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7/28/2019 Structural Design for Low Rise Building
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Supervised by : Dr. Khaled El-Sawy
ID#Name200303838Saeed Mohammad Al Kaabi
200304178Ahmed Obaid Al Dhaheri
200303853Mohammad Owais AL Daraei
200303840Ahmed Abdullah Al Braiki
United Arab Emirates University
College of Engineering
Department of Civil and Environmental Engineering
Industrial Training & Graduation Projects
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-Introduction
-Structural design of :
o Slabs
o Beams
o Stairs
o Columns
o Tie-beams
o Footing
-Environmental, Financial, and Social Impact
-Conclusion
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.fc` Compressive strength of concrete = 28MPa
.fy Yielding strength of steel = 420 MPa
.Ig Moment of inertia
.Ag Area gross
.Ie Effective moment of inertia
.wu Ultimate weight
.Vu Ultimate shear
.Mu Ultimate bending moment
.Mcr Cracking moment .Icr Cracking moment of inertia
.K Effective length factor
.Ec Elasticity of concrete
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The considered low-rise building consists of three blocks
The Majles and Kitchen were designed in GP-I
The Villa is designed in GP-II
The slabs, beams, columns, tie- beams, stairs and footings arestructurally designed.
This structural design in project followed the ACI-318-02M code.
Prokon, AutoCAD and Excel sheets are used in the design.
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Project area = 903 m2
Each floor area of villa
Ground floor = 338 m2
First floor = 261 m2
Each floor consist of several bedrooms, sitting rooms,family hall
Kitchen in ground floor
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What is structural design?
It is finding concrete dimensions and reinforcing steel areas of eachstructural element with insurance of safety and serviceability of themember.
Before designing :
Type of the structural element (i.e., beam, column etc)
The loads carried by the member
The architectural limitations on the member dimensions
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Figure 2:
Architectural plan for first floor.
Figure 1:
Architectural plan for ground floor.
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Figure 3:Typical structural elements
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it is one way
blocks and solid part for R1:
5.3m
5.2m2
short
long
L
L
0.3m
0.43m
nb=22
nr=7
Width of Rib 0.2m
Width of Block 0.38m
No. of Ribs nr 7
No. of Blocks along Ls nb 22
Width of Solid part S1 0.43m
Width of Solid part along Ls (S2) 0.3m
L 5.3m
Ls 5.2m
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m
kN
8.1040.58)1.7(21.4(4.28)1.7(L.L)1.4(D.L)w u
Load Calculation160 160 160420 420
380200380200
Block
Rib
30060
Dimensions in mm
cover)(floorblock)of(weightrib)of(weightslab)topof(weightD.L
m
kN4.380.58)(20.2)(525)10.3
2
0.20.16(25)10.58(0.06D.L
Figure 4: Drawing for ribbed slab (R1)
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Figure 5: Binding moment diagram for R1
Bending Moment for Simply Supported Rib R1
Length (m)
Bending moment
(kN.mm)
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Figure 6: Required steel area for R1
Area of steel is 252mm2 (2#13)
Steel Area R1
Length (m)
Steel Area
(mm2
)
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Shear design for R1
Length (m)
Shear Force
(kN.mm)Figure 7: Shear diagram for R1
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From PROKON software (Vu = 21.1 kn)
stirrupsneed2
VV
20.24KN,2
0.85x47.6
2
V
c
u
c
22.78kN0.85
0.85(47.6)21.1
VVV cus
mmV
dfAS
s
yv450434
22780
3004205.78
Swcs VkNdbfV 4.190300180283
2
3
2max,
kN6.74300180286
1dbf
6
1V wcc
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From PROKON software (Vu = 21.1 kN)
#10@150mmchoose
150mmS
994mm180
3x2x71x420
b
f3A
150mm0.5(300)0.5d
600mm
ofsmallestSmax
w
yv
max
95.25kN30018028
3
1dbf
3
1V
wcs
22.78kNVs
mmS 450
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2#13
2#10
#10@150mm
Check Deflection for R1 Simply supported beam
Minimum thickness of R1 = L/16 = 5.2/16=0.325 m
The actual thickness of R1= 0.36 m > 0.325
It is acceptable
Steel Arrangement
Figure 6:
Final drawing for ribbed slab (R1)
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77.112kN/m8.261.745.051.41.7W1.4WW
8.26kN/mW
45.05kN/m10.53328.55
WWO.Wreaction)(fromWW
kN/m.10.53.53h3kN/mW
3kN/m250.3(0.2x2)W
kN/m.30.20.625bh O.W
8.26kN/m0.58
2.722.07
0.58
Rreaction)rib(fromW
28.55kN/m0.58
7.159.41
0.58
Rreaction)rib(fromW
LDu
L.L
D(wall)partsolidbeamDD.Total
2
D.L(Wall)
partSoild
beam
L.L
L.L
D.LD.L
Beam : DB5
20cm
350cm
Solidpart Solidpart
20cm
20cm
20cm
Beam 30cm
30cm
Wall
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Steel arrangement
Figure 7: Drawing of DB5
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Figure 8: The stair design.
Slope of the stairs
1
28.98380
210tan
Length of the stair
Thickness to control deflection
4.342m=1.23.8=L 22
mmmmL
h 300272
20
11004342
20
min
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Load calculation for the stair
2.8kN/m1.42w
/1.165.108.28.2w
/8.24.12w
/8.22508.04.1w
/5.10253.04.1o.w
Liveload
LD.
coverfloor
step
slab
total
mkN
mkN
mkNhb
mkNhb
stair
beam
mkNLLw
mkNLDw
/8.2.
/4.18cos
1.16.
2
1
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Length (m)
Bending moment
(kN.mm)
Length (m)
Area of steel (mm2)
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From Prokon As=1270mm2 should be between As, min and As,max
Asmin = 0.0033*1400*230=1062.6 mm2
As,max = 0.02125*1400*230=6842.5 mm2 , Asmin < As< Asmax so we will use 7 bars diameter 16
0.02125420600
600
420
280.850.850.75
f600
600
f
`f`0.85(0.75)
0.0033
0.00315
4204
28
0.0033420
1.4
4f
f
f
1.4
ofgreater
yy
c
max
y
c
y
min
1
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Length (m)
V (kN)
Length (m)
As (mm2/mm)
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Shear design
stirrupsforneednosokNVV
kNdbf
V
KNV
c
u
w
c
c
u
,7.1202
4.2411023014006
2885.0
6
'
82.70
3
stirrupsforneednosokNV
V
kNdbf
V
kNV
cu
w
c
c
u
,4.157
2
8.3141030014006
2885.0
6
'
86
3
mmhS
Sthanlessbeshouldmmspacing
90030033
2007
1400
max
max
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Figure 9
The drawing for the stair.
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Column (C1)
212
d
gc
LD
Dd
4933
g
u
cu
3
N.mm8.33x10
1
/2.5IEEI
0.88P
P
bucklingCheck
19.58kNx4.350.3x0.6x25WeightOwn
mm1.35x1012
600x300
12
bhI
968kNP
0.09m0.3x0.30.3hr
29000MPaE4.35m,l,kN/m25
600mmx300mmissectioncrosscolumnAssume
frameunbraced1.5,KAssume
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Column (C1)
1.51K:figure(9)Using
2.78EI/l
EI/l
1.08EI/l
EI/l
beamn
columnu
B
beamn
columnu
A
72.02kN.mhPh
e
M
0.080.030.3
0.015
h
e
3.1P/0.75P1
1
N1019.06)(Kl
EIP
usu
cu
s
5
2
u
2
c
Figure 10
Effective length factor K
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Column (C1)
300mmspacingtieThen
300mm
456mm
300mm
305.6mm
ofsmallesttiesofSpacing
#10issizetieThen
19)#7(choose
1800mm6003000.01AA
0.01use
0.78Ksi895.38MPa/6.
0.60.3
968
A
P
0.2Ksi891.33MPa/6.30.3x0.6x0.
72.02
hA
M
2
gs
min
g
u
g
u
Figure 11
Interaction Diagram
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Figure 12:
The drawing for the column C1.
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Tie-beam (TB1)
The longest tie beam is TB1 which has a length of 7m
which was designed as the critical case. Then, comparedwith the minimum steel area and take the biggest area of
steel:-
In practice, As,min should be greater than the area of steel in
a column that can enough to resist at least 10% of theheavily loaded column load (maximum load is 968 kn)
In practice As,min > 0.5 % Aconcrete of tie beam x-sec
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Length (m)
M (kN.mm)
Length (m)
V (kN)
Figure 13: Moment diagram and shear diagram for TB1.
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For TB1 , As=573 mm2 which is (3 # 16) and compare it with:
10% of the column load (maximum load is 968 kN)
1- As,min =(10% Pu) / (0.9 Fy) =0.1 968 1000 / 0.9 420 = 256 mm2 (2 # 13)
All the tie beams have a cross-section 20cm x 60cm
2- As,min = 0.5 % Aconcrete = 0.005 200 600 = 600mm2 (3 # 16)So, the biggest area is 3 # 16.
For shear (stirrups):
maxs,s
wcmaxs,
svs
sv
VV395.1kN
560200283
2dbf
3
2V
1Check
kN38.57280.164420560
AfydVFind,
957mm0.164
78.52SSo,
S
Asn
fyd
Vs0.164A
#10@280mmchoose
SS
280mmS
989.1mmb
f3A
280mm0.5d
600mm
ofsmallestS
197.6kN560200283
1dbf
3
1V
2Check
max
max
w
yv
max
wcs
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Footing (F1)
94.8kN20.22648.6TB1)(fromPP
291kN40251B11)(fromPP
253.5kN35.5218HB10)(fromPP
LD
LD
LD
2.2m2.2mChoose4.69m150
63.93639.3A
63.93kN)P(P0.1PAssume
AcapacitybearingPPP639.3kN394.8291253.5PP
150kN/m1.5Kg/cmcapacityBearing
F1
2
footing
LDextra
footingextraLD
LD
22
0.95m
0.3m
0.6m
2.2m
2.2m
0.8m
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180.76kN.m)2
0.95(0.952.2227.44)
2
z(LzqM
122.6kN.m)2
0.8
(0.82.2227.44)2
z
(BzqM
227.44kN/m2.22.2
1100.8
A
1.7L1.4Dq
150kN/m136.05kN/m2.22.219.206639.3
APPP
63.93kN19.206kN15)0.4)(240.6(1.50.315)0.4(242.22.2
)h)(lb(1.5)LBh(P
:Check
22ulongu,
11ushortu,
2
chosen
u
22
chosen
extraLD
soilconcretesoilconcreteextra
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12#16)(choose2323.2mm32022000.0033bdA
choose
0.02125420600
600
420
280.850.850.75
f600
600
f
`f`0.85(0.75)
0.0033
0.003154204
28
0.0033420
1.4
4f
f
f
1.4
ofgreater
0.00216
280.85
0.89211
420
280.85
`0.85f`
2R11
f
`0.85f`
0.89MPa891.52kN/m0.08)(0.42.20.9
180.76
bd
MR
entreinforcemLong
2s
minmin
yy
c
max
y
c
y
min
c
n
y
c
2
22u
long
1
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11#16)(choose2178mm30022000.0033A
)choose0.00167(280.85
0.692
11420
280.85
0.69MPa688kN/m0.1)(0.42.20.9
122.6
bd
MR
entreinforcemShort
2s
minmin
2
22u
short
0.95m
0.3m0.6m
2.2m
2.2m
0.8m
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OkVV
1847.09kN
4002320)(300320)(60012
2840.852hd)(bd)(l
12
f4V
750.552kN0.32)0.32)(0.3(0.62.22.2227.44d)d)(b(lBLqV
shearwayTwo
OkVV
527.74kN32022006
280.85Bd
6
f0.85V
265.2kN0.32)(0.952.2227.44d)B(zqV
entreinforcemLong
OkVV
527.74kN32022006
280.85Bd
6
f0.85V
190.14kN0.32)(0.82.2227.44d)B(zqV
entreinforcemshortshearwayOne
cu
c
c
uu
cu
c
c
2uu
cu
c
c
1uu
0.95m
0.3m0.6m
2.2m
2.2m
0.8m
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C1(7#19)columnasentreinforcemsameUse
Safe(Design)PP
2728.8kN(Design)P15.16MPa2.2KsiAg
Pu
figurethisUsing
OK0.010.011600300
1985
A
A
180000mm600300A1100.8kN,P
ColumnNeck
uu
u
g
s
4
gu
Figure 14: Interaction diagram
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Figure 15:
The drawing of the footing F1.
12#162.2m
0.4m
0.85m
12#16
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The environmental aspects and impact of the project should be
controlled by the involved parties to control the bad effects in both
sides either by the environment on the building or on the other way.
- here in UAE, the building usually consumes some of the countrys energy
and water resources and this could be reduced by using green house
technology where solar and wind energies can be used extensively.
- Although the building material considered in this project (i.e., reinforced concrete) isrelatively cheap compared to the steel material, it is, unfortunately, not
environmental friendly
Impact
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Each building in this world represents a unique and creativeidea that is made real by the cooperation of the involved design andconstruction parties.
It is an event that starts by developing the owner ideas on the hands ofengineers to achieve safety, serviceability and creativity of project.
For example, as a social aspects, here in UAE, traditions and thecultural background of the people is totally different from thewestern countries. Thats should be achieved by the architecturaldesign to represent the cultural identity in the scope of the projectdesign beside the modernity.
Impact
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Conclusion and recommendation
The experience gained in the design process is invaluableand represents a major stone in building an efficientstructural designer engineer.
Finally, it is recommended that the design work achievedin this project is originally performed by two groupsusing two different structural systems.
This would provide information enough to compare thecost of each system and get experience with cheapersolutions.