Chain cantenary 2

[gorddrog]

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

 

[MiketheEngineer]

The actual answer is infinity - you will never get it perfectly straight.

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.

 

[BAretired]

Mike is correct. You cannot get it straight, but why would you want to?

BA

 

[rb1957]

that force is applied along the line between the two end points of the chain, no? (it's a long time since i've done this).

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 ??

 

[gorddrog]

Thanks for your answers. The reason I wanted to know the load to straighten the chain was to make sure we left enough clearance between the dock and a breakwall.
Thank You

 

[dik]

See the attached, may be of some help...

Dik

 

[msquared48]

The only way to get the chain straight is for it to be completely vertical.

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

 

[paddingtongreen]

It's quite an interesting question. Completely at rest, there would be 30' vertically down and 70' lying on the sea bed. An increasing force on the boat would move it and the chain would gradually go from infinite slope to 3 in 10, gradually supporting more chain as the hypotenuse lengthens towards, but not reaching, 100'.


Michael.
Timing has a lot to do with the outcome of a rain dance.

 

[rb1957]

"The reason I wanted to know the load to straighten the chain was to make sure we left enough clearance between the dock and a breakwall." ... not sure it works that way ? wouldn't you define a load and see where the dock moves ??

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 ???

 

[BAretired]

The chain weighs 1.41#/'. Its submerged weight is approximately 0.87*1.41 = 1.23#/', assuming steel weighs 489 pcf.

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