My name is Dr. Andrew Assadollahi and I am a Civil Engineering professor at Christian Brothers University (CBU) in Memphis, TN. I am also a licensed Professional Engineer in the State of TN. While some of this content may be helpful to high school students, "Math and Engineering with Dr. A" is a channel designed to help college students with different topics in Engineering and Mathematics.
To maximize your knowledge from these example videos, it is best to WRITE EVERYTHING DOWN. Math (and Engineering) is not a spectator sport! You must participate by writing all of the steps down in a logical manner.
Please feel free to email me any time at aassadol@cbu.edu and check out my website at www.andrewassadollahi.com.
*All videos that show my handwriting were recorded using a 7th gen ipad, Apple Pencil, and the Penultimate app (by Evernote).
2 questions: 1) For base shear, shouldn’t you take trib area to the diaphragm, so only half the wall height? 2) for base shear, why aren’t you taking any pressure from the roof? Per section 28.3.6, you need to take 8psf multiplied by the roof area of the building projected onto a vertical plane normal to the wind.
Hi Jessica, Thank you for your great questions! I split up my response to your question 2 into two parts. 1) For the base shear, think of this as an over-simplified vertical cantilever beam with a uniform load and the base is fixed to the ground. The base shear would be like the horizontal force reaction. All of that wind load on the walls has to go somewhere, and that somewhere is the base. Now, if we were going to investigate or design the roof diaphragm, we would consider pressures on the upper half of the wall height. But in this example, since we were explicitly asked about the base shear, we have to think of the base as an external support that all of the applied loads are transferred to. 2a) The minimum loads specified in Section 28.3.6 (8 psf on the roof and 16 psf on walls) are used when designing structural elements of the main wind for resisting system (MWFRS) and should be applied separately from the actual wind pressures. So, let’s say we are designing a roof beam that is part of the MWFRS. We apply the actual pressures to the building (like we did in the example) and determine the wind effects (W) on that beam (shear, moment, etc). We input these wind effects (W) into the load combinations and design the beam. We then, separately, apply the 16 psf and 8 psf to the building and determine the wind effects again (W) on that beam. We input the effects (W) due to these minimum pressures into the load combinations and check that the beam is still adequate. For the sake of this example, though, I was only intending to demonstrate how to compute base shear due to wind loading using the Envelope Procedure. 2b) Since the roof pressures actually relieve the total horizontal shear due to the windward roof pressure being larger than the leeward wind pressure and both sides of the ridge are experiencing suction, we can neglect the roof wind pressures by Section 28.3.3. i.e. The total horizontal shear is greater than that determined by neglecting the wind forces on the roof. [I probably should have stated that in the video somewhere.] Note, though, that if the windward side of the roof was in compression, we could not neglect the roof pressures since their horizontal components (projected on a vertical plane) would both add to the wall pressures (pointing to the right).
I have made a spreadsheet to calculate Kh,Kz values and came up with a bit different number then Table 26.10-1. Should I not round the calculation and just truncate the number at two places? In this example when h=19.5 ft, Kh = 1.08 and when h=20 ft, Kh = 1.09. I know it is not much but it does change the numbers a little. Exp D h no round rounded formula 15 1.03504 1.04 = (2.41)(15/1935)^(2/11.5) 20 1.08814 1.09 = (2.41)(20/1935)^(2/11.5) 25 1.1312 1.13 = (2.41)(25/1935)^(2/11.5) 30 1.16765 1.17 = (2.41)(30/1935)^(2/11.5) 40 1.22755 1.23 = (2.41)(40/1935)^(2/11.5) 50 1.27613 1.28 = (2.41)(50/1935)^(2/11.5) 60 1.31724 1.32 = (2.41)(60/1935)^(2/11.5) 70 1.35303 1.35 = (2.41)(70/1935)^(2/11.5) 80 1.38482 1.38 = (2.41)(80/1935)^(2/11.5) 90 1.41348 1.41 = (2.41)(90/1935)^(2/11.5) 100 1.43962 1.44 = (2.41)(100/1935)^(2/11.5) 120 1.48599 1.49 = (2.41)(120/1935)^(2/11.5) 140 1.52637 1.53 = (2.41)(140/1935)^(2/11.5) 160 1.56223 1.56 = (2.41)(160/1935)^(2/11.5) 180 1.59456 1.59 = (2.41)(180/1935)^(2/11.5) 200 1.62405 1.62 = (2.41)(200/1935)^(2/11.5) 250 1.68831 1.69 = (2.41)(250/1935)^(2/11.5)
Nice! The equations for Kz beneath Table 26.10-1 would be the most correct to use since these capture the nonlinear nature of this relationship. Thus, using 1.09 (actually 1.088 when rounded to 4 sig figs for h = 20 ft) is the most correct. The advantage of the Table 26.10-1, though, is that you can directly read the values that are presented, and you are also permitted to use linear interpolation with the tabular values....which is helpful for quick calcs and when folks prepare for the PE. Since you created a nice spreadsheet, though, you can continue to use that. If you really want to have some fun with that spreadsheet, put some IF statements for the different z-domains and rig it up to handle all the exposure categories. Awesome stuff!
Thanks for your comment/question. Based on the definition of the "mean roof height" in ASCE 7, for roof angles less than 10 deg, the mean roof height IS PERMITTED to be taken as the roof eave height. So, this is an option but not a requirement. I took it as the actual mean height (20 ft) in part so I can just read the Kh value directly from Table 26.10-1 as 1.08. I will reply to your other comment regarding that value on your other post.
Might be better to plot change of strain vs log(signma') or even log strain vs log(sigma'). The first gives Cc/(1+eo) directly. The second gives two "straight lines' to give preconsolidation pressure (better estimation)
Thanks for your comment. This video is aimed at folks at the first-semester soil mechanics level, which cover this particular topic. I have a few other videos that cover other relationships, including what you mention. See links below: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-1LGJJrJA52o.html ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-K4Itl4T540g.html
@@mathandengineeringwithdr.a9515 - perhaps an interesting story - probably for the pre-consolidation pressure determination - but my mentor studied under Terzaghi and Casagrande at Harvard (my mentor is 92 and still consulting) - anyway Casagrande came up to the Company's office in Toronto as he was consulting on one of our projects - anyway a young chap had done all the preconsolidation constructions on a number of samples and proudly showed them to Casagrande who looked at them and said - "well, this should be about here; and that one there" - but, the lad replied, I did them as per your procedure . . Casagrande looked at him and said - well, I had to develop something so the students would get in the right ball-park! Cheers
Thank you Dr. A for the great video. I'm new to ASCE 7 and it was very helpful. Online hazard tool doesn't show values for places outside the USA. How can we use this method for other countries?
For people who don't know why their graph goes up instead of down, consolidation sinks your specimen, and your dial gauge reads the delta from its original position, and not the actual height relative to the table or the floor or whatever.
Yes. Thanks! Also, depending on how the dial gauge was zeroed, the readings will either decrease or increase in value. Regardless, though, the specimen height decreases during consolidation.
You look very geotechnical engineer 😂😄.. Btw actually I didn't like soil mechanics like other structural subjects because it has a lot of work and formulas 😔
Why you assume the sand is dry then compute gamma dry instead of calculating the gamma wet or wet density as we measure on site with the gauge or ballon test...instead I calculate gamma dry to compare with the proctor [ gamma dry from lab] and calculate the relative desnsity....thank you
Good question. In this problem, we assume the soil above the water table is dry. Also for this problem, we are not given a moisture content for the sand layer.
You can find the point on the beam where V = 0 by setting V(x) equal to zero and solving for "x". This value of "x" will also be where the moment is a maximum. Is that what you are asking?
Why do we compute o' not, at the middle of the clay layer and not at the bottom? None of my books explain why we have to do this. I'm going to take a shot in the dark and say it's because thats where the maximum pore water pressure is during primary consolidation thus resulting in the least amount of primary consolidation
Not the instructor, also a student but I believe this is correct. If a soil is able to drain in two directions you use the middle. You do not use the middle though if it can only drain in one direction
Thanks for your comment. Some of this information is original. Some of it is paraphrased/re-worded from different books and online sources, including "Fundamentals of Geotechnical Engineering" by Das and Sivakugan.
I am also doing the same calculations. My shear diagram would close to zero but my moment diagram won’t. Is there an explanation for that one. There’s a really big excess moment at the end of my strip which I couldn’t explain why.
Hi Carolina! It seems you are correct. It's interesting that no one else has noticed and commented after so many views. Thanks for catching it! I will make a correction video later this week and give you a shoutout if you are OK with it!
Good day. Thank you for your question. "T90" is the time factor that corresponds to 90% primary consolidation (U = 90%). T90 = 0.848 for U = 90%. This value can be found in most Soil Mechanics books and references. Look for a table in a Geotechnical or Soil Mechanics book entitled "U vs. T" or with a similar title.
