Appendix K Design Examples

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Gervase Berry
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1 Appendix K Design Examples

2 Example 1 * Two-Span I-Girder Bridge Continuous for Live Loads AASHTO Type IV I girder Zero Skew (a) Bridge Deck The bridge deck reinforcement using A615 rebars is shown below. Effective Width = 96 * Based on an example prepared as part of Training Classes on AASHTO LRFD Bridge Specifications, ODOT, R.A. Miller K-1

3 Redesign by using A1035 bars with = 100 ksi. A 3,483(12) = 0.90A s (100) s (100) 1.7(7.0)(26) ; A = s 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 are not epoxy coated; hence, the cover is 2.5. The centroid of the bars from the top is 14 ( / 8) + 14(8.5 (2 + 5 / / 8)) x = = 4.12" 28 Crack control reinforcement: γ e = 0.75 f s =Tensile stress in steel reinforcement at the service limit state. d = = 58.4" ρ = A s bd = (26)(58.4) = k = 2ρn + (ρn) 2 ρn = 2(0.0057)(5.718) + ( ) = j = 1 k / 3 = / 3 = f s = M sl A s jd = 2,141(12) = 56.6ksi 60ksi ( ) f s = 56.6ksi s 700γ e d β s f c = = 3.07" s This spacing is too small. Aim for 6 spacing, which is more realistic. Try: 17 top bars: # 12 alternating with # bottom bars: # 12 alternating with # 12 K-2

4 Distances to centroid of bars: ( / 8) + ( / 4) Top Bars: = 2.84" 2 (8.5 (2 + 5 / / 8)) + (8.5 (2 + 5 / / 4)) Bottom Bars: = 5.53" x = = 4.19" 34 >6 O.K. (b) Shear Reinforcement If prestressing steel is ignored, #4 A615 4 o.c. will be needed. Redesign by using A1035 U shaped # 4 stirrups. s = A f d cotθ v y v 0.4(100)(55.63)cot 45 = = 7.8" Controls, say 7 V s Note that s < 74 ; hence, the simplified procedure with β = 2 and θ = 45 o may be used Provide A1035 U shaped #4 7 o.c. K-3

5 Interface Shear Reinforcement Factored horizontal shear, V u = 323 kips V n = ca cv + µ[a vf + P c ]; c = 0.28; µ = 1.0 Although shear reinforcement spacing was previously calculated to be 7 o.c., calculate the spacing of A1035 interface shear reinforcement with limited to 60 ksi. ( ) ( ) V u = φv ni = φ ca cv + µ A vf + P c ; 323 = (20 s) ; s = 57" Say & V n max of 0.2 f ' c A cv = 0.2 7(52 20) = 1456kips 0.8A cv = 0.8 7(52 20) = 5824kips Hence, V n = 323/0.9 = 359 kips as used is o.k. A vf 0.05b v = = in 2 / length Actual A vf = 0.40 / 57 = in 2 / length N.G., Redcue the spacing 0.40 / s = 0.017; s = 23.5" say 23" Use #4 U shaped A1035 interface reinforcement at s = 23. From a durability point of view corrosion resistant A1035 interface shear reinforcement provides advantages. However, practical issues may arise from a placement point of view as the spacing of girder shear reinforcement and that for interface reinforcement are significantly different. K-4

6 Example 2* Simple Span T-Beam Cross Section Elevation The reinforcement using A615 rebars is shown below. The stirrups are symmetrically spaced at 10 o.c. up to 5 from the bearing center line and 16 o.c. up to 25. * Based on an example from LRFD Design of Cast-in-Place Concrete Bridges, 1st Ed., Schneider, E.F. and Bhide, S.B. Portland Cement Association, Skokie, IL, 2006, 156 pages. K-5

7 Redesign by using A1035 bars with a specified yield strength of 100 ksi. 1. Girder Flexure Capacity M u = 2052 k-ft. With 10 # 8 bars in 2 layers, which fit within 22, φm n = 2204 k-ft. Maximum Reinforcement Requirement ( ) c = 3.04 d t 39.5 = < 3 8 Tension controlled, O.K. O.K. Minimum Reinforcement Requirement ( ) 1.2M cr = 1.2 (4691/12) = 469 k-ft Controls 1.33M u = 1.33 (2052) = 2729 k-ft φm n = 2204 k-ft > 1.2M cr Maximum Spacing of Tension Reinforcement ( ) I g = in. 4 f c = My I : = ( )( ) Hence, minimum reinforcement requirements are met. = 1.62ksi f c >80% of f r ; hence, needs to be checked. M service = 1336 k-ft. > M cr = 391 k-ft. Use cracked transformed section properties I cr = in. 4 y - = 6.21 in. (Measured from compression face) d c = / 2(1) = 2.5" f s = n My I = 60 ksi) γ e = 0.75 = ( )( ) = 67.9ksi (Approximately 0.6 = β s = 1+ d c 0.7(h d c ) = (42 2.5) = 1.09 s 700γ e d β s f c = = 2.09" s K-6

8 The actual spacing of 5 # 8 bars, in each layer, is 1+ which is not acceptable. Revise the design by using 12 # 8 bars in 2 layers. Capacity M u = 2052 k-ft. φm n = 2672 k-ft. O.K. Maximum Reinforcement Requirement ( ) c = 3.64 d t 39.5 = < 3 8 Tension controlled, O.K. Minimum Reinforcement Requirement ( ) 1.2M cr = 1.2 (4803/12) = 480 k-ft Controls 1.33M u = 1.33 (2052) = 2729 k-ft φm n = 2672 k-ft > 1.2M cr Maximum Spacing of Tension Reinforcement ( ) I g = in. 4 f c = My I : = ( )( ) ( ) Hence, minimum reinforcement requirements are met. = 1.58ksi f c >80% of f r ; hence, needs to be checked. M service = 1336 k-ft. > M cr = 400 k-ft. Use cracked transformed section properties I cr = in. 4 y - = 6.74 in. (Measured from compression face) d c = / 2(1) = 2.5" = 4.25"> 2.09, f s = n My I = 60 ksi) = ( )( ) = 56.9ksi (Approximately 0.6 = γ e = 0.75 β s = 1+ d c 0.7(h d c ) = (42 2.5) = 1.09 K-7

9 s 700γ e d β s f c = = 3.46" s The actual spacing of 12 # 8 bars, in each layer, is 1+ which is acceptable. Skin Reinforcement ( ) 22 2( ) = 3.4"< 3.46, d e = 39.6 > 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 M f = 278 k-ft Cracked transformed section properties: I cr = in. 4 and y - = 6.74 in. f s = n My (278 12)( ) = 8 = 11.8ksi I cr f min = stress under dead load moment, which is 597 k-ft. > M cr = 400 k-ft. f min = n My (597 12)( ) = 8 = 25.3ksi I cr f f = f min + 8(r / h) = = 15.1ksi 11.8 ksi < 15.1 ksi O.K. Summary: From a strength point of view, 10 #8 A1035 bars (A s = 7.9 in. 2 ) provide adequate flexural capacity. However, the requirements related to spacing of mild reinforcement ( ) result in additional area of steel. The use of A1035 bars as flexural reinforcement saves about 49% in terms of the amount of steel (10 #11 A615 (A s = in. 2 ) vs. 12 #8 A1035 (A s = 9.48 in. 2 )). Girder Shear h 42.0 in. d s = d e 38.5 in. a 3.09 in. d v =d s -a/ in. 0.72h 30.2 in. 0.9d e 34.7 in. Final d v 37.0 in. K-8

10 Assume 1-4 wide support. The critical section is at x = =45 =3.75 Use U shaped # 4 A1035 stirrups Distance (ft.) Point along span V u x x/l kips A v s max = f ' c b v A v 0.4 in. 2 f' c 4 ksi 100 ksi b v 22 in. s max 28.8 in. The simplified procedure for the determination of β and θ may be used if the spacing of the stirrups does not exceed s max = 28.8 in. For the simplified procedure β = 2.0 and θ = 45 o. Distance (ft.) v u Is v u <0.125f' c? s max (in.) x ksi Yes Yes Yes Yes Yes Yes Yes Yes 24 The maximum shear resistance, V n, is given by the lesser of V n =V c +V s +V p and V n =0.25f c b v d v. V n =0.25f c b v d v = 0.25(4)(22)(37) = 814 kips K-9

11 V the critical section = 168 <φv n = 0.9(814) = 733 kips V c = β f ' c b v d v = (2) 4(22)(37) = 103kips At the critical section, V u = 168 kips V u = φ(v c + V p + V s ) 168 = A f d v y v = s s ; s = 17.69" Stirrup layout (symmetrically placed) of U-shaped #4 A1035 stirrups: - Start the first stirrup at 9 from the support. - Provide 3 17 o.c. - Provide o.c. Tensile Capacity of Longitudinal Reinforcement ( ) The following equation needs to be satisfied For this case, the above equation is simplified to (i) Critical Section M u = 633 k-ft & V u = 168 kips For # 4 17 o.c., V s = A f d v y v = = 87.1kips s 17 V s in Eq cannot be V u /φ = 168/0.9 = 187 kips Hence, use V s = 87.1 kips. T = M u φd v + V u φ 0.5V s cotθ = cot(45) = 371kips l db = max( 1.25A f b y ,0.4d b ) = max(, ) = 37.5" f ' c 4 No modification factors are necessary; hence, l d = l db = 37.5" > 12" l d = 38" As seen below, the available distance to develop the bar is 55, which is larger than l d = 38". Therefore, f sx = K-10

12 T provided = A s = ( )100 = 1264kips > T = 371kips O.K. (ii) Midspan M u = 2052 k-ft and V u = 46 kips For # 4 24 o.c., V s = A f d v y v = = 61.7kips s 24 V s in Eq cannot be taken greater than V u /φ = 46/0.9 = 51 kips; hence, use V s = 51 kips. T = M V u + u φd v φ 0.5V s cotθ = cot(45) = 765kips At midspan, there is no concern about development length; hence, f sx =. T provided = A s = ( )100 = 1264kips > T = 765kips O.K. (iii) Face of Bearing M u = 0 and V u = 168 kips (Note: V u is taken as the shear at d v from the face of support.) For # 4 17 o.c., V s = A f d v y v = = 87.1kips s 17 V s in Eq cannot be taken greater than V u /φ = 168/0.9 = 187 kips; hence, use V s = 87.1 kips. T = M u φd v + V u φ 0.5V s cotθ = cot(45) = 143kips K-11

13 l d = 38" As seen below, the available distance to develop the bar is 21. f sx = l d,available l d = = 55.2ksi T provided = A s f sx = ( )55.2 = 698kips > T = 143kips O.K. K-12

Appendix J. Example of Proposed Changes

Appendix J. Example of Proposed Changes Appendix J Example of Proposed Changes J.1 Introduction The proposed changes are illustrated with reference to a 200-ft, single span, Washington DOT WF bridge girder with debonded strands and no skew.