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1、I-1 CHAPTER I DESIGN OF COMPOSITE MEMBERS I1. GENERAL PROVISIONS The available strength of composite sections may be calculated by one of two methods - the plastic stress distribution method or the strain-compatibility method. The composite design tables in the Steel Construction Manual are based on
2、 the plastic stress distribution method. I2. AXIAL MEMBERS Generally, the available compressive strength of a composite member is based on a summation of the strengths of all of the components of the column. The Specification contains several requirements to ensure that the steel and concrete compon
3、ents work together. For tension members, the concrete tensile strength is ignored and only the strength of the steel member and properly connected reinforcing is permitted to be used in the calculation of available tensile strength. Because of concerns about the deformation compatibility of steel an
4、d concrete in resisting shear, either the steel or the reinforced concrete, but not both, are permitted to be used in the calculation of available shear strength. Whether the composite column is an encased column or a filled column, it is important to consider the load path within the composite memb
5、er, and to provide shear transfer mechanisms and appropriate top and bottom details. The design of encased composite compression and tension members is presented in Examples I-3 and I-4. There are no tables in the Manual for the design of these members. The design of filled composite compression and
6、 tension members is presented in Examples I-2 and I-5. The Manual includes tables for the design of filled composite members in compression. I3. FLEXURAL MEMBERS Because a plastic stress distribution is used for both LRFD and ASD designs, the flexural strength of composite beams is generally greater
7、 than that of former ASD designs. Shear connectors, in many cases, have lower horizontal shear strength than was permitted in past LRFD specifications. Designers are encouraged to read the discussion on this subject in Commentary Chapter I. The design of a typical composite beam member is illustrate
8、d in Example I-1. I4. COMBINED AXIAL FORCE AND FLEXURE Design for combined axial force and flexure may be accomplished in any of the three methods outlined in the Commentary. Example I-7 illustrates the plastic-distribution method. To assist in developing this curve, a series of equations is provide
9、d in Figure I-1. These equations define selected points on the interaction curve, without consideration of slenderness effects. Figures I-1a through I-1d outline specific cases, and the applicability of the equations to a cross-section that differs should be carefully considered. As an example, the
10、equations in Figure I-1a are appropriate for the case of side bars located at the centerline, but not for other side bar locations. In contrast, these equations are appropriate for any amount of reinforcing at the extreme reinforcing bar location. In Figure I-1b, the equations are appropriate only f
11、or the case of 4 reinforcing bars at the corners of the encased section. When design cases deviate from those presented the appropriate interaction equations can be derived from first principles. tt SectionStress DistributionPointDefining Equations A C D PLASTIC CAPACITIES FOR RECTANGULAR, ENCASED W
12、-SHAPES BENT ABOUT THE X-X AXIS B area of steel shape area of all continuous reinforcing bars Asysryrcc A s sr cssr PA FAFfA M A A Ah hAA =+ = = = = , 0.85 0 1 2 Ccc CB PfA MM = = , 0.85 () () , , 12 2 0.85 2 0.85 full x-axis plastic section modulus of steel shape = area of continuous reinforcing ba
13、rs at the centerline 2 4 cc D Dsyryrcc s srs rsrsrs csr fA P MZ FZ FZf Z A h ZAAc h h ZZZ = =+ = = = 2 1 2 () () () 2 2 , 12 2 , , 0 0.85 Forbelow the flange 0.852 2 0.852 f B BDsnycnc cnnsn d nn ccsrsyrsrs n cwy w snw n P MMZFZf Zh hZ hht fAAFA h fhtF t Zt h = = = + = + = 1 1 () ()() () ()() 22 , ,
14、 22 Forwithin theflange 0.8522 2 0.852 f ff ff f dd nn ccssrsysyrsrs n cy dd snsnn hth fAAdbAFAdbFA h fhbF b ZZbhh + = = 1 (A) (C) (D) (B) Figure I-1a. W-Shapes, Strong-Axis Anchor Points. SectionStress DistributionPointDefining Equations A E C D PLASTIC CAPACITIES FOR RECTANGULAR, ENCASED W-SHAPES
15、BENT ABOUT THE Y-Y AXIS B area of steel shape area of continuous reinforcing bars Asysryrcc A s sr cssr PA FAFfA M A A Ah hAA =+ = = = = , 0.85 0 1 2 ()() () full y-axis plastic section modulus of steel shape f f sr Esycc EDsEycEc sEsy cEsE Ah PA FfAhb MMZFZf ZZ h b ZZ =+ = = = , , 12 2 0.85 22 0.85
16、 4 1 2 1 Ccc CB PfA MM = = , 0.85 () full y-axis plastic section modulus of steel shape cc D Dsyrsrcc s rsr csr fA P MZ FZ FZf Z h ZAc h h ZZZ = =+ = = = , , 1 2 2 2 0.85 2 0.85 2 4 1 2 () ()() () Forbelow theflange f ffff fyf ff f B BDsnycnc cnnsn nn ccsys n c snsnn P MMZFZf Zh hZ b hh fAAt bFAt b
17、h t Fhtf bb ZZthh = = = + = + =+ , 12 2 2 , , 0 0.85 0.85222 2 420.85 2 22 1 1 () Forabove theflange full y-axis plastic section modulus of steel shape f nn ccsys n c snsy b hh fAAF A h f h ZZ + = = 2 , , 0.852 2 0.85 1 (A) (E) (C) (D) (B) Figure I-1b. W-Shapes, Weak-Axis Anchor Points SectionStress
18、 DistributionPointDefining Equations A C D B PLASTIC CAPACITIES FOR COMPOSITE, FILLED HSS BENT ABOUT THE X-X OR Y-Y AXIS () area of steel shape Asycc A s ci PA FAf M A Ah hr =+ = = = , 2 0.85 0 0.858 1 2 () Ccc CB PAf MM = = , 0.85 () full y-axis plastic section modulus of steel shape cc D Dsycc s c
19、i fA P MZ FZf Z h h Zr = =+ = = , , 1 2 2 3 0.85 2 0.85 0.192 4 1 2 () wy B BDsnycnc snw n cnn cc n c P MMZFZf Zt h Zh h fAh h f ht F = = = = = + 2 , 12 2 2 , 2, 0 0.85 2 0.85 2 0.854 1 1 (A) (C) (D) (B) Figure I-1c. Filled Rectangular or Square HSS, Strong- or Weak Axis Anchor Points SectionStress
20、DistributionPointDefining Equations A C D * is permitted to be used when the composite column is loaded only in axial compression. PLASTIC CAPACITIES FOR COMPOSITE, FILLED ROUND HSS BENT ABOUT ANY AXIS B Asycc A s c PA FfA M Adtt h A =+ = = = , 2 2 0.85* 0 () 4 Ccc CB PfA MM = = , 0.85 () plasticsec
21、tion modulusof steelshape = cc D Dsycc sc c f A P MZ FZf d ZZ h Z = =+ = = , , 1 2 3 3 0.85 2 0.85 6 6 () () () () (in rads) (“thin“ HSS wall assumed) B BsBycBc sBcB cB cs c cscscc c cc sy n P MZFZf d ZZ h Z KK K KKK KfA K Kf h dt KFt hh h = = = = =+ + = = = , 12 33 33 2 , , 2 0 0.85 sin 2 6 sin 2 6
22、 0.02602 0.0848 0.026020.857 0.0848 2 sin 222 (not used, for reference only) = +0.95 Asycc PA Ff A (A) (C) (D) (B) Figure I-1d. Filled Round HSS Anchor Points I-6 Example I-1 Composite Beam Design Given: A series of 45-ft. span composite beams at 10 ft. o/c are carrying the loads shown below. The be
23、ams are ASTM A992 and are unshored. The concrete has fc = 4 ksi. Design a typical floor beam with 3 in. 18 gage composite deck, and 4 in. normal weight concrete above the deck, for fire protection and mass. Select an appropriate beam and determine the required number of shear studs. Solution: Materi
24、al Properties: Concrete fc = 4 ksi Beam Fy = 50 ksi Fu = 65 ksi Manual Table 2-3 Loads: Dead load: Slab = 0.075 kip/ft2 Beam weight = 0.008 kip/ft2 (assumed) Miscellaneous = 0.010 kip/ft2 (ceiling etc.) Live load: Non-reduced = 0.10 kips/ft2 Since each beam is spaced at 10 ft. o.c. Total dead load =
25、 0.093 kip/ft2(10 ft.) = 0.93 kips/ft. Total live load = 0.10 kip/ft2(10ft.) = 1.00 kips/ft. Construction dead load (unshored) = 0.083 kip/ft2(10 ft) = 0.83 kips/ft Construction live load (unshored) = 0.020 kip/ft2(10 ft) = 0.20 kips/ft I-7 Determine the required flexural strength LRFD ASD wu = 1.2(
26、0.93 kip/ft) + 1.6(1.0 kip/ft) = 2.72 kip/ft Mu = 2 2.72 kip/ft(45 ft) 8 = 687 kip-ft. wa = 0.93 kip/ft + 1.0 kip/ft = 1.93 kip/ft Ma = 2 1.93 kip/ft(45 ft) 8 = 489 kip-ft. Use Tables 3-19, 3-20 and 3-21 from the Manual to select an appropriate member Determine beff The effective width of the concre
27、te slab is the sum of the effective widths for each side of the beam centerline, which shall not exceed: (1) one-eighth of the beam span, center to center of supports ( ) 45 ft. 2 8 = 11.3 ft. (2) one-half the distance to center-line of the adjacent beam ( ) 10 ft. 2 2 = 10.0 ft. Controls (3) the di
28、stance to the edge of the slab Not applicable Calculate the moment arm for the concrete force measured from the top of the steel shape, Y2. Assume a = 1.0 in. (Some assumption must be made to start the design process. An assumption of 1.0 in. has proven to be a reasonable starting point in many desi
29、gn problems.) Y2 = tslab a/2 = 7.5 = 7.0 in. Section I.3.1.1a Enter Manual Table 3-19 with the required strength and Y2=7.0 in. Select a beam and neutral axis location that indicates sufficient available strength. Select a W2150 as a trial beam. When PNA location 5 (BFL), this composite shape has an
30、 available strength of: Manual Table 3-19 LRFD ASD bMn= 770 kip-ft 687 kip-ft o.k. Mn/b = 512 kip-ft 489 kip-ft o.k. Manual Table 3-19 Note that the required PNA location for ASD and LRFD differ. This is because the live to dead load ratio in this example is not equal to 3. Thus, the PNA location re
31、quiring the most shear transfer is selected to be acceptable for ASD. It will be conservative for LRFD. I-8 Check the beam deflections and available strength Check the deflection of the beam under construction, considering only the weight of concrete as contributing to the construction dead load. Li
32、mit deflection to a maximum of 2.5 in. to facilitate concrete placement. ()()() ()() 4 33 4 4 5 0.83 kip/ft45 ft1728 in. /ft 5 1,060 in. 384384 29,000 ksi2.5 in. DL req w l I E = From Manual Table 3-20, a W2150 has Ix = 984 in.4, therefore this member does not satisfy the deflection criteria under c
33、onstruction. Using Manual Table 3-20, revise the trial member selection to a W2155, which has Ix = 1140 in.4, as noted in parenthesis below the shape designation. Check selected member strength as an un-shored beam under construction loads assuming adequate lateral bracing through the deck attachmen
34、t to the beam flange. LRFD ASD Calculate the required strength 1.4 DL = 1.4 (0.83 kips/ft) = 1.16 kips/ft 1.2DL+1.6LL = 1.2 (0.83) + 1.6(0.20) = 1.32 klf Mu (unshored) = 2 1.31 kip/ft(45 ft) 8 =331 kip-ft The design strength for a W2155 is 473 kip-ft 331 kip-ft o.k. Calculate the required strength D
35、L+LL = 0.83 + 0.20 = 1.03 kips.ft Ma (unshored) = 2 1.03 kips/ft(45 ft) 8 = 260 kip-ft The allowable strength for a W21 55 is 314 kip-ft 260 kip-ft o.k. For a W2155 with Y2=7.0 in, the member has sufficient available strength when the PNA is at location 6 and n Q= 292 kips. Manual Table 3-19 LRFD AS
36、D bMn=767 kip-ft 687 kip-ft o.k. Mn/b = 510 kip-ft 489 kip-ft o.k. Manual Table 3-19 Check a a = 0.85 n c Q f b = 292 kips 0.85(4 ksi)(10 ft.)(12 in./ft.) = 0.716 in. 0.716 in. 61.2 kips o.k. Va =() 45ft 1.93kip/ft 2 = 43.4 kips Vn/ = 156 kips 43.4 kips o.k. Manual Table 3-3 Determine the required n
37、umber of shear stud connectors Using perpendicular deck with one -in. diameter weak stud per rib in normal weight 4 ksi concrete. Qn = 17.2 kips/stud n n Q Q = 292 kips 17.2 kips = 17, on each side of the beam. Total number of shear connectors; use 2(17) = 34 shear connectors. Check the spacing of s
38、hear connectors Since each flute is 12 in., use one stud every flute, starting at each support, and proceed for 17 studs on each end of the span. 6dstud 336 kips o.k. Pn /c = 236 kips 236 kips 224 kips o.k. Manual Table 4-14 Supporting Calculations The available strength of this filled composite sec
39、tion can be most easily determined by using Table 4-14 of the Manual. Alternatively, the available strength can be determined by direct application of the Specification requirements, as illustrated below. Material Properties: HSS1063/8 Fy= 46 ksi Fu = 58 ksi Concrete fc = 5 ksi () 1.5 1.5 14553,900
40、ksi cc Ewf= Manual Table 2-3 I-11 Geometric Properties: HSS1063/8 t = 0.375 in. b = 10.0 in. h = 6.0 in. Concrete: The concrete area is calculated as follows r = 2t = 2(0.375 in. ) = 0.75 in. (outside radius) bf = b-2r = 10 2(0.75) = 8.50 in. hf = h-2r = 6 2(0.75) = 4.50 in. Ac = bfhf+ (r-t)2 + 2bf(
41、r-t) + 2hf(r-t) = (8.50 in.)(4.50 in.) + (0.375 in.)2 + 2(8.50 in.)(0.375 in.) + 2(4.50 in.)(0.375 in.) = 48.4 in.2 () ()() 2 2 33 1 1222 42()()8 22 121289223 c rtrtb hbhh Irt =+ For this shape, buckling will take place about the weak axis, thus () () () () 1 1 2 2 62 0.3755.25 in. 104 0.3758.5 in.
42、64 0.3754.5 in. 0.375 in. 0.750.3750.375 in. h b h b rt = = = = = () ()() 33 4 2 2 4 (8.5)(5.25)2(0.375)(4.50)8 2 0.375 121289 0.3754 0.3754.5 2 223 = 111 in. c I =+ + HSS106a: As = 10.4 in.2 Is = 61.8 in.4 h/t = 25.7 Manual Table 1-11 Manual Table 1-11 Limitations: 1) Normal weight concrete 10 ksi
43、fc 3 ksi fc = 5 ksi o.k. 2) Not Applicable. 3) The cross-sectional area of the steel HSS shall comprise at least one percent of the total composite cross section. 10.4 in.2 (0.01)(48.6 in.2+ 10.4 in.2) = 0.590 in.2 o.k. 4) The maximum b/t ratio for a rectangular HSS used as a composite column shall
44、be equal to 2.26 y E F. b/t = 25.7 0.90 Therefore use 0.90 EIeff = EsIs + EsIsr + C3EcIc = (29,000 ksi)(61.8 in.4) + (0.90)(3,900 ksi)( 111 in.4) = 2,180,000 kip-in.2 User note: K value is from Chapter C and for this case K = 1.0. Pe = 2(EIeff)/(KL)2 = 2(2,180,000 kip-in.2) / (1.0)(14ft)(12in./ft)2
45、= 762 kips o e P P = 684 kips 762 kips = 0.898 0.898 336 kips o.k. c = 2.00 Pn/c = 470 kips / 2.00 = 235 kips 235 kips 224 kips o.k. Section I2.1b I-13 Example I-3 Encased Composite Column in Axial Compression Given: Determine if a 14 ft tall W1045 steel section encased in a 24 in.24in. concrete col
46、umn with fc = 5 ksi, is adequate to support a dead load of 350 kips and a live load of 1050 kips in axial compression. The concrete section has 8-#8 longitudinal reinforcing bars and #4 transverse ties 12in. o/c., The column is pinned at both ends and the load is applied directly to the concrete enc
47、asement. Solution: Material Properties: Column W1045 Fy = 50 ksi Fu = 65 ksi Concrete fc = 5 ksi Ec = 3,900 ksi (145 pcf concrete) Reinforcement Fyst = 60 ksi Manual Table 2-3 Geometric Properties: W1045: As = 13.3 in.2 Iy = 53.4 in.4 Reinforcing steel: Asr = 6.32 in.2 (the area of 1-#8 bar is 0.79 in.2, per ACI) Isr = 44 224 (0.50) 86(0.79)(9.5)428 in. 44 r Ad +=+= Concrete: Ac = Acg - As - Asr = 576 in.2- 13.3 in.2 - 6.3
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