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1、 Appendix K Design Examples K-1 Example 1* Two-Span I-Girder Bridge Continuous for Live Loads (a) Bridge Deck The bridge deck reinforcement using A615 rebars is shown below. * Based on an example prepared as part of “Training Classes on AASHTO LRFD Bridge Specifications,” ODOT, R.A. Miller AASHTO Ty
2、pe IV I girder Zero Skew Effective Width = 96” K-2 Redesign by using A1035 bars with fy = 100 ksi. 3,483(12) = 0.90As(100) 58.25 As(100) 1.7(7.0)(26) ; As= 8.36in2 Provide 28 # 5 bars (14 on the top with 2.5” cover and 14 on the bottom with 2 5/8” cover). The bars will be at 8” o.c. Note: A1035 bars
3、 are not epoxy coated; hence, the cover is 2.5”. The centroid of the bars from the top is x = 14 (2.5+ 0.55/ 8)+14(8.5 (2+5/ 8+ 0.55/ 8) 28 = 4.12“ Crack control reinforcement: e = 0.75 fs =Tensile stress in steel reinforcement at the service limit state. d = 62.5 4.12 = 58.4“ = As bd = 28 0.31 (26)
4、(58.4) = 0.0057 k =2n+ (n)2n =2(0.0057)(5.718)+ (0.0057 5.718)2 0.0057 5.718 = 0.225 j =1 k / 3=1 0.225/ 3= 0.925 fs= Msl Asjd = 2,141(12) 28 0.31(0.92558.4) = 56.6ksi 60ksi fs= 56.6ksi s 700 e sfs 2dc= 700 0.75 1.067 56.6 2 2.81= 3.07“ This spacing is too small. Aim for 6” spacing, which is more re
5、alistic. Try: 17 top bars: # 5 12” alternating with # 6 12” 17 bottom bars: # 5 12” alternating with # 6 12” K-3 Distances to centroid of bars: Top Bars: (2.5+ 0.55/ 8)+ (2.5+ 0.5 3/ 4) 2 = 2.84“ Bottom Bars: (8.5 (2+5/ 8+ 0.55/ 8)+ (8.5 (2+5/ 8+ 0.5 3/ 4) 2 = 5.53“ x = 17 2.84+17 5.53 34 = 4.19“ 6”
6、 O.K. (b) Shear Reinforcement If prestressing steel is ignored, #4 A615 stirrups 4” o.c. will be needed. Redesign by using A1035 U shaped # 4 stirrups. s = Avfydvcot Vs = 0.4(100)(55.63)cot45 284.6 = 7.8“ Controls, say 7” 5.8.2.5 5.8.2.9 5.8.2.7 Provide A1035 U shaped #4 stirrups 7” o.c. Note that s
7、 1.2Mcr Hence, minimum reinforcement requirements are met. Maximum Spacing of Tension Reinforcement (5.7.3.4) Ig = 262874 in.4 fc= My I = (133612)(4215.42) 262874 =1.62ksi 5.4.2.6: fc80% of fr; hence, 5.7.3.4 needs to be checked. Mservice = 1336 k-ft. Mcr = 391 k-ft. Use cracked transformed section
8、properties Icr = 63200 in.4 y- = 6.21 in. (Measured from compression face) dc =1.5+ 0.5+1/ 2(1) = 2.5“ fs= n My I = 29000 57 4000 (133612)(42 2.5 6.21) 63200 = 67.9ksi (Approximately 0.6fy = 0.6100 = 60 ksi) e = 0.75 s=1+ dc 0.7(h dc) =1+ 2.5 0.7(42 2.5) =1.09 s 700 e sfs 2dc= 700 0.75 1.09 67.9 2 2
9、.5= 2.09“ K-7 The actual spacing of 5 # 8 bars, in each layer, is 1+ 22 2(1.5+ 0.5)51 4 = 4.25“ 2.09”, which is not acceptable. Revise the design by using 12 # 8 bars in 2 layers. Capacity Mu = 2052 k-ft. Mn = 2672 k-ft. O.K. Maximum Reinforcement Requirement (5.7.2.1) c dt = 3.64 39.5 = 0.109 1.2Mc
10、r Hence, minimum reinforcement requirements are met. Maximum Spacing of Tension Reinforcement (5.7.3.4) Ig = 267731 in.4 fc= My I = (133612)(4215.56) 267731 =1.58ksi 5.4.2.6: fc80% of fr; hence, 5.7.3.4 needs to be checked. Mservice = 1336 k-ft. Mcr = 400 k-ft. Use cracked transformed section proper
11、ties Icr = 74218 in.4 y- = 6.74 in. (Measured from compression face) dc =1.5+ 0.5+1/ 2(1) = 2.5“ fs= n My I = 29000 57 4000 (133612)(42 2.5 6.74) 74218 = 56.9ksi (Approximately 0.6fy = 0.6100 = 60 ksi) e = 0.75 s=1+ dc 0.7(h dc) =1+ 2.5 0.7(42 2.5) =1.09 K-8 s 700 e sfs 2dc= 700 0.75 1.0956.9 2 2.5=
12、 3.46“ The actual spacing of 12 # 8 bars, in each layer, is 1+ 22 2(1.5+ 0.5) 61 5 = 3.4“ 36”; skin reinforcement needs to be provided. However, for consistency with the original example, skin reinforcement is not provided in the redesign with A1035 reinforcing bars. Fatigue Limit State Mf = 278 k-f
13、t Cracked transformed section properties: Icr = 74218 in.4 and y- = 6.74 in. fs= n My Icr = 8 (27812)(42 2.5 6.74) 74218 =11.8ksi fmin = stress under dead load moment, which is 597 k-ft. Mcr = 400 k-ft. fmin= n My Icr = 8 (597 12)(42 2.5 6.74) 74218 = 25.3ksi f f = 21 0.33fmin+8(r / h) = 21 0.33 25.
14、3+8 0.3=15.1ksi 11.8 ksi 12“ld= 38“ As seen below, the available distance to develop the bar is 55”, which is larger than ld= 38“. Therefore, fsx = fy K-11 Tprovided= Asfy= (16 0.79)100 =1264kips T = 371kipsO.K. (ii) Midspan Mu = 2052 k-ft and Vu = 46 kips For # 4 stirrups 24” o.c., Vs= Avfydv s = 0
15、.4100 37 24 = 61.7kips Vs in Eq. 5.8.3.5-1 cannot be taken greater than Vu/ = 46/0.9 = 51 kips; hence, use Vs = 51 kips. T = Mu dv + Vu 0.5Vs cot= 205212 0.9 37 + 46 0.9 0.551 cot(45) = 765kips At midspan, there is no concern about development length; hence, fsx=fy. Tprovided= Asfy= (16 0.79)100 =12
16、64kips T = 765kipsO.K. (iii) Face of Bearing Mu = 0 and Vu = 168 kips (Note: Vu is taken as the shear at dv from the face of support.) For # 4 stirrups 17” o.c., Vs= Avfydv s = 0.4100 37 17 = 87.1kips Vs in Eq. 5.8.3.5-1 cannot be taken greater than Vu/ = 168/0.9 = 187 kips; hence, use Vs = 87.1 kips. T = Mu dv + Vu 0.5Vs cot= 0+ 168 0.9 0.587.1 cot(45) =143kips K-12 ld = 38“ As seen below, the available distance to develop the bar is 21”. fsx= ld,available ld fy= 21 38 100 = 55.2ksi Tprovided= Asfsx= (16 0.79)55.2 = 698kips T =143kipsO.K.