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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 âˆ’17 * 20 âˆ’10 * 7 âˆ’ 10 * 4 = 0

RÎ±y = 280/14 =20kN R_{ey} = 37 20 17kN.

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

R_{ay} *7 âˆ’ H * 2âˆ’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

R_{ay} H * y_{B} =0; y_{B} = 80/44.50 = 1.798 m

Similarly, y_{D} = 68/44.50 =1.528m

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 B,C and D one could calculate tension in different segments of the cable.

**Segment bc **

Applying equations of equilibrium,

âˆ‘F_{x} = 0 â‡’ T_{ab} cos Î¸_{ab} = T_{bc} cosÎ¸_{bc}

See Fig.31.4c

**Segment cd **

See Fig.31.4d.

See Fig.31.4e.

Segment de

The tension T_{de} 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 is located below left support by 1 m. Evaluate the reactions and the maximum and minimum values of tension in the cable.

**Fig. 31.6 Example 31.3**

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,

10(4x - x^{2}) = 10x^{2}/3 â‡’ x =25.359 m

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 cosÎ¸ = H . The maximum cable tension occurs at B and the minimum cable tension occurs at C where dy/dx = Î¸ = 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 y_{C}.

Taking moment of all the forces about support B ,

Ray =1/10 [350 + 300 +100_{y}c] (1)

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

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

Ray*3-Ha*2 =0

â‡’ H_{a}= 1.5R_{ay} (2)

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

âˆ‘M_{c} = 0 R_{ay} * 7-H_{a} * y_{c} -50*4 =0

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

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

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

y_{c}^{2}+1.833y_{c} -17 =0 (4)

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

y_{C} = 3.307m.

Substituting the value of in equation (1),

R_{ay} = 98.07 kN

From equation (2),

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

Now the vertical reaction atD,Rdy calculated by taking moment of all the forces about A,

R_{dy} * 10 âˆ’ 100 * 7 + 100 * 3.307 âˆ’ 50 * 3 = 0

R_{dy =}51.93 kN.

Taking moment of all the forces right of C about C , and noting that âˆ‘Mc =0,

R_{dy} 3 = H_{d}* y_{c}â‡’H_{d}=47.109 kN.

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30 videos|72 docs|65 tests

### Test: Cables Subjected To A Distributed Load

- Test | 15 ques | 30 min
### Test: Cables Subjected To Its Own Weight

- Test | 15 ques | 30 min
### Test: Bending Moment Diagram

- Test | 15 ques | 30 min
### Test: Shear Stress Diagram

- Test | 15 ques | 30 min

- Test: Cables
- Test | 15 ques | 30 min
- Cables
- Video | 02:32 min