Calculate the minimum central dip needed.

Subject : Structural Analysis 1

Topic : Cable & Suspension Bridge

Difficulty : Low

$Support \hspace{1mm} reaction \hspace{1mm} calculation: $

$\sum M_a=0$

$28.5*130*{\frac{130}{2}}-V_b*130=0$

$V_b=1852.5KN$

$\sum F_y=0$

$V_a-28.5*130+1852.5=0$

$V_a=1852.5KN$

$Horizontal \hspace{1mm} thrust:$

$Consider \hspace{1mm} part \hspace{1mm} CA$

$BM_c=0$

$1852.5*65-28.5*65*{\frac{65}{2}}-H*d$

$H*d=60206.25$

$H={\frac{60206.25}{d}} KN$

$T_{max}=5000KN (given)$

$T_{max}=\sqrt{\left(V\right)^2 + \left(H\right)^2}$

$5000=\sqrt{\left(1852.5\right)^2 + \left({\frac{60206.25}{d}}\right)^2}$

$d=12.96m$