United Arab Emirates University
Collage of Engineering
Graduation Project Unit
Bakheet Ahmad Al

Mansoori 200101695
Khalfan Ahmad Al

Mansoori 200101697
Saeed Nasser Al

Ahbabi 200101684
Saif Rashid Al

Mansoori 200101629
First Semester
2006

2007
Contents
•
Introduction
•
Background theory
•
Structural systems
•
Design calculation
•
Cost calculation
•
Results and discussion
•
Conclusion and recommendations
Introduction
•
Al
Jazira
Mohammed
Bin
Zayed
Stadium
is
a
multi

use
stadium
in
Al
Jazira
Club
located
in
Abu
Dhabi
.
•
The
stadium’s
original
capacity
was
15
,
000
seats
but
it
is
currently
going
through
an
expansion
stage
which
will
increase
the
stadiums
capacity
to
an
all
seated
40
,
000
ultra

modern
air
conditioned
sporting
arena
.
•
The
expansion
program
includes
two
residential
towers
to
be
built
beside
the
stadium
and
three
phases
the
first
two
phases
are
carried
out
using
reinforced
concrete
and
pre

cast
concrete
structures
whereas
the
third
phase
is
designed
using
structural
steel
systems
.
Introduction
•
This
project
focuses
analysis
and
design
of
alternative
structural
systems
to
the
future
phase
three
of
Al
Jazira
stadium
.
•
The
actual
structural
system
to
be
used
in
construction
of
phase
three
consists
of
an
overhang
steel
roof
truss
supported
by
a
framed
column,
which
is
in
turn
supported
at
the
top
of
phase
two
.
GP
1
overview
•
We design two alternative
structural systems are proposed in
the current project.
•
First system
, is a simple truss
system which consists of overhang
steel roof trusses supported by
trussed columns.
•
Second system
is a modification of
the former one by extending the
height of the trussed column above
the roof level to allow for
installation of cables that are
attached from the other end to the
mid

points of the roof trusses.
Graduation Project (II)
•
calculating
the
structural
loads
acting
on
system
(
2
),
modeling
and
analysis
of
both
system
(
1
)
and
system
(
2
)
using
SAP
2000
software
.
•
Designing
and
detailing
the
analyzed
systems
according
to
the
LRFD
version
of
the
AISC
code
.
•
Developed
Excel
spreadsheets
to
facilitate
the
design
of
the
high
numbers
of
structural
elements
and
connections
included
in
the
proposed
systems
.
Structural
details
were
presented
using
AutoCAD
.
Background Theory
•
Steel
•
Types of loads
•
Load combination
•
SAP
2000
Steel
1.
The aim
of using steel is to reduce the dead weight.
2.
Composed of
Iron
, Small amount of
Carbon
(<
2
%) and other chemical
components (such as
Manganese, Copper, Nickel, Silicon, and Aluminum
)
3.
The aim
of the chemical components is to improve
strength
,
toughness
,
hardness
,
ductility
, and
corrosion resistance
.
4.
Increasing the
Carbon content
leads to an increase in
strength
and
hardness
, but decreases ductility and toughness.
5.
Strength and Ductility
are the most
important characteristics
of structural
steel.
6.
Ductility
is the ability of the member to undergo large deformations without
fracture.
7.
The
economical production
method began around the
middle of the
19
th
century
, by heating iron in contact with charcoal.
8.
A more
advanced process
was introduced by
Sir Henry Bessemer
of England
in
1855
(the
Bessemer process
).
Steel
•
Advantages of construction steel:

High strength

to

weight ratio.

Ductility (large deformation before failure).

Flexibility in structural forms.

Long lifetime if properly maintained.

Recyclable material, environmental friendly.
•
Disadvantage of construction steel:

Buckling susceptibility.

Fireproofing cost, to prevent transmission of heat and the associated large
reduction in strength.

High maintenance cost, e.g. paint coating.

Higher construction and maintenance cost in some parts of the world.
Steel Specification
•
The type of steel that was used in this project is (High strength Low

alloy) A
572
•
Grade
60
, Fy =
60
ksi , Fu =
75
ksi.
•
It gives a high resistance with acceptable cross section dimensions, especially
for large span structures.
•
Circular cross sections were used in this project.
Types of load
•
Dead load
•
Live load
•
Wind load
•
Live roof load
Load combination
•
1.4
D
•
1.2
D +
1.6
L +
0.5
(L
r
or S or R)
•
1.2
D +
1.6
(L
r
or S or R) + (
0.5
L or
0.8
W)
•
1.2
D +
1.3
W +
0.5
L +
0.5
(L
r
or S or R)
•
1.2
D
±
1.0
E +
0.5
L +
0.2
S
•
0.9
D
±
(
1.3
W or
1.0
E)
Where:
D = Dead load
L = Live load
W = Wind load
E = Earthquake load
L
r
= Roof Live Load
SAP
2000
•
SAP
2000
is a
finite element

based
software that represents the most sophisticated
and user

friendly release of the SAP series of computer programs.
•
Powerful
graphical user interface
unmatched in terms of ease

of

use and
productivity.
•
This program features powerful and completely
integrated modules
for design of
both
steel and reinforced
concrete structures.
•
The program provides an
interactive environment
in which the user can study the
stress conditions, make appropriate changes, such as member size revisions, and
update the design without re

analyzing the structure.
Structural systems
Structural system(
1
):
it is consist of trusses only, and divided into three
parts, column, shed and rakar as present in below figure:
Structural system(
2
):
it is consists of trusses and cable also have a
same divination of system(
1
), However, it should
be noted that the column height was increased
to allow for the installation of the cable element
connected to the middle of the overhang as
shown in following figure.
Modeling, analysis and design procedure for
both system
•
The geometry of the proposed systems was generated in
Auto Cad.
•
the coordinates corresponding to each nodal point were
identified.
•
Such information along with joints are members
numbers, tables of cross section dimensions, applied
loads and load combinations are entered into the
SAP
2000
program.
•
The cable properties should be take it from cables
factory otherwise, is not available in SAP
2000
program.
Design calculations
Tension members design
Design Requirements for tensions members:

Pu
Pn
Strength can be determined based on
3
potential failure modes
in our project:


Yielding of the Gross Section.

Fracture of the Net Section.

Stiffness for Tension Members.
Where:
Ø
is the resistance reduction factor:

=
0.90
for yielding failure.
=
0.75
for fracture failure.
P
n
is the nominal strength of the tension member.
P
u
is the factored tensile force.
Tension members design
Strength can be determined based on
3
potential failure
modes in our project:


Yielding of the Gross Section.

Fracture of the Net Section.

Stiffness for Tension Members.
Yielding of the Gross Section:

Tension members design
t
P
n
=
0.9
F
y
A
g
Where:
F
y
is the yield stress of steel used,
A
g
is the gross area of the tension member cross

section
(A
g
=
π
* (D
out
–
D
in
)
2
/
4
).
Where:
F
u
is the ultimate tensile stress of steel.
A
e
is the effective net area at the critical section.
Tension members design
Fracture of the Net Section
:

ØP
n
=
0.75
F
u
A
e
A
e
= A
n
U
Where:
A
n
is the net area at the critical section
=
A
g
–
[
(D
out
–
D
in
) * t
Gusset Plate
],
U is a reduction factor due to the shear lag effect when not all
the x

sectional area is directly connected to the joint
(U =
1
for our project).
Excel Spreadsheet For Design of Tension
members
Compression members design
Pu
Pn
Design Requirements:

Where:
Ø
is the resistance reduction factor:
=
0.85
.
P
n
is the nominal strength of the tension member.
P
u
is the factored tensile force.
Where:
F
cr
critical buckling stress.
A
g
gross area of the member .
g
cr
n
A
F
P
85
.
0
Compression members design
First of all we have to calculate the slenderness coefficient:

E
F
r
KL
y
c
/
Where:
λ
c
= slenderness coefficient
Fy
= yield stress (ksi)
E
= modulus of elasticity (ksi)
K
= effective buckling length factor
L
= laterally unbraced length of member (in)
r
= governing radius of gyration about the axis of buckling (in)
Compression members design
Compare to the value of
1.5
, If is greater than
1.5
then the critical elastic
buckling stress is given by
:

y
c
cr
F
F
.
877
.
0
2
And if is less than
1.5
then the critical inelastic buckling stress is given
by:

y
cr
F
F
c
.
)
658
.
0
(
2
Compression members design
Local buckling
can be avoided only if the
width

to

thickness ratio
(b/t) of
each element in the cross section of the column does not exceed a
specific value called
λ
c
.
The following figure summarizes the two cases of elastic and inelastic
overall buckling:
Figure : Elastic and inelastic buckling of columns
Excel Spreadsheet For Design of Tension
members
Zero members
•
Members doesn’t carry any load.
•
check the stiffness.
•
L/r
<
200
Where:

L : the length of the member.

R : the radius of the member.
Zero members
Welded connections design
•
Use SMAW process, and E
80
electrodes.
•
F
exx
=
80
Ksi
•
t
(G.PL)
=
0.5
in
•
F
y
=
60
Ksi
•
F
u
=
75
Ksi
Welded connections design
Strength of Longitudinal weld:
•
Weld fracture:
ΦR
n
/in =
0.75
(
0.6
F
exx
)(
0.707
S
w
)
•
Shear rapture of the member:
ΦR
n
/in =
0.75
* (
0.6
F
u
) *t
member
•
Shear rapture of gusset plate:
Φ
R
n
/in =
0.75
(
0.6
F
u
)* t
(G.PL)
Welded connections design
Bolted connection design
•
In the construction site either welding or bolting could be used;
however, bolting is more recommended for quality control, ease and
safety reasons.
Bolted connection design:
There is two typical types of high strength
bolts are commonly used, A
325
and A
490
.
Fv
(ksi)
Bolt type
48
A
325
N
60
A
325
X
60
A
490
N
75
A
490
X
Bolted connection design
•
High strength bolts can be used to form two
main types of connections,
Shear failuer in Bolts
and Bearing Failure at Bolt Holes
depending on
the load transfer mechanism of each connection
type.
•
In
Shear failuer in Bolts
:
R
n
≥ P
u
R
n
=
0.75
F
v
A
gv
N
b
N
s
Bearing Failure at Bolt Holes
:
At the external bolts:

Since Le =
2
in >
1.5
d
t = smaller of:
1

t =
0.5
in (gusset plate)
2

t = (
2
*
3
/
8
) in (splice plate)
ØP
n
/bolts =
0.75
(
2.4
)(d)(t)Fu
At the internal bolts:

Since S =
3
d
so,
ØP
n
/bolts =
0.75
(
2.4
)(d)(t)Fu
The factored bearing resistance of the connection:
ØPn
= [(number of external bolts * ØPn/bolts
for the external
) + (number of
internal bolts * ØPn/bolts
for the internal
)]
Beam design
•
The design of beam which stand on
trusses joint of system and carried the
concrete slab.
•
The all beams are a same in the design for
both systems.
22.08
ft
w
•
The beam is rotated by
31
dgree around the
globule axis.
•
It is effected by previews types of load.
•
The beam should be design to resist the
moment and shear.
•
The moment which govern the design should be
the maximum moment of moment due by
different types of load combination.
•
The beam should be design to resist the
moment and shear.
•
The moment which govern the design
should be the maximum moment of
moment due by different types of load
combination.
•
Find:
Max. M
x
Max. M
y
Beam section selecting:
•
Assume: L
b
=L=
22.08
ft
•
steel type used is Fy=
50
ksi
•
C
b
=
1.14
•
From LRFD manual

beam sections charts
We find the beam section according Max.M
u
and beam length.
•
Then find
corresponding design
moment
p
nx
n
n
b
M
M
Cb
of
smallest
x
M
x
M
*
•
Design moment in y
direction
•
Moment check:
Fy
Sy
Fy
Zy
of
smallest
y
M
n
*
*
5
.
1
*
9
.
0
*
*
9
.
0
1
.
0
.
.
y
M
My
Max
x
M
Mx
Max
n
n
•
Shear check:
max
max
2
*
5
.
0
3
.
1
2
.
1
V
Vn
L
W
V
combo
load
W
W
W
LX
WX
DX
Base connection
The design for connection between
the trusses system and concrete slab:
Base Plate Dimensions:

a =
2
Ø
b
b = L
e
L
min
= D
out
+
2
a +
2
b
Where:

D
out
: outer diameter of member.
Ø
b
: diameter of bolt.
Base connection
y
c
c
y
F
f
b
a
t
b
a
b
a
L
f
t
L
F
Mu
Mn
'
'
2
27
.
2
)
(
2
)
(
)
(
*
*
02
.
1
)
2
/
(
*
*
*
9
.
0

Check bearing stress on the concrete slab below the base plate:

'
2
02
.
1
c
Z
f
L
R

For determine the base plate thickness:

Where:

t
= base plate thickness
R
z
= compression force.
F
y
= area of bolt.
= diameter of bolt
.
'
c
f
Base connection
Anchor bolts:


The bolt is subjected to tension and shear:

For shear:

V
V
V
V
b
V
F
f
F
F
nA
V
f
*
75
.
0
Where:

V
= shear force.
A
b
= area of bolt.
n
= number of bolt.
Base connection
Anchor bolts:


The bolt is subjected to tension and shear:

For tension:

t
t
t
t
V
V
t
V
t
V
b
u
t
F
f
F
F
f
f
F
f
F
f
nA
T
f
)
90
(
75
.
0
90
90
9
.
1
117
)
9
.
1
117
(
75
.
0
9
.
1
117
90
9
.
1
117
Where:

T
u
= tension force.
A
b
= area of bolt.
n
= number of bolt.
systems cost
•
Systems weight calculation:
1

calculated the volume for each
member.
V=(D
out
–
D
in
)
*
L
member
2

calculated the weight for each
member. W= V*Specific weight of steel
•
Cost Calculated:
the cost of steel =
6500
Dhs/ton
the total cost = total weight(ton)*cost / ton
system(
1
) cost =
6.30371
ton*
6500
Dhs/ton
=
40.974
Dhs
system(
2
) cost =
6.41431
ton*
6500
Dhs =
41.693
Dhs.
Results and Discussion
•
System
1
is better than system
2
because its economically.
•
The trusses must be welded in the factory.
•
The transportation process must be under the observation of the Consultant to
avoid any damages to the parts.
conclusion & Recommendation
•
Analysis and preliminary design proposed systems by
used SAP
2000
•
Tension and compression members manually design.
•
Welded and bolted connection design.
•
Zero members design.
•
Beam design.
•
Base connection design.
•
Cost calculation for both system.
•
At the end of GP
2
we recommend our collage to give the
student different courses of structural soft war .
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