Transcript Power_Point_ - An-Najah National University

An-najah National University
Faculty of Engineering
Civil Engineering Department
Graduation project
Design of the Library of Palestine
Technical University - Tulkarm
Prepared By:
Omar Jamal Sha’ban
Abdallah Marwan Al-Tibi
Mohammad Mufeed Shishan
Chapters
o Chapter One : Introduction
o Chapter Two : Preliminary Design
o Chapter Three : Three Dimensional
Structural Analysis and Design
General

This is a graduation project that introduces analysis and
design of reinforced concrete structure. This structure is the
building for the library of Palestine Technical University (PTU) in
Tulkarem city – Palestine.

The building will be analyzed and designed using the primary
principles of structural analysis and design by using one
dimensional structural analysis and also using the most modern
analysis of structures which is the three dimensional structural
analysis and design.
Figure 1.1: Architectural 3D model of the building
Note: The red lines in the previous figure are related to expansion joints

The building is divided into three parts which are separated by
expansion joints. Figure 1.1 shows a three dimensional image of the
building.

Part A in the building is composed of four stories and is used as
a library. The first story is an underground (basement) story whose area
is about 743.0 m2 and its height is 3.90 m. The other three stories is an
over ground stories, and the area of each one is about 1546 m2 and its
height is 4.42 m. There is a steel dome in the last story at the roof.

Part B is used as offices, video conference and arcade. And it is
composed of an over ground story whose area is about 472.0 m2 and its
height is 4.42 m.

Part C has not been designed.
Codes and Standards
The structures are designed using practice code and
specifications that control the design process and variables.
The following codes and standards are used in this study:
 ACI 318-08 : American Concrete Institute provisions for
 UBC-97
 IBC-2009
 ASTM
reinforced concrete structural design.
: Uniform Building Code provisions for seismic load
parameters determination.
: International Building Code, which is used here for
live loads.
: For material specifications.
Loads and Load combinations
Loads
1) Gravity loads:
Live load:
It comes from the people, machines and any movable objects in the
buildings. The amount of live load depends on the type of the structure. In this
project the live load is:
7.0 kN/m2 (for Part A)
3.0 kN/m2 (for Part B)
Dead load:
it is consisting of own weight of the structure and any permanent
components. The super imposed dead load is 3.5 kN/m2.
2) Lateral loads:
Seismic loads:
The structure is located in Tulkarm area which is classified as
zone 2A according to Palestine seismic zones.
The UBC97 code seismic parameters are as follows:
 The seismic zone factor, Z= 0.15 .
 The soil is very dense soil and soft rock, so the soil type is Sc .
 The importance factor, I = 1.0 .
 The ductility factor, R = 5.5 .
 The seismic coefficient, Ca = 0.18 .
 The seismic coefficient, Cv = 0.25 .
Load combinations
The ACI318-08 load combinations are used and they are
summarized as follows:




U1 = 1.4D
U2 = 1.2D + 1.6L + 1.6H
U3 = 1.2D + 1.0E + 1.0L
U4 = 0.9D + 1.0E + 1.6H
Where:
D : is dead load
L : is live load
H : is weight and pressure load of soil
E : is earthquake load
Materials
Structural materials

Concrete:
Concrete strength for all concrete parts is B350 ( f’c = 280 Kg\cm2 , 28 MPa )
Modulus of elasticity equals 2.5*105 Kg\cm2 , 2.5*104 MPa
Unit weight is 25 kN\m3

Steel:
Modulus of elasticity equals 2.04*106 Kg\cm2 , 2.04*105 MPa
For steel reinforcement, is 4200 Kg\cm2 , 420 MPa
Non-structural materials
They are mainly, blocks, plasters, tiles, filling, mortar and masonry.
Building Structural Systems
 The structural system in the building parts ( A and B) shall be as
follows:
In part A, the four floors are designed as ribbed slab with
main hidden beams carried by columns.
In part B, the floor is designed as one way ribbed slab in X
and in Y direction with hidden beams.
 Basement walls are used around the ground floor, and all the
exterior walls are composed of concrete, masonry and blocks.
 The masonry stones are fixed to the reinforced concrete walls
using concrete mortar.
Design of Slabs
Ribbed slab analysis and design (Part A)
 The thickness of the slab:
hmin = Ln/33 = 890-50/33 = 25.45 cm Use h=30 cm
(since the loads are heavy).
Where Ln : The maximum spacing between the columns.
The direct design method will be used to check
the preliminary dimensions of the slab and beams.
So, one frame will be taken in calculations to
represent the whole slabs as approximation.
The ribbed slab can be divided to frames in
each direction . Here , calculations are made for
Frame A shown in figure ( 2-1)
Frame bending moment also shown in figure (2-2)
Figure 2.1 Plan of the slab in the
basement floor and frame A
is hatched
Figure 2.2 Moments in Frame A
Figure 2.3 Percentages and values of the moments on the strips of span # 2
For column strip: As for negative moment = 4Ø20 2Ø20 / layer
For column strip: As for positive moment = 2Ø16
For beam in the Frame D= 45 cm, B= 100 cm: As for negative moment = 6Ø20
For beam in the Frame D= 45 cm, B= 100 cm: As for positive moment = 5Ø20
For middle strip: As for negative moment = 2Ø14
For middle strip: As for positive moment = 2Ø14
Figure 2.4 Percentages and values of the moments on the strips of span # 1
For column strip: As for negative moment = 4Ø18 2Ø18 / layer & 2Ø14
For column strip: As for positive moment = 2Ø20
For beam in the Frame: As for negative moment = 5Ø20 & 5Ø20
For beam in the Frame: As for positive moment = 5Ø20
For middle strip: As for negative moment = 2Ø14
For middle strip: As for positive moment = 2Ø14
Ribbed slab analysis and design (Part B)
 The thickness of the slab:
hmin = l/21 = 800/21 = 38 cm  Use h=40 cm
Design of Rib 5 in part B
Rib load Diagram (in kN/m)
Rib Bending Moment Diagram (in kN.m)
 Max. neg. moment= 16.01 kN.m/rib
𝜌 = 0.00235
𝜌min = 0.00333
As = 2Ø12
 Max. +ve moment= 13.64 kN.m/rib
𝜌 = 0.000537
𝜌min = 0.00333
As = 2Ø12
 Design of column # 15
The ultimate load on the column is found using tributary area and
number of stories, and the design load can be calculated using the following
equation:
 Tributary area=59.085 m2
 Number of stories= 5
 Wu=24 kN/m2
 Pu=59.08x5x24 = 7089.6 kN
 Pn req = 7089.6 / 0.65 = 10907.0 kN
 Assume 𝜌 = 0.01
 Pd= Ф Pn = Ф *λ {0.85* fc(Ag-As) + As* fy}
= 1090.70x1000=0.8 {0.85x 280x(-) + 0.01x Agx4200}
 Ag= 1090.70*1000 / 222.1 = 4910.85 cm2
 Square column  b=h =70. cm  use (80x80) the area of column to account
for additional moments and lateral forces
Chapter Three
Three Dimensional Structural Analysis and Design
* General
This chapter is a three dimensional analysis and design, which
introduces the final and practical structural analysis and design of the
structural elements with the practical structural drawings that are ready
for construction.
Structural analysis comprises of set of physical and mathematical laws
required to study and predict the behaviour of structures under a given
set of actions. The structural analysis of the model is aimed to
determine the external reactions at the supports and the internal forces
like bending moments, shear forces, and normal forces for the different
members. Theses internal member forces are used to design the cross
section of three elements.
Property/Stiffness Modification Factors
 Basic calculations
Two way slab system:
Part A
 Membrane f11 modifier = (A1 / A3)= 0.1165/0.301 = 0.387
 Membrane f22 modifier = (A1 / A3)= 0.1165/0.301 = 0.387
 Membrane f12 modifier = (A1 / A3)= 0.1165/0.301 = 0.387
 Bending m11 modifier = 0.25*( I1 / I3) = 0.25*(2.2108*10-3/5.087*10-3) =






0.108
Bending m22 modifier = 0.25*( I1 / I3) = 0.25*(2.2108*10-3/5.087*10-3) =
0.108
Bending m12 modifier = (C ribbed / C solid) = (6.2968*10-4/11.739*10-3) =
0.0536
Shear v13 modifier = (A1 / A3)= 0.1165/0.301 = 0.387
Shear v23 modifier = (A1 / A3)= 0.1165/0.301 = 0.387
Mass m modifier = (M two way rib / M solid) = (9.08/11.25) = 0.80
Weight w modifier = (9.81*M two way rib / 9.81*M solid)= (9.08/11.25) = 0.80
Part B
One way slab system (y-direction):
 Membrane f11modifier = (A2/A3) = 0.044/0.22 = 0.2
 Membrane f22 modifier = (A1/A3) = 0.092/0.22 = 0.418
 Membrane f12 modifier = (A2/A3) = 0.044/0.22 = 0.2
 Bending m11 modifier = 0.25*( I2/ I3) = 0.25*(2.346710-5 / 2.93310-3) =






0.002
Bending m22 modifier = 0.25*( I1/ I3) = 0.25*(2.346710-5 / 2.93310-3) =
0.139
Bending m12 modifier = 0.25*( I2/ I3) = 0.25*(2.346710-5 / 2.93310-3) =
0.002
Shear v13 modifier = (A2/A3) = 0.044/0.22 = 0.2
Shear v23 modifier = (A1/A3) = 0.092/0.22 = 0.4 18
Mass m modifier = (M 1 way rib / M solid) = (0.7/1)= 0.7
Weight w modifier = (9.81*M 1 way rib/ 9.81*M solid) = (0.7/1) = 0.7
One way slab system (x-direction):
 Membrane f11 modifier = (A1/A3) = 0.092/0.22 = 0.418
 Membrane f22 modifier= (A2/A3) = 0.044/0.22 = 0.2
 Membrane f12 modifier = (A2/A3) = 0.044/0.22 = 0.2
 Bending m11 modifier = 0.25*( I1/ I3) = 0.25*(2.346710-5 / 2.93310-3) =






0.139
Bending m22 modifier = 0.25*( I2/ I3) = 0.25*(2.346710-5 / 2.93310-3) =
0.002
Bending m12 modifier = 0.25*( I2/ I3) = 0.25*(2.346710-5 / 2.93310-3) =
0.002
Shear v13 modifier = (A1/A3) = 0.092/0.22 = 0.418
Shear v23 modifier = (A2/A3) = 0.044/0.22 = 0.2
Mass m modifier = (M 1 way rib / M solid) = (0.7/1) = 0.7
Weight w modifier = (9.81*M 1 way rib / 9.81*M solid) = (0.7/1)= 0.7
Structural Model Verification of Part A
Check Equilibrium
The total building dead load = 77047.1 kN
The total building live load = 33627.72 kN
From SAP2000: Total dead load = 74204.27 kN
Total live load = 33109.0 kN
 Error % in dead load= 3.68 %< 5%
 Error % in live load = 1.54 %< 5%
ok.
ok.
Check Compatibility
Structural Model Verification of Part B
 Check Equilibrium
 The total building dead load = 7761.8 kN
 The total building live load = 1421.16 kN
 From SAP2000: Total dead load = 7402.87 kN

Total live load = 1394.19 kN
 Error % in dead load = 4.6 %< 5%
 Error % in live load = 1.9 %< 5%
ok.
ok.
Check Compatibility
Design of slabs
Design of slabs of Part A
Basement floor
Rib 1
For negative moment
use 2 Ф 22
For positive moment
use 2 Ф 14
Rib 2
For negative moment use 2 Ф 20
For positive moment
use 2 Ф 14
*Analysis and Design of Beams:
Design of columns
Analysis and Design of Footings
 footings which used in this project can be classified into the
following types :1) Isolated footing: they have rectangular, square, or circular
shape. This type of footing is used for small loads, and/or large
soil allowable bearing capacity.
2) Wall footing: it is a continuous footing along the length of the
wall.
LEVEL ONE
LEVEL TWO
Design Of Shear Walls:
 For Part A:
 Shear wall (SW 1):
 Vertical reinforcement:
Pu = 1531.8 kN, Mu = 666.1 kN
ρmin = 0.0012 , h = 11.80 m , b = 0.3 m.
As min = 0.0012 X 1180 X 30 = 42.48 cm2
As min for each face = 42.48/2 = 21.24 cm2
From SAP2000:
Use 1 Ф16 / 20 cm
 Horizontal reinforcement:
ρmin = 0.002 , h = 1 m , b = 0.2 m.
As min = 0.002 X 100 X 20 = 4 cm2
As min for each face = 4/2 = 2 cm2
From SAP2000:
Use 1 Ф18 / 25 cm
As a result, for all shear walls:
Vertical reinforcement:
Use 1 Ф16 / 20cm.
Horizontal reinforcement:
Use 1 Ф18 / 25 cm
Design of Stairs
Dimensions:
 Thickness of landing= Ln /20 = 195/20 = 9.75 cm. Thus 15 cm
thickness is suitable.
 Thickness of flight= Ln /24 = 285/24 = 12 cm. Thus 15 cm
thickness is suitable.
 Loads:
 SID Load= 3.5 kN/m2 , LL=5 kN/m2
Design of flight:
Mu = 17 kN.m As = 3.1 cm2 / m  use 1 Ф 14 / 20 cm
Design of landing:
Mu = 8 kN.m As = 2.7 cm2 / m  use 1 Ф 14 / 25 cm
Reinforcement Of Stairs
Thanks for listening
An-najah National University
Faculty of Engineering
Civil Engineering Department
Graduation project
Design of the Library of Palestine
Technical University - Tulkarm
Prepared By:
Omar Jamal Sha’ban
Abdallah Marwan Al-Tibi
Mohammad Mufeed Shishan