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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!

This is OneNote and I am using the Wacom One Tablet. Thanks for asking!

Hey! It should be just basic SOH CAH TOA. CAH is cos(theta) = A / A(plane). Rearranging this, we are left with A(plane) = A / cos(theta). Hope that's okay!

I see now. Thank you!@@simple_civil

P=N cos(theeta). Am i correct sir

@@sabugeorge5001 P=N / cos(theta)! Thanks for asking

I love reading comments like this! I'm glad I could make things easier to understand. I wish you the best in your courses

Thanks for watching. Definitely a long solving process!

No problem, glad I could help!

Thanks for the support!

No worries! The strips are bounded by the axis and underneath the curve . So for Iy, we are going from 0 to 1, and we don't need to do any 'subtracting of the 2' since the equation already gives us what we are looking for. In other words, our dA is already bounded in the area we are interested in. Then for Ix, it is different. We initially have dA bounded between the curve and the y axis from 0 to 2, which does not represent the area we are integrating. This is why we take 1 - (y / 2 ) ^ 1/4 to get dA where it needs to be. Definitely hard to explain in a comment, but I hope this helps!

@@simple_civil Oh i see now ,Understood thanks a lot man🫡

@@hylexsenpai5282 Glad I could help <3

No problem, glad I could help!

Hey! To start, the right hand rule is applicable to our sign conventions here. Our thumb points in the positive direction of our axis, and the moment reactions were both drawn in the positive convention (following our grip around the axis). If we look at only F1, we would have moment reactions about the y and x axis. The support allows no moment reaction to develop on the y axis, and the z axis is parallel to our force, so there will be no moment reaction there either. Now, solving for the moment at x (Max), F1 would create a clockwise moment about that axis. This means our reaction moment will be counterclockwise. This also means our assumption to drawing it this way was correct! Hope this helps!

@@simple_civil Sorry but I'm still not one hundred percent -- if we point our thumb in the direction of the positive y-axis, are our fingers, before curled, the direction of the force (for F1 this would mean that the fingers are pointing down still with the thumb pointing along the positive y-axis and for FBC and FBD are pointing up) and the curling defines the direction? Because if I do that, the curling due to FBC / FBD about the positive y-axis would still be counterclockwise. I appreciate you taking the time to help.

Ill give you a breakdown for each problem: 1) There is no internal hinge in the problem, therefore, we set ec = 0. 2) There is 1 internal hinge. m' is the number of members connected by the hinge. Therefore, we have ec = 4 - 1, which gives us 3. Hope this helps!

Hmm, that would be a good idea for a video! Down the road if I start a Intro to Physics Series I could do some stuff on units.

Hey there, thanks for the question! Basically, when the axis is given at the centroid (or center), that means we do not need to consider the Ad^2 part (since d will equal 0). The global and local axis aligned. Hope that helps, let me know!

Yes you would! The trick is at the end, when using the flexure formula, you need to use the modular ratio to convert the bakelite back into PVC to get the equivalent stress in the PVC. Love reading comments like this, glad I could help!

I'm so happy to hear that! Thanks for watching!