Nzs 3101 06 rc sl 001

Nzs 3101 06 rc sl 001 Nzs 3101 06 rc sl 001 Nzs 3101 06 rc sl 001 Nzs 3101 06 rc sl 001 Nzs 3101 06 rc sl 001 Nzs 3101 06 rc sl 001 Nzs 3101 06 rc sl 001 Nzs 3101 06 rc sl 001 Nzs 3101 06 rc sl 001 Nzs 3101 06 rc sl 001 Nzs 3101 06 rc sl 001 Nzs 3101 06 rc sl 001 Nzs 3101 06 rc sl 001 Nzs 3101 06 rc sl 001 Nzs 3101 06 rc sl 001 Nzs 3101 06 rc sl 001 Nzs 3101 06 rc sl 001 Nzs 3101 06 rc sl 001 Nzs 3101 06 rc sl 001 Nzs 3101 06 rc sl 001 Nzs 3101 06 rc sl 001 Nzs 3101 06 rc sl 001 Nzs 3101 06 rc sl 001 Nzs 3101 06 rc sl 001 Nzs 3101 06 rc sl 001 Nzs 3101 06 rc sl 001 Nzs 3101 06 rc sl 001 Nzs 3101 06 rc sl 001 Nzs 3101 06 rc sl 001 Nzs 3101 06 rc sl 001 Nzs 3101 06 rc sl 001 Nzs 3101 06 rc sl 001 Nzs 3101 06 rc sl 001 Nzs 3101 06 rc sl 001 Nzs 3101 06 rc sl 001

EXAMPLE NZS 3101-06 RC-SL-001

Slab Flexural Design

P ROBLEM D ESCRIPTION

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 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) The computational model uses a finite element mesh, automatically generated by SAFE The maximum element size is specified as 1.0 meter 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 NZS

3101-06 load combination factors, 1.2 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 After completing analysis, design is performed using the NZS 3101-06 code by SAFE and also by hand computation Table 1 shows the comparison of the design reinforcements computed using the two methods

EXAMPLE NZS 3101-06 RC-SL-001 - 2

G EOMETRY , P ROPERTIES AND L OADING

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

T ECHNICAL F EATURES OF SAFE T ESTED

¾ Calculation of flexural reinforcement

¾ Application of minimum flexural reinforcement

R ESULTS C OMPARISON

Table 1 shows the comparison of the SAFE total factored moments in the design strip with the moments obtained by the hand computation method Table 1 also shows the comparison of design reinforcements

Table 1 Comparison of Design Moments and Reinforcements

Reinforcement Area (sq-cm) Load

Level Method

Moment

Medium

,min

C OMPUTER F ILE : NZS3101-06RC-SL-001.FDB

C ONCLUSION

The SAFE results show an exact comparison with the independent results

EXAMPLE NZS 3101-06 RC-SL-001 - 4

H AND C ALCULATION

The following quantities are computed for the load combination:

φb = 0.85

b = 1000 mm

α1=0.85 for f′c ≤55MPa

β1=0.85 for f′c ≤30,

ε ε

= +

c b

c y s

f E = 70.7547

amax = 0.75β1 c b= 45.106 mm

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

w = (1.2w d + 1.5w t ) b

8

2 1

wl

M u =

w y

b d f

⎧ ′

=

⎪

⎪

⎪⎩

min

372 09 4

max

1 4 380 43

c w y

s ,

f

f A

sq-mm

= 380.43 sq-mm

COMB 100

w d = 4.0 kPa

w t = 5.0 kPa

w = 12.3 kN/m

M * = 24.6 kN-m The depth of the compression block is given by:

* 2

1

2

c b

M

α φ

′ = 9.436 mm < amax

The area of tensile steel reinforcement is then given by:

*

2

s

b y

M A

a

f d

φ

=

⎛ −

⎞ = 523.067 sq-mm > A s,min

A s = 5.23067 sq-cm