Boom: Shear and Moment diagrams

[akpalican]

I've been curious about drawing a shear and bending moment diagram for the outer boom of a digger I saw online:

1.) How do you draw the shear and bending moment diagram when you no longer assume that the beam is weightless? Based on the image I attached; is it appropriate to use a point load at the outer boom and inner boom CGs instead of using a distributed load for the boom itself plus a distributed load for the components within each boom?

2.) Looking at the attached image again, would it even be a good idea to use a shear and moment diagram to analyze the booms if the auger wasn't contacting the ground? (Again, I think the part that confuses me most is how to apply the weights when analyzing this "beam")

My reference:

https://faculty.unlv.edu/jagadeep/M...g Shear Force and Bending Moment Diagrams.pdf

 

[drawoh]

akpalican,

You account for the weight of the beam by adding a distributed load. If you treat your auger as the external load, the vertical component of your ram, and your pivot are your reactions. If you are not supported by the ram, you need to apply a moment at the pivot. I see no means of doing this in the photo.

If this were my problem, I would use double integration method to solve this, which means shear and moment diagrams.

--
JHG

 

[drawoh]

akpalican,

If the auger is not contacting the ground, then the ram is supporting the mass of the beam and of the auger.

Is this a student post?

--
JHG

 

[akpalican]

drawoh,

Thank you for your reply. I understand what you're saying and how to analyze it, however, the weight of the beam itself is confusing me. Statics and Mechanics of Materials textbooks neglect the weight of the beam in their analysis. The "beam" I'm trying to figure out is the "outer boom". Each of these booms have components inside of them, like a cylinder to extend it, hydraulic hoses, and whatever else is in there. That's what led me to question if you can find each component's weight and CG, then find the overall CG of the boom as an assembly and apply the resultant weight there for the shear and moment diagrams.

 

[chicopee]

Superimpose the distributed weight of the boom unto the load diagram showing the reaction loads of the auger, of the cylinder at the boom pin connection, and the boom pin connection to the machine structure. Of course, you have to know the vertical reactions at the auger and at the two pins connection and the distributed weight of the boom. From that load diagram you can develop the shear and moment diagrams. You can not dismiss the horizontal forces from the auger because it will not be quiet perpendicular to the boom axis and those horizontal reactions at the pin connections. Then when you have all of that information, you can work on the stresses.

 

[dhengr]

Akpalican:
It would really be helpful if you had drawings of the entire boom, cross sections of the various boom sections, all the major details on the boom. How is the boom rigged, how does it telescope, etc. You have a fair start on some of the loads and reactions, but for the most part the major components are supported at specific points, and that’s where their weights are applied. You have to draw free body diagrams (FBD’s) of each boom section independently, and then combine them for the complete boom configuration. Of course, there are a number of different load conditions, and several max. & min. boom extensions and orientations. Each section of the boom is a cantilever, from the bending and torsional standpoint, and it is supported by the outer end of the previous boom section. The first boom section is supported back at the pin on the truck vert. pivot structure, then again at the outer end of the boom lifting cylinder, and beyond that it cantilevers to its tip. But, from your picture I can’t tell where the end of the first section is.

You didn’t answer an important question, is this a student problem, what’s this inquiry for? Show us some more photos and some boom drawings and details. Who designed and built the crane? Get some details from them.

 

[desertfox]

Hi akpalican

To answer your question directly yes you can place a point load at the centre of gravity of both inner and outer boom, it really depends on how accurate you want your analysis to be.

Another way to analyse it would be to consider it weightless and draw all the shear force and bending moment diagrams, then do a separate bending moment and shear force diagram considering only the mass of the two booms, you then can add the two separate bending moment and shear force diagrams together to obtain the resultant.

Go to the site below and look at the simple beam supported with a udl, that udl could well be the mass of a boom spread over it's length, if you know how to draw bending moment and shear force diagrams for this beam then you know how to include self weight.

http://www.roymech.co.uk/Useful_Tables/Form/Stress_Strain.html

 

[rb1957]

isn't the auger driving into the ground ? shouldn't the force the auger is applying be in the opposite direction.

isn't the hydraulic cyclinder pulling against the boom ?

this is a bad FBD of the boom, some forces acting on the boom (weight) are in the right direction, some forces look to be in the wrong direction.

another day in paradise, or is paradise one day closer ?

 

[rb1957]

thinking about it again ... the hydraulic cylinder force should be reversed, the ground reaction can stay ... the boom weigt and the hydraulic cylinder help the auger dig into the ground.

another day in paradise, or is paradise one day closer ?

 

[akpalican]

rb1957,

This is an arbitrary FBD. My REAL question was how to apply the weight of the booms when drawing a shear and moment diagram.

 

[akpalican]

dhengr,

No, this is not a student problem. There was a debate about which was more accurate: applying the weight of the boom at each boom assembly's CG or applying a distributed load along each boom. This prompted me to go back through some textbooks but I hadn't seen an example where it actually considered the weight of the "beam", hence my question. Thank you for your insight.

 

[rb1957]

you can distribute the weight, or apply it as a concentrated force at the CG

another day in paradise, or is paradise one day closer ?

 

[desertfox]

Hi akpalican

follow the link in my post

 

[drawoh]

akpalican,

Your textbooks are making the assumption that the mass of the beam is negligible, or unimportant. They are showing you how to do the analysis. They are not doing actual design. Sometimes the mass of the beam is not signifiant. In your case, it probably is. It looks like a distributed load to me.

--
JHG

 

[dhengr]

Akpalican:
I doubt that there are any textbooks that suggested that you can ignore the dead load of a beam. They may have mentioned that often the affect of the beam weight is small in comparison the affect of the applied live load. Or, to explain a particular problem derivation, they may have left the uniform dead load out, for the time being, but they were not suggesting you can forget/ignore the dead load (self weight). The boom sections are cantilevers and have a fairly uniform wt./ft., over some lengths. And, you aught to know how to calc. plot the shear, moment and deflection of a uniformly loaded canti. beam, if you’re an engineer. Dust off your Statics, Engineering Mechanics and Strength of Materials text books and do a little studying. Then you superimpose the same canti. beam with its various concentrated loadings on the above uniformly loaded beam for a combined/total load condition. There is no particular advantage to applying the boom weight at its C.G. This approach is less accurate except for finding the fixed end moment and shear, and doesn’t show very good engineering judgement, if used. The example you found from UNLV is showing a sign convention; and then how to draw the shear diagram; and since the moment at any location is the area under the shear diagram, how to draw the moment diagram. Now, assume a beam weight of 50lbs./ft. and draw that shear diag. and then the moment diag. Then combine the shear and moment diags. from the two problems and you’ll have the total load shears and moments. You’ll see that the reactions, shears and moments (their values) don’t change much, but they are not to be ignored, and the uniform load does change the shapes of the diags. and that’s important sometimes. Do the second steps, immediately above, with the beam weight (50lbs./ft.) concentrated at its C.G. and see what happens. You’ll learn something by doing this yourself.

 

[akpalican]

Thanks to all of you who were helpful without being condescending. These forums are suppose to be a tool to learn by asking questions, even if they seem stupid to other people.

 

[Dougt115]

I would use the point load method due to the nature of the booms. The CG of the booms will stay in the same place relative to each boom section but the distributed load will create more of a headache as the booms contract into each other.

If you use the point load system your loading will be a flat between the points and a step at each load equal to and in the direction of the load. The moment will be a sloped line between the point loads. If you use the distributed loads then you will have a combination of steps, at the auger, hydraulic and truck, and a sloped line in between to account for the distributed loading. The Moment diagram will be an exponential curve for each of the booms (where the booms overlap so will your exponential curves this is why it gets difficult).

Your free body diagram is all dependent on whether the auger is drilling down or being pulled out. And the angle of the boom will have an impact on your loads too. Be sure to use both free body diagrams you don't want any surprises.

 

[CCox]

akpalican,

there are some crusty engineers on this site. Don't let it bother you.

 

[akpalican]

Dougt115,

Thanks for all of the insight, I really appreciate it. I figured that both a distributed load across the booms and a point load at the CG would work but I wasn't sure if one method was more common. But, you explained it very well. Thanks again!

 

[akpalican]

CCox,

No kidding! Definitely not what I expected from this forum but luckily, there are quite a few good engineers on here too.