Balanced Moment

Can you provide the link to the thread that is from?

You are quoting from this post.

The negative moment and positive moment are precisely equal when the cantilever is L/√8. For a uniform load of unity, the negative moment is C²/2 and the positive moment is L²/8 - M^neg/2 = L²/8 - C²/4. Check it out.

Until now still cloudy for me sir. How about if we remove the right/left side cantilever and the remaining is one cantilever, how to take equal moment on simple span and cantilever acting uniform load of unity on it?

Thanks.

The details are in the following sketch if not clear enough let me have your mail via PM and I send you the JPEG file.

The same method can be used in other configurations.

hey how youve uploaded the jpeg or picture file to this forum. pls tell the method

On the reply box a photo camera symbol in green (7th from the left in the command bar) is available. You choose it and follow the indications.

Sorry, post #2 is not correct. I goofed there, but not in the original thread.

We are dealing with a beam of span L with a cantilever of unknown length at each end. What we are trying to do is find out how long each cantilever must be in order that the negative moment over each support is numerically equal to the positive moment at midspan. The load is uniform over the span and both cantilevers. For simplicity, say the load is unity.

The cantilever moment is C²/2 where C is the cantilever length.

Without negative moment, the positive moment at midspan would be L²/8 (the simple beam moment under uniform load of 1 kN/m). Because there is a negative moment at each end, the positive moment at midspan is reduced by the full negative moment.

The midspan moment becomes L²/8 - M^neg. This is numerically equal to the negative moment.

So L²/8 - M^neg = M^neg

or L²/8 = 2*M^neg

or M^neg = L²/16

But M^neg = C²/2 = L²/16

so C² = L²/8

and C = L/√8

I hope this clarifies the situation. Sorry about post #2.

It is OK I thought about only one cantilever i.e; a non symetric problem. I use the equation for symetrical cantilervers to determine the position of the shelf supports for books.

In simplicity.....in the field...... Cantilever = 1/3 of the span for equal sq. ft roof loads.

How ever, I am not a graduate engineer..........so...no math included.

In the last fifty years I have built many industrial/commercial structures using this concept for glue lam beam design.

I live in earthquake country.

Mr. Guy

Hello Mr. Guy,

Your rule of thumb comes very close to the theoretical solution. L/√8 is equal to 0.353L, not significantly different from L/3. Your practice over the last fifty years conforms closely to the theoretical solution.

In an actual situation, of course, the snow load could be distributed in an unusual way. If any of the snow load from the two cantilevers is removed (by wind or other means), the positive moment on the beam increases above L²/16 . And for this reason, it is good practice to allow for unbalanced loading (and in fact some of the building codes require it), so it might be good practice to size your glulam beams for a positive moment of, say wL²/12 instead of wL²/16.

A little over design will never get you into trouble and the extra cost is negligible.

Correct

Hi everybody,

I am the original poster taking quote from ba/ael post. Actually my first question was regarding to both ends have cantilever beams. I jumped only into second question, (the reason was I hardly understand this "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.) Asking if how about we remove one of the cantilever beams? to seek more explanation. Reading and analyzing (I don't know if I have this one) I understand already that notion.

And this is the one… let Y₁ = simple span moment = wL_e²/8

Y₂ = cantilever moment = wC²/2

Y₃ = Y₁ + Y_{2(average of negative moment)} = wL2/8 = simple span moment w/o the effect of cantilever moment.

Target Y₁=Y₂

Thus; ½(C²/2+ C²/2) + wC²/2 = wL²/8; from this

C²/2 = L²/16

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