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Elastic Stresses in Unshored Composite Section The elastic stresses at any location shall be the sum of stresses caused by appropriate loads applied separately Steel beam Permanent loads applied before the slab has hardened, are carried by the steel section. Short-term composite section Transient 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.

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Elastic Stresses ( ) The procedure shown in this picture is only valid if the neutral axis is not in the concrete. Use iterations otherwise.

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Elastic Stresses ( ) 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.

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Hybrid Sections , The web yield strength must be: 1.20 f yf ≥ fyw≥ 0.70 f yf and f yw ≥ 36 ksi The hybrid girder reduction factor = R h Where, =2 D n t w / A fn D n = larger of distance from elastic NA to inside flange face A fn = flange area on the side of NA corresponding to D n f n = yield stress corresponding to A fn

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Additional sections – Variable web depth members – Stiffness – Flange stresses and bending moments – Minimum negative flexure concrete deck rft – Net section fracture

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Web Bend-Buckling Resistance ( ) For webs without longitudinal stiffeners, the nominal bend buckling resistance shall be taken as: When the section is composite and in positive flexure R b =1.0 When the section has one or more longitudinal stiffeners, and D/t w ≤ 0.95 (E k /F yc ) 0.5 then R b = 1.0 When 2D c /t w ≤ 5.7 (E / F yc ) 0.5 then R b = 1.0

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Web Bend-Buckling Reduction ( ) If the previous conditions are not met then:

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Calculating the depth D c and D cp (App. D6.3) For composite sections in positive flexure, the depth of the web in compression in the elastic range D c, 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 compressive In lieu, you can use

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Calculating the depth D c and D cp (App. D6.3) For composite sections in positive flexure, the depth of the web in compression at the plastic moment D cp shall be taken as follows for the case of PNA in the web:

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6.10 I-shaped Steel Girder Design Proportioning the section (6.10.2) Webs without longitudinal stiffeners must be limited to D/t w ≤ 150 Webs with longitudinal stiffeners must be limited to D/t w ≤ 300 Compression and tension flanges must be proportioned such that:

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Slender Noncompact Compact Moment Curvature MpMp MyMy Section Behavior

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6.10 I-Shaped Steel Girder Design Strength limit state Composite sections in positive flexure ( ) Classified as compact section if: Flange yield stress (F yf ) ≤ 70 ksi where, D cp is the depth of the web in compression at the plastic moment Classified as non-compact section if requirement not met Compact section designed using Section Non-compact section designed using Section

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Flexural Resistance Composite Sections in Positive Flexure Compact sections At the strength limit state, the section must satisfy If D p ≤ 0.1 D t, then M n = M p Otherwise, M n = M p (1.07 – 0.7 D p /D t ) Where, D p = distance from top of deck to the N.A. of the composite section at the plastic moment. D t = total depth of composite section For continuous spans, M n = 1.3 M y. This limit allows for better design with respect to moment redistributions.

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Flexural Resistance Composite Sections in Positive Flexure Non-Compact sections ( ) At the strength limit state: The compression flange must satisfy f bu ≤ f F nc The tension flange must satisfy f bu + f l /3 ≤ f F nt Nominal flexural resistance F nc = R b R h F yc Nominal flexural resistance F nt = R h F yt Where, R b = web bend buckling reduction factor R h = hybrid section reduction factor

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Ductility requirement. Compact and non-compact sections shall satisfy D p ≤ 0.42 D t This requirement intends to protect the concrete deck from premature crushing. The D p /D t 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 Flexural Resistance Composite Sections in Positive Flexure

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6.10 I-Shaped Steel Girder Design Composite Sections in Negative Flexure and Non- composite Sections ( ) Sections with F yf ≤ 70 ksi Web satisfies the non-compact slenderness limit Where, D c = 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 6.8 If you use 6.8, moment capacity limited to M y If use App. A., get greater moment capacity than M y

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Flexural Resistance Composite Sections in Negative Flexure and Non-Composite Section Discretely braced flanges in compression Discretely braced flanges in tension Continuously braced flanges: f bu ≤ f R h F yf Compression flange flexural resistance = F nc shall be taken as the smaller of the local buckling resistance and the lateral torsional buckling resistance. Tension flange flexural resistance = F nt = R h F yt

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Flange Local buckling or Lateral Torsional Buckling Resistance F n or M n LbLb Inelastic Buckling (non-compact) Elastic Buckling (Slender) LpLp F max or M max Inelastic Buckling (Compact) pf LrLr rf f F yr or M r

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Flexural Resistance Composite Sections in Negative Flexure and Non-Composite Section F nc Compression flange flexural resistance – local buckling

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F nc Compression flange flexural resistance Lateral torsional buckling

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Lateral Torsional Buckling

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Unstiffened Web Buckling in Shear D/t w Web plastification in shear Inelastic web buckling Elastic web buckling

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Shear Resistance – Unstiffened webs At the strength limit state, the webs must satisfy: V u ≤ v V n Nominal resistance of unstiffened webs: V n = V cr = C V p where, V p = 0.58 F yw D t w C = ratio of the shear buckling resistance to shear yield strength k = 5 for unstiffened webs

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Tension Field Action d0d0 D Beam Action Tension Field Action

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Shear resistance – Stiffened Webs Members with stiffened webs have interior and end panels. The interior panels must be such that Without longitudinal stiffeners and with a transverse stiffener spacing (d o ) < 3D With one or more longitudinal stiffeners and transverse stiffener spacing (d o ) < 1.5 D The transverse stiffener distance for end panels with or without longitudinal stiffeners must be d o < 1.5 D The nominal shear resistance of end panel is V n = C (0.58 F yw D t w ) For this case – k is obtained using equation shown on next page and d o = distance to stiffener

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Shear Resistance of Interior Panels of Stiffened Webs

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Transverse Stiffener Spacing Interior panel End panel D

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Types of Stiffeners D Bearing Stiffener Transverse Intermediate Stiffener Longitudinal Stiffener

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Design of Stiffeners Transverse Intermediate Stiffeners Consist of plates of angles bolted or welded to either one or both sides of the web Transverse stiffeners may be used as connection plates for diaphragms or cross-frames When they are not used as connection plates, then they shall tight fit the compression flange, but need not be in bearing with tension flange When they are used as connection plates, they should be welded or bolted to both top and bottom flanges The 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 t w or more than 6 tw.

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Transverse Intermediate Stiffeners Less than 4 t w or more than 6t w Single Plate Double Plate Angle

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Design of Stiffeners Projecting width of transverse stiffeners must satisfy: b t ≥ d/30 and b f /4 ≤ b t ≤ 16 t p The transverse stiffener’s moment of inertia must satisfy: I t ≥ d o t w 3 J where, J = required ratio of the rigidity of one transverse stiffener to that of the web plate = 2.5 (D/d o ) 2 – 2.0 ≥ 2.5 I t = stiffener m.o.i. about edge in contact with web for single stiffeners and about mid thickness for pairs. Transverse stiffeners in web panels with longitudinal stiffeners must also satisfy:

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Design of Stiffeners The stiffener strength must be greater than that required for TFA to develop. Therefore, the area requirement is: If this equation gives A s 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

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Design of Stiffeners Bearing 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. D6.5.2 Web crippling – Art. D6.5.3 Bearing 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.

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Design of Stiffeners To 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 0.75D It is not D because of the restraint offered by the top and bottom flanges.

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Design of Stiffeners Interior panel End panel D btbt tptp 9tw9tw 9tw9tw 9tw9tw

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General 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.

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Strength of Shear Studs Cross-sectional are of the stud in square inches Minimum tensile strength of the stud (usually 60 ksi) 0.85

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Placement A 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.

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Miscellaneous Rules Minimum length = 4 x stud diameter Minimum longitudinal spacing = 4 x stud diameter Minimum transverse spacing = 4 x stud diameter Maximum longitudinal spacing = 8 x slab thickness Minimum lateral cover = 1". Minimum vertical cover = 2”. Minimum penetration into deck = 2”

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