Examples of cables | Additional Study Material for Mechanical Engineering PDF Download

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

Rαy = 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 − 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 

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,

∑F_x = 0 ⇒ T_ab cos θ_ab = T_bc cosθ_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 - x²) = 10x²/3 ⇒ 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 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 y_C.

Taking moment of all the forces about support B ,

Ray =1/10 [350 + 300 +100_yc]             (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

⇒ H_a= 1.5R_ay            (2)

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

∑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_c²+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 − 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 ∑Mc =0,

R_dy 3 = H_d* y_c⇒H_d=47.109  kN.