EUROCODE 2 WORKED EXAMPLES - Federbeton

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EC2 – worked examples 6-64 n1 , n2 Table of Content numbers of tendons above and below section A-A, respectively: considering the anchor plate as rigid a value of 0.5 may be assumed; γ1 =1.1 supplementary partial safety factor against overstressing by overpumping. Bursting force shall be resisted by an area of reinforcement steel of: As,bs = Nbs / fyd = 19.61 cm 2 distributed within lbs/3 to lbs ,(i.e. from 0.40 m to 1.20 m) from the anchor plate. Thus the effective area on a meter length, may be found by the following: Asbs , Asbs , . s× s b .43 ( . .4) = 19 61 = = 57.0 cm 2 × 0 120−0 2 /m 2 3 l bs that may be provided with ties having diameter of 22 mm and spacing both transversally and longitudinally of 250 mm (see Fig. 53); in fact φ22/25×25 corresponds to 60.82 cm 2 /m 2 . Fig. 53 Bursting reinforcement arrangement Verification to spalling force The spalling force may be calculated with the equivalent prism analogy. As for bursting verification, only the longitudinal direction is considered; furthermore, spalling effects arise if upper tendon are tensioned firstly (the eccentricity leads to tension stresses). Thus, a section with a breadth of 0.43 m and a height of 1.50 m has to be verified for one tendon tensioning. The length of the prism for end anchored tendon, is equal to the overall height of the section, i.e. lsl = 1.50 m. Considering an eccentricity for upper prestressing tendon of 0.35 m, the extreme stresses at the end of prism length are calculated by means of the beam theory; for a prestressing force FSd = 4134.0 kN they result (negative if compressive): σtop ⎫⎪ ⎛ 1 035 . × 6 ⎞ ⎧⎪ −15. 38 MPa ⎬ = FSd ⎜− ∓ ⎟ = 2 ⎨ σ .43 . .43 . bott om ⎭⎪ ⎝ 0 × 150 0 × 150 ⎠ ⎩⎪ + 256 . MPa The section along which no shear force results, is placed at 0.428 m from slab bottom fibre (see Fig 54) and the moment for equilibrium turns out:
EC2 – worked examples 6-65 Table of Content 2 3 0 214 0 10 2 3 Msl = σbott om × . × .43 × = 33.61 kNm Fig. 54 Calculation scheme for spalling Assuming zsl = 0.5× lsl and bsl = 0.43 m, the maximum spalling force turns out: Nsl = Msl / zsl = 44.81 kN Disregarding any concrete tensile resistance, the amount of reinforcement is: As = Nsl / fyd = 1.031 cm 2 placed parallel to the end face in its close vicinity.
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EUROCODE 2 WORKED EXAMPLES
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Copyright: European Concrete Platfo
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Foreword to Commentary to Eurocode
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EC2 - worked examples summary EXAMP
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EC2 - worked examples 2-2 EXAMPLE 2
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EC2 - Worked examples 4-2 Moreover:
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EC2 - Worked examples 4-4 EXAMPLE 4
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EC2 - Worked examples 4-6 We have:
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EC2 - worked examples 6-2 MRd3 must
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EC2 - worked examples 6-6 Determina
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EC2 - worked examples 6-8 • VEd =
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EC2 - worked examples 6-10 EXAMPLE
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EC2 - worked examples 6-12 EXAMPLE
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Page 74 and 75: EC2 - worked examples 6-46 Table of
Page 78 and 79: EC2 - worked examples 6-50 where φ
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Page 98 and 99: EC2 - worked examples 7-4 y = 395mm
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Page 102 and 103: EC2 - worked examples 7-8 EXAMPLE 7
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Page 110 and 111: EC2 - worked examples 7-16 0.1687
Page 114 and 115: EC2 - worked examples 7-20 1⎡ λ
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EUROCODE 2 WORKED EXAMPLES - Federbeton

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