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A bridge cable is suspended from towers 80m apart and carries a load of 30KN/m on the entire span. If the maximum sag is 8m, calculate the maximum tension in the cable.

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 Answer
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Figure 1


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

$\sum M_a=0$

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

$v_b=1200KN$

$\sum F_y=0$

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

$V_a=1200KN$

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

$BM_c=0$

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

$H=3000KN$

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

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

$T_{max}=3231.09KN$


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

Figure  2


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

$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 $

Figure 3

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

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

$=2815.46KN$

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

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

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

$=201.85KN$

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