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Structure Analysis I Chapter 5
Transcript of Structure Analysis I - site.iugaza.edu.ps
Structure Analysis IChapter 5
Cables & Arches
Cables
• Cables are often used in engineering structure to
support and or transmit loads from one member to
another
Structural Analysis IDr. Mohammed Arafa
Cables subjected to concentrated
loads
Example 1Determine the tension in cables and what is the dimension h ?
( ) ( )kNT
TT
M
CD
CDCD
A
79.6
0)4(8)2(35.52
0
54
53
=
=−−+
=
o
BC
BC
BCBCy
BCBCx
T
TF
TF
3.32
82.4
0sin8)(79.60
0cos)(79.60
54
53
=
=
=+−→=
=−→=
o
BA
BA
BABA
o
y
BABA
o
x
T
TF
TF
8.53
90.6
0sin3)3.32(sin82.40
0cos)3.32(cos82.40
=
=
=+−−→=
=−→=
( ) mh 74.28.53tan2 ==
Cables subjected to a Uniform
Distributed Load
Cables subjected to a Uniform
Distributed Load
Analysis Procedure
( ) ( )
( ) ( )0
0
0 cos cos 0
0 sin sin 0
0 cos sin 02
Dividing each eq. byΔx and taking the limit as Δx 0
and hence y 0, 0 and T 0
x
y
o
F T T T
F T w x T T
xM w x T y T x
= − + + + =
= − − + + + =
= − + =
→
→ → →
Analysis Procedure
Structural Analysis IDr. Mohammed Arafa
( )
( )0
cos0 (1)
sin(2)
tan (3)
d T
dx
d Tw
dx
dy
dx
=
=
=
Analysis Procedure
( )
( )
H
0
0
at x=0 T=F
Integrate eqs. 1 where T=F at x=0
cos (4)
Integrate eqs. 2 where Tsin =0 at x=0
sin (5)
dividing 4 by 5
tan =
H
H
H
T F
T w x
w xdy
dx F
=
=
=
Analysis Procedure
2
0 0
2
0
2
2
2
,2
H H
H
w x w xdyy
dx F F
w Lat x L y h F
h
hy x
L
= =
= = =
=
Analysis Procedure
( )
max
2 2
max
0
222
max 0 0
cos
cos
is at where is maximum
( )
In our Case at x=L
12
H
H
H
H
T F
FT
T
at x L
T F V
V w L
LT F w L w L
h
=
=
=
= +
=
= + = +
Analysis Procedure
( )
2
2
2
0
2 2
22 2 2
max max 0
2
cos
H
HH
H H
hy x
L
w LF
h
FT F V
T F V F w L
=
=
= = +
= + = +
Summary
Suspension Bridge
Cable Stayed Bridge
Example 2Determine the tension of the cable at points A, B, C
Assume the girder weight is 850 lb/ft
Structural Analysis IDr. Mohammed Arafa
2 2 2 2
max
500 307500 Ib=7.5k
2 2
7.031 7.5 10.280
A B
A B
WLV V
T T T H V
= = = =
= = = + = + =
Example 3Determine the tension of the cable at points A, B
( )2
500 157031.25
2 8HF = =
2
0
2H
w LF
h=
HF
AVBV
HF
Example 4
The beams AB and BC are supported by the cable that
has a parabolic shape. Determine the tension in the cable
at points D and E.
Example 5
Referring to the FBD of member BC, Fig. b,
Then from the FBD of member AB, Fig. a,
Solve for F and B
120 , 10
H y
H yF kN B kN= =
Draw the shear and moment diagrams for members
AB and BC.
Example 5 -continue
100
35.035.020.0
-5.0-5.0
SFD
BMD
Example 6: Determine the maximum and minimum tension in the
parabolic cable and the force in each of the hangers. The girder is
subjected to the uniform load and is pin connected at B.
Draw the shear and moment diagrams for the pin connected girders
AB and BC.
FH
FH
FH
FH
0
5 5 2.5 0.5 0 (1)
0
20 20 10 8 0 (2)
Solve(1) and (2) 0 & 25
A
y H
C
y H
y H
M
B F
M
B F
B F kN
=
− − =
=
− − =
= =
( )
( )
2
0
2
0
0
min
22
max 0
22
2
2025
2 8
1 /
For the main Cabl
25
25 1 20 32.02
e
H
H
H
w LF
h
w
w kN m
T F kN
T F w L
kN
=
=
=
= =
= +
= + =
0
Force in each ha
2.5 1 2.5 2
g
.5
n er
T w kN= = =
Example 7: The cable AB is subjected to a uniform loading of
200kN/m. If the weight of the cable is neglected and the slope angles at
points A and B are 30 and 60, respectively, determine the curve that
defines the cable shape and the maximum tension developed in the
cable.
( )
( )0 0 1
Integrate the equations
(1)
(2)
cos0 cos
sinsin
H
d TT F
dx
d Tw T w x C
dx
= =
= = +
1
0
Substitute at 0
sin 30 tan 3
30cos cos30
eq.(2) will be:
sin tan
0
(3)30
A
H H
H
A H
F FT T
T
x
C T
F
F
w x
=
=
=
= =
+
=
=
=
( )
( )( )
( ) ( )( )( )
0
2
tan 30tan
10.2 tan 30
60
1tan 60 0.2 15 tan 30
Divide
2.6
1 10.2 tan 30 0.2 2.
(3) by (1)
Substitute
6 0.5772.60
0
at 15
(Equation of thecablecurv.0385 0.577 e)
H
H
H
H
H
H
H
H
H
w x F
F
dyx F
dx F
FF
F kN
dyx F x
dx F
x x
x
y
+=
= +
=
= +
=
=
=
= + = +
= +
max
cos
2.6
cos cos
At 0 30
2.63.0
cos3
Substitute in (1)
0
At 15 60
2.65.2
cos60
H
H
A
B
T F
FT
x
T kN
x
T kN T
=
= =
= =
= =
= =
= = =
When the weight of a cable becomes important in the force analysis, the
loading function along the cable will be a function of the arc length s rather
than the projected length x.
Cable Subjected to its own Weight
w = w(s)
To perform a direct integration, it is necessary to replace dy/dx by ds/dx.
one obtains the following relationships
2 2
2 2
1 1
sd dx dy
dy ds ds dy
dx dx dx dx
= +
= − = +
Separating the variables and integrating we obtain
The two constants of integration, say C1 and C2, are found using
the boundary conditions for the curve.
( )( )
2
1/22
2
1
11
H
ds dy
dx dx
dsw s ds
dx F
= +
= +
( )( )1/2
2
2
11
H
dsx
w s dsF
= +
continued on next slide
Example 7
Arches
Arches Types
Example of Fixed Arch
Three Hinge Arches
Example 5
Determine the internal forces at Section D
( )
kNA
kNA
F
kNB
BM
kNB
BM
y
x
x
x
xRightC
y
yA
93
0F
86
0
86
0)20(67)8(60)6(0
67
0)28(60)10(100)40( 0
y
=
=
=
=
=
=+−−=
=
=−−=
86
93
86 cos20
93 sin20
86 sin20
93 cos20
242
242
112.6
58
Example 6
Determine the internal forces at Section D
( )
y
0 25
F 0
25
0 25 (50) (25) 25 (25) 0
25
0
25
A y
y
B xRight
x
x
x
M C kips
A kips
M C
C kips
F
A kips
= =
=
=
= − − =
=
=
=
yF 0
0
0
25
y
x
x
B
F
B kips
=
=
=
=
( )
( )
2
2
2
1
2
25
25
50
252
50
25tan 2 25 0.5
50
26.6
x
o
y x
dyx
dx
dy
dx
−
=
−=
−=
−= = = −
=
25cos 12.5sin 27.95
25 12.5cos 0.0
25(6.25) 12.5(12.5) 0
D
D
D
N
V sin
M
= + =
= − =
= − + =
26.6o =
Example 7
Structural Analysis IDr. Mohammed Arafa
Problem 3
Determine the internal forces at B
( )
( )
( ) ( )
20 1.2 1.5 3.3 0
1.5 3.3 24 1
15 2.4 3 4.5 0
3 4.5 36 2
1 & 2
1.08
13.62
A x y
x y
C x y
x y
y
x
M B B
B B
M B B
B B
Solve
B kN
B kN
= − − =
+ =
= − + =
− =
=
=
