This document discusses structural analysis of cables and arches. It provides examples of determining tensions in cables subjected to concentrated and uniform loads. It also discusses the analysis procedure for cables under uniform loads. Examples are given for calculating tensions at different points of cables supporting bridges. Methods for analyzing fixed and hinged arches are demonstrated through examples finding internal forces at various arch sections.

1. Structure Analysis I
Chapter 5
Structure Analysis I
Chapter 5

2. Cables & Arches

3. Cables
• Cables are often used in engineering structure to
support and or transmit loads from one member to
another
Structural Analysis IDr. Mohammed Arafa

4. Cables subjected to concentrated
loads

5. Example 1
Determine the tension in cables and what is the dimension h ?

6.    
kNT
TT
M
CD
CDCD
A
79.6
0)4(8)2(35.52
0
5
4
5
3



o
BC
BC
BCBCy
BCBCx
T
TF
TF
3.32
82.4
0sin8)(79.60
0cos)(79.60
5
4
5
3










7. 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 

8. Cables subjected to a Uniform
Distributed Load

9. Cables subjected to a Uniform
Distributed Load

10. Analysis Procedure

11.    
   0
0
0 cos cos 0
0 sin sin 0
0 cos sin 0
2
Dividingeach eq. byΔx and taking thelimit as Δx 0
and hence y 0, 0 and T 0
x
y
o
F T T T
F T w x T T
x
M w x T y T x
  
  
 

       
         

       

     



Analysis Procedure
Structural Analysis IDr. Mohammed Arafa

12.  
 
0
cos
0 (1)
sin
(2)
tan (3)
d T
dx
d T
w
dx
dy
dx






Analysis Procedure

13.  
 
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

14. 2
0 0
2
0
2
2
2
,
2
H H
H
w x w xdy
y
dx F F
w L
at x L y h F
h
h
y x
L
  
  
 
Analysis Procedure

15.  
max
2 2
max
0
2
22
max 0 0
cos
cos
is at where is maximum
( )
In our Case at x=L
1
2
H
H
H
H
T F
F
T
T
at x L
T F V
V w L
L
T F w L w L
h






 

 
     
 
Analysis Procedure

16.  
2
2
2
0
2 2
22 2 2
max max 0
2
cos
H
H
H
H H
h
y x
L
w L
F
h
F
T F V
T F V F w L



  
   
Summary

17. Suspension Bridge

18. Cable Stayed Bridge

20. Example 2
Determine the tension of the cable at points A, B, C
Assume the girder weight is 850 lb/ft
Structural Analysis IDr. Mohammed Arafa

25. 2 2 2 2
max
500 30
7500 Ib=7.5k
2 2
7.031 7.5 10.280
A B
A B
WL
V V
T T T H V

   
      
Example 3
Determine the tension of the cable at points A, B
 
2
500 15
7031.25
2 8
HF  

2
0
2
H
w L
F
h


HF
AV BV
HF

26. Example 4

30. Problem 1: 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.

31.  
 
0 0 1
Integrate the equations
(1)
(2)
cos
0 cos
sin
sin
H
d T
T F
dx
d T
w T w x C
dx




  
   


1
0
Substitute at 0
sin30 tan3
30
cos cos30
eq.(2) willbe:
sin tan
0
(3)30
A
H H
H
A H
F F
T T
T
x
C T
F
F
w x






 






32.  
  
     
0
2
tan30
tan
1
0.2 tan30
60
1
tan 60 0.2 15 tan30
Divide
2.6
1 1
0.2 tan30 0.2 2.
(3) by (1)
Substitute
6 0.577
2.60
0
at 15
(Equation of thecablecurv.0385 0.577 e)
H
H
H
H
H
H
H
H
H
w x F
F
dy
x F
dx F
F
F
F kN
dy
x F x
dx F
x x
x
y




 

 



   
 

33. max
cos
2.6
cos cos
At 0 30
2.6
3.0
cos3
Substitute in (1)
0
At 15 60
2.6
5.2
cos60
H
H
A
B
T F
F
T
x
T kN
x
T kN T

 



 
 
 
 
  

34. Problem 2: 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.

35. HF
HF
HF
HF
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

     

     
 



36.  
 
2
0
2
0
0
min
22
max 0
22
2
20
25
2 8
1 /
For the main Cabl
25
25 1 20 32.02
e
H
H
H
w L
F
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    

38. Problem ٣: The beams AB and BC are supported by the cable that
has a parabolic shape. Determine the tension in the cable at points
D, F, and E, and the force in each of the equally spaced hangers

41. Arches

42. Arches Types

44. Example of Fixed Arch

45. Three Hinge Arches

46. 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













47. 86
93
86 cos20
93 sin20
86 sin20
93 cos20
242
242
112.6
58

48. Example 6
Determine the internal forces at Section D

49.  
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
  


    








50. yF 0
0
0
25
y
x
x
B
F
B kips







51.  
 
2
2
2
1
2
25
25
50
25
2
50
25
tan 2 25 0.5
50
26.6
x
o
y x
dy
x
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
 

52. Example 7
Structural Analysis IDr. Mohammed Arafa

56. Problem 3
Determine the internal forces at B