CERM (11th ed.) Ex. Prob. 47.3

CERM (11th ed.) Ex. Prob. 47.3

I think the expression for the moment mp [bar AB] is incorrect and should be (0.04wL2). Can anyone else confirm this?

Also I'm not sure I agree with the overall solution. Wouldn't a more accurate approach be to calculate the slope or rotation at the end of the vertical bar AB and then create and expression for the resulting vertical displacement at point C (it would be a function of the Bar length BC and the rotation at B on Bar AB)? This value could be added to the expression the calculated for the vertical displacement on bar BC.

I think you can get an expression for the rotation at point B [bar AB] by placing a dummy/unit moment there on the Q-System

With the way the solution shows I'm not sure why they would add the deflection of the vertical bar AB to the deflection of the horizontal Bar BC? This is weird because the way I see it the deflection calculated for vertical bar AB is actually a positive global X-Direction displacement?

The second half of the final expression (displacement for Bar BC) I believe is correct and is needed for the calculation. The resulting deflection calculated would be in the global y-direction and correspond to the questions total total vertical displacement.

Any feedback/help would be great.

No one has this review manual I take it? Dang it!

ELASTIC,

Typically questions like this are posted in the Structural (in this case) Exam Prep forum.

http://engineerboards.com/index.php?showforum=38

Thanks for the pointer Ptatohed oking:

Thanks Ptatohed I think I will be living in this area of the forum from here on out. Plenty of interaction going on for my liking.

ELASTIC-1 said:

Thanks Ptatohed I think I will be living in this area of the forum from here on out. Plenty of interaction going on for my liking.

Click to expand...

No problem. I'm not a structures guy but I'll try to look at this problem sometime and let you know if I have any input to your question(s). But, feel free to ask any Transpo, Survey and even Seismic (only because Seismic is still somewhat fresh) questions, anytime.

I'll dig out my copy of the 11th and take a look. It's been a couple years since I cracked this version. I did look at the 12th quickly and it may be the same example problem; but I'll have to look at the 11th to verify first.

Will try to get this looked at by the end of the week.

FYI - I checked for errata for the CERM 11 (there are 2 printings) and, while there is errata for page 47-5, there is no errata for Example 47.3.

http://powertopass.ppi2pass.com/EMARS/errataLogin.jsf

I've dug through my old study materials and I cannot find my 11th edition of the CERM. Am thinking it got loaned out and has yet to be returned. Will keep looking though.

ELASTIC-1 said:

CERM (11th ed.) Ex. Prob. 47.3

I think the expression for the moment mp [bar AB] is incorrect and should be (0.04wL2). Can anyone else confirm this?

Also I'm not sure I agree with the overall solution. Wouldn't a more accurate approach be to calculate the slope or rotation at the end of the vertical bar AB and then create and expression for the resulting vertical displacement at point C (it would be a function of the Bar length BC and the rotation at B on Bar AB)? This value could be added to the expression the calculated for the vertical displacement on bar BC.

I think you can get an expression for the rotation at point B [bar AB] by placing a dummy/unit moment there on the Q-System

With the way the solution shows I'm not sure why they would add the deflection of the vertical bar AB to the deflection of the horizontal Bar BC? This is weird because the way I see it the deflection calculated for vertical bar AB is actually a positive global X-Direction displacement?

The second half of the final expression (displacement for Bar BC) I believe is correct and is needed for the calculation. The resulting deflection calculated would be in the global y-direction and correspond to the questions total total vertical displacement.

Any feedback/help would be great.

Click to expand...

I'll take a crack at this. Just know that I'm no where near an expert on virtual work and barely covered it in college, so I'm going by my quick review of section 47.6-9 right now. One thing to keep in mind is that virtual work is a simplified method for determining deflections, therefore it's hard to compare it to a detailed structural analysis.

As for your first question, the moment mp in the book is right. If you look at the free body diagram for mp of bar AB, you will see that you have a reactant force of 0.2wL acting at 0.2L/2. This gives you a moment, mp = 0.2wL * 0.2L/2 = 0.02wL^2.

The moment equation for virtual work, i.e. Wm, takes into account the rotation of the member, so that is why you don't specifically see them calculate the member's rotation.

As for your third question, they explain in section 47.9 that typical rigid frames resist loading primarily through flexure. You are correct that member AB deflects in the positive x-direction if you perform a true analysis, but it also deflects downward as well. That downward deflection is what the example problem is looking for, hence why it is added to the deflection of bar BC. If they asked for the horizontal deflection at point C, you would have placed a horizontal unit load either at point C or B.

I hope this helps a little, again, I'm not extremely familiar with virtual work other than studying it for the exam. If you have an old structural analysis or mechanics of materials textbook they should be able to help you with the underlying theory if you are interested.

Thanks guys for taking the time to respond to my question.

I simply botched the moment expression and forgot to divide by 2 when multiplying the resultant load times the lever arm.

Somehow in my reading I missed that virtual work takes into account the members rotation too (good to know thanks!).

Ble_PE seems like you're close to understanding what I'm asking here and where I may be disconnected.

Maybe you can help me see/understand how the first integral (for bar AB) in the final expression is actually calculating a true global Y-Displacement or downward deflection?

If we simplified the problem, eliminated bar AB and considered only bar BC as a simple cantilevered beam, with a fixed left end, everything else remaining the same, wouldn't we just take the second integral in the final expression for the answer?

I understand in the simplified example I just gave (or any cantilevered beam) that point on the end of the beam would move both vertically and horizontally (as you were explaining above). The vertical deflection being the much larger of the two (calculated by the expression I think) and the horizontal a result of geometry or bending of the beam. I visualize the horizontal displacement best by thinking of a projection of the deflected beam and how it would not extend to the full horizontal length of the non deflected beam.

What I'm seeing or interpreting is that the first integral in the final expression is calculating displacement in the global X direction (you seemed to agree with me above) which is not what the problem statement requests. In other words the downward deflection is the lesser of the two for the vertical member and not being directly solved for by the expression.

Again any help or feedback would be great.