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A suspension cable of 130m horizontal span is supported at the same level. It is subjected to a UDL of 28.5KN per horizontal meter. If the maximum tension in the cable is limited to 5000KN.

Calculate the minimum central dip needed.


Subject : Structural Analysis 1

Topic : Cable & Suspension Bridge

Difficulty : Low

1 Answer
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Figure


Supportreactioncalculation:

Ma=0

28.51301302Vb130=0

Vb=1852.5KN

Fy=0

Va28.5130+1852.5=0

Va=1852.5KN


Horizontalthrust:

ConsiderpartCA

BMc=0

1852.56528.565652Hd

Hd=60206.25

H=60206.25dKN

Tmax=5000KN(given)

Tmax=(V)2+(H)2

5000=(1852.5)2+(60206.25d)2

d=12.96m

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