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Chapter 3

Q. 3.2

Our firm has been hired to design the rigging for a concert tour. The set designer has presented us with some unique challenges. We need to support three large props in front of a draped, glowing canvas. The senior engineer has worked out with the set designer that we will support the large props from a single cable that does not interfere with the draped canvas.

In order to ensure enough room for the canvas and enough height for the large props, the senior engineer has set the maximum drop of the suspended props to be 15 ft. We initially assume it will be the middle prop, but we must check that.

Our task is to perform preliminary analysis of the cable to determine how far down the props are from the top frame, and to determine the tension in each segment of the cable.

E3.24
E3.25

Step-by-Step

Verified Solution

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.

E3.26
E3.27
E3.28
E3.29
E3.30
E3.31
E3.32
E3.33