Exam 3 will take place on Monday April 20, 6-8 pm in a location TBD. The content of the exam is restricted to Chapters 7 and 8 only (and only the material we have actually covered out of those chapters–Sec. 7.1-7.4, and 8.1 and 8.2)! You can bring 1 sheet of 8.5 x 11 paper with writing on both sides, a pencil, eraser, and calculator. I expect to have no more than 4 problems, and (depending on how complicated they end of being) perhaps fewer than 4.
This table summarizes the relevant digital materials for this exam (click here to access the “Publisher’s Videos”):
| Chapter | Lecture Video | Pulse Pen Recording | Student Solutions (PDF, JPG) | Publisher’s Videos |
|---|---|---|---|---|
| 7. Transverse Shear | Module 16 and Module 17 | Module 16 (no pen recording), Module 17 | Prob. 7-40 (1,2), Prob. 7-50 (1,2) | 6 total, plus Prob. 7-40 |
| 8. Combined Stress | Module 18 (last year’s Module 19) | Module 18 | 7 total |
Note that some of this content is re-purposed from last Spring’s class, since I have not had time to do nice recordings of this year’s lectures. But last year’s are fairly similar, so they still hold a good bit of value. Also, I forgot the Pulse pen one day, so we are missing the audio recording of one lecture. As always, use this post as an open thread for discussion.
7 responses so far ↓
1 tiu2f // Apr 18, 2009 at 12:28 am
I can’t play any of the modules above
2 Tim Ottinger // Apr 18, 2009 at 1:53 pm
Mr. Berger,
I am confused in the Q calculation for problems that ask for the maximum shear stress in the glue necessary to hold the boards together along the seam where they are joined. On page 398, example 7.3, in their Q calculation, they take y’ (bar) as (0.180-.015-.120)m and A’ as (.03)(.150)m. The way I look at it, the shear stress should be evaluated at the joint of the two boards, where the glue is applied, and not in the middle of the board on top, because the glue wouldn’t be applied 0.150m horizontally, it would be only be applied .03m across the top of the vertically standing board.
The way I look at it, y’(bar) should be (.180-.03-.120).
Could you clarify this?
Thanks!
3 berger // Apr 19, 2009 at 10:50 am
Okay, so on Ex. 7.3, I understand what you are saying, but the book’s math is correct. Here’s why. Remember that the Q calculation relies on two quantities:
So the point is that the
parameter does not point to the location of interest. Rather it points to the centroid of the area of interest. Make sense?
4 berger // Apr 19, 2009 at 10:58 am
On the module question…I need to fix this. Done! Try it now, and sorry for the confusion.
5 Plogger // Apr 19, 2009 at 5:52 pm
Is there an old exam uploaded anywhere?
6 Plogger // Apr 19, 2009 at 5:55 pm
Is there an old exam uploaded somewhere?
7 Plogger // Apr 19, 2009 at 6:25 pm
Is there an old test uploaded somewhere?
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