How to Design a Weld For TORSIONAL Loads | Civil PE Exam Prep

[@adelbadiee358]

Great video! thank you Kestävä. Please keep up the good work.

[@Kestava_Engineering]

Thanks, will do!

[@connoreshleman5772]

I was a welder before becoming an engineer and the key to your question at the end is the groove type/angle. SMAW (stick welding) has a hard time getting to the bottom of a 45 degree groove. If you look at the top of the table it shows SMAW again. Generally speaking SAW produces a more consistent stronger weld because it’s more controlled and often done by a computer controlled machine, but it uses a solid shielding material so it can only be done when the weld is flat. PS most contractors I’ve worked with have switched to FCAW or flux core welding as it increases productivity and you can weld more than 10” at a time, fewer starts and stops.

[@sheldon8089]

This is a great video Kestava, very informative, I am a EIT from Canada and I just wanted to say keep up the great work! Cheers

[@Kestava_Engineering]

all the way up in the great white noth eh? jk - thank you for your support Sheldon!

[@sheldon8089]

@@Kestava_Engineering LOL , yes up in the great north, 10 ply bud ! I watch all your videos and they are awesome I just recently bought the design of welded structures book and it is amazing, I use it along with my CISC steel book at the firm I work for, you were right the book is jammed packed with a bunch of great little nuggets of info. Cheers.

[@luismendietaelena9332]

Hi, loving the videos so far. I wanted to ask you, when taking f_v_x you divide Vx/A. Vx is in kips (10kips) and the Area is in in2 (8+8+12 in2) ; however you get k/in as a result. Could you please clarify? Or is just that the formula should be Vx/Length?

[@mohammadfawzysayed5502]

Good job Kestava; One question though, I would have thought you were going to use table 8-9 from the AISC manual, I think it is much faster and should give the same answer, Calculate the resultant force, calculate the applying angle, go to the table and pick the numbers.

[@mohamedsow6093]

I believe there's a typo in the last formula from the manual; it should be 0.928*D*L. That's why your weld size is too big. the final weld size is 5/16"

[@luismendietaelena9332]

@@mohamedsow6093 The formula is 0.928 kips/in/(1/16" of leg). Since he calculated all results as k/in, then his weld size is correct.

[@joshuaevans1703]

how do you know point O is the point where the extreme stresses are located?

[@joeyfalcon3278]

I would say that this is the case as the least amount of steel is around this point. At point O, there is only steel at the left side to help resist the stresses induced by the loads. However, as you move to the left and closer to the center of gravity of the welds along that 8" weld line, you will have steel on all four sides of a given point that helps resist the stress induced by the loads.

[@ropertmores971]

is this for structural pe or this for Am exam ?

[@omar893]

It's a depth problem. I doubt they will bring something that tough in the morning exam.

[@ropertmores971]

because i construction pe

[@Kestava_Engineering]

like Omar said - this one is a structural depth problem - you wont need to worry if you're not structural specific. you should see this on the construction PE. Cheers!