Meriam, Kraige & Bolton — Distributed Forces: Centroids and Centers of Gravity: Problem 5_133

⚡ Mecademy AIENG정역학 · ch5  Problem Statement The Akashi Kaikyō bridge in Japan has a central span of 1991 meters, a sag-to-span ratio of 1 to 10, and a total static loading of 160 kN per lineal meter of horizontal measurement. The weight of both of the main cables is included in this figure and is assumed to be uniformly distributed along the horizontal. Calculate the midspan tension in each of the main cables. If the angle made by the cable with the horizontal at the top of each tower is the same on each side of each tower, determine the total compressive force exerted by each cable on the top of each tower. Problem 5/133 (a) Midspan tension in each of the main cables 1. Formula: For a cable subjected to a load uniformly distributed along the horizontal, the shape is a parabola. The midspan tension (horizontal component) is related to the span , sag , and load by the equation: 2. Substitution: Total load for two cables = . Therefore, load per cable is . Span . Sag-to-span ratio is , so sag . Substituting these values into the formula: 3. Calculation: T 0 C T 0 w T 0 Lhw T = 0 8h wL 2 160 kN/mw= = 2 160 80 kN/m L=1991 m 1:10h= L= 10 1 = 10 1991 199.1 m T = 0 8×199.1 80×1991 2 T = 0 × 8 80 = 199.1 1991 2 10× 199.1 1991×1991 Since : 4. Result: ● Final Conclusion: The midspan tension in each main cable is . (b) Total compressive force on the top of each tower 1. Formula: The total compressive force on the tower from one cable is the sum of the vertical components of the cable t