Which wood cross section can carry highest uniform load

AI Thread Summary
The discussion focuses on determining which wood cross section can support the highest uniform load by calculating the moment of inertia for various shapes. Cross section (D) was initially assumed to be the strongest due to its highest moment of inertia value of 3285.33 in4. However, further analysis revealed that cross section (C) may actually carry the highest load when recalculating with corrected values. Participants debated the appropriateness of seeking help on a take-home quiz, emphasizing the importance of understanding the calculations rather than just finding answers. Ultimately, the conversation highlights the complexity of structural analysis in wood design and the need for careful verification of calculations.
Blugga
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Which wood cross sectioπ caπ carry hïghest uπiform load

Homework Statement


2m69v1g.jpg


Homework Equations


Moment of inertia for rectangle I=(1/12)bh3
τ=(VQ)/(I*b)
σ=(MV)/I

The Attempt at a Solution


I started by finding reactions at B and C
Then drew the sheer and moment diagrams (to save space i made it a link)
http://oi47.tinypic.com/qsmkxh.jpg
To find which cross section can withstand the biggest q, I found the moment of inertia about the z axis (neutral axis) for each figure.
(A)Iz=2*{((1/12)*8*23)+[(2*8)*32]}+2*{((1/12)*8*23)+[(2*8)*12]}
Iz=341.33 in4

(B)Iz=2*{((1/12)*8*23)+[(2*8)*32]}+2*{((1/12)*2*83)}
Iz=469.33 in4

(C)Iz=2*{((1/12)*8*23)+[(2*8)*52]}+2*{((1/12)*2*83)}
Iz=981.33 in4

(D)Iz=2*{((1/12)*8*23)+[(2*8)*92]}+2*{((1/12)*2*83)+[(2*8)*42]}
Iz=3285.33 in4

Because both of the equations I posted for τ and σ are dependent on Inertia, I figured since that when solving for (q) we would have to multiply by Inertia so whichever cross section has the biggest Inertia will also have a bigger (q). Therefore, cross section (D) was the pick. Was I right to assume this?
τmax=[(6.5q)*Q(0)]/[3285.33*2]=65; where Q(0)=208. not sure about that value
∴q=315 lb/ft
σmax=[(34.125lb-ft)(12in/ft)q*10]/3285.33=1800
∴q=1444.10 lb/ft and so the max q can be is 315 lb/ft

This is my first attempt so any help is appreciated.
 
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I can't help you with the math but I can tell you that in the real world D is what is always used when strength is needed.
 
phinds said:
I can't help you with the math but I can tell you that in the real world D is what is always used when strength is needed.

I guess it kinda helps prove my point, but I was kinda looking for a more mathematical response :)
 
Blugga: Is this a take-home test question? I currently cannot match your x1 = 10.5, on your shear diagram, yet. Are you sure x1 is correct? Try again. Therefore, I cannot match your bending stress equation yet.

Your shear stress equation looks fine. Your moments of inertia look fine. How will you know which cross section can carry the highest q if you do not check them? You will just guess?
Blugga said:
Was I right to assume this?
No, I do not think so.
 
It's a sort of take home quiz. The book doesn't go into depth into how to find this sort of info for these shapes. It just explains briefly for 1 or 2 wood boards, and i have 4.

I think you may be right about the X1 being off. I think x1 will just be 10.5ft - 4ft. So x1=6.5ft.
Therefore, Mmax=13.125q.
Also, i guess my reasoning was that from the bending stress alone cross section (D) would be greatest. But i didn't count that the glued joints would make all the difference. I did try to find qmax for each cross section (but with the wrong Mmax) and ended up with cross section (C) carrying the highest q. I'll rework the new figures in the morning. Thanks for the help.
 
nvn said:
Blugga: We usually should not help on quiz or test questions. The PF Rules link at the top of each page says, "Do not ask for answers to exams."

I understand that, but this isn't an exam. It's a take home quiz/assignment. It's worth a very small percentage of the overall grade, about 0.7% and I already had half right so you were only helping with 0.35%. Only reason I asked was because the book was of no help and there was no one else I could ask (TA's and professors never have time). I guess it's best to get things wrong and never figure out how to do them then to bend the "rules" a little... :rolleyes:

But thanks for the help you did provide.
 
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