Aci 318 08 rc sl 001 Aci 318 08 rc sl 001 Aci 318 08 rc sl 001 Aci 318 08 rc sl 001 Aci 318 08 rc sl 001 Aci 318 08 rc sl 001 Aci 318 08 rc sl 001 Aci 318 08 rc sl 001 Aci 318 08 rc sl 001 Aci 318 08 rc sl 001 Aci 318 08 rc sl 001 Aci 318 08 rc sl 001 Aci 318 08 rc sl 001 Aci 318 08 rc sl 001 Aci 318 08 rc sl 001 Aci 318 08 rc sl 001 Aci 318 08 rc sl 001 Aci 318 08 rc sl 001 Aci 318 08 rc sl 001 Aci 318 08 rc sl 001 Aci 318 08 rc sl 001 Aci 318 08 rc sl 001 Aci 318 08 rc sl 001 Aci 318 08 rc sl 001 Aci 318 08 rc sl 001 Aci 318 08 rc sl 001 Aci 318 08 rc sl 001 Aci 318 08 rc sl 001 Aci 318 08 rc sl 001 Aci 318 08 rc sl 001 Aci 318 08 rc sl 001 Aci 318 08 rc sl 001 Aci 318 08 rc sl 001
Trang 1Software Verification
PROGRAM NAME: SAFE
REVISION NO.: 0
EXAMPLE ACI 318-08 RC-SL-001 - 1
EXAMPLE ACI 318-08 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 6 inches thick and spans 12 feet 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 × 1020
k-in) The computational model uses a finite element mesh, automatically generated by SAFE The maximum element size is specified to be 24 inches To obtain factored moments and flexural reinforcement in a design strip, one one-foot-wide strip is defined in the X-direction on the slab, as shown in Figure 1
Free edge
1ft design strip Free edge
Simply supported edge at wall
12 ft span
Y X
Simply supported edge at wall
Free edge
1ft design strip Free edge
Simply supported edge at wall
12 ft span
Y X
Simply supported edge at wall
Figure 1 Plan View of One-Way Slab
One dead load case (DL80) and one live load case (LL100) with uniformly distributed surface loads of magnitudes 80 and 100 psf, respectively, are defined
in the model A load combination (COMB100) is defined using the ACI 318-08 load combination factors, 1.2 for dead loads and 1.6 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 the analysis, design is performed in accordance with ACI 318-08 using SAFE and also by hand computation Table 1 shows the comparison of the moments and design reinforcements computed using the two methods
Trang 2Software Verification
PROGRAM NAME: SAFE
REVISION NO.: 0
EXAMPLE ACI 318-08 RC-SL-001 - 2
G EOMETRY , P ROPERTIES AND L OADING
Depth of tensile reinf d c = 1 in
Clear span l n , l 1 = 144 in
Concrete strength f c = 4,000 psi Yield strength of steel f y = 60,000 psi Concrete unit weight w c = 0 pcf Modulus of elasticity E c = 3,600 ksi Modulus of elasticity E s = 29,000 ksi
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 the design reinforcements
Table 1 Comparison of Design Moments and Reinforcements
Reinforcement Area (sq-in) Load
Level Method
Moment
Medium
,min
A+s = 0.1296 sq-in
C OMPUTER F ILE : ACI318-08RC-SL-001.FDB
C ONCLUSION
The SAFE results show a very close comparison with the independent results
Trang 3Software Verification
PROGRAM NAME: SAFE
REVISION NO.: 0
EXAMPLE ACI 318-08 RC-SL-001 - 3
H AND C ALCULATION
The following quantities are computed for the load combination:
ϕ = 0.9
b = 12 in
As,min = 0.0018bh = 0.1296 sq-in
1
4000
1000
β = − ⎛⎜ ′− ⎞⎟=
c
f
max
0.003 0.003 0.005
a max = β1 c max = 1.59375 in For the load combination, w and M u are calculated as follows:
w = (1.2w d + 1.6w t ) b / 144
8
2 1
wl
M u =
A s = min[A s,min , (4/3) A s,required] = min[0.1296, (4/3)2.11] = 0.1296 sq-in
COMB 100
w d = 80 psf
w t = 100 psf
w = 21.33 lb/in
M u = 55.296 k-in The depth of the compression block is given by:
b f
M d
d a
c
u
ϕ
'
2 85 0
2
−
−
The area of tensile steel reinforcement is then given by:
⎟
⎠
⎞
⎜
⎝
⎛ −
=
2
a d f
M A
y
u s
ϕ
= 0.2114 sq-in > A s,min
A s = 0.2114 sq-in