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Introduction to Bridge Engineering Lecture 4 (II) CONCRETE BRIDGES

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2 Presented By: YASIR IRFAN BADRASHI & QAISER HAYAT Presented To: PROF. DR. AKHTAR NAEEM KHAN &CLASSMATES CONCRETE BRIDGES

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3 Topics to be Presented: Example Problem on: Example Problem on: (i).Concrete Deck Design (ii).Solid Slab Bridge Design (iii).T-Beam Bridge Design CONCRETE BRIDGES

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CONCRETE DECK DESIGN

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5 Use the approximate method of analysis [4.6.2] to design the deck of the reinforced concrete T-Beam bridge section of Fig.E for a HL-93 live load and a PL-2 performance level concrete barrier (Fig.7.45). Use the approximate method of analysis [4.6.2] to design the deck of the reinforced concrete T-Beam bridge section of Fig.E for a HL-93 live load and a PL-2 performance level concrete barrier (Fig.7.45). The T-Beams supporting the deck are 2440 mm on the centers and have a stem width of 350 mm. The deck overhangs the exterior T- Beam approximately 0.4 of the distance between T-Beams. Allow for sacrificial wear of 15mm of concrete surface and for a future wearing surface of 75mm thick bituminous overlay. Use fc’=30 MPa, f y =400Mpa, and compare the selected reinforcement with that obtained by the empirical method [A9.7.2] Problem Statement:

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6 The minimum thickness for concrete deck slabs is 175 mm [A ]. The minimum thickness for concrete deck slabs is 175 mm [A ]. Traditional minimum depths of slabs are based on the deck span length S to control deflection to give [ Table A ] Traditional minimum depths of slabs are based on the deck span length S to control deflection to give [ Table A ] A. DECK THICKNESS Use h s = 190 mm for the structural thickness of the deck. By adding the 15 mm allowance for the sacrificial surface, the dead weight of the deck slab is based on h= 205mm. Because the portion of the deck that overhangs the exterior girder must be designed for a collision load on the barrier, its thickness has been increased by 25mm to h o =230mm

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7 B. WEIGHTS OF THE COMPONENTS [ TABLE A ] For a 1mm width of a transverse strip. For a 1mm width of a transverse strip. Barrier Barrier P b = 2400 x Kg/mm 3 x 9.81 N/Kg x mm 2 = 4.65 N/mm Future Wearing Surface Future Wearing Surface WDW = 2250 x x 9.81 x 75 = 1.66 x N/mm Slab 205mm thick Slab 205mm thick Ws = 2400 x x 9.81 x 205 = 4.83 x N/mm Cantilever Overhanging Cantilever Overhanging W o = 2400 x x 9.81 x 230 = 5.42 x N/mm

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8 C. BENDING MOMENT FORCE EFFECTS – GENERAL An approximate analysis of strips perpendicular to girders is considered acceptable [A9.6.1]. The extreme positive moment in any deck panel between girders shall be taken to apply to all positive moment regions. Similarly, the extreme negative moment over any girder shall be taken to apply to all negative moment regions [A ]. The strips shall be treated as continuous beams with span lengths equal to the center-to-centre distance between girders. The girders are assumed to be rigid [A ] An approximate analysis of strips perpendicular to girders is considered acceptable [A9.6.1]. The extreme positive moment in any deck panel between girders shall be taken to apply to all positive moment regions. Similarly, the extreme negative moment over any girder shall be taken to apply to all negative moment regions [A ]. The strips shall be treated as continuous beams with span lengths equal to the center-to-centre distance between girders. The girders are assumed to be rigid [A ] For ease in applying the load factors, the bending moments will separately be determined for the deck slab, overhang, barrier, future wearing surface, and vehicle live load. For ease in applying the load factors, the bending moments will separately be determined for the deck slab, overhang, barrier, future wearing surface, and vehicle live load.

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9 1. DECK SLAB h = 205 mm, h = 205 mm, Ws = 4.83 x 10 –3 N/mm, S = 2440 mm Placement of the deck slab dead load and results of a moment distribution analysis for negative and positive moments in a 1-mm wide strip is given in figure E7.1-2 Placement of the deck slab dead load and results of a moment distribution analysis for negative and positive moments in a 1-mm wide strip is given in figure E7.1-2 A deck analysis design aid based on influence lines is given in Table A.1 of Appendix A. For a uniform load, the tabulated areas are multiplied by S for Shears and S 2 for moments. A deck analysis design aid based on influence lines is given in Table A.1 of Appendix A. For a uniform load, the tabulated areas are multiplied by S for Shears and S 2 for moments.Table A.1 of Appendix ATable A.1 of Appendix A Fig.E7.1-2: Moment distribution for deck slab dead load.

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10 R200 = Ws (Net area w/o cantilever) S R200 = Ws (Net area w/o cantilever) S = 4.83 x (0.3928) 2440 = 4.63 N/mm = 4.83 x (0.3928) 2440 = 4.63 N/mm M204 = Ws (Net area w/o cantilever) S 2 M204 = Ws (Net area w/o cantilever) S 2 = 4.83 x (0.0772) = 4.83 x (0.0772) = 2220 N mm/mm = 2220 N mm/mm M300 = Ws (Net area w/o cantilever) S 2 M300 = Ws (Net area w/o cantilever) S 2 = 4.83 x ( ) = 4.83 x ( ) = N mm/mm = N mm/mm Comparing the results from the design aid with those from moment distribution shows good agreement. In determining the remainder of the bending moment force effects, the design aid of Table A.1 will be used. Comparing the results from the design aid with those from moment distribution shows good agreement. In determining the remainder of the bending moment force effects, the design aid of Table A.1 will be used.moment distribution Table A.1 moment distribution Table A.1 1. DECK SLAB

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11 2. OVERHANG The parameters are The parameters are h o = 230 mm, W o = 5.42 x N/mm 2 L = 990 mm Placement of the overhang dead load is shown in the figure E By using the design aid Table A.1, the reaction on the exterior T-Beam and the bending moments are: Fig.E7.1-3 Overhang dead load placement

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12 2. OVERHANG R200 = W o (Net area cantilever) L R200 = W o (Net area cantilever) L = 5.42 x ( x 990/2440) 990 = 6.75 N/mm = 5.42 x ( x 990/2440) 990 = 6.75 N/mm M200 = W o (Net area cantilever) L 2 M200 = W o (Net area cantilever) L 2 = 5.42 x ( ) = N mm/mm = 5.42 x ( ) = N mm/mm M204 = W o (Net area cantilever) L 2 M204 = W o (Net area cantilever) L 2 = 5.42 x ( ) = N mm/mm = 5.42 x ( ) = N mm/mm M300 = W o (Net area cantilever) L 2 M300 = W o (Net area cantilever) L 2 = 5.42 x (0.1350) = 717 N mm/mm = 5.42 x (0.1350) = 717 N mm/mm

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13 3. BARRIER The parameters are The parameters are P b = 4.65 N/mm L = 990 – 127 = 863 mm Placement of the center of gravity of the barrier dead load is shown in figure E By using the design aid Table A.1 for the concentrated barrier load, the intensity of the load is multiplied by the influence line ordinate for shears and reactions. For bending moments, the influence line ordinate is multiplied by the cantilever length L. Placement of the center of gravity of the barrier dead load is shown in figure E By using the design aid Table A.1 for the concentrated barrier load, the intensity of the load is multiplied by the influence line ordinate for shears and reactions. For bending moments, the influence line ordinate is multiplied by the cantilever length L. Fig.E7.1-4 Barrier dead load placement

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14 3. BARRIER R200 = P b (Influence line ordinate) R200 = P b (Influence line ordinate) = 4.65( x 863/2440) = 6.74 N/mm = 4.65( x 863/2440) = 6.74 N/mm M200 = P b (Influence line ordinate) L M200 = P b (Influence line ordinate) L = 4.65( ) (863) = N mm/mm = 4.65( ) (863) = N mm/mm M204 = P b (Influence line ordinate) L M204 = P b (Influence line ordinate) L = 4.65 ( ) (863) = N mm/mm = 4.65 ( ) (863) = N mm/mm M300 = P b (Influence line ordinate) L M300 = P b (Influence line ordinate) L = 4.65 (0.2700) (863) = 1083 N mm/mm = 4.65 (0.2700) (863) = 1083 N mm/mm

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15 4. FUTURE WEARING SURFACE F WS = W DW = 1.66 x 10-3 N/mm 2 F WS = W DW = 1.66 x 10-3 N/mm 2 The 75mm bituminous overlay is placed curb to curb as shown in figure E The length of the loaded cantilever is reduced by the base width of the barrier to give The 75mm bituminous overlay is placed curb to curb as shown in figure E The length of the loaded cantilever is reduced by the base width of the barrier to give L = 990 – 380 = 610 mm. Fig. E7.1-5: Future wearing surface dead load placement

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16 If we use the design aid Table A.1, we have R200 = W DW [(Net area cantilever) L + (Net area w/o cantilever) S] R200 = W DW [(Net area cantilever) L + (Net area w/o cantilever) S] = 1.66 x [( x 610/2440) x (0.3928) x 2440)] = 2.76 N/mm M200 = W DW (Net area cantilever) L 2 M200 = W DW (Net area cantilever) L 2 = 1.66 x ( )(610) 2 = -309 N mm/mm = 1.66 x ( )(610) 2 = -309 N mm/mm M204 = W DW [(Net area cantilever) L 2 + (Net area w/o cantilever) S 2 ] M204 = W DW [(Net area cantilever) L 2 + (Net area w/o cantilever) S 2 ] = 1.66 x [( )(610) 2 + (0.0772) ] = 611 N mm/mm = 1.66 x [( )(610) 2 + (0.0772) ] = 611 N mm/mm M300 = W DW [(Net area cantilever) L 2 + (Net area w/o cantilever) S 2 ] M300 = W DW [(Net area cantilever) L 2 + (Net area w/o cantilever) S 2 ] = 1.66 x [(0.1350)(610) 2 + ( ) ] = -975 N mm/mm = 1.66 x [(0.1350)(610) 2 + ( ) ] = -975 N mm/mm 4. FUTURE WEARING SURFACE

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17 D. VEHICULAR LIVE LOAD Where decks are designed using the approximate strip method [A ], and the strips are transverse, they shall be designed for the 145 KN axle of the design truck [A ]. Wheel loads on an axle are assumed to be equal and spaced 1800 mm apart [Fig.A ]. The design truck should be positioned transversely to produce maximum force effects such that the center of any wheel load is not closer than 300mm from the face of the curb for the design of the deck overhang and 600mm from the edge of the 3600 mm wide design lane for the design of all other components [A ] Where decks are designed using the approximate strip method [A ], and the strips are transverse, they shall be designed for the 145 KN axle of the design truck [A ]. Wheel loads on an axle are assumed to be equal and spaced 1800 mm apart [Fig.A ]. The design truck should be positioned transversely to produce maximum force effects such that the center of any wheel load is not closer than 300mm from the face of the curb for the design of the deck overhang and 600mm from the edge of the 3600 mm wide design lane for the design of all other components [A ]

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18 D. VEHICULAR LIVE LOAD The width of equivalent interior transverse strips (mm) over which the wheel loads can be considered distributed longitudinally in CIP concrete decks is given as The width of equivalent interior transverse strips (mm) over which the wheel loads can be considered distributed longitudinally in CIP concrete decks is given as [Table A ] Overhang, X Overhang, X Positive moment, S Positive moment, S Negative moment, S Negative moment, S Where X is the distance from the wheel load to centerline of support and S is the spacing of the T- Beams. Here X=310 mm and S=2440 mm Where X is the distance from the wheel load to centerline of support and S is the spacing of the T- Beams. Here X=310 mm and S=2440 mm(Fig.E7.1-6)

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19 D. VEHICULAR LIVE LOAD Figure E : Distribution of Wheel load on Overhang

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20 D. VEHICULAR LIVE LOAD Tire contact area [A ] shall be assumed as a rectangle with width of 510 mm and length given by Where is the load factor, IM is the dynamic load allowance and P is the Wheel load. Here = 1.75, IM = 33%, P = 72.5 KN.

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21 D. VEHICULAR LIVE LOAD Thus the tire contact area is Thus the tire contact area is 510 x 385mm with the 510mm in the transverse direction as shown in Figure.E7.1-6

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22 D. VEHICULAR LIVE LOAD

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23 D. VEHICULAR LIVE LOAD Figure E : Distribution of Wheel load on Overhang Back

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24 D. VEHICULAR LIVE LOAD 3

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25 D. VEHICULAR LIVE LOAD mm m

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26 D. VEHICULAR LIVE LOAD Fig.E7.1-7: Live load placement for maximum positive moment (a)One loaded lane, m = 1.2 (b) Two loaded lanes, m = 1.0

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27 D. VEHICULAR LIVE LOAD If we use the influence line ordinates from Table A-1, the exterior girder reaction and positive bending moment with one loaded lane (m=1.2) are If we use the influence line ordinates from Table A-1, the exterior girder reaction and positive bending moment with one loaded lane (m=1.2) are

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28 D. VEHICULAR LIVE LOAD For two loaded lanes (m=1.0) Thus, the one loaded lane case governs.

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29 D. VEHICULAR LIVE LOAD 3. MAXIMUM INTERIOR NEGATIVE LIVE LOAD MOMENT. the critical placement of live load for maximum negative moment is at the first interior deck support with one loaded lane (m=1.2) as shown in Fig.E The equivalent transverse strip width is S = (2440) = 1830 mm Using Table A-1, the bending moment at location 300 is

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30 D. VEHICULAR LIVE LOAD 4. MAXIMUM LIVE LOAD REACTION ON EXTERIOR GIRDER

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31 E. STRENGTH LIMIT STATE The gravity load combination can be stated as [Table A ] The gravity load combination can be stated as [Table A ] PP

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32 E. STRENGTH LIMIT STATE

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33 The T-Beam stem width is 350mm, so the design sections will be 175mm on either side of the support centerline used in the analysis. The critical negative moment section is at the interior face of the exterior support as shown in the free body diagram The T-Beam stem width is 350mm, so the design sections will be 175mm on either side of the support centerline used in the analysis. The critical negative moment section is at the interior face of the exterior support as shown in the free body diagram [Fig. E7.1-10] E. STRENGTH LIMIT STATE Back

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34 The values of the loads in Fig E are for a 1- mathematical model strip. The concentrated wheel load is for one loaded lane, that is, The values of the loads in Fig E are for a 1- mathematical model strip. The concentrated wheel load is for one loaded lane, that is, W = 1.2(72500)1400 = N/mm 1. Deck Slab: E. STRENGTH LIMIT STATE s

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35 2. Overhang 3. Barrier E. STRENGTH LIMIT STATE o 200

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36 4. Future Wearing Surface 5. Live Load E. STRENGTH LIMIT STATE

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37 6. Strength-I Limit State E. STRENGTH LIMIT STATE

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38 F. Selection Of Reinforcement The effective concrete depths for positive and negative bending will be different because of the different cover requirements as indicated in this Fig shown.

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39 F. Selection Of Reinforcement u

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40 F. Selection Of Reinforcement

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41 F. Selection Of Reinforcement Maximum reinforcement keeping in view the ductility requirements is limited by [A ] Maximum reinforcement keeping in view the ductility requirements is limited by [A ] Minimum reinforcement [ ] for components containing no prestressing steel is satisfied if Minimum reinforcement [ ] for components containing no prestressing steel is satisfied if

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42 F. Selection Of Reinforcement

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43 F. Selection Of Reinforcement 1.POSITIVE MOMENT REINFORCEMENT :

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44 F. Selection Of Reinforcement Check Ductility Check Ductility Check Moment Strength Check Moment Strength

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45 F. Selection Of Reinforcement 2. Negative Moment Reinforcement Back

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46 F. Selection Of Reinforcement Check Moment Strength Check Moment Strength For transverse top bars, For transverse top bars, Use No. mm. Use No. mm.

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47 F. Selection Of Reinforcement 3. DISTRIBUTION REINFORCEMENT: Secondary reinforcement is placed in the bottom of the slab to distribute the wheel loads in the longitudinal direction of the bridge to the primary reinforcement in the transverse direction. The required area is a percentage of the primary positive moment reinforcement. For primary reinforcement perpendicular to traffic [A ] Where S e is the effective span length [A ]. S e is the distance face to face of stems, that is, S e = = 2090mm

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48 So So Dist.A s = 0.67(Pos.A s )=0.67(0.889) Pos.A sPos.A s = 0.60 mm 2 /mm For longitudinal bottom bars, Use 150 mm, A s = mm 2 /mm F. Selection Of Reinforcement

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49 F. Selection Of Reinforcement 4. SHRINKAGE AND TEMPRATURE REINFORCEMENT. The minimum amount of reinforcement in each direction shall be [A ] Where A g is the gross area of the section for the full 205 mm thickness. For members greater than 150 mm in thickness, the shrinkage and temperature reinforcement is to be distributed equally on both faces. Use 450 mm, Provided A s = mm 2 /mm

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50 G. CONTROL OF CRACKING-GENERAL Cracking is controlled by limiting the tensile stress in the reinforcement under service loads f s to an allowable tensile stress f sa [A ] Cracking is controlled by limiting the tensile stress in the reinforcement under service loads f s to an allowable tensile stress f sa [A ]Where Z = N/mm for severe exposure conditions. d c = Depth of concrete from extreme tension fiber to center of closest bar 50 mm A = Effective concrete tensile area per bar having the same centroid as the reinforcement.

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51 G. CONTROL OF CRACKING-GENERAL M = M DC + M DW M LL c

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52 G. CONTROL OF CRACKING-GENERAL Where = density of concrete = 2400 Kg/m 3. f’c = 30 MPa. So that Use n = 7

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53 G. CONTROL OF CRACKING-GENERAL 1. CHECK OF POSITIVE MOMENT REINFORCEMENT. The service I positive moment at Location 204 is The calculation of the transformed section properties is based on a 1-mm wide doubly reinforced section shown in the Figure E7.1-12

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54 G. CONTROL OF CRACKING-GENERAL Sum of statical moments about the neutral axis yields Sum of statical moments about the neutral axis yields

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55 G. CONTROL OF CRACKING-GENERAL The positive moment tensile reinforcement of No.15 bars at 25mm on centers is located 33 mm from the extreme tension fiber. Therefore, The positive moment tensile reinforcement of No.15 bars at 25mm on centers is located 33 mm from the extreme tension fiber. Therefore, c sa y ys

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56 G. CONTROL OF CRACKING-GENERAL 2.CHECK OF NEGATIVE REINFORCEMENT: The service I negative moment at location is The cross section for the negative moment is shown in Fig.E

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57 G. CONTROL OF CRACKING-GENERAL Balancing the statical moments about the neutral axis gives Balancing the statical moments about the neutral axis gives

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58 G. CONTROL OF CRACKING-GENERAL The negative moment tensile reinforcement of No.15 bars at 225 mm on centers is located 53 mm from the tension face. Therefore dc is the maximum value of 50mm, and The negative moment tensile reinforcement of No.15 bars at 225 mm on centers is located 53 mm from the tension face. Therefore dc is the maximum value of 50mm, and sa

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59 H. FATIGUE LIMIT STATE The investigation for fatigue is not required in concrete decks for multigirder applications [A9.5.3] The investigation for fatigue is not required in concrete decks for multigirder applications [A9.5.3]

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60 I.TRADITIONAL DESIGN FOR INTERIOR SPANS The design sketch in Fig.E summerizes the arrangement of the transverse and longitudinal reinforcement in four layers for the interior spans of the deck. The exterior span and deck overhang have special requirements that must be dealt with separately. The design sketch in Fig.E summerizes the arrangement of the transverse and longitudinal reinforcement in four layers for the interior spans of the deck. The exterior span and deck overhang have special requirements that must be dealt with separately.

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61 J. EMPERICAL DESIGN OF CONCRETE DECK SLABS Research has shown that the primary structural action of the concrete deck is not flexure, but internal arching. The arching creates an internal compression dome. Only a minimum amount of isotropic reinforcement is required for local flexural resistance. Research has shown that the primary structural action of the concrete deck is not flexure, but internal arching. The arching creates an internal compression dome. Only a minimum amount of isotropic reinforcement is required for local flexural resistance.

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62 J. EMPERICAL DESIGN OF CONCRETE DECK SLABS 1. DESIGN CONDITIONS [A ] Design depth excludes the loss due to wear, h=190mm. The following conditions must be satisfied:

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63 J. EMPERICAL DESIGN OF CONCRETE DECK SLABS 2. REINFORCEMENT REQUIREMENTS [A ]

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64 J. EMPERICAL DESIGN OF CONCRETE DECK SLABS 3. EMPERICAL DESIGN SUMMARY while using the empirical design approach there is no need of using any analysis. When the design conditions have been met, the minimum reinforcement in all four layers is predetermined. The design sketch in the Fig.E summarizes the reinforcement arrangement for the interior deck spans.

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65 K. COMPARISON OF REINFORCEMENT QUANTITIES The weight of reinforcement for the traditional and empirical design methods are compared in Table.E7.1-1 for a 1-m wide transverse strip. Significant saving, in this case 74% of the traditionally designed reinforcement is required, can be made by adopting the empirical design method. The weight of reinforcement for the traditional and empirical design methods are compared in Table.E7.1-1 for a 1-m wide transverse strip. Significant saving, in this case 74% of the traditionally designed reinforcement is required, can be made by adopting the empirical design method. (Area = 1m x 14.18m) (Area = 1m x 14.18m)

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66 L. DECK OVERHANG DESIGN The traditional and the empirical methods does not include the design of the deck overhang. The traditional and the empirical methods does not include the design of the deck overhang. The design loads for the deck overhang are applied to a free body diagram of a cantilever that is independent of the deck spans. The design loads for the deck overhang are applied to a free body diagram of a cantilever that is independent of the deck spans. The resulting overhang design can then be incorporated into either the traditional or the empirical design by anchoring the overhang reinforcement into the first deck span. The resulting overhang design can then be incorporated into either the traditional or the empirical design by anchoring the overhang reinforcement into the first deck span.

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67 L. DECK OVERHANG DESIGN Two limit states must be investigated. Two limit states must be investigated. Strength I [A13.6.1] and Extreme Event II [A13.6.2] Strength I [A13.6.1] and Extreme Event II [A13.6.2] The strength limit state considers vertical gravity forces and it seldom governs, unless the cantilever span is very long. The strength limit state considers vertical gravity forces and it seldom governs, unless the cantilever span is very long.

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68 L. DECK OVERHANG DESIGN The extreme event limit state considers horizontal forces caused by the collision of a vehicle with the barrier. The extreme event limit state considers horizontal forces caused by the collision of a vehicle with the barrier. The extreme limit state usually governs the design of the deck overhang. The extreme limit state usually governs the design of the deck overhang.

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69 L. DECK OVERHANG DESIGN 1. STRENGTH I LIMIT STATE: The design negative moment is taken at the exterior face of the support as shown in the Fig.E7.1-6 for the loads given in Fig.E Fig.E7.1-6Fig.E Fig.E7.1-6Fig.E Because the overhang has a single load path and is, therefore, a nonredundant member, then

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70 L. DECK OVERHANG DESIGN

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71 L. DECK OVERHANG DESIGN

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72 L. DECK OVERHANG DESIGN 2. EXTREME EVENT II LIMIT STATE the forces to be transmitted to the deck overhand due to a vehicular collision with the concrete barrier are determined from a strength analysis of the barrier. In this design problem, the barriers are to be designed for a performance level PL-2, which is suitable for “High-speed main line structures on freeways, expressways, highways and areas with a mixture of heavy vehicles and maximum tolerable speeds”

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73 L. DECK OVERHANG DESIGN The maximum edge thickness of the deck overhand is 200mm[A ] and the minimum height of barrier for a PL-2 is 810mm. The maximum edge thickness of the deck overhand is 200mm[A ] and the minimum height of barrier for a PL-2 is 810mm. The transverse and longitudinal forces are distributed over a length of barrier of 1070mm. This length represents the approximate diameter of a truck tire, which is in contact with the wall at the time of impact. The transverse and longitudinal forces are distributed over a length of barrier of 1070mm. This length represents the approximate diameter of a truck tire, which is in contact with the wall at the time of impact. The design philosophy is that if any failures are to occur they should be in the barrier, which can readily be repaired, rather than in the deck overhang. The design philosophy is that if any failures are to occur they should be in the barrier, which can readily be repaired, rather than in the deck overhang. The resistance factors are taken as 1.0 and the vehicle collision load factor is 1.0 The resistance factors are taken as 1.0 and the vehicle collision load factor is 1.0

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74 M. CONCRETE BARRIER STRENGTH All traffic railing systems shall be proven satisfactory through crash testing for a desired performance level [A ]. If a previously tested system is used with only minor modification that do not change its performance, then additional crash testing is not required [A ] All traffic railing systems shall be proven satisfactory through crash testing for a desired performance level [A ]. If a previously tested system is used with only minor modification that do not change its performance, then additional crash testing is not required [A ] The concrete barrier shown in the The concrete barrier shown in the Fig.E (Next Slide) is similar to the profile and reinforcement arrangement to traffic barrier type T5 analyzed by Hirsh(1978) and tested by Buth et al (1990)

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75 M. CONCRETE BARRIER STRENGTH ct Fig. W (Concrete Barrier and connection to deck overhang.)

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76 M. CONCRETE BARRIER STRENGTH …..(E7.1-8)

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77 M. CONCRETE BARRIER STRENGTH t t

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78 M. CONCRETE BARRIER STRENGTH 1. MOMENT STRENGTH OF WALL ABOUT VERTICAL AXIS,M W H. The moment strength about the vertical axis is based on the horizontal reinforcement in the wall. The thickness of the barrier wall varies and it is convenient to divide it for calculation purposes into three segments as shown in Fig. E7.1-18

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79 M. CONCRETE BARRIER STRENGTH

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80 M. CONCRETE BARRIER STRENGTH Neglecting the contribution of compressive reinforcement, the positive and negative bending strengths of segment I are approximately equal and calculated as nI

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81 M. CONCRETE BARRIER STRENGTH For segment II, the moment strengths are slightly different. Considering the moment positive if it produces tension on the straight face, we have For segment II, the moment strengths are slightly different. Considering the moment positive if it produces tension on the straight face, we have n neg n pos n II

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82 M. CONCRETE BARRIER STRENGTH For segment III, the positive and negative bending strengths are equal and For segment III, the positive and negative bending strengths are equal and nIII nIInInIII

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83 M. CONCRETE BARRIER STRENGTH Now considering the wall to have uniform thickness and same area as the actual wall and comparing it with the value of M w H. Now considering the wall to have uniform thickness and same area as the actual wall and comparing it with the value of M w H. This value is close to the one previously calculated and is easier to find

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84 M. CONCRETE BARRIER STRENGTH 2. MOMENT STRENGTH OF WALL ABOUT HORIZONTAL AXIS The moment strength about the horizontal axis is determined from the vertical reinforcement in the wall. The yield lines that cross the vertical reinforcement (Fig.E ) produce only tension in the sloping wall, so that the only negative bending strength need to be calculated. Matching the spacing of the vertical bars in the barrier with the spacing of the bottom bars in the deck, the vertical bars become No.15 at 225mm (As = mm 2 /mm) for the traditional design (Fig.E7.1-14).

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85 M. CONCRETE BARRIER STRENGTH For segment I, the average wall thickness is 175mm and the moment strength about the horizontal axis becomes For segment I, the average wall thickness is 175mm and the moment strength about the horizontal axis becomes At the bottom of the wall the vertical reinforcement at the wider spread is not anchored into the deck overhang. Only the hairpin dowel at a narrower spread is anchored. the effective depth of the hairpin dowel is [Fig.E7.1-17] At the bottom of the wall the vertical reinforcement at the wider spread is not anchored into the deck overhang. Only the hairpin dowel at a narrower spread is anchored. the effective depth of the hairpin dowel is [Fig.E7.1-17] d= = 224 mm

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86 M. CONCRETE BARRIER STRENGTH II+III

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87 M. CONCRETE BARRIER STRENGTH 3. CRITICAL LENTH OF YIELD LINE PATTERN,L C Now with moment strengths and L t =1070mm known, Eq.E7.1-9 yields c ttbw c

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88 M. CONCRETE BARRIER STRENGTH 4. NOMINAL RESISTANCE TO TRANVERSE LOAD,R W LOAD,R W From Eq.E7.1-8, We have w c t bw cc

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89 M. CONCRETE BARRIER STRENGTH 5. SHEAR TRANSFER BETWEEN BARRIER AND DECK The nominal resistance Rw must be transferred acroass a cold joint by shear friction. Free body diagrams of the forces transferred from the barrier to the deck overhang are shown in the Fig.E c

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90 M. CONCRETE BARRIER STRENGTH The nominal shear resistance Vn of the interface plane is given by [A ] The nominal shear resistance Vn of the interface plane is given by [A ] cv n vfc

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91 M. CONCRETE BARRIER STRENGTH The last two factors are for concrete placed against hardened concrete clean and free of laitance, but not intentionally roughened. Therefore for a 1-mm wide design strip The last two factors are for concrete placed against hardened concrete clean and free of laitance, but not intentionally roughened. Therefore for a 1-mm wide design strip ncv vffy

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92 M. CONCRETE BARRIER STRENGTH The minimum cross-sectional area of dowels across the shear plane is [A ] The minimum cross-sectional area of dowels across the shear plane is [A ] vf y v

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93 M. CONCRETE BARRIER STRENGTH The basic development length l hb for a hooked bar with f y = 400 MPa. Is given by [A ] The basic development length l hb for a hooked bar with f y = 400 MPa. Is given by [A ] and shall not be less than 8db or 150mm. For a No.15 bar, db=16mm and which is greater than 8(16) = 128mm and 150mm. The modifications factors of 0.7 for adequate cover and 1.2 for epoxy coated bars [A ] apply, so that the development length l hb is changed to l hb =0.7(1.2)l hb = 0.74(292) = 245mm

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94 M. CONCRETE BARRIER STRENGTH ccw

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95 M. CONCRETE BARRIER STRENGTH The standard 90 o hook with an extension of 12db=12(16)=192mm at the free end of the bar is adequate [A ]

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96 M. CONCRETE BARRIER STRENGTH 6. TOP REINFORCEMENT IN DECK OVERHANG The top reinforcement must resist the negative bending moment over the exterior beam due to the collision and the dead load of the overhang. Based on the strength of the 90o hooks, the collision moment M CT (Fig.E7.1-19) distributed over a wall length of (L c +2H) is

97
97 M. CONCRETE BARRIER STRENGTH The dead load moments were calculated previously for strength I so that for the Extreme Event II limit state, we have The dead load moments were calculated previously for strength I so that for the Extreme Event II limit state, we have u

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98 M. CONCRETE BARRIER STRENGTH Bundling a No.10 bar with No.15 bar at 225mm on centers, the negative moment strength becomes Bundling a No.10 bar with No.15 bar at 225mm on centers, the negative moment strength becomes s n

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99 M. CONCRETE BARRIER STRENGTH this moment strength will be reduced because of the axial tension force this moment strength will be reduced because of the axial tension force T = R w /(L c +2H) By assuming the moment interaction curve between moment and axial tension as a straight line (Fig.E7.1-20]

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100 M. CONCRETE BARRIER STRENGTH u st

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101 M. CONCRETE BARRIER STRENGTH

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102 M. CONCRETE BARRIER STRENGTH

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103 M. CONCRETE BARRIER STRENGTH The development length available for the hook in the overhang before reaching the vertical leg of the hairpin dowel is available l dh = =174mm>155mm

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104 M. CONCRETE BARRIER STRENGTH

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105 M. CONCRETE BARRIER STRENGTH

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106 M. CONCRETE BARRIER STRENGTH

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107 M. CONCRETE BARRIER STRENGTH db

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108 M. CONCRETE BARRIER STRENGTH

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SOLID SLAB BRIDGE DESIGN

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: SOLID SLAB BRIDGE DESIGN PROBLEM STATEMENT: PROBLEM STATEMENT: Design the simply supported solid slab bridge of Fig with a span length of 10670mm center to center of bearing for a HL-93 live load. The roadway width is 13400mm curb to curb. Allow for a future wearing surface of 75mm thick bituminous overlay. Use f c ’=30MPa and f y =400 MPa. Follow the slab bridge outline in Appendix A5.4 and the beam and girder bridge outline in section 5- Appendix A5.3 of the AASHTO (1994) LRFD bridge specifications.

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: SOLID SLAB BRIDGE DESIGN

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112 A.CHECK MINIMUM RECOMMENDED DEPTH [TABLE A ]

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113 B. DETERMINE LIVE LOAD STRIP WIDTH [A ] 1.One-Lane loaded: Multiple presence factor included [C } 11

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114 B. DETERMINE LIVE LOAD STRIP WIDTH [A ]

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115 C. APPLICABILITY OF LIVE LOADS FOR DECKS AND DECK SYSTEMS 1. MAXIMUM SHEAR FORCE – AXLE LOADS [FIG.E7.2-2]

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116 C. APPLICABILITY OF LIVE LOADS FOR DECKS AND DECK SYSTEMS

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117 C. APPLICABILITY OF LIVE LOADS FOR DECKS AND DECK SYSTEMS 1. MAXIMUM BENDING MOMENT AT MIDSPAN- AXLE LOADS [FIG.E7.2-3]

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118 D. SELECTION OF RESISTANCE FACTORS (Table 7.10 [A ]

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119 E. Select load modifiers [A ]

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120 F. SELECT APPLICABLE LOAD COMBINATION (TABLE 3.1 [TABLE A ]) 1. STRENGTH I LIMIT STATE 2. SERVICE I LIMIT STATE 3. FATIGUE LIMIT STATE

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121 G. CALCULATE LIVE LOAD FORCE EFFECTS 1. INTERIOR STRIP.

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122 G. CALCULATE LIVE LOAD FORCE EFFECTS 2. EDGE STRIP [A ]

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123 G. CALCULATE LIVE LOAD FORCE EFFECTS

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124 H. CALCULATE FORCE EFFECTS FROM OTHER loads 1. INTERIOR STRIP, 1-mm WIDE

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125 H. CALCULATE FORCE EFFECTS FROM OTHER loads 2. EDGE STRIP, 1-MM WIDE

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126 I. INVESTIGATE SERVICE LIMIT STATE 1. DURIBILITY

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127 I. INVESTIGATE SERVICE LIMIT STATE a. MOMENT- INTERIOR STRIP sy

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128 I. INVESTIGATE SERVICE LIMIT STATE b. MOMENT-EDGE STRIP

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129 I. INVESTIGATE SERVICE LIMIT STATE 2. CONTROL OF CRACKING a. INTERIOR STRIP ssa r cr c s

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130 I. INVESTIGATE SERVICE LIMIT STATE Location of neutral axis cr

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131 I. INVESTIGATE SERVICE LIMIT STATE STEEL STRESS STEEL STRESS s sy c y sa

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132 I. INVESTIGATE SERVICE LIMIT STATE b. EDGE STRIP ½(10 3 )(x 2 ) = (35 x 10 3 )(510-x) cr

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133 I. INVESTIGATE SERVICE LIMIT STATE STEEL STRESS STEEL STRESS s

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134 I. INVESTIGATE SERVICE LIMIT STATE 3. DEFORMATIONS [A ] e ce cr a a e

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135 I. INVESTIGATE SERVICE LIMIT STATE t g cr e

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136 I. INVESTIGATE SERVICE LIMIT STATE By using I g : [A ]

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137 I. INVESTIGATE SERVICE LIMIT STATE b. LIVE LOAD DEFLECTION: (OPTIONAL)[A ]

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138 I. INVESTIGATE SERVICE LIMIT STATE 4607mm

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139 I. INVESTIGATE SERVICE LIMIT STATE Back

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140 I. INVESTIGATE SERVICE LIMIT STATE

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141 I. INVESTIGATE SERVICE LIMIT STATE DESIGN LANE LOAD DESIGN LANE LOAD Lane

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142 I. INVESTIGATE SERVICE LIMIT STATE The live load deflection estimate of 17mm is conservative because I e was based on the maximum moment at midspan rather than an average I e over the entire span. The live load deflection estimate of 17mm is conservative because I e was based on the maximum moment at midspan rather than an average I e over the entire span. Also, the additional stiffness provided by the concrete barriers has been neglected, as well as the compression reinforcement in the top of the slab. Also, the additional stiffness provided by the concrete barriers has been neglected, as well as the compression reinforcement in the top of the slab. Bridges typically deflect less than the calculations predict and as a result the deflection check has been made optional. Bridges typically deflect less than the calculations predict and as a result the deflection check has been made optional.

143
143 I. INVESTIGATE SERVICE LIMIT STATE 5. Concrete stresses [A ]. As there is no prestressing therefore concrete stresses does not apply.

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144 I. INVESTIGATE SERVICE LIMIT STATE 5. FATIGUE [A5.5.3] Fatigue load should be one truck with 9000-mm axle spacing [A ]. As the rear axle spacing is large, therefore the maximum moment results when the two front axles are on the bridge. as shown in Fig.E7.2-8, the two axle loads are placed on the bridge. No multiple presence factor is applied (m=1). From Fig.E7.2-8

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145 I. INVESTIGATE SERVICE LIMIT STATE

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146 I. INVESTIGATE SERVICE LIMIT STATE a. TENSILE LIVE LOAD STRESSES: One loaded lane, E=4370mm s

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147 I. INVESTIGATE SERVICE LIMIT STATE b. REINFORCING BARS:[A ] min

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148 J. INVESTIGATE STRENGTH LIMIT STATE 1. FLEXURE [A ] RECTANGULAR STRESS DISTRIBUTION [A ] a. INTERIOR STRIP: (2/7)

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149 J. INVESTIGATE STRENGTH LIMIT STATE

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150 J. INVESTIGATE STRENGTH LIMIT STATE

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151 J. INVESTIGATE STRENGTH LIMIT STATE For simple span bridges, temperature gradient effect reduces gravity load effects. Because temperature gradient may not always be there, so assume = 0

152
152 J. INVESTIGATE STRENGTH LIMIT STATE So the strength limit state governs. Use 150 mm for interior strip.

153
153 J. INVESTIGATE STRENGTH LIMIT STATE b. EDGE STRIP

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154 J. INVESTIGATE STRENGTH LIMIT STATE STRENGTH I: Use No. 140mm for edge strip.

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155 J. INVESTIGATE STRENGTH LIMIT STATE 2. SHEAR Slab bridges designed for moment in conformance with AASHTO[A ] maybe considered satisfactory for shear.

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156 K. DISTRIBUTION REINFORCEMENT [A ] The amount of bottom transverse reinforcement maybe taken as a percentage of the main reinforcement required for positive moment as. The amount of bottom transverse reinforcement maybe taken as a percentage of the main reinforcement required for positive moment as.

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157 K. DISTRIBUTION REINFORCEMENT [A ] a. INTERIOR SPAN:

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158 K. DISTRIBUTION REINFORCEMENT [A ] b. EDGE STRIP:

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159 L. SHRINKAGE AND TEMPRATURE REINFORCEMENT Transverse reinforcement in the top of the slab [A5.10.8] Transverse reinforcement in the top of the slab [A5.10.8]

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160 M. DESIGN SKETCH

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161 TABLE A-1 BACK

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162 BACK

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