LebaneseUniversity
Faculty of Engineering II
Final year project
submitted in partial fulfillment of the requirements for the
Diploma of Civil Engineering
by
Catherine HAGE
Ornella NOHRA
Structural design for GRAND STAR hotel
Project supervisor:Mr.Michel MOUAIKEL
GRAND STAR- Hotel
BEIRUT-LEBANON
JULY, 2013
We have taken efforts in this project, the success and final outcome of it required a
lot of guidance and assistance from many individuals and organizations. We are
extremely fortunate to have got this all along the completion of our project work.
Whatever we have done is only due to such guidance and assistance and we will
not forget to extend our sincere thanks to all of them..
We would like to express our gratitude towards our parents and families for their
kind cooperation and encouragement which help us in completion of this final year
project.
We would like to express the deepest appreciation to my professor Dr. Michel
MOUAIKEL who has the attitude and substance of a genius, he continually and
convincingly conveyed a spirit of adventure in regard to research and an
excitement in regard to teaching. Without his guidance and persistent help this
project would not have been accomplished.
We are highly indebted to Mr. Georges MARJ, head of the structural department at
“Dar Al Handasah, Shair and partners” for giving us the opportunity to study one
of their interesting project, though he had busy schedule managing the company
affairs.I also owe my profound gratitude to their engineers for their guidance and
constant supervision, and for providing us necessary informations regarding the
project.
We also wish to extend our thanks for Dr.Assad KALLASSY and Dr. Michel
KHOURY for their care and interest all over these three years of studying.
A special thanks also to those who made this diploma possible, all our doctors at
the faculty of engineering Roumieh for looking forward to make us distinguished
engineers.
My thanks and appreciations also go to my colleague in developing the project and
people who have willingly helped us out with their abilities.
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NOTATIONS
Chapter3:
UBC97: Uniform Building Code 97
C P : Wind ward coefficient or Leeward
coefficient according to ASCE7-05
ASCE7-05: American Society of Civil Engineers 5
Minimum Design Loads for Buildings
and Other Structures
θ: The stability coefficient for PΔ effect
assessment check
AISC : American institute of steel construction
P x : total unfactored gravity load at and above
level x for PΔ effect
ACI: American Building Code Requirements for
structural Concrete
Δ: Seismic story drift for PΔ effect assessment
check
f’c: specified compression strength of concrete
VX: seismic shear force between levels x and x-1
for PΔ effect
fy: specified yield strength of non-prestressed
reinforcement
Z: seismic zone factor according to UBC97
Ig: moment of inertia of gross section
C v & Ca : seismic coefficient according to UBC97
hsx: story height below level for PΔ effect
R: Numerical coefficient representative of inherent
over strength and global ductility capacity of
lateral resisting system according to UBC97
d:distance from extreme compression fiber to
centroid of longitudinal tension
reinforcement
I: Importance Factor
A min : minimum area of flexural reinforcement
W: Total Seismic Dead Load
bw: web width
C t : Numerical Coefficient
H: height of the building
hn: Height of the building
∆d :displacement obtained from the unscaled
dynamic analysis
∆m : maximum inelastic response displacement
under earthquake effect
T A : structural period according to Method A in
UBC97
T B : structural period according to Method B in UBC97
T: Elastic fundamental period of vibration in seconds
of the structure in the direction under consideration
Icr : moment of cracked inertia
V sd : the static base shear calculated taking into
account the requirements of UBC1997
paragraph 1630.10.3
V s :is the design base shear calculated according
to UBC1997-1630.2.1
V d :the unscaled dynamic base shear
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V: Total Base shear
V max : Maximum Total Design Base shear
V min : Minimum Total Design Base shear
LL: Live loads
DL:dead loads
Mx: moment in member in x-direction
My: moment in member in x-direction
K zt : Typographical factor according to ASCE7-05
K d : Direction factor according to ASCE7-05
N: axial force normal to cross section occurring
simultaneously with Vu or Tu; to be
T: torsional moment at section
Chapter 5
ASCE7-05: American Society of Civil Engineers 5
Minimum Design Loads for Buildings and Other
Structures
UBC97: Uniform Building Code 97
V:basic wind speed according to ASCE7-05
K d : wind directionality according to ASCE7-05
I: importance factor according to
ASCE7-05
K zt : topographical factor
Cp : Wind ward coefficient Leeward
coefficient
Chapter 6
W: weight for non-building structures which includes
all dead loads as defined for buildings in
section 1630.1.1.
SDL: Superimposed dead loads
Z : Seismic Zone Factor according to UBC97
Cv: velocity-based ground response coefficient, for a
specific seismic zone and a soil profile, from UBC
Table 16-R
Ca: acceleration-based ground response coefficient,
for a specific seismic zone and a soil profile, from
UBC Table 16-Q.= 0.2
R: Numerical Coefficient representative of inherent
over strength and global ductility capacity of lateral
force resisting system. We have a bearing wall
system .
R requested by the UBC 1997 is 4.5
I: Importance Factor according to UBC97
C t : Numerical Coefficient according to UBC97
T: Elastic fundamental period of vibration in seconds
of the structure in the direction under consideration
T y : structural period according to Method B in UBC97
in y-direction
V y : the Total Static BaseShear in Y direction
Vx : the Total Static Base Shear in X direction
l n : length of clear span in the long direction measured
face to face of supports in slabs without beams and
face to face of beams or other supports in other cases
DL: dead loads
LL: Live loads
Mcr : cracking moment
Bw: web width
𝜌: ratio of non-prestressed tension reinforcement
𝜌𝑚𝑖𝑛 /𝜌𝑚𝑎𝑥 .: minimum/maximum ratio of nonprestressed tension reinforcement
As: area of tension reinforcement
Φ: strength reduction factor
d:distance from extreme compression fiber to
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h n : Height of building
T B : structural period according to Method B in UBC97
T x : structural period according to Method B in
UBC97 in x-direction
F t : whiplash effect force according to UBC97
H: building height
Ux, Uy,Uz: modal displacement given by ETABS in
direction x,y,z respectively
M: building total mass
centroid of tension reinforcement
ӯ=
Ix= moment of inertia is the X direction
Iy=moment of inertia is the Y direction
Xi= the abscissa of the wall from the origin
Yi= the ordinate of the wall from the origin
Xr=the abscissa of the centre of torsion from the origin
Yr= the ordinate of the center of torsion from the origin
Ex= eccentricity in the X direction
Ey= eccentricity in the Y direction
Nu= factored axial load
Mu= factored moment strength at section
δr : : seismic drift in story
Pr: total dead at the specified floor
Chapter 7
ΔS: Design Level Response Displacement, which is
the total drift or total story drift that occurs when
the structure is subjected to the design seismic
forces extracted from Etabs software.
ΔM: Maximum Inelastic Response Displacement,
which is the total drift or total story drift that occurs
when the structure is subjected to the Design Basis
Ground Motion, including estimated elastic and
inelastic contributions to the total deformation
according to UBC97.
R: Numerical Coefficient representative of inherent
over strength and global ductility capacity of lateral
force resisting system. The building is a dual system
(Shear walls with Intermediate Moment Resisting
Frame) R requested by the UBC 1997(Table16-N) is
6.5 (to be on the Safe Side R = 5.5 is adopted).
Fr: seismic shear at the specified floor
E= the earthquake load on an element of the structure resulting
from the combination of the horizontal component, Eh, and the
vertical component, Ev.
Eh= the earthquake load due to the base shear, V, as set forth in
Section 1630.2 or the design lateral force, Fp, as set forth in
Section 1632.
Ev=the load effect resulting from the vertical component of the
earthquake ground motion and is equal to an addition of
0.5CaID to the dead load effect, D, for Strength Design, and
may be taken as zero for Allowable Stress Design.
Em=the estimated maximum earthquake force that can be
developed in the structure as set forth in Section 1630.1.1.
Ω0= the seismic force amplification factor that is required to
account for structural overstrength, as set forth in Section
1630.3.1
Ux, Uy,Uz: modal displacement given by ETABS in
direction x,y,z respectively
Sum Ux, Sum Uy, Sum Uz : modal mass participation
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in a direction given by ETABS
Chapter 8
𝑙𝑢 : unsupported length of a compression member
K: effective length factor
R: radius of gyration equal to 0.3 times the direction
of the rectangular column that stability is
considered.
Ag: gross area of section
Ast: total area of longitudinal reinforcement
d b : nominal bar diameter
Ls: splicing length of deformed bars
L db : devolepment length of deformed bars
Pu: factored axial load
Nu: factored axial load
M 1 : smaller factored end moment on a compression
member positive if member is bent in single curvature
M 2 : larger factored end moment on a compression
member
Pn: nominal axial load strength
Mn: nominal moment strength at section
Vn: nominal shear strength
Vc: nominal shear strength provided by concrete
Vs: nominal shear strength provided by shear
reinforcement
Bw: web width
δ u : lateral design displacement
H w : wall height
c : distance from the extreme compression fiber to
H: thickness of wall
the neutral axis calculated for the factored axial force
As: principal flexural reinforcement
and nominal moment strength
D: distance from extreme compression fiber to P cr: : critical axial load
centroid of tension reinforcement
L: total height of the wall
Av: area of Horizontal shear reinforcement
lw: length of the entire wall or segment of wall
s 2 :spacing of horizontal shear reinforcement
considered in the direction of the shear force
E: modulus of elasticity of concrete
Ah: area of vertical shear reinforcement at spacing
s1
s1:spacing of horizontal shear reinforcement
I m: moment of inertia of section resisting externally
Vu: factored shear load
applied factored loads
Chapter 10
l n :length of clear span in the long direction
measured face to face of supports in slabs
without beams and face to face of beams or
other supports in other cases
Ig: moment of inertia of concrete section
about
centroidal
axis
neglecting
reinforcement
fr: modulus of rupture of concrete
DL: dead loads
Yt: distance of from centroidal axis to
extreme fiber in tension
LL: Live loads
Icr: moment of inertia of cracked section
Mcr : cracking moment
transformed to concrete
E cracked :modulus of elasticity of cracked section
Φ: strength reduction factor
n:modular ratio of elasticity never less then 6; = λ:multiplier of additional long term deflection
Es/Ec
𝜌′ : ratio of non-prestressed compression
Ma: maximal moment in member at stage reinforcement
deflection is computed
Vu : factored shear
h: overall thickness of member
αs: column coefficient 40 interior, 30 edge,
Es: modulus of elasticity of steel
20 corner
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Ec: modulus of elasticity of concrete
As: principal flexural reinforcement
f’c: specified compression strength of concrete
As’: area of compression reinforcement
Ie:effective moment of inertia
Δi adm.: maximum instant deflection
l: span length of beam or one way slab
b0: perimeter of critical section for punching
and shear calculations
d:distance from extreme compression fiber
to centroid of tension reinforcement
d’: distance from extreme compression fiber
to centroid of compression reinforcement
d:distance from extreme compression fiber
to centroid of tension reinforcement
Chapter 11
Mb = balanced moment
M1 = primary moment, internal tendon force
multiplied by its eccentricity from the cross section
centroid
M2 = secondary moment,the deformation held
in place by the internal support will causea
support reaction
Δi :maximum instantaneous deflection
Δ LT :maximum long term deflection
X’,y’= minimum edge distance between
anchorage ends to the closest external surface
in x and y direction
Fpu=
Fpy=
F’ci=
F’c
Acf=
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Table of Contents
Chapter 1 GRAND STAR............................................................................................................................... 19
1.INTRDUCTION ...................................................................................................................................... 19
2.MAJOR CONSTRAINTS .......................................................................................................................... 19
Chapter 2 PROJECT PROCEDURE................................................................................................................. 22
Chapter 3 CONCEPTUAL REPORT ................................................................................................................ 24
1.INTRODUCTION .................................................................................................................................... 24
2.BUILDING DESCRIPTION ....................................................................................................................... 24
3.DESIGN PHILOSOPHY ........................................................................................................................... 25
4.SOIL PROPERTIES .................................................................................................................................. 25
5.DESIGN CRITERIA.................................................................................................................................. 25
5.1 DESIGN CODES FOR PRACTICE AND STANDARDS ......................................................................... 25
5.2 COMPUTER ANALYSIS OF THE BUILDING AND ITS COMPONENTS ............................................... 26
6.MATERIALS ........................................................................................................................................... 28
6.3 Concrete ........................................................................................................................................ 28
6.4 Reinforcing Bars ............................................................................................................................ 28
7. LOADS .................................................................................................................................................. 29
7.1 Super Imposed Dead Loads SDL .................................................................................................... 29
7.2 Live loads ....................................................................................................................................... 30
7.3 Lateral loads ................................................................................................................................ 31
8. CONCRETE PROTECTIVE COVER ........................................................................................................ 35
Chapter 4 STRUCTURAL SYSTEM ................................................................................................................ 37
1.CLASSIFICATION OF STRUCTURAL SYSTEMS ........................................................................................ 37
1.1 BEARING WALLSSYSTEM ............................................................................................................... 37
1.2 BUILDING FRAME SYSTEM ............................................................................................................ 37
1.3 MOMENT RESISTING FRAME ........................................................................................................ 38
1.4 DUAL SYSTEM ................................................................................................................................ 38
2. DETERMINATION OF STRUCTURAL SYSTEM ....................................................................................... 38
Chapter 5 STATIC LOAD ANALYSIS ............................................................................................................. 40
1.INTR0DUCTION .................................................................................................................................... 40
2. STATIC SEISMIC LATERAL LOAD .......................................................................................................... 40
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2.1 STATIC LATERAL FORCE PROCEDURE ............................................................................................ 40
2.2 BASE SHEAR .................................................................................................................................. 41
2.3 TERMS USED TO CALCULATE THE STATIC BASE SHEAR ................................................................ 42
2.4 TERMS USED TO CALCULTE THE LATERAL STORY FORCES ............................................................ 44
2.5 CALCULATION OF THE STATIC SEISMIC BASE SHEAR .................................................................... 44
3. STATIC WIND LATERAL LOAD .............................................................................................................. 46
3.1 INTRODUCTION ............................................................................................................................. 46
3.2 DESIGN PROCEDURE AND PARAMETER’S DETERMINATION ........................................................ 47
3.3 BASE SHEAR CALCULATION .......................................................................................................... 53
Chapter 6 PRELIMINARY DESIGN ............................................................................................................... 57
1.INTR0DUCTION .................................................................................................................................... 57
2.TRANSFER BEAMS PRELIMINARY DESIGN ............................................................................................ 57
3.TRANSFERSLAB PRELIMINARY DESIGN ................................................................................................ 63
3.1 EVALUATION OF THE TRANSFER SLAB LOADING USNG TRIBUTAY AREA METHOD ..................... 63
3.2 LOADS ON TRANSFER SLAB ........................................................................................................... 64
3.3CHECK FOR PUCHING SHEAR ........................................................................................................ 65
3.4 CHECK FOR ONE-WAY SHEAR....................................................................................................... 67
4.COLUMNS PRELIMINARY DESIGN ........................................................................................................ 67
4.1 LOADS ON COLUMNS .................................................................................................................... 67
Calculation of DL ..................................................................................................................................... 69
Calculation of LL ...................................................................................................................................... 69
Calculation of DL ..................................................................................................................................... 75
Calculation of LL ...................................................................................................................................... 75
4.2CHECK FOR COLUMNS ................................................................................................................... 76
5.SHEAR WALL PRELIMINARY DESIGN .................................................................................................... 79
5.1 INTRODUCTION ............................................................................................................................. 79
5.2 CENTER OF TORSION ..................................................................................................................... 79
5.3 CENTER OF MASS .......................................................................................................................... 82
5.4 SHEAR FORCES .............................................................................................................................. 84
6.SLABS PRELIMINARY DESIGN ............................................................................................................... 95
6.1 Ribbed slab preliminary thickness ................................................................................................ 95
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6.2 Solid slab preliminary thickness .................................................................................................... 96
Chapter 7.DYNAMIC ANALYSIS ................................................................................................................... 98
1.INTRIDUCTION ..................................................................................................................................... 98
2.USING ETABS: ....................................................................................................................................... 98
2.1 MODELING PROCEDURE ............................................................................................................... 98
2.2 BUILDING’S GEOMETRY ............................................................................................................... 99
2.3 LOAD COMBINATIONS ................................................................................................................... 100
3.MODAL MASS PARTICIPATION RATIO ................................................................................................ 101
4. SCALING FACTOR............................................................................................................................... 103
5. LATERAL STORY DRIFTS ..................................................................................................................... 107
5.1 CODE REQUIREMENTS ................................................................................................................ 107
5.2 CALCULATION OF THE SEISMIC LATERAL DRIFT ......................................................................... 109
6. CHECK FOR P-DELTA EFFECT ............................................................................................................. 112
Chapter 8.COLUMNS DESIGN.................................................................................................................... 115
1.INTRODUCTION .................................................................................................................................. 115
2.TYPES OF COLUMNS ........................................................................................................................... 116
2.1 SHORT COLUMNS ........................................................................................................................ 116
2.2 SLENDER COLUMNS .................................................................................................................... 116
3. COLUMN DESIGN .............................................................................................................................. 117
3.1 LOADING SELECTION ................................................................................................................... 117
3.2 COLUMN SELECTION ................................................................................................................... 118
3.3REINFORCEMENT ......................................................................................................................... 118
3.4 ANALYSIS ..................................................................................................................................... 122
3.5 S-CONCRETE OUTPUT ................................................................................................................. 125
.......................................................................................................................................................... 128
3.6 SUMMARY RESULT ...................................................................................................................... 129
Chapter 9. SHEAR WALL DESIGN ............................................................................................................... 132
1.INTRODUCTION .................................................................................................................................. 132
1.1 DESIGN FORCES ........................................................................................................................... 132
1.2 IN PLANE EFFECTS ....................................................................................................................... 132
1.3 OUT-OF-PLANE EFFECTS.............................................................................................................. 135
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2. SHEAR WALL DESIGN ........................................................................................................................ 137
Chapter 10. SOLID SLAB DESIGN ............................................................................................................... 143
1.INTRODUCTION .................................................................................................................................. 143
2. PREDIMENSIONING OF THE SOLID SLAB ........................................................................................... 143
3.CHECK FOR DEFLECTION .................................................................................................................... 144
4. CHECK FOR PUNCHING SHEAR .......................................................................................................... 152
5.One-Way Shear Verification .............................................................................................................. 154
6.REINFORCEMENT ............................................................................................................................... 155
Chapter 11. POST-TENSIONED SLAB DESIGN ............................................................................................ 159
1.INTRODUCTION .................................................................................................................................. 159
1.1 HISTORY OF POST-TENSIONING : ................................................................................................ 159
1.2 PRESTRESSING IN PRINCIPLE : ..................................................................................................... 159
2. ADVANTAGES OF POST-TENSIONED FLOORS.................................................................................... 162
3.BONDED VS UNBONDED TENDON SYSTEM : ..................................................................................... 162
3.1 BONDED SYSTEM ........................................................................................................................ 163
3.2 UNBONDED SYSTEM ................................................................................................................... 164
3.3 WHY A BONDED SYSTEM WERE CHOSEN ? ................................................................................. 164
4.PRESTRESS LOSSES : ........................................................................................................................... 165
5. STRESSING STAGES : .......................................................................................................................... 166
6.FLAT PLATE CONCEPT :....................................................................................................................... 169
6.1 LIMITATIONS ............................................................................................................................... 169
6.2 POST-TENSIONED FLAT SLAB BEHAVIOR .................................................................................... 170
2.LOAD BALANCING .......................................................................................................................... 172
7. DESIGN PROCESS ............................................................................................................................... 175
7.1 POST-TENSIONED SLAB’S MINIMUM THICKNESS ....................................................................... 175
7.2 RAM ANALYSIS ............................................................................................................................ 175
Chapter 12 STEEL DECK DESIGN ................................................................................................................ 191
1.INTRODUCTION .................................................................................................................................. 191
2. COMPOSITE ACTION ......................................................................................................................... 192
3. COMPOSITE DECK ADVANTAGES ...................................................................................................... 193
4.INSTALATION OF DECKING ................................................................................................................. 194
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4.1 PLACING OF DECKING ................................................................................................................. 194
4.2 ATTACH MAETAL DECKING TO STRUCTURAL STEEL................................................................... 194
4.3SHEAR CONNECTORS ................................................................................................................... 195
4.4 INSTALATION OF SHEAR CONNECTORS ...................................................................................... 195
4.5 INSTALATION OF CONCRETE ....................................................................................................... 195
4.6 QUALITY CONTROL ...................................................................................................................... 196
5.STEEL DECK CALCULATIONS ............................................................................................................... 196
5.1 STEEL DECK CONCEPT ................................................................................................................. 197
5.2 SECONDARY BEAM CALCULATIONS ........................................................................................... 198
5.3 PRIMARY BEAM CALCULATIONS ................................................................................................ 207
5.4 BOLTED CONNECTIONS .............................................................................................................. 213
Chapter 13.RIBBED SLAB DESIGN.............................................................................................................. 220
1.INTRODUCTION .................................................................................................................................. 220
2.RIBBED SLAB CONCEPT ...................................................................................................................... 220
3.LOAD CALCULATION........................................................................................................................... 222
4.DESIGN FOR BEAMS ........................................................................................................................... 223
4.1 ULTIMATE LINEAR LOADS: .......................................................................................................... 223
4.2 BEAMS WIDTH: ........................................................................................................................... 223
4.2 FLEXURAL REINFORCEMENT ....................................................................................................... 226
4.3 SHEAR REINFORCEMENT............................................................................................................. 229
5.DESIGN FOR SB4 SUBJECTED TO TORSION ........................................................................................ 232
5.1 FLEXURAL REINFORCEMENT ....................................................................................................... 233
5.2 SHEAR REINFORCEMENT............................................................................................................. 234
5.3 TORSIONAL REINFORCEMENT..................................................................................................... 235
6. DESIGN FOR RIBS............................................................................................................................... 236
Chapter14. TRANSFER BEAMS DESIGN ..................................................................................................... 244
1.INTRODUCTION .................................................................................................................................. 244
2. MOMENTS AND SHEARS ON TRANSFER BEAMS ............................................................................... 245
3.DETAILED REINFORCEMENT CALCULATIONS ..................................................................................... 246
3.1 FLEXURAL REINFORCEMENT ....................................................................................................... 246
3.2 SHEAR REINFORCEMENT............................................................................................................. 248
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3.2 LONG TERM DEFLECTION FOR DB3 ............................................................................................. 251
4. SUMMARY RESULTS .......................................................................................................................... 251
Chapter15. TRANSFER BEAMS DESIGN ..................................................................................................... 255
1. INTRODUCTION ................................................................................................................................. 255
2. LOADINGS.......................................................................................................................................... 255
2.1 SOIL LOADINGS ........................................................................................................................... 255
2.1 AXIAL LOADING ........................................................................................................................... 257
3.
DESIGN .......................................................................................................................................... 259
Chapter 16 FOUNDATIONS ....................................................................................................................... 263
1. INTRODUCTION ................................................................................................................................ 263
2. RAFT FOUNDATION ........................................................................................................................... 265
2.1 DEFINITION.................................................................................................................................. 265
2.2 METHODS USED FOR MAT FOUNDATIONS ................................................................................. 266
3. PRE-DIMENSIONING OF RAFT FOUNDATION: ................................................................................... 273
4 .SAFE MODEL FOR MAT FOUNDATION: ............................................................................................. 274
4.1 PUNCHING SHEAR VERIFICATION ............................................................................................... 274
4.2 ONE-WAY SHEAR VERIFICATION ................................................................................................. 282
4.3 SOIL PRESSURE VERIFICATION .................................................................................................... 284
4.3 RAFT ‘S REINFORCEMENT ........................................................................................................... 285
REFERENCES .............................................................................................................................................. 288
TABLE OF FIGURES:
Figure 1: Typicale floor plan .....................................................................................................................................24
Figure 2: Tiling detail ...................................................................................................................................................30
Figure 3: Design response sprectra .......................................................................................................................33
Figure 4: Response spectra provided by ETABS ...............................................................................................33
Figure 5:Bearing wall system ...................................................................................................................................37
Figure 6:Building frame system ..............................................................................................................................37
Figure 7.Moment Resisting Frame .........................................................................................................................38
Figure 8: Position of DB3 ............................................................................................................................................58
Figure 9: Loads on DB3 introduced to RDM6 ....................................................................................................59
Figure 10: Bending moment diagram on DB3 ...................................................................................................60
Figure 11: Bending moment considering the drop.......................................................................................... 61
Figure 12: Shear diagram of DB3 ............................................................................................................................62
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Figure 13: Transfer beam ...........................................................................................................................................63
Figure 14: Transfer slab with its support ............................................................................................................64
Figure 15: Influence surface of each wall in the typical floor plan ........................................................... 68
Figure 16: Position of all transfer beams on GF plan ...................................................................................... 70
Figure 17: DB3 subjected to dead load .................................................................................................................72
Figure 18: Transfer beam subjected to live load ..............................................................................................72
Figure 19: Tributary area on C1 on the basement slabs ............................................................................... 75
Figure 20: position of CDR and CDG with respect to the origin ................................................................. 83
Figure 21: Flowchrat to determine if boundary zones are required........................................................ 90
Figure 22: Boundary zones ........................................................................................................................................91
Figure 23: S-CONSRETE output for wall ..............................................................................................................93
Figure 24: Interaction diagram given by S-CONCRETE I .............................................................................. 94
Figure 25 - Table 9.5.a from the ACI 318-08 ......................................................................................................95
Figure 26: Table 9.5(c) from the ACI318-08 ......................................................................................................96
Figure 27: Basement plan introduced to ETABS ..............................................................................................99
Figure 28: Typical plan introduced to ETABS ....................................................................................................99
Figure 29: Modal Mass Participation Ratio ...................................................................................................... 101
Figure 30: Modes shapes ......................................................................................................................................... 103
Figure 31: Final scaling factor ............................................................................................................................... 105
Figure 32: Dynamic base shear and static base shear before scaling .................................................. 106
Figure 33: Dynamic and static shear convergence ....................................................................................... 106
Figure 34.Story Drift .................................................................................................................................................. 107
Figure 35: Forces and displacements causing p-delta effect .................................................................... 112
Figure 36. Types of compression members ..................................................................................................... 115
Figure 37. Types of columns .................................................................................................................................. 116
Figure 38. Local axis for column in ETABS model......................................................................................... 117
Figure 39.Typical column at typical floors ....................................................................................................... 118
Figure 40. Load -moment diagram for R8-60.90 columns ........................................................................ 124
Figure 41. Walls deformation due to flexion or shear ................................................................................. 134
Figure 42: deformation of a shear wall wit opening .................................................................................... 135
Figure 43. Out-of-plane effects .............................................................................................................................. 136
Figure 44: Reinforcement $ interaction diagram as given by S-CONCERTE ...................................... 138
Figure 45 - Typical floor slab on SAFE ............................................................................................................... 143
Figure 46: Maps of bending moment due to dead load............................................................................... 144
Figure 47: Maps of bending moment due to live load ................................................................................. 145
Figure 48: Deflection due to dead load .............................................................................................................. 145
Figure 49: Deflection due to live load ................................................................................................................ 146
Figure 50: Ultimate reaction given by SAFE .................................................................................................... 152
Figure 51 - Shear in typical slab ........................................................................................................................... 155
Figure 52: M11 for ultimate combination ........................................................................................................ 156
Figure 53: M22 for ultimate combination ........................................................................................................ 156
Figure 54: Bending moment surface for different arrangements of tendons.................................... 171
Figure 55: Balance loading ..................................................................................................................................... 173
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Figure 56: effect of presressing on reactions and moments on a beam............................................... 174
Figure 57: Tendon's cover limitations given by the catalogue ................................................................ 176
Figure 58: the position and the elevation of the tendon along x direction ......................................... 177
Figure 59: the position and the elevation of the tendon along y direction ......................................... 177
Figure 60: strips drwan in x and y direction ................................................................................................... 178
Figure 61: Stress maps at service stage............................................................................................................. 185
Figure 62: Long-term deflection map ................................................................................................................ 186
Figure 63: additional bonded reinforcement in y direction ..................................................................... 188
Figure 64: Additional bonded reinforcement in x direction ..................................................................... 189
Figure 65: Composite slab detail .......................................................................................................................... 191
Figure 66: Metal deck corrugations parallel to the beam .......................................................................... 192
Figure 67: Metal deck corrugations perpendicaular to the beam .......................................................... 193
Figure 68: Position od the steel deck on the ground floor plan .............................................................. 197
Figure 69: Primary beam and secondary beam ............................................................................................. 198
Figure 70: Loading on secondary beams introduced to RDM6 ............................................................... 199
Figure 71: Shear diagram on secondary beam ............................................................................................... 200
Figure 72: Bending moment diagram on secondary beam ....................................................................... 200
Figure 73: Web tear-out block shear .................................................................................................................. 214
Figure 74: Minimum edge distance ..................................................................................................................... 215
Figure 75 - Table 9.5.a from the ACI 318-08 ................................................................................................... 220
Figure 76- Typical plan - Ribbed solution ........................................................................................................ 222
Figure 77:bending moment diagram for the PB ............................................................................................ 225
Figure 78: Linear load on PB2 ............................................................................................................................... 227
Figure 79: Bending moment on PB2 ................................................................................................................... 228
Figure 80- Critical section of calculation of Vu ............................................................................................... 229
Figure 81: Transfer beam on the ground floor ............................................................................................... 244
Figure 82:Envelope ultimate bending moment diagram ........................................................................... 245
Figure 83: Envelope ultimate shear diagram .................................................................................................. 245
Figure 84: Basement wall loading ....................................................................................................................... 255
Figure 85: Ultimate moment diagram on basement wall .......................................................................... 256
Figure 86: Basement wall influence line ........................................................................................................... 258
Figure 87: Ultimate load on the wall .................................................................................................................. 259
Figure 88: Interaction diagram on the basement wall ................................................................................ 260
Figure 89: Different types of foundations ........................................................................................................ 264
Figure 90: Distribution of the bearing pressure on the bottom of the raft......................................... 267
Figure 91: soil bearing pressure for rigid and non-rigid mat foundation ........................................... 267
Figure 92: Bearing pressure as function of soil properties ....................................................................... 268
Figure 93: bed of spring representing soil below the mat......................................................................... 270
Figure 94 - Critical section dimensions b1 and b2 ....................................................................................... 275
Figure 95 - Section properties for shear stress computations ................................................................ 278
Figure 96 - Interior Column ................................................................................................................................... 279
Figure 97 - Edge or corner column...................................................................................................................... 280
Figure 98: punching shear ratio for the raft .................................................................................................... 281
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Figure 99:Shear stress in the raft ......................................................................................................................... 283
Figure 100: Map representing the soil bearing pressure........................................................................... 284
Figure 101: Ultimate bending moment M11 ................................................................................................... 285
Figure 102: Ultimate bending moment M22 ................................................................................................... 285
TABLE OF TABLES:
Table 1: Modifiers assessments for structural elemets in ultimate and service models ................. 28
Table 2: Bar size used in the design .......................................................................................................................29
Table 3: Bar size used in the design .......................................................................................................................29
Table 4: Minimum design live load in buildings ...............................................................................................30
Table 5: Seismic zone factor ......................................................................................................................................44
Table 6:Seismic coefficient Cv ..................................................................................................................................45
Table 7: Occupancy category ....................................................................................................................................45
Table 8: Structural system .........................................................................................................................................45
Table 9: Seismic base shear calculation ...............................................................................................................46
Table 10: Directionality factor .................................................................................................................................48
Table 11: Occupancy category..................................................................................................................................49
Table 12: Importance factor......................................................................................................................................49
Table 13: Nominal heigth and gust speed ...........................................................................................................51
Table 14: Parameters of wind base shear ...........................................................................................................53
Table 15: Wind base shear calculation .................................................................................................................55
Table 16: Loads on DB3 ..............................................................................................................................................59
Table 17: dreq due to bending .................................................................................................................................60
Table 18: Summary of dreq for all transfer beams .......................................................................................... 63
Table 19: Ultimate load due to implanted columns ........................................................................................ 64
Table 20: shear verification .......................................................................................................................................67
Table 21: Total live and dead load due to each wall on TB .......................................................................... 69
Table 22: Position and moment of inertia for each shear wall ................................................................... 80
Table 23: Detremination of center of mass.........................................................................................................82
Table 24: Translational, rotational and bending moment on the shear wall ....................................... 88
Table 25: Verification of lateral drift due to earthquake ........................................................................... 110
Table 26: Verification of P-DELTA effect along x direction ....................................................................... 113
Table 27: Verification of P-DELTA effect in y drection ............................................................................... 113
Table 28 - Lap splice in compression ................................................................................................................. 121
Table 29 - Lap splice in tension ............................................................................................................................ 121
Table 30: Summary of different types of column reinforcement ........................................................... 129
Table 31: Type of each column ............................................................................................................................ 130
Table 32: reinforcement detail of shear walls ................................................................................................ 139
Table 33: Types of all shear walls ........................................................................................................................ 141
Table 34 - Slabs reinforcement ............................................................................................................................. 157
Table 35: limitations of a flat slab ........................................................................................................................ 169
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Table 36: span-depth ratio ..................................................................................................................................... 175
Table 37: Input to the software ............................................................................................................................ 179
Table 38 : Nominal diameter and nominal reinforcement cross section (FRESSINET) ................ 179
Table 39: Load combinations................................................................................................................................. 180
Table 40: Serviceability design requirements ................................................................................................ 181
Table 41: Inital stress verification ....................................................................................................................... 183
Table 42: Service stress verification ................................................................................................................... 184
Table 43- Summary of the beams dimensions ............................................................................................... 226
Table 44: As required for PB2 ............................................................................................................................... 228
Table 45: As chosen for PB2 ................................................................................................................................... 228
Table 46: shear reinforcement .............................................................................................................................. 232
Table 47: Longitudinal reinforcement for all spans ..................................................................................... 234
Table 48: Summary for ribs reinforcement ..................................................................................................... 237
Table 49: Longitudinal reinfoecement of DB3................................................................................................ 248
Table 50:Shear reinforcement for DB3 ............................................................................................................. 250
Table 51: Bending reinforcement for DB1 ....................................................................................................... 252
Table 52: Shear reinforcement for DB1 ............................................................................................................ 252
Table 53: Bending reinforcement for DB4 ....................................................................................................... 252
Table 54: Shear reinforcement for DB4 ............................................................................................................ 252
Table 55: Bending reinforcement for DBA ....................................................................................................... 253
Table 56: Shear reinforcement for DBA ............................................................................................................ 253
Table 57: Dimension and reinforcement of the wall all over it's height .............................................. 261
MANHATTAN II
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CHAPTER 1
GRAND STAR-Hotel
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Chapter 1 GRAND STAR
1.INTRODUCTION
The GRAND STAR-Hotel is a 17-floor superstructure, located in downtown-BEIRUT and
equipped with world-class business facilities.
It’s a brilliant architectural innovation due to its irregular “L” shaped plan, with a very particular
circular feature at its corner, that connects all public spaces.
The building is located on a plain terrain surrounded by two main roads with close buildings
around, green spaces surround this hotel from the other sides.
Grand star hotel has 6 basements for parking and storage, one ground floor for reception, and
16 floors for guests-rooms, suites, restaurants, meeting rooms, health clubs and entertainment
facilities.
This project focuses basically on the structural design of this building without ignoring at any
phase of the project the architectural aspect of this structure that should be conserved leading
in consequence to many structural difficulties that we tried to solve in the best way it could be
executed.
This hotel makes it with no doubt an interesting structural case to study.
2.MAJOR CONSTRAINTS
In hotels, for architectural purposes mainly, vertical elements are only considered to be walls,
with no need for columns. These could be structurally efficient as much as they could present
architectural assets.
The hotel’s guest-rooms are identical, they are positioned on either sides of the hallway.
Including columns within the floor’s structure will provide visible discontinuities along all
partition walls, unlike shear walls.
For this manner, the main partition between rooms and hallways will be based on the shear
walls themselves, because they offer architectural advantages as much as they offer structural
benefits.
Concerning those benefits, shear walls are essential to horizontal loadings such as wind and
earthquake. They are also supporting elements, they carry the slab’s dead and live load form
one story to another.
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The slab, having all those supporting elements, will need a smaller thickness that it could need
if it were supported by columns. Therefore the total dead load of the foundation will decrease,
not forgetting also the reduction of the seismic effect that grows with the building’s weight.
Concerning the basement’s vertical supporting elements, for architectural purposes,columns
form a more suitable solution for parking than shear walls since they offer a larger open space.
Plus, basements are not exposed to the horizontal loadings stated previously, which is why only
columns and cores are considered to be supporting elements.
Having a discontinuous shear wall system requires the presence of transfer beams on
theground floor slab in order to implant those walls.
The columns supporting the vertical feature are also stopped at the ground floor slab, which is
why a transfer slab is in need in their area in order for them to be implanted.
The basement 1 offers a health club that require a large open space, which is why the ground
floor slab above this club is designed to be a steel deck in order to get a big span with no
intermediate supports .
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CHAPTER 2
PROJECT PROCEDURE
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Chapter 2PROJECT PROCEDURE
This report dwells on the recommended structural systems to adopt, the issues to be
considered during the conceptual phases of the project, as well as preliminary design criteria
that sets out the standards and methods that would be used in the design that has been
detailed afterwards.
The aim of the present report is to provide a design of the structural elements of the GRAND
STAR Hotel project. In this study the basic design criteria (codes, loadings, materials…) and the
analysis methods are presented. The basic assumptions of the numerical analysis are also
stated. Based on the design criteria and assumptions data, a rigorous structural analysis is
performed with three dimensional models of the building using the “ETABS” software.
Hereby is an overview on the scope of work followed throughout this project:
1) The architectural plans and sections are overviewed to clarify thehotel’s
architectural concept.
2) According to the type of usage of each part of the floor ,loading plans were
determined and thus a load rundown has been done.
3) Based on the limitations given by the architectural drawings, a preliminary
dimensioning was performed for the each of the vertical elements:columns,
walls, and cores.
4) In addition to the preliminary sections of the vertical elements, beams sections
and slab limits were shown in the preliminary framing plans.
5) Modeling phase: in this phase the hotel is modeled using ETABS.
6) Design phase: the hotel’s response, obtained from the analysis results, led to
the determination of:
- themaximum lateral sway
- the internal forces in the different structural elements, which allowed
the checking of capacity of the proposed columns and walls by the usage
of a specialized other computer software “S-concrete”
-the transfer of data to other software (Safe, …) which will allow the
checking of the proposed foundation and slabs dimensions.
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CHAPTER 3
CONCEPTUAL REPORT
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Chapter 3CONCEPTUAL REPORT
1.INTRODUCTION
The structure, as previously noted, is a 18 story building with 6 basements .
It has an overall height of 63 m. The location of the building on the island is in Beirut.
2.BUILDING DESCRIPTION
The architecture of the stories is divided as following:
-
The basements are underground parking spots having a height of 2.7 m and an area
of 4200 m2, except for the basement 1.
-
The ground floor contains the entrance and the reception halls. It has a height of 3.5
m and an area of 1800 m2 indoor, and an area of 2400 m2 outdoor.
-
The second floor is the first typical floor, all stories above this level are typical. They
are dedicated to the guest-rooms and suites with a height of 3.5 m each and an area
of 1800 m2.
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Figure 1: Typicale floor plan
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3.DESIGN PHILOSOPHY
The building is in reinforced concrete. The vertical elements will be in high strength reinforced
concrete.
The lateral stability is insured by the system of cores and shear walls system.
A significant factor in the design of the basements and the foundations is the overall self-weight
of the building and its applied finishes. It is important to achieve workable and economic design
that the building self weight is kept to a minimum where uplift issues are not critical. Based on
the above, light weight partitions have been selected for the Project at Hand.
4.SOIL PROPERTIES
Due to geotechnical data, a single strata soil has been detected having the following roperties:
-
-
allowable bearing capacity is 10 Kg/cm2
Unit weight =20 KN/m3
Subgrade modulus=20000 KN/m3
Friction Angle= 25˚
Cohesion =0 MPa
5.DESIGN CRITERIA
5.1 DESIGN CODES FOR PRACTICE AND STANDARDS
-
-
Earthquake Loads
According to UBC 1997 – Uniform building code – USA – Chapter 16: Division III to V
Wind Loads
According to ASCE 7 “American Society of Civil Engineers”.
Reinforced Concrete Design
The design will be performed in accordance to UBC 1997 (Uniform Building Code),
Chapter 19 / Building Code Requirements for structural Concrete (ACI 318-95 & ACI 31802) and commentary (ACI 318-95 & ACI 318R-02)
Steel structure
According to AISC: American institute of steel construction.
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5.2 COMPUTER ANALYSIS OF THE BUILDING AND ITS COMPONENTS
5.2.1 SOFTWARES USED
Advanced computer software, based on finite elements methods (FEM), were used to evaluate the
exerted forces on the structure, as well as to calculate the building response.
After assessing the straining stresses on the structural system due to the applied gravitational and
lateral forces, the slabs, beams, foundations have been resized according to the developed stresses and
strains while Columns and walls has been investigated accordingly.
The used softwares are as follow:
-
CSI Series: Etabs and Safe
RDM 6
S-Concrete
Spread Excel sheets
Robot
5.2.2 SOFTWARES USED
The analysis project building was performed using the ETABS and SAFE software.
The following assumptions were made:
•
•
•
•
Slabs and walls are represented by shell finite elements (triangles and quadrangles),
the raft by thick shell, all beams and columns by bar elements.
The raft foundation is considered to be a two way grid. The supporting soil is
represented by elastic supports.
Both Static Equivalent Force Method and Dynamic Spectral Method were used to
estimate the building response.
The seismic accelerations in X & Y direction along with accidental eccentricities of
5% led to four seismic case of loadings:
- Seismic X+ eccentricity Y
- Seismic X- eccentricity Y
- Seismic Y+ eccentricity X
- Seismic Y- eccentricity X
The elements defining each model are defined as follows:
Walls
As mentioned previously, the walls have the main function in resisting lateral forces induced by the
lateral forces and reducing the building sway. They were also used as bearing walls supporting
gravitational loads. Walls were designed as 2D shell elements.
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Slabs
These are to be modelled as 2D shell element using finer meshing elements, and this ensures that the
vertical loads are correctly distributed to the vertical elements.
Columns
The reinforced concrete columns were mainly used to resist gravitational loads. They are to be modelled
as 1D stick elements.
Beams
Transfer beams are modeled as 1D stick elements.
Base Constraints
The building is supposed fixed at foundation level.
5.2.3 CRACKED PROPERTIES
-
For the computation of drift:
I cr = 1*I g for Vertical Elements (Cores and Columns)
I cr = 0.36*I g for Horizontal Elements (Slabs)
I cr = 0.5*I g for Horizontal Elements (Beams)
-
(ACI 318-95 Article 10.11.1)
For the design of Structural Members:
The design will consider the values obtained using I cr = 0.5*I g for all Structural Members, then
the above mentioned structural members will be verified adopting the following inertias
I cr = 0.7*I g for Vertical Elements (Cores and Walls)
I cr = 0.25*I g for Horizontal Elements (Reinforced Concrete Slabs)
I cr = 0.35*I g for Horizontal Elements (Beams)
(ACI 318-95 Article 10.11.1)
UBC 1910.11.1
In ETABS model, the modifiers were set as shown in the table below :
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Table 1: Modifiers assessments for structural elemets in ultimate and service models
6.MATERIALS
Materials used in the construction of the building are mainly concrete and steel. The required
strength and characteristics of those materials are given in the following paragraphs.
6.3 Concrete
F’c= 40MPa for horizontal elements (slabs – beams – raft foundation)
F’c= 50MPa for vertical elements (column - shear walls – cores)
Since vertical elements are subjected to a compression higher than horizontal elements, and
since the concrete is strong in compression and weak under tension , the concrete strength f’c
of the vertical elements should be larger .
6.4 Reinforcing Bars
According to ACI 318-08 chapter 3 titled “Materials” section 3.5.1, reinforcement shall be
deformed reinforcement except for spiral where plain reinforcement can be used.
The yield strength of reinforcement is taken to be fy = 420 MPa for all members.
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The bars used in the design are given in the following table:
Table 2: Bar size used in the design
Bar Size
Diameter (mm)
Area (cm2)
10
0.785
12
1.13
14
1.54
16
2.01
20
3.14
25
4.91
32
8.04
Table 3: Bar size used in the design
7. LOADS
This part is dedicated to the determination of loads applied on the structure. Loads are divided
into two major parts: Dead loads and Live loads.
Dead loads are permanent loads which values are precisely known, they include Super Imposed
Dead Loads SDL and Self Weight SW. SDL are permanent loads other than the structure itself,
whereas Self Weight is the load of the structure itself.
Live loads are variable loads which values are not precisely known but are estimated according
to the use and final destination of the structure.
7.1 Super Imposed Dead Loads SDL
SDL includes the following loads:
-
Partitions
Plaster
Tiling
7.1.1 Partitions
Partition walls are non-load bearing walls constructed in order to separate rooms in each story.
In this building, partition walls between rooms are the shear walls themselves, as for the
partition within the room they are made with plaster boards
Plaster board is a panel made of gypsum plaster pressed between two thick sheets of paper. It
is used to make interior walls. This construction became prevalent as a speedier and lighter
alternative to traditional partition walls.Their weights is 23 kg/m2
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7.1.2 Tiling
Tiles are mainly made of three layers:
-
Tile blocks of 2 cm thickness and of1900 Kg/m3 density
Mortar layer of 2 cm thickness and of 1800 Kg/m3 density
Sandy layer of 5 cm thickness and of 1300 Kg/m3 density
Figure 2: Tiling detail
Tiling’s weight is given by:
W = 1900 × 0.02 + 1800 × 0.02 + 1300 × 0.05
= 127 Kg/m2
Than the total super-imposed dead load is: 150 kg/m2
7.2 Live loads
Live loads are variable loads that depend on the usage of each floor plan.
Based on ASCE 7-05 for minimum design live load in buildings( table 4.1) :
Table 4: Minimum design live load in buildings
An average of 300 kg/m2 is taken between LL(private room) and LL(public room)
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7.3 Lateral loads
There are a number of international design codes that allow a building to be designed to resist
an earthquake, and these are often augmented by local codes where appropriate. It should be
noted that all seismic design codes exist primarily to safeguard against major structural failure
and loss of life, and not to limit damage or maintain function, and all are based on a statistical
approach. The Uniform Building Code (UBC) is considered by some to have a good track record
for producing buildings that offer reasonable life safety when compared to other codes, such as
the Japanese or Eurocode. Though of American origin, it has been used widely throughout the
world including the Middle East.
Structures designed to the UBC should be able to:
-
Resist a minor level of earthquake without damage
Resist a moderate level of earthquake without structural damage but with some
non-structural damage, e.g. to window frames, doors.
Resist a major level earthquake without collapse but with some structural and nonstructural damage.
In the UBC 1997 the Earthquake Hazard, or design ground motion spectrum, corresponds to a
10% probability to be exceeded in 50 years. This is equivalent to an earthquake that has a
return period of 475 years. We propose to design the building using the UBC 1997. Chapter 16,
Division IV Earthquake Design in volume 2 of the UBC provides specifications for earthquake
resistant design.
The UBC 1997 considers 5 seismic zones (1, 2A, 2B, 3 and 4) and assigns Seismic Zone Factors
(Z) that vary between 0.075 and 0.4. Although these factors are akin to the Peak Ground
Acceleration (PGA), the UBC refrains from using the PGA terminology for the Z factors.
The UBC 1997 requires the soil profile to be determined based on the average soil properties
for the top 30m of the soil profile.
The lateral forces due to earthquake will be calculated considering Lebanon in Seismic Zone 2B.
7.3.1 SEISMIC STATIC ANALYSIS
-Z: Seismic Zone Factor
Z = 0.2 Zone 2B
Table 16-I (UBC 97)
-Soil Profile Type depends on soil
investigation
-C v seismic coefficient
(S c has been adopted)
Table 16-J (UBC 97)
C v = 0.32
Table 16-R (UBC 97)
-C a : Seismic coefficient
C a = 0.24
Table 16-Q (UBC 97)
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-R: Numerical coefficient representative of
inherent over strength and global ductility
capacity of lateral resisting system
R = 4.5
Table 16-N (UBC 97)
-I: Importance Factor
I= 1
Table 16-K (UBC 97)
-W: Total Seismic Dead Load
-C t : Numerical Coefficient
C t =0.048
Section 1630.2.2 UBC 97
-h n : Height of the building
Structural Period (section 1630.2.2 UBC97)
Method A:
3�
4
𝑇𝐴 = 𝐶𝑡 × (ℎ𝑛 )
Method B:
𝑇𝐵 = 2𝜋 × �
�∑ 𝑤𝑖 ×𝜕𝑖 2 �
(max T B that can be used is 1.4 T A )
𝑔×(∑ 𝑓𝑖 ×𝜕𝑖 )
Where T: Elastic fundamental period of vibration in seconds of the structure in the direction
Under consideration
The Total Base Shear:𝑉 =
𝐶𝑣𝐼
𝑅𝑇
×𝑊
The Total Design Base Shear not more than: 𝑉𝑀𝑎𝑥 =
2.5×𝐶𝑎 ×𝐼
𝑅
×𝑊
The Total Design Base Shear not less than: 𝑉𝑀𝑖𝑛 = 0.11 × 𝐶𝑎 × 𝐼 × 𝑊
7.3.2 SEISMIC DYNAMIC ANALYSIS
The Total Design Base Shear should not be less than the Total Design Base Shear obtained from
the Seismic Static Analysis (UBC 1997 Section 1631.5.4.).
If this Total Design Base Shear is below the Total Base Shear obtained from the static analysis,
all results are to be multiplied by scaling factor Cx and Cy for directions X and Y respectively.
The vertical component of the earthquake has also been taken into consideration.
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Figure 3: Design response sprectra
Figure 4: Response spectra provided by ETABS
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7.3.3 WIND ANALYSIS
The building has been verified according to ASCE7-05, Chapter 6, Minimum Design Loads for
Buildings and Other Structures. This paragraph summarizes the design parameters and
standards according to which the wind pressures on the building have been assessed:
-Basic wind Speedadopted
140Km/h
Section 6.5.4 (ASCE7)
-Exposure type
Type C is chosen
Section 6.5.6 (ASCE7)
-I: Importance Factor
I= 1.15 for Categorie2
Section 6.5.5 and Table 1-1
(ASCE7)
-K zt :Topographic factor
K zt =1 no hills
Section 6.5.7, and fig.6-3 (ASCE7)
-K d :Direction factor
K d =0.85 buildings
Section 6.5.4 and table 6-4
(ASCE7)
-C P : Wind ward coefficient
C P = 0.8
Section 6.5.4
-C P : Leeward coefficient
C P = 0.5 for L/B =1
Section 6.5.4
-H: height of the building
To be on the safe side, the design will be based on 50 years occurrence in computing along
wind accelerations.
The maximum drift under 50 year wind load is limited to 1/500 of building height. The inter story
drift to be limited to 1/500 of the height of the store but based on the 10 year wind speed.
7.3.4 PΔ EFFECT ASSESSMENT
Seismic forces cause a structure to deflect laterally. As a result, moments of second order are
induced in the structural members due to the displaced gravity loads. This second order effect
is known as the PΔ effect.
According to UBC section 1630.1.3, P𝛥 effects shall be evaluated using the design seismic
forces producing the design level response displacements (Δ S ). In general, PΔ effects need not
to be considered when the stability coefficient (θ), defined as the ratio of second order
moments to primary moments, is less than or equal to 0.1.
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The stability coefficient (θ) for a given story can be computed from the following equation:
𝜃=
Where:
-
𝑃𝑥 × ∆
𝑉𝑥 × ℎ𝑠𝑥
P x : total unfactored gravity load at and above level x
Δ: Seismic story drift
V X : seismic shear force between levels x and x-1
h sx : story height below level
8. CONCRETE PROTECTIVECOVER
For cast in-situ slabs and flat beams
For walls not exposed to earth and weather
For foundations, and basement wall (exterior face )
For faces of walls , columns , beams , or slabs
exposed to water or weather
For drop beams (interior faces ) not exposed to weather
For columns (interior faces ) not exposed to weather
30 mm
30 mm
50mm
50mm
40mm
40mm
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CHAPTER 4
STRUCTURAL SYSTEM
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Chapter 4 STRUCTURAL SYSTEM
1.CLASSIFICATION OF STRUCTURAL SYSTEMS
UBC section 1926.6 details five major categories of buildings of building types distinguished by
the method used to resist the lateral forces. These consist of bearing walls, building frames,
moment- resisting frames, dual systems, and cantilevered columns. These categories are
further subdivided into the types of construction material used. UBC table 16 – N lists the
different framing systems, with the height verification and response modification factors for
each, together with the restrictions imposed on the use of different building types in specific
seismic zones.
1.1 BEARING WALLSSYSTEM
In a bearing wall system, shear walls or braced frames
providesupport for all or most of the gravity loads and for
resisting all lateral loads. In general, a bearing wall system has
comparably lower value for R since the system lacks
redundancy and has a poor inelastic response capacity. The
lateral support members also carry gravity loads and their
failure result in failure of the gravity load carrying capacity. In
seismic zones 3 and 4, the concrete and masonry shear walls
are required to be specially detailed to satisfy UBC sections
1921 and 2106.
Figure 5:Bearing wall system
1.2 BUILDING FRAME SYSTEM
A building frame system has separate systems to provide support for lateral forces and gravity
loads. A frame provides support for essentially all gravity loads
with independent shear walls or braced frames resisting all
lateral forces. The gravity load supporting frame does not
request special ductile detailing, but it is required to satisfy the
deformation compatibility requirements of UBC Section
1633.2.4 and this imposes a practical limitation on the height of
a building frame system. Failure of the lateral support members
will not result in collapseof the building since the frame
continue to support gravity loads.
Figure 6:Building frame system
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1.3 MOMENT RESISTING FRAME
Moment resisting frames are specially detailed to provide good
ductility and support for both lateral and gravity loads by flexural
action. Inn seismic zones 3 and 4, the moment frames are
required to be specially detailed to satisfy UBC sections 1921 or
2211. A high degree of redundancy can be provided and the
system has an excellent ine4lastic response capacity. Large
lateral displacements may be developed while the gravity load
carrying capacity remains intact.
Figure 7.Moment Resisting Frame
1.4 DUAL SYSTEM
A dual system has a comparably higher value of R since a secondary lateral support system is
available to assist the primary non-bearing lateral support system. Non-bearing walls or bracing
supply the primary lateral support for gravity loads and acting as a backup lateral force system.
The moment resisting frame must be designed to independently resist at least 25% of the base
shear and, in addition the two systems shall be designed to resist the total base shear in
proportion to their relative rigidities.
2.DETERMINATION OF STRUCTURAL SYSTEM
Since the columns do not have a regular position and assessment in the towers the dual system
is not very effective in as a secondary lateral support system, therefore shear wall bearing
system will be adopted.
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CHAPTER 5
STATIC LOAD ANALYSIS
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Chapter 5 STATIC LOAD ANALYSIS
1.INTR0DUCTION
The main goal of earthquake-resistant design is to maintain a structure with sufficient strength
and ductility to assure life safety, i.e., to prevent collapse under the most intense earthquake
expected at a site during the life of a structure. In most structures that are subjected to
moderate-to-strong earthquakes, economical earthquake-resistant design is achieved by
allowing yielding to take place in some structural members.
This section is dedicated to the static load analysis according to the static lateral force
procedure exposed in the Uniform Building Code UBC 97,Section 1630.2 for the seismic lateral
load, and according to the American Society of Civil Engineers ASCEfor the wind static lateral
load.
Eventually a comparison will be made in order to choose which gives the critical base shear
2. STATIC SEISMIC LATERAL LOAD
2.1 STATIC LATERAL FORCE PROCEDURE
The concept employed in equivalent static lateral force procedures is to place static loads on a
structure with magnitudes and direction that closely approximate the effects of dynamic
loading caused by earthquakes.
Concentrated lateral forces due to dynamic loading tend to occur at floor and ceiling/roof
levels in buildings, where concentration of mass is the highest. Furthermore, concentrated
lateral forces tend to be larger at higher elevations in a structure. Thus, the greatest lateral
displacements and the largest lateral forces often occur at the top level of a structure
(particularly for tall buildings). These effects are modeled in equivalent static lateral
force procedures of the UBC by placing a force at each story level in a structure.
where:
-
V = base shear force associated with ground motion at the base of the structure
- F x = lateral story force applied at each story level of the structure
- F t = additional lateral force applied at the top level of the structure (per the UBC)
In general, the distribution of lateral story forces is associated with the first (fundamental)
mode of vibration of a cantilevered structure. (In this case, a typical structure is idealized as a
vertical cantilever rigidly attached to the ground).
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The effects of higher modes of vibration are approximated in the UBC by considering an
additional lateral force, F t , applied to the top level of a structure.
The summation of the lateral story forces (plus the additional lateral force at the top, F t )
must be equivalent to the base shear (V) force applied to the structure due to seismic ground
motion.
2.2BASE SHEAR
Base shear is an estimate of the maximum expected lateral force that will occur due to seismic
ground motion at the base of a structure.
Calculations of base shear (V) depend on:
-
soil conditions at the site
proximity to potential sources of seismic activity (such as geological faults)
probability of significant seismic ground motion
the level of ductility and overstrength associated with various structural
configurations and the total weight of the structure
the fundamental (natural) period of vibration of the structure when subjected to
dynamic loading
The UBC addresses the probability of significant seismic activity in various locations
categorizing geographic regions as Seismic Zones 0 through 4 .
Seismic Zone 0: no seismic activity is expectedto occur.
Seismic Zone 4: high probability of significant seismic activity.
The equivalent static force procedure in the Uniform Building Code (UBC
1630.2)specifies the following formula for calculating base shear (V):
-
V = Cv I W / R T (UBC Equation 30-4)
The UBC also specifies the following upper and lower bounds for V:
-
Upper bound: V < 2.5 C a I W / R (UBC Equation 30-5)
Lower bound: V > 0.11 C a I W / R(UBC Equation 30-6)
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An additional lower bound applies in seismic zone 4:
-
V > 0.8 Z N v I W / R(UBC Equation 30-7)
The upper bound value for base shear tends to govern for relatively stiff structures that exhibit
a small (short) fundamental period of vibration (T). The lower bound values for base shear tend
to govern for relatively flexible structures that exhibit a large (long) fundamental period of
vibration (T)
2.3 TERMS USED TO CALCULATE THE STATIC BASE SHEAR
The terms used to calculate base shear (V) in UBC Equation 30-4 are defined as follows:
 W = total seismic dead load of the structure (dead loads plus applicable
portions of some storage loads and snow loads, as specified in UBC 1630.1.1)
 I = importance factor (UBC Table 16-K)
The importance factor is essentially an extra safety adjustment used to increase the calculated
load on a structure based on its occupancy and/or function. Essential facilities (such as
hospitals, fire and police stations, etc.) and facilities that house toxic or explosive substances
have higher seismic importance factors (I = 1.25) than other structures (I = 1.0).
Higher importance factors are intended to insure that structural integrity is not compromised
and important facilities remain operational during emergencies and natural disasters. Based on
typical occupancy classifications for most structures, buildings are frequently designed using an
importance factor of unity (I = 1.0).
Designers should note that the seismic importance factor (I) is not identical to the
importance factor for wind (I w ). Futhermore, the UBC provides two distinct seismic
importance factors - one (I) for the design of typical structural systems/assemblies
such as shear walls and diaphragms, and another (I p ) for the design of critical
elements that are attached to structures and may need to resist large concentrated
seismic forces (see UBC Section 1632).
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 T = C t (h n )3/4(UBC Equation 30-8)
T = fundamental (natural) period of vibration for a structure
The UBC provides the previous simplified method for estimating T based on the height of the
structure (h n )
Where:
-
C t is according to section 1630.2.2
h n = height of the top level of a structure (m)
 R= ductility and overstrength factor ( UBC Table 16-N)
The R factor is intended to account for inelastic structural behavior and the ability of a
structure to displace/deform and dissipate energy without failing. Since all R factors specified in
UBC Table 16-N are greater than unity (R > 1.0), the R factor effectively reduces the calculated
base shear (V) by varying amounts depending on the ductility of a structure. In general, ductile
structural systems should have higher R factors than brittle structural systems. Typical values
of R for many low-rise wood structures are:
-
R = 5.5 for light frame wood buildings with shear walls that support gravity loads
and simultaneously resist lateral loads in structures less than 4 stories high
-
R = 4.5 for light frame wood buildings with shear walls that support gravity loads
and simultaneously resist lateral loads in structures more than 3 stories high
The following additional R factors also apply to wood structures, but are associated with less
commonly used structural systems:
-
R = 6.5 for light frame wood buildings less than 4 stories high in which the frame system
supports gravity loads independently of the shear panels that resist lateral loads
R = 5.0 for light frame wood buildings more than 3 stories high in which the frame system
supports gravity loads independently of the shear panels that resist lateral loads
R = 5.6 for heavy timber braced frames in which the frame system supports gravity
loads independently of the bracing that resists lateral loads
R = 2.8 for heavy timber braced frames in which bracing supports gravity
loads and simultaneously resists lateral loads
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C v and C a : seismic coefficients associated with structural sensitivity to the velocity and acceleration
(respectively) of seismic ground motion. They are based on the geographic location of the
structure (seismic zone) and soil conditions at the site. Their value are specified in UBC Tables
16-R and 16-Q.
2.4 TERMS USED TO CALCULTE THE LATERAL STORY FORCES
Lateral forces that counteract the base shear, V, are assumed to act at each story level of the
structure. The magnitude of each story force ,F x , is determined from the following formula:
Where:
- h x is the height from the base of the structure to level x
- w x is the portion of the building weight assumed to be “lumped” at level x.
w x typically includes the total weight of the floor or ceiling/roof system at level x,plus
half the weight of the vertical elements (walls; columns) located immediately below
level x and half the weight of the vertical elements located immediately above level x.
F t is an additional lateral force assumed to act at the top of a structure. This force is intended to
approximate the effects of higher modes of structural vibration. The magnitude of F t is
determined based on the natural (fundamental) period of vibration of the structure, T:
Ft = 0
F t = 0.07 T V
F t = 0.25 V
when T < 0.7s
when 0.7s < T < 3.57s
when T > 3.57s
2.5 CALCULATION OF THE STATIC SEISMIC BASE SHEAR
-
Theseismic zone factor Z:
Table 5: Seismic zone factor
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-
The seismic coefficient Cv:
Table 6:Seismic coefficient Cv
-
The occupancy category I:
Table 7: Occupancy category
-
The coefficient R:
Table 8: Structural system
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The following table resume the base shear’s calculations:
Table 9: Seismic base shear calculation
3. STATIC WIND LATERAL LOAD
3.1 INTRODUCTION
This method assumes the quasi-steady approximation. It approximates the peak pressures
on building surfaces by the product of the gust dynamic wind pressure and the mean pressure
coefficients. The mean pressure coefficients are measured in the wind-tunnel or by full-scale
tests and are given by pbar/qz(bar).
The implied assumption is that the pressures on the building surface (external and internal)
follow faithfully the variations in upwind velocity. Thus, it is assumed that a peak value of wind
speed is accompanied by a peak value of pressure or load on the structure. The quasi-steady
model has been found to be fairly reliable for wind loading on small structures.
In static analysis, gust wind speed Vz is used to calculate the forces, pressures and moments on
the structure.
Buildings shall be designed to resist the wind effects in accordance with the requirements of
ASCE 7-05 Chapter 6 specified in the standards listed in section 1604 of UBC97.
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ASCE Code specifies three methods for wind analysis procedure:
1.Method 1-Simplified Procedure as specified in Section 6.4 for buildings meeting
the requirements specified therein
2.Method 2-Analytical Procedure as specified in Section 6.5 for buildings meeting
the requirements specified therein
3. Method 3-Wind Tunnel Procedure as specified in Section 6.6
3.2 DESIGN PROCEDURE AND PARAMETER’S DETERMINATION
According to ASCE – section 6.5.3 , the design procedure is given below :
-
The basic wind speed V and wind directionality factor Kd shall be determined in
accordance with Section 6.5.4.
An importance factor I shall be determined in accordancewith Section 6.5.5.
An exposure category or exposure categories and velocity pressure exposure coefficient
K, or Kj, as applicable, shall be determined for each wind direction in accordance
withSection 6.5.6.
A topographic factor Kzt,shall be determined in accordancewith Section 6.5.7.
A gust effect factor G or G j, as applicable, shall be determined in accordance with
Section 6.5.8.
An enclosure classification shall be determined in accordance with Section 6.5.9.
Internal pressure coefficient GCp, shall be determined in accordance with Section
6.5.11.1.
External pressure coefficients C, or GCDforce coefficients Cj , as applicable, shall
bedetermined in accordance with Section 6.5.1 1.2 or 6.5.1 1.3, respectively.
Velocity pressure q, or qj, as applicable, shall be determined in accordance with Section
6.5.10.
Design wind load p or F shall be determined in accordance with Sections 6.5.12, 6.5.13,
6.5.14, and 6.5.15, as applicable.
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3.2.1 WIND DIRECTIONALITY FACTOR Kd
The wind directionalityfactor, Kd, shall be determined from Table 6-4:
Table 10: Directionality factor
3.2.2 IMPORTANCE FACTOR I
An importance factor, Ifor the buildingor other structure shall be determined from Table 6-1
basedon building and structure categories listed in Table 1- 1
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Table 11: Occupancy category
Table 12: Importance factor
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3.2.3 EXPOSURE
For each wind direction considered, the upwindexposure category shall be based on ground
surface roughnessthat is determined from natural topography, vegetation, and constructed
facilities.
Since the building is located in a flat and unobstructed area , an exposure C is to be considered
3.2.4 VELOCITY PRESSURE EXPOSURE COEFFICIENT Kz
Kz= velocity pressure exposure coefficient evaluated at height z
Or using the following formulas :
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Where:
-
zg =nominal height of the atmospheric boundary layer used in this standard.
Values appear inTable 6-2
α= gust-speed power law exponent from Table 6-2
Table 13: Nominal height and gust speed
3.2.5 TOPOGRAPHIC FACTOR Kzt
The wind speed-up effect shall beincluded in the calculation of design wind loads by using the
factor Kzt.
Since site conditions and locations of structures do not meet all theconditions specified
in Section 6.5.7.1 then Kzt= 1.0
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3.2.6 GUST FACTOR G
According to section 6.5.8.1 Rigid Structures:for rigid structures as defined in Section6.2, the
gust-effect factor shall be taken as 0.85
3.2.7 INTERNAL AND EXTERNAL PRESSURE COEFFICIENT cp and Gcp
Those factor are determined according to the following sections of ASCE_7-05 :
6.5.11.1 Internal Pressure Coefficient. Internal pressure coefficients,GCp;, shall be determined
from Fig. 6-5 based on buildingenclosure classifications determined from Section 6.5.9.
6.5.11.1.1 Reduction Factor for Large Volume Buildings,Ri. For a partially enclosed building
containing a single, unpartitionedlarge volume, the internal pressure coefficient, GCp, , shall
be multiplied by the following reduction factor, Ri:
Where:
- AOx = total area of openings in the building envelope (walls androof, in ft2)
- Vi= unpartitioned internal volume, in ft3
6.5.11.2 External Pressure Coefficients.
6.5.11.2.1 Main Wind-Force Resisting Systems. Externalpressure coefficients for MWFRSs Cp
are given in Figs. 6-6,6-7, and 6-8. Combined gust effect factor and external
pressurecoefficients, GCpj, are given in Fig. 6-10 for low-rise buildings.
The pressure coefficient values and gust effect factor in Fig. 6- 10shall not be separated.
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3.3 BASE SHEAR CALCULATION
The velocity pressure and the main wind force are calculated below :
According to ASCE_7-05 , section 6.5.10:
Velocity pressure, qz , evaluated atheight z shall be calculated by the following equation:
qz= 0.00256 Kz Kzt Kd V2 I (lb/ft2)
[In SI: qz= 0.613 Kz Kzt Kd V2 I (N/m^2) ;V in m/s]
Where:
- Kd is the wind directionality factor defined in Section6.5.4.4,
- Kz is the velocity pressure exposure coefficient definedin Section 6.5.6.6,
- Kzt is the topographic factor defined in Section 6.5.7.2,
- qh, is the velocity pressure calculated using Eq. 6-15 at mean roof height h.
Using the following formula :
Table 14: Parameters of wind base shear
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According to ASCE_7-05 , section 6.5.12.2:
6.5.12.2.1 Rigid Buildings of All Heights. Design wind pressures for the MWFRS of buildings of
all heights shall be determined by the following equation:
Where:
- q = qz for windward walls evaluated at height z above themound u
- q = qh,for leeward walls, side walls, and roofs, evaluatedat height h
- qi= qh, for windward walls, side walls, leeward walls, androofs of enclosed buildings and
for negative internalpressure evaluation in partially enclosed buildings
-
qi= qz for positive internal pressure evaluation in partiallyenclosed buildings where
height z is defined as the levelof the highest opening in the building that could affectthe
positive internal pressure. For buildings sited inwind-borne debris regions, glazing that is
not impactresistant or protected with an impact resistant covering,shall be treated as an
opening in accordance withSection 6.5.9.3. For positive internal pressure evaluation,q;
may conservatively be evaluated at height h(q; = q11)
-
G = gust effect factor from Section 6.5.8
Cp = external pressure coefficient from Fig. 6-6 or 6-8
(GCpi) = internal pressure coefficient from Fig. 6-5
q and qi shall be evaluated using exposure defined in Section6.5.6.3. Pressure shall be applied
simultaneously on windwardand leeward walls and on roof surfaces as defined in Figs. 6-6
and6-8.
The following image is extracted from an excel sheet that is established in order to calculate the
base shear due to wind:
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Table 15: Wind base shear calculation
Since the base shear due to wind is smaller than the base shear due to the earthquake , than the design
due to lateral forces will be evaluated from the earthquake’s base shear .
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CHAPTER 6
PRELIMINARY DESIGN
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Chapter 6 PRELIMINARY DESIGN
1.INTR0DUCTION
Using immediately the architectural data for the modeling phase will cause problems since the
range of the structural element dimensions is not quite defined yet, that is why a preliminary
design of the structure is required before the beginning of the modeling phase. This design will
be useful for the modeling inputs and can save time when changes in geometry had to be done
at first. So the preliminary design gives a good idea on the design result and is a major step to
start with.
Preliminary design of the structure includes the following:
-
Transfer beam design
Transfer slab design
Columns design
Shear walls design
Slabs design
Raft design
2.TRANSFER BEAMS PRELIMINARY DESIGN
The depth of the transfer beams is calculated to resist the bending and shearcriterions.
Calculating each depth concerning each criterion, and taking the critical one is what was done.
The transfer beams are subjected to the loads below:
- The implanted walls
- The ground floor solid slab
- Its own weight
Three steps are recommended to calculate the required depth of a transfer beam:
 Step1: finding the required depth due to bending without considering the drop’s weight
 Step2:
- adding the drop’s weight after finding the required depth
- checking if the reinforcement ratio is smaller than ρ max
If the reinforcement is not within the limits, the depth “d” needs to be
changed
 Step3: check the calculated depth calculated for shear
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The detailed procedure of the pre dimensioning of DB3 is stated below:
 Step 1: an excel sheet is established in order to accelerate the work , it provides the
ultimate loads on the transfer beams due to the implanted walls over 18 stories and the
ultimate load due to the ground floor slab.
→ Loads on transfer beams DB3
The following image shows the tributary area of the transfer beam DB3 on the ground floor
slab:
→
→
→
→
→
→
→
→
Figure 8: Position of DB3
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The following table shows the ultimate loads on the transfer beam DB3:
Table 16: Loads on DB3
The transfer beam DB3 is introduced afterwards to the software “RDM6” in order to find its
ultimate maximum bending moment, for which the first required depth is found:
Figure 9: Loads on DB3 introduced to RDM6
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Figure 10: Bending moment diagram on DB3
The required depth “d requried ”that resists the ultimate bending moment, is calculated with the
procedure stated below:
f’c= 300 kg/cm2
fy=4200 kg/cm2
- β = 0.85 − 0.05 ∗ (f′c − 280)/70)
- ρ min = 14.1/f y
- ρ max = 0.85*β *f’ c /f y *(0.003/0.003+0.004)
- ρ chosen =ρ max /2 < 0.85*β *f’ c /f y *(0.003/0.003+0.005)
- R= ρ chosen *f y *(1-0.5* ρ chosen *m)
- bd2=M/(φ*R)
- Findingd requried.
d required due to bending:
β=
ρ min=
ρ max=
ρ chosen=
m=
R=
b(cm)=
Mu(T.m)=
d required (cm)=
Table 17: dreq due to bending
0.8357
0.0034
0.0217
0.0190
16.471
67.393
140
1540
134.66
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 Step 2: adding the drop’s self-weight to the previous loadings in order to find the
ultimate
moment
DL drop= 2.5*1.2*1.4 = 4.2 T/ml
Ultimate DL = 1.2 * DL = 5.04 T/ml
The new ultimate moment diagram is the following:
Figure 11: Bending moment considering the drop
Maximum bending moment =1596 T.m.
The reinforcement ratio due to that moment is calculated with the following procedure:
- m=
- R=
- ρ=
fy
0.85×f′c
Mu
Φ×b×d2
1
m
(Kg/cm2 )
(1 − �1 −
2×m×R
fy
)
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For M u = 1560 T.mρ=1.83% <ρ max =2.17%
The calculated depth needs not to be changed, but it should be verified for the shear criterion
R
→ Step 3: check d required due to bending for shear
The ultimate shear diagram is the following:
Figure 12: Shear diagram of DB3
The ultimate shear force on the transfer beams is resisted by both concrete and stirrups.
Assuming the number and diameter of stirrups at first leads to find the required depth of the
section to resist the remaining shear.
V u (T)=ØV c +ØV s
Where :
Assuming�
-
ØV c = Ø * 0.17 * √𝐹′𝑐 * b w * d
ØV s = Ø * Av * F y * d / S
Av = 14T14 = cm2
and d required =135 cm
𝑆 = 10 𝑐𝑚
ØV c = Ø * 0.17 * √𝐹′𝑐 * b w * d = 132 T
ØV s = Ø * Av * F y * d / S
= 1196T
Ultimate applied shear: Vu = 1114T 1114 < 132+1196 = 1328 T
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The required depth for DB3 is 135 cm.
The same procedure is repeated on all the other transfer beams, results are stated below:
Table 18: Summary of dreq for all transfer beams
3.TRANSFERSLAB PRELIMINARY DESIGN
For architectural purposes in the basements , the columns supporting the circular slab will be
implanted on the ground floor slab .
In this section, the pre-dimensioning of this transfer slab will be discussed
3.1 EVALUATION OF THE TRANSFER SLAB LOADING USNG TRIBUTAY AREA METHOD
Traditionally we may evaluate the loading on the transfer slab by tributary area approach ,
ignoring the flexibility of the transfer slabs , and it’s influence in the load distribution . Also
the transfer slab may be analyzed independently , separated from the rest of the structure and
ignoring the flexibility of the supporting columns and their influence on the bending in the
transfer slab and beam.
Figure 13: Transfer beam
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This approach will provide no variation in the point loads on the transfer slab .in the figure
above , the two points loads at the middle are identical regardless of the supporting conditions
below.
This approach is commonly used and is considered more conservative. The point load on the
transfer slab are internal forces in the columns above and their magnitude is affected by
relative stiffness of all structural elements.
However treating the transfer slab independently and ignoring the rest of the structure is a
more conservative analytical approach .
3.2 LOADS ON TRANSFER SLAB
The following figure shows the transfer slab with its supports , edge and interior beams :
Figure 14: Transfer slab with its support
Ultimate loads due to the implanted columns on the transfer slab:
Table 19: Ultimate load due to implanted columns
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The preliminary transfer slab thickness is defined using the shear criterion since they are the
most critical. A verification on the thickness Is made according to the punching shear and oneway shear.
Since beams are laying on the supporting elements , the punching shear calculation will only
be verified under the implanted columns .
3.3CHECK FOR PUCHING SHEAR
For the implanted column IC19 having the largest ultimate axial load , the punching shear is
evaluated:
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For h=50cm punching is unverified so increasing the thickness is required
After iterations, a thickness of 100 cm for the transfer slab is adopted for punching shear
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3.4 CHECK FOR ONE-WAY SHEAR
V u (T)=ØV c +ØV s
Where :
-
V c = 0.17 * √𝐹′𝑐 * b w * d
V s = Av * F y * d / S
Table 20: shear verification
Vu > ØVc shear reinforcement is needed
For spacing of 15 cm , the shear reinforcement needed is
Av = 24.5 cm2/ml 2T25 @ 30cm each 15 cm
For workability reasons a T16 is used instead of T25 which will increase the transfer slab’s
thickness
Assume 2T16 @ 30cm each 15 cm will induce a 175cm thick slab
4.COLUMNS PRELIMINARY DESIGN
4.1 LOADS ON COLUMNS
The columns preliminary design is performed manually using the influence surface of each
column over the six basements, added to the reactions of the transfer beams on the ground
floor.
Those transfer beams are carrying the loads of the implanted shear walls, which is why a hand
calculation of the wall’s loads on those beams is required to find the reactions on columns. In
addition to those loads the transfer beams are also carrying the ground floor’s slab loads.
The following picture shows the influence surface of each wall in the typical floor plan:
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Figure 15: Influence surface of each wall in the typical floor plan
Walls having same dimensions, and same tributary areas are identically labeled.
Service dead and live loads due to walls over the 18 stories on the transfer beams are
computed using an excel sheet.
As for the input, the section of the wall, its height, its tributary area with its corresponding
super imposed dead load and live load, are required.
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As for the output, the concentrated service live and dead load due to the walls on all stories
on the beams are given.
Those following formulas are used in the excel sheet on each wall:
DL = DLIn�luence surface + Wall weight
LL = LLIn�luence surface
Calculation of DL
DLin�luence surface = In�luence surface × (SDL + 2.5 × slab thickness)
Wall weight = 2.5 × a × b × height
Calculation of LL
LLIn�luence surface = In�luence surface × LL
The following table is extracted from the excel sheet, and shows the total live and dead load
due to each wall on the transfer beams.
Table 21: Total live and dead load due to each wall on TB
W1
W2
W3
W4
W5
W6
W10
Total
DL(T)
975.431
1040.23
970.538
110.14
551.7
470.588
970.538
TOTAL LL
(T)
125.005
134.358
150.287
167.937
87.575
87.575
150.287
Those concentrated loads are then divided by their correspondent wall length to get finally
the linear load on the transfer beam.
The following figures show the transfer beams on the ground floor plan, and the tributary area
that the transfer beam DB3 is carrying.
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Figure 16: Position of all transfer beams on GF plan
The following table shows the total loads on the beam “DB3”, that are introduced to the software
“RDM6” in order to find the reactions on each column
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SPAN 1
LOADS DUE TO THE IMPLATED
WALLS
WALL LENGTH (m)
CONCENTRATED DEAD LOADS
(SERVICE) (T)
CONCENTRATED LIVE LOADS
(SERVICE) (T)
LINEAR DEAD LOADS (SERVICE)
(T/m)
LINEAR LIVE LOADS (SERVICE)
(T/m)
LOADS DUE TO THE GF SLAB
SLAB THICKNESS (cm)
SDL (T/m2)
LL (T/m2)
TRIBUTARY LENGTH (m)
LOAD SHAPES
LINEAR DEAD LOADS (SERVICE)
(T/m)
LINEAR LIVE LOADS (SERVICE)
(T/m)
LOAD DUE TO THE DROP
DROP BEAM HEIGH (cm)
DL OF THE DROP BEAM (SERVICE)
(T/m)
SPAN 2
W3
W6
W6
W4
4.9
2.9
2.9
4.9
970.54
470.59
470.59
1100.14
150.29
87.58
87.58
167.94
198.07
162.27
162.27
224.52
30.67
30.20
30.20
34.27
30
30
30
0.15
0.15
0.15
0.3
0.3
0.3
8.8
8.8-->0
0-->8.8
LINEAR TRIANGULAR TRIANGULAR
30
0.15
0.3
8.8
LINEAR
7.92
7.92-->0
0-->7.92
7.92
2.64
2.64-->0
0-->2.64
2.64
150
150
150
150
5.25
5.25
5.25
5.25
Using the software “RDM 6” has helped to compute the reaction on the beam’s supports which
are the service live and dead concentrated loads on the columns.
The following image shows the transfer beam “DB3”with its subjected dead loads as seen from
the table above as it is introduced to the software “RDM6”.
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Figure 17: DB3 subjected to dead load
The reaction due to dead loads are the following:
SERVICE DEAD LOAD REACTION ON C5
SERVICE DEAD LOAD REACTION ON C6
SERVICE DEAD LOAD REACTION ON C7
663.86 T
1525.97 T
758.09 T
The following image shows the transfer beam “DB3”with its subjected dead loads as seen
from the table above as it is introduced to the software “RDM6”.
Figure 18: Transfer beam subjected to live load
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The reaction due to live loads are the following:
SERVICE LIVE LOAD REACTION ON C5
SERVICE LIVE LOAD REACTION ON C6
SERVICE LIVE LOAD REACTION ON C7
106.21 T
263.5 T
119.62 T
The same work was done on all columns, and the following tables give the service reaction due
to live and dead load on each of them.
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In addition to the loads calculated above, columns are supporting the basement slabs.
Calculation of DL
DLin�luence surface = In�luence surface × (SDL + 2.5 × slab thickness)
Col weight = 2.5 × a × b × height
Calculation of LL
LLIn�luence surface = In�luence surface × LL
The ultimate load of the column at a story x is the sum of the ultimate loads at all the stories
above that level.
The following image shows the tributary areas of the column C1 on the basement slabs:
Figure 19: Tributary area on C1 on the basement slabs
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4.2CHECK FOR COLUMNS
4.2.1 LIMITATIONS
The columns design will be performed according to the chapter 10 of the ACI-318-08 code. The
following formulas are to be used:
According to ACI 10.3.6: Design axial strength ΦPn of compression members shall not be taken
greater than ΦPn,max computed by the following equation:
-
-
For non prestressed members with spiral reinforcement:
𝜙𝑃𝑛,𝑚𝑎𝑥 = 0.85 𝜙 (0.85 𝑓 ′ 𝑐 (𝐴𝑔 − 𝐴𝑠𝑡) + 𝑓𝑦 𝐴𝑠𝑡)
For non prestressed members with tie reinforcement:
𝜙𝑃𝑛,𝑚𝑎𝑥 = 0.8 𝜙 (0.85 𝑓 ′ 𝑐 (𝐴𝑔 − 𝐴𝑠𝑡) + 𝑓𝑦 𝐴𝑠𝑡)
The strength reduction factor Φ is taken as:
-
Φ = 0.75 for spiral reinforcement
Φ = 0.65 for other reinforcement
According to ACI 10.9.1.
The area for longitudinal reinforcement shall not be less than 0.01 Ag nor more than 0.08 Ag,
which means that the reinforcement ratio ρ should be between 1% and 8%
According to ACI 10.9.2.
Minimum number of reinforcement longitudinal bars shall be 4 in rectangular or circular ties
The axial force carried by each column according to its influence surface is calculated as follows:
Pu = 1.2DL + 1.6LL
DL = δ(DLIn�luence surface + Column weight)
LL = δ(LLIn�luence surface )
δ is a magnification factor taken as:
-
δ = 1.2 for interior columns
δ = 1.3 for edge columns
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-
δ = 1.2 for corner columns
Checking if the column’s dimensions and reinforcements are within the limits is computed using
an excel sheet, that required for the input the section of the column, its height, its
magnification factor, its tributary area with its corresponding super imposed dead load and live
load, the basement slab’s thickness and the reactions coming from the transfer beams
calculated previously.
As for the output, the reinforcement ratio 𝝆𝒈 is given, this ratio needs to be within the limits, or
else the column’s dimensions need to be changed.
Detailed calculations for C1 are stated below:
Vertical reinforcement will be computed using the formula:
∅ × 𝑷𝒏 = 𝟎. 𝟖 × ∅ × 𝑨𝒈 × �𝟎. 𝟖𝟓 × 𝒇′ 𝒄 + 𝝆𝒈 × �𝒇𝒚 − 𝟎. 𝟖𝟓 × 𝒇′ 𝒄 ��
4.2.2 Computing Pu= ∅𝑷𝒏 :
-
δ = 1.2 for interior columns
b=0.825 m, h=0.7m Ag=0.825x0.7=0.56 m2
Tributary area =13 m2 for the BAS1 and 48 m2 from the BAS2 till the BAS6.
Height of the column H=2.7m
For basement 1 :
-
DL(T/m2) =2.5 x Slab’s thickness =2.5 x 0.3 =0.75 T/m2
SDL=0.15(T/m2)
Total dead load (T/m2) =0.75 + 0.15 =0.9 T/m2
- LL=0.5 T/m2.
- Total DL (T) =Tributary area x DL + 2.5 x Ag x H
= 13 x 0.9 + 2.5 x 0.56 x 2.7
= 15.48 T
- Total LL(T)=Tributary area x LL
=6.5T
- DL due to DB1=534.22T
- LL due to DB1=71.14T
- Pu (BAS1) = 1.2 x DL + 1.6 x LL
= 1.2 (15.48+534.22) +1.6 (6.5+71.14)
=783.864T
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Frombasement 2 till basement 6 :
-
DL(T/m2) =2.5 x Slab’s thickness =2.5 x 0.3 =0.75 T/m2
SDL=0.15(T/m2)
Total dead load (T/m2) =0.75 + 0.15 =0.9 T/m2
- LL=0.5 T/m2.
- Total DL (T) =Tributary area x DL + 2.5 x Ag x H
= 48 x 0.9 + 2.5 x 0.56 x 2.7
= 46.98 T
- Total LL(T)=Tributary area x LL
=24T
- Pu (BAS2 till BAS6) = 1.2 x DL + 1.6 x LL
= 1.2 (46.98) +1.6 (24)
=94.776T
Total Pu on C1 in the BAS6= Pu (BAS1) +5x Pu (BAS2 till BAS6)
=783.864+ 5 x 94.776
=1257.74 T
δPu=1.2 x 1257.74 =1509.29 T
4.2.3 COMPUTING 𝝆𝒈
δPu=∅ × 𝑷𝒏 = 𝟎. 𝟖 × ∅ × 𝑨𝒈 × �𝟎. 𝟖𝟓 × 𝒇′ 𝒄 + 𝝆𝒈 × �𝒇𝒚 − 𝟎. 𝟖𝟓 × 𝒇′ 𝒄 ��
𝝆𝒈 = 1.04 %
The reinforcement ratio is within its limits, the dimensions of C1 need not to be changed.
The transfer slab is modeled using the software “ROBOT” because it needs the finite element
method to be solved. Loads on that slab are due to the implanted columns over the 18 floors.
Those loads are performed manually using the influence surface of each column on all the
stories.
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5.SHEAR WALL PRELIMINARY DESIGN
5.1 INTRODUCTION
Shear walls are designed according to the shear force and moment they could resist.
-
Shear force criteria:
The static base shear is distributed on the lower story on the walls according to their rigidity,
then the considered walls are checked for their thicknesses and reinforcements if they are
acceptable, if those are not in the limits the shear walls dimensions will have to be changed.
-
Moment criteria:
The static base shear is distributed on all the stories, then each value is divided into a
translational force and rotational force on each wall. The total force on a wall is their sum on
the considered one.
The translational force depends on the rigidity of the shear wall, and the rotational one
depends on the distance between the wall and the center of torsion, and the eccentricity
between the center of mass and the center of torsion.
5.2 CENTER OF TORSION
The center of torsion is to be calculated at first to be able to distribute the shear force since
that distribution depends on the distance from the center of mass of the building and its center
of torsion.
The following table describes the calculation of that eccentricity which is also established using
an Excel sheet.
The center of torsion’s coordinates are given by those formulas:
∑I x
� = xi i
X
∑ Ixi
Where:
-
�=
Y
∑ Iyi yi
∑ Iyi
Ix is the rigidity of wall, or its moment of inertia is the X direction
Iy is the rigidity of wall, or its moment of inertia is the Y direction
Xi is the abscissa of the wall from the origin
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-
Yi is the ordinate of the wall from the origin
The rigidity of the shear walls is given by those formulas:
bh3
Ix =
12
hb3
Iy =
12
Where :
-
b is the dimension on the wall parallel to the moment of inertia’s direction
h is the dimension on the wall perpendicular to the moment of inertia’s direction
The following table shows the positions and moments of inertia of each wall using an Excel
sheet in order to find at last the center of torsion’s coordinates.
Table 22: Position and moment of inertia for each shear wall
Elements
WALL
w1
w2
w3
w4
w5
w6
w7
w8
w9
w10
w11
w12
w13
w14
w15
w16
w17
w18
w19
w20
Position from origin
X(m)
0.325
5.1
16.2
20.975
9.65
11.65
0.325
2.775
8.1
13.2
18.525
20.975
9.65
11.65
0.325
2.775
8.1
13.2
18.525
20.975
Y(m)
35.1125
35.325
35.325
35.1125
30.925
30.925
26.525
26.525
26.525
26.525
26.525
26.525
22.125
22.125
17.725
17.725
17.725
17.725
17.725
17.725
Moment of inertia
Ix (m4)
0.0154
0.0116
0.0116
0.0154
3.9721
3.9721
0.0721
0.0055
0.0038
0.0038
0.0055
0.0721
3.9721
3.9721
0.0721
0.0055
0.0038
0.0038
0.0055
0.0721
Iy (m4)
0.0167
14.6868
14.6868
0.0167
0.0041
0.0041
0.0252
1.5993
0.5081
0.5081
1.5993
0.0252
0.0041
0.0041
0.0252
1.5993
0.5081
0.5081
1.5993
0.0252
Center of torsion
xi*Ixi
0.005005
0.05916
0.18792
0.323015
38.330765
46.274965
0.0234325
0.0152625
0.03078
0.05016
0.1018875
1.5122975
38.330765
46.274965
0.0234325
0.0152625
0.03078
0.05016
0.1018875
1.5122975
yi*Iyi
0.5863788
518.81121
518.81121
0.5863788
0.1267925
0.1267925
0.66843
42.421433
13.477353
13.477353
42.421433
0.66843
0.0907125
0.0907125
0.44667
28.347593
9.0060725
9.0060725
28.347593
0.44667
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w21
w22
w23
w24
w25
w26
w27
w28
w29
w30
w31
w32
w33
w34
w35
w36
w37
w38
w39
w40
w41
w42
w43
w44
w45
w46
w47
w48
w49
w50
w51
w52
w53
w54
w55
w56
w57
w58
w59
w60
w61
9.65
11.65
0.325
2.775
8.1
13.2
18.525
20.975
9.65
11.65
0.325
2.775
8.1
12.3
13.75
18.525
20.975
19.375
19.45
19.6625
23.7625
28.2501
28.2501
28.2501
32.7375
37.05
37.05
37.05
37.025
37.05
37.05
41.45
41.45
45.85
45.85
45.85
45.85
45.85
45.85
50.25
50.25
13.4125
13.4125
8.925
8.925
8.925
8.925
8.925
8.925
4.4375
4.4375
0.3375
0.125
0.05
-0.4847
0.125
0.125
0.3375
-11.225
-16.55
-19
-9.675
-11.225
-16.55
-19
-9.675
-19
-16.55
-11.225
-6.125
-0.8
1.65
-9.675
-7.675
-19
-16.55
-11.225
-6.125
0.8
1.65
-9.675
-7.675
3.9721
3.9721
0.0721
0.0055
0.0038
0.0038
0.0055
0.0721
3.9721
3.9721
0.0167
0.0055
0.0155
0.2908
0.0023
0.0055
0.0167
0.0155
1.5993
0.0154
0.004
0.5081
1.5993
0.0252
0.004
0.0252
1.5993
0.5081
0.6097
1.5993
0.0252
0.0041
0.0041
0.0252
1.5993
0.5081
0.5081
1.5993
0.0252
0.0041
0.0041
0.0041
0.0041
0.0252
1.5993
0.5081
0.5081
1.5993
0.0252
0.0041
0.0041
0.0154
1.5993
0.813
0.163
0.1215
1.5993
0.0154
0.813
0.0055
0.0167
3.6451
0.0038
0.0055
0.0721
3.6453
0.0721
0.0055
0.0038
0.0065
0.0055
0.0721
3.9721
3.9721
0.0721
0.0055
0.0038
0.0038
0.0055
0.0721
3.9721
3.9721
38.330765
46.274965
0.0234325
4.4380575
0.03078
0.05016
29.6270325
1.5122975
38.330765
46.274965
0.0054275
0.0152625
0.12555
3.57684
0.031625
0.1018875
0.3502825
0.3003125
31.106385
0.3028025
0.09505
14.35387581
45.18038493
0.71190252
0.13095
0.93366
59.254065
18.825105
22.5741425
0.203775
2.671305
0.169945
0.169945
1.15542
73.327905
23.296385
23.296385
73.327905
1.15542
0.206025
0.206025
0.0549913
0.0549913
0.22491
14.273753
4.5347925
4.5347925
14.273753
0.22491
0.0181938
0.0181938
0.0051975
0.1999125
0.04065
-0.079006
0.0151875
0.1999125
0.0051975
-9.125925
-0.091025
-0.3173
-35.26634
-0.042655
-0.091025
-1.3699
-35.26828
-1.3699
-0.091025
-0.042655
-0.039813
-0.0044
0.118965
-38.43007
-30.48587
-1.3699
-0.091025
-0.042655
-0.023275
0.0044
0.118965
-38.43007
-30.48587
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w62
w63
w64
w65
w66
w67
w68
w69
w70
w71
w72
w73
w74
w75
w76
w77
w78
w79
w80
54.4375
54.65
54.65
54.4375
19.1001
18.9001
18.6548
22.3876
25.5751
26.9126
28.2501
26.8876
25.5251
29.7626
31.2751
29.7626
31.2751
29.7626
31.2751
-19
-14.225
-3.125
1.65
-4.875
-5.7415
-7.2165
-7.725
-6.225
-5.125
-3.575
0.675
0.425
0.6
0.375
-2.275
-2.275
-7.725
-7.525
0.0154
14.6868
14.6868
0.0154
0.0016
0.0736
0.23
0.0695
0.3906
0.0053
17.0465
0.0055
0.0128
0.0207
0.0182
0.0061
0.0333
0.0284
0.0182
92.4561
0.0167
0.0116
0.0116
0.0167
0.0086
0.0032
0.0625
12.392
0.0056
0.3349
0.0198
0.3565
0.0018
0.7587
0.002
0.5058
0.0025
0.843
0.002
86.3415
0.8383375
-0.3173
802.63362
-0.16501
802.63362
-0.03625
0.8383375
0.027555
0.03056016 -0.041925
1.39104736 -0.018373
4.290604
-0.451031
1.5559382
-95.7282
9.98963406
-0.03486
0.14263678 -1.716363
481.5653297 -0.070785
0.1478818 0.2406375
0.32672128 0.000765
0.61608582
0.45522
0.56920682
0.00075
0.18155186 -1.150695
1.04146083 -0.005688
0.84525784 -6.512175
0.56920682
-0.01505
2885.516409 -244.7741
The center of torsion’s coordinates with respect to the origin shown below are:
-
Xr = 31.20 m
Yr = -2.83 m
5.3 CENTER OF MASS
The calculations of the center of mass of the typical story are stated below:
Table 23: Detremination of center of mass
Areas
A1
A2
A3
A4
A5
A6
Xi
11.18
10.68
22.16
25.38
33.96
45.64
Yi
17.74
-8.66
-2.49
-13.65
-8.90
-8.15
Ai
791.75
284.95
32.46
137.34
105.81
408.45
1760.75
xi*Ai
8847.79
3041.83
719.44
3484.89
3593.65
18640.46
38328.08
yi*Ai
14043.65
-2467.97
-80.75
-1874.63
-941.73
-3328.84
5349.73
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The center of mass’s coordinates with respect to the origin are :
-
� = ∑ Aixi =21.77m
X
∑
�=
Y
Ai
∑ Aiyi
∑ Ai
= 3.04m
This gives an eccentricity between the center of mass and center of torsion of:
-
Ex = 9.44m
Ey =5.87m
Which is large, we can conclude that there is a large rotational force.
The following plan shows the position of the center of mass and center of rigidity of the typical
plan:
Figure 20: position of CDR and CDG with respect to the origin
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5.4 SHEAR FORCES
The shear forces are divided into translational forces and rotational forces as stated before.
The translational force Ftri,that is applied on the shear wall number i is given by:
Ftr,i =
Ii
V
∑I
The rotational force Fri, that is applied on the shear wall number i is given by:
Fr,i =
Where J is:
Ii V e (Xr − Xi)
J
J = � Ixi (xr − xi )2 + � Iyi (yr − yi )2
The final shear force on a certain shear wall is the sum of both translational and rotational
forces
Fon shear wall = Ftr + Fr
3.4.1 LIMITATIONS
This design of the shear wall will be performed according to the ACI 318-08 chapter 11.
Its thickness shall be verified according to the following:
According to ACI section 11.9.3:
The nominal shear Vn at any horizontal section shall not be taken greater than the following
value Vn = 0.83 √f’chd where h is the wall’s thickness and d is the effective depth defined as d =
0.8 lw where lw is the shear wall’s length.
If the applied shear force is less than Vn, the thickness is OK, if it’s not the shear wall’s thickness
should be increased, and if the reinforcement is too large the properties of the wall should be
changed as well.
The shear strength “Vn” is divided into concrete strength ”Vc” and reinforcement strength ”Vs”.
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 Concrete strength Vc :
The concrete strength can be calculated using the simplified or the detailed method.
→ Simplified Method
According to ACI section 11.9.5.:
for walls subjected to axial compression Vc shall not be taken greater than:
𝑉𝑐 = 0.17 𝜆 �𝑓′𝑐 ℎ 𝑑
Where λ is the lightweight concrete factor taken as 1 for normal weight concrete.
According to ACI section 11.2.2.3:
for members subjected to significant axial Tension,
𝑉𝑐 = 0.17 × �1 +
0.29 × 𝑁𝑢
� × �𝑓′𝑐 × 𝑏𝑤 × 𝑑
𝐴𝑔
but not less than zero, where Nu is negative for tension. Nu /Ag shall be expressed in MPa.
→ Detailed Method
According to ACI section 11.9.6.:
Vc shall be permitted to be taken the lesser of the values computed from the following
equations:
-
𝑁𝑢 𝑑
(11-27)
-
𝑉𝑐1 = 0.27 𝜆 �𝑓′𝑐 ℎ 𝑑 +
-
Mu is the moment
Nu is the axial force applied on the shear wall
lw is its overall length.
Where
𝑉𝑐2 = (0.05 𝜆 �𝑓′𝑐 +
4 𝑙𝑤
𝑁
𝑢
𝑙𝑤( 0.1 𝜆 �𝑓′ 𝑐+0.2 ℎ 𝑙𝑤
𝑀𝑢 𝑙𝑤
−
𝑉𝑢 2
)hd
Nu is positive for compression and negative for tension.
(11-28)
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If (Mu/Vu – lw/2) is negative Eq. (11-28) shall not apply.
 Reinforcement shear strength Vs :
According to ACI section 11.9.9:
Where Vu exceeds Vc horizontal shear reinforcement shall be provided
The reinforcement shear strength Vs is the applied shear Vu minus the concrete strength Vc:
→ Horizontal shear reinforcement
ϕVs = Vu − ϕVc
According to ACI section 11.9.9.1
Vs =
Av fy d
s
(11-29)
Where Av is the horizontal shear reinforcement area and s is the spacing.
According to ACI section 11.9.9.2.:
Ratio of horizontal shear reinforcement area to gross concrete area of vertical section, ρt, shall
not be less than 0.0025.
According to ACI section 11.9.9.3.
Spacing of horizontal shear reinforcement shall not exceed the smallest of:
- Lw/5
- 3h
- 450 mm
→ Vertical shear reinforcement
According to ACI section 11.9.9.4:
Ratio of vertical shear reinforcement area to gross concrete area of horizontal section, ρl,shall
not be less than the larger of
ℎ
- 𝜌𝑙 = 0.0025 + 0.5(2.5 − 𝑤 )(𝜌𝑡 − 0.0025) (11-30)
-
0.0025 .
𝑙𝑤
The value of ρcalculated by Eq. (11-30) need not be greater than ρt required by 11.9.9.1.
Lwis the overall length of the wall, and hwis the overall height of the wall.
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According to ACI 318-08 section 11.9.9.5
Spacing of vertical shear reinforcement shall not exceed the smallest of:
- Lw/3
- 3h
- 450 mm
If the shear reinforcement is not within the limits, the dimensions of the corresponding wall
need to be changed.
5.4.2 PRE DIMENSIONING OF WALL 64
The detailed calculation of the wall 64 is stated below, it is the most critical wall for it has a
large moment of inertia which gives it a large translational force, and it also has the largest
distance between its center and the center of rotation.
 Distribution of base shear on all stories
The distribution of the shear force on all stories is stated below:
Fx = (V − Ft ) ×
hx × W x
∑ hx × W x
And since all floors have the same weight, as they are all typical, the distribution of the shear
force formula will then be:
Fx = (V − Ft ) ×
hx
∑ hx
 Translational and rotational forces on W64
The translational force Ftr w64 that is applied on the shear wall 64:
Ftr,w64 =
Iw64
V
∑I
The rotational force Fr, w64 that is applied on the shear wall 64:
Fr,w64 =
Iiw64 V e (Xr − Xw64)
J
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Where J is:
J = � Ixi (xr − xi )2 + � Iyi (yr − yi )2 = 8187327
Having previously calculated :
-
Ix=14.68 m4
ΣIx=92.45 m4
Ft=140.16 T
Static base shear V=1834.8 T as calculated in chapter 5
V-Ft=1 694.7 T
The following table shows the distribution of the shear force on wall64, into translational and
rotational forces, and the moment due to those shear force on the GF.
The pre dimensioning of this wall is to be done according to those values.
Table 24: Translational, rotational and bending moment on the shear wall
Story
hi(m)
18th
63
V /story
(T)
318.547
F translation (T)
F rotational (T)
F(total) T
50.602
-12.646
37.955
Moment ON GF
(T.m)
2391.179
17th
59.5
168.477
26.763
-6.689
20.074
1194.418
16th
56
158.567
25.189
-6.295
18.893
1058.031
15th
52.5
148.656
23.614
-5.902
17.713
929.910
14th
49
138.746
22.040
-5.508
16.532
810.055
13th
45.5
128.836
20.466
-5.115
15.351
698.466
12th
42
118.925
18.891
-4.721
14.170
595.142
11th
38.5
109.015
17.317
-4.328
12.989
500.085
10th
35
99.104
15.743
-3.934
11.808
413.293
9th
31.5
89.194
14.169
-3.541
10.628
334.768
8th
28
79.283
12.594
-3.148
9.447
264.508
7th
24.5
69.373
11.020
-2.754
8.266
202.514
6th
21
59.463
9.446
-2.361
7.085
148.786
5th
17.5
49.552
7.871
-1.967
5.904
103.323
4th
14
39.642
6.297
-1.574
4.723
66.127
3th
10.5
29.731
4.723
-1.180
3.543
37.196
2th
7
19.821
3.149
-0.787
2.362
16.532
1th
3.5
9.910
1.574
-0.393
1.181
4.133
TOTAL
598.5
1834.842
218.623
9768.465
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Total shear force on wall64 on the ground floor level is=218.623 T
Total moment on wall64 on the ground floor level is= 9768.5T.m
 Check the wall’s reinforcement using the software S-Concrete
Since the wall is subjected to a shear, a moment and an axial force all together, hand
calculation of its reinforcement will not be that accurate since its interaction diagram can’t be
taken into consideration , which is why, those reduced values are introduced into the “S
concrete” software along with the wall’s dimensions to check if its reinforcement is within the
limits.
→ Boundary elements calculations:
Flowchart of the process to determine if boundary elements are required:
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Figure 21: Flowchrat to determine if boundary zones are required
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According to UBC-97Section 1921.6.6.4
shear wall boundary zones are not required where:
- Pu<= 0.10Agf’c
for symmetrical walls
Pu<= 0.05Agf’c
for unsymmetrical walls
- Mu/(VuLw) < = 1.0
Short wall or h w /l w < 1.0 for one story wall
- Vu <= 3 Acv (f’c)1/2 [0.25 Acv (f’c)1/2 ] and Mu/(VuLw) < = 3.0
Figure 22: Boundary zones
Check if the boundary zones are required for W64:
From the tributary area calculations Pu w64 = 1902.2 T
Ag x f’c= (9.55 x 0.25) x 50 x 100= 11937.5
Pu/( Ag x f’c) = 0.159 >0.1
Boundary elements are required.
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Calculation of the boundary elements dimensions:
UBC Section 1921.6.6.3imposes a limit on the design axial force above which the wall is no
longer considered effective in resisting lateral loads. This limit is given by:
Pu<0.35P0
Where : P0 : nominal axial loads strength at zero eccentricity
𝑃0 = �0.85 × 𝑓 ′ 𝑐 × �𝐴𝑔 − 𝐴𝑠𝑡 � + 𝑓𝑦 × 𝐴𝑠𝑡 �
where :
Ag=gross area of section.
Ast=area of vertical reinforcement.
As calculated before Pu =1902.2 T
-
Assume 1% steel
𝑃0 = �0.85 × 50 × �9.55 × 0.25 −
=11048 T
9.55 × 0.25
9.55 × 0.25
� + 420 × �
�� × 100
100
100
Pu =1902.2 T <0.35 × 𝑃0 = 0.35 × 11048 = 3866.8 𝑇
So, the wall W64 is considered effective in resisting lateral loads.
Shear walls and portions of shear walls not meeting the conditions of Section 1921.6.6.4 and
having Pu< 0.35Po shall have boundary zones at each end a distance varying linearly from:
0.25Lw for Pu = 0.35 Po
0.15Lw for Pu = 0.15 Po
The boundary zone shall have minimum length of 0.15 lw and shall be detailed in accordance
with Section 1921.6.6.6.
The value of lw is found using the interpolation below:
0.25−0.15
=
0.25−𝑥
0.35−0.15 0.35−0.17
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x=0.16 =>Bondary length=0.16xlw=0.16x955cm=145 cm
At last the reduced values on the wall 64 are the following and they are introduced into the
software S concrete, along with the wall’s dimensions and boundary areas:
-
Vu = 1.1 x 218.623 = 240.48 T
Mu = 1.1 x 9768.5 = 1745.35 T.m
Pu = 1920.2 T
The following figure shows the S-concrete output of the wall 64:
Figure 23: S-CONSRETE output for wall
As we can see the wall’s section needs not to be changed, the reinforcements are within the
limits, so are the spacings of the vertical and horizontal reinforcements.
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The S-concrete software also shows the interaction diagram which is shown below:
Figure 24: Interaction diagram given by S-CONCRETE
I
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6.SLABS PRELIMINARY DESIGN
This sectionshows thepreliminary design of the typical floor slab with two differentconcepts :
-
Ribbed slab
Solid slab
6.1 Ribbed slab preliminary thickness
According to ACI -318-05-9.5.2.1:
Minimum thickness stipulated in Table 9.5(a) shall apply for one-way construction not
supporting or attached to partitions or other construction likely to be damaged by large
deflections, unless computation of deflection indicates a lesser thickness can be used without
adverse effects.
Figure 25 - Table 9.5.a from the ACI 318-08
Longest span for both end continuous :
for beams : 8.8m
for ribs : 8.8m
The slab thickness is calculated for a both end continuous span and a check for deflection will
be needed for the one end continuous span .
The adopted slab thickness is: ℎ =
𝑙
21
=
880
21
= 41.9 = 42𝑐𝑚
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6.2 Solid slab preliminary thickness
Slab thickness is preliminary determined using the following table taken from ACI 318-08
section 9.5. This section is about control of deflection, the slab thickness is chosen according to
the maximum span length in order to minimize deflection.
Figure 26: Table 9.5(c) from the ACI318-08
The maximum span length is 8.55 m which means
Thickness =
ln 8.55
=
= 28.5 𝑐𝑚
30
30
All slabs are designed as 30 cm thick flat slabs and deflection will be checked in zones where
span is greater than 855cm .
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CHAPTER 7
DYNAMIC ANALYSIS
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Chapter 7.DYNAMIC ANALYSIS
1.INTRODUCTION
After finishing the pre dimensioning of the structure’s vertical and horizontal elements under
static loads, the modeling phase, in which the seismic modal analysis is taken into
consideration, takes place.
The model is established using the software” ETABS”
2.USING ETABS:
2.1 MODELING PROCEDURE
In order to model the building on the “ETABS” software, the following steps were followed:
-
The story data is defined (number of stories and their correspondent height)
The different slabs are imported from DXF CAD files to the story data
Materials are defined ( Concrete with f’c = 50 MPa and Concrete with f’c = 30 MPa)
Sections are defined and sized( Frame elements, shell elements..)
Assign frame and shell section modifiers to take into consideration the cracked inertias.
Assign pier labels
Assign area and bar mesh
Assign support conditions
Assign loads (Self weight, dead load, super imposed dead load..)
Assign wind loads parameters as per ASCE 7
Define earthquake parameters as per UBC 97 for the equivalent static force method.
Define Response Spectrum Functions as per UBC 97
Define static and dynamic load combinations
Run the model for the first time
Scale the response spectrum functions
Run the model again in order to take results.
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2.2 BUILDING’S GEOMETRY
Figure 28: Typical plan introduced to ETABS
Figure 27: Basement plan introduced to ETABS
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2.3 LOAD COMBINATIONS
ACCORDING TO UBC 97-SECTION 1612.2.1: Basic load combinations.
Where Load and Resistance Factor Design (Strength Design) is used, structures and all portions
thereof shall resist the most critical effects from the following combinations of factored loads:
 1.4D
(12-1)
 1.2D + 1.6L + 0.5 (Lror S)
(12-2)
 1.2D + 1.6 (Lror S) + (f1L or 0.8W)
(12-3)
 1.2D + 1.3W + f1L + 0.5 (Lror S) (12-4)
 1.2D + 1.0E + (f1L + f2S) (12-5)
 0.9D ± (1.0E or 1.3W)
(12-6)
WHERE:
- f1 = 1.0 for floors in places of public assembly, for live loads in excess of 100 psf
(4.9 kN/m2), and for garage live load.
= 0.5 for other live loads.
- f2 = 0.7 for roof configurations (such as saw tooth) that do not shed snow off
the structure.
= 0.2 for other roof configurations.
EXCEPTIONS:
1. Factored load combinations for concrete perSection 1909.2 where load
combinationsdo not include seismic forces.
2. Factored load combinations of this section multiplied by 1.1 for concrete and masonry
where load combinations include seismic forces.
3. Where other factored load
ACCORDING TO UBC 97-SECTION 1630.1: Earthquake loads.
Structures shall be designed for ground motion producing structural response and seismic
forces in any horizontal direction. The following earthquake loads shallbe used in the load
combinations set forth in Section 1612:
E = ρEh + Ev(30-1)
Em= ΩoEh(30-2)
WHERE:
- E = the earthquake load on an element of the structure resulting from the combination
of the horizontal component, Eh, and the vertical component, Ev.
- Eh = the earthquake load due to the base shear, V, as set forth in Section 1630.2 or the
design lateral force, Fp, as setforth in Section 1632.
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-
Em= the estimated maximum earthquake force that can be developed in the structure
as set forth in Section 1630.1.1.
Ev= the load effect resulting from the vertical component of the earthquake ground
motion and is equal to an additionof 0.5CaID to the dead load effect, D, for Strength
Design, and may be taken as zero for Allowable Stress Design.
Ωo= the seismic force amplification factor that is required to account for structural
overstrength, as set forth in Section1630.3.1.
ρ = Reliability/Redundancy Factor
When calculating drift, or when the structure is located in Seismic Zone 0, 1 or 2, ρshall be
taken equal to 1.
3.MODAL MASS PARTICIPATION RATIO
UBC-97 1613.5.2 Numbers of modes
The requirement of Section 1613.4.1 that all significant modes be included may be satisfied by
demonstrating that for all modes considered, at least 90 percent of the participating mass of the
structure is included in the calculation of response for each principle horizontal direction.
The addition of modes that are taken into this project can be interrupted if the total modal
mass ∑Mi is greater than 90% of the total vibrating mass M of the system, in which case, the
effects of unsuccessful modes can be neglected.
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Figure 29: Modal Mass Participation Ratio
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The percentage of the participating mass, after 20 modes, is 97.21% in X-direction and 97.52%
in Y-direction.
The mass participating ratios along all modes can give an close idea of the building’s
behavior, if it’s being translated or rotated, when it’s being subjected to the correspondent
mode.
As the first 2 modes show, the participating ratio along one direction a way larger than the
one in the opposite direction, this concludes that the building’s displacements are governed
by a translation along the direction that has the bigger ratio.
The following figures clearly show the translational and the rotational displacements affecting
various modes
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Figure 30: Modes shapes
4. SCALING FACTOR
Currently a three-dimensional dynamic analysis is required for a large number of different types
of structural systems that are constructed in Seismic Zones 2, 3 and 4 [1]. The lateral force
requirements suggest several methods that can be used to determine the distribution of
seismic forces within a structure. However, these guidelines are not unique and need further
interpretations.
The current code allows the results obtained from a dynamic analysis to be normalized so that
the maximum dynamic base shear is equal to the base shear obtained from a simple twodimensional static load analysis. Most members of the profession realize that there is no
theoretical foundation for this approach. However, for the purpose of selecting the magnitude
of the dynamic loading that will satisfy the code requirements; this approach can be accepted,
in a modified form, until a more rational method is adopted.
The calculation of the “design base shears” is simple and the variables are defined in the code.
It is of interest to note, however, that the basic magnitude of the seismic loads has not changed
significantly from previous codes. The major change is that “dynamic methods of analysis” must
be used in the “principal directions” of the structure. The present code does not state how to
define the principal directions for a three dimensional structure of arbitrary geometric shape.
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Since the design base shear can be different in each direction, this “scaled spectra” approach
can produce a different input motion for each direction, for both regular and irregular
structures.
Therefore, the current code dynamic analysis approach can result in a structural design which is
relatively “weak” in one direction. The method of dynamic analysis proposed in this chapter
results in a structural design that has equal resistance in all directions.
In addition, the maximum possible design base shear, which is defined by the present code, is
approximately 35 percent of the weight of the structure. For many structures, it is less than 10
percent. It is generally recognized that this force level is small when compared to measured
earthquake forces. Therefore, the use of this design base shear requires that substantial
ductility be designed into the structure.
The definition of an irregular structure, the scaling of the dynamic base shears to the static base
shears for each direction X&Y, the application of accidental torsional loads and the treatment of
orthogonal loading effects are areas which are not clearly defined in the current building code.
The purpose of this section is to present one method of three dimensional seismic analyses that
will satisfy the Lateral Force Requirements of the code. The method is based on the response
spectral shapes defined in the code and previously published and accepted computational
procedures.
According toUBC-97 1631.5.4 Reduction of Elastic Response Parameters for Design
Elastic Response Parameters may be reduced for purpose of design in accordance with the
following items, with the limitation that in no case shall the base shear be less than the Elastic
Response Base Shear divided by the value of R(which is the ductility factor indicated in table 16N UBC)
Scaling factor >1/R
Scaling factor of “ETABS” software:
Essentially ETABS assumes the response spectrum as unitless and that the scale factor 9.81
converts the UBC 97 spectrum into acceleration (m/sec2)
Since the dynamic base shear must not be less than the design base shear obtained from the
static equivalent analysis, scaling factors Cx and Cy are used in the X and Y directions
respectively. Refer to UBC 1997 Section 1631.5.4.
𝐶𝑋 =
𝐶𝑌 =
𝑉𝑠𝑡𝑎𝑡𝑖𝑐
𝑉𝑑𝑦𝑛𝑥
𝑉𝑠𝑡𝑎𝑡𝑖𝑐
𝑉𝑑𝑦𝑛𝑦
× 9.81 = 2.265 ≥ 1/R =0.22
× 9.81 = 2.226 ≥ 1/R =0.22
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The following table shows the scaling factor for the spectrum in the X direction in order to
converge the dynamic base shear to the static base shear.
The following figures show the initial and final scaling factor for the spectrum in the X
direction:
Figure 31: initial scaling factor
Figure 31: Final scaling factor
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The following table is extracted from the “etabs” software and shows the dynamic and static
base shears at the ground floor before the scaling procedure:
Figure 32: Dynamic base shear and static base shear before scaling
The following table is extracted from the “etabs” software and shows the dynamic and static base
shearsconvergence at the ground floor level after scaling:
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Figure 33: Dynamic and static shear convergence
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5. LATERAL STORY DRIFTS
STORY DRIFT is the lateral displacement of one level relative to the level above or below:
Where:
Figure 34.Story Drift
h=height between stories
H = height of building
d= displacement at the top of building
d’=story drift = differential displacement
between stories
5.1 CODE REQUIREMENTS
5.1.1 Drift requirements:
According to UBC.1630.9.2:
The maximum Inelastic Response Displacement shall be computed as follows:
Where:
∆𝑀 = 0.7 × 𝑅 × ∆𝑆
-
ΔM: Maximum Inelastic Response Displacement, which is the total drift or total story
drift that occurs when the structure is subjected to the Design Basis Ground Motion,
including estimated elastic and inelastic contributions to the total deformation.
-
ΔS: Design Level Response Displacement, which is the total drift or total story drift
that occurs when the structure is subjected to the design seismic forces extracted
from Etabs software.
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-
R: Numerical Coefficient representative of inherent over strength and global ductility
capacity of lateral force resisting system. The building is a bearing wall system, the
requested factor “R” by the UBC 1997(Table16-N) is 4.5
According to UBC.1630.10.2the calculated story drift shall not exceed 0.025 times the story
height for structures having a fundamental period of less than 0.7 second. For structures having
a fundamental period of 0.7 second or greater, the calculated story drift shall not exceed 0.020
times the story height.
Since the structure has a fundamental period greater than 0.7 the second criteria isapplicable.
5.1.2 Torsional moment requirements:
According to UBC.1630.7: Horizontal Torsional Moments.Provisions shall be made for the
increased shears resulting from horizontal torsion where diaphragms are not flexible. The most
severe load combination for each element shall be considered for design.
The torsional design moment at a given story shall be the moment resulting from eccentricities
between applied design lateral forces at levels above that story and the vertical –resisting
elements in that story plus an accidental torsion.
The accidental torsional moment shall be determined by assuming the mass at each level to be
displaced from the calculated center of mass in each direction a distance equal to 5 percent of
the building dimension at the level perpendicular to the direction of the force under
consideration.
Where torsional irregularity exists, as defined in Table 16-M of UBC, effects shall be accounted
for by increasing the accidental torsion at each level by an amplification factor, A x , determined
from the following formula:
𝛿𝑚𝑎𝑥 2
𝐴𝑥 = �
�
1.2 𝛿𝑎𝑣𝑔
Where:
-
δ avg : average of displacement at the extreme points of the structure at level x
δ max : maximum displacement at level x
The value of Ax needs not to exceed 3.0.
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5.1.3 Determination of ΔS and ΔM
According to UBC 1630.9.1Dynamic analysis may be performed using finite element software
like Etabs. Where allowable Stress Design is used and where drift is being computed.
The resulting deformation, denoted as ΔS, shall be determined at critical locations in the
structure.
According to UBC-1630.9.2:The Maximum Inelastic Response Displacement, ΔM, shall be
computed as follows:
∆M = 0.7 × R × ∆S
Where R=ductility factor=4.5 (table 16-N)ΔM=3.15 x ΔS.
5.2 CALCULATION OF THE SEISMIC LATERAL DRIFT
The allowable drift is the same in both directions; it only depends on the height of each story.
The following table summarizes the results obtained, it shows the drifts as they are obtained
from ETABS and the allowable drifts and a verification is made.
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Table 25: Verification of lateral drift due to earthquake
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6. CHECK FOR P-DELTA EFFECT
Figure 35: Forces and displacements causing p-delta effect
The resulting member forces and moments and the story drifts induced by P-∆ effects shall be
considered in the evaluation of overall structural frame stability and shall evaluated using the
forces producing the displacements ∆S.
P-∆ needs not to be considered when the ratio of secondary moment to primary moment does
not exceed 0.10.In seismic zones 3 and 4, P-∆ need not to be considered when the story drift
ratio does not exceed 0.02/R.
The seismic zone in this project is 2b, than the first criterion is adopted. It is calculated due the
critical case which is the seismic cases Spec X & Spec Y as shown as below:
δ r × Pr
he × Fr
≤ 0.1
Where:
δ r : : seismic drift in story
Pr: total dead at the specified floor
Fr: seismic shear at the specified floor
He: Height of the specified floor
The following tables show the calculation verification for P-∆ effect along the X and Ydirections:
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Table 26: Verification of P-DELTA effect along x direction
Table 27: Verification of P-DELTA effect in y drection
As those tables show there is no need to introduce the effect of P-Delta to the model.
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CHAPTER 8
COLUMNS DESIGN
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Chapter 8.COLUMNS DESIGN
1.INTRODUCTION
Compression members are structural elements subjected mainly to compression force and
bending moments. We can define three types of compression members:
- Members reinforced with longitudinal bars and lateral ties.
- Members reinforced with longitudinal bars and continuous spirals.
- Composite compression members reinforced with structural shapes: pipes, tubes,...
with or without additional longitudinal steel and various types of lateral
reinforcement.
Figure 36. Types of compression members
Columns are structural compression members which transmit loads from the upper floors to
the lower levels and then to the soil through the foundations. Since columns are compression
elements, failure of one column in a critical location can cause the progressive collapse of
adjoining floors, and in turn, even the collapse of the entire structure.
ACI Code 9.3.2 specifies a strength reduction factor in the design of columns.
For columns subjected to:
-
Axial tension, and axial tension with flexure
Φ = 0.9
Axial compression and axial compression with flexure.
• Members with spiral reinforcement confirming to 10.9.3, Φ =0.75
• Other reinforced members
Φ = 0.70
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2.TYPES OF COLUMNS
Columns are classified into two categories:
- short columns
- slender columns.
2.1 SHORT COLUMNS
The ultimate applied load at a given eccentricity is only governed by the strength of the
materials and the dimensions of the cross section and it is of three kinds:
- Spiral Column:more efficient for e/h < 0.1 their deformation is lager and they have a
more ductile aspect but forming and spiral is expensive.
-
Tied Column: bars in four faces are used when e/h < 0.2 and for biaxial bending.
-
Tied Column:Bars in 2 faces (furthest from axis of bending). It is most efficient when
e/h > 0. Note that a rectangular shape increases efficiency.
2.2 SLENDER COLUMNS
The ultimate applied load is also influenced by slenderness, which produces additional bending
because of transverse deformation.
Figure 37. Types of columns
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3.COLUMN DESIGN
3.1LOADING SELECTION
The design loads are to be taken at the lowest part of the column in each story. In ETABS
model, loads are defined as followed:
-
P:
V2:
V3:
T:
M2:
M3:
axial force negative for compression and positive for tension
shear force in the 1-2 plane
shear force in the 1-3 plane
axial torque
the bending moment in the 1-3 plane (about the 2-axis)
the bending moment in the 1-2 plane (about the 3-axis)
The column major direction is the same as the local 2-axis direction (which is also the same as
the minor axis). Loads acting in the major direction cause M3 bending and V2 shear.
Figure 38. Local axis for column in ETABS model
The columns were checked using the “ S-concrete” software . Their loads under all dynamics and static
combinations are imported from the “ETABS” model to the “ S-concrete” noting the difference between
their local axis’ .
E-TABS LABELS
P
t
S-CONCRETE LABELS
N
T
V3
Vz
M2
V2
M3
My
Vy
Mz
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3.2 COLUMN SELECTION
A detailed calculation will be made for one column “C82”, then summary tables are established for all
the others .
The selection was done according to similar column sections defined in the project. The repeated
column section with maximum forces and moment applied were evaluated according to ACI & UBC. The
chosen columns are mentioned in figures below:
Figure 39.Typical column at typical floors
3.3REINFORCEMENT
Columns will be designed as short columns since the slenderness effects are permitted to be
neglected.
According to ACI 318-08 section 10.10.1 slenderness effects shall be permitted to be neglected
in the following cases:
 For compression members not braced against side-sway:
K lu
≤ 22
r
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 For compression members braced against side-sway:
K lu
M1
≤ 34 − 12 � � ≤ 40
r
M2
Where:
- M 1 /M 2 is positive if the column is bent in single curvature and negative if the member is
bent in double curvature
- Lu is the column height
- r is the radius of gyration given by:
Where:
-
I
A
r=�
I is the moment of inertia
A is the column section
K is the effective length factor taken as 1 which is the worst case while assuming that
the column is pinned at its ends
According to ACI 318-08 section 10.10.5, it shall be permitted to assume a column in a structure
is non-sway if the increase in column end moments due to second-order effects does not exceed
5% of the first-order end moments.
It also shall be permitted to assume a story within a structure non-sway if:
∑ Pu ∆o
≤ 0.05
Q=
Vus lc
Where:
- Σ P u and V us are the total factored vertical load and the horizontal story shear
respectively
- Δ o is the first-order relative lateral deflection between the top and the bottom of
the considered story due to V us which means that it’s the story drift
- Lc length of compression member in frame measured from center to center of the
joints in the frame
3.3.1Vertical Reinforcement
Vertical reinforcement will be computed using the formula:
∅ × 𝑷𝒏 = 𝟎. 𝟖 × ∅ × 𝑨𝒈 × �𝟎. 𝟖𝟓 × 𝒇′ 𝒄 + 𝝆𝒈 × �𝒇𝒚 − 𝟎. 𝟖𝟓 × 𝒇′ 𝒄 ��
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REINFORCEMENT LIMITATION
The percentage of longitudinal reinforcement must be:
1% < ρ < 8%
ACI Code 10.9.1
For cross-sections larger than required for loading, minimum reinforcement may be computed
for the reduced effective area – A g , 0.5A g ; (ACI 10.8.4). Note that the provided strength from
reduced area ad resulting A st must be adequate for loading.
3.3.2 Ties
 Ties in column:
- are used to hold the vertical bars in position
- provide lateral support so that individual bar could have the tendency to buckle
only between the tie supports
- do not contribute to the strength of columns
- are placed at a sufficiently close spacing provide confinement and increase the strain
at which concrete crushes to values well above the maximum of 0.003
 Diameter limitations:
ACI code 7.10.5.1 specifies that all non prestressed bars shall be enclosed by lateral ties, at least
No. 3 (T10 )for longitudinal bars No.10 (T32) and smaller, and at least No.4 for No.11, No.14,
No.18 and bundled longitudinal bars.
 Vertical spacing limitations:
ACI Code 7.10.5.2 specifies that vertical spacing of ties shall not exceed 16 longitudinal bar
diameters, 48 tie bar, or least dimension of the compression member.
The spacing of the lateral ties shall not exceed the minimum between:
- 16 Φl (longitudinal reinforcement diameter)
- 48 Φt (tie reinforcement diameter)
- Minimum dimension of column
ACI Code 7.10.5.3specifies that Ties shall be arranged such that every corner and alternate
longitudinal bar shall have lateral support provided by the corner of a tie with an included angle
of not more that 135 deg and no bar shall be farther than 150mm clear on each side along the
tie from such a laterally supported bar.
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3.3.3 Splicing of Bars in Compression
Splicing of deformed bars in compression is defined in ACI code section 12.16.1, where
compressive lap splices length shall be 0.07 f y db for f y of 420 MPA or less, but not less than
300 mm.
0.071 f y db > 300 𝑚𝑚
𝑓𝑜𝑟 𝑓𝑦 < 420 𝑀𝑃𝑎
The following table shows the lap splices for T12 to T25 obtained from the previous formula.
Diameter
12
14
16
20
25
Lap splice in mm
357.84
417.48
477.12
596.4
745.5
Table 28 - Lap splice in compression
Lap splice in tension are given by the ACI 318-08 section 12.2.2. and 12.15.1.:
ls = 1.3 ld
ld = �
Diameter
12
14
16
20
25
fy
1 2
�d
1.1�f ′ c c
Lap splice in mm
261
355
464
725
1132
Table 29 - Lap splice in tension
Where c is the concrete cover in mm and d is the bar diameter in mm.
The ACI Code (12.3.2) specifies that when bars of different size are lap spliced in compression,
splice length shall be the larger of:
- Splice length of smaller bars
-
Development length of the larger bar Ldb computed as follow :
Ldb = db f y / 4( f 'c )0.5 > max (0.04 f y db ,200 mm)
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3.4 ANALYSIS
Hereby the characteristic of column “C 82” chosen:
Assumed Section: 120x60cm
Since this column extends over18 floors, we attend to minimize the reinforcement used while
going up in the column height. The analysis was proceeded for the first two floors, and another
for the upper floors.
𝑃𝑢 = −687 𝑇
𝑀1 = 10.57 𝑇. 𝑚
𝑀2 = 73.57 𝑇. 𝑚
𝐾= 1
𝑙𝑢 = 3.5 𝑚
3.4.1 Slenderness Verification
The columns slenderness effects should be verified according to:
M1
K lu
≤ 34 − 12 � � ≤ 40
r
M2
Slenderness effect is checked in all columns of that story. The columns can then be designed as
short columns.
𝑀1
34 – 12 ×
= 34 − 12 × (10.57/73.57) = 32.27 < 40
𝑀2
1 × 3.5
𝐾 × 𝑙𝑢
=
= 9.71 < 33
𝑟
0.3 × 1.2
𝐾 × 𝑙𝑢
1 × 3.5
=
= 19.44 < 33
𝑟
0.3 × 0.6
Then the column is short.
3.4.2Design for short column
𝑷𝒏 = 𝟎. 𝟖𝟓 × 𝒇ʹ𝒄 × �𝑨𝒈 − 𝑨𝒔𝒕 � + 𝒇𝒚 × 𝑨𝒔𝒕
ACI 10-3.6 specifies for tied columns a design strength of:
∅ × 𝑷𝒏 = 𝟎. 𝟖 × ∅ × (𝟎. 𝟖𝟓 × 𝒇ʹ 𝒄 × �𝑨𝒈 − 𝑨𝒔𝒕 � + 𝒇𝒚 × 𝑨𝒔𝒕 )
𝑤ℎ𝑒𝑟𝑒: 𝜌𝑔 =
𝐴𝑠
𝐴𝑔
∅ × 𝑷𝒏 = 𝟎. 𝟖 × ∅ × 𝑨𝒈 × �𝟎. 𝟖𝟓 × 𝒇ʹ 𝒄 + 𝝆𝒈 × �𝒇𝒚 − 𝟎. 𝟖𝟓 × 𝒇ʹ 𝒄 ��
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-
A s = total area of longitudinal reinforcement
A g = gross area of section
3.4.3 Vertical Reinforcement
Vertical reinforcement will be computed using the formula:
∅ × 𝑷𝒏 = 𝟎. 𝟖 × ∅ × 𝑨𝒈 × �𝟎. 𝟖𝟓 × 𝒇′ 𝒄 + 𝝆𝒈 × �𝒇𝒚 − 𝟎. 𝟖𝟓 × 𝒇′ 𝒄 ��
-
Then,
Column’s Section: a =120 cm b =60 cm
𝑓′𝑐 = 50 𝑀𝑃𝑎
𝑓𝑦 = 420 𝑀𝑃𝑎
𝑃𝑢 = −687 𝑇
𝑃𝑛 =
𝑃𝑢
𝑀𝑛 =
𝑒=
=
∅
𝑀𝑢
𝑀𝑛
𝑃𝑛
∅
=
−687
𝐴𝑔 = 𝑎 × 𝑏 = 1.2 × 0.6 = 0.72 𝑚2
= −1057 𝑇
0.65
73.57
=
0.65
113.2
1057
= 113.2 𝑇. 𝑚
= 0.107𝑚 = 10.7 𝑐𝑚
−1057𝑇 × 0.107 𝑚
∅ × 𝑃𝑛 × 𝑒
=
= −170.172 𝑇�𝑚2 = −0.245𝐾𝑠𝑖
𝐴𝑔 × ℎ
0.72 × 0.6
∅ × 𝑃𝑛
−1057.29
=
= −1468 𝑇�𝑚2 = −2.11𝐾𝑠𝑖
𝐴𝑔
0.72
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By the diagram shown below, we obtain:
Figure 40. Load -moment diagram for R8-60.90 columns
The section is safeρ = ρ min =1%
As = ρ *b*h = 72 cm2
3.4.4 Ties Design:
-
Use ties Ø10
Maximum Vertical Spacing:
S max = min(16db , 48dtb , least dimension)= min(16x25,48x10,600)
S max = 400 mm
 Use φ 10 ties @ 20 cm
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3.5 S-CONCRETE OUTPUT
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3.6 SUMMARY RESULT
All column have been introduced to the software 6 types of reinforcement were found
The following tables shows the different types of column reinforcement
Table 30: Summary of different types of column reinforcement
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Table 31: Type of each column
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CHAPTER 9
SHEAR WALL DESIGN
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Chapter 9. SHEAR WALL DESIGN
1.INTRODUCTION
Shear walls are vertical elements in the lateral-force-resisting system that transmits lateral forces from
the diaphragm above to the diaphragm below, or to the foundation. Shear walls may also be bearing
walls in the gravity-load system, or they may be components in a dual system framed so as to resist only
lateral loads.
1.1 DESIGN FORCES
Walls may be subjected to both vertical (gravity) and horizontal (wind or earthquake) forces. The
horizontal forces are both in plane and out-of-plane. When considered under their in-plane loads, walls
are called shear walls; when considered under their out-of-plane loads, they are called normal walls.
Walls will be designed to withstand all vertical loads and horizontal forces, both parallel to and normal
to the flat surface, with due allowance for the effect of any eccentric loading or overturning forces
generated. Any wall, whether or not intended as part of the lateral-force-resisting system, is subjected
to lateral forces unless it is isolated on three sides (both ends and top), in which case it is classified as
non-structural. Any wall that is not isolated will participate in shear resistance to horizontal forces
parallel to the wall, since it tends to deform under stress when the surrounding framework deforms.
1.2 IN PLANE EFFECTS
Horizontal forces at any floor or roof level are generally transferred to the ground (foundation)
by using the strength and rigidity of shear walls (and partitions). A shear wall may be
considered analogous to a cantilever plate girder standing on end in a vertical plane, where the
wall performs the function of a plate girder web, the pilasters or floor diaphragms function as
web stiffeners, and the integral reinforcement of the vertical boundaries functions as flanges.
Axial, flexural, and shear forces must be considered in the design of shear walls. The tensile
forces on shear wall elements resulting from the combination of seismic uplift forces and
seismic overturning moments must be resisted by anchorage into the foundation medium
unless the uplift can be counteracted by gravity loads (e.g., 0.90 of dead load) mobilized from
neighbouring elements.
A shear wall may be constructed of materials such as concrete, wood, unit masonry, or metal in
various forms. Design procedures for such materials as cast in- place reinforced concrete and
reinforced unit masonry are well known, and present no problem to the designer once the
loading and reaction system is determined. Other materials frequently used to support vertical
loads from floors and roofs have well- established vertical- load- carrying characteristics, but
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have required tests to demonstrate their ability to resist lateral forces. Various types of wood
sheathing and metal siding fall into this category. Where a shear wall is made up of units such
as plywood, gypsum, wallboard, tilt-up concrete units, or metal panel units, its characteristics
are, to a large degree, dependent upon the attachments of one unit to another, and to the
supporting members.
1.2.1 Rigidity Analysis
For a building with rigid diaphragms, there is a torsional moment, and a rigidity analysis is
required. It is necessary to make a logical and consistent distribution of story shears to each
wall. An exact determination of wall rigidities is very difficult, but is not necessary, because only
relative rigidities are needed. Approximate methods in which the deflections of portions of
walls are combined usually are adequate.
1 .2.2 Wall Deflections
The rigidity of a wall is usually defined as the force required causing a unit deflection. Rigidity is
expressed in kips per inch. The deflection of a concrete shear wall is the sum of the shear and
flexural deflections (see Figure 1). In the case of a solid wall with no openings, the
computations of deflection are quite simple; however, where the shear wall has openings, as
for doors and windows, the computations for deflection and rigidity are much more complex.
An exact analysis, considering angular rotation of elements, rib shortening, etc., is very timeconsuming. For this reason, several short-cut approximate methods have been developed.
These do not always give consistent or satisfactory results. A conservative approach and
judgment must be used.
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Figure 41. Walls deformation due to flexion or shear
1.2.3 Foundation Effects
The rotation at the foundation can greatly influence the overall rigidity of a shear wall because
of the very rigid nature of the shear wall itself; however, the rotational influence on relative
rigidities of walls for purposes of horizontal force distribution may not be as significant.
Considering the complexities of soil behavior, a quantitative evaluation of the foundation
rotation is generally not practical, but a qualitative evaluation will be provided.
1.2.4 Framework Effects
The relative rigidity of concrete or unit masonry walls with nominal openings is usually much
greater than that of the building framework; therefore, the walls tend to resist essentially all or
a major part of the lateral force.
1.2.5 Effect of Openings
The effect of openings on the ability of shear walls to resist lateral forces must be considered. If
openings are very small, their effect on the overall state of stress in a shear wall is minor. Large
openings have a more pronounced effect, and if large enough, result in a system in which
typical frame action predominates (see figure 2). Openings commonly occur in regularly spaced
vertical rows throughout the height of the wall, and the connection between the wall sections
is provided by either connecting beams (or spandrels) which form a part of the wall, or floor
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slabs, or a combination of both. If the openings do not line up vertically and/or horizontally, the
complexity of the analysis is greatly increased. In most cases, a rigorous analysis of a wall with
openings is not required. “Strut and Tie” procedures that depict shear walls as consisting of
compression struts and tension ties are useful tools for the evaluation of shear walls with
openings. In the design of a wall with openings, the deformations must be visualized in order to
establish some approximate method for analyzing the stress distribution to the wall. The major
points that must be considered are the lengthening and shortening of the extreme sides
(boundaries) due to deep beam action, the stress concentration at the corner junctions of the
horizontal and vertical components between openings, and the shear and diagonal tension in
both the horizontal and vertical components.
Figure 42: deformation of a shear wall wit opening
1.3 OUT-OF-PLANE EFFECTS
1.3.1 Lateral Forces
Walls and partitions must safely resist horizontal seismic forces normal to their flat surface
(Figure 3, part a). At the same time, they must resist moments and shears induced by relative
deflections of the diaphragms above and below (Figure 3, part b). The normal force on a wall is
a function of its weight.
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Figure 43. Out-of-plane effects
For cantilevered walls (as parapet), the design force will be applied to the wall in both
inward and outward directions.
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1.3.2 Wall Behavior
Walls distribute normal forces vertically to the horizontal resisting elements above or below.
They may also distribute normal forces to frames, or other walls or frames. A wall may be either
continuous or discontinuous across its supports.
2.SHEAR WALL DESIGN
The walls are designed using the “ S-concrete” software . Their loads under all dynamics and
static combinations are imported from the “ETABS” model to the “ S-concrete” noting the
difference between their local axis’ .
E-TABS LABELS
P
t
S-CONCRETE LABELS
N
T
V3
Vz
M2
V2
M3
My
Vy
Mz
Thefollowing image is extracted from the “S-concrete” software and shows the imported loads
of the pier4. It will be designed by this software under the worst combination that it evaluates
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The following images shows the reinforcement and the interaction diagram of the pier4
Figure 44: Reinforcement $ interaction diagram as given by S-CONCERTE
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All walls have been introduced to the software,based on the limitations noted in chapter
“ PRELIMINARY DESIGN” 6 types of reinforcement were found .
The following tables show the different types of wall reinforcements:
Table 32: reinforcement detail of shear walls
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Table 33: Types of all shear walls
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CHAPTER 10
SOLID SLAB DESIGN
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Chapter 10. SOLID SLAB DESIGN
1.INTRODUCTION
In this chapter the structural design of the typical flat slab is discussed
2. PREDIMENSIONING OF THE SOLID SLAB
Slab preliminary thickness was previously determined according to ACI 318-08 section 9.5.
As seen in chapter 6 section 6.2 , a 30cm thick slab were chosen .
Since the circular area is supported by columns and has a span larger than the one for which
the thickness has been calculated , a verification for deflection and punching shear should be
elaborated .
Since the other part is supported by walls , no need for punching shear verification .
To enter the slab contour , loads and sections properties an import from ETABS to SAFE is done
.
The following picture represents the typical floor plan as introduced into SAFE:
Figure 45 - Typical floor slab on SAFE
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3.CHECK FOR DEFLECTION
Since the circular part has a span larger than the span for which the thickness has been chosen,
a check for deflection for this part is required.
Since SAFE give the short term deflection , long term deflection will be evaluated by both hand
and excel sheet calculation
Figure 46: Maps of bending moment due to dead load
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Figure 47: Maps of bending moment due to live load
Figure 48: Deflection due to dead load
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Figure 49: Deflection due to live load
Screenshots that represent:
- Moment due to dead load
- Moment due to live load
- Deflection due to dead load
- Deflection due to live load
were taken from “SAFE” as seen in the figures above.
The slab is reinforced with the min reinforcing, a check for deflection with this area of steel is
computed.
As min = 0.0018*b*h = 0.0018 *100*30 = 5.4 cm2=> 12@200
→ Manual check for deflection
DATA:
- F’c= 30Mpa
- Mdl= 5.7 t.m
- Mll= 1.94 t.m
- Δ D.L =1.34 cm
- Δ L.L = 0.46 cm
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- Span = 17.5 m
- % of sustained load from LL = 50%
- Rectangular section = (1m)*(0.3m)
- As = A’s = 15.84 cm^2 => 12@150 mesh + 12@150 add
CALCULATION :
-
𝐹𝑟 = 0.7�𝑓 ′ 𝑐 = 0.7 ∗ √30 = 3.83 𝑀𝑝𝑎
𝐸𝑐 = 4700 ∗ �𝑓 ′ 𝑐 = 25742.96 𝑀𝑝𝑎
𝐸𝑠 = 200000𝑀𝑝𝑎
𝑛=
𝐸𝑠
= 7.769
𝐸𝑐
𝑏ℎ3
𝐼𝑔 =
𝑦𝑡 =
12
ℎ
2
= 100 ∗
= 15𝑐𝑚
(30)3
12
Eq- 9.9
8.5.1
= 225000 𝑐𝑚4
Location of neutral axis :
𝑏
∗ 𝑥 2 + (𝑛𝐴𝑠 + (𝑛 − 1)𝐴′ 𝑠)𝑥 − (𝑛𝐴𝑠𝑑 + (𝑛 − 1)𝐴′ 𝑠 𝑑 ′ ) = 0
2
50𝑥 2 + 230.5 x − 3851.62 = 0
𝐼𝑐𝑟 =
𝑥 = 6.76 𝑐𝑚
3
𝑏𝑥
+ 𝑛𝐴𝑠 (𝑑 − 𝑥)2 + (𝑛 − 1)𝐴′ 𝑠(𝑥 − 𝑑 ′ )2 = 61042.2 𝑐𝑚4
3
𝐹𝑟
= 5.745 𝑡. 𝑚
𝑦𝑡
𝑀𝑠𝑢𝑠 = 𝑀𝑑𝑙 + 0.5 𝑀𝑙𝑙 = 6.67 𝑡. 𝑚
𝑀𝑐𝑟 = 𝐼𝑔 ∗
Effective moment of inertia :
- Under dead load only :
-
-
𝐼𝑒(𝑑𝑙) = �
𝑀𝑐𝑟 3
𝑀𝑑𝑙
𝑀𝑐𝑟 3
� ∗ 𝐼𝑔 + �1 − �
𝑀𝑑𝑙
� � ∗ 𝐼𝑐𝑟 = 𝐼𝑔 = 225000 𝑐𝑚4 ( cz Mcr>Mdl)
Under sustained load :
𝑀𝑐𝑟 3
𝑀𝑐𝑟 3
𝐼𝑒(𝑠𝑢𝑠) = �
� ∗ 𝐼𝑔 + �1 − �
� � ∗ 𝐼𝑐𝑟 = 165809 𝑐𝑚4
𝑀𝑠𝑢𝑠
𝑀𝑠𝑢𝑠
Under total load (dl+ll) :
𝑀𝑐𝑟 3
𝑀𝑐𝑟 3
𝐼𝑒(𝑡𝑜𝑡𝑎𝑙) = �
� ∗ 𝐼𝑔 + �1 − �
� � ∗ 𝐼𝑐𝑟 = 130756.5 𝑐𝑚4
𝑀𝑡
𝑀𝑡
Short term deflection :
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-
Deflection due to dead load = ∆𝑑𝑙 ∗
𝐼𝑔
𝐼𝑒𝑑𝑙
= 1.34 𝑐𝑚
Deflection due to (live+ dead ) load = ∆𝑡𝑙 ∗
𝐼𝑔
= 3.097 𝑐𝑚
𝐼𝑒𝑡𝑙
𝐼𝑔
Deflection due to sustained load = ∆𝑠𝑢𝑠𝑙 ∗
𝐼𝑒𝑠𝑢𝑠
= 2.13 𝑐𝑚
Deflection due to live load = ∆𝑡𝑜𝑡𝑎𝑙 − ∆𝑑𝑙 = 3.097 − 1.34 = 1.757 𝑐𝑚
Additional long term deflection :
𝜁
= 2.13 𝑎𝑓𝑡𝑒𝑟 2 𝑦𝑒𝑎𝑟𝑠
1 + 50𝜌′
∆𝑙𝑜𝑛𝑔 = 𝜆 ∗ ∆𝑠𝑢𝑠 = 4.93 𝑐𝑚
𝜆=
→ Check for deflection using excel sheet
Since we obtained the same values in the excel sheet , the sheet will be adopted for check of
deflection .
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The long term deflection for floor supporting non- structural element likely to me damaged by
large deflection is checked with respect to the maximum permissible deflections given by ACI
318-08 table 9.5 b
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To limit the long term deflection , two way can be adopted :
-
Put required reinforcement
Increase slab thickness
Required reinforcement for deflection : 16@200mesh+25@200 add
So we will increase the thickness to 40cm in order to decrease ρ :
As min = 7.2 cm2 =>14@200
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Than deflection is checked for : 14@200 mesh + 16@200 add
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4. CHECK FOR PUNCHING SHEAR
As we can see in the figure above, the slab is supported by walls where punching shear have no
effect. Only for the edge columns punching shear will be checked.
Values of reactions on columns (Vu) taken from SAFE are tabulated in the figure below:
Figure 50: Ultimate reaction given by SAFE
An excel sheet based on ACI-chapter 11 is used to check punching shear on edge column . The
following formulas were used in the sheet :
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The most critical column reaction is verified in order to see if punching shear reinforcement is required.
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Since the ratio Vu/ΦVc obtained in the excel sheet is <1 than no need for shear reinforcement.
5.One-Way Shear Verification
The slab should be designed so that no shear reinforcement should be required.
According to ACI section 11.9.5, the concrete shear strength is given by:
Where:
-
ϕVc = 0.17 λ ϕ �f′cbw d
λ is the lightweight concrete factor taken as 1
b w is the width taken as 1 m for slabs
Φ is equal to 0.85
The concrete strength: f’c = 30 MPa
The effective depth d = 36 cm
ϕVc = 29T
The following figure shows the ultimate shear diagram on the slab given by “SAFE”:
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Figure 51 - Shear in typical slab
If Vu < Φ Vc then the slab doesn’t need shear reinforcement and the concrete section can
resist the applied shear.
Max Vu =25T < 29 T
So a 40cm circular slab to be adopted and a 30 cm slab for the rest of the slab
6.REINFORCEMENT
For the reinforcement design, we choose a grid reinforcement and we add additional
reinforcement on the areas where more reinforcement is needed.
The following figure shows the repartitions of the ultimate bending moment on the slab:
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Figure 52: M11 for ultimate combination
Figure 53: M22 for ultimate combination
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The following table shows the slab’s mesh reinforcement in both direction :
Story
Typical slab
Circular slab
Grid X Direction
Bottom: T12@15
Top: T12@15
Bottom : T14@15
Top : T14@15
Table 34 - Slabs reinforcement
Grid Y Direction
Bottom: T12@15
Top: T12@15
Bottom : T14@15
Top : T14@15
T16@200 is to be added for check of deflection in the circular part
The grid reinforcement is placed according to those maps and additional reinforcement is
sometimes necessary.
Drawings for those results are obtained and present on the plans attached to the present
document.
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CHAPTER 11
POST-TENSIONED SLAB DESIGN
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Chapter 11. POST-TENSIONED SLAB DESIGN
1.INTRODUCTION
1.1 HISTORY OF POST-TENSIONING :
Prestressed concrete was first used by P.H.Jackson of San Francisco in1886. Although the
thought process was the first step , the steel available at that time was of limited strength .
The modern development of prestressed concrete, however, is usually attributed to Eugene
Freyssinet of France . In 1928, Freyssinet began to use high-strength steel wire for prestressing
concrete.
By 1939, Freysinet has designed conical wedges for anchoring the wires at the ends of
prestressed members and special jacks for wire stressing.
In 1940, GustaveMagnel of Belgium developed a system of curved, multi-wire tendons in
flexible rectangular ducts.
Post-tensioned concrete has been used for more than 40 years in the United- state ina wide
variety of construction projects.
First used primarily in bridge construction, applications for post-tensioning now extend far
beyond bridges to include:
- Tanks
- office buildings
- hotels
- parking structures
- shear walls
- etc …
1.2 PRESTRESSING IN PRINCIPLE :
According to “Design Guide Post-Tensioned Concrete Floors “ :
the principle of prestressing is explained below :
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2. ADVANTAGES OF POST-TENSIONED FLOORS
The primary advantages of post-tensioned floors over conventional reinforced concrete in-situ
floor, may be summarized as follows:
-
-
Longer spans and increased architectural design flexibility
Reductions in concrete and reinforcing steel quantities
Smaller and lighter structural members resulting :
→ Lower overall building heights
→ Smaller columns
→ Reduced foundation loads
→ Reduction in lateral force-resisting systems due to reduced building
dead load
Reduced floor cracking due to pre-compression in the slabs and beams
Deflections control due to the balancing of a significant portion of dead load
Rapid construction
Better water resistance
Monolithic connections between slabs , beams and columns with improved
structural integrity through continuous tendons
These advantages can result in significant savings in overall costs. There are also some
situations where the height of the building is limited, in which the reduced story height has
allowed additional story to be constructed within the building envelope.
A common concern regarding post-tensioned slab systems is the perceived inability or difficulty
in modifying the slab for future openings.
It is thoughts that the presence of the post-tensioning precludes or at least complicates this
procedure.
3.BONDED VS UNBONDED TENDON SYSTEM :
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Post-tensioned floors can be constructed using either bonded or unbonded tendons.
The following points may be made in favor of each.
3.1 BONDED SYSTEM
For a bonded system the post-tensioned strands are installed in galvanized steel or plastic ducts
that are cast into the concrete section at the required profile and form a voided path through
which the strands can be installed.
The ducts can be either circular- or oval-shaped and can vary in size toaccommodate a varying
number of steel strands within each duct. At the ends a combined anchorage casting is
provided which anchors all of the strands within the duct.
The anchorage transfers the force from the stressing jack into the concrete. Once the strands
have been stressed , the void around the strands is filled with a cementitious grout, which fully
bonds the strands to the concrete. The duct and thestrands contained within are collectively
called a tendon.
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The main features of a bonded system are summarized below:
- There is less reliance on the anchorages once the duct has been grouted
- The full strength of the strand can be utilized at the ultimate limit state (due to strain
compatibility with the concrete) and hence there is generally a lower requirement for
the use of unstressed reinforcement
- The prestressing tendons can contribute to the concrete shear capacity
- Due to the concentrated arrangement of the strands within the ducts a high force can
be applied to a small concrete section
- Accidental damage to a tendon results in a local loss of the prestress force only and does
not affect the full length of the tendon
3.2 UNBONDED SYSTEM
In an unbonded system the individual steel strands are encapsulated in a polyurethane sheath
and the voids between the sheath and the strand are filled with a rust-inhibiting grease.
The sheath and grease are applied under factory conditions and the completed tendon is
lectronically tested to ensure that the process has been carried out successfully.
The individual tendons are anchored at each end with anchorage castings. The tendons are cast
into the concrete section and are jacked to apply the required prestress force once the
concrete has achieved the required strength.
The main features of an unbonded system are summarized below :
- The tendon can be prefabricated off site
- The installation process on site can be quicker due to prefabrication and the reduced
site operations
- The smaller tendon diameter and reduced cover requirements allow the eccentricity
from the neutral axis to be increased thus resulting in a lower force requirement
- The tendons are flexible and can be curved easily in the horizontal direction to
accommodate curved buildings or divert around openings in the slab
- The force loss due to friction is lower than for bonded tendons due to the action of the
grease
- The force in an unbonded tendon does not increase significantly above that of the
prestressing load
- The ultimate flexural capacity of sections with unbounded tendons is less than that with
bonded tendons but much greater deflections will take place before yielding of the steel
- Tendons can be replaced (usually with a smaller diameter)
- Abroken tendon causes prestress to be lost for the full length of that tendon
- Careful attention is required in design to ensure against progressive collapse
3.3 WHY A BONDED SYSTEM WERE CHOSEN ?
A bonded system will be adopted cause of the number of advantages listed below:
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-
higher flexural capacity
good flexural crack distribution
good corrosion protection
flexibility for later cutting of penetrations
easier demolition
However there are some disadvantages such as :
- additional operation for grouting
- more labor intensive installation
However, the main reason why bonded tendons are preferred relates to the overall cost of the
structure and not just of the post-tensioning.
With unbonded tendons it is usual to have a layer of conventional reinforcement for crack
control. Using bonded tendons there is no such requirement and therefore the overall price of
bonded post-tensioning and associated reinforcement is less than for unbonded tendons.
For unbonded tendons the post-tensioning price may be less, but the overall cost of reinforcing
materials is greater.
4.PRESTRESS LOSSES :
From the time that a post-tensioning tendon is stressed, to itsfinal state many years after
stressing, various losses take placewhich reduce the tension in the tendon. These losses are
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grouped into two categories, namely short-term and longtermlosses.
The short-term losses include:
- Friction losses in the tendon
- Wedge set or ‘draw-in’
- Elastic shortening of the structure.
These losses take place during stressing and anchoring of thetendon.
The long-term losses include:
- Shrinkage of the concrete
- Creep of the concrete including the effect of the prestress
- Relaxation of the steel tendon.
Although these losses occur over a period of up to ten ormore years, the bulk occurs in the first
two years followingstressing. The loss in prestress force following stressing can
be significant (between 10% and 50% of the initial jackingforce at transfer and between 20%
and 60% after all losses) .
5.STRESSING STAGES:
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6.FLAT PLATE CONCEPT :
6.1 LIMITATIONS
According to “ Guide for design of post-tensioned buildings “ , a flat plate will be adopted as a
two-way framing system since spans are less than 9m.
Table 35: limitations of a flat slab
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For a prestressed floor, without primary reinforcement, to be considered as a flat slab the
following criteria have to be applied :
- Pre-compression is normally applied in two orthogonal directions:
Such a floor with no, or moderate, crack formation performs as a homogeneous elastic plate
with its inherent two-way behavior. The actual tendon location at a givenpoint in a floor system
is not critical to the floor’s two way behavior since axial compression, which is the main
component of prestressing, is commonly applied tothe floor at its perimeter.
Flat slab behaviour is, of course, possible with pre-compression applied in one direction only.
However in that situation it must be fully reinforced in the direction notprestressed.
- Aspect ratio (length to width) of any panel should not be greater than 2.0:
This applies to solid flat slabs, supported on orthogonal rows of columns.
For aspect ratios greater than 2.0 the middle section will tend to act as a one-way spanning
slab.
- Stiffness ratios in two directions:
The ratio of the stiffness of the slab in two orthogonal directions should not be
disproportionate. This is more likely to occur with non-uniform cross-sections such asribs.
For square panels this ratio should not exceed 4.0, otherwise the slab is more likely to behave
as one-way spanning.
- Number of panels:
Where the number of panels is less than three in either direction the use of the empirical
coefficient method, for obtaining moments and forces, is not applicable. In suchsituations a
more rigorous analysis should be carried out .
6.2 POST-TENSIONED FLAT SLAB BEHAVIOR
Tests and applications have demonstrated that a posttensioned flat slab behaves as a flat plate
almost regardless of tendon arrangement . The effects of the tendons are, of course, critical to
the behavior as they exert loads on the slab as well as provide reinforcement.
The tendons exert vertical loads on the slab known as equivalent loads, and these loads may be
considered like any other dead or live load. The objective is to applyprestress to reduce or
reverse the effects of gravity in a uniform manner.
Although the shape of the equivalent bending moment diagram from prestress is not the same
as that from uniformly distributed loading such as self-weight, it is possible, with careful placing
of the prestressing tendons, to achieve a reasonable match .
It should be noted that this will cause the peaks of resulting moments to appear in odd places.
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It is helpful to the understanding of post-tensioned flat slabs to forget the arbitrary column
strip, middle strip and nominal percentage tables which have long been familiar to the designer
of reinforced concrete floors. Instead, the mechanics of the action of the tendons will be
examined first.The following figure shows the bending moment surfaces for different
arrangements of tendons:
Figure 54: Bending moment surface for different arrangements of tendons
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2.LOAD BALANCING
The ‘load balancing’ approach is an even more powerful tool for examining the behavior of
two-way spanning systems than it is for one-way spanning members. By the balanced load
approach, attention is focused on the loads exerted on the floor by the tendons, perpendicular
to the plane of the floor.
Asfor one-way floors, this typically means a uniform load exerted upward along the major
portion of the central length of a tendon span, and statically equivalent downward load exerted
over the short length of reverse curvature. In order to apply an essentially uniform upward load
over the entire floor panel these tendons should be uniformly distributed, and the downward
loads from the tendons should react against another structural element. The additional
element could be a beam or wall in the case of one-way floors, or columns in a two-way
system.
However, a look at a plan view of a flat slab reveals that columns provide an upward reaction
for only a very small area. Thus, to maintain static rationality a second set of tendons
perpendicular to the above tendons must provide an upward load to resist the downward load
from the first set. Remembering that the downward load of the uniformly distributed tendons
occurs over a relatively narrow width under the reverse curvatures and that the only available
exterior reaction, the column, is also relatively narrow, it indicates that the second set of
tendons should be in narrow strips or bands passing over the columns.
Methods of accomplishing this two-part tendon system to obtain a nearly uniform upward load
may be obtained by a combination of spreading the tendons uniformly across the width of the
slab and/or banding them over the column lines.
Examples are shown below.
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The equivalent post-tensioning loading is normally referred to as the balanced loading.
The balanced loading is an externally applied set of loads that generally includes larges forces at
the anchorage locations , vertical forces related to the vertical curvature in the tendon profile ,
and friction forces parallel to the tendon .
It assumes that the tendon has been removed from the member and replaced with an
equivalent set of loads .
The balanced loading is an externally applied set of loads that generally includes large forces at
the anchorage location:
-
vertical forces related to the vertical curvature in the tendon profile
friction forces parallel to the tendon .
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Figure 55: Balance loading
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The balanced actions in a structure represent :
-
internal member forces
internal moment
Mb = M1 + M2
where:
-
Mb = balanced moment
M1 = primary moment
= internal tendon force multiplied by its eccentricity from the cross section centroid
- M2 = secondary moment
= the deformation held in place by the internal support will causea Support reaction.
The internal moment that result in the structure due to these reaction is called secondary
moment .
Figure 56: effect of prestressing on reactions and moments on a beam
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7. DESIGN PROCESS
7.1 POST-TENSIONED SLAB’S MINIMUM THICKNESS
The slab thickness must meet two primary functional requirements:
- structural strength
- deflection
Vibration shouldalso be considered where there are only a few panels. The selection of
thickness or type (e.g. plate without drops, plate with drops, coffered or waffle, ribbed or even
beam and slab) is also influenced by concrete strength and loading. There arelikely to be several
alternative solutions to the same problem and a preliminary costing exercise may be necessary
in order to choose the most economical.
Knowing the span and imposed loading requirements, the following table can be used to
choose a suitable span-depth ratio for the section type being considered:
Table 36: span-depth ratio
Span-depth ratio = 40 = 8.8/h => h=22 cm
We will start with 20 cm
7.2 RAM ANALYSIS
According to “FREYSSINET PRESTRESSING the system of the inventor of prestressed concrete“ ,
tendon’s cover and spacing requirement will be shown below :
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Figure 57: Tendon's cover limitations given by the catalogue
Having a=390mm and b= 190 mm
X> 390 +30 = 420 mm
Y> 190+30 = 220 mm
X’>0.5*420+50-10 = 250 mm
Y’> 0.5*220+50-10 = 150 mm
In order to respect this limitation, we use the following tendon’s elevations :
-
100cm at supports
40cm at mid-span
The hallway path has a smaller span relatively to the adjacent areas. It acts as a one-way slab
along X supported on the adjacent walls. So no need for tendons in the Y direction.
Tendons along Y directions are uniformily distributed to take out the effect of the excessive
deflection along the smallest span.
Tendons along X direction are placed adjacent to the opening.
The following figures show the tendon’s position and elevation:
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Figure 58: the position and the elevation of the tendon along x direction
Figure 59: the position and the elevation of the tendon along y direction
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The design of postentioned slabs is different than the solid slabs. For the last ones, a definition
of column strips and middle strips is required for the software to design, but for the
postentioned slabs only one strip is required, it lenghthes from a midspan to the adjacent one.
Different sections along the strip are made by the program, in order to find the average effort
on that strip
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Figure 60: strips drwan in x and y direction
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Table 37: Input to the software
Fpy = 0.9*Fpu
Min radius is limited to 3m to limit losses due to excessive curvature of tendons.
From “FREYSSINET Prestressing“ catalog ( Annex) , we get the propriety of tendons :
Table 38 : Nominal diameter and nominal reinforcement cross section (FRESSINET)
Considering that the jacking stress = 0.75*grade = 0.75*1860 = 1395 N/mm2
→ Losses
- Initial losses =10%
- Losses due to creep and shrinkage = 15%
- Long term losses = 0.15*0.1
- Long term force in the tendon = 0.75*0.9*0.85*fpu = 1067 N/mm2
→ Load combination :
And based on the explanations above, the load cases used are :
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Table 39: Load combinations
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1 PT = force in tendon after all losses = 0.75*0.85*0.9*1860T
→ Total long term deflection :
Since the pre-stressed flexural member are classified as class U , the section is considered
as un-cracked so the calculation of long term deflection is based on the Ig (gross area ) and
multiplied by the shrinkage and creep coefficient which is equal to 2
According to ACI-chapter 18 :
Table 40: Serviceability design requirements
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According to “Post-tensioned “ concrete floors Design handbook”:
3 = 1 (initial )+2 (long term)
→ Stress verification :
According to ACI – chapter 18,
the prestressed flexural member are classified as class U if :
Maximum allowable stresses in concrete immediately after transfer is given by :
Maximum allowable stresses in concrete at service loads is given by :
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The following tables show the stresses limit according to the ACI code and the stresses obtained
in the slab at transfer and service stages.
Table 41: Inital stress verification
Minimum capacity bottom stress = 0.6 f’ ci = 0.6*25 = 15 Mpa
Maximum capacity top stress
= 0.25*√𝑓′𝑐𝑖 = 0.25*√25 = 1.246 Mpa
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Table 42: Service stress verification
Minimum capacity bottom stress = 0.6 f’ c = 0.6*35 = 21 Mpa
Maximum capacity top stress
= 0.5*√𝑓′𝑐 = 0.5*√35 = 2.947 Mpa
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The previous figure show that the stresses are within the limits
The following figure show the stresses at the service stage:
Figure 61: Stress maps at service stage
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.
→ Deflection verification :
According to ACI- section 9.5.4.1 :
The figure below shows the long-term deflection Map:
Figure 62: Long-term deflection map
A comparision between the long term deflection that appeared in the slab and the code
limitation a according to ACI 318-08 table 9.5 b is done. The maximum long-term deflections
are within the limits.
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For punching reason , a beam connecting al the edge columns is placedalong the circular part .
→ Additional bonded reinforcement :
Minimum area of bondedreinforcement is obtained According to ACI – chapter 18 :
Where the stresses in tensile zones exceed 0.17√𝑓′𝑐 additional bonded reinforcement is
required ,and in compressive zones a reinforcement of As = 0.00075 Acf is distributed.
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Figure 63: additional bonded reinforcement in y direction
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Figure 64: Additional bonded reinforcement in x direction
:
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CHAPTER 12
STEEL DECK DESIGN
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Chapter 12 STEEL DECK DESIGN
1.INTRODUCTION
For some time steel formed plate and concrete have been used compositely in building floor
systems.
Composite steel deck floors consist of a profiled steel deck with a concrete topping. Included in
the concrete is some light welded mesh reinforcement which acts to control cracking, to resist
longitudinal shear and, in the case of fire, to act as tensile reinforcement.
Indentations in the profiled deck allow the concrete and steel to bond and share
load. Composite action between the supporting beams and the concrete is created by welding
shear studs through the deck onto the top flange of the beam.
Figure 65: Composite slab detail
During the placement of concrete, the steel deck resists the vertical loads associated with the
weight of the concrete and the construction loads.
As the concrete cures a chemical and mechanical bond forms collectively between thesurface
of the steel deck and the concrete.
The mechanical bond is achieved via embossed features on the surface of the steel deck.
Those deformations on the ribs allow for a stronger bond between the concrete and the
decking.Wholes on the surface of the steel deck are to allowthe positioning of the shear studs.
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2.COMPOSITE ACTION
Composite construction refers to two load-carrying structural members that are integrally
connected and deflect as a single unit.
Usually the formed metal deck is used as a permanent form to carry the fresh concrete and
serve as a working platform. After hardening of the concrete, the metal deck and concrete slab
act compositely to carry the applied live loads if the metal deck is provided with embossments
to provide the shear connection.
A natural consequence was to develop composite action for the steel beams over which the
formed metal deck was placed.
When the metal deck corrugations are parallel to the beam, they do not interfere with the
steel-concrete interaction and the condition is similar to a haunchedslab for which the
Specification provisions are applicable. However, when the metal deck is placed perpendicular
to the steel beams and shear connectors are placed in the ribs of the corrugations, the behavior
of the composite system may differ substantially from the expected behavior.
Figure 66: Metal deck corrugations parallel to the beam
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Figure 67: Metal deck corrugations perpendicaular to the beam
The strength of a composite diaphragm (deck + concrete) depends on the shear bond and the
complimentary behavior of steel and concrete in such an assembly. The steel functions as the
tension element at the bottom of the beam while the concrete functions as the compression
element at the top of the beam.
3.COMPOSITE DECK ADVANTAGES
Connecting the concrete to the steel beams can have several advantages:
- It is typical to have reduced structural steel frame cost.
- Weight of the structural steel frame may be decreased which may reduce
foundation costs
- Shallower beams may be usedwhich might reduce building height
- Increased span lengths are possible
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-
Reduced live load deflections
Stiffer floors
4.INSTALATION OF DECKING
4.1 PLACING OF DECKING
Metal decking is placed on the structural steel at predetermined points.
4.2 ATTACH MAETAL DECKING TO STRUCTURAL STEEL
-
As an alternative to welding,powder actuated tools may be used use to attach
metal decking to structural steel.
Powder actuated toolsuse the expanding gases from a powder load, or booster, to
drive a fastener into the steel beam.
A nail-like fastener is driven through the metal deck.
Thepowderactuated tool, powder load, and fastener must be matched to the
thickness of the structural steel beam flange.
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4.3SHEAR CONNECTORS
-
Depending on the welding process used, the tip of the shear connector may be
placed in a ceramic ferrule (arc shield) during welding to retain the weld.
Shear connectors create a strong bond betweenthe steel beam and the concrete
floor slab which is poured on top of the metal decking.
This bond allows the concrete slab to work with the steel beams to reduce live load
Deflection.
4.4 INSTALATION OF SHEAR CONNECTORS
-
The electrical arc process is commonly used for stud welding
An arc is drawn between the stud and te base metal
The stud is plunged into the molten steel which is contained by the ceramic ferrule.
The metal solidifies and the weld is complete.
The ferrules are removed before the concrete is poured.
4.5 INSTALATION OF CONCRETE
-
-
Concrete is installed by a concrete contractor on top of the composite metal
decking , shear connectors, and welded wire fabric or rebar grid (crack control
reinforcing)
Pumping is a typical installation method for concrete being placed on metal decking.
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4.6 QUALITY CONTROL
-
The shear connectors used in composite construction require specific
inspections and quality control.
Testing procedures are specified in the contract documents or by a local
building authority
5.STEEL DECK CALCULATIONS
The following figure shows the position of the steel deck on the ground floor plan.
As stated before it has been done in order to give a large space in the basement 1 with no need
for intermediate vertical supports, since a steel deck can offer larger spans than a concrete
deck.
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Figure 68: Position od the steel deck on the ground floor plan
5.1 STEEL DECK CONCEPT
The design of the steel structure slab is discussed in this part.
The floor plan has been divided into Primary I shaped beams and secondary I shaped beams
with a metal decking on top.
After placing those structural elements into place and connecting them, a 15cm concrete slab
will be casted on top of the metal decking.
The primary beams have a span of 17.6 m, and the secondary beams a span of 8.8 m.
The total area of the deck is 460 m2
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The following section states the calculations of the primary beams, secondary beams, as well as
their connections.
Figure 69: Primary beam and secondary beam
5.2 SECONDARY BEAM CALCULATIONS
→ WEIGHT CALCULATION
Loads on secondary beams are due to the concrete’s own weight, the super imposed live and
dead load of the slab, and the I shaped beam’s own weight.
- The concrete slab has an average thickness of 15 cm.
- The secondary beams are placed each 150 cm.
SDL=0.15 T/ m2
- LL=0.3T/ m2
- Assume the self-weight of the I beam is equal to 5% of the slab’s DL
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Loads are the following:
- Concrete slab’s own weight:
DL=2.5 T/m3 x 0.15 m =0.375 T/m2
Linear DL =0.375 T/ m2 x 1.5m =0.56 T/ m
- Super imposed dead load:
SDL=0.15 T/m3 x1.5 m =0.225 T/m
- Live load:
LL=0.3 T/m3 x1.5 m =0.45 T/m
- Secondary beam’s self -weight:
W=5%x 0.56 =0.0208 T/m
TOTAL LOADS:
W TOTAL =0.56+0.225+0.45+0.225=1.263 T/m
This linear load is carried by the secondary beam that has 3 spans.
In order to get the moment and shear diagram, the beam and its loads were introduced to the
software RDM 6.
The following figures show the loading, the moment and shear diagram obtained:
Figure 70: Loading on secondary beams introduced to RDM6
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Figure 72: Bending moment diagram on secondary beam
Figure 71: Shear diagram on secondary beam
Max positive moment= 7.825 T.m
Max negative moment= -9.78 T.m
Max shear = 6.7 T
→ ADMISSIBLE STRESS IN FLEXURE:
According to the (AISC F. P.5-45):
ALLOWABLE STRESS: STRONG AXIS BENDING OF I-SHAPED MEMEBERS AND CHANNELS.
MEMBERS WITH COMPACT SECTION:
For members with compact sections, symmetrical about and loaded in, the plane of their minor
axis the allowable stress is: Fb=0.66 Fy
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Fy=50ksiFb=0.66 x 50 =33 ksi
→ Calculation of the required section due to bending:
-
M max = 9.78 T.m =70.72 k.ft
-
S req =
𝑀𝑚𝑎𝑥
𝐹𝑏
=
70.72 𝑘.𝑓𝑡∗12
33 𝑘𝑠𝑖
=25.71 in3
Choiceof the most convenient section:
The table “allowable stress design selection table“gives the convenient W sections according
to S req . The section on top of each group is the lightest one therefore it is the most
economical.Our choice will be based on this criterion.
According to the table, the section W14x22 is chosen, it has the following properties:
S x = 29 in3
- Lc = 4.1 ft
- Lu=4.7 ft
- M R = 70 k.ft
The actual unbraced length in plane of bending “L b ” is found from the bending moment
diagram.
M=0 for x = 7.1m and x= 11.2 m
- L b for negative moment = 11.2-7.1 = 4.1m = 13.41 ft
- L b positive moment = 7.1 m = 23.29 ft
Verification if the assumption of a compact section is valid:
If lb≤Lc
=>Fbcompression tension = 0.66 Fy
Lc<lb≤ Lu =>Fbcompression tension = 0.60 Fy
lb> Lu
=>Fb tension = 0.6 Fy
Fb compression < 0.60 Fy
Lb>Lu => no need for moment redistribution
=>assumption thatFb =0,66Fyis not verified
 1st iteration to find the proper section
Assume Fb = 0.45 Fy = 0.45 *50 = 22.5 Ksi
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S req =
𝑀𝑚𝑎𝑥
𝐹𝑏
=
70.72 𝑘.𝑓𝑡∗12
22,5 𝑘𝑠𝑖
=37.7 in3
Choiceofthe most convenient section :
The table “allowable stress design selection table“ gives the convenient W sections according
to S req . The section on top of each group is the lightest one therefore it is the most
economical one.Our choice will be based on this criterion.
According to the table, the section W12x30 is chosen , it has the following properties :
S x = 38.6 in3
- Lc = 5.8 ft
- Lu=7.8 ft
- M R = 106 k.ft
- d/Af = 4.3
- r t = 1.73 in
According to table 6 page 5-121 ,cb =1
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According to AISC – chapter F – page 5-46, 47
𝑙𝑏
When �102 ∗ 103 ∗ 𝑐𝑏/𝐹𝑦≤ ≤�510 ∗ 103 ∗ 𝑐𝑏/𝐹𝑦
𝑟𝑡
2
When
𝑙𝑏
𝑟𝑡
𝐹𝑏 = � −
3
𝑙 2
𝐹𝑦∗ �𝑟𝑡�
1530∗103 ∗𝑐𝑏
≥�510 ∗ 103 ∗ 𝑐𝑏/𝐹𝑦
For any value of
𝑙𝑏
𝑟𝑡
170∗103 ∗𝑐𝑏
𝐹𝑏 =
:
𝐹𝑏 =
𝑙
( )^2
𝑟𝑡
12∗103 ∗𝑐𝑏
𝑙𝑏
12
= 13.45 ∗
= 93.29
𝑟𝑡
1.73
𝑙∗𝑑/𝐴𝑓
� ∗ 𝐹𝑌 ≤ 0.6𝐹𝑦
(F1-6)
(F1-7)
≤ 0.60 𝐹𝑦
(F1-8)
≤ 0.60 𝐹𝑦
�102 ∗ 103 ∗ 𝑐𝑏/𝐹𝑦 = �102 ∗ 103 ∗ 1/50 = 45.166
�510 ∗ 103 ∗ 𝑐𝑏/𝐹𝑦 = �510 ∗ 103 ∗ 1/50 = 100.99
𝑙𝑏
45.166< < 100.99  Compare (F1-6) and (F1-8)
𝑟𝑡
2
(F1-6) :𝐹𝑏 = � −
3
(F1-8) :𝐹𝑏 =
𝑙 2
𝐹𝑦∗ �𝑟𝑡�
1530∗103 ∗𝑐𝑏
12∗103 ∗𝑐𝑏
𝑙∗𝑑/𝐴𝑓
=
2
� ∗ 𝐹𝑌 = �3 −
12∗103 ∗1
13.45∗4.3∗12
50∗ (93.29)2
1530∗103 ∗1
� ∗ 50 = 19.11 𝐾𝑠𝑖 < 0.6 𝐹𝑦 = 30𝐾𝑠𝑖
= 17.3 𝐾𝑠𝑖 < 30 𝐾𝑠𝑖
Fb max = max ( 19.11 Ksi ; 17.3 Ksi ) = 19.11 Ksi< 0.45 Fy = 22.5 Ksi
The assumption that Fb =0,45 Fy is not verified.
 2nd iteration to find the proper section
Assume Fb = 0.3 Fy = 15 Ksi
S req =
𝑀𝑚𝑎𝑥
𝐹𝑏
=
70.72 𝑘.𝑓𝑡∗12
15 𝑘𝑠𝑖
=56.57 in3
Choice of the most convenient section:
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The table “allowable stress design selection table “ gives the convenient W sections according
to S req . The section on top of each group is the lightest one therefore it is the most
economical one.Our choice will be based on this criterion.
According to the table, the section W18x35 is chosen, it has the following properties :
S x = 57.6 in3
- Lc = 4.8 ft
- Lu=5.6 ft
- d/Af = 6.94
- r t = 1.49 in
12
𝑙𝑏
= 13.45 ∗
= 108.32
𝑟𝑡
1.49
�102 ∗ 103 ∗ 𝑐𝑏/𝐹𝑦 = �102 ∗ 103 ∗ 1/50 = 45.166
�510 ∗ 103 ∗ 𝑐𝑏/𝐹𝑦 = �510 ∗ 103 ∗ 1/50 = 100.99
𝑙𝑏
𝑟𝑡
> 100.99
 Compare (F1-7) and (F1-8)
(F1-7) :𝐹𝑏 =
(F1-8) :𝐹𝑏 =
170∗103 ∗𝑐𝑏
𝑙
( )^2
𝑟𝑡
12∗103 ∗𝑐𝑏
𝑙∗𝑑/𝐴𝑓
=
=
170∗103 ∗1
(108.32)^2
12∗103 ∗1
= 14.5 𝐾𝑠𝑖 < 30𝐾𝑠𝑖
13.45∗6.94∗12
= 10.7 𝐾𝑠𝑖 < 30 𝐾𝑠𝑖
Fb max = max ( 14.5 Ksi ; 10.7 Ksi ) = 14.5 Ksi< 0.3 Fy = 15 Ksi
The assumption that Fb =0,3 Fy is not verified.
 3rd iteration to find the proper section :
Assume Fb = 0.2 Fy = 10 Ksi
S req =
𝑀𝑚𝑎𝑥
𝐹𝑏
=
70.72 𝑘.𝑓𝑡∗12
10 𝑘𝑠𝑖
=84.86 in3
Choiceof the most convenient section :
The table “allowable stress design selection table “ gives the convenient W sections according
to S req . The section on top of each group is the lightest one therefore it is the most
economical one.Our choice will be based on this criterion.
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According to the table, the section W18x50 is chosen, it has the following properties :
S x = 89.9 in3
- Lc = 6.7ft
- Lu=7.9 ft
- d/Af = 4.21
- r t = 1.94 in
𝑙𝑏
12
= 13.45 ∗
= 83.2
𝑟𝑡
1.94
�102 ∗ 103 ∗ 𝑐𝑏/𝐹𝑦 = �102 ∗ 103 ∗ 1/50 = 45.166
�510 ∗ 103 ∗ 𝑐𝑏/𝐹𝑦 = �510 ∗ 103 ∗ 1/50 = 100.99
𝑙𝑏
45.166< < 100.99  Compare (F1-6) and (F1-8)
𝑟𝑡
2
(F1-6) :𝐹𝑏 = � −
3
(F1-8) :𝐹𝑏 =
𝑙 2
𝐹𝑦∗ �𝑟𝑡�
1530∗103 ∗𝑐𝑏
12∗103 ∗𝑐𝑏
𝑙∗𝑑/𝐴𝑓
=
2
� ∗ 𝐹𝑌 = �3 −
12∗103 ∗1
13.45∗4.21∗12
50∗ (83.2)2
� ∗ 50 = 22 𝐾𝑠𝑖 < 0.6 𝐹𝑦 = 30𝐾𝑠𝑖
1530∗103 ∗1
= 17.66 𝐾𝑠𝑖 < 30 𝐾𝑠𝑖
Fb max = max( 22 Ksi ; 17.66 Ksi ) = 22 Ksi> 0.2 Fy = 10 Ksi
The assumption that Fb =0,2 Fy is verified
This W shape is chosen to resist the maximum negative moment, a verification for the positive
moment should be done as well.
Max positive moment= 7.825 T.m = 56.58 k.ft
- Lb= 7.1m = 23.29 ft
- Cb=1
-
𝑙𝑏
𝑟𝑡
=
23.29∗12
1.94
= 144
�102 ∗ 103 ∗ 𝑐𝑏/𝐹𝑦 = �102 ∗ 103 ∗ 1/50 = 45.166
�510 ∗ 103 ∗ 𝑐𝑏/𝐹𝑦 = �510 ∗ 103 ∗ 1/50 = 100.99
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𝑙𝑏
𝑟𝑡
> 100.99  Compare (F1-7) and (F1-8)
(F1-7) :𝐹𝑏 =
(F1-8) :𝐹𝑏 =
170∗103 ∗𝑐𝑏
𝑙
( )^2
𝑟𝑡
12∗103 ∗𝑐𝑏
𝑙∗𝑑/𝐴𝑓
=
=
170∗103 ∗1
(144)^2
12∗103 ∗1
= 8.19𝐾𝑠𝑖 < 30𝐾𝑠𝑖
23.29∗4.21∗12
= 10.02𝐾𝑠𝑖 < 30 𝐾𝑠𝑖
Fb max = max( 10.02Ksi ; 8.19Ksi ) = 10.02Ksi≈ 10 Ksi
The assumption on the positive moment is also verified.
The final I section chosen is W18x50 , it has the following geometrical properties :
-
Depth d=17.99 in
Web’s thickness t w = 0.355 in
Flange’s width b f = 7.5 in
Flange’s thickness t f = o,57 in
T = 15.5 in
The allowable bearing stress of the chosen section for the secondary beams is 0.2Fy <<<0.66FY ,
to allow this section to approach this upper limit, stiffeners should be placed 6ft (<Lc) away
from each other all over the beam’s length .
→ SHEAR VERIFICATION:
According to AISC – page 5-49 :
“F4. ALLOWABLE SHEAR STRESS “
For
ℎ
𝑡𝑤
≤ 380/�𝐹𝑦 , the allowable shear stress is : Fv= 0.4 Fy
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For W18x50
ℎ
𝑡𝑤
=
15.5
0.355
= 43.66 ≤
380
√50
= 53.74Fv = 0.4 Fy = 20 Ksi
V allowable = d*t w *0.4Fy = 17.99*0.355*20
= 127 K
Max shear = 6.7 T = 14K
127k >14k  the section’s web can resist the applicable shear.
5.3 PRIMARY BEAM CALCULATIONS
→ WEIGHT CALCULATION
Loads on primary beams are due to the concrete’s own weight, the super imposed live and
dead load of the slab, and the I shaped beam’s own weight itself.
- The concrete slab has an average thickness of 15 cm
- The primary beams are placed each 880 cm
SDL=0.15 T/ m2
- LL=0.3T/ m2
- Assume the self weight of the I beam is equal to 5% of the slab’s DL
Loads are the following:
- Concrete slab’s own weight:
DL=2.5 T/m3 x 0.15 m =0.375 T/m2
Linear DL =0.375 T/ m2 x 8.8m =3.3T/ m
- Super imposed dead load:
SDL=0.15 T/m3 x8.8 m =1.32 T/m
- Live load:
LL=0.3 T/m3 x8.8 m = 2.64T/m
- Secondary beam’s self weight:
W=5%x 3.3=0.066 T/m
TOTAL LOADS:
W TOTAL =3.3+1.32+2.64+0.066 = 7.326 T/m
This linear load is carried by the primary beam that has 1 span.
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-
Maximum moment M max = wl2/8 = 283.66 T.m = 2051 k.ft
Maximum shear V max = wl/2 = 64.46 T = 142 K
→ ADMISSIBLE STRESS IN FLEXURE:
According to the (AISC F. P.5-45)
ALLOWABLE STRESS: STRONG AXIS BENDING OF I-SHAPED MEMEBERS AND CHANNELS.
MEMBERS WITH COMPACT SECTION:
For members with compact sections, symmetrical about and loaded in, the plane of their minor
axis the allowable stress is:
Fb=0.66 Fy
Fy=50ksi Fb=0.66 x 50 =33 ksi
→ Calculation of the required section due to bending:
M max = 283.66 T.m = 2051 k.ft
S req =
𝑀𝑚𝑎𝑥 2051 𝑘.𝑓𝑡∗12
𝐹𝑏
=
33 𝑘𝑠𝑖
=745.8 in3
Choiceof the most convenient section :
The table “allowable stress design selection table “ gives the convenient W sections according
to S req
The section on top of each group is the lightest one therefore it is the most economical one.
Our choice will be based on this criteria .
According to the table, the section W44x198 is chosen, it has the following properties :
S x = 776 in3
- Lc = 10.6 ft
- Lu=11.2 ft
The actual unbraced length in plane of bending “L b ” is found from the bending moment
diagram.
L b forpositive moment = 17.6 m = 57.74 ft
Verification if the assumption of a compact section is valid :
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If lb≤Lc
=>Fbcompression tension = 0.66 Fy
Lc<lb≤ Lu =>Fbcompression tension = 0.60 Fy
lb> Lu
=>Fb tension = 0.6 Fy
Fb compression < 0.60 Fy
Lb>Lu => no need for moment redistribution
=>assumption thatFb =0,66Fyis not verified
 1st iteration to find the proper section:
Assume Fb = 0.45 Fy = 0.45 *50 = 22.5 Ksi
S req =
𝑀𝑚𝑎𝑥
𝐹𝑏
=
2051 𝑘.𝑓𝑡∗12
22,5 𝑘𝑠𝑖
=1094 in3
Choiceof the most convenient section :
The table “allowable stress design selection table “ gives the convenient W sections according
to S req . The section on top of each group is the lightest one therefore it is the most
economical one.Our choice will be based on this criteria.
According to the table, the section W44x285 is chosen, it has the following properties:
S x = 1120 in3
- Lc = 10.6 ft
- Lu=15.9 ft
- d/Af = 2.1
- r t = 2.95 in
According to table 6 page 5-121, cb =1
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According to AISC – chapter F – page 5-46, 47
𝑙𝑏
When �102 ∗ 103 ∗ 𝑐𝑏/𝐹𝑦≤ ≤�510 ∗ 103 ∗ 𝑐𝑏/𝐹𝑦
𝑟𝑡
2
When
𝑙𝑏
𝑟𝑡
𝐹𝑏 = � −
3
1530∗103 ∗𝑐𝑏
≥�510 ∗ 103 ∗ 𝑐𝑏/𝐹𝑦
For any value of
𝑙𝑏
𝑟𝑡
:
𝐹𝑏 =
𝐹𝑏 =
𝑙𝑏
12
= 57.74 ∗
= 222.7
𝑟𝑡
2.95
𝑙 2
𝐹𝑦∗ �𝑟𝑡�
170∗103 ∗𝑐𝑏
𝑙
( )^2
𝑟𝑡
12∗103 ∗𝑐𝑏
𝑙∗𝑑/𝐴𝑓
� ∗ 𝐹𝑌 ≤ 0.6𝐹𝑦
≤ 0.60 𝐹𝑦
≤ 0.60 𝐹𝑦
(F1-6)
(F1-7)
(F1-8)
�102 ∗ 103 ∗ 𝑐𝑏/𝐹𝑦 = �102 ∗ 103 ∗ 1/50 = 45.166
�510 ∗ 103 ∗ 𝑐𝑏/𝐹𝑦 = �510 ∗ 103 ∗ 1/50 = 100.99
𝑙𝑏
𝑟𝑡
> 100.99  Compare (F1-7) and (F1-8)
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(F1-7) :𝐹𝑏 =
(F1-8) :𝐹𝑏 =
170∗103 ∗𝑐𝑏
𝑙
( )^2
𝑟𝑡
12∗103 ∗𝑐𝑏
𝑙∗𝑑/𝐴𝑓
=
=
170∗103 ∗1
(222.7)^2
12∗103 ∗1
57.74∗2.1∗12
= 3.34𝐾𝑠𝑖 < 30𝐾𝑠𝑖
= 8.25 𝐾𝑠𝑖 < 30 𝐾𝑠𝑖
Fb max = max ( 8.25 Ksi ; 3.34 Ksi ) = 8.25 Ksi< 0.45 Fy = 22.5 Ksi
The assumption that Fb =0,45Fy is not verified
 2nd iteration to find the proper section:
Assume Fb = 0.2 Fy = 10 Ksi
S req =
𝑀𝑚𝑎𝑥
𝐹𝑏
=
2051 𝑘.𝑓𝑡∗12
10 𝑘𝑠𝑖
=2461.2 in3
Choiceof the most convenient section :
The table “allowable stress design selection table “ gives the convenient W sections according
to S req . The section on top of each group is the lightest one therefore it is the most economical
one.Our choice will be based on this criteria.
According to the table, the section W40x655 is chosen, it has the following properties:
S x = 2590 in3
- Lc = 15.1ft
- Lu=45.7 ft
- d/Af =0.73
- r t = 4.43 in
12
𝑙𝑏
= 57.74 ∗
= 156.4
4.43
𝑟𝑡
�102 ∗ 103 ∗ 𝑐𝑏/𝐹𝑦 = �102 ∗ 103 ∗ 1/50 = 45.166
�510 ∗ 103 ∗ 𝑐𝑏/𝐹𝑦 = �510 ∗ 103 ∗ 1/50 = 100.99
𝑙𝑏
𝑟𝑡
> 100.99
(F1-7) :𝐹𝑏 =
 Compare (F1-7) and (F1-8)
170∗103 ∗𝑐𝑏
𝑙
( )^2
𝑟𝑡
=
170∗103 ∗1
(156.4)^2
= 6.95 𝐾𝑠𝑖 < 30𝐾𝑠𝑖
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(F1-8) :𝐹𝑏 =
12∗103 ∗𝑐𝑏
𝑙∗𝑑/𝐴𝑓
=
12∗103 ∗1
57.74∗0.73∗12
= 23.72 𝐾𝑠𝑖 < 30 𝐾𝑠𝑖
Fb max = max ( 23.72 Ksi ; 6.95 Ksi ) = 23.72 Ksi< 0.2 Fy = 10 Ksi
The assumption thatFb =0,2Fy is verified.
The final I section chosen is W40x655, it has the following geometrical properties :
Depth d=43.62 in
Web’s thickness t w = 1.97 in
Flange’s width b f = 16.87 in
Flange’s thickness t f = 3.54 in
T = 33.25 in
-
The allowable bearing stress of the chosen section for the primary beam is 0.2Fy <<< 0.66FY, to
allow this section to approach this upper limit, stiffeners should be placed 15ft (<Lc) away from
each other all over the beam’s length .
→ SHEAR VERIFICATION:
According to AISC – page 5-49 :
“F4. ALLOWABLE SHEAR STRESS “
ℎ
For
𝑡𝑤
≤ 380/�𝐹𝑦, the allowable shear stress is: Fv= 0.4 Fy
For W40x655:
ℎ
𝑡𝑤
=
33.25
1.97
= 16.87 ≤
380
√50
= 53.74Fv = 0.4 Fy = 20 Ksi
V allowable = d*t w *0.4Fy = 43.62*1.97*20
= 1718 K
Max shear = 142 K
1718 k >142 k  the section’s web can resist the applicable shear
→ CHECK FOR DEFLECTION:
According to AISC– ( L3.1 , L3.2 ) ( p.5-180):
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Although deflection, rather than stress, is sometimes the criterion of satisfactory design , there
is no single scale by which the limit of tolerable deflection can be defined. Where limitations on
flexibility are desirable, they are often dictated by the nature of collateral building components,
such as plastered walls and ceilings , rather than by considerations of human comfort and
safety.
The admissible amount of movement varies with the type of component .
The most satisfactory solution must rest upon the sound judgment of qualified engineers.As a
guide the following rules are suggested:
1. The depth of fully stressed beams and girders in floors should if practicable be not less
than ( Fy/800) times the span . If members of less depth are used , the unit stress in
bending should be decreased in the same ratio as the depth is decreased from that
recommended above
dreq≥FyKsi/800 *span
2. The depth of fully stressed roof purlins should if practicable be less than (Fy/800) times
the span . Except in the case of flat roofs
dreq≥FyKsi/1000 *span
For the primary beams :
Span = 57.74 ft
Fy/800 *span = 50/800 *57.74 *12 = 43.30 in
d( W40x655) = 43.62 in > 43.3 in  the deflection is controlled.
5.4 BOLTED CONNECTIONS
-
Top flange of secondary beams is copped 4 in deep because the flange’s thickness
tfis 3.54 in.
Reaction is equal to R= 14K
Bolts A490
Slip critical type connection
Bolt diameter Ø =5/8 in
Section of bolt :
A b = π*Ø2 /4 = 0.3068 in2
According to AISC table J3.2 (p.5-73)
The allowable shear Fv for standard size holes = 21Ksi
The bolts connecting the secondary beams with primary beams are in double shear
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Number of bolts needed:
# = R/ ( Fv*2*Ab)
= 14/(21*2*0.3068)
= 1.02  2 Bolts should be used.
It’s recommended to increase the number of
bolts when the flange is copped to resist the
“block shear”.
Figure 73: Web tear-out block shear
→ Minimum spacing :
According to AISC section J3.8 (p.5-75)
The distance between centers of standard oversized or slotted fastener holes shall not be less
than 22/3 times the nominal diameter of the fastener nor less than that required by the following
paragraph , if applicable .
Along a line of transmitted forces , the distance between centers of holes s shall be not less than
3d when Fp is determined by Equations (J3-1) and (J3-2). Otherwise the distance between
centers of holes shall not be less than the following :
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S ≥ 2P/(Fu*t) + d/2
(J3-5)
Where
P =force transmitted by one fastener to the critical connected part in kips
Fu = specified minimum tensile strength of the critical connected part in Ksi
T = thickness of the critical connected part
d = bolt diameter
spacing = Max 22/3 d = 2.083 in
3d = 1.875 in
2P/(Fu*t) + d/2 = 0.57 in
 Spacing = 2 in
→ Minimum edge distance :
According to AISC table J3.5 (p.5-76) : “Centre of standard hole to edge of connected part”
Nominal bolt diameter = 5/8 in  minimum edge distance to gaz cut edges = lv = 7/8 in
minimum edge distance to rolled edges = lh = 7/8 in
Figure 74: Minimum edge distance
→ Web tear out :
Allowable block shear R BS = 0.30 Av Fu + 0.50 At Fu
Av = {lv + (n-1) s –(n-1) d –d/2} t
Where :
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- d = hole diameter= Ø + 1/8 = 6/8 in
- n = number of bolts
Av = {7/8 + (2-1) 2 –(2-1) *6/8 –(6/8)/2} 0.355
= 0.621 in2
At = (lh -d/2) t
= 0.177 in2
 R BS = 0.30 *0.621*150 + 0.50 *0.177*150
= 41.23 Kips > Reaction = 14 K
No tear out effect in the web.
→ Angle’s dimensions :
L1
= length along the secondary beam
≥ ½ + 2*distance to rolled edge
≥ ½ + 2*7/8 = 2.25 in
L1 = 2.25 in
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According to AISC table (p.4-137) : “ Assembling clearness “
L2
= length along the primary beam
= max
t + H2 + C1 + distance to rolled edge = 1.97 +(1+1/4) + 1 + 7/8 = 5.09 in
K +C3+distance to rolled edge
= 1 + 9/16 + 7/8 = 2.43 in
L2 = 5.09 in
Since K = 1in was assumed it should be verified
L2 should be larger than K + bolt diameter + distance to rolled edge
K + bolt diameter + distance to rolled edge = 1 + 5/8 + 7/8 = 2.5 in < L2 = 5.09 in  L2 is
acceptable
→ Angle’s thickness :
Assume t = ½ in
According to AISC tables for sections:choose L 5 x 3 x 1/2
In the direction of the force:
Edge distance = 2 P / (Fu*t )
= 2*(14/(2*2)) / (150*1/2)
= 0.093 in < ½ in the assumed thickness.
The final chosen angle is L 5 x 3 x 1/2
→ Angle’s height :
According to AISC ( F4 p.5-45 ) (J4 p.5-77) :
Fv = 0.4 Fy on gross area
0.3 Fu on net area
In order to find the required height of the angle, a verification based on the allowable shear
was made.
Since the connection has two angles, each is subjected to half of the reaction.
Reaction = 14K as calculated before
Calculation for one angle :
R/2 = Vall = 0.4*Fy*h*t
= 10h
 h > 0.7 in
0.3*Fu*( h-n*d) t = 10.5*(h-3/2)  h > 2.16 in
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The minimum height to be considered is 2.16 in
One last verification should be made to check if this height is enough for positioning the bolts
with respect to their spacing.
H’ = 2*(Ø + 1/8) + 2*(distance to rolled edge ) + spacing
= 5 in
Take a L 5 x 3 x ½ x 2 for the connection angles
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CHAPTER 13
RIBBED SLAB DESIGN
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Chapter 13.RIBBED SLAB DESIGN
1.INTRODUCTION
This chapter includes the design of the typical floor as a ribbed slab.
The typical floor plan has been divided into beams and ribs with hollow core blocks in
between
2.RIBBED SLAB CONCEPT
The typical floor plan has been divided into ribs positioned perpendicularly to primary beams
to ensure a good transfer of loading.
Thickness selection :
The slab thickness is determined from the ACI 318-05 table 9.5.a.
ACI -318-05- 9.5.2.1 —“ Minimum thickness stipulated in Table 9.5(a) shall apply for one-way
construction not supporting or attached to partitions or other constructionlikely to be damaged
by large deflections, unless computationof deflection indicates a lesser thickness canbe used
without adverse effects.”
Figure 75 - Table 9.5.a from the ACI 318-08
Longest span for both end continuous :
- for beams : 8.8m
- for ribs : 8.8m
Slab thickness:ℎ =
𝑙
21
=
880
21
= 41.9 = 42𝑐𝑚
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Note that for every one end continuous span, deflection verification will be held:
The following image shows the dimensions of the hollow core block used:
At first, primary beams which are supposed to carry the larger loads,are placed in a way to get
the smallest span in order to decrease their deflections. Secondary beams (including ribs) are
placed in a perpendicular direction to the primary beams.
Note that this distinction between primary and secondary beams does not affect their
reinforcement calculation, it is only held for conceptual purposes. A calculation of the load
carried by every element and its transfer to the other structural elements is required.
Hollow core blocks, are distributed between ribs .
The ribs width is taken 20cm, taking into consideration the large slab thickness.
Since the circular part of the slab has a span of 17m, and designing it as a ribbed slab would
demand a large thickness and a large reinforcement area, plus it would need the inclusion of a
column in its middle which is impossible for architectural puposes, designing it as a solid slab
was more convenient.
The following figure contains a plan that show all the ribs and the beams:
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Figure 76- Typical plan - Ribbed solution
3.LOAD CALCULATION
As calculated in “Chapter 3-section 7 “ :
-
LL = 0.3 T/m2
SDL = 0.15 T/m2
→ Weight in ribbed zone:
In one meter length, 5 hollow core blocks can be placed:
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2 5×0.34×0.2×(
V ribbed /m =
0.35+0.31
)
2
0.2+0.35
= 0.204 m3 /m2
V conc /m2 =0.42 − 0.204 = 0.216 m3 /m2
Weight in ribbed zone(T/m2) = 0.216 × 2500 + 0.204 × 1200 = 0.785 T/m2
W u (T/m2) =1.2(0.785 + 0.15) + 1.6 × 0.3 = 1.6 T/m2
4.DESIGN FOR BEAMS
The detailed calculations of the Primary beam PB4 are discussed in thissection, a table will
resume the reinforcing results for all other members.
4.1ULTIMATE LINEAR LOADS:
All beams have a height of 42 cm, since no drops are allowed for architectural reasons.
 Detailed calculations are done below for the primary beam PB4:
PB4 is subjected to its own weight, the ribs weight, the super imposed dead load, and the live
load applied on the slab.
A preliminary width is assumed at first for the beam to take into consideration its own weight,
before calculating its proper width by iterations.
Assuming b=120 cm
→ Linear loads on beam PB4 :
- Own weight : 2.5*b*h = 2.5 * 1.2*0.42 = 1.26 T/ml
- Ribs weight : tributary length * 0.785 = 0.785*7.6 = 5.97 T/ml
- SDL
: 0.15 * tributary length = 0.15*8.8 = 1.32 T/ml
- LL
: 0.3* tributary length = 0.3*8.8 = 2.64 T/ml
Ultimate linear load = 1.2*(1.26+5.966+1.32)+1.6*(2.64) = 14.5 T/ml
4.2 BEAMS WIDTH:
A computation of the convenient beam’s width should be done at first, to make sure if the
assumed one is enough to take all efforts.
The following procedure explains the method used to find the preliminary width of a beam.
It will be adopted for all other beams.
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 β 1 is a factor relating depth of equivalent rectangular compressive stress block to
neutral axis depth:
 If f’c is bigger than 28 Mpa:
f′ c−280
β 1 = 0.85 − 0.05 �
 If f’c is less than 28 Mpa:
70
� (kg/cm2)
β 1 = 0.85
 m=

fy
2
0.85×f′c
(Kg/cm )
ρ max =0.85 × β1 ×
14.1
 ρ min =
fy
(kg/cm2)
f′c
fy
×(
0.003
0.003+0.004
)
At this stage ρ chosen is found.
This ratio ranges from ρ min to ρ max . In order to control the deflection, it is taken as:
 ρ chosen =
ρmax
2
Once the reinforcement ratio is chosen, the factor R that depends from the reinforcement ratio
chosen ρ chosen , and from the yielding stress of the steel fy is to be calculated.
 R = ρchosen × fy × (1 − 0.5 × ρchosen × m) (Kg/cm2)
The nominal moment Mn is:
 Mn = R × b × d2
Mn is also obtained by dividing the ultimate moment Mu by Φ
 Mn =
Mu
 b=
Mu
ϕ
Tthe width b of the beam is:
ϕ×R×d2
Where d is the distance from extreme compression fiber to the centroid of longitudinal
reinforcement.
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Afterwards a comparison of the first assumed width and the calculated one “b” is made:
- If b< assumed width, the section is okey, it can always be decreased.
- If b>assumed width, the beam’s section needs to be changed.
Detailed calculations are done below for the primary beam PB4:
In order to find “b”,M u should be computed, which is why the beam properties, along with it
ultimate loadings are introduced to the software “RDM 6” in order to get the ultimate moment
and shear forces diagrams.
The following pictures are extracted from the “RDM 6” software, they show the ultimate
bending moment and shear forces diagrams for PB2
Figure 77:bending moment diagram for the PB
The maximum positive moment acting on this beam is 13.4 T.m
The negative moment on support is taken to be 0.2 M max>0 =0.2*13.4 = 2.68 T.
The shear at supports is 19.94
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→ Calculation of “b” the beam’s width:
The first assumed width that was taken in order to take into consideration the beam’s own
weight was 120 cm.
The maximum positive moment13.4T.m is used for the verification.
The results obtained from the procedure stated above are the following:
 m = 16.47

R =42.22 Kg/cm2
 ρ chosen = 0.0111
 b = 70 cm <120 cm
The width of the beam PB2is set to be 120 cm.
Same procedure is followed for all the other beams.
The following table shows the results obtained for all the beams:
Beam ID
PB2
PB4
PB11
PB12
PB13
Width(cm) depth(cm) Nb of spans
72.5
42
1
120
42
1
75
42
1
25
42
1
30
42
2
Table 43- Summary of the beams dimensions
4.2 FLEXURAL REINFORCEMENT
When this step of preliminary design is accomplished, the calculation of flexure and shear
reinforcement takes place, the long term deflection will be also checked.
Designing a beam’s longitudinal reinforcement requires finding its bending moments on
supports and spans. Introducing each beam’s input into the software ”RDM6” as its section, its
spans, supports and ultimate loads helped us find all ultimate bending moments and ultimate
shears diagrams.
After finding the ultimate moment Mu, the following steps are done for each beam in order to
find its reinforcement:
-
𝑚=
𝑓𝑦
0.85×𝑓′𝑐
ρmax = 0.85 × β1 ×
ρmin =
14.1
fy
f′c
fy
×(
0.003
0.003+0.004
)
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-
Mu
R=
Φ×b×d2
ρ=
m
(Kg/cm2 )
The reinforcement ratio is:
-
1
(1 − �1 −
if ρ <ρ min , ρ = ρ min
2×m×R
fy
)
if ρ >ρ max , compression reinforcement is required or increasing the beams dimensions is
needed.
Finally the area of steel required for the applied moment is:
-
As = ρ × b × d
Detailed calculations are done below for the primary beam PB2:
The following figures extracted from the software “RDM 6” show the ultimate loading on PB4
which is equals to 14.5 T/ml, and its ultimate bending moment.
Figure 78: Linear load on PB2
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Figure 79: Bending moment on PB2
PB2’s longitudinal reinforcement is calculated according to the procedure stated above, results
are the following:
Table 44: As required for PB2
The following table show the results of reinforcement obtained:
Table 45: As chosen for PB2
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4.3 SHEAR REINFORCEMENT
The critical section of calculation of the shear is situated at a distance of “d” from the edge of
the support.
Figure 80- Critical section of calculation
Vu
of
According to ACI318-M08
11.1, the design of cross
sections subjected to shear
shall be based on:
ΦVn ≥ Vu
Where:
-
Vu is the factored shear force at the section considered
Vn is the nominal shear strength computed by:
Vn = Vc + Vs
Where:
-
Vc is the nominal shear strength provided by concrete
Vs is nominal shear strength provided by shear reinforcement
According to ACI-318 08 11.2.1.1Vc for members subjected to flexure and shear is equal to:
Vc = 0.17λ�f′c bw d
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→ Shear reinforcement categories:
-Category 1:
If V u ≤ Φ.V c /2 no need for transversal reinforcement
-Category 2:
If Φ.V c /2 ≤ V u ≤ Φ.V c minimal transversal reinforcement is used :
min Φ.V s = Φ.A v .f y .d/S max = Φ.3.5.b w .d
Smax =
Avfyt
0.35bw
Smax< d/2< 60 cm.
Noting that for solid slabs, foundations and ribs no need for transversal reinforcement.
-Category 3:
If Φ.V c ≤ V u ≤ (Φ.V c + min Φ.V s )minimal transversal reinforcement is used :
Same conditions are the categorie 2, but without exceptions
-Category 4:
If (Φ.V c + min Φ.V s ) ≤ V u ≤ (Φ.V c + Φx4x0.265.√f' c .b w .d) calculating ΦVS.
ΦVs = Vu − ΦVc
ΦVsmax = Φ x 8 x 0.265 �f′c bw d
If Vs>Vs max the concrete strength f’c should be increased or the section’s dimensions should be
increased as well in order to decrease Vs
Choosing a quantity of shear reinforcement Av and computing the correspondent spacing using
the following formula:
S=
ΦAv fy d
ΦVs
Smax< d/2< 60 cm.
Noting that the shear reinforcement along the span is constant, as its spacing is variable.
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-Category 5: If (Φ.V c + Φx4x0.265.√f' c .b w .d)≤V u ≤(Φ.V c +Φx8x0.265.√f' c .b w .d)
S=
ΦAv fy d
ΦVs
Smax< d/4< 30 cm.
Detailed calculations are done below for the primary beam PB2:
The following figures extracted from the software “RDM 6” shows the ultimate shear force
diagram on PB2
PB2’s shear reinforcement is calculated according to the procedure stated above, results are
the following:
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Table 46: shear reinforcement
5.DESIGN FOR SB4 SUBJECTED TO TORSION
→ Linear loads on beam SB4 :
- Own weight : 2.5*b*h = 2.5 * 0.9 *0.42 = 0.945 T/ml
- Ribs weight : tributary length * 0.785 = 1.75 * 0.785 = 1.37 T/ml
- SDL
: 0.15 * tributary length = 0.15*1.75 = 0.26 T/ml
- LL
: 0.3* tributary length = 0.3*1.75 = 0.525 T/ml
Ultimate linear load = 1.2*(0.945+1.37+0.26)+1.6*(0.525)= 3.93 T/ml
The beam is introduced to the software “RDM 6” in order to get its ultimate bending moment
and shear diagrams.
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5.1 FLEXURAL REINFORCEMENT
The following image is extracted from the software “RDM6” and shows the ultimate bending
moment of the secondary beam SB4.
SB4 has the following properties:
- Number of spans: 4 (each of 8.8 m)
- b=90 cm
- h=42 cm
The following figure is extracted from the software “RDM 6” and shows the ultimate
moment diagram on SB4:
The beam is symmetrical with respect to its middle support.
Max moment on the first span = 2.3 T.m
Max moment on the second span = 1.08 T.m
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Results for longitudinal reinforcement of all spans are the following:
Table 47: Longitudinal reinforcement for all spans
5.2 SHEAR REINFORCEMENT
The following image is extracted from the software “RDM6” and shows the ultimate shear force
of SB3:
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The beam is symmetrical with respect to its middle support.
Max ultimate shear on the first span V u = 2.1 T
Concrete shear strength:
Vc = 2 x 0.625 x �𝑓′𝑐 x b w x d = 2 x 0.265 x √300 x 90 x 37.8/1000 = 31.22 T
φVc = 0.75 x 31.22 =23.422 T
V u <φVc/2no need for transversal reinforcement, but a minimum distribution of stirrups is
recommended.
A V = 5Ф8@20cm for all spans.
5.3 TORSIONAL REINFORCEMENT
The secondary beam SB3is subjected to a torsional moment because the cantilevered ribs are
causing a twisting moment along the length of that supporting beam.
The moment of torsion is equal to the weight in the ribbed zone times the distance between
the linear load’s center of gravity to the secondary beam axis.
Weight in ribbed zone(T/m2) = 0.785 T/m2 as calculated before
The rib’s flange = 55 cm
Ultimate linear load on ribs:
W u =1.2 (Weight in ribbed zone +SDL)x rib’s flange + 1.6 x LL x rib’s flange
=1.2(0.785 +0.15)*0.55 +1.6 x 0.3
= 1.1 T/m
Tu =
Wu x L2
2
=
1.1 x 1.95 ^2
2
= 2.08 𝑇. 𝑚
According to ACI-section 11.6.1
Threshold torsion:It shall be permitted to neglect torsion effects if the factoredtorsional moment
Tu is less than:
(a) For no-prestressed members:
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(b) For prestressed members:
(c) For non-prestressed members subjected toan axial tensile or compressive force:
Where:
-
Tu = factored torsional moment at section (N.mm)
fc′ = specified compressive strength of concrete (MPa)
Acp = area enclosed by outside perimeter of concrete cross section (mm2)
Pcp= outside perimeter of concrete cross section (mm)
→ Check if torsion may be neglected:
A cp = 90 x42 = 3780 cm2=378000 mm2
P cp = 2 x(90+42) = 264 cm =2600 mm2
𝐴𝑐𝑝 ^2
T cr = Ø x 0.083 x�𝑓′𝑐 x (
𝑃𝑐𝑝
T u = 2.08 T.m < T cr = 2.2 T.m
378000 ^2
) = 0.85 x 0.083 x√30 x (
2600
) =2.2 T.m
The torsion is neglected.
6. DESIGN FOR RIBS
The following tables show the summary results for all ribs.
Their flexural and shear reinforcement was made with the same procedures stated above.
N.B : some ribs, located near the openings are subjected to number of concentrated loads
equivalent to reactions of ribs in the perpendicular direction.
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Table 48: Summary for ribs reinforcement
N1
Span
1
2
3
4
Length
(m)
8.565
8.80
8.80
8.565
1
2
3
4
5
Length
(m)
0.725
1.2
1.2
1.2
0.725
Span
Vu (T)
1
2
3
4
4.6
4.6
4.6
4.6
Support
Moment (T.m)
As req (cm^2)
As chosen
L (cm)
4.94
2.5
2.5
4.94
4.79
2.41
2.41
4.79
2T20
2T14
2T14
2T20
1000
900
1000
900
Moment (T.m)
As req (cm^2)
As chosen
L (cm)
0.741
7.03
5
7.03
0.741
2
5.51
3.84
5.51
2.34
2T12
2T20
2T16
2T20
2T12
800
700
700
700
800
Av chosen
(cm^2)
1.01
1.01
1.01
1.01
Av
Spacing (cm)
2T8
2T8
2T8
2T8
15
15
15
15
Moment (T.m)
As req (cm^2)
As chosen
L (cm)
5.08
1.293
2.05
4.15
4.93
2.34
2.34
4.02
2T20
2T12
2T12
2T16
1000
1000
1000
900
Moment (T.m)
As req (cm^2)
As chosen
L (cm)
0.762
6.68
4.21
6.15
0.6225
2.34
5.22
3.21
4.78
2.34
2T12
2T20
2T16
2T20
2T12
800
600
700
700
800
Av
Spacing (cm)
2T8
2T8
2T8
2T8
15
15
15
15
N2
Span
1
2
3
4
Length
(m)
8.565
8.8
8.8
8.565
1
2
3
4
5
Length
(m)
0.725
1.2
1.2
1.2
0.725
Span
Vu (T)
1
2
3
4
4.5
4.5
3.7
3.7
Support
Av chosen
(cm^2)
1.01
1.01
1.01
1.01
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N3
Span
1
2
Length
(m)
8
3.325
1
2
3
Length
(m)
0.4
1.2
0.25
Span
Vu (T)
1
2
4.2
4.2
Support
Moment (T.m)
As req (cm^2)
As chosen
L (cm)
4.63
1
4.5
2.34
2T20
2T12
900
300
Moment (T.m)
As req (cm^2)
As chosen
L (cm)
0.639
5.33
0.2
2.34
4.11
2.34
2T12
2T16
2T12
600
450
300
Av chosen
(cm^2)
1.01
1.01
Av
Spacing (cm)
2T8
2T8
15
15
Moment (T.m)
As req (cm^2)
As chosen
L (cm)
4.63
1
4.5
2.34
2T12
2T12
350
400
Moment (T.m)
As req (cm^2)
As chosen
L (cm)
0.639
5.33
0.2
2.34
4.11
2.34
2T12
2T14
2T12
300
300
300
N4
Span
1
2
Length
(m)
3.325
3.325
1
2
3
Length
(m)
0.25
1.2
0.25
Span
Vu (T)
1
2
4.2
4.2
Support
Av chosen
(cm^2)
1.01
1.01
Av
Spacing (cm)
2T8
2T8
15
15
Moment (T.m)
As req (cm^2)
As chosen
L (cm)
1.046
2.34
2T12
350
Moment (T.m)
0.16
0.16
As req (cm^2)
2.34
2.34
As chosen
2T12
2T12
L (cm)
350
350
Av
Spacing (cm)
2T8
15
N5
1
Length
(m)
2.1
Support
1
3
Length
(m)
0.25
0.725
Span
Vu (T)
1
1.36
Span
Av chosen
(cm^2)
1.01
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N6
Span
1
Length
(m)
4.7
Moment (T.m)
1.26
As req (cm^2)
2.34
As chosen
2T12
L (cm)
500
Support
1
3
Length
(m)
0.725
0.24
Moment (T.m)
0.19
0.19
As req (cm^2)
2.34
2.34
As chosen
2T12
2T12
L (cm)
500
500
Span
Vu (T)
1
1.23
Av chosen
(cm^2)
1.01
Av
Spacing (cm)
2T8
15
Moment (T.m)
As req (cm^2)
As chosen
L (cm)
0.02
0.55
2.34
2.34
2T12
2T12
600
600
Moment (T.m)
As req (cm^2)
As chosen
L (cm)
0.01
0.63
0.08
2.34
2.34
2.34
2T12
2T12
2T12
600
600
600
N7
Span
1
2
Length
(m)
2.5
3.3
1
2
3
Length
(m)
0.2
1.2
0.2
Span
Vu (T)
1
2
1.2
1.2
Support
Av chosen
(cm^2)
1.01
1.01
Av
Spacing (cm)
2T8
2T8
15
15
Moment (T.m)
As req (cm^2)
As chosen
L (cm)
0.5
0.5
2.34
2.34
2T12
2T12
350
400
Moment (T.m)
As req (cm^2)
As chosen
L (cm)
0.07
0.83
0.07
2.34
2.34
2.34
2T12
2T12
2T12
300
300
300
Av
Spacing (cm)
2T8
15
2T8
15
N8
Span
1
2
Length
(m)
3.325
3.325
1
2
3
Length
(m)
0.25
1.2
0.25
Span
Vu (T)
1
1.24
Av chosen
(cm^2)
1.01
1.24
1.01
Support
2
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N9
Span
1
Length
(m)
2.5
Moment (T.m)
0.74
As req (cm^2)
2.34
As chosen
2T12
L (cm)
300
Support
1
3
Length
(m)
0.25
0.725
Moment (T.m)
0.1
0.1
As req (cm^2)
2.34
2.34
As chosen
2T12
2T12
L (cm)
300
300
Span
Vu (T)
1
1.36
Av chosen
(cm^2)
1.01
Av
Spacing (cm)
2T8
15
Moment (T.m)
As req (cm^2)
As chosen
L (cm)
4.94
2.5
2.5
4.94
4.79
2.41
2.41
4.79
2T20
2T14
2T14
2T20
1000
900
1000
900
Moment (T.m)
As req (cm^2)
As chosen
L (cm)
0.741
7.03
5
7.03
0.741
2
5.51
3.84
5.51
2.34
2T12
2T20
2T16
2T20
2T12
800
700
700
700
800
N11
Span
1
2
3
4
Length
(m)
8.565
8.8
8.8
8.565
1
2
3
4
5
Length
(m)
0.725
1.2
1.2
1.2
0.725
Span
Vu (T)
1
2
3
4
4.6
4.6
4.6
4.6
Support
Av chosen
(cm^2)
1.01
1.01
1.01
1.01
Av
Spacing (cm)
2T8
2T8
2T8
2T8
15
15
15
15
Moment (T.m)
As req (cm^2)
As chosen
L (cm)
2.34
2.54
4.93
2.34
2.41
4.79
2T12
2T14
2T20
600
1000
900
Moment (T.m)
As req (cm^2)
As chosen
L (cm)
0.351
4.906
7.054
0.74
2
4.71
6.78
2.34
2T12
2T16
2T25
2T12
550
550
650
800
N13
Span
1
2
3
Support
1
2
3
4
Length
(m)
5.75
8.8
8.565
Length
(m)
0.3
1.2
1.2
0.725
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Span
Vu (T)
1
2
3
3.64
4.1
4.1
Av chosen
(cm^2)
1.01
1.01
1.01
Av
Spacing (cm)
2T8
2T8
2T8
15
15
15
Moment (T.m)
As req (cm^2)
As chosen
L (cm)
10.4
5.71
5.74
10.4
10.56
10.56
10.56
10.56
9T14
9T14
9T14
9T14
900
950
950
950
Moment (T.m)
As req (cm^2)
As chosen
L (cm)
1.56
16
11.4
0.7
1.6
10.56
12.13
10.64
12.13
10.56
9T14
9T14
9T14
9T14
9T14
500
900
950
950
550
SB1-SB8
Span
1
2
3
4
Length
(m)
8.565
8.8
8.8
8.565
1
2
3
4
5
Length
(m)
0.725
1.2
1.2
1.2
0.725
Span
Vu (T)
1
2
3
4
10.5
8.3
8.3
10.5
Support
Av chosen
(cm^2)
251
251
251
251
Av
Spacing (cm)
5T8
5T8
5T8
5T8
15
15
15
15
Moment (T.m)
As req (cm^2)
As chosen
L (cm)
11.97
7
6.77
14.9
10
10
10
11.3
9T12
9T12
9T12
9T12
900
950
950
950
Moment (T.m)
As req (cm^2)
As chosen
L (cm)
1.795
18.7
13.5
20
2.235
16
14.3
10.2
15.3
10
9T12
9T12 +9T10
9T12
9T12+9T10
9T12
500
900 / 650
950
900/650
550
SB2-SB3SB7
Span
1
2
3
4
Support
1
2
3
4
5
Length
(m)
8.565
8.8
8.8
8.565
Length
(m)
0.725
1.2
1.2
1.2
0.725
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Span
Vu (T)
1
2
3
4
12.15
12.15
10.78
12.7
Av chosen
(cm^2)
251
251
251
251
Av
Spacing (cm)
5T8
5T8
5T8
5T8
15
15
15
15
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CHAPTER 14
TRANSFER BEAMS DESIGN
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Chapter14. TRANSFER BEAMS DESIGN
1.INTRODUCTION
The main difficulty in this project is the fact that the shear walls of 17 floors rely on transfer
beams that transmit their loads to basement’s columns, receiving as a result a significant
vertical load and a big moment at their fixed ends.
The main concern that is studied in this chapter in order to make these beams
structurallyefficient in this complex structure is their reinforcement, since their pre
dimensioning was already done.
The ground floor’s transfer beams allow to carry the loads coming from the implanted walls,
and the ground floor slab itself.
Moments on the transfer beams were extracted from the “E-tabs” model, not forgetting the
envelope combination that was created in order to find the maximum ultimate moment on
spans and supports.
The figure below shows the transfer beams of the ground floor plan:
Figure 81: Transfer beam on the ground floor
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2.MOMENTS AND SHEARS ON TRANSFER BEAMS
The following figure is extracted from the software “ETABS” and shows the envelope moment
diagram for beam DB3.
The envelope combination takes into consideration the largest positive and negative moment.
Figure 82:Envelope ultimate bending moment diagram
The following figure is extracted from the software “ETABS” and shows the envelope shear
diagram for beam DB3.
Figure 83: Envelope ultimate shear diagram
As the diagram shows, the maximum bending moment received by the beam from the
envelope combination are:
SPAN 1:
-
Max positive moment in span: M=178.58 T.m
Max negative moment at the first support: M=-60.25 T.m
Max negative moment in span: M=-63.49 T.m
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-
Max negative moment at the second support: M=-225.07 T.m
Max shear force at the first support: T=-255.76 T
Max shear force at the second support: T=467.50 T
SPAN 2:
-
Max positive moment in span: M=156.69 T.m
Max negative moment at the second support: M=-312.259 T.m
Max negative moment in span: M=-62.68 T.m
Max negative moment at the third support: M=-57.91 T.m
The bending reinforcement is based on those moments.
Max shear force at the second support: T=-486.45 T
Max shear force at the third support: T= 231.74 T
3.DETAILEDREINFORCEMENT CALCULATIONS
A detailed reinforcement calculation of the transfer beam DB3 is discussed in this section, all
other transfer beams are reinforced with the same procedure.
A summary table will show the results obtained for all other transfer beams.
3.1 FLEXURAL REINFORCEMENT
The detailed calculation of the transfer beam DB3’s longitudinal reinforcement is stated below.
The following procedure describes the steps for longitudinal reinforcing, and an excel sheet was
established in order to accelerate the work.
-
𝑚=
𝑓𝑦
0.85×𝑓′𝑐
ρmax = 0.85 × β1 ×
ρmin =
14.1
fy
Mu
R=
Φ×b×d2
ρ=
m
f′c
fy
×(
0.003
0.003+0.004
)
(Kg/cm2 )
The reinforcement ratio is:
-
1
(1 − �1 −
if ρ <ρ min , ρ = ρ min
2×m×R
fy
)
if ρ >ρ max , compression reinforcement is required, or increasing the beam’s dimensions
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Finally the area of steel required for the applied moment is:
As = ρ × b × d
-
→ Bending reinforcement for DB3:
Ultimate moment= 178.58
𝑚=
𝑓𝑦
=
Dimensions
Breadth of section (b)
Thickness of section (t)
Concrete clear cover (dc)
Depth of the section (d)
Materials
=
=
=
=
140
150
4
135
cm
cm
cm
cm
Steel yield strength (fy)
=
4200
Kg/cm2
Concrete cylinder strength (f'c)
=
300
Kg/cm2
4200
0.85×𝑓′𝑐 0.85∗300
ρmax = 0.85 × β1 ×
ρmin =
R=
ρ=
14.1
fy
Mu
=0.0035
Φ×b×d2
1
m
=
T.m
=16.47
f′c
fy
×(
178.58∗10^5
0.9∗140∗135^2
(1 − �1 −
2×m×R
fy
0.003
0.003+0.004
) = 0.85*0.85*
R
300
4200
∗(
0.003
0.003+0.004
) =0.02211
R
=7.77 kg/cm2
)=
1
16.47
ρ < ρmin therefore ρ = ρmin
(1 − �1 −
2∗16.47∗5.82
4200
) = 0.0018
As = ρ × b × d = 0.0035 x 140 x 135 = 66.15 cm2  As=14 T 25
The following table shows the reinforcement due to all other bending moments on the first and
second span, taking into consideration the envelope combination for DB3:
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Table 49: Longitudinal reinfoecement of DB3
3.2 SHEAR REINFORCEMENT
The following procedure describes the steps for shear reinforcing, and an excel sheet was
established in order to accelerate the work.
According to ACI318-M08 11.1, the design of cross sections subjected to shear shall be based
on:
ΦVn ≥ Vu
Where:
-
Vu is the factored shear force at the section considered
Vn is the nominal shear strength computed by:
Vn = Vc + Vs
Where:
-
Vc is the nominal shear strength provided by concrete
Vs is nominal shear strength provided by shear reinforcement
According to ACI-318 08 11.2.1.1Vc for members subjected to flexure and shear is equal to:
Vc = 0.17λ�f′c bw d
→ Shear reinforcement categories:
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-Category 1:
If V u ≤ Φ.V c /2 no need for transversal reinforcement
-Category 2:
If Φ.V c /2 ≤ V u ≤ Φ.V c minimal transversal reinforcement is used :
min Φ.V s = Φ.A v .f y .d/S max = Φ.3.5.b w .d
Smax =
Avfyt
0.35bw
Smax< d/2< 60 cm.
Noting that for solid slabs, foundations and ribs no need for transversal reinforcement.
-Category 3:
If Φ.V c ≤ V u ≤ (Φ.V c + min Φ.V s )minimal transversal reinforcement is used :
Same conditions are the categorie 2, but without exceptions
-Category 4:
If (Φ.V c + min Φ.V s ) ≤ V u ≤ (Φ.V c + Φx4x0.265.√f' c .b w .d) calculating ΦVS.
ΦVs = Vu − ΦVc
ΦVsmax = Φ x 8 x 0.265 �f′c bw d
If Vs>Vs max the concrete strength f’c should be increased or the section’s dimensions should be
increased as well in order to decrease Vs
Choosing a quantity of shear reinforcement Av and computing the correspondent spacing using
the following formula:
S=
ΦAv fy d
ΦVs
Smax< d/2< 60 cm.
Noting that the shear reinforcement along the span is constant, as its spacing is variable.
-Category 5: If (Φ.V c + Φx4x0.265.√f' c .b w .d)≤V u ≤(Φ.V c +Φx8x0.265.√f' c .b w .d)
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S=
ΦAv fy d
ΦVs
Smax< d/4< 30 cm.
Table 50:Shear reinforcement for DB3
Dimensions & Shear
Ultimate shear force(Vu)
Breadth of section (b w )
Depth of section (d)
Spacing of stirrups (s)
Number of legs (n)
=
=
=
=
=
231.7
140
135
10
14
Materials
Stirrups steel yield strength (fy)
Concrete cylinder strength (f'c)
Strength reduction factor φ
= 2400
= 300
= 0.75
T
cm
cm
cm
Kg/cm2
Kg/cm2
Concrete shear strength:
Vc = 2 x 0.625 x �𝑓′𝑐 x b w x d = 2 x 0.265 x √300 x 140 x 135/1000 = 175 T
φVc = 0.75 x 175 =130.12 T
Min φVs=φ x 3.5 x bw x d = 49.61 T
Categorie 4
𝑠=
φ Av Fyd d
(𝑉𝑢−φVc)
𝐴𝑣 =
𝐴𝑣 =
𝑠(𝑉𝑢−φVc)
φ Fyd d
10(231.74 − 131.12)
= 4.18 𝑐𝑚2
0.75 ∗ 2400 ∗ 135
The number of longitudinal bars chosen is 14, therefore 14 stirrups T12 are used:
4.18
Required stirrup area per leg =
14
= 0.3 𝑐𝑚2 Av=Ø8@10
The following table shows the reinforcement due to all other shear forces on the first and
second span, taking into consideration the envelope combination for DB3:
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3.2 LONG TERM DEFLECTION FOR DB3
According to the ACI 318-11 table 9.5.a:
Minimum thickness of non-prestressedbeams or one-way slabs unless deflections are calculated:
The transfer beam DB3 has a span of 10 m, therefore the minimum thickness unless deflections
are calculated is for one end continuous:
𝑙
18.5
=
1000
18.5
= 54𝑐𝑚
And having its thickness = 150 cm, no need to check its deflection.
4.SUMMARY RESULTS
All other transfer beams are reinforced with the same procedure, the following tables show the
reinforcing results:
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Table 51: Bending reinforcement for DB1
Table 52: Shear reinforcement for DB1
Table 53: Bending reinforcement for DB4
Table 54: Shear reinforcement for DB4
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Table 55: Bending reinforcement for DBA
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Table 56: Shear reinforcement for DBA
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CHAPTER 15
BASEMENT WALL DESIGN
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Chapter15. BASEMENT WALL DESIGN
1.INTRODUCTION
Basement walls are calculated to resist lateral loads due to earth and water pressures.They are
subjected to triangular soil and water loadings, and axial forces coming from the slab.
Therefore, the design will be performed as a beam with 100 cm width and a length equal to the
wall’s height which is from basement 6 to the ground floor.
2.LOADINGS
As the section of the basement wall shows, this beam will be subjected to composite flexure
(N,M) and the interaction diagram must be drawn.
2.1 SOIL LOADINGS
Figure 84: Basement wall loading
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The soil has the following properties:
Soil Properties
Unit weight in KN/m3
20
Φ=
25
Earth pressure coeff Ko
0.577
K0=1-sinΦ= 1-sin(25) =0.577
The soil pressure at the lowest point of the wall is given by the following:
P = K o γ h = 0.577 x 20
KN
m3
𝑥 18 𝑚 = 207.72 KN/m2 = 20.72 T/m2
Taking the wall’s width equals to 1m, the load at the lowest point is given by the following:
F=P x 1m =207.72
Finding Mu(T.m) on each span:
KN
m
= 20.77 T/m
This figure is extracted from the software “RDM6” and shows the ultimate loading due to soil
on the basement wall, noting that the pinned supports are the basement slabs and the fixed
support is the raft foundation.
Ultimate load on the lowest point of the wall is equal to:
W= 1.6 x 207.7 =332.32 KN/m
This load varies linearly along the length of the wall, it is equal to zero at the GF slab:
Figure 85: Ultimate moment diagram on basement wall
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Max Moments on spans:
-
Mu max on span1
Mu max on span2
Mu max on span3
Mu max on span4
Mu max on span5
Mu max on span6
=7.6 KN.m =0.76T.m
=30.8 KN.m =3.08 T.m
=51.1 KN.m =5.11 T.m
=72.54 KN.m =7.254 T.m
=92.25 KN.m =9.225 T.m
=117.6 KN.m =11.76 T.m
Moments on supports:
-
Muon support 1 =1.52 KN.m =0.152T.m
Muon support 2 =38 KN.m =3.8T.m
Muon support 3 =81.9KN.m =8.19T.m
Muon support 4 =123.6 KN.m =12.36T.m
Muon support 5 =164.8 KN.m =16.48T.m
Muon support 6 =209 KN.m =20.9T.m
Muon support 7 =239 KN.m =23.9T.m
2.1 AXIAL LOADING
The axial loading is equal to the slab weight according to the influence surface of the wall and
the wall’s weight.
The influence area varies from one wall to another, the design of basement wall will be where
it is the most subjected to an axial load.
The following picture shows the most critical one and the other sections will be similar to
this one. In other words, the design will be performed per linear meter and the resulting
reinforcement will be displayed the same for all basement walls.
The most critical influence line is shown in the following picture:
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Figure 86: Basement wall influence line
IL = 5.6 m
On the basement 1 wall we have the following loads:
-
-
DL = IL × 1 x (slab weight + SDL on slab) + Wall weight
= 5.6 × (2.5 × 0.3 + 0.15) + 2.5 × 0.3 × 1 × 2.7
= 𝟕. 𝟎𝟔 𝐓/ml
LL = IS × (LL on slab)
= 5.6 x 1 × 0.35
=𝟏. 𝟗𝟔 𝐓/ml
The factored load is:
Wu = 1.2DL + 1.6 LL
N u =11. 6 T/mL
The following table shows the axial force on each basement wall:
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BAS 1
BAS 2
BAS 3
BAS 4
BAS 5
BAS 6
DL (T/m)
7.06
14.12
LL(T/m)
Wu(T/m)
1.96
11.608
4.76
24.56
21.18
7.56
37.512
25.24
10.36
46.864
35.3
42.36
13.16
15.96
63.416
76.368
Figure 87: Ultimate load on the wall
3. DESIGN
This wall is introduced as a beam to the software“ S-CONCRETE”, it is subjected to an axial force
and a bending moment, which is why an interaction diagram must be drawn in order to locate
our (N,M) point, to check if it is within the limits.
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The Bas 6 results are the following:
-
Vertical reinforcement: T 20 @ 15
Horizontal reinforcement: T 12 @ 15
Figure 88: Interaction diagram on the basement wall
The same procedure is repeated for all the basement walls, the following table shows the
dimensions and reinforcement of the wall all over its height.
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Table 57: Dimension and reinforcement of the wall all over it's height
Support 1
Span BAS 1
Support 2
Span BAS 2
Support 3
Span BAS 3
Support 4
Span BAS 4
Support 5
Span BAS 5
Support 6
Span BAS 6
Support 7
h(cm)
Mu(T.m)
30
30
30
30
30
0.152
30
5.11
40
12.36
40
7.25
40
16.48
40
9.22
40
20.9
40
11.76
40
23.9
0.76
Wu(T/m)
T14@15
11.608
3.8
3.08
BARS
T14@15
T16@15
24.56
8.19
T16@15
T16@15
37.512
T16@15
T20@15
46.864
T20@15
T20@15
63.416
T20@15
T20@15
76.368
T20@15
T20@15
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CHAPTER 16
FOUNDATIONS
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Chapter 16 FOUNDATIONS
1. INTRODUCTION
The substructure or foundationis the part of a structure that is usually placed belowthe
surface of the ground.Footings and other foundation units transfer the loads from the structure
to the soil or rock supporting the structure because the soil is generally much weaker than the
concrete columns & walls thatmust be supported, the contact area between the soil & the
footing is much larger thanthat between the supported member & the footing.
Foundation types:
-
Strip footings or wall footings display essentially one-dimensional action, cantilevering
out on each side of the wall.
Spread Footings are pads that distribute the column load to an area of soil around the
column. These distribute the load in two directions. Sometimes spread footing have
pedestals, are stepped, or are tapered to save materials.
A pile cap transmits the column load to a series of piles, which in turn, transmit the load
to a strong layer at some depth below the surface “hard strata”.
Combined footings transmit the loads from two or more columns to the soil. Such a
footing is often used when one column is close to a property line.
A mat or raft foundation transfers the loads from all the columns in a building to the
underlying soil. Mat foundations are used when very weak soils are encountered.
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Figure 89: Different types of foundations
The choice of foundation’s type is selected in consultation with the geotechnical engineer.
Factors to be considered are:
- The soil strength
- The soil type
- The variability of the soil type over the area and with increasing depth
- The susceptibility of the soil and the building to deflections
The two essential requirements in the design of foundation are :
- total settlement of the structure < tolerably small amount
- differential settlement of the various parts of the structure be eliminated as nearly
as possible.
With respect to possible structural damage, the elimination of differential settlement, i.e.,
different amounts of settlement within the same structure, is even more important than
limitations on uniform overall settlement.
To limit settlements as indicated, it is necessary to:
- Transmit the load of the structure to a soil stratum of sufficient strength
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-
Spread the load over a sufficiently large area of that stratum to minimize bearing
pressure
2. RAFT FOUNDATION
2.1 DEFINITION
A shallow single Foundation unit that supports all columns & walls of a structure or parts of a
structure may be called a raft foundation or mat foundation. It is usually designed as l flat slab.
Raft foundation is required when dealing with:
-
multi-story buildings
overhead water tanks, chimneys, etc
in submerged structure
multi-story structures with basement and in retaining walls, etc
Foundation engineering often consider mats when dealing with any of the following conditions:
-
The structural loads are so high
the soil conditions so poor that spread footings would be exceptionally large
area(spread footings)> (building area/3)
mat foundation
-
The soil is very erratic & prone to excessive differential settlements. The structural
continuity & flexural strength of a mat will bridge over these irregularities.
-
The structural loads are erratic, and thus increase the likelihood of excessive differential
settlement again, the structural continuity and flexural strength of the mat will absorb
these irregularities
-
Lateral loads are not uniformly distributed through the structure and thus may cause
differential horizontal movement in spread footing or pile caps. The continuity of a mat
will resist such movements
-
The uplift loads are larger than spread footings can accommodate.The greater weight
and continuity of a mat may provide sufficient resistance
-
The bottom of the structure is located below the ground table, so waterproofing is an
important concern. Because mats are monolithic, they are much easier to waterproof.
The weight of the mat also helps resist hydrostatic uplift forces from the groundwater.
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In this project, due to the heavy loadings, the earthquake design and some of the previous
provisions the mat foundation might be used.
2.2 METHODS USED FOR MAT FOUNDATIONS
Various methods have been used for the mat foundations.They can be divided into two
categories:
- RIGID METHOD
- NON-RIGID METHOD
2.2.1 RIGID METHOD
The simplest approach to structural design of mats is the rigid method. It assumes the mat is
much more rigid than the underlying soils, which means any distortion in the mat are too small
to significantly impact the distribution of bearing pressure depends only on the applied loads
and the weight of mat, and either uniform across the bottom of the mat (if the normal acts
through the centroid and no moment load is present) or varies linearly a cross the mat (if
eccentric or moment loads are present) as shown in figure below:
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Figure 90: Distribution of the bearing pressure on the bottom of the raft
For analysis purposes, the mat becomes an inverted and simply loaded two-way slab, which
means the shears, moments, and deflection may be easily computed using the principles of the
structural mechanics. The engineer can then select the appropriate mat thickness and
reinforcement.
Although this type of analysis is appropriate for spread footings, it doesn't accurately model
mat foundations becomes the width-to-thickness ratio is much greater in mats and the
assumption of rigidity is no longer valid.
Portions of a mat beneath columns and bearing walls settle more than the portions with loss
load, which means the bearing pressure will be greater beneath the heavily-loaded zones
Figure 91: soil bearing pressure for rigid and non-rigid mat foundation
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This redistribution of bearing pressure is most pronounced when the ground is stiff compared
to the mat as shown in the following figure, but is present to some degree in all soils
Figure 92: Bearing pressure as function of soil properties
Because the rigid method does not consider this redistribution of bearing pressure, it doesn't
produce reliable estimates of the shear, moments, and deformations in the mat. In addition,
even if the mat was perfectly rigid, the simplified bearing pressure distribution are not correctin reality; the bearing pressure is greater on the edges and smaller in the center.
2.2.2 NON- RIGID METHOD
To become the in accuracies of the rigid method by using analyses that consider deformations
in the mat and their influence on the bearing pressure distribution. These are called non-rigid
methods, and produce more accurate values of mat deformations and stresses, unfortunately
non-rigid analyses also are more difficult to implement because they require consideration of
soil-structure interaction and because the bearing pressure distribution is not as simple.
→ Coefficient of subgrade reaction :
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Because non-rigid method consider the effects of local mat deformations on the distribution
of bearing pressure, it is necessary to define the relation slip between settlement and bearing
pressure. This is usually done using the coefficient of subgrade reaction, Ks (also known as the
modulus of subgrade reaction, or the subgrade modulus)
The coefficient K s has units of force length cubed.
The interaction between the mat and the underlying soil may there be represented as a"bed of
springs" each with a stiffness Ks per unit area, as shown in fig .
Portions of the mat that experience more settlement produce more compression in the
"springs," which represents the higher bearing pressure, whereas portions that settle less don't
compress the springs as for and thus have less bearing pressure. The sum of these spring forces
must equal the applied structural loads plus the weight of the mat.
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Figure 93: bed of spring representing soil below the mat
This method of describing bearing pressure is called a soil-structure interaction analysis
because the bearing pressure depends on the mat deformations, and the mat deformations
depends on the bearing pressure.
 Determination of the coefficient of subgrade reaction :
Most mat foundation designs are currently developed using either the Winkler method or the
pseudo-coupled method, both of which depend on our ability to define the coefficient of
subgrade reaction, Ks.
Unfortunately, this task is not as simple as it might first appear because Ks is not a fundamental
soil property. Its magnitude also depends on many other factors, including the following:
- The width of the loaded area:
A wide mat will settlement more than a narrow one with the same q because
it mobilizes the soil to a greater depth, therefore, each has a different (ks )
-
The shape of the loaded area :
The stresses below long narrow loaded areas are different from those below square
loaded areas therefore, ks will differ
-
The depth of loaded area below the ground surface :
At greater depths, the change in stress in the soil due to q is a smaller percentage of
the initial stress, so the settlement is also smaller and ks is greater
-
The position on the mat :
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To model the soil accurately, ks need to be larger near the edges of the mat and smaller
near the center
-
Time :
Much of the settlement of mats on deep compressible soils will be due to
consolidation and thus may occur over a period of several years. Therefore, it may be
necessary to consider both short-term and long-term cases.
Actually, there is no single ks value, even if we could define these factors because the q-δ
relationship is nonlinear and because neither method accounts for interaction between the
springs.
Various techniques of measuring or computing ks were tried :
- Some rely on plate load test to measure ks in situ:
However, the test results must be adjusted to compensate for the differences in width, shape,
and depth of the plate and the mat. Plate load tests include dubious assumption that the soils
within the shallowzone of influence below the plate are comparable to those in the much
deeper zone below the mat. Therefore, plate load test generally do not provide good estimates
of ks for mat foundation design
- Others have derived relationships between ks and the soils modulus of elasticity,
E(Vesic&Saxena, 1970). Although these relationships provide some insight, they too are limited
-
Another method consists of computing the average mat settlement using the
techniques of settlement and expressing the results in the form of ks using the following
equation
 Structural design :
The structural design of mat foundations must satisfy:
- strength requirements
- serviceability requirements
This requires two separate analyses, as follows:
- Evaluate the strength requirements result from the load combinations.The mat
must have :
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a. sufficient thickness, T
b. required reinforcement to resist these loads
c.
As with spread footings, T should be large enough that no shear reinforcement is needed.
-
Evaluating mat deformations using the unfactored loads. These deformations are
the result of :
a. concentrated loading at the column locations, possible non-uniformities
in the mat
b. variations in the soil stiffness
In effect, these deformations are the equivalent of differential settlement.
If they are excessive, then the mat must be made stiffer by increasing its thickness.
Winkler method is used (i.e., when all “springs” have the same Ks) and the geometry of the
problem can be represented in two-dimensions.
Mat foundation is designed with the aid of a computer using the finite element method (FEM).
This method divides the mat into hundreds or perhaps thousands of elements. Each element
has certain defined dimensions, a specified stiffness and strength (which may be defined in
terms of concrete and steel properties) and is connected to the adjacent elements in a specified
way.
The mat elements are connected to the ground through a series of “springs,” which are
defined using the coefficient of subgrade reaction. Typically, one spring is located at each
corner of each element.
The loads on the mat include the externally applied column loads, applied line loads, applied
area loads, and the weight of the mat itself. These loads press the mat downward, and this
downward movement is resisted by the soil “springs.”
If the results of the analysis are not acceptable, the design is modified accordingly and
re-analysed.
This type of finite element analysis does not consider the stiffness of the superstructure. In
other words, it assumes the superstructure is perfectly flexible and offers no resistance to
deformations in the mat which is conservative.
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3.PRE-DIMENSIONING OF RAFT FOUNDATION:
There is two methods used for the pre-dimensioning of a mat foundation :
- Simplified method :
10 cm depth is assumed for each building’s floor than H(mat) = 230cm
- Detailed method :
Calculate the required depth of an isolated footing laying under a column
having an average Pu from the existing once and check for shear is required for
columns having a higher Pu.
Since building above the ground floor lays on a part of the raft , this part will be subjected to a
bigger load than the one supporting only the basement which is why a raft subjected to
different load will be designed.
The calculation of the footing depth is shown in the tables below, and is based on the following
formulas :
Footing dimension :
𝑠𝑒𝑟𝑣𝑖𝑐𝑒 𝑙𝑜𝑎𝑑
𝑎𝑙𝑙𝑜𝑤𝑎𝑏𝑙𝑒 𝑏𝑒𝑎𝑟𝑖𝑛𝑔 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦
𝐴=𝐵= �
Part subjected to bigger load :
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Part subjected to smaller load:
4.SAFE MODEL FOR MAT FOUNDATION:
Loads applied on the raft are imported from the “ETABS” model, and the raft is modeled as a
slab lying on an elastic soil, represented by springs,using the software “SAFE”.
The elastic modulus of subgrade reaction Ks of the soil based on geotechnical informationis
equal to 20000 KN/m^3.
The primary thickness introduced is the one found by the pre-dimensioning method, it is to be
increased each time one of the following criteria is unchecked :
- Two-way shear
- One-way shear
- Soil pressure
4.1 PUNCHING SHEAR VERIFICATION
According to ACI 318-08 chapter 11 and PCA notes:
The unbalanced moment at a slab-column connection must be transferred from the slab to the
column by eccentricity of shear and by flexure.
Studies of moment transfer between slabs and square columns found that 0.6 Mu is transferred
by flexure across the perimeter of the critical section b o and 0.4 Mu by eccentricity of shear
about the centroid of the critical section.
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For a rectangular column, the portion of moment transferred by flexure γ f Mu increases as the
dimension of the column that is parallel to the applied moment increases. The fraction of
unbalanced moment transferred by flexure is:
𝛾𝑓 =
1
2
𝑏
1+ � 1
3 𝑏
2
And the fraction of unbalanced moment transferred by eccentricity of shear is:
𝛾𝑣 = 1 − 𝛾𝑓
Where b1 and b2 are the dimensions of the perimeter of the critical section, with b1 parallel to
the direction of analysis.
The following picture illustrates the critical section dimensions b1 and b2:
Figure 94 - Critical section dimensions b1 and b2
The unbalanced moment transferred by eccentricity of shear is γ v M u where M u is the
unbalanced moment at the centroid of the critical section.
Assuming that shear stress resulting from moment transfer by eccentricity of shear varies
linearly about the centroid of the critical section, the factored shear stresses at the faces of the
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critical section due to the direct shear Vu and the unbalanced moment transferred by
eccentricity of shear γ v M u are:
Vu1 =
Where:
-
Vu2 =
Vu γv Mu c
+
Ac
J
Vu γv Mu c′
−
Ac
J
Ac is the area of concrete section resisting shear transfer equal to the perimeter b o
multiplied by the effective depth d.
J is the property of critical section analogous to polar moment of inertia of segments
forming area Ac.
c and c’ are distances from centroid axis of critical section to the perimeter of the
critical section in the direction of analysis.
Expressions for Ac, c, c’, J/c and J/c’ are given in the following figures.
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Figure 95 - Section properties for shear stress computations
The maximum shear stress Vu1 shall not exceed ΦVn where ΦVn shall be determined from the
following:
For slabs without shear reinforcement: ΦVn = ΦVc
WhereVc is the smallest between:
-
2
Vc = 0.17 �1 + � λ√f′cbo d
Vc = 0.083 �
β
αs d
bo
+ 2� λ√f′cbo d
Vc = 0.33 λ√f′cbo d
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Where:
-
β is the ratio of long side to short side of the column, concentrated load or reaction
area.
αs = 40 for interior columns,
=30 for edge columns,
=20 for corner columns.
If Vu <ΦVn, thickness is verified.
If not, shear reinforcement should be provided with the following section:
Av =
The spacing limit are shown in the figure below:
Vs s
fy d
Figure 96 - Interior Column
The following picture gives the spacing limits for edge and corner columns.
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Figure 97 - Edge or corner column
To check the punching shear on the raft, an envelope of the ultimate combinations is added, to
take into consideration the worst case.
The shear force that causes the punching shear is:
F=Vu –q u *A
Where :
-
q u = ultimate soil pressure
A = area of the critical punching perimeter
Vu = ultimate shear force acting on the raft
The following figure shows the ration of ΦVc/Vu, if the ratio is less than one then no need
for punching shear reinforcement .
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Figure 98: punching shear ratio for the raft
Since the punching shear ratio is higher than one for both parts, than the raft’s thickness must
be increased.
The following figure shows ΦVc/Vu for a thickness of 250cm for the high loaded part and
100cm for the low loaded part.
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Since the punching shear ratio is less than one for both parts, than the raft’s thickness is
verified.
4.2ONE-WAY SHEAR VERIFICATION
The raft should be designed so that no shear reinforcement should be required.
According to ACI section 11.9.5.the concrete shear strength is given by :
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ϕVc = 0.17 λ ϕ �f′cbw d
Where:
-
λ is the lightweight concrete factor taken as 1
b w is the width taken as 1 m for slabs and rafts
Φ is equal to 0.85
The concrete strength: f’c = 30 MPa
The following figure shows the shear stress in the raft as given by “SAFE”:
Figure 99:Shear stress in the raft
 For the thicker part :
The effective depth d = 235 cmϕVc = 194T
The software gives us the applied shear force Vu at a distance equal to d from the column.
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Max Vu =150T < 194 T
 For the thinner part :
The effective depth d = 85cmϕVc = 70T
Max Vu =10T < 70 T
Since the one-way shear is verified for both parts, the raft’s thickness will remain the same.
4.3 SOIL PRESSURE VERIFICATION
Based on geotechnical investigations, the soil allowable bearing pressure is :
10 Kg/cm2 = 100 T/m2
The following picture shows the soil pressure for the envelope of service combinations:
Figure 100: Map representing the soil bearing pressure
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Since the values obtained by the software are smaller than the allowable ones, this criteria is
checked.
4.3 RAFT ‘S REINFORCEMENT
The following figures show the ultimate bending moment maps for both direction :
Figure 101: Ultimate bending moment M11
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Figure 102: Ultimate bending moment M22
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Choosing a mesh of the minimum required reinforcement was the first step to reinforce the
raft.
According to ACI-08:
-
For the thickness of 250 cm:
As min = ρ min x b x h = 45 cm2/ml  2 Layers of T25@15cm
-
For the thickness of 100 cm:
As min = ρ min x b x h = 18 cm2/ml  1 Layer of T20@15cm
Reinforcing with respect to this mesh required additional bottom steel reinforcement in a
major part of the raft, which is why an additional layer of mesh is distributed on the bottom of
the raft.
Reinforcement:
-
For the thickness of 250 cm:
Top reinforcement: 2 Layers of T25@15cm
Bottom reinforcement: 3 Layers of T25@15cm
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-
For the thickness of 100 cm:
Top reinforcement: 1 Layers of T20 @15cm
Bottom reinforcement: 2 Layers of T20@15cm
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REFERENCES
Our references are divided into three categories
A. Codes:




UBC 1997
ASCE STANDARD 7-05
ACI 318-99 & ACI 318-02
AISC 1989
B. Books:



GUIDE FOR DESIGN OF POST-TENSIONED BUILDINGS ( PTI DC20.9-11)
POST-TENSIONED CONCRETE FLOORS DESIGN HANDBOOK (SECOND
EDITION)
FREYSSINET PRESTRESSING ( THE SYSTEM OF THE INVENTOR OF
PRESTRESSED CONCRETE )
C. Websites:

www.google.com
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