 kips kips = 4 kips The available strength in axial ompression an be determined diretly from the at KL = 14 ft as: φ P n = 353 kips 353 kips > 336 kips o.k. P n /Ω = 36 kips 36 kips > 4 kips o.k. Table 4-14 Supporting Calulations The available strength of this filled omposite setion an be most easily determined by using Table 4-14 of the. Alternatively, the available strength an be determined by diret appliation of the Speifiation requirements, as illustrated below. Material Properties: HSS10 6 3/8 F y = 46 ksi F u = 58 ksi 1.5 ' Conrete f = 5 ksi ( ) 1.5 E = w f = = 3,900 ksi Table -3

6 I-11 Geometri Properties: HSS10 6 3/8 t = in. b = 10.0 in. h = 6.0 in. Conrete: The onrete area is alulated as follows Table 1-11 r = t = (0.375 in. ) = 0.75 in. (outside radius) b f = b-r = 10 (0.75) = 8.50 in. h f = h-r = 6 (0.75) = 4.50 in. A = b f h f + π(r-t) + b f (r-t) + h f (r-t) = (8.50 in.)(4.50 in.) + π(0.375 in.) + (8.50 in.)(0.375 in.) + (4.50 in.)(0.375 in.) 3 3 ( ) = 48.4 in. 4 ( r t) ( r t) bh 1 1 ( b)( h) π 8 π h 4 I = + + r t π 3π For this shape, bukling will take plae about the weak axis, thus h1 = 6 ( 0.375) = 5.5 in. b1 = 10 4( 0.375) = 8.5 in. h = 6 4( 0.375) = 4.5 in. b = in. ( r t) = = in. 3 3 (8.5)(5.5) (0.375)(4.50) 4 π 8 I = + + ( 0.375) π π ( 0.375) 4.5 4( 0.375) + + 3π = 111 in. HSS10 6 a: A s = 10.4 in. I s = 61.8 in. 4 h/t = 5.7 Limitations: 1) Normal weight onrete 10 ksi > f > 3 ksi f = 5 ksi o.k. Table 1-11 Setion I1. ) Not Appliable. 3) The ross-setional area of the steel HSS shall omprise at least one perent of the total omposite ross setion. Setion I.a 10.4 in. > (0.01)(48.6 in in. ) = in. o.k. 4) The maximum b/t ratio for a retangular HSS used as a omposite olumn shall be equal to.6 EF y. b/t = 5.7<.6 EF y =.6 9,000ksi 46ksi = 56.7 o.k. User note: For all retangular HSS setions found in the the b/t ratios do not exeed.6 EF y. 5) Not Appliable.

7 I-1 Calulate the available ompressive strength C = 0.85 for retangular setions P o = A s F y + A sr F yr + C A f = (10.4 in. )(46 ksi) (48.4 in. )(5 ksi) = 684 kips Se. I.b Eqn. I-13 A s C 3 = = A + As Therefore use in. 48.4in in. = > 0.90 Eqn. I-15 EI eff = E s I s + E s I sr + C 3 E I = (9,000 ksi)(61.8 in. 4 ) + (0.90)(3,900 ksi)( 111 in. 4 ) =,180,000 kip-in. Eqn. I-14 User note: K value is from Chapter C and for this ase K = 1.0. P e = π (EI eff )/(KL) = π (,180,000 kip-in. ) / ((1.0)(14ft)(1in./ft)) = 76 kips Po P e = 684 kips 76 kips = Eqn. I <.5 Therefore use Eqn. I- to solve P n P n = P o P P o e = (684 kips) ( ) = 470 kips Setion I.1b Eqn. I-6 φ = 0.75 φ P n = 0.75(470kips) = 353 kips 353 kips > 336 kips o.k. Ω =.00 P n /Ω = 470 kips /.00 = 35 kips 35 kips > 4 kips o.k. Setion I.1b

8 I-13 Example I-3 Enased Composite Column in Axial Compression Given: Determine if a 14 ft tall W10 45 steel setion enased in a 4 in. 4in. onrete olumn with f = 5 ksi, is adequate to support a dead load of 350 kips and a live load of 1050 kips in axial ompression. The onrete setion has 8-#8 longitudinal reinforing bars and #4 transverse 1in. o/., The olumn is pinned at both ends and the load is applied diretly to the onrete enasement. Solution: Material Properties: Column W10 45 F y = 50 ksi F u = 65 ksi Conrete f = 5 ksi E = 3,900 ksi (145 pf onrete) Table -3 Reinforement F yst = 60 ksi Geometri Properties: W10 45: A s = 13.3 in. I y = 53.4 in. 4 Table 1-1 Reinforing steel: A sr = 6.3 in. (the area of 1-#8 bar is 0.79 in., per ACI) 4 4 πr π(0.50) 4 I sr = + Ad = 8 + 6(0.79)(9.5) = 48 in. 4 4 Conrete: A = A g - A s - A sr = 576 in in in. = 556 in. I = I g - I s - I sr = 7,00 in. 4 Note: The weak axis moment of inertia is used as part of a slenderness hek.

9 I-14 Limitations: 1) Normal weight onrete 10 ksi > f > 3 ksi f = 5 ksi o.k. Setion I1. ) F yst < 75 ksi F yst = 60 ksi o.k. 3) The ross-setional area of the steel ore shall omprise at least one perent of the total omposite ross setion. Setion I.1a 13.3 in. > (0.01)(576 in. ) = 5.76in. o.k. 4) Conrete enasement of the steel ore shall be reinfored with ontinuous longitudinal bars and lateral ties or spirals. The minimum transverse reinforement shall be at least in. of tie spaing. 0.0 in. /1 in. = in. /in. > in. /in. o.k. 5) The minimum reinforement ratio for ontinuous longitudinal reinforing, ρ sr, shall be ρ sr = A 6.3in. sr A g = 576in. = > o.k. Eqn. I-1 Calulate the total required ompressive strength P u = 1.(350 kips) + 1.6(1050 kips) = 100 kips P a = 350 kips kips = 1400 kips Calulate the available ompressive strength P o = A s F y + A sr F yr A f = (13.3 in. )(50 ksi) + (6.3 in. )(60 ksi) +0.85(556 in. )(5 ksi) = 3410 kips As C 1 = A + A = in. 556 in in. + = 0.15 s Eqn. I- Eqn. I-5 EI eff = E s I s + 0.5E s I sr + C 1 E I = (9,000 ksi)(53.4 in. 4 ) + 0.5(9,000 ksi)(48 in. 4 ) + (0.15)(3,900 in. 4 )(7,00) =3,700,000 kip-in. Eqn. I-4 User note: K value is from Chapter C and for this ase K = 1.0. P e = π (EI eff )/(KL) = π (3,700,000 kip-in. ) / ((1.0)(14in.)(1in./ft)) = 8,90 kips Po P e = 3, 410 kips 8, 90 kips = <.5 Therefore use Eqn. I- to solve P n Eqn. I-3 Setion I.1b

10 I-15 P n = P o φ = 0.75 Po P e = (3,410 kips) ( ) = 870 kips Ω =.00 Eqn. I-6 Setion I.1b φ P n = 0.75(870kips) = 150 kips 150 kips > 100 kips o.k. P n /Ω = 870 kips /.00 =1440 kips 1440 kips > 1400 kips o.k. Beause the entire load in the olumn was applied diretly to the onrete, aommodations must be made to transfer an appropriate portion of the axial fore to the steel olumn. This fore is transferred as a shear fore at the interfae between the two materials. Determine the number and spaing of -in. diameter headed shear studs to transfer the axial fore. Solution: Material Properties: Con. f = 5 ksi E = 4070 ksi Shear Studs F u = 65 ksi Table 3-1 Geometri Properties: W10 45 A st = 13.3 in. A flange = 4.97 in. A web = 3.36 in. d st = 10.1 in. Shear Studs A s = in. Con. d = 4 in. Table 1-1 Calulate the shear fore to be transferred Pu,100 V = = =,800 kips φ 0.75 ( ) V = P a Ω = 1, 400 =,800 kips V = V(A s F y /P o ) = 800 kips ((13.3 in. )(50 ksi) / (3410 kips)) = 546 kips Calulate the nominal strength of one in. diameter shear stud onnetor Eqn. I-10

11 I-16 Q n = 0.5A s ' fe < A s F u ' 0.5A s fe = 0.5(0.196 in. ) (5ksi)(3900ksi) = 13.7 kips A s F u = (0.196 in. )(65 ksi) = 1.7 kips Eqn. I-1 Therefore use 1.7 kips. Calulate the number of shear studs required to transfer the total fore, V V / Q n = 546 kips / 1.7 kips = 43 An even number of studs are required to be plaed symmetrially on two faes. Therefore use studs minimum per flange Determine the spaing for the shear studs The maximum stud spaing is 16 in. Setion I.1f The available olumn length is 14ft (1 in./ft) = 168 in. and the maximum spaing is = 168 in./(+1) = 7.3 in. Therefore, on the flanges, use single 7 in. Stud plaement is to start 10.5 in. from one end. Determine the length of the studs for the flanges; d dst - 3 in. = 4 in in. - 3 in. = 3.95 in. Therefore use 3 ½ in. in length for the flanges. Note: The subtration of 3 in. is to ensure suffiient over. Summary: use -in. diameter shear stud onnetors as shown on eah flange, 7 in.