45793867 Design of Super Structure

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A. Design of Superstructure1.0 Design DataFig 1.BRIDGE CROSS-SECTIONPage 11.1. Materials and its Properties:M25Fe415Characteristics Strengthof Concrete fck = 25 MPaPermissible direct compressive stress, c= 6.2 MPaPermissible flexural compressive stress, cbc = 8.3 MPaMaximum Permissible shear stress, max ( 0.07*fck) = 1.75 MPaFig 1.BRIDGE CROSS-SECTIONMaximum Permissible shear stress, max ( 0.07 fck) 1.75 MPaBasic Permissible Stresses of Reinforcing Bars as per IRC : 21-1987, Section III:Permissible Flexural Tensile stress, st= 200 MPaPermissible direct compressive stress, co= 170 MPaSelf weight of materials as per IRC : 6-2000:Concrete (cement-Reinforced) = 24 kN/m3Fig 1.BRIDGE CROSS-SECTIONMacadam (binder premix) = 22 kN/m31.2. Geometrical Properties:Effective Span of Bridge = 24.00 mTotal length of span = 24.56 mNumbers of span = 2Width of expantion Joint = 40 mmTotal length of Bridge = 49.2 mFig 1.BRIDGE CROSS-SECTIONg gNos. of longitudinal Girder = 3Spacing of Girder = 2.4 mRib width of main girder = 400 mmOverall depth of main girder = 2000 mmDepth of kerb above deck slab = 225 mmNos. of cross girder = 6Spacing of cross girder = 4.8 mRib width of cross girder = 300 mmFig 1.BRIDGE CROSS-SECTIONRib width of cross girder = 300 mmOverall depth of cross girder = 1500 mmDeck slab thickness = 220 mmDeck slab thickness at edge = 150 mmThickness of wearing coat = 80 mmFillet size (horizontal) = 150 mmFillet size (vertical) = 150 mmBridge Width:Fig 1.BRIDGE CROSS-SECTIONCarriageway width = 6 mFootpath width = 0.45 mKerb width Outer = 0.15 mKerb width Inner = 0 mTotal Width of Deck Slab = 7.2 mTotal depth of Kerb Outer = 0.375 mTotal depth of Kerb Inner = 0 mFig 1.BRIDGE CROSS-SECTION Fig 1.BRIDGE CROSS-SECTIONPage 12.0 Design of Slabh il l b i d i d b i id h h dcbc 3280Fig 1.BRIDGE CROSS-SECTION57KN350KN37.5KNIRC Class AA Track Loading IRC Class A Loading IRC Class AA Wheel Loading Fig. 2.a, 2.b & 2.cPage 22.1 Design of Cantilever slab: The cantilever slab is designed by effective width method.300 mm at junction with rib150 mm at free end0.5 kN/m (assumed)Impact factor = 54 % (for IRC class A loading)25 % (for IRC class AA loading)Thickness of slab =Self weight of Railing =cbc 3280Fig 1.BRIDGE CROSS-SECTION57KN350KN37.5KNIRC Class AA Track Loading IRC Class A Loading IRC Class AA Wheel Loading Fig. 2.a, 2.b & 2.c( g)Dead Load Bending Moment and Shear Force:S.No. Item Width Depth Unit Wt1 Railing/Parapet0.5 kN1.525-0.100=0.9 m 0.45 kN.m2 Kerb (outer) 0.2 0.225 24 1.08 kN1.525-0.100=0.9 m 0.97 kN.m3 Kerb (inner) 0 0 24 0 kN0.15+0.175/2=0.2375 m 0 kN.m4 Wearing Coat 0.4 0.08 22 0.704 kN0.150/2= 0.075 m 0.05 kN.mAssumedLoad / m run (kN) Moment Distance cbc 3280Fig 1.BRIDGE CROSS-SECTION57KN350KN37.5KNIRC Class AA Track Loading IRC Class A Loading IRC Class AA Wheel Loading Fig. 2.a, 2.b & 2.c4 Wearing Coat 0.4 0.08 22 0.704 kN0. 50/0.075 m 0.05 kN.m5 Slab 1 0.15 24 3.6 kN 1.525/2= 0.5 m 1.8 kN.m1 0.075 24 1.8 kN 1.525/3= 0.3333 m 0.6 kN.mTotal kN kN.m= 7.684 kN= 3.875 kN.m7.684Dead load Shear force at theface of ribDead load Bending Moment at the face of rib3.875cbc 3280Fig 1.BRIDGE CROSS-SECTION57KN350KN37.5KNIRC Class AA Track Loading IRC Class A Loading IRC Class AA Wheel Loading Fig. 2.a, 2.b & 2.cLive Load Bending Moment and Shear Force:cbc 3280Fig 1.BRIDGE CROSS-SECTION57KN350KN37.5KNIRC Class AA Track Loading IRC Class A Loading IRC Class AA Wheel Loading Fig. 2.a, 2.b & 2.cIRC Class AA will not operate on the cantilever slab that shown in fig 2.b & 2.c above and Class A Loading is to be considered and the load will be as shown in fig 2.a above.Effective width of dispersion be is computed by equation be = 1.2X+ bwHereX= 0.125 mbw= 0.41 mcbc 3280Fig 1.BRIDGE CROSS-SECTION57KN350KN37.5KNIRC Class AA Track Loading IRC Class A Loading IRC Class AA Wheel Loading Fig. 2.a, 2.b & 2.cbw= 0.41 mHencebe= 0.56 mIRC Class A Loading Load = 28.5 kNLive Load per m width including impact = 76.339 kNMaximum Moment due to live load = 9.5424 kNmAverage thickness of cantilever slab = 225 mmTaking pedestrain load (LL) = 5.0 kN/m2Eff ti idth f l b 0 45cbc 3280Fig 1.BRIDGE CROSS-SECTION57KN350KN37.5KNIRC Class AA Track Loading IRC Class A Loading IRC Class AA Wheel Loading Fig. 2.a, 2.b & 2.cEffective width of slab = 0.45 mCantilever length of slab = 1 mMaximum Bending moment = 1.406 kN.mShear force at the face of slab = 2.250 kNTotal Design Shear Force = 86.3 kNTotal Design Bending Moment = 14.82 kN.mDesign of Section:Modular Ratio, m = = 11.245cbc 3280Fig 1.BRIDGE CROSS-SECTION57KN350KN37.5KNIRC Class AA Track Loading IRC Class A Loading IRC Class AA Wheel Loading Fig. 2.a, 2.b & 2.cPage 2Neutral axis factor, k = = 0.3182Lever arm factor, j = = 0.8939cbc 3280stcbcm cbcm +31 kstj k 21RbMd jMst..=d jMst. . =bddtan M- V =vco c k k . .2 1=5 . 0 7 . 0 14 . 11 = d k1 25 . 0 5 . 02 + = k bdA s= =2kco4 . 02 1 = K KcvPage 3Moment of resistance coefficient, R = = 1.1804Therefore, required effective depth of slab=d = = 112.06 mmEffective depth of slab, provided = 254 mm > d reqO.K.cbc 3280stcbcm cbcm +31 kstj k 21RbMd jMst..=d jMst. . =bddtan M- V =vco c k k . .2 1=5 . 0 7 . 0 14 . 11 = d k1 25 . 0 5 . 02 + = k bdA s= =2kco4 . 02 1 = K Kcvp , p qArea of steel required, Ast = 326.42 mm2Provide 10 mm bars @ 200 = 393 mm2> required, Ok.Distribution Steel:Distribution steel is to be provided for 0.3 times live load moment plus 0.2 times deadload moment.mm c/c, giving area of steel = cbc 3280stcbcm cbcm +31 kstj k 21RbMd jMst..=d jMst. . =bddtan M- V =vco c k k . .2 1=5 . 0 7 . 0 14 . 11 = d k1 25 . 0 5 . 02 + = k bdA s= =2kco4 . 02 1 = K Kcvload moment.Moment = 4.06 kN.m Effective depth 244 mmArea of steel required, Ast = 93.057 mm2Half reinforcement is to be provided at top and half at bottom.Provide 10 mm bars 200 mm c/c at both top and bottom, giving area of t l 392 5 mm2> required OKcbc 3280stcbcm cbcm +31 kstj k 21RbMd jMst..=d jMst. . =bddtan M- V =vco c k k . .2 1=5 . 0 7 . 0 14 . 11 = d k1 25 . 0 5 . 02 + = k bdA s= =2kco4 . 02 1 = K Kcvsteel = 392.5 mm2 > required, OK.Check for min. area of Steel:Min. area of steel @ 0.12 % = 360 mm2< Provided. O.K.Design for Shear:Dead load shear = 7.68 kNLive load Shear = 2.250 kNcbc 3280stcbcm cbcm +31 kstj k 21RbMd jMst..=d jMst. . =bddtan M- V =vco c k k . .2 1=5 . 0 7 . 0 14 . 11 = d k1 25 . 0 5 . 02 + = k bdA s= =2kco4 . 02 1 = K KcvTotal = 9.93 kNtan = 0.150Shear stress, = 0.005 N/mm2Percentage area of tension steel, pt = 0.13 %Allowable shear stress as per code is given bycbc 3280stcbcm cbcm +31 kstj k 21RbMd jMst..=d jMst. . =bddtan M- V =vco c k k . .2 1=5 . 0 7 . 0 14 . 11 = d k1 25 . 0 5 . 02 + = k bdA s= =2kco4 . 02 1 = K Kcv( d being in m) = 0.986(where ) = 0.500446 1.00Ad tcbc 3280stcbcm cbcm +31 kstj k 21RbMd jMst..=d jMst. . =bddtan M- V =vco c k k . .2 1=5 . 0 7 . 0 14 . 11 = d k1 25 . 0 5 . 02 + = k bdA s= =2kco4 . 02 1 = K KcvAdopt 1Value ot = for M25 grade of concrete from code = 0.4 N/mm2Allowable shear stress = 0.3944 N/mm2 > , Hence Safecbc 3280stcbcm cbcm +31 kstj k 21RbMd jMst..=d jMst. . =bddtan M- V =vco c k k . .2 1=5 . 0 7 . 0 14 . 11 = d k1 25 . 0 5 . 02 + = k bdA s= =2kco4 . 02 1 = K KcvPage 32.2 Design of Interior Panels: The slab panel is designed by Pigeauds method.DPage 4BCShort span of slab, Bs = 2 mLong span of slab, Ls = 4.5 mAFig : 3 Bridge PlanCalculation of Bending momentsa) Due to Dead load:Self weight of wearing coat = 1.76 kN/m2Self weight of deck slab = 5.28 kN/m2Total = 7.04 kN/m2Since the slab is supported on all four sides and is continuous, Piegauds curves are used to calculate bending moments.Ratio k = Bs/Ls = 0 44 Ratio, k = Bs/Ls = 0.44As the panel is loaded with UDL,u/Bs = 1v/Ls = 1Where, u & v are the dimensions of the loaded area.From the Pigeauds curve,m1 = 0.0457m2 = 0.0086Total dead load W = 63.36 kNMoment along short span, M1 = W (m1 +0.15m2) = 2.98 kN-mMoment along long span, M2 = W (0.15m1 +m2) = 0.98 kN-mConsidering effects of continuity, 0.8Moment along short span, M1 = 2.38 kN-mMoment along long span, M2 = 0.78 kN-mb) Due to Live load: Class AA Tracked Vehicle b) Due to Live load: Class AA Tracked VehicleFor maximum bending moment one wheel is placed at the center of panel.Tyre contact length along short span, x = 0.85 mTyre contact length along long span, y = 3.6 mLoaded length, u = 1.034 mLoaded width, v = 3.766 mWheel load, W = 350 kNRatio, k = Bs/Ls = 0.44/B 0 517 N u/Bs = 0.517v/Ls = 0.837From the Pigeauds curve,m1 = 0.0813m2 = 0.0147Moment along short span,= 29.227 kN-mMoment along long span,= 9.413 kN-mM1= W(m1+0.15m2)M2= W(0.15m1+m2)W1=350kNFig: 4Page 4Bending moment including impact and continuity,M1 = 29.227 kN-mM2 = 9.413 kN-mFig: 4Page 5c) Due to Live load: Class AA Wheeled VehicleCase-I: When two loads of 37.5 kN each and four loads of 62.5kN are placed such that two loads of 62.5kN lies at center line of pannel.Tyre contact width (along short span), = 0.30 mTyre contact length (along long span), = 0.15 mDisperced width along short span, u = 0.510 mDisperced width along long span, V = 0.380 mK = 0 44 K = 0.4462.5kN62.5kN62.5kN62.5kNW1 W4W2 W5YX XBending moment due to load W1: 62.5 kNRatio, k = Bs/Ls = 0.44Fig: 537.5kNW337.5kNW6YRatio, k Bs/Ls 0.44u/Bs = 0.255v/Ls = 0.084From the Pigeauds curve,m1 = 0.1965m2 = 0.1383Moment along short span,= 13.578 kN-mMoment along long spanM1= W(m1+0.15m2)Moment along long span,= 10.486 kN-mBending moment including impact and continuity,M1 = 13.578 kN-mM2 = 10.486 kN-mBending moment due to load W2: 62.5 kN Wheel load is placed unsymmetrical wrt the X-XIntensity of loading, q = 322.4