Analysis by General Cable Theorem:

Because there is more than one applied load, this is the most efficient approximate analysis method.
Equivalent beam:

+\circlearrowright \sum{M_E} = 0 = A_y(20  ft)  –  2  k(18  ft)  –  1.5  k (13  ft)  –  2  k(3  ft)

A_y = 3.08  k

Find the internal moments:

+\circlearrowright \sum{M_{cut}} = 3.08  k (2  ft)  –  M_B

M_B = 6.16  k  •  ft

+\circlearrowright \sum{M_{cut}} = 3.08  k (7  ft)  –  2  k(5  ft)  –  M_C

M_C = 11.56  k  •  ft

+\circlearrowright \sum{M_{cut}} = 3.08  k(17  ft)  –  2  k(15  ft)  –  1.5  k(10  ft)  –  M_D M_D = 7.36  k  •  ft

Cable geometry:

From similar triangles:

\frac{h^′_B}{13  ft} = \frac{2  ft}{20  ft}          h^′_B = 1.30  ft

 

\frac{h^′_C}{13  ft} = \frac{7  ft}{20  ft}          h^′_C = 4.55  ft

 

\frac{h^′_D}{13  ft} = \frac{17  ft}{20  ft}          h^′_D = 11.05  ft

Calculate the sag at C:

s_C = 15  ft  –  4.55  ft = 10.45  ft

General Cable Theorem:
Horizontal component of cable times sag at C is moment from equivalent beam.

10.45  ft  •  H = 11.56  k  •  ft

H = 1.106  k

Find the vertical positions of the other loads:
General Cable Theorem:

s_B  •  H  –  s_B(1.106  k) ≡ 6.16  k  •  ft

S_B = 5.57  ft

s_D  •  H = s_D(1.106  k) ≡7.36  k  •  ft

s_D = 6.65  ft

Vertical positions:

h_B = s_B + h^′_B = 5.57  ft + 1.30  ft

h_B = 6.87  ft

h_D = s_D + h^′_D = 6.65  ft + 11.05  ft

h_D = 17.7  ft

This configuration results in point D being 2.7 ft too low, since the maximum drop of any of the suspended props is 15 ft.

Calculate the resulting cable forces anyway for comparison:
Lengths of cable segments:

l_{AB} = \sqrt{(2  ft)²  +  (6.87  ft)²} = 7.16  ft

 

l_{BC} = \sqrt{(5  ft)²  +  (15  ft  –  6.87  ft)²} = 9.54  ft

 

l_{CD} = \sqrt{(10  ft)²  +  (17.7  ft  –  15.0  ft)²} = 10.36  ft

 

l_{DE} = \sqrt{(3  ft)²  +  (17.7  ft  –  13.0  ft)²} = 5.58  ft

Cable forces:
Since the horizontal component is known for all segments,

F_{cable} = H (\frac{l_{seagment}}{w_{seagment}})

 

F_{AB} = 1.106  k(\frac{7.16  ft}{2  ft}) = 3.96  k

 

F_{BC} = 1.106  k(\frac{9.54  ft}{5  ft}) = 2.11  k

 

F_{CD} = 1.106  k(\frac{10.36  ft}{10  ft}) = 1.15  k

 

F_{DE} = 1.106  k(\frac{5.58  ft}{3  ft}) = 2.06  k

Summary:

Analysis with Maximum Sag at D:

Cable geometry:
Since the start and end points are not changed,

h^′_B = 1.30  ft

 

h^′_C = 4.55  ft

 

h^′_D = 11.05  ft

Calculate the sag at D:

s_D = 15  ft  –  11.05  ft = 3.95  ft

General Cable Theorem:
Horizontal component of cable times sag at D is moment from equivalent beam.

3.95  ft  •  H ≡ 7.36  k  •  ft

H = 1.863  k

Find the vertical positions of the other loads:
General Cable Theorem:

s_B  •  H = s_B  (1.863  k) ≡ 6.16  k  •  ft

s_B = 3.31  ft

s_C  •  H = s_C  (1.863  k) ≡ 7.36  k  •  ft

s_C = 6.21  ft

Vertical positions:

h_B = s_B + h^′_B = 3.31  ft + 1.30  ft

h_B = 4.61  ft

h_C = s_C + h_C^′ = 6.21  ft + 4.55  ft

h_C = 10.76  ft

Lengths of cable segments:

l_{AB} = \sqrt{(2  ft)²  +  (4.61  ft)²} = 5.03  ft

 

l_{BC} = \sqrt{(5  ft)²  +  (10.76  ft  –  4.61  ft)²} = 7.63  ft

 

l_{CD} = \sqrt{(10  ft)²  +  (15  ft  –  10.76  ft)²} = 10.86  ft

 

l_{DE} = \sqrt{(3  ft)²  +  (13  ft  –  15  ft)²} = 3.61  ft

Cable forces:

F_{AB} = 1.863  k(\frac{5.03  ft}{2  ft}) = 4.69  k

 

F_{BC} = 1.863  k(\frac{7.93  ft}{5  ft}) = 2.95  k

 

F_{CD} = 1.863  k(\frac{10.86  ft}{10  ft}) = 2.02  k

 

F_{DE} = 1.863  k(\frac{3.61  ft}{3  ft}) = 2.24  k

Summary:

Conclusion:
The configuration that meets the concert designer’s requirements has point D hanging at 15 ft. The peak cable force in this configuration is 4.69 k.

Observations:
The final configuration results in a 79% increase in tension in one of the cable segments and an increase of 0.73 k in another segment compared with the initially assumed configuration.
∴ The choice of the sag can have a significant effect on the resulting axial forces in the cable segments.