Complex beam bending problem 2

[SamCrossley]

Hi

Wondering if anyone could point me in the right direction for solving this. I am looking work out the approximate reaction forces (and eventually bending moments) for a trailer design I'm putting together and have approximated it to a simply supported beam with multiple reactions (2 x wheels, 1 x tow hitch) and multiple loads. I've so far tried removing the R2 reaction and solving through superposition, but I'm struggling using beam theory to calculate the deflection at its position (x=1.724).

For section EF I have:
M(x) = -798.758392x + 1211.009551

Which integrating gives:

EI (dv/dx) = -399.379196x^2 + 1211.009551x + C1
EIv = -133.1263987x^3 + 605.504775 x^2 + C1x + C2

But I'm unsure what boundary conditions to apply to this section to get C1 and C2.


Does anyone know of a good way to solve this easily without having to go into FEA methods?

EI = 8890

Thanks

 

[BAretired]

Yes. It is quite straightforward, but I haven't done it yet. A simple 2D Frame program will give you the answer, but another way is to remove R2 and solve the statically determinate structure.

Calculate the deflection at x=1.724m for the determinate structure, then calculate deflection at the same point for a unit load. From this, calculate the magnitude of R2 and its effect on R1 and R3.

BA

 

[SamCrossley]

Thanks for your reply

I've removed R2 and solved for the reactions (R1' = 1975.958392N , R3' = 427.4916081N) but I cant find the deflection at x=1.724.

I have M(x)= -798.758392x + 1211.009551 for the section containing x=1.724, but integrating this leaves me with unknown constants. Is there a way to determine where the maximum deflection for the entire system is (i.e. dv/dx = 0) to easily find these constants? Or is there a better way to find the deflection?

Thanks

 

[delagina]

many practicing structural engineers dont do this by hand anymore. you may have better luck asking engineering students.

 

[msquared48]

Personally, I use Beam Pro and have an answer in less than 5 minutes - 4 of that I'm in the bathroom.

Mike McCann
MMC Engineering

 

[BAretired]

There are many ways to find the deflection. I don't know which one you consider best.

The problem is indeterminate to the first degree which is why you remove one reaction. You could have removed R1 or R3, but R2 seemed easier. After removing R2, you are left with a simple span beam from D to I (R1 to R3) with cantilever to the left of D.

The loads on the cantilever produce a negative moment at point D which I'll call M_D. M_D = P1*(a+b) where P1=490.5kN and a, b are the distance from each load to D. Considering only the loads on the cantilever, the moment varies linearly from D to I and the deflection at Point F is a function of M_D. By inspection, it is upward. Its magnitude may be found using moment-area principles or from any text which includes a beam diagram with moment applied to one end of a simple beam.

For each of the loads inside the span, deflection at F is given in the Steel Handbook as well as many other texts.

You add the deflection at F from M_D and each of the loads inside the span. That is the deflection of the determinate beam.

Now place a unit load at point F and calculate the deflection at F. Reaction R2 which was removed is numerically equal to the deflection due to all loads divided by deflection due to a unit load. In other words, R2 is the reaction required to bring the net deflection to zero.

The values of R1 and R3 will be changed by R2*a/L and R2*b/L where a, b are distance from the opposite support.

Another method is to draw the Shear Force and Bending Moment diagrams for the determinate beam, then calculate deflection at F using Moment-Area principles and finally solve for R2 as suggested above.



BA

 

[civeng80]

30 years ago, I guess this would have been a bit tedious to solve (like BAret suggested), but today it can be solved with about a minute of computer punching.

 

[civeng80]

Sam, Google continuous beams and you could get a free calculator to solve this.

 

[bridgebuster]

ASCE Steel Tools has a free excel sheet.Works well, don't recall if it includes cantilevers. In transit right now, can upload on Thursday.

 

[paddingtongreen]

It is a simple enough moment distribution problem, especially if loads E and G are moved to the supports, after all it is a trailer so the loads are not that well defined.

Michael.
"Science adjusts its views based on what's observed. Faith is the denial of observation so that belief can be preserved." ~ Tim Minchin

 

[msquared48]

BA:

The jobs not finished until the paperwork is done...

By the way, civeng80 vindicated me...

Mike McCann
MMC Engineering

 

[bridgebuster]

Attached is the spread sheet that's on AISC Steel Toools

 

[IDS]

So did we get an agreed answer?

I get:
R1 = 1535.49 kN
R2 = 572.67 kN
R3 = 295.28 kN

Spreadsheet file attached.


Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[IDS]

Forgot to attach it:

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[dik]

I use a spreadsheet by Yakov... a contributor to this site... slick, fast, and few bells or whistles...

Dik

 

[rb1957]

there are many free calculators available. personally, i would solve it by hand ('cause i dislike canned s/ware). a possible issue with the way the beam is being modelled is the reactions points ... in reality these are more like springs rather than infinitely stiff reaction points and as this is a redundant (or hyper-static) problem this will affect the results. i'd've thought (quite possibly wrongly) that assuming stiffer supports would be unconservative for beam bending moment (it maximises that redundant reaction).

Quando Omni Flunkus Moritati

 

[aggman]

For a statically indeterminate beam like this I would use moment distribution. It is very straight forward to solve this way. You can google (Moment Distribution by Hardy Cross) and you will get a bunch of information and descriptions of how to use the method. This is of course assuming you do not have some simple beam or frame program to solve it quicker.

I did want to add the following thought though. In my world trailer suspension systems are load distributing. If this trailer is using a tandem suspension package than the problem is statically determinate because the load is distributed equally to the two, three, etc axles through the spring hangers (more or less). Just solve for the trailer suspension reaction to the center of the axle group, balance the load to each axle, then carry the forces up to the frame through the spring hangers. If the trailer is not using suspension support then it's still statically determinate because the trailer will lift one axle off the ground when you go up and down hills so it has to span the load to one axle or the other.