Is 456 00 rc sl 001

Is 456 00 rc sl 001 Is 456 00 rc sl 001 Is 456 00 rc sl 001 Is 456 00 rc sl 001 Is 456 00 rc sl 001 Is 456 00 rc sl 001 Is 456 00 rc sl 001 Is 456 00 rc sl 001 Is 456 00 rc sl 001 Is 456 00 rc sl 001 Is 456 00 rc sl 001 Is 456 00 rc sl 001 Is 456 00 rc sl 001 Is 456 00 rc sl 001 Is 456 00 rc sl 001 Is 456 00 rc sl 001 Is 456 00 rc sl 001 Is 456 00 rc sl 001 Is 456 00 rc sl 001 Is 456 00 rc sl 001 Is 456 00 rc sl 001 Is 456 00 rc sl 001 Is 456 00 rc sl 001 Is 456 00 rc sl 001 Is 456 00 rc sl 001 Is 456 00 rc sl 001 Is 456 00 rc sl 001 Is 456 00 rc sl 001 Is 456 00 rc sl 001 Is 456 00 rc sl 001 Is 456 00 rc sl 001 Is 456 00 rc sl 001 Is 456 00 rc sl 001 Is 456 00 rc sl 001 Is 456 00 rc sl 001 Is 456 00 rc sl 001

Software Verification

PROGRAM NAME: SAFE

REVISION NO.: 0

EXAMPLE IS 456-00 RC-SL-001 - 1

EXAMPLE IS 456-00 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 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 IS 456-00 load combination factors, 1.5 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 was performed using the IS 456-00 code by SAFE and also by hand computation Table 1 shows the comparison of the design reinforcements computed using the two methods

Software Verification

REVISION NO.: 0

EXAMPLE IS 456-00 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 (kN-m) A s + A s

Medium

,min

C OMPUTER F ILE : IS456-00RC-SL-001.FDB

C ONCLUSION

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

Software Verification

PROGRAM NAME: SAFE

REVISION NO.: 0

EXAMPLE IS 456-00 RC-SL-001 - 3

H AND C ALCULATION

The following quantities are computed for the load combination:

γs = 1.15

γc = 1.50

α = 0.36

β = 0.42

b = 1000 mm For the load combination, w and M are calculated as follows:

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

2 1

8

wl

M =

,min

0.85

s

y

f

=

= 230.978 sq-mm

COMB 100

w d = 4.0 kPa

w t = 5.0 kPa

w = 13.5 kN/m

M = 27.0 kN-m

max

250

165 415

85

y

y

y u,

y

y

y

f

x

f d

≤

⎧

⎪

⎪

⎩

≤

≤

= max

u,

x

d 0.466

Software Verification

REVISION NO.: 0

EXAMPLE IS 456-00 RC-SL-001 - 4

The depth of the compression block is given by:

ck

u

f bd

M m

α

2

=

= 0.16

2

x d

β β

= = 0.172497 < x u,max

d

The area of tensile steel reinforcement is then given by:

z d 1 x u

d

β

⎧

⎫

⎬ = 115.9439 mm

u s

M A

= = 582.178 sq-mm > A s,min

A s = 5.818 sq-cm