Flitch Beam - Design example of flitch beam
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- Опубліковано 30 вер 2024
- In this video we have to design for a flitch beam due to depth limitation of the supporting new beam.
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Can you tell me how to calculate the req'd. inertia? thank you
Can you give us a quick explanation why you choose two 1/4" steel plates instead of one 1/2" steel plate. I would think an (LVL - 1/2"steel - LVL) beam would also work.
Probably because he needed the added bending strength of the additional lvl and 3 would help evenly sandwich the plates
useless if you skip a the part about how you get required inertia
This is the most Proper, Process and Design Calculations
Dear Sir, I have watched your video and found it is quite useful for my study. I love the way you explained thats why i message. just wondering if you have time for my question. I am doing calculating about flitch beam, in which steel plate is by I shape, and considering the dowel connections. I am not a civil engineering so I would like to learn from you. Could you please kind enough to give me some clues on this field. Thank you so much and looking for your consideration!
ok so a couple things so far. you have the wrong value in for W as you don't multiply by the length of the beam you multiply by the trib, and when you are solving out for b, I have no idea how you came up with 0.36, with the numbers you have provided it should come out as 0.29. do us a favour if you can and show the math.
Thanks for the video
Thank you for pointing that out. The trib width is 10.66ft. 55psf*10.66ft=587 plf. I used 595 so the numbers will be off by a little bit. As for the width of steel it is 0.29. I must have plugged something wrong in the calculator. However the process of the design of the flitch beam is correct. For future videos I will be more precise with the results. Again thank you.
Exactly correct -the wrong formula was used. The video should be corrected. A 50 PSF TL applied to a long beam (100') with a very narrow tributary length (2') should generate a low 100 PLF (2' x 50 psf). But by this incorrect formula multiplying 50 PSF times the beam length of 100' the result would be 5000 plf which would be tremendously off.
w should equal to 55*10.75' you times the length again
you forgot the shear transfer at the throughbolts from the wood to steel - pretty important
Didn’t forget I just didn’t have the time to put it in this video. I will do another video showing the shear transfer. Thank you though.
What are you waiting for? Have been 5 years ? Come on, kidding, awesome, thanks for sharing@@FGJEngineering
thanks man, great video
Really helpful!
Hi Paul. I can't seem to come up with the Ireq'd of 275 or 300 in the example.. if w=591.25 and L=12.75 and E=1700000(?) .. I get a number of 2.25 for I req'd. What could I be doing wrong?
Sorry, had to correct units.. got it now
@@lesdupre466 I'm still not getting 275 or 300 with the correct or incorrect w. Also, i see you used 1.7 x 10^6 for Ew but he used 2.0 x 10^6, which one is correct?
@3:07 min you multiplied 55 x 12.75 by my calculation that is 701.25 not 592.25. Did I miss something?
chvydrptop he was supposed to multiply it
by 10.75
Nice video, but isnt it pounds per square feet. Not pounds-square-feet. I dont doubt his result.
You’re correct sir. For future videos I’ll say it correctly. Very respectfully, Felix
@@FGJEngineering yes, but more important to know what lbs per square foot means so that it is not just plug and play equations but rather engineering reality that you are communicating.
How do you calculate the required Inertia?
Hey Paul, for this example we have a simple span beam so to find the required inertia it would be I=((5wL^3)/(384E))*240. The 240 is for D+L. So if your looking for deflection of L/360 then substitute the 240 with 360. Please let me know if I was clear.
is it I=((5wL^3? or L^4?
Peter D'Angelo it's L^3. In the deflection equation we are solving for the required inertia. If we say deflection is L/240 and we solve for inertia, the L^4 becomes L^3
Thanks for the video. How can required Inertia for Dead+Live (55psf) be less than for Live (40psf) only?
+John Endacott it has to do with the deflection criteria.... L/240 for dead and live and L/360 for live... in this case even though live load is 40psf the live load criteria makes the required inertia large