since the curve is a catenary, y = a cosh(x/a)
we have the arc length of half of the curve is
∫[0,20] √(1 + y'^2) dx = 22.7
y' = sinh(x/a), so √(1 + sinh^2(x/a)) = cosh(x/a)
∫[0,20] cosh(x/a) dx = 22.7
a sinh(x/a) [0,20] = a sinh(20/a) = 22.7
a = 22.6589
so the curve is
y = 22.6589 cosh(x/22.6589)
y(20) = 32.0736
So the maximum sag is 32.0736 - 22.6589 = 9.4147 m
A cable 45.4 m long is carrying a uniformly distributed load
along its span. If the cable is strung between two posts at the same level, 40 m apart, compute the smallest value that the cable may cable may sag.
1 answer
1 answer