Cables (Part - 2) Civil Engineering (CE) Notes | EduRev

Structural Analysis

Civil Engineering (CE) : Cables (Part - 2) Civil Engineering (CE) Notes | EduRev

The document Cables (Part - 2) Civil Engineering (CE) Notes | EduRev is a part of the Civil Engineering (CE) Course Structural Analysis.
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Example 31.1
Determine reaction components at A and B, tension in the cable and the sag yB , and yE of the cable shown in Fig. 31.4a. Neglect the self weight of the cable in the analysis.

Cables (Part - 2) Civil Engineering (CE) Notes | EduRev

Cables (Part - 2) Civil Engineering (CE) Notes | EduRevCables (Part - 2) Civil Engineering (CE) Notes | EduRev

Cables (Part - 2) Civil Engineering (CE) Notes | EduRevCables (Part - 2) Civil Engineering (CE) Notes | EduRev

Since there are no horizontal loads, horizontal reactions at A and B should be the same. Taking moment about E, yields

Ray ×14 - 17×20− 10×7 − 10×4 = 0

Cables (Part - 2) Civil Engineering (CE) Notes | EduRev

Now horizontal reaction H may be evaluated taking moment about point C of all forces left of C.

Ray ×7−H×2 -17×3 = 0

H = 44.5 kN

Taking moment about B of all the forces left of B and setting MB = 0, we get

Cables (Part - 2) Civil Engineering (CE) Notes | EduRev

Similarly, Cables (Part - 2) Civil Engineering (CE) Notes | EduRev

To determine the tension in the cable in the segment AB , consider the equilibrium of joint A (vide Fig.31.4b).

Cables (Part - 2) Civil Engineering (CE) Notes | EduRev

The tension Tab may also be obtained as 

Cables (Part - 2) Civil Engineering (CE) Notes | EduRev

Now considering equilibrium of joint BC and D one could calculate tension in different segments of the cable.

Segment bc
Applying equations of equilibrium,

∑ Fx = 0 ⇒ Tab cosθab =Tbc cosθbc

Cables (Part - 2) Civil Engineering (CE) Notes | EduRev

See Fig.31.4c

Segment cd

Cables (Part - 2) Civil Engineering (CE) Notes | EduRev

See Fig.31.4d.
See Fig.31.4e.

Segment de

Cables (Part - 2) Civil Engineering (CE) Notes | EduRev

The tension Tdc may also be obtained as

Cables (Part - 2) Civil Engineering (CE) Notes | EduRev

Example 31.2
A cable of uniform cross section is used to span a distance of 40m as shown in Fig 31.5. The cable is subjected to uniformly distributed load of 10 kN/m. run. The left support is below the right support by 2 m and the lowest point on the cable C is located below left support by 1 m. Evaluate the reactions and the maximum and minimum values of tension in the cable.

Cables (Part - 2) Civil Engineering (CE) Notes | EduRev

Cables (Part - 2) Civil Engineering (CE) Notes | EduRev

Assume the lowest point C to be at distance of x m from B . Let us place our origin of the co-ordinate system xy at C . Using equation 31.5, one could write,

Cables (Part - 2) Civil Engineering (CE) Notes | EduRev                              (1)

Cables (Part - 2) Civil Engineering (CE) Notes | EduRev                            (2)

where ya and yb be the y co-ordinates of supports A and B respectively. From equations 1 and 2, one could evaluate the value of x . Thus,

Cables (Part - 2) Civil Engineering (CE) Notes | EduRev
From equation 2, the horizontal reaction can be determined.

Cables (Part - 2) Civil Engineering (CE) Notes | EduRev

Now taking moment about A of all the forces acting on the cable, yields

Cables (Part - 2) Civil Engineering (CE) Notes | EduRev

Writing equation of moment equilibrium at point B , yields

Cables (Part - 2) Civil Engineering (CE) Notes | EduRev

Tension in the cable at supports A and B are

Cables (Part - 2) Civil Engineering (CE) Notes | EduRev

The tension in the cable is maximum where the slope is maximum as T cosθ = H. The maximum cable tension occurs at B and the minimum cable tension occurs at C where dy/dx = θ = 0 and TC = H =1071.81 kN

Example 31.3
A cable of uniform cross section is used to support the loading shown in Fig 31.6. Determine the reactions at two supports and the unknown sag yC.

Taking moment of all the forces about support B ,

Cables (Part - 2) Civil Engineering (CE) Notes | EduRev                  (1)

Ray = 65+ 10yc

Taking moment about B of all the forces left of B and setting MB = 0, we get,

Ray × 3 - Ha × 2 = 0

⇒ Ha = 1.5 Ray                           (2)

Taking moment about C of all the forces left of C and setting MC = 0 , we get

∑MC = 0     Ray × 7 - Ha × yC - 50 × 4 = 0

Substituting the value of Ha in terms of Ray in the above equation,

7 Ray - 1.5 Ray yC - 200 = 0                          (3)

Using equation (1), the above equation may be written as,

Cables (Part - 2) Civil Engineering (CE) Notes | EduRev             (4)

Solving the above quadratic equation, yC can be evaluated. Hence,

yC = 3.307m.

Substituting the value of yC in equation (1),

Ray = 98.07 kN

From equation (2),

Ha = 1.5Ray = 147.05 kN

Now the vertical reaction at D, Rdy is calculated by taking moment of all the forces about A ,

Rdy × 10 -100 × 7 +100 × 3.307 - 50 × 3 = 0

Rdy = 51.93 kN.

Taking moment of all the forces right of C about C, and noting that ∑ MC = 0,

Rdy × 3 = Hd × yc   ⇒ Hd = 47.109 kN.

Summary

In this lesson, the cable is defined as the structure in pure tension having the funicular shape of the load. The procedures to analyse cables carrying concentrated load and uniformly distributed loads are developed. A few numerical examples are solved to show the application of these methods to actual problems. 

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