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For the cable loaded as shown in figure ME06-02 2m 2m 1.5m 2m 2m 3 kN 8 kN Figure ME06-02 6a. Find the tension in cable CD. b. 8.625 kN d. 3.546 kN a. 6.786 kN C. 4.815 kN 6b. Find the tension in cable AB. a. 5.80 kN C. 3.20 kN b. 6.90 kN d. 7.40 kN 6c. Find OBc. a. 25.60° C. 15.80° b. 32.27° d. 46.30°

For the cable loaded as shown in figure ME06-02 2m 2m 1.5m 2m 2m 3 kN 8 kN Figure ME06-02 6a. Find the tension in cable CD. b. 8.625 kN d. 3.546 kN a. 6.786 kN C. 4.815 kN 6b. Find the tension in cable AB. a. 5.80 kN C. 3.20 kN b. 6.90 kN d. 7.40 kN 6c. Find OBc. a. 25.60° C. 15.80° b. 32.27° d. 46.30°

Structural Analysis
Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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For the cable loaded as shown in figure ME06-02 2m 2m 1.5m 2m 2m 3 kN 8kN Figure ME06-02 6a. Find the tension in cable CD. a. 6.786 kN C. 4.815 kN b. 8.625 kN d. 3.546 kN 6b. Find the tension in cable AB. a. 5.80 kN C. 3.20 kN b. 6.90 kN d. 7.40 kN 6c. Find OBc. a. 25.60° C. 15.80° b. 32.27° d. 46.30°
Transcribed Image Text:For the cable loaded as shown in figure ME06-02 2m 2m 1.5m 2m 2m 3 kN 8kN Figure ME06-02 6a. Find the tension in cable CD. a. 6.786 kN C. 4.815 kN b. 8.625 kN d. 3.546 kN 6b. Find the tension in cable AB. a. 5.80 kN C. 3.20 kN b. 6.90 kN d. 7.40 kN 6c. Find OBc. a. 25.60° C. 15.80° b. 32.27° d. 46.30°
SUSPENDED CABLE - CATENARY note: A suspendedcable is assumed to be A CATENARY when the uniformly distributed load GENERAL CABLE THEOREM is assumed to be acting ALONG THE LENGTH If the weight of the cable is negligible, OF THE CABLE the horizontal component of tension at any point of the cable is constant. T2 M = Hy – key formula M= moment at any point of the cable due to vertical forces left/night of said point. H = honzontal component of tension at any point of the cable Y = bertical distance between line connecting the supports and the point considered y2 X1 X2 y = ccosh formula 1 or x = cinvcosh T= wy fomula 2 y? = c? + s? → fomula 3
Transcribed Image Text:SUSPENDED CABLE - CATENARY note: A suspendedcable is assumed to be A CATENARY when the uniformly distributed load GENERAL CABLE THEOREM is assumed to be acting ALONG THE LENGTH If the weight of the cable is negligible, OF THE CABLE the horizontal component of tension at any point of the cable is constant. T2 M = Hy – key formula M= moment at any point of the cable due to vertical forces left/night of said point. H = honzontal component of tension at any point of the cable Y = bertical distance between line connecting the supports and the point considered y2 X1 X2 y = ccosh formula 1 or x = cinvcosh T= wy fomula 2 y? = c? + s? → fomula 3
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