Example on Design of Slab Bridge Design Data and Specifications Chapter 5 SUPERSTRUCTURES Superstructure consists of 10m slab, 36m box girder and 10m T-girder all simply supported. Only the design of Slab Bridge will be used for illustration. Roadway Grade = 1660.00 m, amsl HWM = 1643.56 - Roadway grade dictates elevation of superstructure and not minimum free board requirement. I. Slab II. T-Girder III.Box-Girder Clear span = 10m Clear span 10m Clear span = 36m Road way width = 7.32m Road way width = 7.32m Road way width = 7.32m Curb width = 0.8m Curb width = 0.80m Curb width = 0.80m -Materials Concrete: Class A concrete: Cylinder strength f c = 28MPa [A5.4.2.1] [A5.4.2.4] Steel: f y = 400MPa E s = 200GPa Design method is Load and Resistance Factor Design (LRDF) AAiT, Department of Civil & Environmental Engineering Page 1
Reference: AASHTO LRFD Bridge Design Specifications, SI units, 2 nd Edition, 2005. Slab Bridge Design 1. Depth Determination [A2.5.2.6.3] Minimum recommended depth for slabs with main reinforcement parallel to traffic is Where S is the span, S=c/c of supports clear span + d, S=10+0.4/2+0.43/2=10.415m Use D = 540 mm, d= 540- F/2-25 = 499mm S=10.415m Clear span + d = 10000 + 499 = 10.499m Ok! (Cover) 2. Live Load Strip Width [Art.4.6.2.3] a) Interior Strip i) One lane loaded: multiple presence factor included [C.4.6.2.3] L 1 is smaller of 10415 or 18000. W 1 is the smaller of 8920 or 9000 L 1 = 10415 W 1 = 8920 ii) Multiple lanes loaded AAiT, Department of Civil & Environmental Engineering Page 2
W=Actual edge to edge width = 8920mm N L = Int(clear roadway width/3600) Use E=3256.63mm Equivalent concentrated and distributed loads Truck: P 1 =35/3.2566=10.75; P 2 = 145/3.2566 = 44.52 Tandem: P 3 =110/3.2566 = 33.78 Lane: w = 9.3/3.2566 = 2.856 b) Edge Strip Longitudinal edge strip width for a line of wheels [Art.4.6.2.1.4] E= distance from edge to face of barrier + 300+1/4* strip width E= 800 + 300+3256.63/4 = 1914.08mm > 1800mm E=1800mm 3. Influence Lines for Shear Force and Bending Moment Slab bridges shall be designed for all vehicular live loads specified in AASHTO Art 3.6.1.2, including the lane load [Art.3.6.1.3.3] a) Inter Strip i) Maximum Shear Force This governs AAiT, Department of Civil & Environmental Engineering Page 3
Impact factor = 1+IM/100 = 1+33/100 = 1.33, not applied to lane load [Art.3.6.2.1] V LL + IM =1.33*72.52+14.87 = 111.32 ii) Maximum bending Moment Truck: M Tr = 44.52(0.703+2.553) + 10.75(0.103) = 146.06 knm Tandom: M Ta = 33.78(2.304*2) =155.66 knm this governs Lane: MLn = 2.856*(1/2)*2.604*10.415 =38.73kNm M LL + Im = 1.33*155.66+38.73 = 245.76kNm b) Edge Strip Because E= 1800mm, one lane loaded with a multiple presence factor of 1.2 will be critical 4. Select resistance factor, φ [Art. 5.5.4.2.1] Strength Limit States (RC) φ Flexure & Tension 0.90 Shear & Torsion 0.90 Axial Compression 0.75 Bearing On concrete 0.70 Compression in strut and tie model 0.70 5. Select Load Modifiers, η 1 Strength service fatigue i) Ductility η 0 0.95 1.0 1.0 [Art. 1.3.3] ii) Redundancy η R 1.05 1.0 1.0 [Art. 1.3.4] iii) Importance η I 1.05 1.0 1.0 [Art. 1.3.5] AAiT, Department of Civil & Environmental Engineering Page 4
η 0 = η R = η I = 1.0 6. Select Applicable Load Combinations [Table 3.4.1-1] Strength I U=η (1.25DC + 1.50DW + 1.75(LL+1M)+1.0FR+γ TG TG Service I U=1.0(DC+DW) +1.0(LL+IM) + 0.3(WS+WL+1.0FR Fatigue U=0.75*(LL+IM) 7. Dead Load Force Effects a) Interior Strip:- Consider a 1m Strip, ρ con =2400 kg/m3 [Table 3.5.1-1] W DC = (2400*9.81)* 10-3 kn/m 3 * 0.54 m = 12.71kN/m 2 W DW = (2250*9.81)* 10-3 kn/m 3 * 0.075m = 1.66kN/m 2 75mm bituminous wearing surface, ρ bit = 2250kg/m3 [Table 3.5.1-1] V DC = ½ * 12.71*10.415 = 66.21kN/m V DW = ½ * 1.66*10.415 = 8.64kN/m b) Edge Strip: V DC = ½* 16.06*10.415 = 83.63kN/m 8. Investigate Service Limit State AAiT, Department of Civil & Environmental Engineering Page 5
i) Durability: Cover for main reinforcement steel for [Art. 5.12] deck surface subjected to tire wear = 60mm bottom of cast in-place slab = 25 mm η D = η R = η I = 1.0 η = 1.0 a) Moment Interior Strip M=1.0(172.34 + 22.51 + 245.76) = 440.61 knm Reinforcement: Assume j=0.875 and f s = 0.6 f y = 0.6*400 = 240 b) Moment Edge strip: M=1.0(217.76 + 0 + 533.56) = 751.32kNm ii) Control of Cracking [Art.5.7.3.4] Components shall be so proportioned that the tensile stress in the mild steel Reinforcement at the service limit state, f s, does not exceed f sa AAiT, Department of Civil & Environmental Engineering Page 6
Z crack width parameter (N/mm) = 23000N/mm for severe exposure d c depth of concrete measured from extreme tension fiber to center of bar located closest there to. Clear cover used to compute d c 50mm a) Interior strip 190 <394.6 Okay! b) Edge Strip 140<418.98 Okay A- Area of concrete having the same controid as the principal tensile reinforcement and bounded by the surfaces of the cross-section and a line parallel to the neutral axis divided by the number of bars (mm2), clear cover here also 50mm The concrete is considered cracked if tensile stress in concrete 80% of the modulus of rupture, [Art. 5.7.3.4&5.4.2.6] a) Interior Strip check concrete tensile stress against 0.8fr Mint = 440.61kNm/m AAiT, Department of Civil & Environmental Engineering Page 7
Now, steel stress should be calculated for elastic cracked section. The moment of inertia of the composite transformed section should be used for the stress calculation N=7, na sprove = 7*4232.88 = 29630.16mm 2 Equivalent concrete area Determine x from ½*1000*x 2 = 29630.16(499-x) x=144.87mm Now I cr = 1/3*1000*144.87 3 + 29630.1(499-144.86) 2 = 4.729*10 9 mm 4 /m. Steel stress over n, f s /n = M(d-x)/I cr =(440.61*10 6 *354.13)/(4.729*10 9 ) = 32.99MPa f s =7*32.99=230.93MPa 0.6fy Now, f sa can be computed: b) Edge Strip M edge = 751.32KNm/m f s = 230.93 f sa = 240Mpa OK! AAiT, Department of Civil & Environmental Engineering Page 8
½*750*X2 = 7*4882.93(749-x) x = 219.655mm<250mm Curb height I cr = 1/3*750*(219.655) 3 + 7*4882.93(749-219.655) 2 = 12.227*10 9 mm 4 f s /n = M(d-x)/Icr = 751.32x10 6 *(749-219.655)/12.227x10 9 ) = 32.53Mpa f s = 7*32.53Mpa=227.71Mpa f s <f sa Ok! iii) Deformations Deflection and camber calculations shall consider dead load, live load, erection loads, concrete creep and shrinkage. [Art. 5.7.3.6.2] Immediate (instantaneous) deflections may be computed taking the moment of inertia as either the effective moment of inertia, Ie or the gross moment of inertia, I g The long-term deformation (due to creep and shrinkage) may be taken as the immediate deflection multiplied by the following factor 3.0-1.2(A s/as) 1.6 if immediate deflection is calculated using I e. if immediate deflection is calculated using I g. AAiT, Department of Civil & Environmental Engineering Page 9
a) Dead Load Camber: Total dead load of the bridge and the whole bridge cross-section is considered W DC = 12.71*8.62+(2.53+0.59+0.23)*1.8*2=121.62KN/m W DW =1.66*7.32 M a actual maximum moment (Nmm) f r modulus of rupture y t distance from N.A to extreme tension fiber (mm) f r = 0.63 = 0.63 = 3.33Mpa, Location of N.A, Since the section does not crack under DL, Ig should be used Chamber 4*4.53=18.12mm upward AAiT, Department of Civil & Environmental Engineering Page 10
b) Live Load Deflection (Optional) [Art. 2.5.2.6.2] Use design truck alone or design lane plus 25% of truck load. [Art. 3.6.1.3.2] When design truck is used alone, it should be placed so that the distance between its resultant and the nearest wheels is bisected by span centerline. All design lanes should be loaded. M DC+DW+LL+IM = 1813.79+1.33*146.06*3.2566*2*1.0 = 3079.04KNm>M cr Multiple presence factor Design TruckLoad First load, P=385.7KN,a=8.78,b=1.635m,X=4.48m Second load, P=385.7,a = x = 4.48m, b = 5.935m Third load, P=93.1kKN,a=10.235,b=0.18m,X=5.935 (Δ LL+IM ) 1 =1.75+3.83+0.003=5.583<<13mm Ok! Design Lane Load +25% of design Truck Load: W=9.3*2*1=18.6kN/m AAiT, Department of Civil & Environmental Engineering Page 11
Δ LL+IM =1.33+1.48+2.79mm<<13.02mm Ok! TandemLoad Single concentrated tandem load at mid-span (spaced at zero meter) P=1.33*220*2*1 = 585.2KN With average I e over the entire span used instead of Ie at section of maximum moment as done here, smaller deflection would result. The contribution of compression steel is also neglected. For these reasons, live load deflections are made optional in AASHTO. 9. Investigate Fatigue Limit State. U=0.75(LL+IM), IM=15% Fatigue load shall be one design truck with 9m axle spacing. Maximum moment results when the two from axles are on the span and the rear axle is out of span. a) Tensile Live Load stress: One lane loaded E=4298.2mm f s max = 7*5.58 = 39.05 Mpa b) Reinforcing Bars: The stress range in straight reinforcement bars resulting from fatigue load combination shall not exceed. f f =145-0.33f min +55(r/h) AAiT, Department of Civil & Environmental Engineering Page 12
f f -is stress range f min -minimum LL stress, where there is stress reversal=0 for our case r/h=0.3 f f =145-0.33(0)+55(0.3)=161.5Mpa f max <f f ok! 40.66<161.5 ok! 10. Investigate Strength Limit State i) Flexure: Equivalent Rectangular stress Distribution [Art. 5.7.2.2.2] a) Interior strip Mu=ηΣγiQ I =1.05[1.25M DC + 1.5M DW + 1.75M LL + IM +γ TG M TG ] For simple span bridges, temperature gradient effect reduce gravity load effects. Because temperature gradient may not always be there, assume γtg=0 Mu=1.05 [1.25(172.34) + 1.5(22.51) + 1.75(245.76)] = 713.23kNm/m Mu = φasf y d(1-0.588 ρfy/f c) D=540-32/2-25 = 499mm Ρ = 0.0086 >ρ min = 0.03*f c/fy = 0.03*28/400=0.0021 As = 0.0086*1000*499=4291.4mm 2 Use b) Edge Strip Mu=ηΣγiQ I =1.05[1.25(217.76) +0+ 1.75(533.56)+0) = 1266.22KN/m D=540 + 250 32/2-25 = 749mm ρ=0.00904>ρ min As =ρbd = 0.0086*750*749=5082.19mm 2 AAiT, Department of Civil & Environmental Engineering Page 13
ii) Sheaf Slab bridges designed in conformance with AASHO, Art 4.6.2.3 may be considered satisfactory for shear. Art. 4.6.2.3 deals with approximate method of analysis of slab bridges using equivalent strip method. But if longitudinal tubes are placed in the slab as in pre stressed concrete, and create voids and reduce the cross section, the shear resistance must be checked. iii) Distribution Reinforcement: The amount of bottom transverse reinforcement may be taken as a percentage of the main reinforcement required for positive moment as a) Interior strip: Transverse reinforcement = 0.175*4347.34mm 2 = 745.6mm 2 m b) Edge strip: Transverse reinforcement = 0.1715 * 5063.8 mm 2 = 868.44mm 2 m AAiT, Department of Civil & Environmental Engineering Page 14
iv) Shrinkage& Temperature Reinforcement: Reinforcement for shrinkage & temperature stresses shall be provided near surfaces of concrete exposed to daily temperature changes. The steel shall be distributed equally on both sides a) Interior Strip: m, transverse. 2.1 LIMIT STATES GENERAL Bridges shall be designed for specified limit states to achieve the objectives of constructibility, safety, and serviceability, with due regard to issues of inspectibility, economy, and aesthetics, as specified in Chapters 3 11. Regardless of the type of analysis used, Equation 2.1 shall be satisfied for all specified force effects and combinations thereof. Equation 2.1 below is the basis of the LRFD methodology. Each component and connection shall satisfy Equation 2.1 for each limit state, unless otherwise specified. For service and extreme event limit states, resistance factors shall be taken as 1.0, except for bolts, for which the provisions of Chapter 8: Bridge Details apply. All limit states shall be considered of equal importance. i i Q i R n = R f (2.1) AAiT, Department of Civil & Environmental Engineering Page 15
Where: for loads for which a maximum value of i is appropriate: i = D R I 0.95 (2.2) for loads for which a minimum value of i is appropriate: i = 1 1.0 (2.3) D R I Where: i = load modifier: a factor relating to ductility, redundancy, and operational importance i = load factor: a statistically based multiplier applied to force effects Q i = force effect = resistance factor: a statistically based multiplier applied to nominal resistance (see chapters 5,6,7 8, 10 and 12). R n = nominal resistance D = a factor relating to ductility, as specified below R = a factor relating to redundancy as specified below I = a factor relating to operational importance as specified below R f = factored resistance: R n Ductility, redundancy, and operational importance are significant aspects affecting the margin of safety of bridges. Whereas the first two directly relate to physical strength, the last concerns the consequences of the bridge being out of service. The grouping of these aspects on the load side of Equation 2.1 is, therefore, arbitrary. However, it constitutes a first effort at codification. In the absence of more precise information, each effect, except that for fatigue and fracture, is estimated as ±5 percent, accumulated geometrically, a clearly subjective approach. With time, improved quantification of ductility, redundancy, and operational importance, and their interaction and system synergy, shall be attained, possibly leading to a rearrangement of Equation 2.1, in which these effects may appear on either side of the equation or on both sides. AAiT, Department of Civil & Environmental Engineering Page 16