BRIDGE DESIGN - Purdue Engineering

BRIDGE DESIGN - Purdue Engineering

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 Short-term composite section Permanent loads applied before the slab has hardened, are carried by the steel 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. Elastic Stresses (6.10.1.1) The procedure shown in this picture is only valid if the neutral axis is not in the concrete. Use iterations otherwise. b Myt f ct nI tr t b/n t yt yb Original section Transformed section f sb

My b I tr Elastic Stresses (6.10.1.1) Effective Width (Interior) According to AASHTO-LRFD 4.6.2.6.1, 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. Hybrid Sections 6.10.3, 6.10.1.10 The web yield strength must be: 1.20 f fyw 0.70 f yf

yf and fyw 36 ksi The hybrid girder reduction factor = Rh 12 (3 3 ) Rh 12 2 Where, =2 Dn tw / Afn Dn = larger of distance from elastic NA to inside flange face Afn = flange area on the side of NA corresponding to Dn fn = yield stress corresponding to Afn Additional sections

6.10.1.4 Variable web depth members 6.10.1.5 Stiffness 6.10.1.6 Flange stresses and bending moments 6.10.1.7 Minimum negative flexure concrete deck rft. 6.10.1.8 Net section fracture Web Bend-Buckling Resistance (6.10.1.9) For webs without longitudinal stiffeners, the nominal bend buckling resistance shall be taken as: Fcrw 0.9 E k D

tw 2 where, k bend buckling coefficient 9 Dc / D 2 where, Dc depth of web in compression in elastic range When the section is composite and in positive flexure R b=1.0 When the section has one or more longitudinal stiffeners, and D/tw 0.95 (E k /Fyc)0.5 then Rb = 1.0 When 2Dc/tw 5.7 (E / Fyc)0.5 then Rb = 1.0 Web Bend-Buckling Reduction (6.10.1.10) If the previous conditions are not met then:

2 Dc awc Rb 1 rw 1.0 1200 300 awc tw where, rw 5.7 and awc E Fyc 2 Dc t w b fc t fc Calculating 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 compressive Dc f DC1 f DC 2 fWS f LL IM tf f DC1 f DC 2 fWS f LL IM csteel c3n cn In lieu, you can use fc Dc d t fc 0 f c ft where, d depth of steel sec tion f c and ft are the compression and tension flange stresses Calculating 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: ' F A F A 0 . 85 f D yt t yc c c As Fyr Ar Dcp 1 2 Fyw Aw 6.10 I-shaped Steel Girder Design Proportioning the section (6.10.2) Webs without longitudinal stiffeners must be limited to

D/tw 150 Webs with longitudinal stiffeners must be limited to D/tw 300 Compression and tension flanges must be proportioned such that: b D / 6 f bf 2t f 12.0 t f 1.1 tw 0.1 I yc I yt 10 Section Behavior

Moment Mp My Compact Noncompact Slender Curvature 6.10 I-Shaped Steel Girder Design Strength limit state 6.10.6 Composite sections in positive flexure (6.10.6.2.2) Classified as compact section if: Flange yield stress (F ) 70 ksi yf 2 Dcp tw where, Dcp 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 6.10.7.1 Non-compact section designed using Section 6.10.7.2 E 3.76 Fyc 6.10.7 Flexural Resistance Composite Sections in Positive Flexure Compact sections At the strength limit state, the section must satisfy 1 M u fl S xt f M n 3 If D 0.1 D , then M = M p t n p Otherwise, 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 section For continuous spans, Mn = 1.3 My. This limit allows for better design with respect to moment redistributions. 6.10.7 Flexural Resistance Composite Sections in Positive Flexure Non-Compact sections (6.10.7.2) At the strength limit state: The compression flange must satisfy fbu f Fnc The tension flange must satisfy fbu + fl/3 f Fnt Nominal flexural resistance Fnc = Rb Rh Fyc

Nominal flexural resistance Fnt= Rh Fyt Where, Rb = web bend buckling reduction factor Rh = hybrid section reduction factor 6.10.7 Flexural Resistance Composite Sections in Positive Flexure Ductility requirement. Compact and non-compact sections shall satisfy Dp 0.42 Dt This 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.

6.10 I-Shaped Steel Girder Design Composite Sections in Negative Flexure and Noncomposite Sections (6.10.6.2.2) Sections with Fyf 70 ksi Web satisfies the non-compact slenderness limit 2 Dc E 5.7 tw Fyc Where, 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 6.8 If you use 6.8, moment capacity limited to M y

If use App. A., get greater moment capacity than M y 6.10.8 Flexural Resistance Composite Sections in Negative Flexure and Non-Composite Section Discretely braced flanges in compression fbu 1 fl f Fnc 3 Discretely braced flanges in tension fbu 1 fl f Fnt 3

Continuously braced flanges: fbu f Rh Fyf Compression 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 Flange Local buckling or Lateral Torsional Buckling Resistance Fn or Mn Fmax or Mmax Inelastic Buckling Inelastic Buckling (Compact) (non-compact) Fyr or Mr Elastic Buckling

(Slender) Lp Lr Lb pf rf f 6.10.8 Flexural Resistance Composite Sections in Negative Flexure and Non-Composite Section Fnc Compression flange flexural resistance local buckling f b fc 2 t fc pf 0.38 E Fyc

When, f pf Fnc Rb Rh Fyc When, f rf Fnc 1 Fyr 0.7 Fyc rf 0.56 E Fyr Fyr f pf 1 Rh Fyc rf pf Rb Rh Fyc Fnc Compression flange flexural resistance

Lateral torsional buckling Lb E Fyc L p 1.0 rt When, Lb L p rf rt Fnc Rb Rh Fyc Fyr Lb L p Fnc Cb 1 1 Rb Rh Fyc Rb Rh Fyc Rh Fyc Lr L p Fnc Fcr Rb Rh Fyc When, Lb Lr When, Lb Lr Where, 2 2

f f Cb 1.75 1.05 1 0.3 1 2.3 f2 f2 Fcr rt E Fyr Cb Rb 2 E 2 Lb rt b fc Dt 12 1 c w 3 b fc t fc

Lateral Torsional Buckling Lateral support Lb Unstiffened Web Buckling in Shear Web plastification in shear Vn V p 0.58Fyw D.t w Inelastic web buckling Vn 1.48tw2 EFyw Elastic web buckling 3 4.55t w E Vn D 2.46 E Fy

3.07 E Fy D/tw 6.10.9 Shear Resistance Unstiffened webs At the strength limit state, the webs must satisfy: Vu v Vn Nominal resistance of unstiffened webs: Vn = Vcr = C Vp where, Vp = 0.58 Fyw D tw C = ratio of the shear buckling resistance to shear yield strength k = 5 for unstiffened webs D Ek If , 1.12

; tw Fyw If , If , 1.12 then C 1.0 Ek D Ek 1.12 1.40 ; then C D Fyw tw Fyw tw D Ek 1.40 ; tw Fyw

then C 1.57 D tw 2 Ek Fyw Ek Fyw Tension Field Action Vn Vcr VTFA Beam Action Tension Field Action D d0

6.10.9 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 (do) < 3D With one or more longitudinal stiffeners and transverse stiffener spacing (do) < 1.5 D The transverse stiffener distance for end panels with or without longitudinal stiffeners must be do < 1.5 D The nominal shear resistance of end panel is Vn = C (0.58 Fyw D tw) For this case k is obtained using equation shown on next page and do = distance to stiffener

Shear Resistance of Interior Panels of Stiffened Webs If the sec tion is proportioned such that : 2 D tw b fc t fc b ft t ft 0.87 (1 C ) Vn 0.58 Fyw D tw C 2 do 1 D where, d o transverse stiffener spacing k shear buckling coefficient 5 If not , then Vn 0.58 Fyw

5 do D 2 0.87 (1 C ) D tw C 2 do do 1 D D

2.5 Transverse Stiffener Spacing D Interior panel End panel d o 1.5D d o 3D do 1.5 D Types of Stiffeners Transverse Intermediate Stiffener

Longitudinal Stiffener D do 1.5 D d o 1.5D Bearing Stiffener 6.10.11 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 tw or more than 6 tw. Transverse Intermediate Stiffeners Single Plate Angle Double Plate Less than 4 tw or more than 6tw 6.10.11 Design of Stiffeners Projecting width of transverse stiffeners must satisfy: bt 2.0 + d/30 and

bf/4 bt 16 tp The transverse stiffeners moment of inertia must satisfy: It d o t w3 J where, J = required ratio of the rigidity of one transverse stiffener to that of the web plate = 2.5 (D/do)2 2.0 2.5 It = 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: It bt D Il bl 3.0 d o 6.10.11 Design of Stiffeners The stiffener strength must be greater than that required for TFA to develop. Therefore, the area requirement is: Fyw 2 Vu D As 0.15 B (1 C ) tw 18

t V F crs w v n where, Fcrs elastic local buckling stress Fcrs 0.31E 2 Fys bt tp and , B 1.0 for stiffener pairs B 1.8 for sin gle angle stiffener B 2.4 for sin gle plate stiffener 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

6.10.11 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. 6.10.11 Design of Stiffeners To prevent local buckling before yielding, the following

should be satisfied. E bt 0.48t p Fys The factored bearing resistance for the fitted ends of bearing stiffeners shall be taken as: Rsb n 1.4 Apn Fys 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. 6.10.11 Design of Stiffeners D Interior panel d o 3D tp

End panel d o 1.5 D bt 9tw 9tw 9tw 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. Strength of Shear Studs Qr sc Qn 0.85 Cross-sectional are of the stud in square inches Qn 0.5 Asc ' f c Ec Asc Fu Minimum tensile strength of the stud (usually 60 ksi) 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. 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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