A r - Near East University

Transcription

DETERMINATION OF THE REACTIONS OF A FIXED ENDED
DOUBLE ARCH BY AN ANALYTICAL METHOD
A Master Thesis
m
Civil Engineering Department
Near East University
By
Mohamed Atwan
December, 1997
/ f1111,11W~Jt1w1111111
NEU
Approval of the Graduate School of Natural and Applied Sciences
~.y-----,
P~
Dr. Omit HASSAN
Director
I certify that this thesis satisfies all the requirements as a thesis for the degree of Master
-
of Science.
Chairman of the Department
We certify that we have read this thesis and that in our opinion it is fully adequate, in
scope and quality, as a thesis for the degree of Master of Science in Civil Engineering.
Asst.
OKAY
~ad
Supervisor
Examining Committee in Charge:
Asst.
Prof. Dr. Fuad
OKAY:~···ifl/~
Assoc. Prof. Dr. Huseyin GOK<;EKUS.... ..
··
....
"JJ"fAA·}J_·······················
. ..
Prof. Dr. Burhanettin ALTAN (Committee Chairman).-r"l/·?/
···············.....
III
DETERMINATION OF THE REACTIONS OF A FIXED ENDED
DOUBLE ARCH BY AN ANALYTICAL METHOD
ATWAN, Mohamed
Faculty of Engineering
Department of Civil Engineering, M.S. Thesis
Supervisor: Asst ._ Prof. Dr. Fuad Okay
7 4 pages, December 1997
ABSTRACT
The support reactions at the fixed ends of a space structure consisting of
four, equally spaced, quarter circular arches are investigated; the structure is
loaded by a single force perpendicular to the base of the structure.
18 equations for the 18 unknowns are obtained by Castigliano's
theorem. A model problem is numerically solved with the help of the
software Excel by Microsoft.
Key words : Castigliano' s theorem, strain energy, matrix equations,
superposition, reciprocal theorem.
IV
ANKESTRE MESNETLi
<;iFT KEMER PROBLEMiNiN
ANALiTiK METODLA
cozuvu
ATWAN, Mohamed
Muhendislik Fakultesi
Insaat Muhendisligi Bolumii, Yuksek Lisans Tezi
Tez Yoneticisi: Yard. D09. Dr. Fuad OKAY
74 Sayfa, Arahk 1997
OZET
Bu cahsmada dort ceyrek cembersel kemerin esit olarak yerlestirilmesi
ile elde edilen bir uzay tasiyici sistemin ankastre mesnetlerindeki
tepkiler
incelenmistir, Yapi kemerlerin birine, yapmm zeminine dik olarak etkiyen
tekil bir kuvvetle yukludur.
Problemin 18 bilinmiyeni icin gereken denklemler Castigliano teoremi
ile elde edilmistir, Omek bir problem Microsoft firmasinca tiretilmis olan
Excel isimli bir program yardimi ile ntimerik olarak cozulrnustur.
Anahtar Kelimeleri: Castigliano Teoremi, Sekil Degistirme Enerjisi, Matris
Denklemleri, Karsilik Teoremi, superpozisyon.
V
ACKNOWLEDGEMENTS
The author is deeply indebted to the chairman of civil engmeenng
department Assoc. Prof. Dr. Huseyin Gokcekus who is the founder and the
nurturer of the department at Near East University. Dr. Gokcekus has
provided excellent motivation and continuous support for this study. His
support did not diminish even while he was busy organizing an international
conference on water problems in the Mediterranean countries.
I am also fully thanking my supervisor Assb. Prof. Dr. Fuad Okay, for
building and creating my deep gratitude, and for his support, remarkable
suggestions and comments through this study, as well.
Finally, the encouragement and financial support provided by my
family is highly appreciated.
VI
TABLE
OF CONTENTS
Page
ABSTRACT
----------------------------
OZET
_ --------------------------
--------------------------------------------------------------
ACKNOWLEDGEMENTS
LIST OF FIGURES
----------------------------------------------
---------------------------------------------------------
III
IV
V
VII
CHAPTER
I : INTRODUCTION
CHAPTER
II: DEFINITION OF THE PROBLEM -----------------------
CHAPTER
III : CONSTRUCTING THE INTEGRATIONS
CHAPTER
IV: OBTAINING THE EQUATIONS AND SOLUTION -- 57
CHAPTER
V : CONCLUSION
RE FE REN CES
----------------------------------------
1
6
----------- 17
----------------------------------------
----------------------------------------------------------------
73
74
VII
LIST OF FIGURES
Page
Figure 1: Two arches coinciding each other perpendicularly
4
Figure 2: The reactions and the internal forces in leg A ------------------
8
Figure 3: The reactions and the internal forces in leg B -----------------
11
Figure 4: The reactions and the internal forces in leg C -----------------
12
Figure 5: The reactions and the internal forces in leg D
before the load P -----------------------------------------------
14
Figure 6: The reactions and the internal forces in leg D
after the load P
------------------------------------------------
15
I. INTRODUCTION
In this study, the forces and the moments of a main frame of a dome
where an external concentrated load is applied normal to it, are analyzed by
Castigliano' s theorem. The main frame is chosen as two arches coinciding each
other perpendicularly.
Domes are frequently used as roofs over large circular floor areas for
assembly halls, gymnasiums, field houses, mosques; and other buildings. They
are very strong structures, since they are shells, and they are constructed of selfsupporting reinforced concrete shells.
A dome exerts outward thrusts continuously around its perimeter. These
are resisted by a tension ring. The dome and the tension ring are usually
supported on columns spaced around the perimeter and braced to provide lateral
stability for the structure. Also, a dome is an integral or self-contained unit. Its
perimeter may be supported directly by the foundation, which carries the
vertical and horizontal thrusts of the ribs. The simplest form of dome would be
2
generated by revolving a solid arch about a vertical axis through its center. The
spaces between ribs may be spanned by purlins to support the roof deck [l].
Any plane-curved bar or rib, properly supported at its ends and so loaded
that it acts primarily in direct compression, may be called an arch. It is assumed
that the plane of curvature of the rib is also a plane of symmetry for each cross
section and that external forces applied to the arch act only in this plane. Under
such conditions, deformation will also take place in the plane of symmetry, and
the problem of analysis becomes a two-dimensional one. If the cross sections of
the rib are unsymmetrical with respect to the plane of curvature or if loads are
applied normal to this plane, there will be twisting of the loaded rib, and under
such conditions it is not properly considered to be an arch [2].
A ribbed arch is one of a series of arches providing structural support for
the roof deck which spans the areas between arches and is continuous from one
end of the roofed area to the other.
In contrast to rigid frames, arches are so proportioned that the stresses
produced by the loads are primarily compressive and the shears and flexural
stresses are relatively small. Flexural stresses must be considered, however, in
the design of the arch section, splices, and supports.
A reinforced concrete roof arch, of constant dimensions in the
longitudinal direction of a building , may be connected to heavy footings at the
ends of the transverse span by means of dowels so that the arch itself maybe
3
considered to have two fixed supports .In the analysis of such an arch, only a
typical slice of a unit width in the longitudinal direction needs to be considered
[3].
In this study the problem of two arches coinciding each other
perpendicularly, which are shown in figure 1, is solved by the help of
Castigliano's theorem. Since the problem is a space (three dimensional)
problem, each fixed support brings six unknown reactions.
There are four fixed supports in our structure, thus we have twenty four
unknown reactions correspondingly. Any six of them can be obtained from the
equations of equilibrium conditions. We choose these six unknowns to be the
six reactions in one leg. The remaining 18 unknowns are obtained by the
deformation conditions at the other three legs.
The displacement
o J at a coordinate j
of a structural that behaves linear
or nonlinear due to the effect of external applied loads and of temperature
variation, shrinkage, or other environmental causes can be obtained from
0. J
=
oU
*
( 1.1)
oF1
where U* is the complementary energy expressed in terms of the forces
F. [ 4].
J
4
ri~ure 1. I WO orcnes coincioir1~ eocn otner rer~enoirnlorlt
5
In the special
case when the structure
is linear
elastic
and the
deformations are caused by external forces only, the complementary energy U*
is equal to the strain energy U by using the conservation of energy, and equation
(1.1) becomes
<5. = oU
1
oF.J
(1.2)
This equation known as Castigliano's theorem. It must be remembered that its
use is limited to the calculation of displacement in linear elastic structures
caused by applied loads [ 4]. In our problem the loads are kept such small that
the deformation will be linear elastic.
6
II. DEFINITION OF THE PROBLEM
As it is shown in figure 1, all the forces and the moments are taken into
consideration in analyzing the dome. It is known that, a dome is an integral or
self-contained unit. Thus, the analysis will be focused on two arches coinciding
each other perpendicularly, which they are represented by four legs, and each
leg has a fixed support at its end. In the analysis of such an arch, only a typical
slice of a unit width in the longitudinal direction needs to be considered.
A. Equilibrium of the structure
From the equations of equilibrium
conditions,
six equations will be
obtained at any leg. In our problem leg D is chosen as follows:
A
+B
X
D
X
+C
+D
=0
X
X
X
-A
- B
- C
X
X
X
(2.1)
7
=
Ay + By + Cy + Dy
0
(2.2)
Dy=-Ay-By-Cy
(2.3)
Having the moment at joint D will result in'.
:Z:A,(=0
MDx + M Ax+ MBx +MCx -Az r+Cz r = 0
M
Dx
= -M
Ax
- M
Bx
- M
Cx
+A
z
r-
C r
(2.4)
z
IMy=O
M~ +MAy +MBy +MC); +Azr+Bz 2r+Czr-Pu=O
M
I>;;
=-M
Ay
-M
By
-M
Cy
-A r-2B
z
z
--cz r+Pu
(2.5)
M +M +M +M +A r-C r-A r-B '2r-C r=O
Dz
Az
Bz
Cz x
x
y
y
y
M =-M -M
-M
-Ar+Cr+Ar+2Br+Cr
Dz
Az
Bz
Cz x
x
y
y
y
(26)
·
8
/,:
/<..,/'
/?
//
fi~ure L. 1ne reactions ono tne internal forces in le~ A.
9
B. Internal Forces in Arches
Now, four legs will be analyzed separately at any section or in other words
at any angle 8, where 8 ranges from O to rc/2. M, , T and Me will be found, where
M, is the radial moment, T is moment of twisting and Me is the angular moment,
acting on an arbitrary cross-section.
From figure 2, which shows leg A, Me ,T and M, will be found as follows:
Taking the moment about z-axis at the section
M sin
r
e + T cos e + M Az
+A
x
r (1 - cos 8) = 0
(2.7)
And taking the moment about y-axis at the same section
-Mr cos 8 + Tsin 8 + M Ay - Axr sin 8 = 0
(2.8)
Similarly, having the moment about x-axis at the same section
M8+MA
X
+A rsinB-A
y
Z
r(l-cosB)=O
(2.9)
Equation (2.9), will result in:
M
e =-M
Ax
-A
y
rsinB+A r(l-cosB)
z
(2.10)
And Equation (2. 7), will result in:
Mr sine= -TcosBM
-TcosB-
r
M Az - Axr(l-
M Az - Axr(l-
= -------=-==--------
sm e
.
cos e')
cos e')
(2.11)
10
Substituting
T = -M
Mr
in equation (2.8), will result in:
cosB- M
Az
Ay
sine+ A r(I-cosB)
(2.12) ·
X
Substituting the value of T in (2.11 ), will result in:
Mr
= - M Az
sin
e+
M Ay cos
e-
AX r sin
e
(2.13)
Here, the same steps will be followed in leg B, which shown in figure 3 as
follows:
z:
M
y:
-M() + MB
x:
- M cos() + T sin () + MB + B r sin () = 0
r
. X
y
r
sinB+TcosB+MB
z
-B
y
r(l-cosB)=O
- B r sin 8 + B r(I-
y
X
Z
cos())= 0
which will result in:
M8
= MBy
-B
X
rsinB+ B r(l-cosB)
(2.14)
Z
T = - MB cos B - MB sin B - B r(l - cos B)
Z
M
r
= -MB
Z
(2.15)
y
X
sinB+ MB X cosB+ By rsinB
(2.16)
The same procedure will be done in leg C, which shown in figure 4 as
follows:
z
y
Mr cos O - Z'sin B + Mey - Cxrsine
X
M
B
+M
Cx
+ C r sin B + C r(l - cos B)
y
z
= 0
=
0
11
~i~ure
1 1ne reactions ono tne internal fore es in le~ b.
12
(z
ri~ure
t 1ne reoclion) rm~ lne internal form in le~ (.
13
which will result in:
= - MC X
Me
- C r sine - C r(l - cose)
y
Z
r(I-cosB)
(2.18)
cosB- MC Z sinB+ CX rsinB
(2.19)
T= MC sine-MC
y
M
r
= -MC y
Z
cose-c
(2.17)
X
For leg D, two cases will be analyzed, because of the load P, which is
applied normal to it at a distance u, from its edge.
When 8 ranges between O and a, as shown in figure 5, following the
I-
same steps as before, it will result in:
z
M r sin B + T cos ·B + MD z + D y r(I - cos B)
X
M cos B - T sin B + MD
r
X
=0
+ D y r sin B = 0
y
which results in:
M BI
= -M
Dy+ D xrsine+ D2r(I-cosB)
(2.20)
where M BI is the angular moment ranging from O to a.
T
M
= -MD
r
z
= -MD
cosB+ MD sine+ D r(I-cosB)
X
y
(2.21)
sinB+ MD X cosB-D y rsinB
(2.22)
I
Z
14
~i~ure ~. lne readions ano lne internal forces in le~ ~ oefore lne loao ~.
15
fi~~re 0. 1ne reoc\ion~ ono tne in\ernol !me~ in le~ ~ ol\er \ne looa ~.
16
II-
When
e
ranges between a and rc/2, which shown in figure 6,
'allowing the same steps will result in:
z
Mr sine+ TcosB+ MDz + Dyr(l- cos e')
=0
TsinB + M Dx + DYrsinB = 0
x
Mr cosB-
y
M82 + MDy -DxrsinB-D2r(l-cosB)+
P[r(l-cosB)-u]
which results in:
M82 = -MDy + DxrsinB+ D2r(l-cosB)-
where
MB2
the
is
angular
T= -MD Z cosB+ MDX sinB+Dy r(l-cosB)
M r = -MD
Z
sinB+ MD cosB-D
X
Now, the equations of
y
Me , Mr
be used in the next chapter.
rsinB
P[r(l- cosB)- u]
moment
ranging
(2.23)
from
a
to
rc/2.
(2.24)
(2.25)
and T are obtained for each leg, and they will
I/
III. CONSTRUCTING THE INTEGRATIONS
As it is explained in the first chapter, Castigliano's theorem will be used
for solving this problem, and it is known that, the deflection 5 . is found as:
J
oU
(3.1)
8 j = oF ·
}
And it is known that, the strain energy U is equal to:
U=-
1
2 i
3
I
=1
M2
-'
Eli
(3.2)
dx.
z
The effect of normal forces and shear forces are small enough
that we
can easily neglect them.
By applying equation (3 .1 ), andequation (3 .2) the deflection in each leg
will be found as:
(3.3)
where
M.
l
is the moment
corresponding moment of inertia.
on an arbitrary axis, and I.
l
is the
18
Some of the integrals which will be used in the following equations:
.
f
Sin () COS() d () = ~
f
c os O
f
s i ri P d()
2
()
2
tr I 2
f
ao
cos2
=sin()
= - c o e O
()d()
0
tr I 2
f
sin2
()
ae
0
=~+sin
2
=~-sin
2
2()
4
2()
4
-
tr
4
-
= tr
4
where
sin O = 0
sin rc/2
=
1
cos O = 1
cos rc/2 = 0
Since the distance traveled 's' is a circular arc, it is obvious that
s
=
r e' and ds = r de
By using equation (3 .3 ), the deflection in leg A will be found in terms of
one of the six reactions. And from equations (2.10), (2.12) and (2.13) we will
get:
19
oM e
oA x
oT
oA X
- 0
- r(l
- cosB)
oM
___ r_= -rsinB
oA X
Applying each term in equation (3.3) will result in:
f
M
A EI
e
e
oM
oA
e
ds
=
0
(3.4a)
X
f _I_ or
A GJ oA ds
X
1r I
=
f
2 [-MA
Z
cosB- MA sine+ A r(l- cosB)]
y
X
r(l-cosB)rdB
GJ
0
(3.4b)
f
M
_r
oM
__ r ds
A EI r oA X
e- A
=
n I 2 [ - MA sin B + MA cos
Z
y
f
0
Elr
X
r sine]
.
(-rsmB) r dB
20
M
2
= _r_[M
EI
Az 4
r
A rtr]
tr-~+
(3.4c)
x 4
2
From equations (2.10), (2.12) and (2.13) we will get
oM ()
oA y
oT
oA y
oM
----'-r_
oA
=
-rsin()
- 0
= 0
y
Me oM e
f -A
Ele
ds
oAY
1r/2[-MA
=
-A
f
x
rsinB+A
y
r2
EI ()
=-[M
Ax
f _!_ oT
A GJ oA
y
ds = 0
Mr oMr
A EI ~ds=O
r
z
(-rsinB)rdB
Elg
0
f
r(l-cosB)]
y
+A
r it
--A
Y 4
r
-]
z 2
(3.5a)
(3.5b)
(3.5c)
21
oM
---=()::....-=
oA
z
oT
oA
r(l-cosB)
=
O
z
dMr = 0
oA
z
MB oM()
J-
ds
A Eie 0A2
1r!2[-MA
= J
X
-A
O
Jr
Ax
()
f _I_ oT
(1--)-A
2
ds=O
A GJ 0A2
A
rsinB+A r(l-cosB)J
Z
r(l- cosB)rdB
EI()
r2
=-[M
EI
f
y
M
oM
EI
oAz
_r __ rds=O
r
r
31r
-+A r(--2)]
Y 2
z 4
(3.6a)
(3.6.b)
(3.6c)
22
Equations (2.13), (2.13) and (2.13) will give
oM e
oM Ax
oT
oM
-
- 1
= 0
Ax
oM r
oM Ax
= 0
Jr/ 2 [-MA.x -A rsinB+ A r(l-cose)]
y
z
- I
O
r2
EI
=-[M
o
EI(}
ff
-+A
Ax 2r
f_I._ oT
ds=O
AGJ oM Ax
M
ff
+A (1--)]
Y
z
2
~n~e
(3. 7a)
(3.7b)
oM
f-r
r ds=O
AE!r oM Ax
(3.7c)
23
oM fJ
oM
=
0
Ay
----o T
oM
fJ
.
= -
Sln
Ay
oM
___ .:.._r_
oM Ay
=
C O S
fJ
M
!JM
__f}___
f} ds
A Elf} !JM Ay
f
=0
(3.8a)
f _I__
I} T ds
A GJ !JM Ay
1r/2 [-MA
= f
z
0
r2
MA
GJ
2r
=-[--z
f
A
M
_r
+M
oM
cosfJ-M A sinfJ+A r(l-cosfJ)]
y
x
( - sin fJ )rd f}
GJ
7r
--~]
Ay 4r
A
(3.8b)
2
r ds
=. oMAy
1r / 2 [-M Az sinfJ +MA
= f
0
y
Elr
cosfJ- Axr sinfJ]
(cosfJ) rd O
24
-M
2
- _r_
A
Az + M
..!!__~
2r
Ay 4r
2 ]
- EI [
r
(3.8c)
oM B
= 0
oM A z
oT
----=
-
cosB
oM A z
oM
__
.,..:.r_
= - sin B
oM Az
M() oM ()
oM
ds= 0
()
Az
f EI
A
f I_ oT
(3.9a)
ds
A GJoMAy
;rr/2 [-MAzcosB-MA
f
=
GJ
2
GJ
M
f-r
(-cosB)rdB
lY
0
r
=-[M
sinB+Axr(l-cosB)]
M Ay
ff
ff
-+--+A
Az 4r
oM
(--1)]
x 4
2r
(3.9b)
r ds
AEir oM Az
ff
sin e' + MA cost) - A r sin BJ
I 2 [ - MA
= f
z
r2
=-[M
EI
r
x
y
0
( - sin e' )rd()
Eir
ff
M Ay
----+A
Az 4r
2r
ff
-]
x 4
(3.9c)
25
Similarly, The same steps will be followed to find the deflection in leg B.
oM [_
oB
X
oT
oB
= 0
=
o
X
oM e
dB
= -rsinB
X
f
M
oM
_r __ rds=O
B EI r ss X
(3.10a)
(3.10b)
,r I 2 [MB
=
f
0
y
- B rsinB+ B r(l- cosB)]
2
x
(-rsinB)rdB
EIB
(3.10c)
26
oM
oB
[.
r
sin
e
y
--oT
oB
=
=
-r(l
- cosB)
y
e
dM
oB
= o
y
J
B
M
oM
_r __ r ds
EI
r
ssy
st I 2 [ - MB sinB + MB cosB + B r sinB]
z
x
y
(rsinB) rdB
0
E!r
= J
r2
=-[-M
EI
Tr
M Bx
-+--+B
Bz 4
r
f _I__ oT
2
rst
-]
(3 .11 a)
Y 4
ds
BGJ oBY
,r/2
[-M BzcosB-M
J
=
y
0
r2
=-[M
GJ
J
M
r(l-cosB)]
BxsinB-B
(-r{l-cosB})rdB
GJ
Tr
M Bx
(1--)+--+B
Bz
4
2
Y
31r
4
r(--2)]
(3.llb)
oM
__f}___~s=O
B Elg oBY
(3.1 l c)
27
oM
__ r =0
oB z
8T = 0
oB z
oM
oB_e
=
r(I - cosB)
z
Applying each term in equation (3 .3) will result in:
f
M
oM
_r --LJs=O
EI oB
B
r
z
f __!_~S=
(3.12a)
0
(3.12b)
B GJ 0B2
M8;oM8
f -
ds
B Eie oBz
1r I 2 [ MB - B r sin B + B r(l - cos B)]
=
f
y
X
0
r2
=-[M
EI (}
Z
r(l- cosB)rdB
EJB
1r
(--1)-B
By 2
31r
r
-+B
x 2
z
r(--2)]
4
oM r
oM Bx = COS (3
oT = - sin (3
oM Bx
oM (3
oM Bx = 0
(3.12c)
28
f
M
oM
et ,
oM Bx
_r_
B
1r
=
=
r ds
I 2 [ -M Bz sin 6 + M Bx cos6 + B r sin 6]
f
y
( cos 6) rd 6
0
st ,
2 -M
B
r
Bz M
ff
y
EI [ 2r + Bx 4r + 2]
r
f _I__
oT
ds
c:
B GJ oMBx
ffl2
[-MB
= f
(3.13a)
cosB-MB
z
X
GJ
0
r2 M Bz
=-[--+M
GJ 2r
sinB-B
r(l-cosB)]
y
( - sin B)
ff By
-+-]
Bx 4r
2
M
oM
_f}_
e ds
B EIB oMBx
f
=0
rde
(3.13b)
(3.13c)
From equations (2.14 ), (2.15) and (2.16) we will get
oM r
= 0
oM By
oT
=
0
=
1
oM By
oM
e
oM By
M
f-r
oM
rds=O
B Elr oMBy
(3.14a)
29
f I_
oT ds=O
B GJoMBy
f-
(3'.14b)
M0 oM0
ds
B Eie oMBy
tr/2
= I
[MB -B
y
rsinO+B r(l-cosO)]
z
X
0
r2
=-[M
EI
()
(l)rd e
EIO
tr
--B
By 2r
x
tr
+B (--1)]
(3.14c)
z 2
oM
r
oM
= - sin B
Bz
oT
oM
= -Cose
Bz
oMe
oM
= 0
Bz
f
B
=
M
_r
st;
oM
r ds
oM Bz
tr I 2 [ - MB sin O + MB cos e + B r sin OJ
f
z
x
y
(-sinO)rdO
0
Elr
r2
=-[M
EI
r
tr M Bx
-----B
Bz 4r
2r
tr
-]
(3.15a)
Y 4
f I_
oT ds
B GJ IJM Bz
tr/2 [-MB2cos0-MBxsin0-B
= f
0
GJ
r(l-cosO)]
y
(""'"cosO)rdO
30
(3. l 5b)
M
f _fl
B EI()
oM
()
ds
=0
(3.15c)
oM Bz
By following the same steps, the same thing will be done for finding the
deflection in leg C. From equations (2.17), (2.18) and (2.19) we will get
oM r
oC
-
.
- r sm {)
X
oT
= - r (1
oC
-
COS {))
X
oM ()
=
oC
O
X
Applying each term in equation (3.3), will result in:
f
M oM
___!_ __I__cJs
C EI r
scX
,r/2 [-M0 cosB-MCz-sinB+C rsinBJ
y
x
(rsinB) rdB
0
Elr
= f
>
31
=
2 -M
r
Cy
EI [
2
- M Cz
r
J
_!_~s
rst
Jr
4 + Cx 4]
(3.16a)
c GJ ocx
;r/2 [MC sinB-MC2cosB-Cxr(l-cosB)]
=
y
J
0
r
[-r(l-
cosB)]rdB
GJ
2 -M.
Cy
=-[
GJ
2
M
oM
+M
J _f}_ __ o ds
c EI8 ocx
1r
(1--)+C
Cz
4
3:r
r(--2)]
x 4
(3.16b)
=0
(3.16c)
oM
t:
=
0
oCY
_!I_=
0
oCY
oM
----'8~= - r sine
!JC
y
M oM
J _r ---1:,Js = 0
C EI r !JCy
J
(3.17a)
_!_~s=O
(3.17b)
C GJ oCY
M8oM8
J---ds
C EIB oCY
:rr/2 (-MC
= f
0
X
-C
y
rsin8-C
EIB
Z
r(l-cos8)]
(-rsinB)rdB
32
r2
=-[M
EI
o
rst
+C -+C
Cx
Y 4
r
-]
(3.17c)
z 2
oM
=0
__ r
oC
z
or
oC
=0
z
oM B = -r(l - cosB)
ac2
I
Mr oMr
C EI oC ds
r
z
=O
(3.18a)
(3.18b)
MB oM8
f---ds
c E18 ac2
1r
=
I 2 [ - MC - C r sin B - C r(l - cos B)]
X
y
Z
f
0
EIB
{-r(l-cosB)}rdB
(3.18c)
33
oM r
=
0
=
0
oM Cx
!} T
oM
cx
dM B
= -
1
oM Cx
Applying each term in equation (3.3) will result in
f
M
oM
_r
r ds = 0
C Elr oMCx
f __!_ oT
C GJ oM
Cx
(3.19a)
(3. l 9b)
ds=O
MB oM8
f -
ds
C EIB oMCx
,r/2
=
f
-C
[-MC
0
r2
=-[M
EI (}
r(l-cosB)]
rsinB-C
y
x
z
(-l)rdB
EIB
Jr
-+C
Cx 2r
Jr
Y
+C (--1)]
z 2
(3.19c)
)
34
oM r
=
-cos()
=
sin ()
oM Cy
oT
oM Cy
oM ()
su Cy
= 0
f
C
M
_r
oM
r
et; oMCy
ds
Jr / 2 [ - MC cos() - M Cz sin () + Cx r sin fJ]
·
y
(-cosfJ)rd()
O
EI r.
= f
(3.20a)
f _I_ o T
ds
C GJ oMCy
=
1r/2
f
[MC
sinfJ-MC
y
0
r2
=-[M
GJ
x
( sin fJ) rd()
GJ
1r
MCz
-----]
Cy 4r
2r
M
oM
_§_
() ds
C EI() oMCy
f
r(l-cosfJ)]
cosfJ-C
z
=0
Cx
2
(3.20b)
(3.20c)
35
oM r
oM Cz
or
- - sine
= _
cos
e
oMCz
_oM__..:::..e_
oM Cz
=
o
f
M
_r
oM
r ds
C Elr oMCz
1r I 2 [ -
= f
MC cos B - MC sin B + C r sin BJ
z
y
x
(-sinB)rdB
Elr
0
2 M
r
Cy
=-[--+M
EI
2r
r
tr
--C
CZ 4r
tr
x 4
-]
(3.21a)
f _I_
t3 T ds
C GJ oMCz
tr / 2 [MC sin B - M CZ cos B - Cx r( 1 - cos B)]
y
(-cosB)rdB
GJ
0
= f
(3.21b)
(3.21c)
36
From leg D four equations are obtained as explained in the second
chapter, and they are as follows:
M·
r
= - MD Z sin () + MD X cos()
.
T = -MD
D r sin ()
y
cos()+ MD sin()+ D r(I- cos e')
Z
y
X
. y + DX rsin()+ DZ r(l-
M()I = -MD
cos e')
M ()2 = M ()I - P[ r - r cos () - u]
From figure (2.4) and figure (2.5) we will integrate M
r
and T from O to
rc/2, but for M Bl it will be integrated from O to a, and for M ()2 it will be
integrated from a to rc/2.
Substituting
the values of D ,D ,D ,MD
X
y
Z
X
,MD
y
and MD
Z
in the
previous equations, we will have:
M
r
= -(
MA
Z
- MB .,... MC - A r + C r + A r + 2rB + C r) sin e'
Z
Z
XX
y
y
y
+(-M Ax -M Bx -MCx + A2r-C2r)cosB
- ( -A
T=-(MA
Z
y
(3.22)
- B - C )r sin e'
y
y
-MB
Z
-MC
Z
-A
r+C r+A
XX
y
r+2rB
y
+C y r)cosB
+ ( - M Ax - M Bx - M Cx + A2r - C 2r) sin e'
+ ( - A - B - C )r(l - cos B)
y
M
(}J.
=-{-M
Ay
y
y
-M
-M
By
Cy
-A z r-2rB z -Cr+Ai)+(-A
z
+(P-A -B -C )r(l-ca;{})
z
z
z
(3.23)
X
-B -C )rsinB
X
X
(3.24)
37
Me2 = MBl - P[ r - r cos e - u]
(3.25)
We know from leg D, that
f
Me oMe
---ds=
D EI e
where
f
O EI e
o•
o•
a Mei oMBl
1rl2 Me2 0Me2
ds+ f
o•
a
--
EI e
(3.26)
ds
o•
means that, leg D will be differentiated with respect to the
unknown reaction •.
From equation (3.25), it is given that:
Me2
= MBl
-P[r-rcose-u]
and
Thus, we have
f
Me oMe
--ds=
D EI e o•
n/2 Mel oMBl
f
O
---dsEI e
o•
n/2 P[r-rcose-u]
f
O
EI e
oMBl
o•
ds
(3.27)
=a-b
)
38
I - The equations in leg D containing the unknown reactions of leg A.
From equation (3 .22), we will have
oM
__ r
oA
= rsinB
X
f
M
oM
_r __ r ds
EI
D
rr/2
= f
r
oA
X
[-MD
2
sinB+ MD cosB-D rsinB]
x
0
(rsinB)rdB
y
".
r2
=-[-M
EI
r
tr
M Dx
-+---D
Dz 4
2
rst
-]
(3.28a)
Y 4
From equation (3.23) it is found that:
--oT = rcosB
oA
X
f _I_ oT
ds
D GJ oAx
rr/2
= J
0
[-MD2cosB+MDxsinB+D
r(l-cosB)]
y
(rcos8) rde
GJ
(3.28b)
39
From equation (3.24), we will get
dv!
__§J._ds
,r/2 flr-rcosB-u]
f
EI
a
at
8
X
,r I 2 flr-rcosB-u](-rsmfJ)rdB
f
=
EI
a
_ -Pr
-
2
cos
a
Eie
e
r
[r--cosa-u]
2
(b)
M8 oM8
f---ds=a-b
DE!(} iJAX
r2
=-[M
EI (}
Dy
-D
rtr
--D
x 4
r
r
-+Pcosa(r--cosa-u)]
z 2
2
It is found from equation (3.22) that:
iJM
__ r =0
iJA
y
(3.28c)
40
f Mr oMr
D EI oA ds
y
r
=0
(3.29a)
oT
--=-r
oA
y
f _I__ oT
ds
D GJ oAY
;r/2[-MD
cosB+MD
Z
= f
X
y
r(l-cosB)]
GJ
0
r2
=-[M
GJ
sinB+D
Dz
-M
Dx
-D
(-r)rdB
Jr
r(--1)]
y 2
(3.28b)
From equation (3.24), it is found that:
oM191
--=0
oA
y
Me oMBI
f
ds
D Elf) oAY
=0
(3.29c)
From equation (3.22),we will get
oM
__ r =rcose
oA
z
M oM
_r __ rds
D EI r oA z
=f
=
1r I 2 [ - M Dz sin fJ + M Dx cos fJ - D r sin B]
f
y
(rcosfJ) rdB
EI
.
0
r
)
41
2 -M
r [
Dz
M
tt
EI
2
+ Dx
4-
r
D
r]
y
2
(3.30a)
Similarly, from equation (3.23) it is found that:
oT
-=rsm
oA
z
. O
= f
oT
_!'__
ds
D GJ oAz
,r/2 [-MD
= f
sinB+D r(l-cosB)]
cosB+MD
z
x
0
(r sinB)rdB
y
GJ
2 -M
= L: [
GJ
,r + D !::_]
Dz + M
Dx 4
2
(3.30b)
Y 2
From equation (3 .24), we will get
oM
_ _.:;.B_l = r cose
oA
z
n: I 2
=
f
0
,r I 2
=
f
0
r2
=-[-M
EI
r
Tr
+D -+D r(l--)]
Dy
x 2
z
4
e
,r I 2 P[r-rcosB-u]
f
a
=
EI8
(a)
d M BI
oA
ds
z
2
,rfl P[ r - r cosB - u] (
B) dt)
rcos r
a
EIB
r2 p
= -[r(lEIB
.
sma)-r{-
,r
4
a
sin2a
-(-+--)}-u(l-sma)]
2
4
.
(b)
42
M6oM
f -D
EI
e
e
~ds=a-b
z
2
= _r EI
[- M
e
+ D !_ + D r(l - ~) - P {r(l - sin a)
- z
z
4
Dy
-+--sin2a)} -u(l-sma.
2
4
1r
(a
-r{-4
(3.30c)
)} ]
From equation (3.22), we will have
oM
__ r_ =-cose
=:
M
=f-r
=. =:
D
TC
=
oM
rds
I 2 [ - M Dz sine+ M Dx cos e - D r sin e']
f
Y
(-cos&)rde
Elr
0
(3.31a)
From equation (3.23), it is found that:
or
. e
---=-sin
oMAx
f
=
D
T
ds
oT
GJ oMA
X
tr/2
I
=
[-M
0
2 M
Dz
=-[--M
r
GJ
2r
Dz
cosB+ M Dx sine+ Dyr(l- cosB)](-sin&)rde
GJ
1C
---]
Dx 4r
D
Y
2
(3.31 b)
43
_oMBl
___.c.___
=0
=:
M
f
(}
oM
Bl ds
D EI(} oM Ax
=0
(3.31c)
oM
__ r_=O
=;
M
oM
f-r
rds=O
(3.32a)
D Eir oM Ay
And, from equation (3.23) it is given that:
oT =O
oMAy
_I_
f
oT ds=O
D GJoMAy
(3.32b)
oM(}l
--=1
=;
tr/2
f
O
M(}l oM(}l
--
EI(} oM Ay
ds
44
,r; 2 [ - MDy
+ D r sin B + D r(l - cosB)]
f
X
Z
(l) rdB
0
r
EIB
2
tr
=-[-M
El
-+D
Dy 2r
o
,r/2
P[r-rcosB-u]
f
El
a
e
tr
x
+D (--1)]
(a)
z 2
oM Bl ds
oM A
y
J
Jr 2 P[ r - r cos B - u] (1) rd o
a
Elg
r2 P
;r
Elg
2
= -[(-
u ;r
.
- a)-(1-sma)-
-(-
r 2
-a)]
(b)
M
oM
g
g ds=a-b=
D Elg oMAy
f
r2
;r
= -[-M
EI
;r
-+D
Dy 2r
o
+D (--1)
x
z 2
(3.32c)
- P { ( ;r - a) - (1- sin a) - ~ ( ;r - a)}]
2
r 2
It is found from equation (3.22) that:
oM
__
-«;
f
r_
M
_r
=sine
oM
r ds
D Elr oMAz
;r I 2 [ - MD sin B + MD cos B - D
=
f
0
z
x
Elr
· y
r
r
sin BJ
( sin B)
rd B
45
r2
=-[-M
EI
;rr
M Dx
-+---D
Dz 4r
2r
r
;rr
-]
(3.33a)
Y 4
Similarly from equation (3.23),we will get
oT
cos()
---=
oM Az
f _!_
o T ds
GJ oM Az
D
[-MDzcosB+MDxsinB+D
;rr/2
f
=
GJ
0
r2
=-[-M
;rr
M Dx
-+--+D
Dz 4r
2r
GJ
r(l-cosB)]
y
(c o s e") rd()
;rr
(1--)]
Y
4
(3.33b)
And from equation (3.24), it is found that:
oMf)l
--=0
=:
Me
.~ oM e
fD Elf)
oM
Az
ds= 0
(3.33c)
II. The equations in leg D containing the unknown reactions of leg B.
oM
__ r =0
oB
X
f
Mr
D EI
r
s»,
oB
ds
X
=
0
(3.34a)
46
Similarly, from equation (3.23), it is found that:
oT =
oB
O
X
f __!_ oT
ds= 0
(3.34b)
D GJ oBx
And from equation (3 .24 ), we will get
oM
__ B_I
oB
sin B
X
MBI oMBI
tr/2
f
--
£18
0
r2
=--[M
EI ()
= -r
Dy
-D
ds
oBx
rst
--D
x 4
r
r
-+Pcosa(r--cosa-u)]
z2
(3.34c)
2
oM
__ r
oB
y
su
M
f-r
D
= -r sine'
E!
__ r ds
r
asy
)
47
M
2
Tr _ _k+D
=-r-[M
EI
Dz 4
r
2
(3.35a)
r1r]
Y 4
And from equation (3.23), we will get
--oT = -r(cosB+
oB
1)
y
f I__ o T
ds
D GJ oBY
r2
=-[M
GJ
1r
(-+1)-M
Dz 4
3
--D
Dx 2
rst
~]
y 4
(3.35b)
oMB1=0
oB
y
M
oM
I __fl __ e ds = o
(3.35c)
DEie oBY
oM
__ r =0
oB
z
I Mr
D EI
r
=.
oB
ds
z
=
0
(3.36a)
And from equation (3.23), it is found that:
oT
oB z
=O
)
48
(3.36b)
oM
__ B_l
oB
z
= r(l + cosB)
,r I 2 Mel oMBI
J -
EI8 0B2
0
,r
3r
(-+l)+D
-+D
Dy 2
x 2
r2
=-[-M
EI
ds
e
rtt
z 4
-· ]
(a)
,r I 2 P[r- rcosB-u] oMB1
J
ssz
E18
a
r2P
= -[r{(-
EI8
ds
,r
.
- a)+ (1-sma)}2
.
,r a sin2a
r{(l-sma) + (- - {- +--}
)}
4
2
4
-u{(,r -a)+(l-sina)}]
2
M8oM8
J---ds=a-b
nEie oBz
2
r
=-[-M
EI ()
;r
3r
rr:
;r
(-+l)+D
-+D
--P{r{(--a)+(l-sma)}
Dy 2
x 2
z 4
2
.
-r{(l-sma)
;r
+(-4
oM
r_
= -cos()
oMBx
f
D
M
_r
oM
=. =;
r ds
.
a sin 2a
;r
.
{-+--}
)}-u{(- -a)+ (1-sma)}}]
2
4
2
__
(b)
(3.36c)
49
(3.37a)
t3 T
. B
---=
dM
-sin
Bx
f _!_ »r
D GJoM
Bx
ds
r2 M Dz
tr Dy
---]
=-[--M
GJ 2r
Dx 4r
2
(3.37b)
Similarly, from equation (3.24) will be found
oMBl
--=O
oMBx
M
f __fl__
dM
B ds = 0
D EJB oMBx
(3.37c)
And, equation (3.22) will result in:
oM
---'r_
=0
oMBy
M
f .u:
oM
r ds=O
D E!r oMBy
(3.38a)
Similarly, equation (3.23) will give
oT
=O
oMBy
I__!_
oT
D GJ oMBy
ds=O
(3.38b)
50
From equation (3 .24 ), we will find that:
cJMBI
=
--=-=-
1
cJMBy
tr
I 2 M BI cJ M BI
f -
0
ds
EIB cJM By
r2
;rr
= -EI-[-MDY-2r
;rr
+Dx +Dz(2-
I)]
(a)
B
;rr/2
J
P[r-rcosB-u]
ds
a
EIB
r2 P
;rr
Ef 8
2
= -[(-
f
cJMBI
M
cJM By
.
- a)-(1-sma)-
cJM
_§_
e
ds
=a-
u ;rr
-(-r
2
a)]
(b)
b
D EIB cJM By
r2
= -EI-[ - M Dy
;rr
;rr
;rr
-2r + D x +Dz ( 2 - l) - p { ( 2 - a)
(}
(3.38c)
-(1- sin a)- ~(;rr - a)} J
r
2
cJM
_ __;r_ =sin(}
cJMBz
f
M
_r
oM
r
ds
D Elr oMBz
(3.39a)
)
51
Similarly, equation (3.23) will result in:
oT
--=cosB
oMBz
f
T oT ds
D GJ oMBz_
r2
=-[-M
GJ
Tr
M Dx
-+--+D
Dz 4r
2r
Tr
(1--)J
Y
4
(3.39b)
And, equation (3.24) will give
oM81
--=0
oMBz
f
M
__j}__
oM
B ds = 0
D EIB oMBz
(3.39c)
III. The equations in leg D containing the unknown reactions of leg C.
oM
__ r =-rsinB
oCX
f
M
_r
D EI r
su
__ r
scX
ds
(3.40a)
)
52
oT
-- = -rcosB
oC
X
I .I_ oT
D GJ
ocx
r2
=-[M
GJ
ds
Tr
MDx
-----D
Dz 4
2
y
r(l--)]
Tr
(3.40b)
4
oM
--=6-=--=1 -r sin
oC
B
X
MB oM6
---ds
I
D EIB oCX
r2
= --[M
EI
Dy
o
-D
sr
r
r
- -D -+ Pcosa(r- -cosax 4
z 2
2
u)]
(3.40c)
oM
__ r =0
oC
y
f
M
oM
_r __ rds=O
D EI r oC y
(3.41a)
And, equation (3.23) will give
oT
--=-r
oC
y
I .I_ oT
ds
D GJ oCY
)
53
r2
= -GJ [ M Dz -
ff
M Dx - Dyr(
2 - I)]
(3.41b)
Similarly, equation (3.24) will result in:
oMB1
-~=O
oC
y
M oM
_§__ __ eds= 0
D EI() oCY
f
(3.41c)
oM
= -rcosB
__ r
oC
z
f
M
_r
D EI r
ot«
__ r
scz
ds
r2 MD
ff
r
=-[--2-M
-+D
-]
EI
2
Dx 4
Y 2
r
(3.42a)
ar
--
oC
.
=-rs1n 8
z
f __!_ oT
ds
D GJ ac2
r2 MDz
=-G-J[-2--M
ff
Dx 4-Dy
r
2]
(3.42b)
oM
---=-8-=-=
1 r cos B
oC
z
)
Y.D.U
FEN BiLiMLERi ENSTiTUSU
TEZ SINA VI TUTANAK FORMU
Gonderen: .. .Insaat Mflhendisligi.
Gonderilen: Enstitu Mudurlugu
.. EABD Baskanhgi
Enstitu Anabilim Dahrmz YUKSEK LiSANS/8 ! I I · 1tA Programi ogrencisi
......... Mohamed ATW AN
tez cahsmalanm sonuclandmnis ve kurulan juri
onunde tezini savunmustur. Smav tutanagi asagidadir.
29/12/1997
Tarih
Enstitu Anabilim Dali Baskam
SINAV TUTANAGI:
Jurimiz 29/12/1997 Tarihinde toplanmis, yukanda
adi
ogrencinin
gecen
'Determination of the Reactions of a Fixed Ended Double Arch by an
Analytical Method'
konulu tezi incelenmis ve yapilan sozlu smav sonunda
OYBiRLidi,g l J
VJ ile asagidaki karan almistir.
~Ba~anh
D Basansiz
D
Diizeltme (Yuksek Lisans 3 ay/Doktora 6 ay)
Tez Smavi Jurisi
Unvam, Adi Soyadi
Baskan
Prof. Dr. Burhanettin ALTAN
Tez Damsrnam
Dye
Imza
Asst. Prof. Dr. Fuad OKAY
Assoc. Prof. Dr. Hiiseyin GOK<;EKU~
Dye
Dye
Uzatma alan veya basansiz olan ogrenciler icin juri raporu eklenmelidir. Juri raporunu
tum juri uyeleri ve/veya yalmzca juri baskam irnzalamahdir.
Gonderen: FBE
Gonderilen: OiDB
Yukarida adi gecen ogrenci Smav Tutanaginda belirtildigi iizere mezun olmaya
HAK KAZANMI~TIR/Kl~.aSi.rlAIAAIISf.flR.
Geregini rica ederim.
~----;
Tarih
Enstitii Muduru
54
r2
=-[-M
EI
Dy
e
+D
r
;rr
-+D r(l--)x 2
z
4
(3 .42c)
P{r(l-sina)-r(;rr
-{a+
4
2
sin2a})-u(l-sina)}]
4
t3M
---Cr_ = - cos e
t3MCx
M
f
_r
t3M
r ds
D Elr t3MCx
r2
M Dz
=-[---M
EI
2r
;rr
-+-]
Dx 4r
r
Dy
2
(3.43a)
. B
t3T
--=-sm
t3MCx
f _!_
t3T ds
D GJ t3MCx
=
2 M
D
Dz
;rr
Y
-G-J [-2-r- - M Dx -4r - -2-]
r
(3.43b)
t3 M BI
---"--=-
=
0
t3M Cx
M
t3M
_f}_
B ds= 0
D EIB t3MCx
f
(3.43c)
55
oM
-----'r-=0
oMCy
M
oM
f-r
rds=O
EI
oMC
D
r
.y
(3.44a)
And, equation (3 .23) will result in:
oT =O
=;
f
.I_
oT
ds=O
(3.44b)
D GJ oMCy
oMBI
-~=I
=;
MB oM8
f ds
D EI8 oMCy
r2
ff
= -EI-[-M
.
ff
DY-2r +Dx +Dz(2-l)
()
(3.44c)
- P { ( ff
-
a) - (1 - sin a) - ~ ( ff
2
r
-
2
oM
--'-roMCz
f
M
_r
D n,
= sin B
oM
r ds
oMCz
a)}]
56
r2
=-[-M
EI
r
Jr
M Dx
-+---D
Dz 4r
2r
Jr
-]
(3.45a)
Y 4
oT
--=cose
oMCz
f __!_ oT
ds
D GJ oMCz
r2
Jr
M Dx
= -G-J[-M Dz -4r =t::
Jr
Dy(l-4)]
(3.45b)
Finally, equation (3.24) will give
oMB1
--=0
oMCz
(3.45c)
Thus, all the equations in leg D, containing the unknown reactions of the
other three legs, are obtained.
57
IV. OBTAINING THE EQUATIONS AND SOLUTION
It is known
from Castigliano 's theorem
and from the fact that
deformations at fixed ends should be zero that:
M oM
T oT
f _r __ r ds+
f --ds+
A EI r o •
A GJ o •
Mr oMr
f ---ds+
D EI r o •
Me oMe
f ---ds+
A EI e o •
T oT
f --ds+
D GJ o •
Me oMe
f ---ds=O
D EI e o •
(4.1)
where 8• refers to the 6 unknowns of leg A. And it is worthy to mention
that, equation ( 4.1) is also applicable for leg B and C.
By using equation ( 4.1) for leg A, B and C, 18 equations will be obtained.
And we arrive that:
sr
M oM
T
f _r __ r ds+
f ---ds+
A Eir oAX
A GJ oAX
Mr oMr
+f ---ds+f
D ".
oAX
T at
---ds+f
D GJ oAX
Me oMe
f ---ds
A Eie oAX
Me oMe
---ds=O
D Eie oAX
(4.2)
58
Substituting the value of each term, will give us:
r2
1r M Ay
EI [MAz 4--2-+
r
31r
r2
+ A r(--2)]+0+-[-M
x 4
EI
r2
+-[-M
GJ
tt
MDx
-+---D
Dz 4
2
r
1r
r-]
y 4
(4.3)
Jr MDx
-+--+D
Dz 4
r2
1r
M Ay
GJ[M A/4-l)--2-
rn
Ax 4]+
2
y
Jr
r(l--)]
4
1r
r
r
-D r- -D -+ Pcosa(r- -cosa - u)]
y
x 4
z2
2
r2
+--[MD
EI(}
Substituting
the values of
=0
Dx ,DY ,D2 ,MDx ,M Dy and MDz
m
equation (4.3), it can be written as follows:
M
1r/2 - 1 1r/2
+-]+M
Az GJ
EI
1
1
1
[ ---J
Ay - 2GJ- 2EI - EI n
[
r
r
1r/2 Jr- 2 1r/4
+A [r(-'-+--+-)]+M
x EI
GJ EI11
r
M
r
-r
+A [-]+B
Y GJ
Jr/4 1r/4
[-+-]+
Bz EI
GJ
u
Jr/4 m4
[-+-]+C
Cz EI
GJ
x
r
1r/4
EI
1r;A 1r/4
-+-)]
GJ EI
[r( -
r
1r/4
[r(--Y
EI
r
(}
1r/4+1
-r
10
)]+C [-0
]+
J
y J
1 1
1
1 1
M [ -(-+-)]+M
[ -(-+-)]
Ax 2 EI
GJ
Bx 2 EI
r
r
1 1
1
r/2 r/2
+ Mc) - 2 ( EI + GJ)] + Az [ EI + GJ r
r
-l
r/2 n2
r/2
CZ[- EI - GJ- iI ]+MBy[EI
r
(}
3r.i2
+B [ -]+B
z EI
(}
rm4
[-Jx EI
(}
u
(}
1
GJ
r/2
EI ] +
(}
-1
]+MCy[EI ]
(}
P[-u+r,2-cosa(r-rcosa,2-u)]
EI
(}
To simplify this equation, some techniques are used as follows:
(4.4)
59
h
It is supposed that: b
= /3, where h and b are the length and the width of a
cross sectional area of the leg respectively. So, the moment of inertia I e, and
the radial moment
Ir, can be written as follows [5].
(4.4a)
(4.4b)
Also, for the same cross sectional area, it is known that, the torsional
constants J is given by the following formula [ 4].
J
= bh
3 1
h
h4
[ - - 0.21- (1- -)
]
3
b
12b4
3 1
h
J = bh [- -0.21-(13
b
J=/3
2
h4
12
-)]12b4 12
4
I() [_!.-0.21/J(l-.[!__)]12
3
12
1
/3 4
Let y =12/32 [--0.21/J (1--)]
3
12
so J = r I()
( 4.4c)
(4.4d)
(4.4e)
(4.4t)
(4.4g)
And, modulus of rigidity is given by:
G=
E
2(1 + v)
(4.4h)
where, Eis the modulus of elasticity, and vis poison ratio [5].
Substituting these values in equation ( 4.4), it will give the first equation.
60
A [r{',z-/2 +2(n--2)(l+v)
x /12
r
+n-}]+A [-2r(l+v)]
4
r
Y
+ A [-r- + r(l + v) _ ~] + M [-__!_ {-1- + 2(1 + v)}]
z 2 p2
r
2
Ax 2 p2
r
+M
Ay
[-(l+v) __ 1__ 1]+M
[2(n-/2-l)(l+v)
r
Az
r
2/12
+~]
2/J2
+B (n-]+B [r{ -n- _ 2(n-/4+1)(1+ v)}]+B [-3r]
X 4
Y
r
Z
2
4/J2
+M
[-_!_{_1_+2(1+v)}]+M [-l]+M
[~+n-(l+v)]
2 /J2
y
By
Bz 4/J2
2y
+C [r{ -n- _ n-(1+ v) + n-}]+C [-2r(l+ v)]
X
4/J2
2y
4
Y
r
Bx
+C [--=!.:_ _ r(l + v) _ ~] + M [ _ _!_ {-1- + 2(1 + v)}]
z 2/12
r
2
Cx 2 /J2
r
1r
+ MC [ -1] + MC [-2+
Y
z 4/J
n-(1 + v)
]
2y
-1
1
2
+ cosa - - cos a]
2
2
= Pr[-
(4.5)
Following the same steps, the remaining equations will be obtained as in
the same way.
A [-2r(l + v)] + A (n- + rn-(l + v)] + A [-r _ 2r(l + v)] +
X
r
Y4
r
z2
r
M
[l+2(l+v)]+M
r
Ax
+M
[2(1 + v)] + M
Bx
r
[-2(l+v)]+B
Az
r
[2r(l+n-/2)(1+v)]
y
r
[-2(1 + v)] + C [2r(l + v)] + C (n-(1 + v)]
Bz
r
X
r
y
r
+C [2r(l + v)] + M [2(1 + v)] + M [-2(1 + v)]
z
r
Cx
r
Cz
r
=0
(4.6)
61
r(l + v)
A [ -+r
x 2JJ2
r
[-1-+
2
Ax
2
r
;r
x 2
;r
[--2X
4/3
+B [r(l+-)]+MB
Z
4.
Ay
3r(l+ v)]
r
y 2/J2
;r(l + v)
2y
;r-2 )
[l]
2y
4132
(
z 4 JJ2
(1 + v)] +B [-r ]+ B [~-
Az 2/J2
M
r
y 2
[l-1r _ _!!__1r(l+v)]+M
+r;r(l+v)]+M
2y
+M
2r(l + v)] + A [ -+r
rat
-- r] + A [-r
--
[1]+
]+MB
Y
[-1- + (1 + v)] + C [ __ r__ r(l + v) _ _c] + C [ _ 2r(l + v)]
Bz 2132
r
x 2132
r
2
Y
r
+C [-r;r + r;r _ r;r(l + v)] + M
z 4132
4
1
+Mc [-2+
z 2/J
2y
(1 + v)
r
A [-=!.__(l+v)]+A
x 2/J2
r
+M
Cx
a
]=Pr[-+
[ -;r _ ;r(l + v)] + M [l]
2y
Cy
4132
sin2a
2
.
-smacosa]
(4.7)
4
[l+2(1+v)]+A
Y
r
[l-1r _ _!!__1r(l+v)]
z
2 4/J2
2r
[~ {2 +-1-+ 2(1 + v)}]+ M
[---=!_- (1 + v)] +
Ax 4r
p2
y
Az 2r/J2
ry
B [-l-+3(l+v)]+M
y 2p2
r
[~{-l-+2(1+v)}]+M
Bx 4r /32
r
[---=!__(l+v)]
Bz 2r/32
+C [-1- + (1 + v)] + C [ 2(1 + v)] + C [ ;r {-1- + 2(1 + v)}]+
x 2132
M
r
Y
r
z 4 /32
[~ {-1-+ 2(1 + v)}]+ M [---=!_- (1 + v)]
Cx 4r p2
y
Cz 2r /32
ry
r
=0
rr
(4.S)
62
A [-(l+v)
x
y
+M
Az
[
__ l__
2/32
(1 + v)
ry
+C [-l)+C
X
Z
l)+A [l]+M
(,z-(I+v)+_1r_+~]
z
Ay
2ry
4rf32
2r
1
---]+B
2r /32
[l)+MC
x
[-l]+B
Tr
[-+l]+M
z 2
Tr
[-]
By 2r
(4.9)
[~]=P[sina-acosa]
)l 2r
1r
2(1r/2- 1)(1 + v)
- 2(1 + v)
1
(1 + v)
A [-2+
]+A [
] + A [-2+
]
x 2/3
r
Y
r
z 2/J
r
- 1
(1 + v)
- 1
(1 + v)
1r(l + v)
1r
+ MAx [--2 ] + MA [--2 +
] + M Az [
+ --2]
2r/J
ry
Y 2rf3
ry
2rf3
rr
-Tr
+B [-y 4/32
+M
2(1r/4+1)(1+v)
r
[ --+Tr
Bz 4rf32
- 1
C [--
z 2/32
]+M
-1
[--Bx 2rf32
1r(l + v)] C [ - Tr
+
-2ry
x 4/32
(1 + v)
r
]+M
-1
[--Cx 2r/J2
(l+v)
rr
( 4.10)
1r(l + v)] C - 2(1 + v)
+
[
)+
2y
y
r
(1 + v)
rr
]+M
r1r
[-]+B
x2
[
Tr
1r(l + v)
--+
Cz 4r/32
r1r
A [-]+A
x4
-r
[-]+MA
z2
y
rat
C [-]+C
x4
-r
[-]+MC
z2
1
[-l)=Pr[--+cosa--cos
Y
2
[-l]+B
]
z
[-2r]+MB
1
2
2
2ry
y
]=0
[-2]+
( 4.11)
a]
63
A [ -nr _ 2r(n-/4 + 1)(1 + v)] + A [2r(n-/2 + 1)(1 + v)] +
x 4/32
r
Y
r
[--=c_- 3r(l + v)] + M
A
r
z 2/J2
M
[-1- + 3(1 + v)] +
Ax 2/32
r
[---n- _ 2(n-/4 + 1)(1 + v)] + B [-rn-_ + _3r_n-_;_(l_+_v--'-)]
Az 4/32
2/32
r
+M
[ -+1
Bx /32
4(1+v)]
r
y
+
M
[ -n-Bz 2/32
r
n-(l+v)]
r
+
C [ _!!!_ + 2r(n-/4 + 1)(1 + v)] + C [2r(n-/2 + 1)(1 + v)] +
x 4/32
r
Y
r
C [-r- + 3r(l + v)] + M [-1- + 3(1 + v)]
z 2/32
r
Cx 2/32
r
+M
[ -n- _ 2(n-/4 + 1)(1 + v)] = O
r
Cz 4/32
- 3r
A [-]+A
x2
-3r
+C [-]+C
x 2
nn[r(-+l)]+MA
[-+l]+B
z
4
Y2
nn[r(-+l)]+MC
[-+1]=
z
4
·
Y2
+(1- sin a)} -r{(l-
M
3rn[-2r]+B [-]+MB
[n-]
x
z2
y
nrtt
nPu(-+l)-P-+P[r{(--a)
2
4
2
r
[2(1 + v)] + A [ -n- _ n-(l + v)] +
Y
r
z 4 /32
2r
[-=-!._ _ (1 + v)]
[~
+ n-(l + v)] + M
Ax 4r/J2
2ry
Az 2r/J2
+B [-1-+ 4(l+v)]+M
y /32
r
ry
[~+n-(l+v)]+M
Bx 2r/J2
ry
[2]
Bz r/J2
+C [-1- + (1 + v)] + C [ 2(1 + v)] + C [ ~
+ n-(1 + v)]
x 2/J2
r
y r
z 4/J2
2r
+M
(4.13)
sin a)+ (n- - {a+ sin2a} )} - u{(n- - a)+ (1- sin a)} J
4
2
4
2
--2__ _ (1 + v)] + A
A [
x 2 /J2
(4.12)
[-=-!.__- (1 + v)] = 0
[~
+ n-(1 + v)] + M
Cx 4r/J2
2ry
Cz 2r/J2
ry
(4.14)
64
7r
7r
A [-l]+A [l]+MA [-]+B [-2]+B [1r]+MB [-]+C [-1]
x
z
y 2r
x
z
y r
x
+C [1] + MC [ ~]
z
Y 2r
=
( 4.15)
P[sina - a cosa]
A [_!:__+1r(l+v)]+A [-2(1+v)]+A [-1-+(l+v)]+
x 4/32
2y
Y
r
Y 2/32
r
M
[--=.!____(l+v)]+M [~+1r(l+v)]+B
Ax 2r/J2
rr
Az 4r/J2
2ry
+M
[2]+M
Bx r/J2
[~+1r(l+v)]+C
Bz 2r/J2
rr
[-1r _1r(l+v)]
y 2/32
r
[-1r _1r(l+v)]+
x 4/32
2y
( 4.16)
C [-2(1 + v)] + C [ --2_ _ (1 + v)] + M [--=.!___ _ (1 + v)]
y
r
z 2/32
y
Cx 2r/J2
rr
+M
[-"- + 1r(l + v)]
Cz 4r/J2
2ry
=0
A [-r1r +r1r _r1r(l+v)]+A [2r(l+v)]+A [~-~-r(l+v)]
x 4/32
4
2r
y r
z 2/32
2
+M
[-1-+(l+v)]+M
Ax 2/32
y
[-l]+M
Ay
r
[-1r _1r(l+v)]+
Az 4/32
2y
M [ 1
(l+v)
B [ -r1r] + + B [ - rtt + 2r(1r/4+l)(l+v)] + B [-3r]
+
+--]
x 4
y 4/32
r
z 2
Bx 2 /32
r
+M
[-l]+M
By
[ -1r _1r(l+v)]+C [ r1r +r1r +2r(1r-2)(1+v)]
Bz 4/J2
2y
x 2/32 4
r
r
+C [ 2r(l+ v)] + C [ ---+
y r
z 2/32
M
[--2__ (l+v)
~2~
_l]+M
r
r(l+ v)] + M
r
2
[ -+1
Cx 2/32
(l+ v)] +
[ -1r + 2(1-1r/2)(1+v)]
Q2~
= Pr[- -1 + cosa - -1 cos2 a]
2
r
2
r
r
( 4.17)
65
A [-2r(l + v)] + A (;r(l + v)] + A [-2r(l + v)] + M
[2(1+ v)]
X
r
y
r
z
r
Ax
r
[-2(1 + v)] + B [2r(l + Jr/2)(1+ v)] + M
+M
r
Az
y
[-2(1+v)]+C
M
Bz
r
r
Bx
( 4.18)
[2r(l+v)]+C
(Jr +r;r(l+v)]+
x
y
y 4
y
y
C [~+2r(l+v)]+M
[l+2(1+v)]+M
r
z 2
[2(1+ v)] +
Cx
[-2(1+v)]=O
y
r
Cz
A [- _r__ r(l + v) _~]+A [ 2r(l + v)] + A [- r;r + r;r _ r;r(l + v)]
x 2/32
r
2
y
r
z 4/32
4
2r
[~+;r(l+v)]+M
+M
Ax 4/32
[l]+M
2r,
Ay
Bx 4/32
C [~+r(;r-2)+r;r(l+v)]+M
+M
1
(1 + v)
y
[-=!_ _ (1 + v) ] + A
A
x 2/32
r
[-Jr-+
M
Ax 4r/J2
Bx 4r/J2
+M
r
( 4.19)
[~+2r(l+v)]+
Y 2
r
[1]
2y
4/32
Cy
[ 2 ( 1 + v)] + A [ - ;r _ ;r(1 + v) ] +
y
r
z 4/32
2r
[-=-!_- (1 + v)] + B
;r(l + v)] + M
Az 2r/J2
;r(l+v)] + M
2ry
z 2
ry
[ ----1
Bz 2r/J2
+C [1 + 2(1+ v)] + C [ ;r - 1 + ~
y
By
a sin2a
.
] = Pr[-+---smacosa]
2
4
2ry
[ --+Jr
+M
[1]
2y
[;r-l+~+;r(l+v)]+M
Cx 2
2y
[--Cz 2/32
[-r]
x 2
[~+;r(l+v)]+M
z
[--l __ (l+v)]~C [-r--~+r(l+v)]+C
Bz 2 /32
r:
x 2/32 2
r
z 4p2
r
2/32
[r(l+;r/4)]+M
+B [-r-+3r(l+v)]+B
Y 2/32
r
+M
[--l __ (l+v)]+B
Az
4/32
[-1- + 3(1 + v)]
y
2/32
(l+v)] + C [ -+-1
ry
x
2/32
+ ;r(l + v)]
2r
[-=-!_- (1+ v)] = 0
[~ +~
+ ;r(l + v)] + M
Cx 2r 4r/J2
2ry
Cz 2r/J2
,ry
r
(l+v)]
r
(4.20)
66
ff
A [-l]+A [l]+MA
[-]+B
x
z
Y 2r
+C [---2_-l_(l+v)]+C
x 2/32
y
1
(1 + v)
+MC [--2]
z 2r/J
rr
A [~+ff(l+v)]+A
X 4/32
2y
M
[~Ax 2r/32
x
[l]+M
z
=
[ --+ff
Bz 4r/J2
( 4.21)
[-1-+(l+v)]+
r
Y
Z
(1+ v)]+ M
[-ff-+
ry
Az 4r/J2
r
2/32
ff(l+ v)]+
2ry
[~-(l+v)]
rr
Bx 2r/32
ff(l + v)] C [ -ff
+
-+
2ry
x 2/32
2(1- ff/2)(1+ v)]
+
y
[--=!_ _ (1 + v)] + M
[-1 __
Cy 2r/J2
ff
[-]
Y 2r
[-ff-+~+ff(l+v)]
Cy 4r/J2
2r
2ry
[-2(1+v)]+A
C [-2(1 + v)] + C
y
r
z 2/32
+M
ff
[-+l]+MB
z 2
.
P[sma- acosa]
B [-ff
_2(ff/4+l)(l+v)]+M
Y 4/32
r
+M
[-l]+B
y
[~
_ (1 + v)]
Cx 2r/J2
rr
(1+ v)]+ M [~+
r~
Cz 2r/J2
ff(l+ v)]
rr
(4.22)
=0
According to Maxwell's reciprocal theorem, in a linear elastic structure,
the displacements at coordinate i due to a unit force at coordinate j is equal to
the displacement at j due to a unit force acting at
i , which is given by the
equation
where
fu is the displacement at i due to a unit force at j, and ~ds the
displacement at j due to a unit force at i .
67
The displacements in this equation are the flexibility coefficients. For a
structure in which m coordinates are indicated, the flexibility coefficients, when
arranged in a matrix of the order m x m, give the flexibility matrix of the
structure. This matrix must be symmetrical, by virtue of the previous equation
[4].
So, from the 18 equations which are obtained, a symmetrical matrix of the
order l 8x 18 is arranged analytically
in a general form. For solving these
equations numerically, some numerical values must be given. And in our case
we choose
P
= 10.0
KN.
R
=
m.
10;0
a= 60
u
0
= r(l
= -Tr
3
- cos a)= 5.0
m.
0. 3
fJ = 2 .0
V=
Then equation ( 4.4f) becomes:
r = 22 .n
After applying these values to the equations, it can easily be solved as:
[A] [x]
=
[CJ
where [ A] is a symmetrical matrix of the order l 8x 18, which shows the
elements of the equations as shown in the next page.
68
{O
co
..•...
N
co
N
0
0
I
..•...
co
..•...
0
0
0
0
-.::t"
0
0
0
0
0
0
0
co
.•....
0
N
I
0
I
.•....
O'l
0
I
0
I
I
{O
co
..•...
0
co
N
N
0
0
0
0
.•....
0
0
N
{O
N
co
N
I
I
N
0
0
I
I
I
0
co
,..._
.•....
0
LO
0
,..._
.
0
0
CO
..•...
0
,..._
0
co
..•...
{O
LO
co
..•...
0
0
co
0
I
,..._
0
0
0
I
0
0
co
..•...
0
0
0
I
co
co
,..._
.•....
,..._
LO
co
..•...
0
C"')
0
0
{O
LO
{O
O'l
N
LO
I
N
0
0
,..._
co
.•....
co
{O
0
I
0
0
I
I
co
..•...
N
0
0
0
,..._
co
co
.•....
.•....
0
0
N
0
I
0
O'l
O'l
0
0
N
LO
0
LO
,..._
co
.•....
O'l
0
co
..•...
N
co
..•...
LO
,..._
N
N
co
-.::t"
0
.•....
,..._
.•....
0
O'l
N
0
N
N
co
I
I
N
..•...
..•...
I
I
N
co
co
{O
I
{O
co
N
0
..•...
I
I
0
N
co
..•...
,..._
.•....
0
0
0
0
~
N
0
0
0
0
I
0
{O
co
..•...
LO
..•...
I
,..._
0
0
-.::t"
I
co
N
0
LO
..•...
0
0
N
-.::t"
O'l
N
0
I
0
0
,..._
,..._
.•....
O'l
0
-.::t"_
{O
O'l
NI
LO
I
LO
N
0
0
0
0
0
0
..•...
I
I
N
I
LO
0
0
0
O'l
0
..•...
C"')
I
{O
N
0
.•....
I
N
0
I
0I
co
..•...
0
0
0
I
0
I
co
co
..•...
..•...
co
..•...
N
co
0
N
co
..•...
C"')
0
I
,..._
LO
co
LO
0
0
0
I
co
.•....
0
I
I
co
{O
I
N
-.::t"
..•...
0
NI
0
,..._
0
0
0
LO
LO
s::t"
0
I
O'l
N
0
0
I
-.::t"_
0
O'l
0
..•...
I
0
I
co
,..._
..•...
C"')
I
N
co
.•....
0I
N
co
..•...
..•...
I
0
0
0
I
{O
O'l
N°"
I
I
-.::t"
LO
LO
co
,..._·
N
-.::t"
O'l
N°"
LO
0
I
0
0
~
,..._
LO
.•....
{O
O'l
N
0
0
co
N
0
co
..•...
0
0
0
0
0
0
L()
.•....
0
co
,..._
..•...
0
0
LO
,..._
,..._
..•...
O'l
C\i
{O
O'l
N
O'l
N
0
0
LO
0
LO
co
,-..:
.•....
,..._
0
0
-.::t"
LO
0
co
,..._
O'l
co
N
0
N
-.::t"
{O
0
0
0
.•....
I
I
O'l
N
0
I
0
0
{O
co
.•....
N
0
0
I
0
I
N
co
co
..•...
0
I
0
I
0
co
..•...
0
..•...
I
0
0
I
.•....
I
0
O'l
0
N
0
0
I
0
N
{O
co
.•....
0
co
N
0
0
N
N
co
{O
I
co
..•...
O'l
N
0
0
0
{O
.•....
I
0
0
I
co
N
N
co
..•...
0
0
I
0
0
0
N
I
I
co
..•...
0
.•....
s::t"
0
0
0
LO
0
0
0
co
.•....
0
I
N
N
0
0
,..._
0
NI
0
.•....
,..._
0
,..._
LO
co
.•....
0
-.::t"
co
.•....
0
0
I
-.::t"_
0
NI
I
O'l
N
0
co
0
0
0
,..._
0
{O
0
N
,..._
0
C\i
N
O'l
C"')
I
0
0
0
N
I
0
0
I
0
N
0
I
0
.•....
0
I
0
0
LO
0
I
L()
0
0
.•....
co
.•....
0
LO
I
N
co
0
,..._
0
0
LO
{O
0
N
0
co
.•....
{O
co
.•....
0
,..._
0
,..._
0
co
..•...
..•...
,..._
I
,..._
{O
0
I
C"')
I
O'l
0
0
N
0
0
N
-.::t"_
I
0
I
0
0
0
C"')
N
0
0
I
0
co
,..._
..•...
co
.•....
N
0
0
.•....
,..._
co
..•...
LO
0
0
N
co
0
N
0
I
LO
co
.•....
O'l
N
LO
0
co
..•...
..•...
LO
co
.•....
0
N
,..._
O'l
0
co
.•....
I
0
,..._
N
I
{O
O'l
{O
N
LO
.•....
0
I
0
co
.•....
LO
0
0
0
0
0
0
N
co
..,...
0
0
N
N
{O
co
N
0
.•....
I
N
co
co
..•...
{O
0
I
I
..•...
I
I
{O
co
N
0
69
And [x] is a matrix of the order 18xl which shows the unknowns as:
= AX
X
= Ay
2
X
1
X3
=
AZ
x4
=
M Ax
XS=
M Ay
x6
=
X
=B
X
X
7
8
9
MAz
=B
X
y
=B
z
xlO
=
MBx
xll
=
MBy
xl2
xl3
= M Bz
= ex
xl4
=
Cy
XIS= CZ
xl6
=
MCx
xl7
=
MCy
xl8
=
MCz
70
Also [c] is a matrix of the order 18xl which gives the constants as:
c1
= -12.5
c2 = 0.0
c3 = 30.709
c4
= 0.0
cs = 3.424
c6 = 0.0
c7
= -12.5
's = 0.0
= 64.952
Cl0 = 0.0
C9
cl 1
= 3.424
cl2
cl5
= 0.0
= -12.5
= 0.0
= 30.709
cl6
= 0.0
cl 7
= 3.424
cl3
cl4
cl8 = 0.0
71
Then, by the help of a computer program (Microsoft Excel), the matrix is
solved, and the results are found as:
A
A
A
X
y
z
= 0.886596
= 0.42503
= 0.47057
M Ax = -0.98242
M Ay = 3.577007
M Az = -4.57213
B
=1.816587
B
= -0.00959
X
y
B = 1.606615
z
M Bx = 0.125205
M By = 5.564286
M Bz = -0.04756
C = 0.895539
X
C = -0.41242
y
C = 0.441539
z
M Cx = 1.034371
MCy = 3.6332
M Cz = 4.572547
where the forces are given by KN, and the moments are given by KN.m.
72
By substituting the results in the equations, which obtained from statics in
the second chapter, the forces and the moments ofleg D can be found as:
D
X
= -3.59869
D = -0.00302
y
D
z
= 7.481275
MDX =0.113159
M Dy = -4.02789
M Dz = 0.071212
Now, all the forces and the moments of the two arches are found, and the
problem is solved.
73
V. CONCLUSION
In this study, 18 equations by 18 unknowns are obtained for analyzing a
dome problem, by using Castigliano' s theorem.
These equations are constructed analytically in a very general way, that
can be solved by applying any external concentrated load normal to the dome.
And it is worthy to note that all terms that contributed the deformation are taken
into consideration.
By applying the principle of superposition, this problem can be solved for
different external loads [2]. This can be done not only for concentrated loads,
but also for distributed loads as well. This procedure can be performed by
applying the loads as frequent as it is desired. Thus a satisfactory analysis can be
obtained from engineering point of view.
74
REFERENCES
1.
WHITNEY CLARK HUNTINGTON., "Building Construction,"
THIRD EDITION, JOHN WILEY & SONS, INC, 1963
2. S.P.Timoshenko & D.H.Young.,"THEORY of STRUCTURES,"
SECOND EDITION, McGraw-Hill, Inc, 1965
3. C.K. Wang.," INTERMEDIATE STRUCTURAL ANALYSIS,"
International Edition, McGraw-Hill, 1983
4. A. GHALI ,A.M. NEVILE.,"Structural Analysis," Second Edition,
CHAPMAN AND HALL
5. Beer & JOHNSTON.,"Mechanics of Materials," Second Edition,
McGraw- Hill, Singapore, 1992

A r - Near East University

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