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02-02ChapGere.0007

# 02-02ChapGere.0007 - 28,800,000 psi The cable consists of...

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Problem 2.3-15 The main cables of a suspension bridge [see part (a) of the figure] follow a curve that is nearly parabolic because the primary load on the cables is the weight of the bridge deck, which is uniform in intensity along the horizontal. Therefore, let us represent the central region AOB of one of the main cables [see part (b) of the figure] as a parabolic cable supported at points A and B and carrying a uniform load of intensity q along the horizontal. The span of the cable is L , the sag is h , the axial rigidity is EA , and the origin of coordinates is at midspan. (a) Derive the following formula for the elongation of cable AOB shown in part (b) of the figure: d 5 } 8 q h L E 3 A } (1 1 } 1 3 6 L h 2 2 } ) (b) Calculate the elongation d of the central span of one of the main cables of the Golden Gate Bridge, for which the dimensions and properties are L 5 4200 ft, h 5 470 ft, q 5 12,700 lb/ft, and E 5
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Unformatted text preview: 28,800,000 psi. The cable consists of 27,572 parallel wires of diameter 0.196 in. Hint: Determine the tensile force T at any point in the cable from a free-body diagram of part of the cable; then determine the elongation of an element of the cable of length ds ; finally, integrate along the curve of the cable to obtain an equation for the elongation d . Solution 2.3-15 Cable of a suspension bridge 86 CHAPTER 2 Axially Loaded Members B A O L 2 q y (b) (a) x h — L 2 — Equation of parabolic curve: dy dx 5 8 hx L 2 y 5 4 hx 2 L 2 B A O L 2 q y x h — L 2 — D B O L 2 q y x h — D H V B H B F REE-BODY DIAGRAM OF HALF OF CABLE © F horizontal 5 (Eq. 1) © F vertical 5 (Eq. 2) V B 5 qL 2 H B 5 H 5 qL 2 8 h H 5 qL 2 8 h 2 Hh 1 qL 2 ¢ L 4 ≤ 5 © M B 5 0 `~ A-PDF Split DEMO : Purchase from www.A-PDF.com to remove the watermark...
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