written 6.8 years ago by | • modified 2.9 years ago |
The maximum dip in the cable from the left support is 4m. Find the maximum and minimum tension in the cable.
Subject : Structural Analysis 1
Topic : Cable & Suspension Bridge
Difficulty : High
written 6.8 years ago by | • modified 2.9 years ago |
The maximum dip in the cable from the left support is 4m. Find the maximum and minimum tension in the cable.
Subject : Structural Analysis 1
Topic : Cable & Suspension Bridge
Difficulty : High
written 6.8 years ago by | modified 2.7 years ago by |
To find l:
$\frac{l_{1}}{l_{2}}= \frac{h_{1}}{h_{2}}$
$l_{1} = \sqrt{\frac{h_{1}}{h_{2}}} * l_{2}$
$l_{1} = \sqrt{\frac{4}{1.5}} * l_{2}$
$l_{1} = 1.6329 l_{2}$
$l=l_{1}+l_{2}$
$25=1.6329l_{2}+l_{2}$
$l_{2}= 9.495m$
$l=l_{1}+l_{2}$
$25=l_{1}+9.45$
$l_{1}=15.505m$
Support reaction calculation:
$V_{A}=w*l_{1}=10*15.505=155.05KN$
$v_{B}=w*l_{2}=10*9.495=94.95KN$
$H_{A}=\frac{w*l_{1}^2}{2*h_{1}}= \frac{10(15.505)^2}{2*4}=300.51KN$
$H_{B}=\frac{w*l_{2}^2}{2*h_{2}}= \frac{10(9.495)^2}{2*1.5}=300.51KN$
$T_{A}=\sqrt{V_{a}^2+H^2}$
$T_{A}=\sqrt{(155.05)^2+(300.51)^2}$
$T_{A}=338.15KN$
$T_{B}=\sqrt{V_{b}^2+H^2}$
$T_{B}=\sqrt{(94.95)^2+(300.51)^2}$
$T_{B}=315.15KN$
$T_{max}=338.15KN$
$T_{min}=315.15KN$
a line pq 100 mm is inclined30 to hp and 45 to vp its midpoint is in the vp and 20 above the hp draw its projection if its one end p is in third quardrent and other end qis in first quarderent