Mechanics of Materials Blog

Generic Fixed Fixed with Triangular Load

2009-05-31T23:32:00.000-07:00

I'm not sure what has come over me ... but I decided to crank out the making of the fixed-fixed beam with triangular load into a generic solution.

Baby! ...
http://www.associatedcontent.com/article/1802670/generic_solution_for_a_fixed_fixed.html?cat=4

Scary! Fun.

Updated the Superposition Examples Article ...

2009-05-30T18:41:00.000-07:00

I update my article on some examples of the use of the Method of Superposition ... cool.

Here ... http://www.associatedcontent.com/article/1155162/example_solutions_of_indeterminate.html?cat=4

Generic Expression

2009-05-30T14:50:00.001-07:00

... here is an article showing how we can take a specific solution and make the answer `generic'.

http://www.associatedcontent.com/article/1798856/generic_results_of_solving_for_the.html?cat=4

Deflections by Integration - Stat Det Beam

2009-05-30T14:45:00.000-07:00

Okay ... here is the solution for the deflection of a statically determinate beam, simply supported, triangular load starting from zero at one (left) end and ending at the maximum at the other (right) end. The answer is in many textbooks (and manuals). In earlier articles we did this beam with fixed end conditions and with fixed ends with one end displaced.

http://www.associatedcontent.com/article/1797334/deflection_of_a_statically_determinate.html?cat=4

In the solution our boundary conditions are ...

1. Known shear at one end.

Known Shear at the other end used as a check.

2. Known moment at one end.

Known Moment at other ends used as a check.

3. Zero displacement at one end.

4. Zero displacement at the other end.

Cool.

Load Combinations, ASD ...

2009-02-14T15:07:00.001-08:00

here ... http://www.associatedcontent.com/article/1476848/load_combinations_and_load_factors.html?cat=4

Spacing of nails, rebar, etc.

2009-02-13T16:45:00.001-08:00

... this may be helpful ... students seem to get this messed up.

Spacing ... http://www.associatedcontent.com/article/1474217/spacing_of_nails_bolts_and_rebar.html?cat=15

Diaphragm and Shear Wall info ...

2009-02-02T19:11:00.001-08:00

hey, anyone following this blog ... here is a link to the APA design stuff on shear walls and diaphragms. Good stuff. APA is a big player ...

http://www.apawood.org/pdfs/managed/L350.pdf?CFID=304277&CFTOKEN=24478590

Fourth Exam ...

2008-12-01T21:40:00.001-08:00

Review ...

1. Euler Buckling of columns ... Pcr and sigma cr.

2. Eccentricity ... The Secant Formula

3. Lateral Torsional Buckling (basic)

4. Design Equations for Allowable stress / load ... (Steel and Wood ... gnarly!)

5. Plane Stress, Strain, Moh's Circle, and Hooke's Law ...

Secon Exam Review

2008-12-01T21:35:00.000-08:00

Review ... 2nd exam ...

1. Torsion ... maximum stress

2. Angle of twist

3. Round shapes and rectangular (with respect to the above)

4. Statically determinate (series) torsion

5. Statically indeterminate torsion

6. V,M Diagrams ... yes, be able to do them ... also find them for simple cases

7. Flexure formula ... sigma = My/I

8. Extreme fiber stress ... f = Mc/I = M/S

9. MOI, S

10. Shear stress ... tau = VQ/Ib

11. Shear stress for a rectangular section ... tau = (3/2) V/A

12. The transformed section ... n = E1/E2

(That's a lot of stuff!!!)

Review - Test 1

2008-12-01T21:32:00.000-08:00

Review ... for the First Exam,

1. Stress, Strain, E

2. Stress-Strain curve, yield stress, proportional limit, elastic behavior, plastic behavior

3. Axial load ... delta = PL /AE

4. Series systems ... (statically determinate)

5. Parallel systems ... (statically indeterminate)

Test No. 4 ...

2008-11-19T15:50:00.000-08:00

If I haven't said so already, you need to be able to get G from E and v, or get one of any of the variables from the other two, ... via, for example, the equation on p. 501, namely 7-38.

draft Secant Formula Calcs ...

2008-11-15T14:06:00.001-08:00

are here ... http://www.associatedcontent.com/article/1213007/stability_3.html?cat=4

Coming exam ... things to study.

2008-11-14T11:32:00.001-08:00

Here is what I am thinking you should study and be familiar with for the Fourth Exam ...

1. Make sure you understand the idea of weak and strong direction with regard to a column buckling.

2. Me able to calculate the slenderness ratio for a column (with respect to weak and strong directions). Note the slenderness ratio calc is not the same for a wood column vs. a steel column.

3. Related to the above ... understand the `radius of gyration'.

4. Be able to calculate the basic Lateral Torsion Buckling capacity of a beam, given the needed section properties. (Make sure you use the correct I value, etc.)

5. Be able to calculate the maximum flexural stress of a column subject to an eccentric axial load using the Secant formula. Also be able to calculate the corresponding deflection.

6. Given stresses (normal and shear) on faces oriented in one direction, be able to obtain the stresses and strains on faces oriented any other direction, including principal stresses and max shear stress. Be able to sketch the stresses as they act on the rotated shape.

7. Ditto with 6 in regard to strains.

8. Be able to go from planes stresses to strains ... and be able to sketch the strained element.

9. And be able to go from strains to stresses.

10. And with 8 and 9 above ... make sure you understand the difference between plane stress and plane strain ... and which equations to use how, accordingly.

Strain gages will not be on the exam.

Inelastic buckling (to be covered next week might be). (I haven't decided yet.) Pressure vessel stuff won't be. (But that doesn't mean you shouldn't read up on it.)

Strain Gage Equations ...

2008-11-12T20:32:00.000-08:00

So, did I tell you? ... for the equations you might want to use for the strain gage problem ... consider the stuff on p. 524 of the text.

Assignment 34, Problem 5 ...

2008-11-12T20:09:00.000-08:00

Okay, Problem 5 ... we assume the thing is in plane stress. What that means is that ϵ_z is some number, not necessarily zero, and also not known. (The strain gages didn't give it to us.) So, if we want to use the equations on p. 507 (7-54a,b) we don't have anything (any number) to dump in for ϵ_z (7-53c) into the equations for plane stress (7-54a,b). And I think we'll get the same thing as 7-36a,b on the top of p. 501. And then we can proceed.

That is so cooool!

2008-11-12T17:45:00.001-08:00

Okay, I just finished the first 3 probs of Assn 34. What is really cool is that the strain in the z-direction (the bulging or squeezing) ϵ _z is the same regardless of the orientation of the stuff in stuff in the x-y plane. Which makes sense, if you think about it. I get ϵ_z = - 60 x 10^-6 in both cases.

And hint ...

2008-11-12T16:49:00.001-08:00

If you need `G' ... consider equation 7-38 on p. 501.

Assignment 34 is up ...

2008-11-12T16:37:00.000-08:00

For assignment 34 ... use ν (nu) = 0.30 for Prob. 1 - 3 and ν (nu) = 0.35 for Prob. 5.

Assume the element in Prob. 1 - 3 is in `plane stress'.

(I have not worked these yet - so give me a shout if there are any `problems'.)

Bracing Length ...

2008-11-05T21:58:00.000-08:00

Gang, with regard to Assignment 31, I have provided an example at the end of the following link.

In other words, if you previously downloaded this example, you may want to look at what I added to the end.

http://www.associatedcontent.com/article/1180159/basic_lateral_torsional_buckling_example.html?page=2&cat=4

11.4 - 8

2008-11-05T18:33:00.000-08:00

If you sketch a column with ends fixed against rotation, but one end displaced, it looks kind of like half a sine wave ... me thinketh that would be the same as the pinned-pinned case ... Le = L (k = 1.0).

Slenderness Ratios ... (Limiting)

2008-11-03T18:13:00.001-08:00

Limiting Slenderness Ratios ...

1. Wood ... Le/d shall not exceed 50, except during construction shall not exceed 75 ... (where Le/d is the larger of Le1/d1 and Le2/d2).

2. Steel ... kL/r not to exceed 200.

3. Concrete ...

Stability ... Content

2008-11-03T18:10:00.000-08:00

Here ...

Stability,

Part 1 ... http://www.associatedcontent.com/article/1170754/stability_part_1.html?cat=4

Part 2 ... http://www.associatedcontent.com/article/1171012/stability_part_2.html?cat=4

Part 3 ... http://www.associatedcontent.com/article/1173292/stability_part_3.html?cat=4

Let me know if there are any typos, etc.

Doc

AWC TR 14 - Beam Stability

2008-11-01T14:31:00.000-07:00

... next we get into stability. As a sub-topic we may cover beam (in-) stability. Kind of to go along with our course, and also because I am on the Wood Design Standards Committee of the American Wood Council (AWC) ... I/we may be looking at their technical report on beam stability.

So, for kicks and giggles ... http://www.awc.org/pdf/tr14.pdf

Day 27

2008-10-30T15:43:00.000-07:00

... here is using superposition to get the generic solution to the fixed-fixed beam with triangular loading, and our lecture example problem ...

http://www.associatedcontent.com/article/1154680/solution_of_a_statically_indeterminate.html?page=2&cat=4

Here is the solution to the beam with the middle support 1.00 in. low ... and an example of replacing the support at mid-span with another beam.

http://www.associatedcontent.com/article/1155162/example_solutions_of_indeterminate.html?cat=4

Day 26 ...

2008-10-30T15:39:00.000-07:00

Here is the solution to the statically indet. beam with the `zero' far end conditions (Day 26) ...

http://www.associatedcontent.com/article/1147406/beam_deflections_by_method_of_integration.html?cat=4

And here is the solution with the bent and deflected far end conditions ...

http://www.associatedcontent.com/article/1149072/beam_deflections_by_integration_statically.html?cat=4

Don't say I never gave you anything.

Columns ...

2008-10-29T16:59:00.000-07:00

At some point, maybe after the exam, we'll get into columns ...

Here ... http://aitc-glulam.org/pdf/Column/DF-2-1_00.PDF

Assignment 26 ...

2008-10-28T21:53:00.001-07:00

So, for the beam without the truck crashing into the frame ... the equations for V and M will give us

Vmax = 1750 lb at the right support,
M pos max = about 1070 lb-ft between 5 and 6 ft from the left, and
M neg max = 2500 lb-ft at the right support.

And I think in class we got, or at least talked about, the max deflection of about 0.14 in.

... then the truck hits the frame and we re-do the problem with non-zero right end displacement and rotation. So, it's a lot of the same calcs, except that you get (some) different constants of integration.

You will thus end up with different equations for V, M, angle, and v.

With them I get ...

Vmax = 1752 lb at the right end ...
M pos max about 1010 lb-ft near the mid-span, and ...

(gotta leave something for you to do.)

I eventually set up a spreadsheet that allowed me to crank out my answer with any combination of the two constants of integration relating to the conditions at the right end. That way I could test my stuff with what we did in lecture.

I will let you know if my solutions will be available prior to the exam, but they are not available at present.

Assignment 27 ...

2008-10-27T14:52:00.001-07:00

Okay, Assignment 27 is up on the Schedule. I think it can be solved.

Today's lecture ... beam with triangular load fixed at both ends.

2008-10-27T11:09:00.000-07:00

Superposition ... Gang, note that on the board today I sketched a case of a beam fixed at one end and pinned at the other, with that other end cranked on by a moment ... and it was this case that we added to Case 4 to (potentially) solve our example problem. The problem is - Case 6 in the book is not the case I describe above, since it is `missing' the pin. So, we ... need to recognize that Case 4 will also accrue a deflection, and include that in the mix of putting all the constraints together.

... now I remember ...

2008-10-27T11:06:00.000-07:00

Okay - here is what I was going to mention: consider a roof joist or truss spanning, say, 32ft from exterior wall to another. The truss is designed to span the entire 32 ft. The roof gets built, and then we come in and frame in some interior walls. What will happen if the interior walls are framed right up to the bottom of the joists (while the joists are `lightly' loaded) and then (later) we get a heavier load???

Assignment 26 ...

2008-10-26T13:55:00.000-07:00

What I am looking for here are the equations for Shear, Moment, EI theta, EI v, v, and max deflection. But I think it would also be helpful to look at max Shear and max positive and negative moments in each situation.

Note that when looking for max deflection you should probably look at the deviation of the curved shape of the beam from a straight line between the ends.

Assn 26 is up.

2008-10-24T11:57:00.000-07:00

Assignment 26 is posted on the Schedule. Remember to be careful with `units'.

Assn 25, cont.

2008-10-22T17:28:00.000-07:00

Okay, it works out ... like I said it would ... but not without a few units challenges. Watch to make sure you have the inches, ft, k, lb, and all that stuff straight.

Assignment 25 ... slight change.

2008-10-22T13:02:00.000-07:00

I added something to Assignment 25 ... express the maximum deflection that you get in terms of L/_____ ... say, L/395, or L/610, or whatever it is you get. This kind of number might be real useful to the architect of the project. In other words, L/360 might be the bare-bones minimum, but when you report to him (or her) ... L/650 ... he (or she) might say, "That's Golden!"

Schedule Changes ...

2008-10-22T11:24:00.000-07:00

Okay - check out the schedule changes ...

http://www.serioushunter.com/EDU/Engr/350/Schedule.html

I moved the bi-axial, axial+bending, eccentric axial load stuff to be due Monday.

If you are stuck ... come with Q's on Friday ... or, even better, maybe email me before then.

Note: I assigned another Moment-Area-Method problem. Since the lecture was optional - so is this problem (Assn 25). I will count it as extra credit. If you weren't in class and want the extra credit - team up with someone who took good notes - as we got the problem all set up in class. Also due Monday.

Assn 23 - MAM

2008-10-21T08:28:00.000-07:00

Okay, sadly, the book doesn't have many examples on the M-A Method. And what's worse, is the examples that the book has - are not very good examples.

For the first problem, it's like where we left off in lecture. Go back to the first example we did in lecture - it was kind of easy because at the fixed end we knew that the deflection and slope are both zero. Hence, any offset relative to the (left) end ends up being the deflection.

Now, take the second example we did, and the first HW problem. At the left end we don't know the slope. Fine. Leave it unknown. Shop around for a place on the beam where we know what the slope is. (The mid-span, silly.) If you chop the beam of the second example in half at the mid-span and turn it upside down - you have almost exactly the first example. The key ingredient being that at the midspan the slope is zero ...

So, then, use the 2nd M.A. theorem from the midspan out to either end, and you get the deflection directly.

Assignment 24 is up ...

2008-10-20T12:45:00.000-07:00

Be sure that you scroll down the page ... there are 3 problems.

Assignment 23 ... Several things ...

2008-10-19T19:12:00.000-07:00

Several things to note on this assignment: ...

1) ... like we did in class, we can solve for the needed EI, and then the needed I, to get a beam that will meet the deflection criteria. There will probably be a number of beams that are `big enough' (have enough `I'). Generally we will want to select the beam that has the lowest weight per foot.

2) ... once you pick a beam, then you need to calc the actual deflection using the I for the beam you pick. In other words, you will calc a minimum I using the L/360 = 0.533 in. But since the beam you end up picking will likely have more I than the minimum needed, the deflection will be correspondingly less.

3) ... interesting, ... the beam I get for Problems 1 and 2 is also good enough for Problem 3 with regard to deflection. (But the beam I get for Problem 3 does not check out good for flexure, Part c) ...)

Oh, look! ...

2008-10-17T14:15:00.000-07:00

In class today we found we needed some 20.6 in.^4 of MOI for the beam ... and we found that the smallest W section in the Appendix was more than enough. But look at the next page ... an S 6 x 12.5 would work (22 vs. 20.6) ... and then on the next page after that ... a C 8 x 11.5 (32.6 vs. 20.6). And there are other shapes available that are not in the Appendix.

The W 8 x 15 was so much more than what we needed that probably the added self weight would not change our final answer much ... but these other shapes ... that are closer `to the wire' ... maybe we should add in the deflection due to the self weight.

And, note, we would still need to check for shear and moment.

Assignment 22

2008-10-16T11:30:00.000-07:00

Okay ... cool ... it checks out `pretty darn close'. Note a couple things: 1) we're calculating deflections here ... serviceability ... NOT safety. We're talking about how something looks, or feels, or maybe a small crack in sheetrock ... not something collapsing. As such, I think we can give ourselves a bit of liberty in calcs and approximations ... AS LONG AS WE KNOW WHAT WE ARE DOING.

... that's my opinion.

Also, ... I'll post some XC for Assn 22 as well.

Assignment 21 ... `answer' ...

2008-10-16T11:09:00.000-07:00

... wow, that was a lot of integrating.

I get that the 5-1/8 x 15.5 and the 5-1/8 x 16.5 are `good' on all counts (bending, shear, LL defl, and LL+1.5DL deft) ... the only thing that doesn't check perfectly is the self weight, if you use the 16.5 in. depth beam. I get that it weighs 20.55 plf @ 35 pcf. I'm not gonna re-do the whole thing for 0.55 plf. Besides, the 35 is `heavy'. Do the check with 33 pcf, then you'll be good with either size beam.

Assignment 21 ... some help on Parts d and e ...

2008-10-15T22:50:00.000-07:00

Okay ... GULP

Parts a) and b) are kind of like what we did up to and including the last exam ...

Part c) ... you are going to have to get the `w' for Live load only ... integrate to get V, intergrate to get M, integrate to get EI v', and integrate to get EI v ... then check the max value of v against the limit of L/360 ... and this much we have kinda done the last 2 lectures.

... now it gets fun (actually ugly) ...

Part d) ... now you need to determine the `w' for Dead load and self weight. Integrate, integrate, integrate, and integrate to get an expression for the v for Dead load and self weight. What we get would be the `immediate' deflection due to Dead load + self weight. But we are going to allow for `creep' ... this means that after a while the Dead load and self weight (which are `sustained') will cause it to deflect even more ... on the order of 50% more. So, the total, long term deflection due to Dead load and self weight will be the equation for v that you come up with ... multiplied by 1.5.

So, the Total long term deflection is going to be vLL + 1.5 (vDL+sw) ...

I'm thinking this equation will have 8 terms (2 for each of the powers of x, x^3, x^4, and x^5).

Find the max total long term deflection ... and test it against the limit of L/240.

...

Part e) ... see if indeed our use of 20 plf for the self weight is good ... using 35 pcf for wood.

If so, and, if all of the checks a), b), c) and d) check out okay ... then you're done.

If not, gulp, then add 1.5 in. to the depth, and do again.

Assignment 21

2008-10-15T21:20:00.000-07:00

Whoa ... Assignment 21 may not be as easy as one might think I alluded to. I mean, in some sense, I was right in how we can ratio things up and down LL vs DL and DL deflection x 1.5 ... but it gets sticky with the self weight of the beam being uniform, and the other loads being trapezoidal. Don't wait until the last minute to start this problem.

I checked ... for our beam in class today ...

2008-10-13T11:07:00.000-07:00

I checked ... for the simple beam loaded with a triangular loading (zero at one end and max at the other) ... the position along the beam of max moment / zero shear is NOT the same position as that for max deflection / zero slope. (Though they are not far away from each other.) But the matter is not trivial - as we come to grips with the fact that it is not, but close, we will be able to use such to `cheat' on other problems.

More to follow ...

Cavities ...

2008-10-08T16:20:00.001-07:00

Oh, the cavities in the GP wood I-joists ... also real sweet for running electrical, mechanical, plumbing stuff.

You can tell that my main `channel' is structure. A general contractor prolly would have looked at the same pic - and first thought of being able to run stuff through the joists.

Sweet!

Angle of Twist ... Wood Members

2008-10-07T22:25:00.000-07:00

Here is the angle of twist from the Wood Handbook. What's embarrassing is that I have that book right next to me on my bookshelf, and probably have (or had) it also on my computer in digital format ... but found `it' by a Google search on the Internet.

http://www.fpl.fs.fed.us/documnts/fplgtr/fplgtr113/ch08.pdf

Second Exam, cont. ...

2008-10-07T14:43:00.001-07:00

No, shear flow is not on the exam.

Second Exam ...

2008-10-07T14:09:00.000-07:00

Make sure you understand everything we covered, in particular ...

1) area loads (psf) vs. line loads (plf) ...
2) tributary width
3) V, M ... and be able to construct Load/RXN, V, and M diagrams
4) bending stress
5) shear stress (at any location on any section)
6) be able to calculate the self-weight of a beam
7) and don't forget to recall your notes on our lecture(s) on torsion

Zach and W-Shapes ... so cool!

2008-10-07T13:31:00.000-07:00

Zach came in and I told him everything. (No, not really.) But we got snooping around on the Internet and ran into something cool. First - here's the background. Zach asked about W-shapes for steel beams. Follow along. Generally we use steel for somewhat long spans. Long spans suggest that `bending' will be a big item. For bending we want a large value of I (or S). To get a large value we get the metal as far away from the n.a. as possible - hence, a good portion of the metal away from the middle and located in the flanges (top and bottom).

Shear, as we have seen, is carried by the `middle' of the section. If we have a beam that is going to carry a relatively long span, we don't need as `much' in the middle. Hence the `scrawny' webs.

The Zach asked ... what about wood I-beams? Well, I haven't seen wood I-beams. But wood I-JOISTS are quite common ... employing the same thinking.

Then we found something totally cool. Look at the pic in the attached link.

Note the `thin' webs ... with the wood flanges at distance away ...

And not only that ... remember that in general for simply supported members the shear is greatest near the supports??? That means the demand for web material is greatest near the supports, and decreases out toward mid-span??? So, this manufacturer uses un-needed web material somewhere else ... less web material the farther away from the supports ... YEAH, BABY! That's capitalizm, free enterprise, efficiency, at its finest.

http://www.gp.com/build/product.aspx?pid=6383

Correction to Solution of Prob. 2, Day 13 ...

2008-10-05T16:58:00.000-07:00

The section modulus for the 2 x 10 (used flat) is ...

bh^2/6 = 9.25(1.5)^2/6 = 3.469 in.^3 ... not the number shown.

Thus, the bending stress becomes 1038 psi ... not the number shown.

Span Calculator ... wood joists and rafters

2008-10-04T20:12:00.000-07:00

... here is a `calculator' that I have found useful (for wood joists and rafters) ... you don't need to download any software, or sign up for something - it's on an external site (American Wood Council).

http://www.awc.org/calculators/span/calc/timbercalcstyle.asp?species=Douglas+Fir-Larch&size=2x10&grade=Select+Structural&member=Floor+Joists&deflectionlimit=L%2F360&spacing=16&wet=No&incised=No&liveload=50&snowload=-1&deadload=15&submit=Calculate+Maximum+Horizontal+Span

Assignment 17 ... trick question ...

2008-10-04T18:43:00.000-07:00

Gang,

... note that for a shape that is not symmetrical about the n.a. ... (that) there are TWO values for S (for both the new and origian shapes).

(By `new' I mean the one in the assignment ... with the two 2x8s ... and by `old' I mean the one we did in class, with the single 2x10.)

S1 = I/c1 ... and ... S2 = I/c2 ...

... where ...

c1 is the distance from the n.a. to the top fiber, and c2 is the distance from the n.a. to the bottom fiber (or vice versa).

On W Shape (Steel) Beams ...

2008-10-03T23:08:00.000-07:00

W Shape Designations are cool ...

The designation is `Shape' `Nominal Depth' x `Self Weight per Linear Foot of Beam' ...

So, a W 10 x 12 ...

... is an AISC `W' (Wide Flange) shape ...

... has a nominal depth of 10 in. ... (actual depth approx 10.0 in. ... see Appendix E) ...

... and (self) weight per foot of 12 (12 plf).

Hint for 5.10-7

2008-10-03T17:07:00.000-07:00

See p. 289

Book Prob 5.10-7 ...

2008-10-03T17:03:00.000-07:00

Gang,

There are several `right' answers for the shear part of this problem ...

The author uses eq. 5-48a for his answer ...

On p. 350 one might get the impression that eq. 5-50 is usable.

The AISC Manual of Steel Construction (The `Code') ... specifies that we are to use ...

V(allowable) = Fv d t ...

Where d is the `whole' depth of the beam ... not the d1 in eq. 5-50.

So, you can use whichever ... but just DOCUMENT what you are doing.

Doc

Beam Software Ad ...

2008-10-03T11:56:00.001-07:00

Cool, ... one of the google ads shows some software that handles steel beams, glulams ... and some other stuff like `LVL' ...

("What the ... is LVL!???")

("Well, silly, LVL is `Laminated Veneer Lumber.")

We have talked fairly extensively in class about sawn lumber and glulams ... but there are a lot of other wood products out there - that perhaps we'll talk about on Monday (LVL being one of them). And, of course there is steel. And we have already talked about (reinforced) concrete. Those are the big players for beams.

Assignment 17 ...

2008-10-03T11:37:00.001-07:00

Assignment 17 is up ... dealing with Shear Flow, and springing from the example we did in class. I came up with the problem while riding my bike from campus. Remind me not to do that again (think about about m.o.m. and ride a bike at the same time through traffic - it's scary!).

In the problem I say to `round' the spacing to some reasonable number (say, mult. of 1/2 in.). By reasonable I mean ... some `laborer' is gonna have to measure out the locations of the holes, and then start drilling. Also, be careful which direction you round. Think about it - I think you'll see what I am saying.

With regard to fastening the 2 x 8s to one another ... we simply want them to `stay' together. This would not necessarily be a structural connection. I would probably just specify the fastening required for any multiple-piece dimension lumber beam (in the building code).

(I can look it up (the specific requirement) if anyone is interested ... but where I can I generally direct the contractor to (generically) simply follow the Code.)

In the problem I suggest that if our new S is greater than the original, and the new I is greater than the original, then our `new' beam automatically satisfies bending and deflection (assuming the original did). Do you really believe that? Don't take my word for it - think it through - based on what we talked about today.

Shear stress goes like 1 over Area ...

Bending stress goes like 1 over Section Modulus ...

Deflection goes like 1 over MOI ... (or 1 / EI) ...

(We haven't talked about deflection as much yet - but you can look at some of the formulas in the book. For example, the formula for deflection of a simple beam under uniform load is ...

Δ = (5/384) w L^4 / EI ...

Note the EI in the denominator ...

(You should probly memorize the above formula ... you'll need it later on.)

Standard Sizes for Sawn Lumber ...

2008-10-02T21:28:00.000-07:00

Gang, I have talked rather freely about lumber sizes ...

Your book has kind of a combined table ... p. 903, Appendix F. The first 3 sets of rows are what are called `Dimension Lumber'. The last two sets of rows are what are called `Timbers'. Note that for these we have `nominal sizes' and `actual sizes'. And we have discussed this a bit in class. It's a bit strange, but note that a 4 x 8 dim. lumber is 3-1/2 x 7.25 ... while a 6 x 8 timber is 5-1/2 x 7-1/2. In other words, the timber actual sizes are 1/2 in. less than the nominal sizes ... but the dim. lumber sizes are not that easy.

We have also discussed `glulam' ... short for `structural glued laminated timber'. These are generally described in actual sizes.

Around here ... common widths are 3-1/8 in., 5-1/8 in., 6-3/4 in., 8-3/4 in., and sometimes 10-3/4 in. Depths are multiples of 1.5 in. Sometimes we might refer to a `nominal width' for a glulam ... for example, a 6 in. wide (nominal) would be 5-1/8 in. actual.

More info is here ... http://www.aitc-glulam.org/shopcart/Pdf/113-2001.pdf

More to follow ...

Centroids, MOIs...

2008-10-02T11:47:00.000-07:00

Okay, ... for those of you needing practice on centroids, and MOIs, I should assign dozens or so problems, but I won't. I'll let you pick your own at your own discretion.

Assignment 15 Solution, Problem 1, page 1 ...

The calc that I do to get ybar = 3.94 in. ... that's basically the calc that the author does on. p. 832. The equation that I use to get it is basically his 12-7b, except that I call it a `moment' equation ... and the Area is in the numerator of the left side of the equation (instead of the denominator of the right hand side). It's the same thing. In my calculation I am taking moments about the `top' of the section.

Once the new n.a. location is determined, we need to calc the MOI of the composite with respect to this new n.a., and can use the method the author uses in Sec. 12.5 ... with a specific example on p. 840.

Get good with this (centroid and MOI) stuff. You signed up for the class agreeing (in concept) that you understood it from statics. I realize that you may have passed statics without that understanding ... so catch up now. This is real life use of that stuff.

Allowable Stress and Allowable Capacity ...

2008-10-02T11:37:00.000-07:00

Remember, ... the common way to deal with Allowable Stress Design (ASD) is to deal with it at the `stress' (material) level ... for example,

... is fb (stress on the material under design load) ≤ Fb (allowable stress) ? ...

or

the equivalent `Unity Check' ... is fb / Fb ≤ 1.00 ? ...

ALTERNATELY, and increasingly common in real life,

We can do the check at the `member' level ...

So, what we do is, for example with bending, ... let the extreme fiber stress equal the allowable stress, and solve for the allowable moment.

... in general ... fb = M/S

But letting fb = Fb ...

M = Mr = Fb S ... where Mr is the allowable moment resistance.

And we could do the same for shear ...

Let fv = Fv ...

fv = (3/2) V / A

Then ...

Vr = (2/3) A Fv ...

Yeah, Baby!

Assn 16 is up ...

2008-10-01T20:14:00.000-07:00

Assignment 16 is pretty straightforward ... good stuff. (See schedule.) In both of these problems we won't worry about the deflection, or stability (of the compression zone) issues.

(Those come later.)

Wood Beam with `crack'! ...

2008-10-01T14:06:00.000-07:00

... some more blurbing on side cracks in wood beams ... over on the wood blog ...

http://woodengineering.blogspot.com/

Double XC ...

2008-09-30T21:24:00.000-07:00

Okay ... this is wild ... I posted some DOUBLE EXTRA CREDIT for Assignment 15, No. 5 on the Schedule.

Basically what I am having you do ... is, with the steel just at yield, and supposing the concrete isn't crushing ... let's calculate the applied load that we would be allowed to apply to the slab/joist system. I set the problem up in the `Strength Design' approach - which is what we generally use for reinforced concrete. The other calcs in Assignment 15 do not have factors of safety in them. This final exercise does. Cool. I found that we wouldn't be allowed to apply much. So, in real life, I'd either start over in the design, ...

Or add more steel.

Yeah, Baby!

Clarification on Assignment 15, No. 4

2008-09-30T17:05:00.001-07:00

On problem no. 4 ... I am referring to being under the load of the self weight only (as in no. 3).

Note: even though we don't `count' the concrete below the n.a. for strength ... it still weighs what it weighs, and it's still `hanging there' (hopefully) ... so our self weight for the problem doesn't change.

gouged beam, cont.

2008-09-30T16:25:00.001-07:00

... note that the gouge makes the beam unsymmetrical ... so, we basically have two S values. (S stands for `section modulus', in case I forgot to mention it in class ... the term that is.) Unless we distinguish explicitly between the allowable extreme fiber stress as being compressive or tensile, we test it for both.

... steel yield stress ... REBAR ...

2008-09-29T20:53:00.001-07:00

In our example we used a yield stress for the steel of 60,000 psi. This corresponds to Gr. 60 (60 ksi) steel, which is probably the most common deformed bar reinforcement. Gr. 40 and Gr. 75 are not uncommon, and the (soft) metric equivalents of the above. Recall that we were using 36,000 psi for the yield strength in examples earlier in the class ... but those examples dealt with other things, not rebar.

... extreme fiber in tension ...

2008-09-29T20:51:00.001-07:00

Good point ... generally we take y_t to be from the n.a. to the center of the rebar.

However, if you want to calculate it to the extreme edge of the (round) bar itself, another 1/2 of 0.50 in. away, you may do so. Just make it clear what your are doing.

Assignment 15, more XC

2008-09-29T20:46:00.000-07:00

Okay, I added some more extra credit to Assignment 15 ... namely, for the last problem where we are dealing with the steel just at yield in tension (60,000 psi) ... calculate the corresponding compressive stress (extreme fiber) in the concrete. Discuss. If it is well below 3000 psi ... then, good. Not only good, we will be in the linear elastic range ... so all our calcs are valid. If we start getting compressive stresses near the compressive strength, then the concrete behavior isn't linear, and things get more complicated. If the compressive stress in the concrete exceeds the compressive strength, before the steel yields, that is not so good. Generally we want the steel to yield before the concrete crushes.

Assignment 15 is up ... Draft

2008-09-29T11:30:00.001-07:00

...

Assignment 15 is up, in draft form. (See the Schedule.)

Schedule is here: ... http://www.woodengineering.com/EDU/Engr/350/Schedule.html

I'll blog more to follow.

The Unity Check ...

2008-09-28T20:36:00.000-07:00

The `Unity Check' is basically checking the ratio of applied/allowable and seeing if it is (making sure it is) less than 1.00.

The things we won't think about ... (Gouged Beam)

2008-09-28T19:19:00.000-07:00

A couple of the things we won't think about, at this stage, are ...

1) ... dimension lumber, timber, laminating stock, etc. are assigned `grades' based on the pressence and size of knots, and other stuff, that reduce the strength of the wood as compared to `clear' wood (no knots, defects, etc.). And then these grades are assigned design values from which we get allowable stresses. Each grade, we could say, has an allowable knot size on such a such face, or edge, of the member. Then we take into account the relative size of the knot, defect, etc. with respect to the width of the edge (say, the loaded edge) by means of adjustment factors. What I am getting at is this ... if the gouge makes the loaded edge less wide, then the allowed defect has more of an effect than in the original design. So, in a situation like this, we might somehow try and take that into account, qualitatively, or we might demand that the condition be inspected with regard to any knots, etc. in the vicinity in the gouge.

2) ... the other thing is that glulam beams are generally laid up of laminations of varying strength and stiffness. More is generally demanded of the top and bottom (lams) of the beam, than the inner part. If indeed the beam is manufactured with different quality materials top/bottom versus inner, we should really consider a transformed section analysis with the gouge, as more is demanded of the gouged region than what we would get using a single rectangular section.

(We'll cover transformed section analysis probably tomorrow.)

In the consulting situation that `inspired' this example, I found that in taking into consideration the section loss due to the gouge, and the greater demand on the gouged area due to the layup of the beam, that we (THEY) were right on the `knife edge'.

Options, then, were offered as follows:

a) ( ... whoever gouged it needs to) Replace it ...
b) Come up with a structural fix ... (perhaps addition of another reinforcement lam).
c) Do more precise calculations to see exactly which side of the `knife edge' we are on ...

BUT WHY!??? ... I wasn't the one who gouged the beam.

Gouged Beam ...

2008-09-27T22:37:00.001-07:00

Okay, ... I have kind of come up with a strategy / template for the `Gouged Beam' problem. Link to it here ...

http://www.woodengineering.com/EDU/Engr/350/gouged.pdf

I think it makes sense ... (which is kind of scary, since I am the one who came up with it.)

Rupture Stress, Concrete

2008-09-26T13:51:00.000-07:00

Gang, this rupture stress, f_r, that we talked about in class today (also called the `modulus of rupture' for concrete) ... do NOT take the `7.5' to suggest 2-sig fig accuracy.

... f_r = 7.5 √ f'c ...

... where f'c (the 28-day compressive strength) is necessarily in psi, and f_r is in psi also.

This is one of few places in mechanics where our equation is NOT dimensionally consistent.

Question: ... what do you suggest we use for an `allowable' (flexural tension) stress for concrete? Hint: think about our conversation in class today. Think about the results we obtained in class ... the slab-joist system spanning overhead and it's in-ability to carry even its own weight without reinforcement. Think about the fact that we would actually never, never do that.

Essentially, building codes will only `allow' us to use un-reinforced concrete (`plain' concrete)ONLY in conditions where it will never, never be expected to be loaded in flexural tension. Around here - that means in slabs on grade (supported by the ground), or slabs fully supported by some other means. And in some cases footing / foundation elements.

Design Software ...

2008-09-25T10:08:00.000-07:00

Recently I purchased and have started using `Visual ABC' by IES for my engineered wood calcs. I have used Woodworks SIZER for years. I switched for the exercise of learning new software, and, also, because, in the end, I think I'll be able to do more with it (some calcs that I can't necessarily do with SIZER). But they are both good. IES (Integrated Engineering Software) is at ... http://www.iesweb.com ... Visual ABC isn't as easy to climb into and use (as SIZER), but it is good.

For statically indeterminate frames I have used CADRE (1.02) ... it is a very old version - but so easy to climb into and crank out answers with ... I LOVE IT. Cadre (newer versions) are available at http://www.cadreanalyitic.com ... My computer crashed ... I am hoping I can recover my age-old version.

Cadre solves by the method of `Finite Elements' (FEA or Finite Element Analysis) ... we haven't talked about the method much, except for ... the idea of ...

Q: "How big, or small, does an element need to be?"

A: "An element is arbitrarily big, or small. It is as big, or small, as it needs to be ... to analyze our problem.

... for concrete ...

2008-09-24T20:54:00.000-07:00

The ACI Code (Concrete code) provides the following approximation for rectangular (concrete) solids ...

tau_max = T (2a + 2b) / (a^2 b^2) ...

You will find that it essentially underestimates the stress a bit as it is generous in how the concrete is distributed.

...

Rectangular sections, cont.

2008-09-24T14:54:00.000-07:00

... perusing the lit. you'll find that for a rectangular section you will run into equations of the general form ...

... tau max = alpha T / (a^2 b ) ...

where

... a is the dimension of one (particular) side ...

... b is the dimension of the other side ...

... and alpha is a coefficient that is a function of the ratio of a and b, given either in tabular form, or in some kind of a graph (figure).

... torsion stress for a rectangular section ...

2008-09-24T14:27:00.001-07:00

... from the Timber Construction Manual (TCM), 5th edition, p. 114 ... the maximum torsional stress τmax in a rectangular shape section is ...

... τ max = T(3a + 1.8 b) / (a^2 b^2) ...

... where ...

... T is the applied torque ...
... a is the dimension of the wide (typically side) face ...
... b is the dimension of the narrow face ...
... where τ max is experienced at the middle of the two wide faces.

... the wording of p.113 and 114 in the TCM is not limited, per se, to glu-lam ... though that is the context.

Indeed I found one reference that indicates the equation's use for rectangular shapes in general (?).

... more to follow ...

Design Software

2008-09-24T11:38:00.000-07:00

... here is where you can get some wood design software. This particular link will take you to ... the Woodworks SIZER. You can probably find a demonstration version.

http://www.cwc.ca/WoodWorks+Software/index.htm

Go to U.S. Version, software demo downloads, and so on.

You'll probably find that you'll have to agree not to use the software for commercial purposes, that you can't print results, and that you will not have a very robust list of materials to choose from. BUT, from what we have talked about already in class, you should be able understand the language, parameters, and start using the software to check (or do) some of the stuff we are doing in class.

I have used (licensed versions of) Sizer for years in my consulting work. I also helped them on one of their revisions.

I (also) use some other softwares for my consulting ...

More later, enjoy.

Dedication

2008-09-24T11:28:00.000-07:00

This blog is dedicated to the better understanding of Engineering Mechanics of Materials ... primarily the `basic' stuff encountered in a typical 300 level college course. In this blog we (you students, maybe other engineers, and I) can banter back and forth on the subject, share discoveries, and wrestle with this stuff to the point where it has meaning in real life.

Do not use this blog as a substitute for your college textbook (or coming to class). Your M.o.M. text is one that will be worth keeping (as long as you are an engineer).

Please follow the rules of blogging (whatever they are) ... and let's get going.