4.1 Bridge g Deck Design g 4.2 DECK SLAB DESIGN METHODS Interior region of deck 1. Empirical method - described in LRFD 9.7.2 2. Strip method - loads described in LRFD 3.6.1.3.3 analysis described in LRFD 4.6.2 3. Refined methods - Finite Element Method, etc, using loads described in LRFD Chap. 3 S n equal spans @ S Overhangs: LRFD 9.7.2.2 1. Empirical method not applicable to overhangs. 2. Strip method or refined method must be used in the cantilever overhang to account for wheel loads between the railing and the exterior grider, and for collision loads applied to the railing. 4.3 REINFORCING DEFINITIONS Railing reinforcement • • Slab ( "Deck" ) • Horizontal • Longitudinal slab reinforcement [ 2 layers ] Vertical • • • • • • • Railing Overhang Transverse slab reinforcement [ 2 layers ] Interior region 1 4.4 LRFD DESIGN CRITERIA Slab transverse reinforcement (interior region): gravity loads Service limit state Strength limit state Slab transverse reinforcement (overhang): Service limit state Strength limit state Extreme event limit state gravity loads collision (rail) plus reduced gravity loads Slab longitudinal reinforcement (all) Secondary flexure; temperature/shrinkage control Rail horizontal and vertical reinforcement: Extreme event limit state collision (rail) Problem Definition Live Load: 4.5 HL-93 Deck Concrete f’c = 4 ksi wc =150 pcf Mild Steel (Non-Prestressed) fy = 60 ksi Es =29,000 ksi Dimensions Thickness = 8.0 in. Cover = 2.5 in. (Top) = 1.0 in. (Bottom) (LRFD 9.7.11 & 13.7.3.1.2) (LRFD 5.12.3) Future Wearing Surface Allowance: FWS = 30 psf Deck Design 4.6 2 7 Definitions used in empirical design method Interior region of deck S n spans @ S S Effective length: Le Le The effective length is the distance between the flange tips (S− bf) plus the flange overhang ½ (bf − bw): Le = S − ½ (bf + bw) S . maximum 3 .bf bw bf Empirical Method z Based on extensive lab tests z Load resistance mechanisms 4.9 Empirical Method – Design Conditions z Diaphragms at lines of support z Concrete and/or steel girders z CIP and water cured deck z Uniform depth z Effective length to depth ratio z Effective length – 6 to 18 – 13.5 ft..8 » Flexure » Arching Action z FEM verification z Factor of safety z No analysis required z Isotropic reinforcement z Not applicable to overhang design 4.4. . minimum z Minimum overhang to depth ratio » 5 » 3.4 ksi. minimum z Slab thickness – 7.10 Empirical Method – Design Conditions z Core depth – 4.2 / ft. z Reinforcement doubled in end zone if skew exceeds 25° 4.0 in. spacing – As.5 in. (#4 bars @ 13 in.e..0 in. each way: 0.2 / ft.185 in.2 / ft. if barrier is composite z fc’ . o. each way: 0.11 Empirical Method – Design z Bottom Layer..2 / ft.276 in.27 in. (#5 bars @ 13.prov’d = 0.) z G d 60 steel Grade t l z Outermost bar in direction of effective length z Maximum spacing – 18 in.c.18 in. transverse steel outside) • • • • longitudinal steel transverse steel Slab thickness: longitudinal steel Wearing surface Design depth: ts (excluding wearing surface) • • Top cover • • Core depth: tc Bottom cover transverse steel 4 .) z Top Layer. minimum z Deck is composite 4.4. spacing – As.12 Empirical design method details Steel placement: Reinforcement shall be provided in each face of the slab with the outermost layers placed in the direction of the effective length (i.prov’d = 0. 1’ width S n spans @ S 5 .14 » Moving M i load l d analysis l i – Truck axles moved laterally – Multiple presence factors – Dynamic load allowance – Total moment divided strip width » LRFD Table A4.1-1 (used in this design example) 4. and the reinforcing required to connect the railing to the slab.13 Empirical Method – Final Design Strip Method z Continuous beam loaded with truck axle loads z Equivalent strip widths – interior.4. exterior. The deck is subdivided into 1-ft-wide strips perpendicular to the supporting girders. The girders are assumed to provide rigid point supports for the continuous-beam strip. The span length is taken as the center-to-center distance between the girders.15 STRIP METHOD Applicability: The strip method (also called the approximate method in the LRFD Specification) provides for selecting all transverse and longitudinal reinforcing in the slab. and overhang z DL moments on a per foot width basis z LL moments: 4. The strips are treated as continuous beams. 0 6' 16 kips 16 kips vary position and spacing of truck(s) 6’ 1’ width 6 .20 6' 8' 16k16k 16k16k 6' 4' Two trucks 100% of the moment caused by two trucks: m = 1.18 Live loads on slab strip ( continued ) Use whichever produces the greatest moment at each point across width of deck: 16k 16k One truck 120% of the moment caused by one truck: m = 1.4. 16k 16k 16k 16k 16k 16k 6' 8' 6' 16k 16k 6' 4' 6' Two adjacent trucks One truck 16k 16k 6' spacing varies Two trucks in alternate spans (use influence lines for placement) 4. Recall the lane width is 10 feet.16 Dead loads on slab strip thickened slab in overhang main slab railing (parapet) wearing surface S wearing surface added slab thickness main slab thickness railing weight per foot length S 4.17 Live loads on slab strip Live load used in deck design is that caused only by truck axle loads ( lane load not included ). the truck load moments must be increased by the impact factor: MLL = MTruck ( 1 + IM ) = MTruck ( 1 + 0.19 Examples of critical wheel load positions: Maximum negative moment at exterior support ( wheel load in cantilever ): 16 k Multiply moment produced by one truck by m = 1.2.2 16 k 16 k Position of inner wheel relative to support depends on grider spacing.4 S Approximate location of maximum positive moment caused by dead loads S 4.21 Live loads on slab strip ( continued ) The moments computed using the truck load patterns just described are distributed to the 1’-wide deck strip by dividing by each moment value (foot-kips) by equivalent strip widths that are defined in LRFD Table 4.6.0 In all cases.33 ) 4.2 6’ 1’ Maximum positve moment in end segment: Multiply moment produced by one truck by m = 1.20 Live loads on slab strip ( continued ) Examples of critical wheel load positions ( continued ): Maximum negative moment at first interior support : 16k 6’ 16k 6’ 16k 16k 6’ 16k 4’ 16k S < 10 feet S > 10 feet Multiply moment produced by one truck by m = 1.1. 6’ ≈ 0.2 Multiply moment produced by two trucks by m = 1.3-1: Equivalent strip width: SW Mdesign 1’ width Mdesign ( foot − kips / foot ) = MLL ( foot − kips) SW ( feet ) 7 .Live loads on slab strip ( continued ) 4. 64 4. 3 in.90 2.73 2. 4. The critical load at the exterior girder is generally that associated with truck collision into the rail.30 6.04 3.73 3.02 4.40 9.85 10.19 1.79 12.20 3. m = 1.47 2.09 10.24 1.01 6.81 12. 18 in.30 0.79 8.95 1.25 9.54 8.74 1.31 2. 24 in.88 4.14 9.92 9.02 4.23 9.66 2.83 13’-0” 13’-3” 13’-6” 13’-9” 14’-0” 14’-3” 14’-6” 14’-9” 15’-0” 8.63 9.58 2.51 7.46 12.65 6. 2.86 7.65 4.97 3.67 10.03 10. The moments given in this table account for: 1.74 1.79 6.20 2.44 9.45 6.59 11. and the maximum negative at the interior girders ( m = 1.76 7.21 9.4.36 9. AASHTO LRFD Table A4.90 1.07 1.47 Negative Moment .95 1.38 2.86 6. Placement of trucks to produce the maximum positive moment beween girders.77 4.41 1.23 Live loads on slab strip ( continued ) Table may be used for all positive LL moments Must compute negative LL moment in overhang Table may be used for negative LL moments at all interior supports AASHTO LRFD Table A4.07 1.63 3.52 8.2 applied to moments caused by one truck.43 2.76 8.83 1.16 6.33 1.25 2.71 4.38 7.80 10.04 8.18 3.25 1.02 12.89 2.19 3.16 9. 6 in.31 11.88 13.55 9.55 11.72 6.12 4.33 These moments may be combined directly with the dead load moments to determine the the critical moments (per foot width) to be used in designing the deck reinforcing in the interior deck segments.32 1.66 8.43 10.99 4.78 8.99 9. The girder flanges help resist the negative moment at the girders.32 1.1-1 (shown on the next page) gives values of maximum live load moments per foot width that may be used in designing the interior slab segments in slabgirder bridges.36 3.28 0.65 1.16 11.19 1. The dynamic (impact) factor of 1.37 11.99 7.91 11.24 12.94 11.68 4.02 12.94 5.34 1.74 1.67 12. thus reducing the negative moment for which the slab must be designed Maximum negative moment in slab when girder is treated as a point support.74 3.24 Live loads on slab strip ( continued ) Treating the slab strip as a continuous beam with rigid point supports results in overlyconservative negative moments at the centerline of each girder.50 2.30 10.98 4.50 10.58 2. Moment computed at dc when girder is treated as a point support Actual negative moment experienced by slab dc 8 .26 1.28 8.68 2.07 1.14 11.27 10.20 3.18 7.67 4.08 10.07 11.02 9.0 applied to moments caused by two trucks) 3.22 Live loads on slab strip ( continued ) Computer analyses are required to consider all possible live load combinations and positions.1-1 Girder Spacing ft 4’-0” 4’-3” 4’-6” 4’-9” 5’-0” 5’-3” 5’-6” 5’-9” 6’-0” Positive Moment k-ft/ft 4. 9 in.57 7.60 1.57 1. The live load moments at the exterior girder must be computed separately. 12 in.11 2.50 1.29 1.72 10.39 1.88 2. Grider spacing 2.37 8.67 8. k-ft/ft Distance from CL of girder to design section for negative moment 0 in.63 4.59 6.66 4.90 9.38 6.00 2.73 2.31 1.58 6. 18 and 24” from the girder centerline. 10 Future wearing surface = 30 psf = 0. 9. 6.3 ksf M FWS = 0.6: The design section for negative moments in precast I-shaped and T-shaped concrete beams may be taken at one-third the flange width width. 12.2. dc = bf 3 ≤ 15.0" Table A4. 1 x 9 2 = 0. 10 9 .1.27 Strip Method – DL Moments M = z z z DL / LL ratio C = 10 or 12 Self weight = 8(150)/12 = 100 psf = 0.4.24 kip − ft .0 15 0 inches inches.81 kip − ft .26 Limit states » Service: crack control » Fatigue: need not be checked » Strength: factored moments » Extreme event: vehicular collision 4. / ft .2. / ft . Strip Method z 4.25 Live loads on slab strip ( continued ) dc From LRFD 4.6.1 ksf M DL = z wl 2 c 0. Moments at other dc locations can be found by interpolating moment values at adjacent tabulated distances.03 x9 2 = 0.1-1 gives values of negative live load moment at distances dc = 3.6.1.6: This reduction in negative moment replaces the use of reduced span length to compute moments as done in the standard specification. but not exceeding 15. From LRFD Commentary 4. from the centerline of the point support assumed at the centerline of the girder. 81+0. / ft.24 + 1.4.30 Strip Method – Strength LS Moments z Strength Limit State » Negative Interior Moment: Mneg.str Mneg.24+6.str = 1.87 kip-ft.29 = 12.25x0.38 kip-ft.5x0. neg M LLI = 3. 10 . / ft .5x0.76 kip-ft.75x3. » Use 12 in in.81 + 1.71kip .str = -(1.24 + 1. 4.71) = -7.34 kip-ft. » Positive Moment: Mpos = (0. (governs) ≤ 15 in. / ft.29 kip . z Critical section for negative moment » (1/3) bf = 14 in.ft .29) = 7.71) = -4. p / ft.ft .25x0.75x6. (conservative) pos M LLI = 6.81 + 1. / ft . 4.24+3.29 Strip Method – Service LS Moments z Service Limit State: » Negative Interior Moment: Mneg = -(0. » Positive Moment: Mpos.28 Strip Method – LL Moments z Table A4-1 z Span = 9 ft. / ft.81+0. 65 ) * 0. / ft.38 ksi fM neg = 4.7.24 4 = 0. » As = (12/10)(0.K.8 * 0.90 )( 0.65 in.c.021 ≥ 0.controlled section φ = 0.2 / ft.003 = 0.372)(60) = 0.65 Therefore.str = 7.87 kip-ft.547 in.Strip Method – Flexure Design z 4. > Mneg./ft.24 fc' = 0.9 for flexure 4.87 kip-ft.003 = (5. tension . φMn = φ a= Asfy ⎛ a⎞ ⎜d − ⎟ b ⎝ 2⎠ a= (0.19 − (12 ) 2 ⎝ ⎠ ϕMn = 8. / ft.23 kip-ft.ft.7.2/bar) = 0. 0.85 0. ft /ft ⎜ ( 5.31 in.45 ksi > 0.33 Strip Method – Crack Control z Maximum spacing of tension reinforcement » LRFD Article 5. / ft.3.8 * 0.372 in.19 − 0.8 fr = 0.4 applies 11 .005 c 0. » Try No. o.85)(4)(12) As f y 0.4 applies if fMneg > 0.3.38 ksi ⎛ 12 * 8 2 ⎞ ⎜⎜ ⎟⎟ ⎝ 6 ⎠ Therefore. z O.76 * 12 = 0. Article 5.85 fc' b c= a 0.23 kip ki .85 Check Tension/Co mpression controlled section εt = (d t − c ) * 0.31 Mneg.547 = = 0.547 ⎞ ) ⎟ = 8. 4.8fr 0. (0. 5 at 10 in.372 )( 60 ) ⎛ 0.32 Strip Method – Flexure Design φMn = φ φM n = As fy ⎛ a⎞ ⎜d − ⎟ b ⎝ 2⎠ ( 0.str = -7. d c ) 0. As = No. ds = 8 – 2.7 ( 8 . 5 bar)/2 = 2.1) E = 33.5 4.5 – 0. γ e = 0.912 )( 5.77 0 .00597 )( 8 ) + [( 0.265 j = 1 − 0.5” (clear cover) + 0.c.0 = 3.372 = 0.2/ ft.00597 )( 8 ) = 0.000 )( 0. 5 at 10” o.k)ds 3 M Elevation Section εs fs Strain Stress T Resultant Forces Figure 3: Reinforced concrete rectangular beam section at service load Strip Method – Crack Control fs = M As jd s 4. = 0.76 kip-ft.000 w 1.265 / 3 = 0.36 where: M = -4.372 in.31/10*12 = 0.19 ) j =1- c k = 2 ρn + (ρn ) .7 ( h .000 / 3./ft.372 )( 0.Strip Method – Crack Control 4.625/2 = 5. n = modular ratio = Es / Ec = 29.81 =1+ = 1.4 ksi ( 0.( 0.625 (diameter of No.81 ) fs = Stress in reinf.7.75 for Class 2 exposure dc = cover – extreme tension fiber to center of extreme reinf.ρn 2 k = ( 2 )( 0.00597 )( 8 )] .830 = 7. based on cracked section analysis Strip Method – Crack Control 4. βs = 1 + dc 2.830 ksi c ρ= ρ= k 3 c As bd 0.76 * 12 ) = 32.00597 (12 )( 5. Use 8 ≥ 6 OK (LRFD 5.912 2 fs = ( 4.150 )1.81 in.35 Calculate fs z b 1kd 3 s εc kd s fc C ds Neutral Axis jds = (1 .19 in.34 Maximum spacing of tension reinforcement z s≤ 700γ e − 2d c βs fs where .19 ) 12 .57. = 2.5 f ' = ( 33.2. /ft.) Mneg prov’d = 11. R i d maximum Revised i spacing i = 7.2) z z z At bottom In secondary direction Percent of reinforcement for Mpositive 220 ≤ 67%.81 = 3.c.c.) 13 .39 (LRFD 9.77 * 32.47 in.5 kip-ft. S 220 = 75 % > 67 %.67(0. (As prov'd = 0.K. 5 at 8 in.c.) Mpos prov’d = 13. z Reduce spacing to 7 in. i O.2 / ft.2 / ft. where S = 108 . o.75 1. o. prov’d = 0.2 / ft.2 7 2 in. o. are adequate (As.53 in. o.53 in.bf Strip Method – Distribution Reinforcement 4.37 Strip Method – Crack Control s≤ 700γ e β s fs 2d c = 700 * 0. o.3.c.5 67% Governs As = 0.2 / ft.4. Provide #5 @ 12 in.c.53 in. z Similar calculations for Mpositive suggest No.31 in.3 kip-ft.465 in2/ft./ft. 8 .) = 0.7. N.G. z Provided No.310 in.5 ft. o. OK z Therefore.c. (As prov'd = 0.4 2 * 2.6 = 102 in. 4.38 LONGITUDINAL STEEL Secondary longitudinal reinforcement (parallel to the girders ) in the bottom of the slab is required as a percentage of the primary positive-moment transverse reinforcement perpendicular to the girders: Longitudinal steel percentage of primary reinforcement = 220 Le % ≤ 67% where: Le = effective length (feet) is the distance between the flange tips plus the flange overhang: Le = S − (bf + bw) / 24 S temperature steel Le • ° ° • • • ° • • ° • ° • • ° • SbL bw bf longitudinal steel (bottom of slab) S . > 3. for negative interior moments: Provide #5 @ 7 in. = 8. 5 at 10 in. 3. (+ 71%) 14 .922 in.10.ft/ft (governs) M cr = 0.3 * 96 As = = 0.11 24 * 60 Maximum spacing: 3*8=24 in or 18 in (governs) Provide No.40 42’ – 6” = 510 in 8 in Method 1 : Consider full width of deck : As = 1. 4 @ 18 in. 5.2 Mcr or 1.33 Mu z (LRFD 5.2 * 7.575 in.41 Strip Method – Minimum Reinforcement Mr ≥ lesser of 1. Reinf.465 + 0.38 = 16.37 fc′ = 0.11 2 * (510 + 8) * 60 Method 2 : Consider unit width of deck = 12 in 12 in Area = 12 * 8 = 96in 2 8 in Drying perimeter = 2 * (12 + 0) = 24 in 1.2 / ft.33M pos.185] = 0.087. therefore As = 0.9 = 9.7 kip p .str = 1.10.27 in2 / ft.27 = 1.in.87 = 10.11 ≤ As ≤ 0.47 kip .8 − 1 0.276 + 0.74 * ( 1.33M neg. therefore As = 0.2 / ft.3 * 510 * 8 = 0.3.33 * 12. 1. o.ft/ft z Mpos prov’d = 13.2M cr = 1.c./ft. 5.74 ksi 12 * 8 2 ) = 94.60 Eq. Traditional z Total reinforcement per square foot of deck: Empirical method: 2[0.Strip Method – Shrinkage & Temp.9 kip p .42 Empirical vs. (As prov'd = 0.5 kip-ft/ft OK 4.5 kip-ft/ft > 9. = 7. 6 1.37 4 = 0. (.41%) Traditional method: 0.310 + 0.8 − 2 4.) 4.3bh As = 2(b + h )fy Eq.2) Mcr = fr Sc fr = 0.53 + 0.str = 1.085 .5 kip .7.3 kip-ft/ft and Mneg prov’d = 11.33 * 7.47 kip .ft/ft 1.ft. 0 foot from the inside face of the railing: W L = uniform load distributed longitudinally 1.43 DECK CANTILEVER / RAILING ANALYSIS The deck cantilever is designed for the negative moment at the exterior girder caused by whichever of the two load conditions shown below produces the greatest value of this moment: Dead load moment: railing. FWS PLUS Full truck live load (gravity load) Reduced truck gravity load plus lateral load applied to railing OR Critical moment location Critical moment location Deck Cantilever / Railing Analysis ( continued ) 4. may be use if railing is continuous 4.0 ft WL = 1.0 kip/ft 15 .may only be used if railing is continuous Strip method .simplified method: continuous railing WL 1.0 ft dc Two methods for determining W L: Simplified method .45 Deck Cantilever / Railing Analysis ( continued ) Truck gravity load .must be used if railing is discontinuous. deck.4.44 Truck gravity load: The truck live load is applied as a uniformly distributed load W L located 1. expressways and interstate highways with a mixture of trucks and heavy vehicles 4. and very low mixtures of heavy vehicles TL-4 .Test Level Four: Applicable to high-speed highways.Test Level Two: Applicable to local and collector roads with favorable site conditions.5 4.2-1 Railing Test Levels TL-2 Ft Transverse ( KIP ) TL-1 TL-4 TL-6 175 Lt ( IN ) [ Given in FT in table ] 48 48 48 42 96 96 96 Minimum Rail Height H ( IN ) 27 27 27 32 40 54 90 16 .47 RAILING DESIGN Truck load applied to railing: The primary purpose of the railing (also called the barrier or parapet) is to contain and redirect vehicles using the structure in order to: (1) protect the occupants of a vehicle that collides with the railing.strip method: discontinuous rail gap WL 16k 16k 1.0 • X(feet) Pw SW Overhang 4.5 9 18 18 39 41 58 Fv Vertical Down ( KIP ) 4.0 ft SW X SW SW Overhang WL = (inches) = 45 + 10. Design parameters are grouped into six (6) Test Levels in AASHTO 13.Test Level Three: Applicable to high-speed arterial highways with favorable site conditions. The railing and deck overhang are designed to survive a vehicle collision force whose intensity depends on the nature of the highway and the traffic it carries.46 Deck Cantilever / Railing Analysis ( continued ) Truck gravity load .0 ft 1. Four of the most common Test Levels are: TL-1 .7.4.5 4. freeways.0 ft 1.5 27 54 54 116 124 FL Longitudinal ( KIP ) 4.5 18 50 80 80 Designations Table A13. reduced posted speeds. (3) protect persons and vehicles on roadways and other areas under the bridge. and a small number of heavy vehicles TL-3 .Test Level One: Applicable to roads with low speeds and traffic volumes TL-2 .2 . (2) protect other vehicles on the structure at the time of the collision.48 Railing Design ( continued ) Fv FL Lt Ft H ( to top of wall ) Critical section for negative moment in deck overhang Design Forces and TL-3 TL-5A TL-5B 13. 1).1.49 Railing Design ( continued ) The minimum design rail collision load is based on the conditions (the highway type traffic volume and site features) for which bridge will be used (LRFD Table 13.51 Railing Design ( continued ) Railing design model .7.it must equal or exceed design load Ft given in Table A13. Rw is force required to “break” wall .Case 1 ( interior ) : Lc Rw H Lc = Rw Lt + 2 2 ⎛ L t ⎞2 8 H M w ⎜⎜ 2 ⎟⎟ + Mc ⎝ ⎠ ( in ) 45° ⎛ ⎛ L c ⎞2 ⎞ 2H ⎜ ⎟ = 2 L − L ⎜ 8 M w + M c ⎜⎜ H ⎟⎟ ⎟ ( k ) ⎝ ⎠ ⎠ c t ⎝ Rw Rw TCT = VCT = L + 2 H c ( k/in ) Rw H Lc + 2H ( in-k/in ) MCT = MCT VCT H = height of the railing wall ( inches ) Lt = length along which impact force is assumed distributed to wall ( inches ) Mw = average unit ultimate moment resistance of wall about vertical axis ( in-k/in ) Mc = average unit ultimate moment resistance of wall about horizontal axis ( in-k/in ) Lc = width of failure mechanism that offers the least Rw ( inches ) 4. The moments produced at the base of the rail by the load that causes the rail to fail are applied to the deck at the outer edge of the cantilever overhang.50 Railing Design ( continued ) Railing design model .Case 2 ( end ) : Lc = Rw = Lt + 2 ⎛ L t ⎞2 8 H 2 M w ⎜⎜ ⎟⎟ + Mc ⎝ 2 ⎠ Rw H ⎛ 2H ⎛ L ⎞2 ⎞ ⎜ 8 M w + Mc ⎜⎜ c ⎟⎟ ⎟ ⎟ (k) 2 L c − L t ⎜⎝ ⎝ H ⎠ ⎠ TCT = VCT = MCT = Lc ( in ) Rw H Lc + H Rw Lc + H 45° Rw ( k/in ) ( in-k/in ) MCT VCT 17 .2-1 H Vehicle loads used in rail collision Extreme Event Limit State analysis: Rw Rw MCT VCT Reduced wheel load is applied with Rw MSLAB TCT = VCT TCT = VCT MCT MCT 1’ width deck strip Forces per foot deck width caused by horizontal force Rw 4.4. – extreme event limit state z Design Case 2: DL & vert.75 (LL+IM) 4.2) Design Forces and Designations Ft Transverse Force 54 KIP FL Longitudinal Force 18 KIP Fv Vertical Force Down 18 KIP Lt and LL 3.54 Strength of Barrier: ILDOT F-Shape Concrete Barrier 18 .53 Vehicle Impact Forces z Extreme Event Test Vehicle – TL4 (LRFD 13.5 DW + 1.25DC + 1.0.5 FT Lv 18 FT He min (Height of impact above deck) 32 IN H Minimum Height of Barrier 32 IN Safety Barrier 4. & long. 0 – extreme event limit state Typically does not govern for concrete barriers z Design Case 3: Strength I Limit State 1. vehicle impact forces Load & Resistance Factors = 1 1.52 Design Case 1: DL and trans.7.Overhang Design z 4. Vehicle impact forces Load & Resistance Factors = 1.0. 55 4.57 19 .56 Yield Line Case 1 : ⎛L ⎞ ⎛ L ⎞ 8H (M b + Mw ) Lc = ⎜ t ⎟ + ⎜ t ⎟ + Mc ⎝2⎠ ⎝2⎠ 2 ⎛ 2 Rw 1 = ⎜⎜ ⎝ 2Lc − Lt 2 ⎞⎛ ML ⎞ ⎟⎟⎜ 8M b + 8Mw + c c ⎟ ⎜ H ⎟⎠ ⎠⎝ Strength of Barrier – Yield Line Case 2 4.Strength of Barrier – Yield Line Case 1 Strength of Barrier 4. controls ⎝ 2Lc − Lt ⎠⎝ ⎠ Rw for barrier = 61.4 kip ⎠⎝ ⎠ Yield Line Case 2 : ⎛ M + Mw ⎞ ⎛L ⎞ ⎛L ⎞ Lc = ⎜ t ⎟ + ⎜ t ⎟ + H ⎜⎜ b ⎟⎟ = 6.58 Yield Line Case 2 : ⎛ M + Mw ⎛L ⎞ ⎛L ⎞ Lc = ⎜ t ⎟ + ⎜ t ⎟ + H ⎜⎜ b ⎝2⎠ ⎝2⎠ ⎝ Mc 2 ⎛ 2 Rw 2 = ⎜⎜ ⎝ 2Lc − Lt ⎞ ⎟⎟ ⎠ 2 ⎞⎛ ML ⎞ ⎟⎟⎜ M b + Mw + c c ⎟ ⎜ H ⎟⎠ ⎠⎝ Strength of Barrier 4.Strength of Barrier 4.6 kip z 61.60 Yield Line Case 1 : ⎛L ⎞ ⎛ L ⎞ 8H (M b + Mw ) = 13.3 ft ⎝2⎠ ⎝2⎠ ⎝ Mc ⎠ 2 2 ⎛ ⎞⎛ M c Lc ⎞⎟ 2 ⎜ Rw 2 = ⎜⎜ ⎟⎟⎜ M b + Mw + H ⎟ = 61.6 kip.6 kip > Ft = 54 kip OK 20 .7 ft Lc = ⎜ t ⎟ + ⎜ t ⎟ + Mc ⎝2⎠ ⎝2⎠ 2 ⎛ 2 Rw 1 = ⎜⎜ ⎝ 2Lc − Lt 2 ⎞⎛ M c Lc ⎞⎟ ⎜ ⎟⎟⎜ 8M b + 8Mw + H ⎟ = 134.59 Strength of Barrier 4. Flexural Design of Deck 4.63 Yield Line Case 2 » At the inside of barrier – M over Lc – T over Lc+ H 21 .61 P h d M Distribution of Mc and T 4.62 Yield Line Case 1 » At the inside of barrier – M over Lc – T over Lc+2H Distribution of Mc and T 4. 2 / ft.92 ⎞ ⎛ ⎛ 8 0. / ft.94 = 4.5 – 0. 7 at 13 in.0 kip − ft . (Flexural strength of barrier about hor.185 in. / ft.10/0.1 .64 At the inside face of the barrier: z MDC = (8/12)*(0. OK 22 . / ft . 2 ⎠ 2 ⎠ ⎝ ⎝2 z Mn = 15.60)/13*(12) = 0.450)*(1.5)2 / 2 = 0.16 kip / ft. z P = T1 (yield line case 1) = 6. (Empirical design – No.30 kip-ft. z z z z z z z Reinforcement at Top of Deck 4.185x60 = 11. / ft.16/(0. at centroid of deck P h d M Reinforcement at Top of Deck 4.53 53 < M = 14 14. / ft.25 in.06 − ⎟ − 6.150)*(1.5/2 = 5.74x60 = 44.7 kip − in.94⎜ − ⎟ = 179. Mbarrier = (0. a = 4.9 9 = 14.1 kip / ft. / ft . de = 8 – 2.5)2 / 2 = 0. » T = 0.2 / ft. / ft.Flexural Design of Deck 4.66 T+P P h d a M Strains Stresses C Forces 0.6. Mc = 13. o. 2 ⎠ 2 ⎠ ⎝ ⎝2 z z Mn = 2 2. C = 11.06 0 06 +0 0.34 34 + 13 13. 0. alternating with No. / ft.74 in.10 ⎞ ⎛ ⎛ 8 0.10 ⎞ M n = 11.4⎜ 5.65 T+P P h d a M Strains Stresses C Forces a⎞ h⎞ a⎞ ⎛ ⎛ ⎛ ⎛h a⎞ M n = C ⎜ d − ⎟ + P ⎜ d − ⎟ = T1⎜ d − ⎟ − P ⎜ − ⎟ 2⎠ 2⎠ 2⎠ ⎝ ⎝ ⎝ ⎝2 2⎠ As = 0.25 − ⎟ − 6. axis) Design forces for deck (at inside face of barrier): M = MDC + Mbarrier + Mc = 0.34 kip-ft.94⎜ − ⎟ = 30.30 kip-ft.. / ft .30 30 kip-ft kip-ft. / ft . 4 @ 13” o. c = a/0. » As = (0.85 = 0.85 = 0. 4 at 13 o.c.c.1⎜ 5.85x12x4) = 0.12 in.9 kip-ft.20+0.) T1 = T + P T1 = T + P = 0. = 2.3 kip − in.4 kip / ft. ft NG Provide additional No.94 kip / ft.10 in.0 > M = 14.53 kip − ft .06 kip-ft. = 15.c.92 ⎞ M n = 44. 67 » Away from barrier: – Dispersion at 30 to 45 deg 23 .4.