Pre-stressed Bridge Structural Design Calculations to the specifications of Eurocode BS 5400-4: 1990 Bridge Geometry and Materials As regards the bridge Super-structure geometry, the superstructure type is reinforced concrete deck supported on medium span prestressed girders made of continuous for live load with three span, the middle span being 280m, and two end spans at 145m each. The bridge deck total width is 15m in total inclusive of the footway and cycle way on either side of the tramway as represented by the sketch below. Tramway- 5.0 m Footway 5.0m Cycle-way 5.0m Pylons Tier 1 Tier 2 7.5m 7.5m HA and HB Loading for Bridge Deck Taking Carriageway = 15m wide Deck Span = 570 m (center to center for a simple supported medium single span) Designing for a unit width of bridge deck: Number of notional lanes = 3 Notional lane width = 15m/3= 5m Loaded length (L) = effective span = 570m W =336(1/L)0.67 kN/m (per notional lane) W= 336(1/570)0.67 = 4.875 kN/m Knife Edge load = 100kN (per notional lane) a2= 0.0137[bL (40-L) + 3.65(L-20) a2 = 0.0137 [5.0(40-570) +3.65(570-20) = 0.867 It is noteworthy to remember that loaded lengths less that 20m in span have the load proportioned for a standard lane of 3.65m width implying 0.274bL =bL/3.65 Calculating for a unit width of deck: W= (3.16 x a2)/3 For a meter width of deck: W = (31.6 x 0.867)/3.0 = 9.13 kN/m KEL = (100 x 0.867)/3.0 = 28.9 kN Maximum mid span Bending Moment with KEL at mid span: M = (W * L2)/8 + (KEL * L)/4 M = (9.13 * 252)/8 + (28.9 * 25)/4 = 894.0 kNm Considering, γfL = 1.20 (Serviceability limit state - combination 1) γfL = 1.50 (Ultimate limit state - combination 1) Designing for HA moment for a meter width of deck : (Moment at serviceability limit state) M(serviceability) = 894.0 x 1.2 = 1072.8 kNm (Moment at ultimate state) M(ultimate) = 894.0 x 1.5 = 1341 kNm It is authoritative to make note of the employment of γf3 subject to BS 5400 Part3 as well as Part 5 - γf3 is used with the design strength so Mult = 1341 kNm. Consequently from the specifications of BS 5400 Part 4 - γf3 is used with the load effect so therefore; Multimate = 1.1 x 1341 = 1475 kNm. Nominal load per axle = 30units x 9.13kN = 274kN The maximum bending moment will be achieved by using the shortest HB vehicle using a spacing of 6m spacing subject to BS 5400-2:2006, Figure 12. The maximum moment for a simply supported span occurs under the inner axle when the vehicle is positioned such that the mid span bisects the distance between the centroid of the load and the nearest axle. Consequently, with a 25m span and the 6m HB vehicle with equal axle loads, the inner axle is placed at 1.5m from the mid span as recapitulated by the representation below. 1.3m 1.3m 1.5m 1.5m 9.7m 6.7m X RL 12.5m 12.5m RR 25.0m RL = 274(6.7+8.0+14.0+15.3)/25 = 482.24 kN RR = 4 * 274 – 482.24 = 613.76 kN Moment at X = 482 * 11.0 – 274 * 1.3 = 4946 kNm The HB vehicle occupies one lane with HA load in the adjacent lane. Assume for instance that the HB load is carried by a standard lane width of 3.50m. Hence the moment per meter width of deck slab = 4946/3.50 = 1413kNm γfL = 1.10 (Serviceability limit state - combination 1) γfL = 1.30 (Ultimate limit state - combination 1) Design HB moment for a meter width of deck : Msls = 1.1 x 1413 = 1554 kN/m (compared to 1072.8 for HA load) Mult = 1.3 x 1413 = 1837 kN/m (compared to 1341 for HA load) Hence in this case HB load effects would govern although a grillage or finite element type distribution would reduce the HB moment considerably. Pre-stressed Concrete Beam Design to BS 5400 Part 4 Designing a simple supported pre-stressed concrete beam which bears a 300mm thick concrete slab and 100mm of surfacing, together with a nominal live load udl of 20.0 kN/m2 and kel of 28.9 kN/m, nominal wind load of 4kN/m2, and footpath live load of 5kN/m2, . The span of the beam is 25.0m centre to centre of bearings and the beams are spaced at 6.0m intervals. γconc. = 144kN/m3 25 units of HB to be considered at SLS for load combination 1 only (BS 5400 Part t4) The Loading per beam (at 6.0m c/c) Important to note: The loading has been simplified to demonstrate the method of designing the beam (See BS 5400 Pt2, or DB 37/01 for full design loading) Nominal Dead Loads: Slab = 144 × 0.30 × 6.0 = 259.2 kN/m Beam = say Y5 beam = 10.78 kN/m Surfacing = 25 × 0.1 × 6.0 = 15.0 kN/m Nominal Live Load: HA = 20 × 6.0 + 28.9 = 20 kN/m + 28.9kN 25 units HB = 25 × 20 / 4 per wheel = 125.0 kN per wheel Tabulated below are the load factors for serviceability and ultimate limit state subject to BS 5400 Part 2 Table 1: SLS Dead load VfL concrete ULS Comb. 1 Comb. 3 1.0 1.0 Comb. 1 Comb. 3 1.15 1.15 Superimposed Dead load VfL surfacing 1.2 1.2 1.75 1.75 VfL HA 1.2 1.0 1.5 1.25 VfL HB 1.1 - - - VfL - 0.8 - 1.0 Live load Temperature difference Taking the Concrete Grades to be: C40/50 fcu = 50 N/mm2, fci = 40 N/mm2 for the beam and C32/40 fcu = 40 N/mm2 for the slab and considering BS 5400 Part 4 we look at the Section Properties: It is worth to note that the Modular ratio effect for different concrete strengths between beam and slab may be ignored. Property Beam Section Composite Section 449.22×103 599.22×103 456 623 2nd Moment of Area (mm4) 52. 905x109 103x515x109 Modulus at level 1 (mm3) 116.020×106 166.156×106 Modulus at level 2 (mm3) 89.066x 106 242.424 x106 Modulus at level 3 (mm3) - 179.402 x106 Area (mm2) Centroid (mm) Effects of Differential Shrinkage Total shrinkage of insitu concrete = 300 × 10-6 Assuming that 2/3 of the total contraction of the pre-stressed concrete is done prior to the deck slab casting as well as taking the residual shrinkage to be 100 × 10-6 , consequently the disparity shrinkage is 200 × 10-6 Force to restrain differential shrinkage: F = - εdiff * Ecf * Acf * φ F = -200 * 10-6 * 25 × 1000 * 300 * 0.43 = -645 kN Eccentricity acent = 502mm Restraint moment Mcs = -645 × 0.502 = -323.80 kNm Self load of beam and load of deck wedge is sustained by the beam. Upon the curing of the deck slab concrete any further loading , both superimposed and live loads, is sustained by the compound section of the beam as well as the slab. Temperature Difference Effects Employing the temperature differentiations provided in BS: 5400 Part 2 Figure 9 for a basic beam section, we find the Coefficient of thermal expansion = 12 × 10-6 per °C. From BS 5400 Part 4, Table 3: we obtain Ec = 34 kN/mm2 for fcu = 50N/mm2 consequently, controlled temperature stresses per °C = 32 × 103 × 12 × 10-6 = 0.25 N/mm2 Constructive Temperature Difference Force F to restrain temperature strain : 0.25 × 1000 × [ 300 × ( 3.0 + 5.25 ) ] × 10-3 + 0.25 × ( 300 × 250 × 1.5 + 750 × 200 × 1.25 ) × 10-3 = 693.75 kN Moment M about centroid of section to restrain curvature due to temperature strain : 0.25 × 1000 × [ 150 × ( 3.0 × 502 + 5.25 × 527 ) ] × 10-6 + 0.25 × ( 300 × 250 × 1.5 × 344 - 750 × 200 × 1.25 × 556 ) × 10-6 = 261.5 - 26.7 = 234.8 kNm Repeal temperature difference Force F to restrain temperature strain : - 0.25 × [ 1000 × 150 × ( 3.6 + 2.3 ) + 300 × 90 × ( 0.9 + 1.35 ) ] × 10-3 - 0.25 × 300 × ( 200 × 0.45 + 150 × 0.45 ) × 10-3 - 0.25 × 750 × [ 50 × ( 0.9 + 0.15 ) + 240 × ( 1.2 + 2.6 ) ] × 10-3 = - 385.9 - 19.3 - 295.1 = - 700.3 kN. Subsequently, moment M about centroid of section to restrain curvature due to temperature strain : - 0.408 × [ 150000 × ( 3.6 × 502 + 2.3 × 527 ) + 27000 × ( 0.9 × 382 + 1.35 × 397 ) ] × 10-6 - 0.408 × 300 × ( 200 × 0.45 × 270 - 150 × 0.45 × 283 ) × 10-6 + 0.408 × 750 × [ 50 × ( 0.9 × 358 + 0.15 × 366 ) + 240 × ( 1.2 × 503 + 2.6 × 543 ) ] × 10-6 = - 194.5 - 0.6 + 153.8 = - 41.3 kNm Checking for Shear Considering it as critical: V = 29.5 kN/m and hence VEd = 29.5 – 0.14 × 12.3 = 27.8 kN/m VEd = 27.8 × 103/144 × 103 = 0.19 MPa VRd = 0.53 MPa this implies no shear reinforcement necessary Designing for the Aqueduct The following is assumed: a) Flow rate = 30m3/hr b) Bed width = 100m c) Water depth = 1.8m d) Level of full supply = 220m e) Side gradient = 1.5: 1 f) Manning coefficient for pre-stressed concrete = 0.016 g) High flood discharge = 300m3/hr h) High flood level = 248.0 m i) High flood depth = 2.5 m By use of Lacey’s regime of wetted perimeter where Pw = 4.83 √Q = 4.83√300 = 83.66m Allowing the clear span between piers be 8 m and the pier thickness be 1.5 m. Provide 8 coves of 8 m each = 64.0 m 7 piers of 1.5 m each = 10.5 m Hence the waterway between abutments is taken to be 74.5 m subject to the Bed width of canal References BS 5400-4: 1990. Steel, concrete and composite bridges —Part 4: Code Of Practice for Design of Concrete Bridge. UDC 624.21.01:624.012.4:624.014.2 [006.76 (083.75). Retrieved from < http://ultimaswari.files.wordpress.com/2010/03/5400-4-1990.pdf>