This document provides an overview of cables and their analysis. It defines cables as flexible members that can only withstand tension. It lists some common engineering applications of cables. It then outlines the assumptions made in cable analysis, including that cables are flexible, have negligible self-weight, and experience only tension forces. The document explains the procedure for analyzing cables subjected to concentrated loads, including drawing free body diagrams, applying equilibrium equations, and determining tensions and reactions. It provides examples of solving for cable tensions, reactions, slope, and horizontal forces using this procedure.

1. Cables
Presented By,
Miss. Shinde Bharti M. (Assistant Professor)
Department of Civil Engineering
Sanjivani College of Engineering, Kopargaon.
Email- shindebharticivil@sanjivani.org.in
1

2. Cables-
Cables are the flexible members capable of
withstanding only tension , designed to support either
concentrated load or distributed load.
Cables are used in many engineering applications
such as suspension bridges, transmission lines, aerial
tramways, guy wires for high towers etc.
2

3. 3
Assumptions –
1. Cables are Flexible
2. Self weight of the cable is negligible, compared to the
load supported by cable.
3. The internal forces at any point in the cable reduce to
a force of tension directed along the cable.

4. 4
Analysis of Cables-
The procedure for the analysis of cables subjected to concentrated point load
is as explained below,
1. Draw the free body diagram of the entire cable.
2. Now, there will be 4 unknown components and only 3 equations of
equilibrium then identify the point of additional information as position or
slope of the point.
3. Then, cut the cable at that point end draw free body diagram of any one
side section left or right. And write the additional equation to determine
the unknown as,
• If the position is known, take about the point of cut for the new
free body diagram.
• If the slope is known, take for the new free body
diagram.
M 0
H 0 and V 0  

5. 5
4. To find the elevation of a cable at a given point or slope and
tension at that point- cut the cable at that point (where elevation is to be
determined) and draw the free body diagram of any one side of the section
and writing about the point gives elevation and writing
gives tension in the member.
5. For a cable supporting vertical load only the maximum tension occurs in the
steepest portion of the cable.
M 0 H 0 and V 0  

6. 6
Example-
1. Three loads are suspended as shown from the cable ABCDE. Knowing that dc= 3m,
determine a) the components of the reaction at E, b) the maximum tension in the cable.
Ans:
Step 1-
Considering the equilibrium of entire cable and applying the conditions of equilibrium to find
the reactions at support
A E
A E
A E
E
A
H 0 , H H 0
V 0 , V V 9
M 0 , 2x 4 3x8 4x12 V x16 0
V 5KN
V 4KN
   
  
    

 




7. 7
Step2-
Cut the cable at point C and consider the free body diagram of the right side,
C E
E A
2 2
A A A
2 2
E E E
E
M 0 , 4x 4 5x8 H x3 0
H 8KN and H 8KN
R H V 8.944KN
R H V 9.43KN
Maximumtensionincable R 9.43KN
   
  
   
  
 


8. 8
Example-
2. Knowing that dc=2.44m, determine a)the reaction at A, b)the reaction at E.
Ans:
Step 1-
Considering the equilibrium of entire cable and applying the conditions of equilibrium to
find the reactions at support
A E
A E
A
E E
E E
H 0 , H H 0
V 0 , V V 3.56
M 0 ,
1.33x 2.44 0.9x 4.88 1.33x7.32 V x9.76 H x1.83 0
1.83H 9.76V 17.37
   
  

    
  




9. 9
Step2-
Cut the cable at point C and consider the free body diagram of the right side,
C E E
E E
E E
A A
2 2
A A A
2 2
E E E
E
M 0 , 1.33x 2.44 H x 2.44 V x 4.88 0
H 2V 1.33
solving thesimultaneousequations,
H 3.545KN, V 2.437KN
and H 3.545KN, V 1.123KN
R H V 3.71KN
R H V 4.30KN
Maximumtensionincable R 9.43KN
   
   
  
 
   
  
 


10. 10
Example-
3. Determine a) distance dc fir which portion DE of the cable is horizontal, b) the
corresponding reactions at A and E.
Ans:
Step 1-
Considering the equilibrium of entire cable and applying the conditions of equilibrium to
find the reactions at support
A E
A E
A
E E
E E
H 0 , H H 0
V 0 , V V 20
M 0 ,
10x7 5x 4 5x 2 V x10 H x 4 0
4H 10V 100
   
  

    
 




11. 11
11
Step2-
Cut the cable at point C and consider the free body diagram of the right side,
Step 3-
To find the dc- cut the cable at point C and consider the free body diagram of left side
section,
D A A
A A
A A
E E
2 2
A A A
2 2
E E E
A
M 0 , 5x5 5x3 H x 4 V x7 0
4H 7V 40
solving thesimultaneousequations,
H 25KN, V 20KN
and H 25KN, V 0KN
R H V 32KN
R H V 25KN
Maximumtensionincable R 32KN
     
   
  
 
   
  
 

CM 0 , 25xdc 20x 4 5x 2 0
dc 2.8m
    
 


12. 12
Example-
4. Cable ABC supports two loads as shown. Knowing that b=6.4m, determine, a) the
required magnitude of the horizontal force P, b) the corresponding distance a.
Ans:
Step 1-
Considering the equilibrium of entire cable and applying the conditions of equilibrium to
find the reactions at support
A
A
A
H 0 , H P
V 0 , V 1424KN
M 0 ,
Px6.4 623xa 801x6.4 0
6.4P 623a 5126.4
 
 

   
   




13. 13
Step2-
Cut the cable at point B and consider the free body diagram of the right side,
BM 0 , Px 2.74 801x(6.4 a) 0
2.74P 801a 5126.4
solving thesimultaneousequations,
P 1068.27KN, a 2.74m
    
    
  


14. 14
Thank You