Anchor shank strengths

[Neeves]

In Theory, Practice and theory are the same, in practice they are not.

Winter is drawing in for some of you and you will be looking for some mental stimulus to replace your time on the water!

I would value some help from engineers:

The lever arm of a shank can be theoretically represented by the lever arm formula:

We are assuming a load, horizontal to a set anchor with the load applied at 90 degrees to shank length. The load is to be applied in increments, not a snatch load (I'll come to that if this initial thread develops).

Load = width of arm (W) x thickness of plate (T) x thickness of plate (T) x yield stress (Y) all divided by length (L).

To provide a figure in newtons the answer is divided by 1/3 and to convert to kg divide again by 10 (approx). Measurements are in mm and MPa. Length is shank length, to fluke (so the lever length) and width is the 'thinnest' point, where the shank horizontal meets the shank vertical. For yield stress I am using either minimum (which is often quoted on spec sheets of steel) and/or typical if I know it. I am not looking for exactness - not only is this for a yacht its an anchor - there is no great precision needed! Its the same formula for inches and pounds, different constant used.

To calculate where the shank will bend, if it were a straight beam, one would normally use half of length (as that where is bends). So one would incorporate L/2 not L

But a shank is not a straight beam. It is composed usually of a thicker 'vertical' section attached to the flluke and a longer thinner section (horizontal) which is possible thicker than necessary at the end to accommodate the shackle hole.

In order to find the point of maximum stress I took a number of model shanks, I cut them from al plate, and measured where the bend, or fail point, occurred.

I find that for a Delta, type, shank the failure point is not at the mid point but closer to the shackle hole giving 1.43, rather than 2

This then gives and equation of Load = W x T squared x Y

all divided by 3 x 10 x 1.43 (or 43) x L

This would be load in Kg at which the shank would bend

Now I have checked this on a shank at it seems to be about right, but that does not mean I have not made two, or more, mistakes that cancel out!

I have also bent Spade and Fortress shanks and they behave differently.

Because they taper in 2 dimensions they appear to have been much more cleverly designed to allow load to be fairly even along the length of the shank and both when they bend have a very even curve (so there is no 'failure' in the middle or at 1.43). A Fortress should be easier to consider as it is a 'simple' (or very cleverly designed) tapered beam. The Spade is a Delta shank with the geometry smoothed out, but it is tapered but made much more complex as it is a triangular box (An Ultra is a later consideration!)

Can anyone comment on my initial formula

L= W x T squared x Y/43 x L

(I'm happy if someone can conduct the theory to ammend my 1.43 (or cut up their own shanks)!)

and can anyone suggest a formula to cope with the tapered Fortress shank? Does L/2 still work or is it L (or something in between?)

Finally I have a set of calculations for various anchors with Delta type shanks and I happen to have similar calculations by Peter Smith (on the Rocna). My calculations and his are in general agreement (and the Rocna comes out well - so this not to develop into a Rocna knocking thread! Nor any anchor knocking thread - dream on?) Once I have the formula I'll share the information I have, spread sheet, but my calculations are limited to those to which I have access and know the Yield Stress.

Jonathan

 

[Neeves]

In order for me to extend my spread sheet I would value knowledge of the Yield Stress of a Delta anchor. The dimensions are published, or were, on Lewmar's website but the closest I can get to steel quality is something like, Hi-tensile manganese alloy (which tells me nothing). I vaguely recall it being suggested (somewhere on YBW) that the steel was around 600-700 MPa (though the 700 MPa seems a bit high). Can anyone be a bit more exact - If the closest answer is, say 600 - 650 MPa - I can calculate on both figures with a qualification note.

Jonathan

 

[JumbleDuck]

Load = width of arm (W) x thickness of plate (T) x thickness of plate (T) x yield stress (Y) all divided by length (L).

Looks a bit out to me. Starting from

Bending moment at yield = Elastic section modulus x Yield stress

and

Elastic section modulus = second moment of area / maximum distance from neutral axis

I'd get

F x length = sigma_y x (width x thickness^3 / 12) / (thickness / 2)

So

F = sigma_y x width x thickness^2 / (6 x length)

All of which only works for a rectangular shank.

To calculate where the shank will bend, if it were a straight beam, one would normally use half of length (as that where is bends). So one would incorporate L/2 not L

I don't understand where you get that. Failure would be expected at maximum bending moment, which for a given force means at the longest lever arms, not half-way along.

 

[Neeves]

There are 2 lever arms.

There is a very strong vertical lever, the thick bit of the shank attached to the fluke - it bends, or twists - but not much. There is then the horizontal of the lever arm where it does bend, a lot.

So there are two 'ends' to the lever one at the end of the horizontal which is being supported by the vertical section and the whole is then fixed where it joins the fluke.

So my 1.43 factor is an experimental way to defining where the main failure actually occurs, which is in the horizontal of the shank. If the shank is a rectangular the failure point will be at the end (but having tried it - this might be theory but its not what I see in practice, though that might be for a variety of defects in holding a bit of metal in a vice and bending it!) If you look at most bent shanks (that are tapered), none occur at the end (no matter which shank) the maximum bend is a long way from where it joins the fluke. I am sure there is a nice neat way to arrive at 'my' failure' point but in the absence of education I did it experimentally - which might not be correct, hence the request. Really I'm trying to factor in the 'L' of the shank and the variation in width into one simple formula. I'm making the crude assumprion that many shanks are similar to a Delta - and having superimposed a number of shanks one on top of the other, Spade, Delta, Excel, this is largely correct. I've excluded the Manson, I do not have one of the right size to compare and I'm also excluding Bruce for similar reasons (and its geometry is very different in any case).

Appreciate your comments

Jonathan

 

[Neeves]

JumbleDuck

To explain the factor '2'. In theory if a beam is fixed at one end and a load applied to the free end and the load is applied as a point load then the stress is greatest at the fixed end, and the load along the beam increases linearly If the beam is fixed at both ends and the load is applied by moving (or loading) one of those fixed points then the point of maximum stress is in the centre of the beam. Much depends how you consider the function of a shackle (or swivel) but I have assumed, possibly incorrectly that the shank is like a beam with both ends fixed.

If you like, take a bamboo cane, hold each end in each hand and bend - the maximum stress is in the centre of the cane. Hold only one end and the cane will break, greatest stress, near the fixed end.

Much depends on how the beam, or shank is loaded.

 

[GHA]

Little bump in case you missed this thread...
http://www.ybw.com/forums/showthread.php?375505-Another-article-on-anchors
Nylon halyards?

 

[Neeves]

Nudge accepted (and not ignored), not sure why no-one else picked it up (including me! - stupid)

One wants to get it right - so one answer is to go to collective wisdom.

Jonathan

 

[Plevier]

JumbleDuck

To explain the factor '2'. In theory if a beam is fixed at one end and a load applied to the free end and the load is applied as a point load then the stress is greatest at the fixed end, and the load along the beam increases linearly If the beam is fixed at both ends and the load is applied by moving (or loading) one of those fixed points then the point of maximum stress is in the centre of the beam. Much depends how you consider the function of a shackle (or swivel) but I have assumed, possibly incorrectly that the shank is like a beam with both ends fixed.

If you like, take a bamboo cane, hold each end in each hand and bend - the maximum stress is in the centre of the cane. Hold only one end and the cane will break, greatest stress, near the fixed end.

Much depends on how the beam, or shank is loaded.

I couldn't understand your original post, congrats to JumbleDuck!
The anchor shank is definitely a cantilever. A shackle to the end is definitely a hinge joint.
So the shear force is constant at any point along it.
The bending moment increases linearly from the point of force application to the fixed end. So if the shank is of constant section, the fixed end is where it will bend. Like a nail in a plank if you pull the end of the nail sideways.
If it is not of constant section, to work out where it will bend and at what side load you'll have to calculate how the section modulus changes along it and compare with the bending moment at that point, as JumbleDuck indicates. It could be anywhere. Could be quite tricky if the shank section changes non-linearly and particularly if a fabricated hollow section.

 

[JumbleDuck]

To explain the factor '2'. In theory if a beam is fixed at one end and a load applied to the free end and the load is applied as a point load then the stress is greatest at the fixed end, and the load along the beam increases linearly If the beam is fixed at both ends and the load is applied by moving (or loading) one of those fixed points then the point of maximum stress is in the centre of the beam. Much depends how you consider the function of a shackle (or swivel) but I have assumed, possibly incorrectly that the shank is like a beam with both ends fixed.

If you have a beam which is built in at both ends and you displace one of those ends transversely relative to the other (which is what I think you mean by "the beam is fixed at both ends and the load is applied by moving (or loading) one of those fixed points") then the maximum bending moment is at the ends, but of different signs, and changes linearly (constant shear force) along the length, passing through zero in the middle. Maximum stresses are in that case also at the end(s).

That's because a single beam built in at two relatively displaced ends is identical to two cantilevered beams of half the length pinned together at their free ends and similarly displaced.

If you like, take a bamboo cane, hold each end in each hand and bend - the maximum stress is in the centre of the cane. Hold only one end and the cane will break, greatest stress, near the fixed end.

In your cane example you are applying a moment to each end, which will result in a constant bending moment and therefore constant bending stresses (and constant curvature) along the length of the cane. There is in that case no reason to expect failure to occur in the middle.

With all due respect, I think you need to learn quite a lot more about the analysis of beams before taking this any further.

 

[Neeves]

Hollow section - later, much! But it must be possible - or they (Spade and Ultra) were both developed by trial and error - trial and error must have been very hard work!

The shank section, of a Delta type shank, definitely changes non-linearly. Consequently, in the absence of anything else, trial and error seems an option. But on the basis of your and Jumbleducks comments my point loading was the right way to go. I had this nagging feeling over 2 fixed points and still have the same feeling - as to whether a shackle, under effectively point load, is a hinge?

Thanks for the input.

Jonathan

 

[Plevier]

Hollow section - later, much! But it must be possible - or they (Spade and Ultra) were both developed by trial and error - trial and error must have been very hard work!

The shank section, of a Delta type shank, definitely changes non-linearly. Consequently, in the absence of anything else, trial and error seems an option. But on the basis of your and Jumbleducks comments my point loading was the right way to go. I had this nagging feeling over 2 fixed points and still have the same feeling - as to whether a shackle, under effectively point load, is a hinge?

Thanks for the input.

Jonathan

Yes those calculations are possible. Standard engineering.
Yes it is a hinge. It cannot transmit any bending moment can it?
JumbleDuck's post a few minutes ago is also 100% right.

 

[bignick]

I don't understand where you get that. Failure would be expected at maximum bending moment, which for a given force means at the longest lever arms, not half-way along.

+1

I'd start with a diagram showing your structure and loads....
Then we'll all be working on the same load case and can choose our assumptions and simplifications.....

Nick

 

[Delfin]

+1

I'd start with a diagram showing your structure and loads....
Then we'll all be working on the same load case and can choose our assumptions and simplifications.....

Nick

Lots of knowledge here, so I'll direct my question to these experts. On another forum, the manufacturer of a new anchor being sold in the US, Mantus, did static modal stress analysis of different anchor shanks. They compared their mild steel (A36) to shanks made of A514 steel like Bisalloy 80. The purpose of the comparison was to reassure purchasers of a mild steel shank that it would perform about as well as a Delta, etc. They concluded from their analysis that the force to bend the mild steel shank of a Mantus was only 50% less than the force to bend an identical Mantus shank made of A514 (which they don't actually make, but say they will). I don't have the dimensional profile of the Mantus shank, but I can tell you that it is quite thin. My question is whether the chart they prepared below makes any sense. It appears that they are conflating the point at which the different shanks would start to bend to the point they would become permanently deformed (Yield vs. UTS), although I may be confused. So, does this chart strike you as an attempt to confuse rather than inform? Having misplaced my slide rule, I'm unqualified to answer the question myself but your thoughts would be most appreciated.

View attachment 36335

 

[bignick]

My reading of their chart is that unsafe = red, marginal = yellow, and presumably safe would be green (if there were any!).
The chart shows that only the Manson and Mantis hi-strength could (marginally!!) withstand the loads on a 40' boat in 20kts. All anchor shanks would be unsafe in '30kt winds'. Their Mantis 'mild steel' would fail at 300lb load, with a load of 530lb applied......?

All the above based on the fact that their calcs and underlying assumptions are correct....

 

[Neeves]

I posted this in answer to Delfin;s query on the other thread:

I'd go with GHA, the Yield Stress is when the shank will deform permanently the UTS is when it will break. I'd never thought about it but I guess that as soon as you start to apply a load it will start to bend, though this is going to be imperceptible for a long time. Different materials, metals, appear to have different ability to bend, apparently HT steels a much greater ability to bend without permanent deformation than mild steels. But I do not think this has anything to do with the Mantus data. Bending ability, and the ability to be elastic is a bit of a red herring (valuable - but not relevant here).

My take on the Mantus figures, based on the fact they use (declared) A36 Mild Steel with a min YS of 250 MPa and A514 with a min YS of 690 MPa means that the load under which they will bend is in the proportion to the different YS, or 2.76:1. It has never been mentioned any of the dimensions vary - so we are comparing like with like. Their quoted bending forces are 300lb and 600lb which suggests that either it should be 300lb and 828lb or 217lb and 600lb. However they may have been using typical YS. Equally they could be working on a YS of 345 MPa - but they have never suggested this quality of steel (this would then give them bending forces of 300lb and 600lb) However the fact they appear to be manipulating the data for their own anchor does suggest one must be careful in viewing the rest of the data (all of which are competitors to Mantus).

However there could be a whole host of reasons for the way the data is produced, which may be perfectly valid. Certainly the original work and illustrations were excellent and very impressive which does make one wonder how they derived the data.

But a 44lb anchor with a shank able to withstand only 133kg or 96kg does not look very impressive - but shank strength does not appear to be very high on people's priorities. Most people seem to look at price - and maybe slick marketing.

I cannot comment on the Boss nor Danforths but the Delta seems low. I am twitchy of the Boss shank - it does seem thin but the other dimensions may compensate (I've not had the opportunity to get up close and personal) and I never see genuine Danforths. Delta are not renowned for bent shanks, at all, and the figure quoted is not much more than 2 healthy men could impose on the shank (a healthy fit man is supposed to be able to exert his own body weight).

Jonathan

I think one critical factor has been raised by bignick - (which was why I started this thread in the first place!)

viz 'All the above based on the fact that their calcs and underlying assumptions are correct....'

I was trying to pin down the calcs - and ensuring no assumptions were left 'unturned'

Jonathan

 

[Neeves]

The formula I have used is as follows:

The yield force or distortion load is

1/3 Y x b d^2/L in Newtons or 1/30 Y x b x d^2/L in kg (I'm rounding to keep the formula simple)

I am using (^2) for squared as this seems to be what is used above.

This derived as follows:

Section Modulas, Z = b x d^2/6

Stress Max S = W x L/2Z = 3W x L/b x d^2

Load to reach yield Y W=(1/3) Y x b x d^2/L

Deflection of bar y = WL^3/(8 x E x I) = 3/2 W x L^3/ (E x b x d^3)

Tensile stress E = 3/2 W x L^3/ (y x b x d^3)

Where

Z = Section Modulas m^3
S = Max stress, N
y = deflection of bar, m
L = Length between load point and bend point, or end of lever. m
b = breadth or width of bar, m
d = thickness of plate, m
W = force on end of bar N
E = Tensile strength MPa
Y = yield strength MPa

There are 2 ways to measure L, length of lever arm.

Both need an assumption.

L can be the maximum length of the shank (multiplied by a factor to compensate where failure actually occurs)........(1)

L can be the length to the point where failure occurs........(2)

I measured the failure point to be at about L/1.43 where L is the length from the shackle hole to the point where the trailing edge of the shank is welded to the fluke.

Both 'L' should be the same but (1) is a more common measurement (but then assumes all Delta shanks are in proportion).

b, breadth or width of bar is the width of the plate (so not to be confused with plate thickness) at the 'end' of the horizontal of the shank, before it turns 'vertical' - this point is at about L/1.66 for a Delta shank.

So based on L being the measure from shackle point to fluke plate my formula becomes:

W = b x d^2 x Y/43 x L

Or Force (or load) to bend = width of plate x (thickness of plate)^2 x Yield Stress/ 43 x Length

If this needs more clarification with a drawing, let me know - it just takes a bit longer.

Jonathan

Correction, or addition. I am also assuming, which is what I did in the first place, that this is a cantilever arm, fixed at one end and that the load is applied as a hinge (or shackle). The scenario changes if one uses the 'Y' of a swivel - so simple rotating shackle.

And load is to be applied at 90 degrees, so maximum.

 

[Neeves]

My formula in the previous post assumes there is a weak point in the shank.

For a loaded cantilever beam of even dimensions the maximum stress is at the fixed end of the beam and the load at any point along the beam is a simple function of where one takes that point, half way along the beam the stress will be half the stress at the end.

If I now take a tapered beam such that I ensure the stress is the same all along the beam then when I load it the beam should bend in a perfect curve (I know some perfect curves but I'm not sure what a perfect curve on a beam is). With this perfectly tapered beam the beam should fail evenly, it should all bend permanently in the same way at the same time. How do you calculate stress? or is it the same as the cantilever beam formula b x d^2 x Y/L (ignore the 1.43 factor). Presumably to check for perfection, of taper, you would need to check stress at a number of points along the tapered beam to ensure they met the linear development of stress along length (half stress at end occurs half way along beam).

I ask as images of bent Ultra, Spade and Fortress look perfect, or even, curves.

Jonathan

 

[JumbleDuck]

The formula I have used is as follows:

The yield force or distortion load is

1/3 Y x b d^2/L in Newtons or 1/30 Y x b x d^2/L in kg (I'm rounding to keep the formula simple)

I am using (^2) for squared as this seems to be what is used above.

This derived as follows:

Section Modulas, Z = b x d^2/6

OK for a rectangular section being bent around the b direction.

Stress Max S = W x L/2Z = 3W x L/b x d^2

The factor of two is wrong.

Maximum stress = Bending moment / Elastic Section Modulus
S = WL/Z = 6WL/bd^2

Load to reach yield Y W=(1/3) Y x b x d^2/L

W = Ybd^2/6L

Still a factor of 2 out

Deflection of bar y = WL^3/(8 x E x I) = 3/2 W x L^3/ (E x b x d^3)

For a cantilever beam loaded at the end

y = WL^3/3EI and since I = bd^3/12, y = 4 WL^3/Ebd^3

Tensile stress E = 3/2 W x L^3/ (y x b x d^3)

E is the Young's Modulus for the material, not the tensile stress or, as you later describe it, the tensile strength. If you want the maximum tensile strength in the beam as a result of a tip deflection it's easier to combine

y = WL^3/3EI

and

S = WL/Z = WLd/2I (since Z = I/(d/2))

to get

S = 3/2 Eyd/L^2

Where

Z = Section Modulas m^3
S = Max stress, N
y = deflection of bar, m
L = Length between load point and bend point, or end of lever. m
b = breadth or width of bar, m
d = thickness of plate, m
W = force on end of bar N
E = Tensile strength MPa
Y = yield strength MPa

There are 2 ways to measure L, length of lever arm.

Both need an assumption.

L can be the maximum length of the shank (multiplied by a factor to compensate where failure actually occurs)........(1)

L can be the length to the point where failure occurs........(2)

It doesn't need to be fudged like this and in any case depends wholly on how the shank profile along its length. No general factor is likely to be of any use here.

I hate to pour cold water on enthusiasm, but as I wrote before, you really need to spend some time learning beam bending theory. Your confusion over E clearly shows that.

PS. If your loads are in newtons and your lengths are in metres (which they should be) then your stresses have to be in pascals, not MPa, or you're going to need correcting factors of 10^3 and 10^6 all over the place.

 

[JumbleDuck]

If I now take a tapered beam such that I ensure the stress is the same all along the beam then when I load it the beam should bend in a perfect curve (I know some perfect curves but I'm not sure what a perfect curve on a beam is). With this perfectly tapered beam the beam should fail evenly, it should all bend permanently in the same way at the same time. How do you calculate stress? or is it the same as the cantilever beam formula b x d^2 x Y/L (ignore the 1.43 factor). Presumably to check for perfection, of taper, you would need to check stress at a number of points along the tapered beam to ensure they met the linear development of stress along length (half stress at end occurs half way along beam).

I am not at all sure what you mean by "a perfect curve". For a simple tip-loaded cantilever the deflected shape is a parabola. If you want a circular arc then you could use the elastic bending equation in the form

M/I = E/R

and rearrange to get

I = MR/E = WLR/E

showing that you need I to increase linearly along the length of the beam. If the cross section is to remain the same shape, then each dimension would have to increase with the fourth root of proportional distance along the beam. In that case, though, the section modulus would not increase as fast as I so failure would take place near the tip. If your goal is to have the outer elements of the shank reach yield along the whole length simultaneously then you want to make sure that

Y = Wx/Z(x) where x is distance from the tip and Z(x) is section modulus as a function of x

Rearrange to get

Z(x) = Wx/Y

Again assuming a constant shape, that requires dimensions which vary as the cube root of proportional distance along the bar ie

d(x) = r(root) (x/L)^(1/3)
b(x) = b(root) (x/L)^(1/3)

In that case I would vary as

I(x) = I(root) (x/L)^(4/3)

which plugged into

R = EI/M = EI/Wx

give

R = R(root) (x/L)^(1/3)

is that sufficiently perfect a curve?

 

[bignick]

All the units in the table are in pounds (lbs). Nothing to do with stresses (psi) whatsoever.

Suggest Roark as a good place to start reading up on stresses and strains.

(ps. not aimed at Jumbleduck at all, who seems to understand exactly what is going on)