Cantilever beam with horizontal and vertical component point load at free end

[helisphere]

How do you calculate the stress in a cantilevered beam with a point load at the free end where the load has a large horizontal component loading the beam in tension as well as bending from the vertical component? And the beam bends far enough so it's slope will match the slope of the load at some point along the span.

I can see that beyond the spanwise point where the slope of the beam is equal to the slope of the load, the beam will simply be in pure tension but where I am having trouble is calculating the deflection and bending stress in the spanwise portion before the slope matches the load direction...

 

[Rothlisburger]

Can you attach a sketch or photo? If you're doing large deflections, you get non-linearities from the change in geometry which make the deflection calculations difficult to do analytically. If your deflections are not that great compared to the length of your beam, then you can do it by hand. In any case, a picture is worth a thousand words.

 

[helisphere]

Do I have to calculate this numerically and iterate through the geometry changes? Are there any free or at least cheap software applications that calculate this kind of beam loading?

The real problem is a distributed horizontal and vertical loading that varies with x (span) but I am trying to get the concept of how to solve this one in basic principle first.

 

[AlephZero]

Note, in your picture you appear to have two different (independent) nonlinear effects.

One is that the stiffness of the beam changes depending on its deflection.

The other is that the position of the load changes when the tip of the beam moves.

Both of these are standard features in the big commercial finite element packages like Nastran, Abaqus, etc.

Sorry, I don't have any experience with cheap or free FE software. The deflected shape issues, the usual buzzwords and keywords to look for are "geometric nonlinearity" or "large rotations with small strains". For the change in load position, look for "follower forces".

Note, if you are appliyng a distributed load, there are further issues of whether your "vertical" and "horizontal" loads should have fixed directions in space, or rotate to be normal and tangential to the deformed shape of the beam (for example, like a pressure load). A "full feature" nonlinear FE program will have both options available.

 

[1988ajk]

Personally I would convert the force you have in the diagram (top picture) into forces perpendicular to the beam x and y using trigonometry.

the y force you resolved can be used to find the bending stress

the x force you resolved puts the beam in pure tension, this can be used to find the tensile stress = F/A

 

[Rothlisburger]

Actually, both the x and y forces contribute to bending stress. Only the x force contributes to axial stress.

From the attached image, the net bending moment is M = T [b sin(theta) – a cos(theta)]. The axial force is just F = T cos(theta). The max normal stress on the cross section would then be sigma = M c / I + F / A.

The deflection is best done with finite element analysis. Google says there are some free FEA programs out there.

-David

 

[1988ajk]

This has got me intrigued now. I am not saying that Rothlisburger is wrong, I just ask the question;

what value should you use for 'a' as the deflection is unknown?

This is my personal opinion on how i would go about it...

I have attached a very rough picture to this, apologies i don't have a scanner at home.

I have scaled the force T using a free body diagram. From the picture this shows that the force F acts perpendicular to the beam causing it to bend, I always like to scale off in this manner as you can visually see if its correct. The force F creates a simple bending moment M, meaning bending stress can be calculated using the beam bending theory Rothlisburger stated.

However there is an axial stress involved. so to find that do the same method (free body dia) to derive a horizontal (x) component force perpendicular to the beam, sigma = F/Cross Sectional Area

then the total stress within the beam is the bending stress + the axial stress.

I welcome feedback to this as its my personal opinion, and one problem I have yet to come across at work.

Adam

 

[Rothlisburger]

Worst case scenario would be a = 0. Any deflection of of the end would tend to lessen the bending stress (because the T cos(theta) term opposes the T sin(theta) term).

 

[pongo38]

If deflections are small in relation to length then stresses are approximately N/A + or - My/I

 

[AlephZero]

This has got me intrigued now. I am not saying that Rothlisburger is wrong, I just ask the question;

what value should you use for 'a' as the deflection is unknown?

Your first post was the "correct" way to do this, assuming the behaviour of the structure is linear.

But the way I read the OP, it implies the behaviour is not linear. There is no easy way to find the correct deformed shape of the beam by a hand calculation. Once you have the deformed shape, you could find the shear force, bending moments, and stresses by hand, but any computer program that will find the displacements will do the rest "for free".

Some structures only "work" because of the nonlinear behaviour. For example, if you model a fan blade on a large jet engine as a beam, and apply the axial loads caused by the rotation of the fan, and the bending loads caused by the gas pressuire on the blade, the answer from a linear analysis says the blade would bend like a banana and the tip would move forwards 0.5 meters or more out of the front of the engine. That is obviously ridiculous. The actual amount of bending, because of the nonlinear behaviour, is only a few millimeters.

 

[Studiot]

Surely this is just a standard combined stress question, complicitated by the curvature of the cantilever?

To analyse try a differential element, apply the standard formulae and integrate along the (known) curve of the bar.

If the curvature is kept small then the standard formula alone will suffice.

https://www.physicsforums.com/showthread.php?t=416415&highlight=combined+stress

 

[1988ajk]

I think AlephZero summed the situation up perfectly,

However I followed your link Studiot and unfortunately the equation you posted I have never come across. further to this I downloaded the file 'chapter 8'.

The document backs up that my theory was correct where cutting T into very simple x and and y components and adding the axial and bending stresses together, equation (8.4) gives the total stress in the beam.

the notes i downloaded from the links are great reading, many thanks.

 

[1988ajk]

Studiot,

where is this material 'Chapter 8' from?

Just for my records, as there is some usefull information on here that I may need to reference in my future work.

 

[Studiot]

For combined axial and flexural loads, or loads applied at angles or eccentricities to the main axes of the beam, the resultant stress at any point is given by the algebraic sum of the axial and flexural stress at that point.

S = S_axial + S_flexural

S = -or+ P/A -or+ My/I

The axial stress will normally be uniform, whereas the flexural stress will vary with position.

The equation I posted in the referred thread was simply the application of this equation to the problem then to hand.

I am sorry my Mathtype is not currently so I can't prettify the equation.

I highlighted paragraph 8, not chapter 8 of the internet reference.

You should be able to find reference to combined stresses in any decent advanced strenght of materials book - I like Singer, Chapter 9 is devoted to it.

 

[1988ajk]

yes,the method is exactly as I posted in post 7 & 12. I shall check Singer out in the Library, Thanks.

 

[AlephZero]

Surely this is just a standard combined stress question, complicitated by the curvature of the cantilever?

To analyse try a differential element, apply the standard formulae and integrate along the (known) curve of the bar.

Yes, if you know the shape of the bar. But that is a big "if".

If the curvature is kept small then the standard formula alone will suffice.

Not necessarily, because the axial force in the bar produces an extra transverse stiffness term, exactly the same as the transverse stiffness in a guitar string is caused by the tension in the string. This affects the deflected shape of the bar.

Whether this effect is significant depends on the situation, but it some situations it can increase the transverse stiffness by an order of magnitude compared with the standard "beam bending" formulas.

(Of course the elastic transverse stiffness of a guitar string is zero, for all practical purposes, but piano designers do take the elastic stiffness of the metal "strings" into account as well as the stiffness created by the tension, especially for the thick bass "strings").

 

[helisphere]

Simply combining bending and axial loads only works when using the the small deflection assumption. This is obviously not the case here.

In this case, the free end of the beam, outboard of some unknown point, is not bending at all and therefore will develop no internal bending moment and no bending stress, it will only have axial load (tension) and F/A stress. That part is easy but the inboard section of the beam, however, is taking on a curve shape and is developing internal bending stresses. Sure, if the curve is known then the calculations are fairly straightforward as noted above but, the curve is not known.

The obvious thing about this beam is that it's quite flexible relative to the load magnitude. It's not as flexible as a string but it is leaning in that direction and the larger the load, the more the beam would deflect like a string. If it were to go to that extreme, the beam would be completely straight and lined up with the load except for a very small part right at the base attachment. It seems to me that the bending moment gets more and more focused at the root and less distributed spanwise.

I think that I have to solve this one numerically by guessing an initial curve and point of inflection for the beam then calculating the external bending moment based on the load and the internal bending moment based on the curve of the beam. Then either by guessing or by some numerical algorithm keep calculating a new curve and the new internal and external bending moments until they converge and are equal.

Unfortunately I think that might be a little beyond my ability. I am going to keep searching for a free or inexpensive FEA application. If anybody knows of one specifically, please advise.

Thanks for the replies

 

[pongo38]

If the force were compressive, there would be a line of thrust, and the difference between the line of thrust and the bent beam would be the bending moment diagram, to some scale. In compression problems the second order effects (increased bending moment due to deflection) might be worth computing, but, turning the problem round so that the force is tensile, you still have a line of tension not passing through the the encastre support, and whose distance from the bent member is to some scale the bending moment diagram. However, the deflection is less than in the compressive case because of the straightening effect of a tensile force. I think the member is only straight for an infinitesimally short length at the end, not a finite length as suggested earlier.

 

[Studiot]

Simply combining bending and axial loads only works when using the the small deflection assumption. This is obviously not the case here.

It's not obvious to me since I can find no details about the degree of curvature in this thread, which is why my remarks were general. If you were to supply some more details a more precise approcach could be sugested.

An alternative to FE might be to segment the bar as a series of free bodies, which individually are not overbent, apply the formula to achieve a series of simultaneous equations and solve these equations.

 

[raghu.mech]

This is an interesting problem.

The answer is strongly dependent on the physics of the problem. It is important to know what happens to the force as the beam deflects. If the direction remains invariant globally then they are best applied as point loads. If they act like follower forces they are to be applied as element forces - pressure. In large deformation analysis the element loads act on the deformed geometry.

The solution is obtained in most analysis packages by invoking geometric non-linearity option.

(I am presuming that a finite element solution is being sought. After re-reading the posts I am not sure whether my presumption is right.)

 

[helisphere]

Ok, many of you are asking for more details concerning the real problem. I posted it here, simplified, just to see if I might get a few ideas about solutions. Anyway, the real problem I am trying to solve is: can you make a rigid (helicopter) rotor blade out of wood, rigid hub, all bending stress resolved in the wood blade itself.

The horizontal component force is actually a distributed load of F = m*w^2*r (centrifugal force) where the blade is a NACA 63-015 airfoil but for simplicity you can probably just use an elliptical shape of 5 in by 0.75 in since it is a 5 in chord 15% thick. Specific gravity of the wood is 0.45, E is 1.63x10^6psi, compression/tension max stress is 7,200 psi. The lift Force (vertical force) is L= 1/2*rho*v^2*CL*C*r. I was figuring a max lift CL of 1.2, C = 5 in and I was using a 9 ft (108in) span, r for the radius. To find the actual forces you must integrate both equations with respect to r so dF= m*w^2*dr and m is a function of r and dL = 1/2 rho*w^2*r^2*C*CL*dr. Integrate both sides of both equations, blah blah blah, hopefully you get it...

Also, you could use a different wood that its mechanical properties might be better but again, the goal is to see if a rigid rotor blade made from wood is possible and feasible. Wood is natures great composite structure and it has a wonderful strength to weight ratio, especially when you consider that because of how light (and weak) it is you end up with a larger cross section to get the needed strength, typically, and when you consider the larger cross section's larger moment of inertia you get a strength of very high value and a weight of very low value...

Of course you have the option to strengthen the blade root with doublers. You will find that the tip of the blade (at max CL) wants to deflect to about 12 degrees and the root about 8 degrees with the distributed loads specified.

The horizontal load remains horizontal for any deflection but the vertical load technically is always perpendicular to the blade span. Although I think you could just calculate it as always vertical and it would be close enough...

Well it's more of an exercise because I am curious, but what the heck...

 

[lumeal]

i've a question for beam with finite difference scheme. beam with have forth order derivative is equal to distribution load,EI (d^4 w)/〖dx〗^4 =q(x). My problem is, the load is a point load at the center with free support. if in the equation, i remove the distribution load, then it will equal to zero. then how I'm suppose to have the point load in my equation? i can't solve the matrix if all is equal to zero.

 

[Rafa]

Only a question. Do the surface where the forze is applied bend? or this surface does not change. Because to seprate the force in horizontal with Fcos(alpha) and vertical (F sin (alpha) i will think in this surface or not??