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EXAMPLE CSA A23.3-04 RC-SL-001

Slab Flexural Design

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 the 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 was specified to be 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 (1 × 1015

kN/m) To obtain factored moments and flexural reinforcement in a design strip, one one-meter wide strip is defined in the X-direction on the slab as shown in Figure 1

Free edge

1 m design strip Free edge

Simply supported edge at wall

4 m span

Y X

Simply supported edge at wall

Free edge

1 m design strip Free edge

Simply supported edge at wall

4 m span

Y X

Simply supported edge at wall

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 CSA

A23.3-04 load combination factors, 1.25 for dead loads and 1.5 for live loads The model is analyzed for these load cases and load combinations

The slab moment on a strip of unit width is computed analytically The total factored strip moments are compared with the SAFE results After completing the analysis, design is performed using the CSA A23.3-04 code by SAFE and also by hand computation Table 1 show the comparison of the design reinforcements computed using the two methods

EXAMPLE CSA A23.3-04 RC-SL-001 - 2

Depth of tensile reinf d c = 25 mm

Clear span l n , l 1 = 4000 mm

Yield strength of steel f sy = 460 MPa Concrete unit weight w c = 0 N/m3 Modulus of elasticity E c = 25000 MPa Modulus of elasticity E s = 2x106 MPa

¾ Calculation of flexural reinforcement

¾ Application of minimum flexural reinforcement

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

Reinforcement Area (sq-cm) Load

Level Method

Moment

Medium

,min

C OMPUTER F ILE : CSAA23.3-04RC-SL-001.FDB

The SAFE results show a very close comparison with the independent results

EXAMPLE CSA A23.3-04 RC-SL-001 - 4

The following quantities are computed for the load combination:

φ c = 0.65 for concrete

φ s = 0.85 for reinforcement

A s,min = 0.2 c

y

f f

′

b w h = 357.2 sq-mm

b = 1000 mm

α1 = 0.85 – 0.0015f' c ≥ 0.67 = 0.805

β1 = 0.97 – 0.0025f' c ≥ 0.67 = 0.895

c b =

y

f

+ 700

700

d = 75.43 mm

a b = β1 c b = 67.5 mm

For the load combination, w and M * are calculated as follows:

w = (1.25w d + 1.5w t ) b

8

2 1

wl

M u =

A s = min[A s,min , (4/3) A s,required] = min[357.2, (4/3)540.63] = 357.2 sq-mm = 0.22•(150/125)2•0.6•SQRT(30)/460•100•125

= 282.9 sq-mm

COMB 100

w d = 4.0 kPa

w t = 5.0 kPa

w = 12.5 kN/m

M f = 25.0 kN-m The depth of the compression block is given by:

b f

M d

d a

c c

f

φ

α '

2 1

2−

−

The area of tensile steel reinforcement is then given by:

⎟

⎠

⎞

⎜

⎝

⎛ −

=

2

a d f

M A

y s

f s

φ

= 540.630 sq-mm > A s,min

A s = 5.406 sq-cm