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SAFE 2014 0
EXAMPLE Eurocode 2-04 RC-SL-001 Slab Flexural Design PROBLEM DESCRIPTION The purpose of this example is to verify slab flexural design in SAFE. A one-way, simple-span slab supported by walls on two opposite edges is modeled using SAFE. The slab is 150 mm thick and spans 4 meters between walls. To ensure one-way action, Poisson’s ratio is taken to be zero. The computational model uses a finite element mesh, automatically generated by SAFE. The maximum element size is specified as 1.0 meter. The slab is modeled using thin plate elements. The walls are modeled as line supports without rotational stiffnesses and with very large vertical stiffness (11015 kN/m). To obtain factored moments and flexural reinforcement in a design strip, one onemeter wide strip is defined in the X-direction on the slab, as shown in Figure 1.
Simply supported edge at wall
4 m span
Simply supported edge at wall
Free edge 1 m design strip
Y X
Free edge Figure 1 Plan View of One-Way Slab
One dead load case (DL4KPa) and one live load case (LL5KPa) with uniformly distributed surface loads of magnitudes 4 and 5 kN/m2, respectively, are defined in the model. A load combination (COMB5kPa) is defined using the Eurocode 204 load combination factors, 1.35 for dead loads and 1.5 for live loads. The model is analyzed for both load cases and the load combination. The slab moment on a strip of unit width is computed analytically. The total factored strip moments are compared with the SAFE results. These moments are identical. After completing the analysis, design is performed using the Eurocode 2-04 code by SAFE and also by hand computation. Table 1 shows the comparison of the design reinforcements computed by the two methods.
EXAMPLE Eurocode 2-04 RC-SL-001 - 1
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GEOMETRY, PROPERTIES AND LOADING Thickness Depth of tensile reinf. Effective depth Clear span
T, h dc d ln, l1
= = = =
150 25 125 4000
mm mm mm mm
Concrete strength Yield strength of steel Concrete unit weight Modulus of elasticity Modulus of elasticity Poisson’s ratio
fck fsy wc Ec Es
= = = = = =
30 460 0 25000 2x106 0
MPa MPa N/m3 MPa MPa
Dead load Live load
wd wl
= =
4.0 5.0
kPa kPa
TECHNICAL FEATURES OF SAFE TESTED Calculation of flexural reinforcement Application of minimum flexural reinforcement RESULTS COMPARISON Table 1 shows the comparison of the SAFE total factored moments in the design strip the moments obtained by the hand computation method. Table 1 also shows the comparison of the design reinforcements. Table 1 Comparison of Design Moments and Reinforcements
National Annex CEN Default, Norway, Slovenia and Sweden
Finland , Singapore and UK Denmark
EXAMPLE Eurocode 2-04 RC-SL-001 - 2
Reinforcement Area (sq-cm)
Method
Strip Moment (kN-m)
SAFE
25.797
5.400
Calculated
25.800
5.400
SAFE
25.797
5.446
Calculated
25.800
5.446
SAFE
25.797
5.626
As+
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Calculated
25.800
SAFE 2014 0 5.626
A s ,min = 204.642 sq-mm
COMPUTER FILE: Eurocode 2-04 RC-SL-001.FDB CONCLUSION The SAFE results show an acceptable comparison with the independent results.
EXAMPLE Eurocode 2-04 RC-SL-001 - 3
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HAND CALCULATION The following quantities are computed for the load combination:
1.0 for fck ≤ 50 MPa 0.8 for fck ≤ 50 MPa b = 1000 mm
For the load combination, w and M are calculated as follows: w = (1.35wd + 1.5wt) b wl12 M 8
As ,min
0.0013bw d max fctm 0 . 26 bd f yk = 204.642 sq-mm
COMB100 wd = 4.0 kPa wt
= 5.0 kPa
w = 12.9 kN/m M-strip = 25.8 kN-m M-design= 25.8347 kN-m For CEN Default, Norway, Slovenia and Sweden:
m, steel = 1.15 m, concrete = 1.50 αcc = 1.0 The depth of the compression block is given by: m
M bd 2 f cd
EXAMPLE Eurocode 2-04 RC-SL-001 - 4
25.8347 106 = 0.08267 1000 1252 1.0 1.0 30 /1.5
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For reinforcement with fyk 500 MPa, the following values are used: k1 = 0.44 k2 = k4 = 1.25(0.6 + 0.0014/εcu2) = 1.25
is assumed to be 1 k1 x for fck 50 MPa = 0.448 k2 d lim x x mlim 1 = 0.294 d lim 2 d lim
1 1 2m = 0.08640 f bd As cd = 540.024 sq-mm > As,min f yd As = 5.400 sq-cm
For Singapore and UK:
m, steel = 1.15 m, concrete = 1.50 αcc = 0.85: The depth of the compression block is given by:
m
M bd 2 f cd
25.8347 106 = 0.097260 1000 1252 1.0 0.85 30 /1.5
x x mlim 1 = 0.48 d lim 2 d lim
k1 x for fck 50 MPa = 0.60 k2 d lim For reinforcement with fyk 500 MPa, the following values are used:
k1 = 0.40 k2 = (0.6 + 0.0014/εcu2) = 1.00
EXAMPLE Eurocode 2-04 RC-SL-001 - 5
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is assumed to be 1 1 1 2m = 0.10251 f bd As cd = 544.61 sq-mm > As,min f yd As = 5.446 sq-cm
For Finland:
m, steel = 1.15 m, concrete = 1.50 αcc = 0.85: The depth of the compression block is given by: m
M bd 2 f cd
25.8347 106 = 0.097260 1000 1252 1.0 0.85 30 /1.5
x x mlim 1 = 032433 d lim 2 d lim
k1 x for fck 50 MPa = 0.5091 k2 d lim For reinforcement with fyk 500 MPa, the following values are used: k1 = 0.44 k2 = 1.1 k4 = 1.25(0.6 + 0.0014/εcu2) = 1.25
is assumed to be 1 1 1 2m = 0.10251 f bd As cd = 544.61 sq-mm > As,min f yd As = 5.446 sq-cm
EXAMPLE Eurocode 2-04 RC-SL-001 - 6
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For Denmark:
m, steel = 1.20 m, concrete = 1.45 αcc = 1.0 The depth of the compression block is given by: m
M bd 2 f cd
25.8347 106 = 0.0799153 1000 1252 1.0 1.0 30 /1.5
x x mlim 1 = 0.294 d lim 2 d lim
k1 x for fck 50 MPa = 0.448 k2 d lim For reinforcement with fyk 500 MPa, the following values are used: k1 = 0.44 k2 = k4 = 1.25(0.6 + 0.0014/εcu2) = 1.25
is assumed to be 1 1 1 2m = 0.08339 f bd As cd = 562.62 sq-mm > As,min f yd As = 5.626 sq-cm
EXAMPLE Eurocode 2-04 RC-SL-001 - 7
