Tapered beam Deflection 5

[TP29]

Hello,
I'm hoping someone here can help. I have a question concerning my dinghy mast that have been puzzling me for some time now and I have been unable to find any reference book or website that covers this topic. I am a Chartered Engineer, but not a structural one, so know the basics of structural theories but confess that this is beyond me.

If a sailboat mast of uniform section is unstayed then it can be considered as a cantilever beam and I am able to calculate the deflection using simple beam theory (assume for now that the sail applies a UDL to the mast). How do I calculate the deflection, but when the mast section is uniformly tapered (so that the 'I' value decreases linearly from root to tip)? Clearly the deflection will be greater - can I replicate this by simply applying a uniformly increasing load to a non-tapered section? If not, how else can I calculate the deflection? It must be possible to do this using hand calcs rather than FEA, surely. For the sake of detail, I should add that the mast section is a hollow cylinder and that the taper applies in all axes iso that the section remains round. I guess this makes it a high aspect ratio frustum of a cone.

Many thanks in advance for any suggestions, but be warned - if I get a good response I may pose another question!

 

[WillisV]

I am not familiar with any easy closed form simple method of calculating deflection for a continously tapering beam. Most hand methods would top out as far as being reasonable with a few different steps in moment of intertia. A computerized numerical method approach would be best.

 

[broekie]

From the AASHTO Standard Specifications for Structural Supports for Highway Signs, Luminaires, and Traffic Signals page B-6, the formula for the deflection of a tapered tubular cantilevered beam under a uniform load is as follows:

y = w * L^4 / (7.872EI * (db/da)^3.282)

where:

I is the moment of inertia at the free end
db is the diameter at the fixed end
da is the diameter at the free end

Hope this helps. Not sure if I'll be able to handle another question, but give it a try.

 

[kxa]

I guess if you knew the deflection already you could work it backward to find equivaqlent linear load over constant x-section mast (average diameter). The only way I know is to break it up in segemnts and assume each segment has a uniform x-section.

 

[transmissiontowers]

The MOI is not a linear function, it is dependent on the ID^4 and the OD^4. If the mast thickness is constant it is a little easier problem, but if you know the ID as a function of length and the OD as a function of length, you can do a spreadsheet to break the mast into 1 foot pieces and calculate the average OD and ID of that section and calculate the deflection of each 1 foot segment, then add them up to get the total deflection. This method neglects the P-Delta effect where the deflected shape gets more deflection because the top vertical load is not over the base but has a moment arm and increases the deflection. Depending on the vertical tip load the P-Delta effect is usually less than 10%.

_____________________________________
I have been called "A storehouse of worthless information" many times.

 

[prex]

The basic equation for deflections of beams in bending is
y''=-M/EJ.
For a cantilever with UDL, if you take x=0 at the beam tip, you have M=wx²/2
For a round thin hollow section J=[π]tR³. So J doesn't vary linearly: if the thickness is constant then the variation will be with the cube of x, if the section area is constant, it changes with the square of x.
Let's take the first case: you may write
R=R_t+(R_r-R_t)x/L
Now
y''=-wx²/2E[π]t(R_t+(R_r-R_t)x/L)³
This can be integrated analytically on

http://integrals.wolfram.com/index.jsp

but the result is too long to be reported here (contains the inverse of a linear form and a log times a linear form), or can be integrated numerically by using a spreadsheet.
I hope this gives at least a picture of the problem you are facing, that is really solvable by elementary means, though some patience is needed.

prex

http://www.xcalcs.com

Online tools for structural design

 

[UcfSE]

I was going to have you write an equation for moment of inertia in terms of the location along the mast. Then you will have to put that into your equation for d(delta) and integrate to find your deflection. Once you have your equation you can put write it in terms of a definite integral and use a calculator, Mathcad, MATLAB, or perhaps an online solver to find the deflection. Going back to basics is really the only way I know to solve this unless you can find it already given in another source such as those mentioned above or perhaps in Roark's Formulas for Stress and Strain.

 

[rb1957]

M(x) is a function of load, unchanged with the tapered mast.
v(x), being the slope of the beam, is the integral of [M(x)/EI] for a constant I, but in your case I is a function of x, so v(x) = integral[M(x)/EI(x)]
now M(x) is going to be proportional to x^2 (for a UDL)
maybe you can linearise I(x) ... i'd think this would be close enough given your assumption of loading, maybe piece-wise linear.

btw, i think a UDL is very simplistic assumption. the load from the sail is probably more proportional to the sail chord and leech tension (from the vang). and how are you accounting for the shrouds (which restrict deflection of the mast, creating a node in it's displaced shape)

good luck, i think you can solve this by hand methods

 

[SlideRuleEra]

Here is one iterative hand solution. Consider the tapered cylinder as short segments of different (constant) diameters, all stacked one on top of the next to simulate the complete structure. Compute the deflection at the top of the bottom cylinder (cylinder with the largest diameter), here is the catch - compute the slope of the deflection curve at the top of this bottom cylinder. The next cylinder (with the second largest diameter)in not plumb but is "leaning" at the angle just computed. Do the deflection same calcs for each cylinder in turn and do the trig to compute the "offset" of each cylinder because of the cumulatively increasing "lean". The sum of all cylinder deflections and all offsets can be a reasonable approximation.

broekie's formula is sure a lot better.

http://www.SlideRuleEra.net

 

[TP29]

Thnaks guys - really great help! Who would have thought the answer could be found in a standard on road signs! I think I might try splitting the mast into a number of differing sections as per SlideRuleEra, just to check. I agree the loading assumption is very simplistic, it is approximately proportional to the chord length - but I didn't want to comlicate things too much at this stage. No shrouds though, so that bit is simpler.

Anyway, with such good answers its time for the next question...

I was considering stiffening the mast to prevent it bending too much in gusty winds. I know I could use shrouds (stays) to hold the mast rigid or just increase the sectional inertia but I want to do it differently.
Do you know what diamonds wires are? They are just tensioned wires that run from the base of the mast to a point higher up the mast, say 70% of the length. At the midpoint of the wire (so 35% of the mast height) a spreader bar, perpendicular to the mast, holds the wire out so that you have what looks a bit like a suspension bridge on its side. There is one either side of the mast, so that they form a diamond. Simple.
What I can't work out is how to calculate how much stiffer the mast is with these attached. Obviously, wire tension and spreader length will increase the siffness, but how? Is it just a case of the spreader bar applying a point load equal to the tension in the wire (resolved through the diamond angle), which opposes the applied sail force?

 

[rb1957]

now i think you're getting complicated !

diamond wires work (i think) because of their tension. the diamond connects to two points on the mast (i know you know, but it helps explain my subsequent waffle). these two points form the major diagonal of the diamond. consider one point remains fixed, and the other point deflects. this'll strain the two diamond wires connected to it, increasing the tension in one, reducing it in the other (hence the preload in the wires, you wouldn't want one of the wires going slack. also the spreader (the minor diameter of the diamond) is going to rotate with the mast's deflection, which'll strain the wires some.

maybe you can solve this iteratively, tho' i think FEA would be much easier (and non-linear at that). consider the mast without the diamonds, get the deflected shape, the deflection at the apex of the diamond (where it attaches to the mast) and the slope where the spreaders attach. now look at the diamond with these enforced displacements; one corner is fixed, the two spreader points move (oppositely) and the thrid corner deflects (to the side). the lengths of the diamond wires will all change, causing changes in the internal loads in the wires (increasing the tension in some, reducing it in others). i think this'll produce, effectively, a restorative couple to the mast (a moment opposite the the deflection from the loading). apply this couple to the mast, recalculate the displacements and slopes, and iterate !

an interesting problem, good luck

 

[mrMikee]

It's been a long time since I solved something like this by hand and even longer since I wanted to. There's a free FEA evaluation program (linear) from cadreanalytic.com that is actually quite good. The Lite version at $142 has tension only members which might be handy here.

It's not that I want to turn the discussion towards analysis software (that's been talked about often enough) it's just that I don't want anyone to suffer needlessly.

Just an idea.

Regards,
-Mike

 

[transmissiontowers]

In the AASHTO formula y = w * L^4 / (7.872EI * (db/da)^3.282)

the "uniform load" w is not uniform since the mast is tapered, the wind is on the diameter which is tapered if the wind on the mast is the thing that causes the moment and deflection. I'm not a sailor, but isn't the sail connected to the mast with a series of ropes which would put discreet load points on the length of the mast?

_____________________________________
I have been called "A storehouse of worthless information" many times.

 

[GregLocock]

The lateral load up the mast is very non uniform, and the poster who wondered about the standing rigging also had a very good point.



Cheers

Greg Locock

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[TP29]

There are many different mast/sail configurations but my particular mast is not supported by standing rigging. The sail is continuously connected to the mast from top to bottom. A constant chord sail (i.e. rectangular) will actually have an elliptical load distribution due to air pressure leakage at the ends. In reality most sails are triangularish, so even less force is generated at the top due to less sail area. In addition, vang and sheet loads apply additonal loads on the mast, though fore and aft not sideways. However, I was struggling with the simple UDL situation, so I thought I didn't want to complicate matters!

Many thanks for all the words of wisdom, it seems these aren't such trivial problems after all...

 

[TP29]

Thought I'd report back.
The AASHTO formula only works for relatively thick wall thicknesses. When the wall thickness:diameter ratio is less than about 1:8 two things happen. Firstly, the taper has very little effect on deflection when compared to an untapered section (for same fixed end section size). Secondly, reducing the free end daimeter/MOI actually decreases the deflection, giving very curious results. I suppose that teaches me a lesson about understanding the source and limitations of empirical data! Extrapolation is always dodgy...

 

[StephenA]

As a first approximation the I of the diamond wires could be added to the I of the mast. The centroid remains the same. It works so long as the wires remain in tension.

It is not perfect because there is no horizontal shear connection between the points of connection.

 

[rb1957]

this website may help in solving the integration ...

http://integrals.wolfram.com/index.jsp

the loading (UDL or varying DL) leads to a simple enough moment curve, and the MoI expression is straight-forward too, but integrating them probably isn't pretty.

still, given the UDL assumption, i think linearising MoI, at least peicewise, should be accurate enough. for the diamond wires, i think drawing the deformed diamond will illuminate the wire strains/stresses/loads.

 

[broekie]

Sorry, everyone. My mistake on the previous empirical solution to the problem. That equation is for solid round cantilevered beams. From page B-5 of the same manual, here is the equation for a uniform load for a tapered hollow section:

y = w*L^4 / (2E*pi*t*(Rb-Ra)^4) * (3*Ra*(-ln(Rb/Ra) - Ra/Rb + Ra^2/6/Rb^2 + 1/2) + Rb)

For a triangular load,

y = W*L^3 / (6E*pi*t*(Rb-Ra)^5) * (12*Ra^2*ln(Rb/Ra) + Ra*(8*Ra^2/Rb - Ra^3/Rb^2 - 8*Rb) + Rb^2)

where:

w = uniform load
W = total load
t = wall thickness
Ra = radius measured to wall centerline at free end
Rb = radius measured to wall centerline at fixed end

Sorry about the confusion my first answer may have caused. Probably should read the fine print a little more carefully.

 

[bjb]

Design of Welded Structures by Blodget has a method for calculating the deflection of a beam with non-uniform section. It is based on the area-moment method, and you basically divide the beam into segments. Software would probably serve you better, but being able to approximate the deflection by hand allows you to check the software output.