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JG33
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| تاريخ الالتحاق: | الجمعة أغسطس 29th, 2008 |
| الموقع: | USA |
| عدد المواضيع: | 32 |
| Exams Taken: | PD, GS, BD/MM, CD, SP, BP, BT | | Exams Passed: | SP, BP, BT | | Describes Me: | |
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كتب: الاثنين سيبتمبر 15th, 2008 06:25 pm |
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I can't seem to figure out this flashcard problem.
Calculate the allowable axial load for a 10" x 12" wood column, Douglas Fir Structural Select, with an unbraced length of 20 ft. The answer is 82.156K
Can anyone help me????
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sharcitect
Member
| تاريخ الالتحاق: | الاحد أوكتوبر 14th, 2007 |
| الموقع: | NYC |
| عدد المواضيع: | 239 |
| Exams Taken: | | | Exams Passed: | | | Describes Me: | A-R-C-H-I-T-E-C-T |
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كتب: الخميس سيبتمبر 25th, 2008 06:28 pm |
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Hi,
I'll take a crack at this.
But Ballast says this about wood columns on pg. 9-10
"The allowable unit stress in pounds per square inch of cross-sectional area of square or rectangular solid columns is determined according to a complex formula that considers the effective length; whether the wood is visually graded or machine graded; and whether the wood is sawn lumber, round timber piles, or glue-laminated timber. Because of the complexity of the formula, it is unlikely that the test will ask for specific values to be calculated."
Apparently, the flash card people disagree.
Below is how I would go about solving this without knowing the complex formulas that Ballast refers to.
I have just started studying for this one so I would appreciate it if someone could critique this approach and tell me where I went wrong.
Euler's formula is:
P/A = (pi^2 * E) / (l/r)^2
A= 109.25 sq in
E= 1,900,000 lbs for select structural Doug Fir
L = 20 feet or 320 in
r = sqr rt of I/A so you will need to find I
I= bd^3 / 12 (in the smaller dimension) = 11.5 * 9.5^3 / 12 = 821.7 in^4
r= radius of gyration in small dim. would be sqr rt of I/A = sqr rt of 821.7/9.5x11.5 = 2.74 in
Although Ballast says for simple wood columns r is not used but the smallest dimension of the column instead (pg 4-10) so I guess I didn't need to calculate r after all.
I guess that leaves our equation as:
P/A = (pi^2 * 1,900,000 lbs) / (320 in /9.5 in)^2 or P = A*((pi^2 * E) / (l/9.5 in)^2)
I get 1805 kips. Damn.
How about when I take the simple table 9.2a on Ballast pg 9-3 and find compression parallel to grain to be 1700 lbs for select structural doug fir and then use the simpler equation F= P / A or 1700 lbs = P / 109.25 or P = 1857.25 kips. Damn again.
Am I forgetting some size / slenderness factors or am I just way off somewhere else?
Started this answer to the OP trying to help, now I feel more lost than before...
Thanks for any advice.
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UofMArch
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| تاريخ الالتحاق: | الاثنين أغسطس 6th, 2007 |
| الموقع: | Adrian, Michigan USA |
| عدد المواضيع: | 48 |
| Exams Taken: | PD, GS, LF, ME, BD/MM, CD, SP, BP, BT | | Exams Passed: | PD, ME, BD/MM, CD, SP, BP, BT | | Describes Me: | |
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كتب: الاثنين نوفمبر 10th, 2008 02:00 am |
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The one bright spot of this entire lousy weekend I finally figured out this equation. I found out on Friday that my firm is closing next week. So the day before my structures exam I get to file for unemployment! It’s not that the equation will ever show up on the exam but after trying to figure it out for three weeks it was a true elation. I know this is bringing up an old posting but if I was looking for this same answer there is more than likely others with the same question.
Actually I can’t take credit, Kaplan question and answers book has a very similar question.
Use Fc = .3 E/(l/d)2 with E for Douglas Fir Structural Select being 1,600,000 and the actual 10 x 12 being 9-1/2 x 11-1/2 (don’t ask me it came from a chart). You use 9-1/2 as d (the smaller dimension) and the length multiplied by 12 to convert to inches. You get Fc = 752. You then use the P = Fc x A. The area of a 10 x 12 is 109.3 (from chart). Multiply 752 x 109.3 and you get 82,202.7, which is as close and I could get to the 82,156 on the back of the card.
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JG33
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| تاريخ الالتحاق: | الجمعة أغسطس 29th, 2008 |
| الموقع: | USA |
| عدد المواضيع: | 32 |
| Exams Taken: | PD, GS, BD/MM, CD, SP, BP, BT | | Exams Passed: | SP, BP, BT | | Describes Me: | |
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كتب: الاثنين نوفمبر 10th, 2008 04:17 pm |
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Thank you both! That problem haunted me for weeks!!!! I just ran across a similar problem in the ALS last night in the final exam. Same Fc equation trial and error with each answer. 
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sharcitect
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| تاريخ الالتحاق: | الاحد أوكتوبر 14th, 2007 |
| الموقع: | NYC |
| عدد المواضيع: | 239 |
| Exams Taken: | | | Exams Passed: | | | Describes Me: | A-R-C-H-I-T-E-C-T |
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كتب: الاثنين نوفمبر 10th, 2008 10:48 pm |
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Thanks for the solution. I have decided to skip learning about sizing wood columns and concrete beams because they are both too complicated. I will roll the dice that tomorrow these items will not be on my test.
Wish me luck!
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