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**Example 31.1**

Determine reaction components at A and B, tension in the cable and the sag y_{B} , and y_{E} of the cable shown in Fig. 31.4a. Neglect the self weight of the cable in the analysis.

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

R_{ay} ×14 - 17×20− 10×7 − 10×4 = 0

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

R_{ay} ×7−H×2 -17×3 = 0

H = 44.5 kN

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

Similarly,

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

The tension T_{ab} may also be obtained as

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

**Segment bc**

Applying equations of equilibrium,

∑ F_{x} = 0 ⇒ T_{ab} cosθ_{ab} =T_{bc} cosθ_{bc}

See Fig.31.4c

**Segment cd**

See Fig.31.4d.

See Fig.31.4e.

**Segment de**

The tension T_{dc} may also be obtained as

**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.

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,

(1)

(2)

where y_{a} and y_{b} be the y co-ordinates of supports A and B respectively. From equations 1 and 2, one could evaluate the value of x . Thus,

From equation 2, the horizontal reaction can be determined.

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

Writing equation of moment equilibrium at point B , yields

Tension in the cable at supports A and B are

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 T_{C} = 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 y_{C}.

Taking moment of all the forces about support B ,

(1)

R_{ay} = 65+ 10y_{c}

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

R_{ay} × 3 - H_{a} × 2 = 0

⇒ H_{a} = 1.5 R_{ay} (2)

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

∑M_{C} = 0 R_{ay} × 7 - H_{a} × y_{C} - 50 × 4 = 0

Substituting the value of H_{a} in terms of R_{ay} in the above equation,

7 R_{ay} - 1.5 R_{ay} y_{C} - 200 = 0 (3)

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

(4)

Solving the above quadratic equation, y_{C} can be evaluated. Hence,

y_{C} = 3.307m.

Substituting the value of y_{C} in equation (1),

R_{ay} = 98.07 kN

From equation (2),

H_{a} = 1.5R_{ay} = 147.05 kN

Now the vertical reaction at D, R_{dy} is calculated by taking moment of all the forces about A ,

R_{dy }× 10 -100 × 7 +100 × 3.307 - 50 × 3 = 0

R_{dy} = 51.93 kN.

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

R_{dy} × 3 = H_{d} × y_{c} ⇒ H_{d} = 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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