If the cable is supported by saddles with are stayed by wires inclined at 30⁰ to the horizontal , determine the forces acting on the towers. If the same inclination of back stay passes over pulley, determine the forces on the towers.

Subject : Structural Analysis 1

Topic : Cable & Suspension Bridge

Difficulty : High

$1. Support \hspace{1mm} reaction \hspace{1mm} calcualation:$

$\sum M_a=0$

$(30*80*{\frac{80}{2}})-V_b*80=0$

$v_b=1200KN$

$\sum F_y=0$

$V_a-(30*80)+1200=0$

$V_a=1200KN$

$Taking \hspace{1mm} moment \hspace{1mm} about \hspace{1mm} central \hspace{1mm} point \hspace{1mm} C$

$BM_c=0$

$-H*8-((30*40)*{\frac{40}{2}})+V_a*40=0$

$H=3000KN$

$Maximum \hspace{1mm} tension \hspace{1mm} occurs \hspace{1mm} at \hspace{1mm} support $

$T_{max}=\sqrt{\left(V_a\right)^2 + \left(H\right)^2 }= \sqrt{\left(1200\right)^2 + \left(3000\right)^2 }$

$T_{max}=3231.09KN$

$CASE \hspace{1mm} 1: If \hspace{1mm} the \hspace{1mm} cable \hspace{1mm} is \hspace{1mm} supported \hspace{1mm} by \hspace{1mm} saddle, \hspace{1mm} the \hspace{1mm} anchor \hspace{1mm} cable \hspace{1mm} T_1 \hspace{1mm} is \hspace{1mm} given \hspace{1mm} by,$

$T_1cos\alpha=T_{max}cos\theta$

$Also, \hspace{1mm} we \hspace{1mm} know \hspace{1mm} that, $

$H=T_{max}cos\theta$

$\theta=\cos\,inverse[\frac{H}{T_{max}}]$

$\theta=\cos\,inverse[\frac{3000}{3231.1}]$

$\theta=21.80^\circ$

$T_1cos30^\circ=3231.09*cos21.80^\circ$

$T_1 = 3464.13 KN$

Vertical Force on Tower is given by

$= T_1sin\alpha + T_maxsin\theta$

$3464.13sin30^o + 3231.09 sin21.80$

$T_1=2931.98KN$

$CASE \hspace{1mm} 2: If \hspace{1mm} the \hspace{1mm} cable \hspace{1mm} is \hspace{1mm} on \hspace{1mm} pulley \hspace{1mm}, the \hspace{1mm} vertical \hspace{1mm} force \hspace{1mm} on \hspace{1mm} tower \hspace{1mm} is $

$T_{max}(sin\alpha+sin\theta)$

$=3231.09[sin30^\circ+sin21.80^circ)$

$=2815.46KN$

$Horizontal \hspace{1mm} force \hspace{1mm} on \hspace{1mm} tower $

$=T_{max}[cos\theta-cos\alpha]$

$=3231.09[cos21.80^\circ-cos30^\circ]$

$=201.85KN$