Chain cantenary
Chain cantenary
(OP)
Hello
We are securing a dock to the bottom of a bay and would like to know how much force would be required to straighten the chain.
The chain is 100' of 3/8, grade 30 which weighs 1.41 pds\ft,
141pds dry ,the water is 30ft deep and the anchor is a 3000pd block of concrete that sits basically flush with the bottom.
Could someone help us with this calculation.
Thank You
We are securing a dock to the bottom of a bay and would like to know how much force would be required to straighten the chain.
The chain is 100' of 3/8, grade 30 which weighs 1.41 pds\ft,
141pds dry ,the water is 30ft deep and the anchor is a 3000pd block of concrete that sits basically flush with the bottom.
Could someone help us with this calculation.
Thank You






RE: Chain cantenary
For example, if it weighs 1.41 lbs/ft it will take about 3525# to get it within 6'', double that for 3'' and on and on and on.
Just google catenary and you can find the simple formula that describes this.
RE: Chain cantenary
BA
RE: Chain cantenary
and the chain cannot become precisely straight 'cause the weight is acting on the chain, deflecting it. (if it was a weightless chain, it could become straight).
why are you concerned about the load to straighten the chain ? can the dock move that far ??
RE: Chain cantenary
Thank You
RE: Chain cantenary
Dik
RE: Chain cantenary
But then it couldn't take any lateral load in that position, so why use the chain?
Mike McCann
MMC Engineering
Motto: KISS
Motivation: Don't ask
RE: Chain cantenary
Michael.
Timing has a lot to do with the outcome of a rain dance.
RE: Chain cantenary
i think mike got a point ... initially the chain is 30' straight down and 70' lying on the bottom of the lake. now move the dock, what shape is the chain ?? inclined above the bottom of the lake ? i think it'll have something to do with the mud on the bottom of the lake resisting the chain being pulled up ???
RE: Chain cantenary
The diagonal length is 100', the depth is 30', so the horizontal distance is 95.4' if the chain is absolutely straight (discounting any elastic strain as a result of infinite tension). The angle is 17.45 degrees from horizontal.
If we approximate the catenary to a parabola (it will be fairly close). So, M = 1.23(95)^2/8 = 1387#. With a tension of 1000# on the slope or 954# horizontal, the sag is 1387/954 = 1.45'.
The internet has more accurate formulae available for a catenary but I'm too lazy to apply them.
BA