Abstract

1
Abstract

• This graduation project aims to analyze and
  design a building, Northern mountain,
  Nablus-Palestine, consists of a garage, offices
  and residential apartments. The structure is 14
  stories and the area of each floor is 330.11 .
  The total area of the project is 4621.54

2
The project chapters

• In GP1, the following have been done
• Chapter 1 Introduction, which describes the
  structure location, loads, materials, codes and
  standards and the basic structural system of the
  structure.
• Chapter 2 preliminary analysis and design for
  slab systems-one way ribbed slab system.
• Chapter 3 preliminary analysis and design for
  beams using 2D analysis on SAP.
• Chapter 4 design of columns and use live load
  reduction factor.

3
The project chapters

• Chapter 4 redesign of columns
• chapter5 Wind load analysis.
• Chapter 6 Earthquake analysis.
• Chapter 7 3D modeling for the project.
• Chapter 8 Stairs design.
• Chapter 9 Tie beams and ramp design.
• Chapter 10 Design of beams based on SAP2000
  results.
• Chapter 11 Design of slab- Two way solid slab.
• Chapter 12 Analysis and design of footings.

4
Chapter four column
SAP AS for columns this section in it
5

• This table for number of bars

col As(mm2) bar diameter Area of bar of bars
1 12780 32 803.84 16
2 12549 32 803.84 16
3 9670 32 803.84 12
4 57919 32 803.84 12
5 6899 32 803.84 10
6 15555 32 803.84 12
7 24497 32 803.84 12
8 9119 32 803.84 10
9 4225 32 803.84 6
10 4225 32 803.84 6
11 4225 32 803.84 6
6

• Section in column 11

7
Chapter Five Analysis for Shear walls Lateral
loads

• The design wind loads for buildings and other
  structures
• , shall be determined using one of the following
  procedures
• 1- Simplified Procedure for low rise
  building(hlt18m).
• 2- Analytical Procedure.
• 3- Wind Tunnel Procedure.

8
Design procedures

• Basic wind load and many constants shall be
  determined the wind load is obtained from the
  following equation
• P qzGCp
• Where
• - p is the wind load(N/mm2)
• - qz is the wind pressure
• - G is the gust effect factor
• - Cp is exposure coefficient
• qz 0.613KzKzt Kd I (N/m2)
• Where
• - qz is the velocity pressure
• - Kd is the wind directionality
  factor.
• - Kz is the velocity pressure
  exposure coefficient.
• - Kzt is the topographic factor.
• - V is the wind velocity(m/s)
  
• - I is the importance factor

9
from the ACI code the constants are calculated
and found to be as follow

• I 1.0 (for normal buildings).
• Kz 1.33 (box system or shear walls) from table
  6.1 in ACI318-08,
• KZt 1 for the Nablus topography.
• Kd0.85 in Nablus city.
• V 35 mph 15.65 m/s ( in Nablus city)
• G0.85 from ACI tables( in Nablus city)
• 0.8 for wind ward
• Cp 0.50 for lee ward
• 0.70 for side ward
• 0.70 for roof ward

10
from the ACI code the constants are calculated
and found to be as follow

• qz0.6131.1310.851
• 144.2
• P144.20.850.898 N/m2
• check for Pmin
• Pmin500/1.3Cp500/1.30.8 gt P
• so, use P 384.6 N/m2

11
some information about the building

• Fcall (0.3-0.4) Fc 105 Kg/cm2
• Story height 3.3m
• 14 stories
• Residential building
• Fc for tension 0.10Fc for compression 10
  Kg/cm2
• WD8.46 KN/m2
• WL 3KN/m2
• Load combination 0.9D 1.6 L

12
for the lateral loads in XZ-plain
13
check the overturning in the X-direction

• Slab weight 8.4633014 39085.2 KN
• Walls weight (0.32546.228.1) 9736.65 KN
• Walls weight (0.22546.210) 2310 KN
• Wind load on the building 0.38546.210.5
  186.76 KN
• W 0.951131.8 46018.665 KN
• H (186.76/46.5)1.6 8.1 KN/m
• M -(0.58.146.2)(246.2/3) 46018.7(35.1/2)
  
• 801875.9 KN.m
• the building weight is sufficient to resist the
  wind-load in X-direction.( the same in
  Y-direction and it is okay)

14
Check the torsional effect in ( XY-plan)

• Center of mass
• X17 m
• Y 5.314 m
• Center of rigidity
• X 17.11m
• Y 8.98 m

15
the difference between the two centers in the
X-direction is small, but in Y-direction is
large, so a sample calculation will be made on
the shear wall number 2 in the Y-direction.

• e 8.97-5.31 3.66 m
• V 0.38510.546.2 186.76 KN
• MT Ve
• MT 3.66186.76 683.55 KN.m
• J 4355 KN.m2
• F (115.36/2.31) 49.9 50 KN/ 0.5 Kg/cm2 lt 105
  Kg/cm2
• F ( 68.76/2.31) 29.77 KN/ 0.29 Kg/ lt 10.5
  Kg/
• the wind-load effect is very small so it can be
  neglected

16
Chapter Six Analysis for seismic force using
UBC97 code

• Seismic Forces- Methods of analysis
• Equivalent static method
• Dynamic analysis
• Response spectrum analysis
• Time history analysis

17
selection of analysis method

• Static method The static lateral force procedure
  of Section 1630 may be used for the following
  structures
•  a. All structures, regular or irregular, in
  Seismic Zone 1 and in Occupancy Categories 4 and
  5 in Seismic Zone 2.
• b. Regular structures under 73m in height with
  lateral force resistance provided by systems
  listed in Table 7-E.
• c. Irregular structures not more than five
  stories or 20m in height.

18
selection of analysis method

• Dynamic analysis The dynamic lateral-force
  procedure of Section 1631 shall be used for all
  other structures, including the following  
• a. Structures 240 feet (73 152 mm) or more in
  height.
• b. Structures having a stiffness, weight or
  geometric vertical irregularity of Type 1, 2 or
  3,
• c. Structures over five stories or 65 feet (19
  812 mm) in height in Seismic Zones 3 and 4 not
  having the same structural system throughout
  their height
• d. Structures, regular or irregular, located on
  Soil Profile Type SF, that have a period greater
  than 0.7 second. The analysis shall include the
  effects of the soils at the site.

19
Equivalent lateral force method (Static Method)-
UBC 97

• The total design base shear in a given direction
  shall be determined from the following formula
• UBC97
  30-4
• The total design base shear need not exceed the
  following
• UBC97 30-5
• The total design base shear shall not be less
  than the following
• UBC97 30-6

20
Equivalent lateral force method (Static Method)-
UBC 97

• Where
• Z seismic zone factor, Table 7-A
• I importance factor,
• R numerical coefficient representative of the
  inherent over strength and global ductility
  capacity of lateral- force- resisting systems,
  Table 7-E
• Ca acceleration seismic coefficient, Table 7-C
• Cv velocity seismic coefficient, Table 7-D
• W the total dead load and applicable portions
  of other loads listed below
• 1. In storage and warehouse occupancies, a
  minimum of 25 of the floor live load shall be
  applicable
• 2. Design snow loads of 1.5kN/m2 or less need not
  be included. Where design snow loads exceed 1.5
  kN/m2, the design snow load shall be

21
Soil profile type

• Soil Profile Types SA, SB, SC, SD and SE are
  defined in Table 7-B and Soil Profile Type SF
  is defined as soils requiring site-specific
  evaluation.

22
Structural period

• Method A
• The period, T is given by
• hn height of structure in meters
• Ct factor given by
• Ct0.0853 for steel moment resisting frames
• Ct 0.0731 for reinforced concrete moment
  resisting frames and eccentrically braced frames
• Ct0.0488 for all other buildings
• - T is the basic natural period of a simple one
  degree of freedom system which is the time
  required to complete one whole cycle during
  dynamic loading

23
Structural period

• Method B
• The fundamental period T may be calculated using
  the structural properties and deformational
  characteristics of the resisting elements.
• Where
• Mi the mass of the building at the level i
• the deflection of the level i, calculated
  using the applied lateral forces
• Fi the lateral force applied on the level i

24
Site and the building information

• Number of stories 14.
• Story height 3.3 m.
• Materials concrete cylinder compressive
  strength at 28 days, fc 28 ??????
• -Steel yield strength, ????420 ??????
• Soil rock, Sb type in accordance with UBC 97
  provisions.
• All columns are square with side length equals
  to 650 mm.
• All beams are 400 mm section width and 500mm
  total thickness.
• The slab is two way solid slab of 200 mm
  thickness.
• External shear walls thickness is 300 mm, and
  the internal are 200 mm wide
• Shear walls weight 25 KN/m.

25
Site and the building information

• The live load is 3 kN/m2 office and residential
  building.
• The superimposed dead load is 3.36 kN/m2.
• The perimeter wall weight is 21 kN/m.
• Importance factor (I) 1.
• Zone factor (??)0.2, 2B UBC 97
• ????0.20, ????0.2.
• Sizes of all columns in upper floors are kept
  the same.
• The floor diaphragms are assumed to be rigid.
• Lateral-force-resisting system is Duel
  systems(4.1.c)
• Location of building Nablus city .
• Earthquake load as..UBC97

26
Seismic Load

• According to Uniform Building Code (UBC97), the
  seismic coefficients Ca0.20, Cv0.20.

27
The modified plane of the buildingto reduce the
eccentricity
28
weight of the building

• Solid slab own weight slab thickness x unit
  weight of concrete
• ????0.20255 ????/??2
• ??????????29951495 ????/??2
• - Beams weight length of beams x cross section
  area x unit weight of concrete
• beams 400500 ,Area13.42 ??2
• ????????????13.4225335.5
• Notice that the beam depth below slab is used
  which is 0.50-0.20 0.30 m.
• - Columns weight length of columns x cross
  section area x unit weight of concrete columns
  650X650 at all floors ,Area0.4225 ??2,
  ??0.014876 ??4
• ????????????????110.42253.325383.41 ???? (
  columns at edge of walls are neglected )
• -Shear wall weight length of wall x cross
  section x unit weight of concrete
• Wshear walls1295 KN
• - Parameter wall weight length of wall unit
  weight(KN/m)
• Wwall1031.5 KN
• - Superimposed dead load on slab area of slab x
  superimposed dead load/m2
• ??????????????????????????2993.361006.05 ????
• - Total weight of one story 5546.46 ????
• - Total weight of building 5521.461477650.44
  ????

29
The structure period
?Ix27.9075 m4
30
The structure period

• ?Iy6.1763 m4

31
Base Shear in Y-direction

• R 6.5 from UBC97 Table16-N (Duel system 4.1.c).

- The base shear, V, need not exceed
- The base shear shall not be less than
??0.11???????? ??0.110.201??0.022?? gt
0.0217?? So, total base shear will be ??0.022
7756.044 1700.6 KN
32
Base Shear in Y-direction

• since T gt 0.7 sec, there is Ftop.

Ftop need not exceed 0.25V, 0.25 V 425.15
KN Ftop lt 0.25V, so Ft 168.45 KN and the
distribution of the rest of the base shear on the
stories is according to the following equation
Where x is the number of the story. Table 7.1
shows a summary of distribution of forces for the
stories
33
Table 7.1 Distribution of forces to stories
34
center of rigidity
35
calculation of eccentricity due to seismic
forces

• center of mass(CM)
• X17.43 m
• y5.25 m
•  
• center of rigidity (CR)
• X18.187 m
• Y6.8177 m
•  
• emax eo e2
• emin eo - e2
• eo the real eccentricity is the distance between
  the CR( center of stiffness and the CM and center
  of mass). 0e
• eo accidental eccentricity
• e2 accidental eccentricity
• e2 0.05L
• - L the floor dimension perpendicular to the
  direction of the seismic action.
• eo18.187 -17.43 0.757 m
• e2 0.0535.1 1.755 m
• emax1.755 0.757 2.512 m
• emin -0.998 m

36
horizontal distribution of story shear to the
walls

• The distribution of the total seismic load, to
  walls will be in proportion to their rigidities..
  The flexural resistance of walls with respect to
  their weak axes may be neglected in lateral load
  analysis. Table 7.4 summarizes the force
  distribution to the columns and walls in y
  direction.
• the table 7.3 shows (Km, Kp, dm and dp) for the
  shear walls.

• Where
• - ???? The shear force of the structural walls m
  , parallel to the direction of the seismic
  action.
• - ???? The shear force of the structural walls
  p, perpendicular to the direction of the seismic
  action.
• - dm,dp The distance from the centers of gravity
  of the structural walls m and p to the considered
  center of rigidity.
• - ???? The torsion rigidity of the considered
  level.
• ???? The translation rigidity of the considered
  wall.
• K The total translation rigidity of the
  considered level

37
the parallel walls are W1, W2, W3, W4, W5, W6,
W7the perpendicular walls are W1, W2, W3, W4,
W8, W9, W10, W11
38
Table 7.3 Distribution of forces to shear wall
at each story (V is in KN
39
  Table 7.5 the total force on each shear wall
at each story (KN) ( VVmVp)
40
Calculation of the base shear in the other
direction(X)

• The same procedures as in Y-direction

The base shear, V in the y-direction will be R
6.5 from UBC97 Table16-N (Duel system 4.1.c
The base shear, V, need not exceed
- The base shear shall not be less than
??0.11???????? ??0.110.201??0.022?? lt
0.046 ??
So, total base shear will be ??0.046 7756.044
3576.65 KN
41
Calculation of the base shear in the other
direction(X)

• since T lt 0.7 sec, there is no Ftop.
• The distribution of the base shear on the stories
  is according to the following equation

Where x is the number of the story. Table 7.5
shows a summary of distribution of forces for the
stories.
42
calculation do eccentricity due to seismic
forcescenter of mass(CM)

• center of mass(CM)
• X17.43 m
• y5.25 m
•  
• center of rigidity (CR)
• X18.187 m
• Y6.8177 m
•  
• emax eo e2
• emin eo - e2
• eo the real eccentricity is the distance between
  the CR( center of stiffness and the CM and center
  of mass). 0e
• eo accidental eccentricity
• e2 accidental eccentricity
• e2 0.05L
• - L the floor dimension perpendicular to the
  direction of the seismic action.
• eo6.8177-5.25 1.4617 m
• e2 0.0510.5 0.525 m
• emax1.41170.5251.9867 m
• emin 0.8867 m

43
horizontal distribution of story shear to walls

• Table7.7( Km(KN), Kp(KN), dm(m), dp(m)) for the
  shear walls

44
Table 7.8 Distribution of forces to shear wall
at each story (V is in KN)
45
Table 7.9 the total force on each shear wall at
each story(VVmVp) (KN)
46
Tables used in the analysis
47
Tables used in the analysis
48
Tables used in the analysis
49
Chapter 7- 3D modeling

• Structural System used
• In this project, 3D modeling will be made for the
  structure, and two-way slab with beams will be
  used as the structural system, since the project
  will be analyzed and designed to resist the
  earthquakes.
• No analysis will be made in this chapter, only
  the modeling.
• By using 2D modeling on SAP2000 and applying many
  thicknesses to obtain an economical one, it was
  found the following
• Slab thickness 20 cm-
• Beams are 4050 cm-
• and for the modifiers for SAP
• - Beam0.35
• - Slab0.25
• - Column0.70

50
Structural materials

• 1) Concrete
• The main material used in the structural is the
  concrete
• The cylindrical compressive strength after 28
  days,
• fc 24 MPa (B300) for slabs and beams.
• Modulus of elasticity ,E 2.3
• fc 28 MPa (B350) for columns.
• Modulus of elasticity, E2.49
• Unit weight is 25 KN/
• 2) Steel
• The Steel yield strength used for steel
  reinforcement is 420 MPa
• The modulus of elasticity is 2.04 MPa

51
3D-Modeling on SAP 2000
52
Sap2000 results

• Equilibrium check
• The final check in the model is Check local
  equilibrium by comparing the results of moments
  for beams and slabs from structural
  three-dimensional model using SAP 2000 and hand
  calculation.
• the loads from SAP as shown on the following
  table
• Table 7.1 the loads resulted from SAP

Support reactions - Dead (own weight)
49265.037 KN -Superimposed dead 14084.717 KN -
Live 12575.64 - Parameter wall weight 14371.363
KN - Dead superimposed dead parameter wall
77721.117 KN (77650.44 KN by hand
calculations)) The mass and support reactions are
equal to that in hand calculations
53
Sap2000 results

• Periods
• Table 7.2 Structure periods

First mode- movement in y- direction The period
in Y-direction in hand calculations is
1.415seconds (UBC 97) The period in X-direction
in hand calculation 0.665 seconds (UBC 97)
54
Equivalent static method

• in this project, the base shear will be
  calculated by the equivalent static method using
  two values for the period
• 1- the period produced from the hand calculation(
  Tnx 0.665 sec and Tny 1.415 sec)
• 2- and the period produced by SAP 2000 (Tnx1.533
  sec and Tny1.83 sec)

1) The first period- the manual one
Base shear resulted from SAP 2000 using the UBC
97 code.(using the periods that are calculated
manually)
55
2- Base shear according to the period produced by
SAP 2000- Tnx1.533- Tny 1.83
Base shear resulted from SAP 2000 using the UBC
97 code.(using the periods that are calculated by
SAP)
56
Comments on the results

• In order to have the same results of the hand
  calculation and the equivalent static method, the
  assumptions that have been made to the hand
  calculation must meets the equivalent static
  method conditions.
• In hand calculation, we assumed that the building
  is moving 100 in the X-direction to calculate
  the Tnx. and the same assumption for y-direction.
• but SAP 200 results show that the X-mode has 0.36
  model participating mass ratio and Y-mode has
  0.66 model participating mass ratio. this is
  opposite to the assumption, so the equivalent
  static method give results with a high percentage
  of error

57
Response Spectrum Method (UBC 97 code).
58
the main direction of the seismic force is
X-direction( U1)
the main direction of the seismic force is
Y-direction (U2)
the main direction of the seismic force is
Z-direction (U3)
59
Structure periods

• The modes periods and their participating mass
  ratios( for 150 modes)

60
Response Spectrum Method (UBC 97 code).
the response spectrum results for the seismic
force in the three directions are as shown in the
figure below
61
Analysis and design of shear walls

Lw/5
S
3h

• 500 mm

62
Analysis and design of shear walls

• find the concrete strength to resist shear for
  all shear walls
• Find the Vu from SAP, on each shear wall

63
Analysis and design of shear walls

• The shear walls have been organized in groups
• Group number one include shear walls number( 1
  2 ).
• Group number two include shear walls number( 3
  4 ).
• Group number three include shear walls number (
  5,6,7,8 and 9).

0.1 B 0.17.7 0.77
m d min of 4b 40.3 1.2 m
The boundary reinforcement
The web reinforcement
64
Dividing the shear walls in groups
65
Analysis and design of shear walls
66
Analysis and design of shear walls
67
window in the following shear walls(1, 2, 3, 4,
10 and 11), special reinforcement shall be
provided as follow
68
Windows in shear walls
in the earlier mentioned shear walls, use the
following reinforcement Av 0.0015bwS1 Avh0.00
25bwS2 S1(d/3), 500 mm S2(d/3), 500 mm
69
Tie beam design

• the main idea to use tie beam in structure is
  decrease deferential settlement of the footing
  due to loads transfer from other elements of
  structure

Beam ID Section (mm) Effective depth (mm)
Tie beam 500300 440
70
(

Beam ID Section(mm) d(mm) Length (m)
(1-2) 500300 440 5.7

• SAP 2000 results
• Bending moment diagram

Maximum positive moment 21.3 KN.m
Use 3
71
Shear force diagram
Vu/

• Design tie beam for shear
• Ultimate shear load 15 KN
• Ultimate Shear load at critical section 13.9 KN
• Design shear force Vu/ , 0.75
• 13.9/0.75 18.5 KN
• Vc ?/6 bwd (1.0) (300)(440)
  116.4 KN Vn 18.5 KN
• Use 1 cm minimum.

72
Ramp 3D analysis and Design

• Ramp is a flat supporting surface tilted at an
  angle, with one end higher than the other

elements Section(mm)
Beam 500250
slab One way solid (h 25cm)
73
Modification factors for structural elements

• Beam modification factor

Slab modification factor
74
In this figure show the bending moment diagram
for ramp slab.


Design ramp slab Slab thickness 25 cm (one way
solid slab) Design slab (A-B) for flexure
Use area of steel As (min)
660 mm2
Slab ID Longitudinal reinforcement Longitudinal reinforcement
Slab ID Negative moment Positive moment
Slab (A-B) 6 /1 m 6 /1 m
Slab (B-C) 6 /1 m 6 /1 m
Slab (C-D) 6 /1 m 6 /1 m
Slab (D-E) 6 /1 m 6 /1 m
Slab (E-F) 6 /1 m 6 /1 m
Beam Design Bending moment in the beam duo to
load transfer from ramp slab is very small, so
that beam design must be depends on the minimum
design case
Section of beam (5025 cm)
As (min) 0.0033500200 300 mm2
Use 3
75
Design beams
From SAP we design beams and take As and this
picture can show that
76
of beam   As of beam of bars of beam   As of beam of bars
b1 top 578 4 b15 top 798 6
  bottom 578 4   bottom 578 4
b2 top 717 4 b16 top 574 4
  bottom 578 4   bottom 283 2
b3 top 631 5 b17 top 578 4
  bottom 536 4   bottom 530 4
b4 top 661 5 b18 top 578 4
  bottom 578 4   bottom 415 3
b5 top 724 5 b19 top 491 4
  bottom 578 4   bottom 243 2
b6 top 707 5 b20 top 578 4
  bottom 578 4   bottom 278 2
b7 top 261 2 b21 top 174 2
  bottom 130 1   bottom 296 2
b8 top 668 5 b22 top 199 2
  bottom 578 4   bottom 336 3
b9 top 830 6 b23 top 959 7
  bottom 578 2   bottom 606 2
b1o top 662 5 b24 top 611 4
  bottom 576 2   bottom 552 2
b11 top 619 4 b25 top 578 4
  bottom 535 4   bottom 511 4
b12 top 864 6 b26 top 428 3
  bottom 578 4   bottom 212 2
b13 top 710 5 b27 top 745 5
  bottom 522 3   bottom 578 4
b14 top 589 4        
  bottom 412 3        
And this picture can show beam number
From this table we take As top and bottom in each
beam
But Asmin .0033400450 594mm 4f14 so we
need to change every beam which is have number
of bars less than minimum to minimum.
77
And this is picture for As/s for shear
We can see from this picture all values about
.333 so, if we use f10 stirrups then we need 5f10
/m 1f10/20cm.
And this is beams in detail
78
And this cross section in beam 11
79
Design slabs
From SAP we design slabs and take As and this
pictures can show that for x direction
80
bottom face
81
but ASmin 360mm2/m and this more than 215mm on
the bottom face so use Asmin and this is the
details in the x direction
82
(No Transcript)
83
And this for y direction