Cables - 2 | Structural Analysis - Civil Engineering (CE) 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

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.