Three unequal supports, two concentrated loads problem..

[truckdesigner]

Attached is a loading diagram that I have come across for a test gantry we are building.

I have not come across such an abnormal loading set-up before, and cannot find any literature that addresses my problem ie. shear and bending moment diagrams and Max shear and bending moment.

Can somebody please help me?

Regards.

 

[desertfox]

Hi truckdesigner

I am looking at solving your beam, perhaps tomorrow or saturday i might get to post a solution.

desertfox

 

[handex]

I'm with hokie here, stuff this this is very standard and suits spacegass to a tee. No need to waste your time doing moment distribution etc. If you want verification post your output and I'll quickly run the model myself for a check

 

[rb1957]

the on-line calculator will work just as well, it's set up for metric but it'll work just as well with your imperial units. then run your FE to check. then try a textbook solution.

i think you may have been looking for precisely this problem but the general problem (an indeterminate beam) is (should be) in just about every textbook. the 3 moment equation works well (looks a little scary but it is easy to apply), unit force method is an approach i'd use. google "indeterminate beam", "3 moment equation", "unit force method".

 

[ST04]

I believe that (3-Moment equation) method is the fastest and easier for hand calculation.

If you will use plastic analysis there is very simple method related to work or energy (don't remember the exact name). you can solve this problem in a half page or less by that method.

Note (As I understood, the loads are not moving, if they moves you might need to draw the influence line).

Regards

 

[ST04]

For plastic analysis
try to get this book:
Plastic methods for steel and concrete structure - by Stuart S.J.Moy.

The method written in my previous post is (virtual work method).

It allows you to find the collapse load, then you can compare it with the applied loads. or If you want the applied loads to be the collapse load you can find the plastic moment then determine the steel section to be used.

regards

 

[BAretired]

Another way is to remove the reaction at point 3, see how much the beam deflects, then replace R3 and equate the two deflections. See attached.

BA

 

[rb1957]

yeah, that's the unit force method

 

[MechEng2005]

My fourth edition of Hibbler has chapter 12 titled "Deflections of Beams and Shafts" and 4 of the 9 sub-sections starts with the three words "Statically Indeterminate Beams." Assuming this wasn't removed between the 4th and 6th editions, I don't know how much more attention you could ask for in a textbook that is pretty general mechanics.

-- MechEng2005

 

[hokie66]

Superposition was mentioned previously, and BA's approach is one application of the principle. You could also remove the support at point 2, find the deflection at that point and the load required to produce that deflection if applied at point 2, and add the bending and shear diagrams.

 

[handex]

attached is the bending moments and reactions from a SG model. Build your own of course (only takes a few mins) but use it for a check

 

[BAretired]

handex has the correct answer. R3 should be 11.44, not 10.9, but the formula I gave in my last post was okay.

BA

 

[ToadJones]

People are saying its "routine" because they aren't admitting to the fact that they are using software

 

[paddingtongreen]

R3 has been worrying me. Why is it there. It is a liability in that it places more force on R2.

Michael.
Timing has a lot to do with the outcome of a rain dance.

 

[BAretired]

paddington, I agree it doesn't seem to make a lot of sense. It's only advantage seems to be that it reduces the maximum moment on the beam from 42 T-m to 34.5 T-m. Maybe truckdesigner can tell us why they are building it that way.

BA

 

[rb1957]

release the RH support and apply the load ... the RH end of the beam is deflecting upwards ... so the reaction here has to keep the end from deflecting, so you get a -ve reaction (increasing the reactions at the other two points). the OP gave an explanation for this construction ... but it's not something i completely understand

 

[desertfox]

Hi truckdesigner

My soluton as promised, my answers are those of handex.
Also if you convert your beam to english units and use the on line calculator I provided you will obtain similar results, I say similar results because you can't place more than one point load per span, just put a resultant force of 80T on span 1.

desertfox

 

[truckdesigner]

Folks - to say that I am overwhelmed by the responses is an understatement! Thank you all so much for your advice, concern etc.

I got the exact same result as handex with the Space Gass model, so am confident with its accuracy. I am still battling with the hand calculation to be honest, but when I get some time (after hours) I will sit down and nut it out.

I have attached a photo of the frame in question. We are widening the "working area" to four metres so wanted to check for Max. Bending Stress. I have spoken to another Engineer - one with a little more structural experience than I, and he has told me that two "310 UC 118" members welded side by side will be more than enough, so all good. This is a stock Australian Standard beam. "Full penetration weld all around" of course.

Looking forward to my next encounter with this forum.

Regards.