EUROCODE 2 WORKED EXAMPLES - Federbeton

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EC2 – worked examples 7-16 0.1687 ⎛ 0.0223 ⎞ σ s,cr = 0.6⋅3.086⋅ ⋅ ⎜1+ 15⋅ ⎟= 41.78 MPa 0.0223 ⎝ 0.1687 ⎠ 160 ⎛ 41.78 ⎞ ⎛ 0.1687 ⎞ w k = ⋅ 1 3.4 50 0.17 26 0.12 mm 5 ⎜ − ⎟⋅⎜ ⋅ + ⋅ ⋅ ⎟= 2 ⋅10 ⎝ 160 ⎠ ⎝ 0.0223 ⎠ The obtained values are in good agreement with those evaluated within the design. The values from the verification are slightly larger because of the fact that in the considered section the internal drive lever arm is lower than the approximated value 0.9d assumed in the approximated design procedure. In fact, being h0/d the adimensional lever arm in units of effective height d, in the three case we have kw = 1 h0/d = (43.70 – 19.59/3) / 43.70 = 0.85 kw = 2/3 h0/d = (43.70 – 21.46/3) / 43.70 = 0.836 kw = 1/3 h0/d = [(18·160·18.99 + 3·160·13.79 2 /18.99) / (18·160 + 3·160·13.79/18.99) + + 2/3·24.71] / 43.70 ≅ 0.8 Let’s remark that the presence of a compressed reinforcement is highly recommended to make ductile the section in the ultimate limit state. The reinforcement increase the lever arm of the section reducing the difference between the approximated values and those coming from the verification. The approximated method previously discussed can be successfully applied in the design of the ultimate crack state. The obtained results are reported in the Tables 7.1 and 7.2 and they are shown in Figure 7.9. Table 7.3 and Figure 7.10 report numerical values and graphs for the maximal diameter and the required reinforcement expressed as a function of fixed values for σs. Stating a suitable precision for the approximated method, those values are evaluated using the (7.16) (7.14). Table 7.1. Approximated method. Table 7.2. Exact method. w k (mm) A s (mm 2 ) σ s (MPa) h 0/d A s (mm 2 ) w k (mm) σ s (MPa) h 0/d 0.1 11151 140 0.9 11151 0.120 160 0.811 0.2 6903 221 0.9 6903 0.213 238 0.836 0.3 5310 190 0.9 5310 0.306 304 0.85 Table of Content Fig. 7.9. Comparison between the exact and approximated methods.
EC2 – worked examples 7-17 Table of Content Table 7.3. Approximated method – Determination of maximum diameter. wk = 0.1 mm (A) wk = 0.2 mm (B) wk = 0.3 mm (C) σs (MPa) φmax (mm) As (mm 2 ) σs (MPa) φmax (mm) As (mm 2 ) σs (MPa) φmax (mm) As (mm 2 ) 137 30 11111 214 30 7001 280 30 5430 140 26 11151 221 26 6903 290 26 5310 145 20 10486 233 20 6508 309 20 4910 149 16 10205 243 16 6245 325 16 4672 156 10 9750 261 10 5816 355 10 4282 Fig. 7.10. Diagrams for Maximal diameter (φ max) – Metal area (A s ) – Steel tension (σ s).
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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 2-6 Set B (ve
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EC2 - worked examples 2-8 Table of
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EC2 - worked examples 2-10 Verifica
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EC2 - worked examples 2-12 Table 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 6-14 Taking d
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EC2 - worked examples 6-16 ⎛ fck
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EC2 - worked examples 6-20 Fig. 6.1
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EC2 - worked examples 6-22 Assumed
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EC2 - worked examples 6-26 (applied
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EC2 - worked examples 6-30 θ 11 =
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