Example Bridge Deck Design for Assign. 3 Summary

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1 Bridge Design Spring 2012 Example Bridge Deck Design for Assign. 3 Summary 1 / 2 Geometry Material Properties N gird 6 f' c 4 ksi S 8.00 ft b f 0 in f r ksi = 0.37 * SQRT( f' c ) L over 3.25 ft E c 3640 ksi = 1820 * SQRT(f_c) W barrier 1.25 ft f y 60 ksi contin_barrier FALSE cover top 2 in Expos_Fac top 0.75 Loads cover bot 1 in Expos_Fac bot 1 h wear 2 in P wheel -16 k W_out ft = (Ng - 1) * Sg + 2 * Lo W_in ft = W_out - 2 * Wbarrier N_lanes 3 =INT( W_in / 12 ) Flexure Design: OK Overhang Between the Girders h 8 in 8 in with: # 5 3" on the top with: # 5 6" on the top # 5 10" on the bottom # 5 10" on the bottom Ductility: M n k-ft / ft M n OK OK Mu Mu k-ft / ft Ductility: M n OK OK Ductility: Mu M n OK OK Mu T u 4.13 Top Reinforcement Top Reinforcement bar top 5 in bar top 5 s max 42 in s max 13 in OK s top 3 in s top 6 in Bottom Reinforcement Bottom Reinforcement bar bot 5 bar bot 5 s max in s max 12 in OK s bot 10 in s bot 10 in

2 Bridge Design Spring 2012 Example Bridge Deck Design for Assign. 3 Summary 2 / 2 Distribution Reinforcement OK bar tdistrib 5 A s_dist in 2 s distrib 14 in A s_dist_req' in 2 Temp. & Shrinkage Reinforcement OK bar_size 4 A s_t&s 0.13 s 18 in A s_t&s_req 0.11 in 2 Calculations: M n : M n : h 8 in h 8 in bar top 5 in 2 =bar top bar top 5 in 2 =bar top bar bot 5 in 2 =bar bot bar bot 5 in 2 =bar bot s top 3 in =s top s top 6 in =s top s bot 10 in =s bot s bot 10 in =s bot cov top 2 in =cover top cov top 2 in =cover top cov bot 1 in =cover bot cov bot 1 in =cover bot P ax 4.13 k P ax 0 k see sheet Calc_fMn M n_ pos k-ft / ft see sheet Calc_fMn M n_ neg k-ft / ft for details M n_ neg k-ft / ft for details 1.2 M cr 9.47 k-ft / ft 1.2 M cr 9.47 k-ft / ft Crack Control Crack Control h 8 in h 8 in bar top 5 bar top 5 bar bot 5 bar bot 5 s top 3 in s top 6 in s bot 10 in s bot 10 in cov top 2 in cov top 2 in cov bot 1 in cov bot 1 in M serv_pos 0.00 k-ft / ft M serv_pos 5.49 k-ft / ft M serv_neg k-ft / ft M serv_neg k-ft / ft s max_top 42.3 in see sheet Calc_fs_serv s max_top 13.1 in see sheet Calc_fs_serv 18.0 for details s max_bot 12 in for details Distribution Reinforcement S eff 8 ft = S g - b f /2/12 A s_dist / A s = MIN( 0.67, 220 / SQRT( S eff ) / 100 ) A s_bar in 2 A s in 2 =A s_bar * 12 / s bot A s_dist_ req'd in 2 / ft = A s * A s_dist / A s bar_size 5 s 14 in A bar 0.31 in 2 A s_dist in 2 / ft = A bar * 12 / s Temperature & Shrinkage Reinforcement A s_t&s_req'd in 2 / ft = 12 * h / ( 2 * (12 * h) * f y ) A s_t&s_req'd 0.11 in 2 / ft = MAX( 0.11, MIN(0.6, A s_t&s_req'd ) )

3 Loads & Moments 1 / 2 S 8.00 ft L over 3.25 ft b f 0.00 in W barrier 1.25 ft Dead Loads: Overhang Between the Girders h slab_over 8 in h slab 8 in A barrier 2.03 sf 293 h wear 2 in x barrier 0.42 ft w slab_over -100 psf = -150 * h slab_over /12 w slab -100 psf = -150 * h slab /12 w wear psf = -145 * h wear /12 w wear psf = -145 * h wear /12 w barrier plf = -150 * A barrier Live Loads: P wheel -16 k Moments: M- (at Ext. Girder) (0.35 L of Ext. Span) (1st Int. Girder) Dead Loads: I slab ft 2 from sheet "Influ_+M" I slab 5.35 ft 2 from sheet "Influ_-M" M slab_over k-ft / ft = w slab_over /1000 * 1/2 * L over ^2 M slab k-ft / ft = I slab * w slab /1000 M slab k-ft / ft = I slab * w slab /1000 M wear k-ft / ft M wear k-ft / ft = I slab * w wear /1000 M wear k-ft / ft = I slab * w wear /1000 =w wear /1000 * 1/2 * (L over - W barrier )^2 I barrier 1.58 ft from sheet "Influ_+M" I barrier ft from sheet "Influ_-M" M barrier k-ft / ft = w barrier /1000 * (L over - x barrier ) M barrier k-ft / ft = I barrier * w barrier /1000 M barrier k-ft / ft = I barrier * w barrier /1000

4 Loads & Moments 2 / 2 M- (at Ext. Girder) (0.35 L of Ext. Span) (1st Int. Girder) Live Loads: One Two One Two x wheel 1.00 ft = L over - W barrier - 1 Truck Trucks Truck Trucks M wheel k-ft = P wheel * x wheel I truck ft, from Sheet "Influ_+M" I truck from sheet "Influ_-M" w 4.58 ft = ( * x wheel )/12 M truck = P wheel * I truck M truck MPF 1.2 MPF MPF M wheel k-ft / ft = M wheel * MPF /w M truck = M truck * MPF M truck = M truck * MPF M truck 28.4 k-ft 1 Truck Controls M truck k-ft 1 Truck Controls M line k-ft / ft w 6.57 ft = ( * S ) /12 w 6.00 ft = ( * S ) /12 M truck k-ft / ft = MIN( M wheel, M line ) M truck 4.33 k-ft / ft = M truck / w M truck k-ft / ft = M truck / w IM 0.33 IM 0.33 IM 0.33 M truck k-ft / ft = M truck * (1 + IM) M truck 5.76 k-ft / ft =M truck * (1 + IM) M truck k-ft / ft =M truck * (1 + IM) Strength I Load Combination: M U k-ft / ft M U k-ft / ft M U k-ft / ft = 1.25 * (M slab_over + M barrier ) * M wear * M truck = 1.25 * M slab * M wear *M barrier * M truck Extreme Event II Load Combination: M collision k / ft from sheet "Barrier" T collision 4.13 k-ft / ft from sheet "Barrier" M U k-ft / ft = 1.25 * (M slab_over + M barrier ) * M wear * 0.5 * M truck + 1 * M collision T U 4.13 k / ft = 1 * T collision Service I Load Combination: k-ft / ft M serv 5.49 k-ft / ft M serv k-ft / ft = M slab_over * + M wear + M truck = M slab * M wear + M barrier + M truck = M slab * M wear + M barrier + M truck

5 Barrier 1 / 4 Area of barrier: in in in^2 in X Y A x_bar A barrier 293 in^2 x_bar 5.08 in ht of barrier H 32 in Extereme Event II Limit State: Strength of slab should be sufficent to resist force Rw at top of barrrier (from ALBDS, Sect. 13) From Table A13.2-1: TL-4 Ft 54 k transverse force on barrier FL 18 k longitudinal force on barrier Fv 4.5 k vertical force on barrier Lt 3.5 ft length of impact force

6 Barrier 2 / 4 2. Mn about vert. axis M w : = avg of +'ve Mn and -'ve Mn Consider equivalent rectangular section of width b and depth h: h width of section b 32 in = H depth of section h 9.16 in = A barrier / H Avg effective depth, d, of Horiz. Rebar: b Bars on Left: distance from distance from h left face right face Bar = 2 + 5/8 + 1/2 * 4/ Bar = 2 + 5/8 + 1/2 * 4/ Bar = 2 + 5/8 + 1/2 * 4/ Avg Bars on Right: distance from distance from h right face left face Bar = 2 + 5/8 + 1/2 * 4/ = h - d_right Bar = 2 + 5/8 + 1/2 * 4/ Bar = * 5/8 + 1/2 * 4/8 4.5 Bar = 4 + 5/8 + 1/2 * 4/ Avg bar_size 4 A bar 0.20 in 2 A s_left_bars 0.60 in 2 = 3 * A bar A s_right_bars 0.80 in 2 = 4 * A bar (comp. on left) (comp. on right) d 5.47 in d 6.96 in d' 2.88 in d' 3.53 in A s 0.80 in 2 =A s_right_bars A s 0.60 in 2 =A s_left_bars A s ' 0.60 in 2 =A s_left_bars A s ' 0.80 in 2 =A s_right_bars b 32 M+ d 5.47 d' 2.88 A s 0.80 A s ' 0.60 M- d 6.96 d' 3.53 A s 0.60 A s ' 0.80 M n See Sheet "Calc_fMn_2" for calculation details M n M w k-ft =( M n+ + ABS( M n- ) ) / 2

7 Barrier 3 / 4 2. Mn about long. axis M c : only +'ve M (outside surface in compression) consider equivalent rectangular section of width b and depth h: b 12 in h 9.16 in b bar_size 5 s 9 in assume bar spacing A bar 0.31 in 2 same as bottom bars in slab bar in A s 0.41 in 2 = A bar * 12 / s h (comp. on left) M+ d 6.84 in = h - 2-1/2 * 5/8 d' 2.31 in = 2 + 1/2 * 5/8 A s 0.41 in 2 =A s A s ' 0.41 in 2 =A s b 12 M+ d 6.84 d' 2.31 A s 0.41 A s ' 0.41 M- d 0.00 d' 0.00 A s 0.00 A s ' 0.80 M n M n- #DIV/0! M C k-ft / ft

8 Barrier 4 / 4 3. Check Development Length inclined bar from barrier into slab: f' c 4 ksi f y 60 ksi Tension: L avail 9.8 in = (8-1) * 1.4 8" slab, 1" cover, bar angled at 45 degrees for straight bars L db 15 in 12 in 11.6 in = 1.25 * A bar * f y / SQRT(f' c ) 15 in = 0.4 * bar * f y max: 15 in for hooked bar L db 11.9 in 6 in 5 in = 8 * bar 11.9 in =38* bar /SQRT(f' c ) max: 11.9 in Reduce A s by L avail / (L db for straight bars) A s_reduc 0.27 in 2 = A s * L avail / L db Check Development Length of bars in Compression: L db 11.8 in 11.8 in = 0.63 * bar * f y / SQRT(f' c ) 11.3 in = 0.3 * bar *f y max: 11.8 in M C k-ft / ft with no reduction k-ft / ft with A s = 0.18 in k-ft / ft with A s = 0.18 in 2 and A s ' = 0 use M C 8.07 k-ft / ft (conservative) R W : L c ft = Lt / 2 + SQRT( (Lt /2 )^2 + (8 * H/12 * M w ) / M C ) R W 69.1 k = ( 2 / ( 2*L c - Lt ) ) * ( 8 * M w + M C * L c^2 / (H/12) ) OK, Rw > Ft = 54 k M at base of barrier from Rw, per foot M CT k-ft / ft = -(R W * H/12) / L c T 4.13 k / ft = R W / (L c + 2 * H/12)

9 Calc_fMn 1 / 2 h 8 bar top 5 f' c 4 ksi bar bot s top 6 f r 0.74 ksi =0.37 * SQRT( f' c ) s bot 10 cov top 2 f y 60 ksi cov bot 1 P ax 0 M n_ pos M n_ neg Mcr 9.47 A bar_top 0.31 in 2 A bar_bot 0.31 in 2 A s_top in 2 = A bar_top * 12 / s top A s_bot 0.37 in 2 = A bar_bot * 12 / s bot bar_top in bar_bot in d 6.69 in =h - cov bot - bar_bot / 2 d 5.69 in = h - cov top - bar_top / 2 d' 2.31 in = cov top + bar_top / 2 d' 1.31 in = cov bot + bar_bot / 2 A s 0.37 in 2 = A s_bot A s 0.62 in 2 = A s_top A s ' 0.62 in 2 = A s_top A s ' 0.37 in 2 = A s_bot find y c : Calc. y c from F = * f'c * 12 * 1 * y c - (.003 / y c * (y c - d')) * E s * A s ' + f y A s - P = 0 assuming tensile steel yields Use quadratic formula to solve for y c y = (-b +/- sqrt( b^2-4ac) / (2a) a = * f' c * 12 * 1 a = * f' c * 12 * 1 b -32 = A s * f y - P ax - A s ' * * b 5 = A s * f y - P ax - A s ' * * c 125 =A s ' * * * d' c 42 =A s ' * * * d' y c_try 1.49 in y c_try = ( -b - SQRT( b^2-4 * a * c)) / ( 2 * a) y c 1.49 y c 1.18 Internal Forces: a in = y c * 1 a in s ' = / y c * (y c - d') s ' s = / y c * d s

10 Calc_fMn 2 / 2 C k = 0.85 * f' c * a * 12 C fs' 47.6 ksi =MAX(-f y, MIN(f y, s ' * 29000) ) fs' 9.9 ksi =MAX(-f y, MIN(f y, * 29000) ) Fs' 29.5 k = fs' * A s ' Fs' 3.7 = fs' * A s ' fs 60.0 =MAX(-f y, MIN(f y, s * ) ) fs 60.0 =MAX(-f y, MIN(f y, s * ) ) T 22.3 k = A s * fs T 37.2 = A s * fs Sum F Sum F dif 0.00 dif 0.00 M n k-ft M n k-ft = -(C * (h/2 - a/2) + Fs' * (h/2 - d') + T * (h/2 - d) )/12 = (C * (h/2 - a/2) + Fs' * (h/2 - d') + T * (h/2 - d) )/ = IF( s <0.002,0.65,IF( s <0.005, * s,0.9)) 0.90 for Strength Limit States 0.90 for Extreme Event Limit State 0.90 M n Ductility: 1.2 M cr 9.47 k-ft = 1.2 * f r * 1/6 * 12 * h^2 / 12

11 Calc_fs_serv 1 / 2 h 8 in f' c 4 ksi bar top 5 in bar bot 5 in 2 f r 0.48 ksi = 0.24 * SQRT( f' c ) s top 6 in s bot 10 in f y 60 ksi cov top 2 in E c 3640 ksi = 1820 * SQRT(f' c ) cov bot 1 in n 7.97 = / E c M serv_pos 5.49 k-ft / ft M serv_neg k-ft / ft e_bot 0.75 s max_top 13.1 ksi e_top 1 s max_bot 12.1 ksi Check if tension in concrete > 0.8 fr: f ct 0.51 ksi = M serv_pos * 12 / (1/6 * 12 * h^2) f ct 0.61 ksi = -M serv_neg * 12 / (1/6 * 12 * h^2) f ct > 0.8 f r TRUE = IF( f ct > 0.8 * f r, TRUE, FALSE ) f ct > 0.8 f r TRUE = IF( f ct > 0.8 * f r, TRUE, FALSE ) find y c : A top_bar 0.31 in 2 A bot_bar 0.31 in 2 A s_top in 2 = A top_bar * 12 / s top A s_bot in 2 = A bot_bar * 12 / s bot bar_top in bar_bot in d 6.69 in =h - cov bot - bar_bot / 2 d 5.69 in = h - cov top - bar_top / 2 d' 2.31 in = cov top + bar_top / 2 d' 1.31 in = cov bot + bar_bot / 2 A s 0.37 in 2 = A s_bot A s in 2 = A s_top A s ' 0.62 in 2 = A s_top A s ' 0.37 in 2 = A s_bot Calc. y c from F = 0-1/2 * K * y c * E c *y c * 12 - K * (y c - d') * E s * A s ' + K * (d - y c ) * E s * A s = 0 divide through by K * E c, let n = E s / E c -1/2 * 12 y c 2 - n * A s ' * (y c - d') + n * A s * (d - y c ) = 0 Use quadratic formula to solve for y c y = (-b +/- sqrt( b^2-4ac) / (2a) a -6.0 = -6 a -6.0 = -6 b -8 = -n * (A s ' + A s ) b -8 = -n * (A s ' + A s ) c 31 = n * (A s ' * d' + A s * d) c 32 = n * (A s ' * d' + A s * d) y c 1.72 in y c 1.74 = ( -b - SQRT( b^2-4 * a * c)) / ( 2 * a) Sum F k = a * y c^2 + b * y c + c Sum F k = a * y c^2 + b * y c + c

12 Calc_fs_serv 2 / 2 Calc. s max: I trans 95 in 4 I trans 99 in 4 = 1/12 * 12 * y c^ * y c * (y c /2)^2 + n * A s ' * (y c - d')^2 + n * A s * (d - y c )^2 M service 5.49 k-ft / ft M service 6.52 k-ft / ft f ss 27.4 ksi = n * M service * 12 * (d - y c ) / I trans f ss 24.9 ksi = n * M service * 12 * (d - y c ) / I trans e 0.75 = e_bot e 1.00 bar_bot 0.63 in bar_top 0.63 in d c 1.31 in = cov bot + bar_bot / 2 d c 2.31 in = cov top + bar_top / 2 s 1.31 = 1 + d c / (0.7 * (h - d c )) s 1.58 = 1 + d c / (0.7 * (h - d c )) s max_top 12.1 in = 700 * e / ( s * f ss ) - 2 * d c s max_top 13.1 in = 700 * e / ( s * f ss ) - 2 * d c s max_bot 12.1 in = IF( f ct > 0.8 f r, s max_top, 18 ) s max_top 13.1 in = IF( f ct > 0.8 f r, s max_top, 18 )

13 Influ_+M 1 / 2 overhangs 3.25 ft span 8.00 ft I for M due Slab and Barrier: I for M due Wheel Loads: I barrier ft I slab ft 2 one truck two trucks max I x_left_wheel space between trucks 9 ft in ft ft ft ft x_model D_model x I I_1_truck I_2_trucks N edge overhang cg of barrier rail eft-most wheel pos support N N N N n3a N N N N N N N N N N N N N N N N N N N N N N N N N N N N

14 Influ_+M 2 / 2 W_out ft W_in 40 ft min_s 4 ft max_s 24 ft =INT( W_in - ( 2*(overhangs+2) ) ) Slope_L Slope_R _slope E-02 rad 6.70E-02 rad E-01 rad I, ft M, k ft Influence for M at.35l of Ext. Span N E-02 N E-02 N E N E-02 N E-02 N3AL E-02 N3AR E-02 N E-02 N E-02 N E N E N E N E-02 N10L E-02 N10R E-02 N E-02 N E-02 N E-03 N E-04 N E-03 N16 M due 0Wheels at 0.35L 0 of 0Ext. Span0-7.66E-03 N E-03 N E-03 N E N E-03 N E N E-03 N E-03 N E-04 N E-03 N Truck 2.05E-03 N E-03 N Trucks 2.61E-03 N E-03 N E N E-03 N E-04 N E-04

15 Influ_-M 1 / 2 overhangs 3.25 ft span 8.00 ft I for M due Slab and Barrier: I for M due Wheel Loads: I barrier ft I slab ft 2 one truck two trucks max I x_left_wheel space between trucks 4.00 ft in ft ft ft ft X, Model, model x I I_one_truck_two_truck N edge overhang cg of barrier rail eft-most wheel pos support N N N N n3a N N N N N N N N N N N N N N N N N N N N N N N N N N N N N N N

16 Influ_-M 2 / 2 W_out ft W_in 40 ft min_s 4 ft max_s 24 ft =INT( W_in - ( 2*(overhangs+2) ) ) Slope_L Slope_R _slope -1.07E-02 rad 9.31E-03 rad 2.01E-02 rad ks N E-03 N E-03 N E-03 N E-03 N E-03 N3AL E-03 N3AR E-03 N E-03 N E-03 N E N E N E N E-03 N10L E N10R E N E N E-03 N E-03 N E-04 N E-03 N16 0M due Wheels E N E N E-03 N E N E-03 N E N E N E N E-05 N E N E N Truck 8.70E-04 N E N Trucks 8.47E N E N E N E-04 N E-04 N E-05 N E-04 N E-04 I, ft M, k ft Influence for M at 1st Interior Support

Design of a Balanced-Cantilever Bridge

Design of a Balanced-Cantilever Bridge CL (Bridge is symmetric about CL) 0.8 L 0.2 L 0.6 L 0.2 L 0.8 L L = 80 ft Bridge Span = 2.6 L = 2.6 80 = 208 Bridge Width = 30 No. of girders = 6, Width of each girder