PR beam connection--what is the slope formula? 1

[NearBeer]

I work for a small storage rack manufacturer and have the joy of updating our capacity spreadsheets (based on 3rd Ed. LRFD). However, the formulas used have no reference as to where they came from. Some are self explanatory such as those found in the AISC book but others look like they are from a Mechanics of Materials book.

In one of the formulas, it appears that the beam is assumed to have partially restrained connections and that the rotation/slope is (1/2) of ML/EI. In the MoM book, for a cantilevered beam with one end fixed and dist loading, the rotation/slope is ML/EI. Therefore, is the slope of a beam with a dist load and both ends having partially restrained connections 1/2 of that of a cant beam?

If anybody has any clue as to what the slope formula is and can clue me in, I would appreciate it. Some of this engineering stuff is having a hard time coming back to me; I spent 10 yrs in the military and never once used my CE degree!

Thanks for your time and in search for next cold one,

Near Beer

 

[Lion06]

it depends on the moment-rotation curve of the connection. Blodgett has a discussion on this and a rational approach for the moment-rotation curves of plates for PR connections (Section 5.5 and 5.6). If the PR connections are top and bottom angles, there are several papers which address this, but don't give a rational approach. They do however give some tables/charts.

 

[NearBeer]

Section 5.5 and 5.6 of what reference? Thanks for your response.

/R,

Near Beer

 

[EdR]

For a simply supported beam with moments M applied at both ends of the beam the slope at each end is M L/(2 E I)...

Hope that helps

Ed.R.

 

[NearBeer]

EdR: Thank you very much! Can you humor me and give me a specific textbook/reference where that formula can be found? My Mech of Mat. book does not have formulas for beams with both end fixed, just can beams. This exercise is turning into a documentation goat rope; the only books I have to work with are my old MoM book and the 3rd ed. of LRFD...thankfully the company is going to reimburse me for ordering the updated stuff!

Again, many thanks...Near Beer

 

[Lion06]

The reference for my first post is Blodgett's "Design of Welded Structures"

 

[Lion06]

I'm not sure how the slope for an end moment only on a simple beam is all that useful. How would you use that?

 

[msquared48]

To determine the true deflection at the tip of the cantilever, so one can see if material will tend to slide off the end of the storage racks?

Mike McCann
MMC Engineering

 

[EdR]

Gentlemen:

A bunch of you have misread my post.....The formula is for a "simply" supported beam (pinned at one end and rollered at the other) with a concentrated moment (M) applied at each support and has nothing to do with a cantilever (unless you are superimposing a group of load cases for a more complicated loading situation...remember the method of superposition from MoM??)

NearBeer: It is in Gere & Timoshenko and can also be derived by taking a simply supported beam with a moment on one end (a standard case in any text or Roark) and adding the slopes (superposition) from two cases with the moments on different ends.....i.e. M L/(3 E I)+ M L/(6 E I)=M L/(2 E I)

Ed.R.

 

[Lion06]

But my question is what good is the end slope for a simply supported beam with end moments only? What is that telling you in the way of PR connections?

 

[NearBeer]

Good information! I have attached the spreadsheet that I am working off of and someone’s handwritten notes that was used a reference for the spreadsheet so that you can have an idea of what I am doing. Unfortunately, our engineer doesn’t remember who created the spreadsheet or what the sources of the formulas were.
A little background on where this is going: The work we do is mainly manufacturing with some design thrown in there on occasion. So, MSquared48 is correct that slope is important to see if our product will slide off of its support. Also, for the issue I am working on now (how much slope our beam deflects) will have an impact on the sizing of the structural channel to be used. This particular job is going to number in the tens of thousands of units and having to use a C4 over a C3 will ultimately translate into how much weight to ship………………
(…......which in turn affects our profit which in turn may affect my bonus check which affects my restoration of a 65 Mustang Fastback?….I digress rapidly.)
The association (Rack Manufacturers Institute) that our company belongs to has shown, through testing, that the beam to column connection (PR) behaves more or less like a spring. Therefore, the maximum moment that the beam can carry should be reduced because of this spring effect (aka joint rigidity). Another factor that RMI accounts for but is not adequately addressed in AISI/AISC is the nature of perforated web that we routinely manufacture in our C3/C4 columns: typically (2) holes ½” dia that are side by side and 4” o.c. traversing the length of the column.

Now for the unpleasant/tricky stuff:

Referencing the .pdf file of the handwritten notes, it looks like our mystery engineer is reducing the moment carrying capacity of the beam by subtracting out the moment associated with beam/column connection spring constant. Utilizing:

? = M/F
?1 = ?2 – ?3

where M is the moment at the connection and F is the spring constant (determined from RMI testing; see cell F7 in the attached spreadsheet. NOTE: The value for the end moment in F7 is different from C21) and

?2 is the rotation imparted by a distributed load

?3 is the rotation in the beam/column connection.

I am able to follow his calcs up till the point where he converts ?3 into MeL/2EI. Not knowing how mystery engineer arrived at this equation, I decided to seek your help on the forum.

StructuralEIT: If what I have written doesn't asnwer your question, shoot me a PM and we will discuss from there.

 

[NearBeer]

Couldn't get the spreadsheet to attach in the previous post so here it is.

 

[Lion06]

I didn't go through the whole post, but the ML/(2EI) is the simple beam end rotation where M is (2/3) of (wl^2)/8 (or the fixed end moment, (wl^2)/12).
This gives you the second point that you need for the beam line. The first is zero rotation with the fixed end moment (wl^2)/12........ or (2/3)(wl^2)/8.
Once you have the beam line, you have to draw the moment rotation curve for the connection (as I mention above), and where the two intersect is the moment and rotation that you can actually expect to have in your beam/connection.

I'll look through the spreadsheet when I get a chance, but definitely check out Blodgett (Sec. 5.5 and 5.6), he goes through this very thoroughly and makes it easy to understand.

 

[Lion06]

F appears to be the slope of the moment-rotation curve of the connection. I can't see his reference, but that makes sense. Me is an input that is the fixed end moment for a rigid connection. Me/F will give you the end rotation of your particular beam - though I'm not quite sure why Me is an input, it should be calc'd based on other input information.
I'm not seeing anything regarding the theta1, theta2, and theta3 that you reference.

 

[BAretired]

The change in slope from point A to point B in any member is the area under the M/EI diagram between point A and point B.

For a cantilever beam with a point load at the end, the moment at the base is PL. The area under the M/EI diagram is PL^2/2EI, and that is the rotation of the tip of the cantilever since the base of the cantilever is fixed.

Taking it a step further, the deflection of the cantilever at the tip is the rotation times the distance from the rotation to the point in question. Since the rotation is concentrated at the c.g. of the moment diagram, i.e. L/3 from the base, the deflection at the tip is given by the expression PL^2/2EI * 2L/3 = PL^3/3EI...a familiar expression from any standard text.

Using the principals of Beam Analogy, the rotation and deflection of any point of a beam may be determined if we know the shape of the moment diagram under the actual load.


Best regards,

BA

 

[kslee1000]

I think there is a mistake in the equation for the calcultion of rotation Re (I don't know how to do math symbol here, so use "R" as THITA).

Under the moment diagram:

Re1 = Re2 - Re3
Re1 = wL^2/24EI - MeL/2EI

The first term on the right should be wL^3/24EI (moment area with respect to end support), similarily, the second term is the result of Me*L*(L/2). Making sense? Please check.

 

[BAretired]

kslee1000,

I agree with your first term, wL^3/24EI but not your second. Under pure moment across the beam, the area under the M/EI curve is Me*L/EI, half of which goes to each end using beam analogy, i.e. Me*L/2EI.

Best regards,

BA

 

[kslee1000]

Rotation = Moment Area * Arm (with respect to point of interest)

 

[kslee1000]

BA:

You are correct. Thanks for pointing out my mistake. It shoud be Me*L/2EI.

 

[BAretired]

Consider an analogous beam with a uniform load of M/EI. The total load is ML/EI. The analogous beam shear is ML/2EI which is rotation of the real beam. The analogous moment at midspan is ML^2/8EI which is deflection in the real beam under constant moment.

Best regards,

BA