Oh...ok. The AASHTO stuff is a huge hole in my knowledge. I'm clueless here except what's in SERM and what little knowledge I've picked up doing practice problems.Look at the end of Chapter 4 in AASHTO for the Appendix A. It is a table of moments for the deck design based on the span.
Yea...I'll check those tonight when I get home. I started to look at it this morning and got off into the errata of the NCEES sample exam. Wow...that's grown since I last checked it...which was before the exam in April. Guess I should keep up with that more.Now you know how I feel with the IBC, ASCE 7, ACI masonry, PCI, and many others :beerchug:
The AASHTO is enormous, but at least its all in one code. You might flag a few sections of AASHTO, like the effective flange width sections and the table I mentioned. Tab the load combination page as well. I've been tabbing quite a bit of stuff in the larger manuals.
Let me know about those two problems I question above. I could be wrong though.....like the wood problem
Aw shoot, I was adding the three values together instead of multiplying them! :brickwall: Check the errata for the chord forces on those problems you question, the leader was pointing to the wrong line.
Got a few more questions with the same SE I practice exam, maybe you could help/clarify. The solution to 501 shows the L for the column as 13.67 ft. According to ACI, the L should be the clear distance between floors. Wouldn't that be 12ft then, not 13.67?
One other one that I think may be incorrect is the LRFD solution to #525. The section is doubly symmetric, so I think they should use section H1. You would then see if Pr/Pc > 0.2 I get Pr = 32k. The Pc = phi * Fcr * A = 0.9 * 23.7 * 15.6 = 332.75k. The ratio ends up being 0.096 < 0.2, therefore the second equation applies. Pr/(2Pc) + Mr/Mc <1. In their solution, they are finding the stress in the column (P/A) and then use an equation fa/Fa + Mr/Mc This equation lacks the 2 in the denominator of the axial portion. I end up getting 0.814 for the combined ratio compared to their 0.853
I did OK with the afternoon portion, just barely passing. The masonry expansion/contraction questions really threw me for a loop. Such an odd group of questions, and there were like 5 of them!
On 501, I agree it should be 12'...ACI 10.11.3 says it's the clear distance.Aw shoot, I was adding the three values together instead of multiplying them! :brickwall: Check the errata for the chord forces on those problems you question, the leader was pointing to the wrong line.
Got a few more questions with the same SE I practice exam, maybe you could help/clarify. The solution to 501 shows the L for the column as 13.67 ft. According to ACI, the L should be the clear distance between floors. Wouldn't that be 12ft then, not 13.67?
One other one that I think may be incorrect is the LRFD solution to #525. The section is doubly symmetric, so I think they should use section H1. You would then see if Pr/Pc > 0.2 I get Pr = 32k. The Pc = phi * Fcr * A = 0.9 * 23.7 * 15.6 = 332.75k. The ratio ends up being 0.096 < 0.2, therefore the second equation applies. Pr/(2Pc) + Mr/Mc <1. In their solution, they are finding the stress in the column (P/A) and then use an equation fa/Fa + Mr/Mc This equation lacks the 2 in the denominator of the axial portion. I end up getting 0.814 for the combined ratio compared to their 0.853
I did OK with the afternoon portion, just barely passing. The masonry expansion/contraction questions really threw me for a loop. Such an odd group of questions, and there were like 5 of them!
Yeah I got a different answer then them using the P/Pn, which was why I wondered if combining ASD axial stresses and LRFD moments ends up throwing things off a bit.That's what I was thinking on the doubly symetric...tubes, W sections...the WT, Channels would be singly symetric.
On 525, I looked into the P/Pn from the tables, but it didn't give the correct answer. I meant to bring that book to work today to talk with someone over that problem, but I forgot it. I'll still see if I can describe it to him. Let you know what he says.
I thought about splitting my AASHTO up, but decided not to since I need the index to find most of it.
If you are combining ASD and LRFD, you will get the wrong answer...look at the tables. The LRFD values are significantly higher.Yeah I got a different answer then them using the P/Pn, which was why I wondered if combining ASD axial stresses and LRFD moments ends up throwing things off a bit.
In Illinois we aren't supposed to break up the AASHTO manual, but I did it anyways. I'm bringing the enormous empty binder just in case they give me grief. I copied the index and put it at the end of each binder :beerchug:
That's what I was thinking on the doubly symetric...tubes, W sections...the WT, Channels would be singly symetric.
On 525, I looked into the P/Pn from the tables, but it didn't give the correct answer. I meant to bring that book to work today to talk with someone over that problem, but I forgot it. I'll still see if I can describe it to him. Let you know what he says.
I thought about splitting my AASHTO up, but decided not to since I need the index to find most of it.
I'll check into this when I get home...I don't have that here with me.One more question for you guys.... number 538 in the NCEES SE1 book. I understand that the net area is defined in 2.3.3.3.1 as the least of the center to center bar spacing, 6 times nominal wall thickness, or 72 inches. That makes sense, but the area of the steel is shown on page 136 as 0.8 square inches. How do they get that number? It seems to me like it should be a #5 area at 0.31 square inches since we're taking the net width at 4 feet. Do you think I'm missing something or is it a mistake?
honestly I don't think the steel should be included at all in this problem. You can only include the steel in masonry axial design if it is tied. However, I got Pa=12831 for the masonry only and I used .31 for the Ast and got Ast=.31...getting Pa for the steel equal to 1027...total of 13859...not the answer in the book either way. But...if I run their numbers, it doesn't add up in my calculator either way. My way gets me close to the answer. Don't know what to tell you other than I don't agree with it either wayI'll check into this when I get home...I don't have that here with me.
honestly I don't think the steel should be included at all in this problem. You can only include the steel in masonry axial design if it is tied. However, I got Pa=12831 for the masonry only and I used .31 for the Ast and got Ast=.31...getting Pa for the steel equal to 1027...total of 13859...not the answer in the book either way. But...if I run their numbers, it doesn't add up in my calculator either way. My way gets me close to the answer. Don't know what to tell you other than I don't agree with it either way
for the steel thing I was talking about see 2.3.2.2.1 and 2.1.6.5. It never says the steel is tied...but I guess if it doesn't...you are supposed to assume it is tied.
It's just a matter of leaving out the part of the equation that includes the steel. I don't know what to think on that. Yes, though...other than the Ast and the fact that when I punch in their numbers I don't get the same answer, I think they have done the problem correctly...i just have to question whether you should be using the steel.I see what you're saying as far as the code goes for compression steel with the lateral tie requirement. Unfortunately I can't find any other examples that show how to deal with strictly axial loads on reinforced walls using ASD design. I think I'm just going to trust that the example solution is mostly correct, except for the Ast part.
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