Meriam, Kraige & Bolton — Distributed Forces: Centroids and Centers of Gravity: Problem 5_138

⚡ Mecademy AIENG정역학 · ch5  Problem Statement Determine the weight per unit length of the 30-ft steel beam which will produce a maximum tension of 1500 lb in the cable. Additionally, find the minimum tension in the cable and the total length of the cable. Problem 5/138 (a) Weight per unit length of the beam 1. Formula: For a cable supporting a load uniformly distributed along the horizontal, the shape is a parabola . The horizontal span is the sum of horizontal distances from the lowest point to each support: . The vertical heights are related to these distances by: The weight is then: 2. Substitution: From the diagram, , , and . Determine and : Relate maximum tension (at support ) to minimum tension : w w y= x 2T 0 w 2 L L=x + A x B = x B x A h B h A w w= x A 2 2T h 0A L=30 fth = A 6 fth = B 3 ft x A x B = x B x A = 3 6 ≈ 21.4142 x + A x = B 1.4142x + B x = B 2.4142x = B 30 ft x = B ≈ 2.4142 30 12.426 ft x = A 30−12.426=17.574 ft T = max 1500 lbA T 0 3. Calculation: Now calculate the weight per unit length : 4. Result: ● Final Conclusion: The weight per unit length of the beam is . (b) Minimum tension in the cable 1. Formula: The minimum tension in a parabolic cable occurs at its lowest point and is equal to the horizontal component of the tension, . 2. Substitution: From the calculation in sub-question (a): 3. Calculation: 4. Result: ● Final Conclusion: The minimum tension in the cable is . (c) Total length of the cable T = A T = 0 1+ ( x A 2h A ) 2 1500 T = 0 1+ ( 17.574 2