- A Simply Supported Pretensioned Aashto Bridge Girder Of Normalweight Concrete Is To Be Designed The Beam Profile Is Show 1 (60.03 KiB) Viewed 48 times
- A Simply Supported Pretensioned Aashto Bridge Girder Of Normalweight Concrete Is To Be Designed The Beam Profile Is Show 2 (40.46 KiB) Viewed 48 times
- A Simply Supported Pretensioned Aashto Bridge Girder Of Normalweight Concrete Is To Be Designed The Beam Profile Is Show 3 (38.7 KiB) Viewed 48 times
- A Simply Supported Pretensioned Aashto Bridge Girder Of Normalweight Concrete Is To Be Designed The Beam Profile Is Show 4 (31.91 KiB) Viewed 48 times
- A Simply Supported Pretensioned Aashto Bridge Girder Of Normalweight Concrete Is To Be Designed The Beam Profile Is Show 5 (31.57 KiB) Viewed 48 times
12 12 16 11 28 15 36 19 45 17 / 16 7 18 Land Type 1 35 to 45 ft Type II 40 to 50 ft Type III 55 to 80 ft 20 42 42 8 BR 23 54 33 63 42 72 8 10 m 10 + -26 28- Type IV 70 to 100 ft Type V 90 to 120 ft Type VI 110 to 140 ft Section Properties of AASHTO Bridge Girders A Type in. I. in. in. C in. C2 in. ? in. 52 Wo plf S in? in 1 II III 28 36 45 54 63 72 IV 276 369 560 789 1013 1085 22,750 50,979 125,390 260,741 521,180 733,320 15.41 20.17 24.73 29.27 31.04 35.62 12.59 15.83 20.27 24.73 31.96 36.38 82 138 224 330 514 676 288 384 583 822 1055 1130 176 1807 ZSZ 3220 5070 6196 8908 10544 16741 16307 20587 20157 VI
FAG Deng Given : fc', fel 1 Live load dead load (not including cum voright) 7 Beam with constant eccentridity cover didtonce de required, R, Spa Spg, and stand dantels Brom with deflected tendons Bromwikotidle takticky - L fti = for fes bife fei 0.7 foi fcs = -0.15 fe' fai 6 1 = fts alfa VE fel = 0.6fi fos -0.45 FC .6 fti fts - AVE 1 : fei - 0.6 foi fos: -0.45 z. 1 * Make an estimate of the own weight lo # Compute Mo, Me w Mo: 0,178 LI: al LA L M: 01/4 G L ar M ur148 *** ME Mo + M.Qa Md + ML 3. Si, required ME Rfti fes Si, sured. M - RM Rfu - fes Rfi - Sz, required. Si, equired. - RM Se, respied Me - & Me fts - Refei Sz, maquirá. M- its - Refri where Motor de L-L')
- Canding table chain standard, valolle 16cm w ste salat CHON available that satisfies the commodo cents, such 31, attual Sud, Note that simland 51. mimi a torly the same Saschal Samord? beam should be adopted (antes availability requiremenila diletate etter asymetric For the section, chosen + having Ac (n!) calculate (6/81 As 150 (assuming normal weight nike recalculate Mr Mo + Md. and me of deflected tendonu Go back to and check the steps and find Sound Sy maled 6, actual for your chestn"stchen > i equined and similarily Szaluual > Srequired # If not chasse a bigger section - W - Mo - Me and if deflected tenda) y el and back to step on HAVE SECTION IN HAND Ic, Ic/A, 41,92, Siectal Su actual Mo + 1 5. + 2 e 1 r² [ R fri + (ME - Its) (2 - ME - fts) - RC fti 52 [R(-fo] G1-fu) - ( - RG (fritt CE r [R()+(-:)] (1 - fes) - R5 (here
Find do = ₂ - c 22/F1 If de dc required : Design pre coure 1 : e is correct 2 *e is R(1 + (-fes + TA Sz salve for P2 IF de < de regulired Declaration wrong (too big required Use e - e emak = cz - demoised * R(1 + Emas (2/1²) (-fts + Ac = solve for Pi . 0.82fpy fpi frollo emin C 0.74 fpu Thus, the required steel area is Apyran Number of strands required is area of single stand Pi fpi Ap, ro a → round upward to a whole number to get total number of strands that will be used Unless ducts are used round apward sathat equal number of strands in each duct. - Ap, actual Ap. Calculate Pe = RP Spe Pe Ap & make sure fpe fpu > 0.5