Presentation on theme: "Elastic Stresses in Unshored Composite Section"— Presentation transcript:
1Elastic Stresses in Unshored Composite Section The elastic stresses at any location shall be the sum of stresses caused by appropriate loads applied separatelySteel beamPermanent loads applied before the slab has hardened, are carried by the steel section.Short-term composite sectionTransient loads (such as live loads) are assumed to be carried by short-term composite action. The short-term modular ratio, n, should be used.Long-term composite section.Permanent loads applied after the slab has been hardened are carried by the long-term composite section. The long-term modular ratio, 3n, should be used.
2Elastic Stresses ( )The procedure shown in this picture is only valid if the neutral axis is not in the concrete.Use iterations otherwise.
3Elastic Stresses (188.8.131.52) Effective Width (Interior) According to AASHTO-LRFD , the effective width for interior girders is to be taken as the smallest of:One quarter of the effective span length (span length in simply supported beams and distance between permanent load inflection points in continuous beams).Average center-to-center spacing.Twelve times the slab thickness plus the top flange width.
4Hybrid Sections 6.10.3, 184.108.40.206 The web yield strength must be: 1.20 fyf ≥ fyw≥ 0.70 fyf and fyw≥ 36 ksiThe hybrid girder reduction factor = RhWhere, b=2 Dn tw / AfnDn = larger of distance from elastic NA to inside flange faceAfn = flange area on the side of NA corresponding to Dnfn = yield stress corresponding to Afn
5Additional sections 220.127.116.11 – Variable web depth members – Stiffness– Flange stresses and bending moments– Minimum negative flexure concrete deck rft.– Net section fracture
6Web Bend-Buckling Resistance (18.104.22.168) For webs without longitudinal stiffeners, the nominal bend buckling resistance shall be taken as:When the section is composite and in positive flexure Rb=1.0When the section has one or more longitudinal stiffeners, and D/tw≤ 0.95 (E k /Fyc)0.5 then Rb = 1.0When 2Dc/tw ≤ 5.7 (E / Fyc)0.5 then Rb = 1.0
7Web Bend-Buckling Reduction (22.214.171.124) If the previous conditions are not met then:
8Calculating the depth Dc and Dcp (App. D6.3) For composite sections in positive flexure, the depth of the web in compression in the elastic range Dc, shall be the depth over which the algebraic sum of the stresses in the steel, the long-term composite and short term composite section is compressiveIn lieu, you can use
9Calculating the depth Dc and Dcp (App. D6.3) For composite sections in positive flexure, the depth of the web in compression at the plastic moment Dcp shall be taken as follows for the case of PNA in the web:
106.10 I-shaped Steel Girder Design Proportioning the section (6.10.2)Webs without longitudinal stiffeners must be limited toD/tw ≤ 150Webs with longitudinal stiffeners must be limited toD/tw≤ 300Compression and tension flanges must be proportioned such that:
126.10 I-Shaped Steel Girder Design Strength limit stateComposite sections in positive flexure ( )Classified as compact section if:Flange yield stress (Fyf ) ≤ 70 ksiwhere, Dcp is the depth of the web in compression at the plastic momentClassified as non-compact section if requirement not metCompact section designed using SectionNon-compact section designed using Section
136.10.7 Flexural Resistance Composite Sections in Positive Flexure Compact sectionsAt the strength limit state, the section must satisfyIf Dp≤ 0.1 Dt , then Mn = MpOtherwise, Mn = Mp(1.07 – 0.7 Dp/Dt)Where, Dp = distance from top of deck to the N.A. of the composite section at the plastic moment.Dt = total depth of composite sectionFor continuous spans, Mn = 1.3 My. This limit allows for better design with respect to moment redistributions.
146.10.7 Flexural Resistance Composite Sections in Positive Flexure Non-Compact sections ( )At the strength limit state:The compression flange must satisfy fbu ≤ ff FncThe tension flange must satisfy fbu + fl/3 ≤ ff FntNominal flexural resistance Fnc = Rb Rh FycNominal flexural resistance Fnt= Rh FytWhere,Rb = web bend buckling reduction factorRh = hybrid section reduction factor
156.10.7 Flexural Resistance Composite Sections in Positive Flexure Ductility requirement. Compact and non-compact sections shall satisfy Dp ≤ 0.42 DtThis requirement intends to protect the concrete deck from premature crushing. The Dp/Dt ratio is lowered to 0.42 to ensure significant yielding of the bottom flange when the crushing strain is reached at the top of deck.
166.10 I-Shaped Steel Girder Design Composite Sections in Negative Flexure and Non-composite Sections ( )Sections with Fyf ≤ 70 ksiWeb satisfies the non-compact slenderness limitWhere, Dc = depth of web in compression in elastic range.Designed using provisions for compact or non-compact web section specified in App. A.Can be designed conservatively using Section 6.8If you use 6.8, moment capacity limited to MyIf use App. A., get greater moment capacity than My
176.10.8 Flexural Resistance Composite Sections in Negative Flexure and Non-Composite Section Discretely braced flanges in compressionDiscretely braced flanges in tensionContinuously braced flanges: fbu≤ ff Rh FyfCompression flange flexural resistance = Fnc shall be taken as the smaller of the local buckling resistance and the lateral torsional buckling resistance.Tension flange flexural resistance = Fnt = Rh Fyt
18Flange Local buckling or Lateral Torsional Buckling Resistance Fn or MnInelastic Buckling(Compact)Inelastic Buckling(non-compact)Fmax or MmaxFyr or MrElastic Buckling(Slender)LpLrLblpflrflf
196.10.8 Flexural Resistance Composite Sections in Negative Flexure and Non-Composite Section Fnc Compression flange flexural resistance – local buckling
22Unstiffened Web Buckling in Shear D/twWeb plastification in shearInelastic web bucklingElastic web buckling
236.10.9 Shear Resistance – Unstiffened webs At the strength limit state, the webs must satisfy:Vu ≤ fv VnNominal resistance of unstiffened webs:Vn = Vcr = C Vpwhere, Vp = 0.58 Fyw D twC = ratio of the shear buckling resistance to shear yield strengthk = 5 for unstiffened webs
24Tension Field ActionBeam ActionTension Field ActionDgd0
256.10.9 Shear resistance – Stiffened Webs Members with stiffened webs have interior and end panels.The interior panels must be such thatWithout longitudinal stiffeners and with a transverse stiffener spacing (do) < 3DWith one or more longitudinal stiffeners and transverse stiffener spacing (do) < 1.5 DThe transverse stiffener distance for end panels with or without longitudinal stiffeners must be do < 1.5 DThe nominal shear resistance of end panel isVn = C (0.58 Fyw D tw)For this case – k is obtained using equation shown on next page and do = distance to stiffener
26Shear Resistance of Interior Panels of Stiffened Webs
28Types of Stiffeners Transverse Longitudinal Intermediate D Bearing
296.10.11 Design of Stiffeners Transverse Intermediate Stiffeners Consist of plates of angles bolted or welded to either one or both sides of the webTransverse stiffeners may be used as connection plates for diaphragms or cross-framesWhen they are not used as connection plates, then they shall tight fit the compression flange, but need not be in bearing with tension flangeWhen they are used as connection plates, they should be welded or bolted to both top and bottom flangesThe distance between the end of the web-to-stiffener weld and the near edge of the adjacent web-to-flange weld shall not be less than 4 tw or more than 6 tw.
30Transverse Intermediate Stiffeners Single PlateAngleDouble PlateLess than 4 tw or more than 6tw
31Design of StiffenersProjecting width of transverse stiffeners must satisfy:bt ≥ d/30and bf/4 ≤ bt ≤ 16 tpThe transverse stiffener’s moment of inertia must satisfy:It ≥ do tw3 Jwhere, J = required ratio of the rigidity of one transverse stiffener to that of the web plate = 2.5 (D/do)2 – 2.0 ≥ 2.5It = stiffener m.o.i. about edge in contact with web forsingle stiffeners and about mid thickness for pairs.Transverse stiffeners in web panels with longitudinal stiffeners must also satisfy:
32Design of StiffenersThe stiffener strength must be greater than that required for TFA to develop. Therefore, the area requirement is:If this equation gives As negative, it means that the web alone is strong enough to develop the TFA forces. The stiffener must be proportions for m.o.i. and width alone
33Design of StiffenersBearing Stiffeners must be placed on the web of built-up sections at all bearing locations. Either bearing stiffeners will be provided or the web will be checked for the limit states of:Web yielding – Art. D6.5.2Web crippling – Art. D6.5.3Bearing stiffeners will consist of one or more plates or angles welded or bolted to both sides of the web. The stiffeners will extend the full depth of the web and as closely as practical to the outer edges of the flanges.The stiffeners shall be either mille to bear against the flange or attached by full penetration welds.
34Design of StiffenersTo prevent local buckling before yielding, the following should be satisfied.The factored bearing resistance for the fitted ends of bearing stiffeners shall be taken as:The axial resistance shall be determined per column provisions. The effective column length is 0.75DIt is not D because of the restraint offered by the top and bottom flanges.
35Design of StiffenersInterior panelEndpanelDbttp9tw
36General Considerations Shear studs are needed to transfer the horizontal shear that is developed between the concrete slab and steel beam.AASHTO-LRFD requires that full transfer (i.e. full composite action) must be achieved.Shear studs are placed throughout both simple and continuous spans.Two limit states must be considered: fatigue and shear. Fatigue is discussed later.
37Strength of Shear Studs 0.85Cross-sectional are of the stud in square inchesMinimum tensile strength of the stud (usually 60 ksi)
38PlacementA sufficient number of shear studs should be placed between a point of zero moment and adjacent points of maximum moment.It is permissible to evenly distribute the shear studs along the length they are needed in (between point of inflection and point of maximum moment), since the studs have the necessary ductility to accommodate the redistribution that will take place.
39Miscellaneous Rules Minimum length = 4 x stud diameter Minimum longitudinal spacing = 4 x stud diameterMinimum transverse spacing = 4 x stud diameterMaximum longitudinal spacing = 8 x slab thicknessMinimum lateral cover = 1".Minimum vertical cover = 2”.Minimum penetration into deck = 2”