7_Slabs
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Transcript of 7_Slabs
11/16/2014
1
CE 432 1st Semester 14/15
Dr. Nadim Shbeeb
Slabs
One Way Slabs
CE 432 1st Semester 14/15
Dr. Nadim Shbeeb
Introduction• Reinforced concrete building construction
commonly has floor slabs, beams, girders and
columns continuously placed to form a
monolithic system.
11/16/2014
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CE 432 1st Semester 14/15
Dr. Nadim Shbeeb
Introduction• B1,B2, and B3 are supported on the Girders
• B4,B5, and B6 are supported on the Columns
• G1,G2, and G3 are supported on the Columns
CE 432 1st Semester 14/15
Dr. Nadim Shbeeb
Types of Slabs• One-Way Slabs
– Solid Slabs
– Ribbed Slabs (Ribbed Joist floor)
2SpanShort
Span Long>
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CE 432 1st Semester 14/15
Dr. Nadim Shbeeb
Types of Slabs
• Two-Way Slabs
– Solid Slabs
– Ribbed Slabs (Ribbed Joist floor)
CE 432 1st Semester 14/15
Dr. Nadim Shbeeb
Types of Slabs
• Flat Slabs
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CE 432 1st Semester 14/15
Dr. Nadim Shbeeb
Example S1
• A one-way single span reinforced concrete
slab has a simple span of 5.0 m. The service
live load is 15 kN/m2, the dead load is 10
kN/m2 (including self weight) and 28 MPa
concrete is specified for use with steel with a
yield stress equal to 420 MPa. Design the slab,
following the provisions of the ACI code.
CE 432 1st Semester 14/15
Dr. Nadim Shbeeb
Example S1• First determine the thickness of the slab:
– From Table 9.5a of the ACI code
– Assume a 1.0 m thickness strip (b=1.0 m)
m 250.020
5
20===
lh
mm 217.5d and mm 250h use thus250 mm 88.163
28
420*0123857.0588.011000*420*0123857.0*9.010*112
0123857.06.0
m-kN 5.1128
kN/m 3615*6.110*2.1
26
max
2
==<=⇒
−=
=
==
=+=
d
d
lwM
w
uu
u
ρ
11/16/2014
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CE 432 1st Semester 14/15
Dr. Nadim Shbeeb
Example S12mm 89.26935.217*1000*0123857.0 ==sA
Use 5φ25
m-kN 70.181
0113.01000*186
51.3694
90.00980.096.50
96.505.217003.0
mm 50.96cmm 31.431000*28*85.0
420*37.2454a
mm 5.2175.1220250
min
=
>==
=⇒=
−=
=⇒==
=−−=
u
t
M
d
ρρ
φε
Using ρ formula As=1451.88 mm2 3 φ25
Mu=113.84 kN-m
CE 432 1st Semester 14/15
Dr. Nadim Shbeeb
Example S2
• A reinforced concrete slab is built integrally
with its supports and consists of two equal
spans, each with clear span of 4.5 m. The
service live load is 5 kN/m2, and 28 MPa
concrete (Density =23.5 kN/m3) is specified for
use with steel with a yield stress equal to 420
MPa. Design the slab, following the provisions
of the ACI code.
11/16/2014
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CE 432 1st Semester 14/15
Dr. Nadim Shbeeb
Example S2
• First determine the thickness of the slab:
– From Table 9.5a of the ACI code
– Use h=165 mm, thus
• Calculate the moments using the ACI coefficients listed in section 8.3.3 of the ACI code
m 161.028
5.4
28===
lh
2
2
kN/m 656.125*6.188.3*2.1
kN/m 88.3165.*5.23
=+=
==
uw
DL
CE 432 1st Semester 14/15
Dr. Nadim Shbeeb
Example S28.3.3 — As an alternate to frame analysis, the following approximate
moments and shears shall be permitted for design of continuous
beams and one-way slabs (slabs reinforced to resist flexural stresses
in only one direction), provided (a) through (e) are satisfied:
(a) There are two or more spans;
(b) Spans are approximately equal, with the larger of two
adjacent spans not greater than the shorter by more than 20
percent;
(c) Loads are uniformly distributed;
(d) Un-factored live load, L, does not exceed three times un-
factored dead load, D; and
(e) Members are prismatic.
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CE 432 1st Semester 14/15
Dr. Nadim Shbeeb
Example S2
kN.m 679.105.4*656.12*24
1
24 :supportexterior At
kN.m 306.185.4*656.12*14
1
14 :midspanAt
kN.m 476.285.4*656.12*9
1
9 :supportinterior At
22
22
22
===−
===+
===−
n
n
n
wlM
wlM
wlM
24
1−
14
1+
9
1−
CE 432 1st Semester 14/15
Dr. Nadim Shbeeb
Example S2
mm 115d thusmm 165h Use
mm 40.70
)59.01(
0181.0
'
_
0.005
==
=
−
=
=
c
y
y
Interioru
f
fbf
Md
ρφρ
ρ
mEach 145 Use
mm 733.693
)36.2(85.0
85.0
2
'2''
φ
ρ
φ
φφρ
==⇒
−−=
bdA
df
Mfbd
b
f
f
f
s
y
ucc
y
c
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CE 432 1st Semester 14/15
Dr. Nadim Shbeeb
Example S2
m-kN 44.29
0071.01000*108
69.769
90.0173.098.15
98.15108003.0
mm 98.151000*28*85.0
420*69.769
85.0
1c
mm 108750165
_
min
=
>==
=⇒=
−=
==
=−−=
Interioru
t
M
d
ρρ
φε
Similarly
m)per 143 (Use mm 360
m)per 143 (Use mm 437
2_
2_
φ
φ
=
=
Exteroirs
midspans
A
A
CE 432 1st Semester 14/15
Dr. Nadim Shbeeb
Example S2
Shrinkage bars=0.0018bh=297 mm2 use 1φ20
c
unu
u
V
dwlw
V
φ2
1kN 38.31108.*656.12
2
5.4*656.12*15.1
215.1
<=−=
−=
Shear Consideration
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CE 432 1st Semester 14/15
Dr. Nadim Shbeeb
Reinforcement Details
L2L1
L1/5 L1/3 L2/3 L2/3
L2L1
L1/5 L1/3 L2/3 L2/3
L1/7 L1/5 L2/4 L2/4
Straight bars
Bent bars
CE 432 1st Semester 14/15
Dr. Nadim Shbeeb
Ribbed Slab
40 cm
36 cm
2 cm
14, 18 ,24 , 32 cm
7 cm
9 cm
3.5 cm
2.5 cm
12-15 cm20 cm 20 cm
Main
Reinforcement
Shrinkage and
Temperature
ب�ط
مونه
حصمه
قصاره
11/16/2014
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CE 432 1st Semester 14/15
Dr. Nadim Shbeeb
Ribbed Slab
Design as T-beam
52 cm
12 cm
7 cm
14,18, 24,32 cm
Review ACI provisions 8.13
CE 432 1st Semester 14/15
Dr. Nadim Shbeeb
3.0 m
6.5 m
0.5 m
0.5 m
0.5 m
0.5 m
Design the Ribbed slab below according
to the ACI code provisions fy=420 MPa and fc’=21 MPa
B1
B3
B2
B4
11/16/2014
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CE 432 1st Semester 14/15
Dr. Nadim Shbeeb
Ribbed Slab
15 cm20 cm 20 cm
2 cm
18cm
7 cm
9 cm
3.5 cm
2.5 cmب�ط
مونه
حصمه
قصاره
cm 75.1816
3
16===
lhTo estimate bw assume:
Cover =25 mm,
Stirrups φ8,
Main reinforcement 2 φ20,
distance between reinforcement=30mm mmmm
bw
150 Use136
30)20825(*2
=
+++=
CE 432 1st Semester 14/15
Dr. Nadim Shbeeb
Ribbed Slab
15 cm20 cm 20 cm
2 cm
18cm
7 cm
9 cm
3.5 cm
2.5 cmب�ط
مونه
حصمه
قصاره
2kN/m 34.10.55
0.73kN/m 73.024*18.0*)19.015.0(
2
1===+
To estimate DL:
Rib :
ب�ط : 2kN/m 625.025*025.0 =
:مونه 2kN/m 77.022*035.0 =
حصمه :2kN/m 8.120*09.0 =
قصاره : 2kN/m 44.022*020.0 =
Block:2kN/m 36115
20
1
550
1.*.
.*
.=
hf:2kN/m 68124070 .*. =
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CE 432 1st Semester 14/15
Dr. Nadim Shbeeb
Ribbed Slab
2kN/m 64.150.2*6.137.10*2.1 =+=⇒ uw
DLSlab=1.8+.625+.77+.44+1.36+1.68=8.02 kN/m2
Assume Partitions DL=2.35 kN/m2
DLTotal=10.37 kN/m2
LL=2.0 kN/m2
Thus load per rib is
kN/m 60864.15550 .*.wu ==⇒
CE 432 1st Semester 14/15
Dr. Nadim Shbeeb
Ribbed Slab
kN.m 17.138
5.3*81.8
8
22
===lw
M uu
Design as T-beam
26
mm 563.193
)2
70215(*420*9.0
10*17.13=
−
=sA
Assume stress block depth equal to flange thickness
fha <=××
×=
>==
mm 28.85502185.0
420563.193
006.0215*150
563.193minρρ
11/16/2014
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CE 432 1st Semester 14/15
Dr. Nadim Shbeeb
Ribbed SlabUse 2φ14
u
t
1
M kN.m 26.24
)2
17.13215(42088.3079.0
00955.0215*150
88.307
005.00386.049.15
49.15215003.0
mm 49.1585.0
17.13ac
mm 17.13550*21*85.0
420*88.307
>=
−×××=
==
>=−
=
===
==
u
w
M
a
ρ
ε
β
CE 432 1st Semester 14/15
Dr. Nadim Shbeeb
Ribbed Slab
2 cm
18 cm
7 cm
9 cm
3.5 cm
2.5 cm
15 cm20 cm 20 cm
2φ14 1φ8
ب�ط
مونه
حصمه
قصاره
Design for Shear as before
Design B3 & B4 using wu=15.63*1.5=23.4 kN/m
Design Columns with axial Compression load of
23.4*7/2=81.9 kN/m