Download presentation

Presentation is loading. Please wait.

Published byDylan Ruiz Modified over 3 years ago

1
1 The following is an example of calculations performed to determine the time dependent camber values and prestress losses for a BT84 precast concrete beam with a span length of 150 feet. The beam has 56 strands including 28 deflected strands. The assumed beam spacing is 6-0. An 8 deck with 2 of build-up was assumed. Prestressed Concrete Beam Camber – BT84

2
2 Compute upward deflection at release of strand Assume that at release that the modulus of elasticity is 0.70 times the final value. EC = (0.70) 33000K1wC1.5 fC (AASHTO 5.4.2.4-1) = (0.70)(33000)(1.0)(0.145)1.5(7.00) = 3374 ksi I = 710000 in4 A = 772 in4 yb = 42.37 y C = 4.68 ye = 22.37 w = 0.06925 k/in

3
3 Prestressed Concrete Beam Camber – BT84 Compute initial stress loss Total jacking force = (31 kips/strand)(56 strands) = 1736 kips Start with 15 ksi loss in prestress at release Strand stress at release = 31/0.153 – 15.0 = 188 ksi Prestress force at release = (188)(0.153)(56) = 1611 kips Prestress moment at midspan at release = (1611)(42.37 – 4.68) = 60735 inch kips

4
4 Prestressed Concrete Beam Camber – BT84 Maximum moment from beam weight = (0.06925)(1800)2/8 = 28046 inch kips Net maximum moment = 60735 – 28046 = 32689 inch kips fcgp = Concrete stress at strand cg = 1611 + (32689)(42.37 – 4.68) = 3.82 ksi 772 710000 Strain in concrete at strand cg = (3.82)/(3374) = 0.0011322 Stress loss in strand = (0.0011322)(28500) = 32.27 ksi Recompute initial stress loss in strand assuming 30 ksi initial stress loss

5
5 Prestressed Concrete Beam Camber – BT84 Assume 30 ksi loss in prestress at release Strand stress at release = 31/0.153 – 30.0 = 173 ksi Prestress force at release = (173)(0.153)(56) = 1482 kips Prestress moment at midspan at release = (1482)(42.37 – 4.68) = 55856 in. kips Maximum moment from beam weight = (0.06925)(1800)2/8 = 28046 inch kips Net maximum moment = 55856 – 28046 = 27810 inch kips fcgp = Concrete stress at strand cg = 1482 + (27810)(42.37 – 4.68) = 3.39 ksi 772 710000 Strain in concrete at strand cg = (3.39)/(3374) = 0.001005 Stress loss in strand = (0.001005)(28500) = 28.64 ksi Assume an initial loss of 29 ksi

6
6 Prestressed Concrete Beam Camber – BT84 Use 174 ksi stress in strand, 1491 kips, and a strain value of 0.001005 Use the Moment-Area method to determine the upward deflection of the beam at release. Compute the moment in the beam induced by the prestressed strands. See the following drawings that are used in deflection calculations.

7
7 Prestressed Concrete Beam Camber – BT84

8
8 Use the Moment Area method to determine the deflection at midspan from the prestress force at release. The deflection is equal to the moment of the area of the M/EI diagram between the end of the beam and midspan about the end support. The moment of the area between midspan and the end of the beam about the end of the beam = (720)(0.000012448)(720/2) + (360/2)(0.000023459)(720+360/4) + (0.000023459 – 0.000012448)(720/2)(2/3)(720) = 8.55 This is the upward deflection caused by the load from the prestress strand not including the weight of the beam.

9
9 Prestressed Concrete Beam Camber – BT84 The downward deflection of the beam from self weight = 5wL 4 384EI w = 0.06925 kips/inch, L = 1800 inches, E = 3374 ksi, I = 710000 in4 = (5)(0.06925)(1800)4 = 3.95 inches (384)(3374)(710000) The net upward deflection at midspan at release = 8.55-3.95 = 4.60 inches

10
10 Prestressed Concrete Beam Camber – BT84 Compute the upward deflection 3 months after release. The creep coefficient is determined from AASHTO formula 5.4.2.3.2-1, Ψ(t, ti) = 1.9 ks khc kf ktd ti -0.118 k s = 1.45 – 0.13(V/S) 1.0 V/S is the volume to surface ratio = 772 = 2.67 288.8 ks = 1.45 – 0.13(2.67) = 1.10

11
11 Prestressed Concrete Beam Camber – BT84 khc = 1.56 – 0.008H, H = 70 (From AASHTO Figure 5.4.2.3.3-1) khc = 1.56 – (0.008)(70) = 1.0 kf = 5 = 5 = 0.625 1+fc 1 + 7.0 ktd = t = 90 = 0.732 61 – 4fc + t 61 – (4)(7.0) + 90 ti = 1.0 days

12
12 Prestressed Concrete Beam Camber – BT84 Ψ(t, ti) = (1.90)(1.10)(1.0)(0.625)(0.732)(1.0) -0.118 = 0.956 Creep deflection = (0.956)(4.60) = 4.40 inches Compute strand stress loss due to creep. ΔfpCR = Ep fcgp Ψ(t, ti) Kid (5.9.5.4.2b-1) Eci Kid = 1. 1 + EpAps [1+ Age 2 pg ][1 + 0.7Ψ b (t f, t i )] EciAg Ig

13
13 Prestressed Concrete Beam Camber – BT84 Ep = 28500 ksi Aps = (56)(0.153) = 8.57 in2 Eci = 3374 ksi Ag = 772 in2 epg = 37.69 in Ig = 710000 in4 Ψb = 0.956 fcgp = 3.39 ksi

14
14 Prestressed Concrete Beam Camber – BT84 Kid = 1 = 0.715 1 + (28500)(8.57) [1+ (772)(37.69) 2 ][1 + (0.7)(0.956)] (3374)(772) 710000 ΔfpCR = (28500)(3.39)(0.956)(0.715) = 19.57 ksi (3374) Compute strand stress loss from shrinkage fpSR = εbidEpKid εbid = ks khs kf ktd 0.48x10 -3 (AASHTO 5.9.5.4.2a, 5.4.2.3.3-1)

15
15 Prestressed Concrete Beam Camber – BT84 k s = 1.45 – 0.13(2.67) = 1.10 k hs = 2.00 – 0.014H, H=70, khs = 2.0 – (0.014)(70) = 1.02 k f = 5 = 5 = 0.625 1+fc 1 + 7.0 k td = t = 90 = 0.732 61 – 4fc + t 61 – (4)(7.0) + 90 ε bid = (1.10)(1.02)(0.625)(0.732)(0.48x10 -3 ) = 0.0002464 f pSR = (0.0002464)(28500)(0.715) = 5.02 ksi Total prestress loss at 90 days from creep and shrinkage = 19.57 + 5.02 = 24.59 ksi

16
16 Prestressed Concrete Beam Camber – BT84 Downward deflection from creep and shrinkage loss = (8.55)(24.59/174) = 1.21 inches Strand stress loss from relaxation (AASHTO C5.9.5.4.2c-1) = f pR1 = f pt log(24t) [(f pt /f py ) -0.55] [1- 3(Δf pSR + Δf pCR ) ]K id K L log(24t 1 ) fpt f pt = stress after transfer = 174ksi t = 90 days K L = 45 t i = 1.0 days f py = (0.90)(270) = 243 ksi Δf pSR = Stress loss in strand from shrinkage = 5.02 ksi Δf pCR = Stress loss in strand from creep = 19.57 ksi K id = 0.715

17
17 Prestressed Concrete Beam Camber – BT84 f pR1 = (174) log [(24)(90)] [(174/243) – 0.55] [1 – 3(5.02 + 19.57)](0.715) = 0.639 ksi (45) log [(24)(1.0)] 174 Downward deflection due to relaxation in strand = (0.639)(8.55) = 0.03 inches (174) Total downward deflection due to strand stress loss = 1.21 + 0.03 = 1.24 Total upward deflection at 3 months = 4.60 + 4.40 – 1.24 = 7.76 inches Compute downward deflection due to deck = 5wL 4 384EI

18
18 Prestressed Concrete Beam Camber – BT84 Assumed beam spacing = 6-0 Deck thickness = 8 Assumed build-up = 2 Sectional area of deck = (0.667)(6.0) + (0.167)(4.0) = 4.67 ft2 Deck weight per foot = (0.150)(4.67) = 0.700 kips/ft = 0.05833 kips/inch L = 150 feet = 1800 inches EC = 33000 K1 wc 1.5 fc = (33000)(1.0)(0.145) 1.5 7.00 = 4821 ksi w = 0.0583 kips/inch

19
19 Prestressed Concrete Beam Camber – BT84 I = 710,000 in 4 Δ = (5)(0.0583)(1800) 4 = 2.33 inches (384)(4821)(710000) Compute strand stress gain from the deck load Moment from deck = (0.0583)(1800) 2 = 23611 inch kips (8) Concrete stress = (23611)(37.69) = 1.253 ksi (710000) Strain in concrete = 1.253 = 0.0002599 4821 Stress gain in strand = (28500)(0.0002599) = 7.41 ksi Upward deflection from strand stress gain = (8.55)(7.41) = 0.36 inches (174)

20
20 Prestressed Concrete Beam Camber – BT84 Determine deflections and strand stress losses after deck is in place Compute the transformed moment of inertia ( I ) for the beam with the deck in place. Assume deck concrete strength = 4.00 ksi. E beam = 4821 ksi E deck = 33000K1WC 1.5 f C = (33000)(1.0)(0.145) 1.5 (4.00) = 3644 ksi n = E beam = 4821/3644 = 1.32 E deck Effective deck width = 6.00/1.32 = 4.55 feet = 54.6 inches

21
21 Prestressed Concrete Beam Camber – BT84 Distance to composite centroid from bottom of beam = (772)(42.37) + (54.60)(8.00)(88.00) = 58.86 inches 772 + (54.60)(8.00) Composite Moment of Inertia = IC = 710,000 + (772)(58.86 – 42.37)2 + (54.60)(8.00)3 + (8.00)(54.60)(88.00 – 58.86)2 = 12 1,293,156 in4

22
22 Prestressed Concrete Beam Camber – BT84 Total transformed area = 772 + (8)(54.6) = 1209 in 2 Strand cg to composite girder cg = 58.86 – 4.68 = 54.18 inches Compute creep deflection five years beyond release of prestress The creep factor to be applied to deflection at release = Ψ b (t f, t i ) - Ψ b (t d, t i ) The creep factor to be used for deflection from deck = Ψ b (t f, t d )K df Ψ(t f, t i ) = 1.9 k s k hc k f k td t i -0.118 k s = 1.10 k hc = 1.0 k f = 0.625

23
23 Prestressed Concrete Beam Camber – BT84 k td = t = 1825 = 0.98 61 – 4f c + t 61 – (4)(7.0) + 1825 t i = 1 Ψ(t f, t i ) = (1.9)(1.10)(1.0)(0.625)(0.98)(1.0) -0.118 = 1.28 Ψ(t d, t i ) = 1.9 k s k hc k f k td t i -0.118 k s = 1.10 k hc = 1.0 k k = 0.625 k td = t = 90 = 0.73 61 – 4f c + t 61 – (4)(7.0) + 90

24
24 Prestressed Concrete Beam Camber – BT84 ti = 1 Ψ(t d, t i ) = (1.9)(1.10)(1.0)(0.625)(0.73)(1.0) -0.118 = 0.95 Ψ b (t f, t i ) - Ψ b (t d, t i ) = 1.28 – 0.95 = 0.33 Compute the creep applied to deck deflection at five years beyond release of prestress. Base this deflection on an elastic deflection assuming a full composite moment of inertia at time of application of deck load. = 5wL 4 384EI = (5)(0.05833)(1800) 4 = 1.28 inches (384)(4821)(1293156)

25
25 Prestressed Concrete Beam Camber – BT84 Apply creep factor Ψ(t f, t d ) = 1.9 k s k hc k f k td t i -0.118 k s = 1.45 – 0.13(V/S) V = 773 in 3 /in, S = 289 in 2 /in, V/S = 773/289 = 2.67 k s = 1.45 – 0.13(2.67) = 1.10 k hc = 1.0 k f = 0.625 k td = t = 1825 = 0.982 61 – 4f c + t 61 – (4)(7.0) + 1825

26
26 Prestressed Concrete Beam Camber – BT84 ti = 90 days Ψ(tf, td) = (1.90)(1.10)(1.0)(0.625)(0.982)(90)-0.118 = 0.754 Creep deflection from deck = (0.754)(1.28) = 0.96 inches Compute upward deflection of beam from creep applied to upward at release of prestress. As with deck deflection, base this deflection on an elastic deflection assuming a full composite moment of inertia at time of release. Deflection at release based on non composite moment of inertia = 4.60. Deflection if beam were composite = (4.60)(710000) = 2.52 (1293156) Creep deflection from release of prestress = (0.33)(2.52) = 0.83

27
27 Prestressed Concrete Beam Camber – BT84 Compute downward deflection from strand stress loss due to creep and shrinkage. From AASHTO 5.9.5.4.3a-1, the strand stress loss from shrinkage after deck placement = f pSD = ε bdf E p K df K df = 1. 1 + E p A ps [1+ A c e 2 pc ][1 + 0.7Ψ b (t f, t i )] E ci A c I c Ψ b (t f, t i ) = 1.9 k s k hc k f k td t i -0.118 ti = 90 k s = 1.10 khs = 1.02 kf = 0.625

28
28 Prestressed Concrete Beam Camber – BT84 k td = t = 1825 = 0.982 61 – 4f c + t 61 – (4)(7.0) + 1825 Ψ b (t f, t i ) = (1.9)(1.1)(1.02)(0.625)(0.982)(90) -0.118 = 0.77 K df = 1 = 0.743 1 + (28500)(8.57) [1+ (1209)(54.18) 2 ][1 + (0.7)(0.77)] (3375)(1209) (1293156) ε bdf = k s k hs k f k td 0.48x10 -3 = (1.1)(1.02)(0.625)(0.982)(0.48x10 -3 ) = 0.0003305 f pSD = (0.0003305)(28500)(0.743) = 7.00 ksi

29
29 Prestressed Concrete Beam Camber – BT84 Compute strand stress loss from creep from deck pour to 5 years after release f pCD = E p f cgp [Ψ b (t f, t i ) - Ψ b (t d, t i )]K df + E p f cd Ψ b (t f, t d )K df E ci E c Ψ b (t d, t i ) = (1.90)(1.10)(1.0)(0.625)(0.732)(1.0 -0.118 ) = 0.956 Ψ b (t f, t i ) = 1.9 k s k hc k f k td ti- 0.118 t i = 1 k s = 1.0 k hc = 1.0 k f = 0.625

30
30 Prestressed Concrete Beam Camber – BT84 k td = t = 1825 = 0.982 61 – 4f c + t 61 – (4)(7.0) + 1825 Ψ b (t f, t i ) = (1.9)(1.10)(1.0)(0.625)(0.982)(1) -0.118 = 1.283 Ψ b (t f, t d ) = 1.9 k s k hc k f k td t i -0.118 k td = t = 1735 = 0.981 61 – 4f c + t 61 – (4)(7.0) + 1735 Ψ b (t f, t d ) = (1.9)(1.1)(1.0)(0.625)(0.981)(90) -0.118 = 0.754

31
31 Prestressed Concrete Beam Camber – BT84 Ψ b (t f, t d ) = (1.9)(1.1)(1.0)(0.625)(0.981)(90) -0.118 = 0.754 E p = 28500 ksi E ci = 3374 ksi E c = 4821 ksi f cgp = 3.39 ksi f cd = Change in concrete stress at strand cg due to losses between transfer and deck placement and stress loss from deck placement Total strand losses between transfer and deck placement = 19.57 + 5.02 + 0.639 = 25.23 ksi

32
32 Prestressed Concrete Beam Camber – BT84 Concrete stress loss from prestress loss = 25.23 + (25.23)(0.153)(56)(42.37-4.68)(42.37-4.68) = 0.465 ksi 772 710000 Stress loss in concrete at strand cg from deck placement = My I M = wL 2 = (0.05833)(1800) 2 = 23624 inch kips 8 8 y = 37.69 inches, I = 710000 in 4 Stress loss in concrete from deck = (23624)(37.69) = 1.25 ksi (710000) fcd = 0.465 + 1.25 = 1.71 ksi

33
33 Prestressed Concrete Beam Camber – BT84 K df = 1. 1 + E p A ps [1+ A c e 2 pc ][1 + 0.7Ψ b (t f, t i )] E ci A c Ic K df = 1 = 0.701 1 + (28500)(8.57) [1+ (1209)(54.18)2][1 + (0.7)(1.283)] (3374)(1209) 1293156 f pCD = (28500) (3.39)[1.283 – 0.956](0.701) + (28500) (1.71)(0.754)(0.701) = 11.91 ksi (3374) (4821) Total stress loss from deck poor to 5 years = 7.00 + 11.91 = 18.91 ksi

34
34 Prestressed Concrete Beam Camber – BT84 On the non composite section, this stress loss would result in the following deflection: = (18.91)(8.55) = 0.93 inches 174 With the composite section, = (0.93)(710000) = 0.51 inches (1293156) The total upward deflection between the time of the deck pour and 5 years = -0.96 + 0.83 -0.51 = -0.64 inches

35
35 Prestressed Concrete Beam Camber – BT84 Compute shortening 2 weeks after transfer of prestress Compute creep coefficient at 2 weeks (14 days) Ψ(t, t i ) = 1.9 k s k hc k f k td t i -0.118 ks = 1.45 – 0.13(2.67) = 1.10 khc = 1.56 – 0.008H, H = 70 (From AASHTO Figure 5.4.2.3.3-1) khc = 1.56 – (0.008)(70) = 1.0 kf = 5 = 5 = 0.625 1+f c 1 + 7.0 k td = t = 14 = 0.30 61 – 4f c + t 61 – (4)(7.0) + 14

36
36 Prestressed Concrete Beam Camber – BT84 t i = 1.0 Ψ(t, t i ) = (1.90)(1.10)(1.0)(0.625)(0.30)(1.00) -0.118 = 0.39 Shortening at release = (1491)(1800) = 1.03 inches (772)(3374) Shortening from creep = (1.03)(0.39) = 0.40 inches Compute shrinkage coefficient at 2 weeks (14 days) ε sh = k s k hs k f k td 0.48x10 -3 k s = 1.45 – 0.13(2.67) = 1.10 k hs = 2.00 – 0.014H, H=70, k hs = 2.0 – (0.014)(70) = 1.02

37
37 Prestressed Concrete Beam Camber – BT84 k f = 5 = 5 = 0.625 1+f c 1 + 7.0 k td = t = 14 = 0.30 61 – 4f c + t 61 – (4)(7.0) + 14 ε sh = (1.10)(1.02)(0.625)(0.30)(0.48x10 -3 ) = 0.00010098 Shortening from shrinkage = (1800)(0.00010098) = 0.18 inches Neglect effects of strand relaxation Total shortening at 2 weeks = 1.03 + 0.40 + 0.18 = 1.61 inches

38
38 Prestressed Concrete Beam Camber – BT84 Compute total stress loss at 27 years The total loss in strands up to the time of deck pour (90 days) = 29 + 19.57 + 5.02 + 0.64 = 54.23 ksi Strand stress gain from deck pour = 7.41 ksi Strand stress loss after deck pour = 54.23 – 7.41 = 46.82 ksi Strand stress loss due to shrinkage between time of deck pour and 27 years = f pSD = ε bdf E p K df K df = 1. 1 + E p A ps [1+ A c e 2 pc ][1 + 0.7Ψ b (t f, t i )] E ci A c I c

39
39 Prestressed Concrete Beam Camber – BT84 Ψ b (t f, t i ) = 1.9 k s k hc k f k td t i -0.118 k s = 1.45 – 0.13(V/S) V = 773 in 3 /in, S = 289 in 2 /in, V/S = 773/289 = 2.67 k s = 1.45 – 0.13(2.67) = 1.10 k hs = 1.02 k f = 0.625 k td = t = 9855 = 1.00 61 – 4f c + t 61 – (4)(7.0) + 9855 t i = 90.0

40
40 Prestressed Concrete Beam Camber – BT84 Ψ b (t f, t i ) = (1.90)(1.10)(1.02)(0.625)(1.00)(1) -0.118 = 1.33 K df = 1 = 0.805 1 + (28500)(8.57) [1+ (1209)(54.18) 2 ][1 + (0.7)(0.78)] (4821)(1209) 1293156 ε bdf = k s k hs k f k td 0.48x10 -3 ε bdf = (1.1)(1.02)(0.625)(1.00)(0.48x10 -3 ) = 0.000336 f pSD = (0.000336)(28500)(0.805) = 7.71 ksi Compute loss in strand from creep from time of deck pour to 27 years f pCD = E p f cgp [Ψ b (t f, t i ) - Ψ b (t d, t i )]K df + E p f cd Ψ b (t f, t d )K df E ci E c

41
41 Prestressed Concrete Beam Camber – BT84 E p = 28500 ksi E ci = 4821 ksi f cgp = 3.39 ksi Ψ b (t f, t i ) = 1.9 k s k hc k f k td t i -0.118 Ψ b (t f, t i ) = (1.90)(1.10)(1.02)(0.625)(1.00)(1)-0.118 = 1.33 Ψ b (t d, t i ) = 1.9 k s khc k f k td t i -0.118 k td = t = 90 = 0.73 61 – 4f c + t 61 – (4)(7.0) + 90

42
42 Prestressed Concrete Beam Camber – BT84 Ψ b (t d, t i ) = (1.90)(1.1)(1.0)(0.625)(0.73)(1.0) -0.118 = 0.95 Ψ b (t f, t d ) = 1.9 k s k hc k f k td t i -0.118 t i = 1 k td = t = 9765 = 1.00 61 – 4f c + t 61 – (4)(7.0) + 9765 Ψ b (t f, t d ) = (1.90)(1.1)(1.0)(0.625)(1.00)(1) -0.118 = 1.31 Compute f cd Strand stress loss between time of transfer and deck pour = 46.82 ksi

43
43 Prestressed Concrete Beam Camber – BT84 Prestress force loss = (46.82)(56)(0.153) = 401 kips Moment loss = (401)(37.69) = 15114 inch kips Stress loss in concrete = 401 + (15114)(37.69) = 1.32 ksi 772 (710000) Moment from deck = 23611 inch kips Concrete stress loss at strand cg = (23611)(37.69) = 1.25 ksi (710000) f cd = 1.32 + 1.25 = 2.57 ksi f pCD = E p f cgp [Ψ b (t f, t i ) - Ψ b (t d, t i )]K df + E p f cd Ψ b (t f, t d )K df E ci E c

44
44 Prestressed Concrete Beam Camber – BT84 f pCD = (28500)(3.39)[1.33 – 0.95](0.777) + (28500)(2.57)(1.31)(0.805) = 24.78 ksi (3374) (4821) Total prestress loss = 46.82 + 7.71 + 24.78 = 79.31 ksi Compute strand stress gain from deck shrinkage f pSS = E p f cdf K df [1+0.7Ψ b (t f, t d )] E c f cdf = ε ddf A d E cd ( 1 - epc ed ) [1+0.7 Ψ d (t f, t d )] A c I c ε ddf = k s k hs k f k td 0.48x10 -3 k s = 1.45 – 0.13(V/S) 1.0

45
45 Prestressed Concrete Beam Camber – BT84 V = (72.00)(8.00) = 576.0 in 3 /in S = (72.00)(2) + (8.00)(2) = 160.0 in 2 /in V/S = 436.80/125.20 = 3.60 k s = 1.45 – 0.13(3.60) = 0.982 k hs = 1.02 k f = 5 = 5 = 1.0 1+ f ci 1 + 4.0 k td = t = 9765 = 0.67 61 – 4f c + t 61 – (4)(4.0) + 9765 ε ddf = (0.982)(1.02)(1.0)(1.0)(0.48x10 -3 ) = 0.000478

46
46 Prestressed Concrete Beam Camber – BT84 A d = (72.00)(8.00) = 576 in 2 E cd = 33000K1WC 1.5 fC = (33000)(1.0)(0.145) 1.5 (4.00) = 3644 ksi Ψ b (t f, t d ) = 1.9 k s k hc k f k td t i -0.118 Ψ b (t f, t d ) = (1.9)(1.1)(1.0)(0.625)(1.0)(1.0) -0.118 = 1.31 Ψ d (t f, t d ) = (1.90)(0.982)(1.2)(1.0)(1.0)(1.0) -0.118 = 1.90 A c = 1209 in 2 I c = 1293156 in 4

47
47 Prestressed Concrete Beam Camber – BT84 e pc = 54.18 inches e d = 84 + 8/2 – 58.86 = 29.14 inches f cdf = (0.0004704)(576)(3644)( 1 - (54.18)(29.14) ) = -0.167 [1+(0.7)(1.90)] (1209) (1293156) f pSS = 28500 (-0.167)(0.805) [1+(0.7)(1.31)] = -1.52 ksi 4821 Total Stress Loss @ 27 Years = 79.31-1.52 = 77.79 ksi

48
48 Prestressed Concrete Beam Camber – BT84

Similar presentations

Presentation is loading. Please wait....

OK

Add Governors Discretionary (1G) Grants Chapter 6.

Add Governors Discretionary (1G) Grants Chapter 6.

© 2018 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on javascript events w3school Ppt on machine translation online Ppt on agriculture insurance in india Ppt on regional trade agreements definition Ppt on bluetooth network security Ppt on wind and the sun Ppt on indian entertainment and media industry Ppt on business environment nature concept and significance of study Ppt on national parks in india Ppt on external gear pump