Estimating Effect of Cantilever on Span

If there are cantilevered ends (at 0M and 6M), then the six meter beam is not simply supported at the 1M and 5M points.

You can assume whatever you want, but such assumptions are not valid structurally speaking.

A cantilevered beam cannot also be a simple beam. Two different approaches. The supports at 1M and 5M are probably non-bearing or what seems to be cantilevered at 0M and 6M, not so.

Perhaps this beam is not a structural member at all, but merely for show. By this I mean, it is not carrying any real structural load.

Thanks eriew. I think we are at cross purposes though. There is no other support except at 1m and 5m, the 1m sticking out is the cantilever.

Thank you very much ba/ael. That's a good, solid, useful answer. I can see the sense in how you worked it and when I have a few free moments I might play around with a bit of paper and a calculator and see whether or not a rule of thumb comes out of it.

Gas is my area so I don't play much with structural figures but I do like to know how everything holds together and it was bugging me not being able to come up with an answer to a fairly simple discussion.

I'm not sure why a rule of thumb is required as the calculations are pretty simple as is. If you move the supports to 1.5m and 4.5m respectively, then the negative moment is 1.125w over each support.

The span portion of the beam becomes 3m with a simple span moment of 1.125w. Combining all loads, the mid-span moment is 0 for the continuous beam. This is what you would expect because if you cut the beam at mid-span, each half is balanced over its own support with a double cantilever of 1.5m each side. For this case L_e = 0.

That's true ba/ael. The only reason I chase rules of thumb is that they are useful to picture things that you don't work with every day. In my field, for example, I know that the energy produced from burning LPG or NG is much the same if you look at the amount of air used. I don't need to remember all the figures (although I do, because it's my business). I know that when 1 cubic metre of air is mixed nicely with gas it'll produce a bit less than 4MJ.

I haven't had a chance to play with your figures yet but already I think the point is obvious that the effect of the cantilever disappears rapidly as you reduce it. I'll remember that but I won't remember the calculation to prove it unless I have cause to use it regularly. I suppose the point is that somewhere in the past I would have known how to calculate this but the only structural stuff I know now is basic rules such as if you increase the depth of a beam, it's strength goes up by the square of the increase but the deflection goes down by the cube. Sort of stuff that sticks because once you know it, it makes sense.

Okay, nutwood. You probably know a heck of a lot more about structural engineering than I know about the burning of LPG or NG. And I agree with you that it is useful to have a few rules of thumb to help get a handle on unfamiliar situations. Occasionally, however, rules of thumb can get you in hot water...but then, you won't have that problem as you are not designing any high rise buildings based on them.

Best regards

That's just it. You don't use this stuff to design buildings on but when when a discussion arises over lunch, as did the prompt for this post, you have an idea about what's what. When there's actually dollars and lives at stake you go a bit deeper. Having said that, the converse is also true, that when there's dollars and lives at stake, it's also good to be able to look at a design and say "that's BS". I've done it a couple of times in my life and it's really good when someone comes back, mopping the egg of their face, and say's "whoops,I missed a zero".

It's a belief of mine that the ability to look at something and tell whether it is right or wrong is becoming a lost art.

Hello Nutwood

I tried to find a case in Roark to cover this but surprisingly there isn't one (not in mine anyway, which is a copy of I think Issue 5). .

So working it out from scratch I make max bending moment (at centre) 17160N.m. BM at the supports -5721N.m. BM = zero somewhere in between but I don't suppose you're interested in just where.

Without the end cantilevers (ie simple supports, 4m span, same load per metre) max BM = 22880N.m. To give same max BM as above, 17160N.m, estimated length = 3.464m.

Agrees with ba/ael, I like his idea of using superposition, I didn't think of that.

Cheers.....Codey

Codey,

"BM = zero somewhere in between but I don't suppose you're interested in just where".

You have already determined just where in your next paragraph. The points of zero moment, otherwise known as points of inflection occur 0.268m from each support:

i.e. (4.0 - 3.464)/2 = 0.268

Best regards,

According to my coy of Blodget, for a beam of length L between supports with A overhang carrying a uniform load of W:

M(support)=(W*A^2)/2

Def(center)=((W*L^2)/(384*E*I))*(5*L^2-24A^2)

Perhaps this image might help.

Draw the bending moment diagrams by considering the cantilevers as pure cantilevers and the beam in the middle as simply supported. qL²/2 and qL²/8 (or PL/2 and PL/8 if you use overall load)

Take the bending moment curve that you have for the simply supported portion and "hang" it to the bending moment for the cantilevers. You now have two cross over points with the horizontal axis which are the points of inflexion of the moment diagram. This gives a remarkably good approximation to the distance between points of inflexion (3m about), even when you are no good at art. It constitutes a rough check which I think was what you were after. You can make it more precise by doing a more accurate graph of the parabolic shapes.

This works very well for the symmetrical loads/cantilevers but requires a little more graphic thought for unsymmetrical loads/cantilevers. Also it helps being a statically determinate problem but don't let that put you off using graphical methods for more complex problems.

It would be interesting to see how your rule of thumb might vary for varying ratios of cantilever. Perhaps a rule from that could be established. Thinking about it graphically, the greater the centre span in proportion to the outer cantilever, the steeper the graph and so the less the "balanced cantilever" rule applies.

The deflections are quite another kettle of fish.

Well, thank you to all who have helped with my simple little question although, with the best will in the world, I can't picture omw7's image. My brain obviously doesn't work that way!

My thoughts are that with my beam there is a support point at about the 1.24m mark where the bending moment over the supports is the same as at the centre of the beam. If that's rounded to 1.2 m, we now have 3.6 m for the centre span so it's possible to conclude that having aprox 1.5 times the cantilever behind it, is a "balanced" structure. As you move away from this in either direction, the bending moments start to differ from each other at an exponential rate.

Does the above make sense or am I typing rubbish? Also would it follow that if the beam was supported at the 1.24m mark (and the 4.76m of course), the deflection at the end of the beam would be the same as at the centre

The "balanced" moment occurs when the cantilever moment is equal to one half the simple span moment. That is, when C²/2 = L²/16, C and L being the cantilever and span length respectively.

Solving this equation, C = L/√8 = 0.3536L

Also, 2C + L = 6m

So L = 3.515m and C = 1.24m, confirming your thoughts expressed in your last post.

Deflection, however will not be the same at the end of the cantilever and the mid-span of the beam. But that is another story.

This has been an interesting discussion for me. I now have a mental picture of the stresses in a cantilever.

I note that it's been said twice now that the deflections are another story. Can deflection be attacked in this sort of manner or does the nature of the beam start to have an effect? I know in simple spans deflection behaves itself and conforms to basic rules.

With regard to the above example, I'd imagine that with my 6m beam supported at the 1.24m mark, the end of the cantilever would droop more than the centre of the beam. Perhaps that's why, in the original case under discussion, the beam was supported 1m in from the ends?

Deflection can be calculated quite easily for any situation. In the present case, let us assume that the only load on the beam is the uniform load on the two cantilevers. Then, the deflection at the ends of the two cantilevers will be equal, but the deflection at midspan of the beam will be negative, that is, it will rise up. The rotation of the beam will be equal and downward at the end of each cantilevers.

If we now superimpose a uniform load on the span portion of the beam, the mid-span will deflect downward and the ends of the cantilevers will deflect upward, but can the rotations be equal? The slope of the beam at midspan must be zero (by symmetry). In other words, the beam is horizontal at midspan. Can the slope of the ends of the cantilevers be zero too? Well, for some ratio of C to L, the answer is yes! But if it is, the deflections are obviously not equal. If you don't believe me, draw a sketch of the deflected shape...with zero slop at midspan and zero slope at the ends of the cantilevers. You should see that the end of each cantilever must be higher than the midspan of the beam.

In order for the deflections to be equal, the shape of the deflection curve must be symmetrical. But it is not! The slope at midspan is zero. The slope at the end of each cantilever is downward (the ends of each cantilever droop downward). Thus, the deflection at the end of the cantilever is not, in general, the same as the deflection at midspan of the beam.

In general, if equivalent span is defined as the span having the same simple span moment, then it will not necessarily (and probably infrequently) satisfy the condition of having the same deflection.

Thank you ba/ael. I can picture what you are saying. The only bit I found puzzling was the last sentence. You seem to be saying that deflection sometimes will be equal in an "equivalent span" situation. I would have thought that the ends of the cantilever would always have a greater deflection than the span centre under this condition.

I hope you don't mind this drawn out discussion? It's interesting to me but I don't want to presume on your good nature too much.

Perhaps I worded it badly. What I meant to say was that the equivalent simple span beam will have the same midspan bending moment as the continuous beam but will not have the same deflection because the points of inflection do not necessarily have zero deflection.

I was not trying to compare the deflection at the midspan to the deflection at the ends of the cantilevers. As a matter of interest however, when the supports are 1.24m from each end, midspan deflection appears to be slightly greater than that of the cantilevers.

Ah-ha, I see what you mean. It's very interesting that the deflection is greater at mid span. With the bending moments being the same at all three points of stress, I'd would have put money on the deflections being either equal or greater at the ends.