Software Verification

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1 EXAMPLE 17 Crack Width Analyi The crack width, wk, i calculated uing the methodology decribed in the Eurocode EN :2004, Section 7.3.4, which make ue of the following expreion: (1) w = ( ),max ε ε (eq. 7.8) k r m cm r,max i the maximum crack pacing εm εcm i the mean train in the reinforcement under the relevant combination of load, including the effect of impoed deformation and taking into account the effect of tenion tiffening. Only the additional tenile train beyond the tate of zero train of the concrete at the ame level i conidered. i the mean train in the concrete between crack (2) εm εcm may be calculated from the expreion fct,eff σ kt ( 1+αeρp.eff ) ρ p,eff σ εm ε cm = E 0.6 (eq. 7.9) E E σ αe i the tre in the tenion reinforcement auming a cracked ection. For pretenioned member, σ may be replaced by σ, the tre variation in pretreing tendon from the tate of zero train of the concrete at the ame level. i the ratio Ec / Ecm ρp,eff i A / Ac,eff Ap and Ac,eff ; Ap i the area of tendon within Ac,eff, and Ac,eff i the area of tenion concrete urrounding the reinforcing. EXAMPLE 17-1

2 kt i a factor dependent on the duration of the load kt = 0.6 for hort term loading kt = 0.4 for long-term loading (3) In ituation bonded reinforcement i fixed at reaonably cloe center within the tenion zone [pacing 5(c + φ / 2)], the maximum final crack pacing may be calculated from r,max = k3c + k1k2k4φ / ρp,eff (eq. 7.11) φ i the bar diameter. Where a mixture of bar diameter i ued in a ection, an equivalent diameter, φeq, hould be ued. For a ection with n1 bar of diameter φ1 and n2 bar of diameter φ2, the following equation hould be ued: 2 2 n1φ 1+ n2φ2 φ eq = (eq. 7.12) n φ + n φ c i the cover to the longitudinal reinforcement k1 i a coefficient that take into account the bond propertie of the bonded reinforcement: = 0.8 for high bond bar = 1.6 for bar with an effectively plain urface (e.g., pretreing tendon) k2 i a coefficient that take into account the ditribution of train: = 0.5 for bending = 1.0 for pure tenion k3 and k4 are recommended a 3.4 and repectively. See the National Annex for more information. For cae of eccentric tenion or for local area, intermediate value of k2 hould be ued that may be calculated from the relation: k2 = (ε1 + ε2) / 2ε1 (eq. 7.13) ε1 i the greater and ε2 i the leer tenile train at the boundarie of the ection conidered, aeed on the bai of a cracked ection. EXAMPLE 17-2

3 PROBLEM DESCRIPTION The purpoe of thi example i to verify that the crack width calculation performed by i conitent with the methodology decribed above. Hand calculation uing the Eurocode EN :2004, Section are hown below a well a a comparion of the and hand calculated reult. A one-way, imply upported lab i modeled in. The modeled lab i 254 mm thick by 914 mm wide and pan 9,754 mm, a hown in Figure 17-1, and i the ame lab ued to validate the Eurocode PT deign (ee deign verification example Eurocode 2-04 PT-SL-001). To tet the crack width calculation, even #5 longitudinal bar have been added to the lab. The total area of mild teel reinforcement i 1,400mm 2. Currently, will account for ome of the PT effect. account for the PT effect on the moment and reinforcing tree but the tendon area are not conidered effective to reit cracking. Pretreing tendon, Ap Mild Steel, A 229 mm 254 mm Length, L = 9754 mm 914 mm 25 mm Elevation Section Figure 17-1 One-Way Slab EXAMPLE 17-3

4 A 254-mm-wide deign trip i centered along the length of the lab and ha been defined a an A-Strip. B-trip have been placed at each end of the pan, perpendicular to Strip-A (the B-Strip are neceary to define the tendon profile). A tendon with two trand, each having an area of 99 mm 2, wa added to the A- Strip. The load are a follow: Load: Dead = elf weight GEOMETRY, PROPERTIES AND LOADING Thickne T, h = 254 mm Effective depth d = 229 mm Clear pan L = 9754 mm Concrete trength f 'c = 30 MPa Yield trength of teel fy = 400 MPa Pretreing, ultimate fpu = 1862 MPa Pretreing, effective fe = 1210 MPa Area of Pretre (ingle trand) Ap = 198 mm 2 Concrete unit weight wc = KN/m 3 Modulu of elaticity Ec = N/mm 3 Modulu of elaticity E = 200,000 N/mm 3 Poion ratio ν = 0 Dead load wd = elf KN/m 2 TECHNICAL FEATURES OF TESTED Calculation of the reported crack width. RESULTS COMPARISON Table 1 how the comparion of the crack width to thoe calculated by hand. EXAMPLE 17-4

5 Table 1 Comparion of Reult FEATURE TESTED INDEPENDENT RESULTS RESULTS DIFFERENCE Crack Width (mm) 0.151mm 0.161mm 6.62% COMPUTER FILE: S17.FDB CONCLUSION The reult how an acceptable comparion with the independent reult. EXAMPLE 17-5

6 HAND CALCULATIONS: Deign Parameter: Mild Steel Reinforcing f c = 30MPa fy = 400MPa Pot-Tenioning fpu = 1862 MPa fpy = 1675 MPa Streing Lo = 186 MPa Long-Term Lo = 94 MPa fi = 1490 MPa fe = 1210 MPa Pretreing tendon, Ap Mild Steel, A 229 mm 254 mm Length, L = 9754 mm 914 mm 25 mm Elevation Section Load: Dead, elf-wt = m x kn/m 3 = kn/m 2 (D) ω =5.984 kn/m 2 x m = kn/m Ultimate Moment, 2 wl1 M U = = x (9.754) 2 /8 = 65.0 kn-m 8 Reinforcing teel tre, 2 σ = 207N / mm (calculated but not reported by ) EXAMPLE 17-6

7 Check of Concrete Stree at Midpan: Figure 17-1 Setting ued for thi example EXAMPLE 17-7

8 Calculation of Crack Width: w = ( ε ε ) k r,max m cm fct,eff σ kt ( 1+αeρp,eff ) ρ p,eff σ εm ε cm = 0.6 E E, p,eff c.eff 2 2 ( ) ρ = A / A = 1.53mm / mm / 60mm / mm ρ p,eff = N / mm 206 N / mm 0.4 ( 1+ 8( 0.026) ) εm ε cm = εm εcm = r,max = kc 3 + kkk φ/ ρp,eff = 3.4( 19.0mm) + 0.8( 0.5)( 0.425) 15.8mm / = 168mm Total crack width, wk r,max( εm εcm ) = = 168mm ( ) = 0.151mm EXAMPLE 17-8