Lecture 9 part V

Lecture 9 part V - PART V FLEXIBLE CABLES 49 FLEXIBLE...

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49 PART V FLEXIBLE CABLES
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50 F LEXIBLE CABLES In previous introduction, cables were assumed to be straight two-force members capable of transmitting only axial tensile forces. When transverse loads are applied to a cable, it cannot remain straight but sags . Sag is defined as the difference in elevation between the lowest point on the cable and a support . The span of the cable is defined as the horizontal distance between supports. In the following, we assume the cables are perfectly flexible and inextensible. The bending resistance of a cable is neglected and the resultant internal force on any cross section must act along a tangent to the cable at that cross section . The central question for the cable problem is solve for the cable force T in an arbitrary section in terms of other known parameters. Cables subjected to concentrated loads If the concentrated loads are much larger than the weight of the cable, the weight of the cable can be neglected in the analysis and the segments of the cable can be considered as straight two-force cables.
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51 Fig. 8-12 Flexible cable supporting a system of concentrated loads Equations for FBD in Fig. 8-12 b ): 11 2 2 33 00 Ay MB a P x P x P x =⇒ = () 1 y B Px a =+ + 123 yy y FA B P P P + − − − = A PPPB =++− xx x FB A = ⇒− = (9) Equations for FBD in Fig. 8-12c): 0 x F 22 44 cos cos cos cos x x A TT T T B θ θθθ == = = = (10) The maximum tension force will occur in the segment with the largest angle of inclination . Such a segment must be adjacent to one of the two supports. The equilibrium equations 0 x F = for pins at supports A and B are
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52 11 4 4 cos cos xx AT BT θ == sin sin yy (11) Then 22 14 x yx y TA A TB B =+ tan tan x x A B A B θθ −− (12) If the
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Lecture 9 part V - PART V FLEXIBLE CABLES 49 FLEXIBLE...

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