# cantilever beam with horizontal and vertical component point load at free end

by helisphere
Tags: beam, cantilever, component, free, horizontal, load, point, vertical
 P: 22 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...
 P: 6 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.
 P: 22 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.
P: 5,375

## cantilever beam with horizontal and vertical component point load at free end

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 dont 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.
 P: 23 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
 P: 6 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 Attached Thumbnails
 P: 23 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 dont 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 Attached Thumbnails
 P: 6 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).
 P: 667 If deflections are small in relation to length then stresses are approximately N/A + or - My/I
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 Quote by 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?
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.
 P: 5,462 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. http://www.physicsforums.com/showthr...ombined+stress
 P: 23 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.
 P: 23 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.
 P: 5,462 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 = Saxial + Sflexural 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.
 P: 23 yes,the method is exactly as I posted in post 7 & 12. I shall check Singer out in the Library, Thanks.
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 Quote by 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.
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").