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I am so glad I could help. I wish you all the best in your studies!

Good question! I think it comes down to the type of system you are analyzing. If external forces are applied to a pully for example, then they must satisfy equilibrium. But, if the pully is supported by a frame in this scenario, then the frame members must also satisfy equilibrium. For this to happen, forces exerted by the pully, and reactions in the member must be considered. Let me know if this helps!

Thanks for your support, it means a lot and I'm glad I can help! And of course I am open to topic recommendations, so feel free to let me know.

@@simple_civil I am glad and privileged to be offered help. Statics is not my strong suit but I hope with all my heart I pass. Here's the list: 1) Equilibrium of particles 2) Force system resultants 3) Equilibrium of a Rigid body 4) Structural analysis The topics are very broad, and I happen to only need help with specific questions. If you are open to having a look, I will be very happy to email them to you.

@@ansarjawahir2404 Thank you for the recommendations. I have some of these concepts on the channel, but can definitely work to add additional problems to my catalogue. Stay tuned

Good question! I just worked with the dimensions I was given. If we treat the bracket as the hypotenuse of a triangle, we have a 1.5m base and 2m height. This is equivalent to the 3 4 5 triangle i draw later in the problem. From there, if you wanted the angle with respect to x, you can do tan-1 (3/4). Hope that helps!

Hey! Unfortunately I have not covered any other problems for strain rosettes, I have done other videos on on principle stresses that cover very similar solving procedures. Check em out in the playlist here! ruclips.net/video/79XX7_RtLBU/видео.html

Cheers, glad I could help!

Glad I could help my friend!

I think the simplest answer is because 'in-plane' refers to the stresses developed at the principle angle. For 'max stress', you can simply plug in using the given stresses. However 'max in plane stress' must be derived using principle stresses, which are acting on the principle plane. Hope this helps!

Glad I could help! Thanks for watching <3

Glad I could help, thanks for watching!

Glad I could help <3

Love seeing comments like this, I'm glad my content has been helpful and I appreciate the recommendation! By the way, this is CSA A23.3, and typically European and American code is very similar. Differences tend to be in the code notation and how information is organized, but the principles used in all codes are the same. Thanks again for your comment.

Good question! Basically, your neutral axis is the transition point in your section between tensile stress and compressive stress. This means our y value for tensile stress is the total depth of the tensile region. This values is based on our tensile region which is from the base of our section, to the neutral axis. Thus, y bar is used! Hope this helps.

No problem, you can find the link to the playlist below. Currently working on making more videos so stay tuned! ruclips.net/p/PLu2TVt_O6kVXCcGHzQFpK_pBdyJpQDtEw

Haha well I am glad I could fill the gap!

Good question! For the given triangle, the centroid is not exactly at 1/2. If you search up 'right triangle centroid' in google images, you will see that we need to move 1/3 from the left of the triangle to reach its centroid. Hope this helps.

​@@simple_civil yeah it does thanks. I really enjoyed your class,thumbs up man.

@@edwinanthony2383 Glad I could help my friend

Haha, Parallel axis theorem definitely gets harder with exams! But as long as you have strong fundamentals, you can solve any problem! Thanks for watching

Thank you so much! I appreciate the support

An example that comes to the top of my head would be having an an axially loaded member, fastened together in the middle by a bolted connection, but the connection between the two members is at an angle instead of perfectly perpendicular to the load. I'm sure there are more examples in textbooks, but I also assume that the most likely used for these formulas would be for material testing purposes. Hope this helps!

Bit confused by this. The hypotenuse length is not really relevant to the problem and the dimension of the hypotenuse is not mentioned explicitly. Sorry!

Thanks for the support, I really appreciate it!

Give it a try! It is a similar procedure, just now use the y axis as your reference axis!

@@simple_civil I did got like 10.3x10^9

Glad I could help!

@@simple_civil the evaluation was today, i think the mass geometry exercice, from the mechanical engineering mechanics I course is full correct. Thank you a lot

@@GB-eq4db That is amazing! Glad I could be there to help!

Thanks for the support, love from Canada!

You can do either, but typically you convert the concrete to steel. It is much easier as the slab is a simple rectangle, while the beam has multiple individual rectangular sections. Additionally, for analysis, you would take the tributary area of that slab above the beam, and use that as your untransformed b value. hope that helps!

Thanks! I use OBS for my recordings, it is a free software

Nevermind. :) It is actually really just as simple doing it from the right as it is from the left. I just have the roller support as the constant rather than the pin support. (And I just need to keep track of where I'm measuring x from. But I got the right answer, so it works! Thanks!!)

@@sceneryj Haha glad you were able to test both ways. At the end of the day, whatever is intuitive for you, go for it. Glad you enjoyed the video as well

Wow! I am so glad I was able to help you and your classmates! Thanks again for the support my friend

Hey! thanks for the support by the way. Like I said in the video, it is best to break down a four member joint into a FBD to imagine the reactions taken by each. I believe in the case you are asking for, we would have four y components and three x components in the members cumulatively. So in this case, it is hard to say without analyzing other joints to determine some unknowns in that joint first. Hope this helps!

Glad I could help!

Glad I could help !

No problem! Glad I could help my friend

Hey! Its a special triangle (also called a 3 4 5 triangle). You can use 1.5, 2 and 2.5, but its just common to write it as 3 4 5 when you have the equivalent relationship. Hope that helps

I believe the first part of the equation, you end up with the wrong sign. -(-65-(-125))/2 sin (2 x 145) should give approximately a positive 28.2 and the 75cos (2 x 145) should give approximately 25.6. Hope that helps!

Hey! I did this video a while ago, but Ill see if I can help. From the FBD (free body diagram) I have, Fkd is drawn in the positive x (pointed to the right with respect to the joint) and the negative y (pointed downward with respect to the joint). So the signs in the equations seem to be right based on what is drawn. Then when Fkd is solved, we end up with a negative value, meaning the Fkd drawing needs to be flipped and pointed towards the joint. Then, based on our convention, we are pushing the joint. Thus, Fkd is a compressive force. Hope this helps!