NUMERICAL ON THE DETERMINATION OF TENSILE FORCES WHEN CABLE SUBJECTED TO THE CONCENTRATED LOAD

NUMERICAL ON THE DETERMINATION OF TENSILE FORCES WHEN CABLE SUBJECTED TO THE CONCENTRATED LOAD

A cable supported at its ends with span 40m apart carries a load of 20KN,10KN and 12KN at the distances of 10m, 20m and 30m respectively from left support. If the vertical distance of the point where the 10KN load is carried is 13m below the level of end supports. Determine 

1. Support reactions 

2. Tension forces at different parts of the cable

3. Total length of cable

Step 1: Determination of the support reactions

V a + V b = 20 + 10 + 12= 42KN…. (1)

Taking moment about support A

V b (40) –12 (30) -10 (20) - 20 (10) = 0…. (2)

Solving Eq 2

V b= 19KN

Substitute the value of V b in Eq 1

V a= 23KN

Taking moment about D

V a (20) – H (13) – 20(10) = 0

23 (20) – H (13) – 20(10) = 0…. (3)

Solving Eq 3

Therefore, H = 20KN

Step 2: Determination of dip distance Yc and Ye

Bending Moment about C 

V a (10) – H (Yc) = 0 (Since the moment is zero in cables)

23(10) -20(Yc) = 0

Yc = 11.5m

Similarly Bending moment about E

Vb (10) – H (Ye) = 0

19(10) – 20(Ye) = 0

Ye = 9.5m

Step 3: Calculation of tension forces

At node C                                                

Tan Ꝋ1 = Yc /10

Tan Ꝋ1 = 11.5 /10

Ꝋ1 = Tan-1(1.15)

Ꝋ1= 49⁰

Tan Ꝋ2 = (Yd – Yc)/10

Tan Ꝋ2 = (13-11.5)/10

Ꝋ2 = Tan-1(0.15)

Ꝋ2= 8.53⁰

Applying Sin rule

T1/ Sin (90 - Ꝋ2) = T2 / Sin (90 + Ꝋ1) = 20 / Sin (180- Ꝋ1+ Ꝋ2)

Equating any two terms 

T2 / Sin (90 + 49) = 20 / Sin (180- 49+ 8.53)

T2 = 20.21KN

Similarly 

T1/ Sin (90 - Ꝋ2) = 20 / Sin (180- Ꝋ1+ Ꝋ2)

T1/ Sin (90 – 8.53) = 20 / Sin (180- 49+ 8.53)

T1 = 30.47KN

At Node D

Tan Ꝋ3= (Yd -Ye) /10

Tan Ꝋ3 = (13-9.5) /10

Ꝋ3 = Tan-1(0.35)

Ꝋ3= 19.3⁰

Applying Sin rule

T2/ Sin (90 +Ꝋ3) = T3 / Sin (90 + Ꝋ2) = 10 / Sin (180- Ꝋ2- Ꝋ3)

Equating any two terms 

T3 / Sin (90 + 8.53) = 10 / Sin (180- 8.53- 19.3)

T3 = 21.18KN

At Node E 

Tan Ꝋ4= (Ye) /10

Tan Ꝋ4 = (9.5) /10

Ꝋ4 = Tan-1(0.95)

Ꝋ3= 43.53⁰

Applying Sin rule

T3/ Sin (90 +Ꝋ4) = T4 / Sin (90 - Ꝋ3) = 12 / Sin (180- Ꝋ4+Ꝋ3)

Equating any two terms 

T4 / Sin (90 – 19.3) = 12 / Sin (180- 43.53+19.3)

T4 = 27.59KN

Step 4:

Determination of length of cable

Total deflected length of cable = AC + CD + DE + EB

AC = 10 Sec Ꝋ1

AC = 10 Sec 49 

AC = 15.24m

CD = 10 Sec Ꝋ2

CD = 10 Sec 8.53

CD = 10.11m

DE = 10Sec Ꝋ3

DE = 10Sec 19.3

DE = 10.6m

EB = 10Sec Ꝋ4

EB = 10Sec 43.53

EB = 13.53m

Total deflected length of cable = 15.24+10.11+10.6+13.53

                                        = 49.74m