Statics: Lesson 69 - Moment of Inertia, Composite Shape Method
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Thank you dad
THE FK is wrong with you?!
LOL fancy seeing you here I’ve seen you on mma vids. Small world indeed
Thanks for carrying me through.. glad i wont have to do this after today...you helped me cross the finish line
I'm pretty sure you've had to do it multiple times till now, myself coming back to this video thrice a year lmao
Amazing explanations Jeff! Very much appreciated.
You are the patron saint of statics thank you Jeff
The moment of inertia for the cirlce in the second line should be 38.63 not 36.63. Dr.H says its 38.63 but writes down 36.63
the final answer should be 469.37 in ^4 if you round those totals to 2 decimals.
Another small problem is: he rotated the whole shape (and switched up b & h places)...
But forgot to use the same "switch method" for the triangle....
The correct value for triangle alone is -124.5024m^4
@@samuelchuonyo7022 No he did the triangle correctly.
@@samuelchuonyo7022 The whole object is suppose to be rotated like this when solving for Iyy the triangle is treated the same as the rest.
38.63…. Amazing how your mind can hear something different than you just read a loud to yourself…
thank you, that deviation bugged me😄
DR. Hanson ,thank you for another awesome lecture on the Moments of Inertia and the Composite shape method. These topics are a combination of Physics and Statics.
These videos are quite informative and really helped me understand the topics. I passed my last test thanks to you!!
It's Saturday Dec 11, 2021, 5:24pm, I've been working on the same statics problem for 2 hours. I don't know what's wrong. I come to youtube and I find you, Mr. Hanson. Thanks!! Just Thanks!
Haha same. My final is in 90 minutes and I was stuck understanding the moment of inertia too.
Just want to mention the correct answer is 469.366 in^4. I say this because Jeff accidentally wrote down -36.63 in^4 despite saying -38.63in^4. Other than that amazing video thanks jeff
This guy is freakin dope
You make this fun Jeff. You are my slime. Many thanks!
Thank you so much Sir! May GOD BLESS YOU MORE 😇 💕 you're such a great teacher😊
If I were your child/student though I'm not I'm so very very proud of you😊😇
bless up bless up!
bless up ! Bless up!
Thank god I found this man before my exams god blesses
You’re the boss thank you
You are the reason im about to pass my final. A true God-send
Wow you really saved the day haha. This makes so much sense. You have a gift!
Great professor. I understood it completely !!!
Kudos👍👍👍 to you, Mr Hanson.
Can you teach some of my other classes as well? I have to retake a geologic hydrology class next semester and could really use some help lmao. Thank you for all of your help though seriously, I will recommend you whenever possible to those in need.
Hey jeff!
Love the video, however i believe you have made a mistake near the end with the equation for the triangle. When completing your calculations you set up the formula for the rectangle so that you could swap base and height - you did this correctly when putting in the values to your formula.
However, you did not swap the formula for the triangle yet you still switched the base and height values which i believe is incorrect?
If I'm wrong please feel free to explain this to me as i'd love to know!
Hi mate there is no mistake.. the formula of the rectangle was firstly in the way that is used for Iyy but the formula for the triangle is other way around (Ixx).. for the circle it does not matter anyway(base = height).. in order to have them in the same way he changed the rect. Formula to Ixx too and he turned the whole shape 90 degree in his imagination and calculated the values accordingly.. i hope this helps
how dare you question jeff
Nice Lecture
is there a way i can have the same book?
i really love this channel!
thank you. what about the moment of inertia at left edge?
good job
am i crazy or is the x value in the table for shape 3 6 not 7
he did correct. the triangle he used to subtract is not the one intersect with the circle but it is on the right upper corner that doesn't exist. Look carefully
The Iy'=Iy+Ad^2 is the parallel axis theorem, right? For my professor, the Iy is on left and Iy' is on the right with the formula as Iy=Iy'+Adx^2. Which is right? Thanks so much for these videos! They're super easy to follow and comprehend. Thanks again!
I believe the one on the left is correct.
if i may ask,which book is he using
we love you jefff
broski i rate u fam
im confused on the last part. For the triangle equation, should the base be 3" and the height be 6" ? it seems Dr. Hanson had the opposite
I am wondering the same thing
Im confused as to why you calculated x bar when the question is about y
THE GOATTTTTTTTTTTTTT!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
why you are teaching so coooool???🙌🙌😍😍😍
wow so amazing, thx
I would want to ask the difference between bh³/3 and bh³/12 for the moment of inertia about the x axis.
b*h^3/3 is for the 2nd moment of area of a rectangular shape, about the edge that has length b.
b*h^3/12 is for the 2nd moment of area of a rectangular shape, about its center, through an axis that is parallel to length b
Through the parallel axis theorem, also known as Steiner's Theorem, you can show that b*h^3/3 is related to b*h^3/12:
I_edge = I_center + A*d^2
I_edge = b*h^3/12 + b*h*(h/2)^2
I_edge = b*h^3/12 + b*h^3/4
I_edge = b*h^3/12 + 3*b*h^3/12
I_edge = b*h^3/3
thanks sir
Love from India 🇮🇳
Thanks god for sending you here !
Huge help!!!!
why is the centroid axix of the triangle 1/3 not 1/2 of 3 ? n is it 1/3 of all triangles ? so confused .
Because the mass isn't equally concentrated on both sides of the midpoint of the base. The centroid is therefore "weighted" closer to the altitude edge of the right triangle.
Let the triangle be the area bounded by the x-axis and the equation:
y = -h/b * x + h
Each area element, has a width dx, and a height of y, and thus dA = y*dx
The first moment of area (which I'll call mu) is integral r*dA. About the y-axis, r=x. Substitute in dA and get:
mu = integral r*dA =
mu = integral x*y*dx
mu = integral x*(-h/b * x + h) dx, from x=0 to x = b
Carry out the integral and get:
mu = b^2*h/6
Divide mu by the area of the triangle, and get c, the distance from the y-axis to the centroid:
c = mu/A
c = b^2*h/6 / (1/2*b*h)
c = b/3
Ty sir Jeff
thnxxx so much ur a great teacher
You are the best 😘
May i please ask which type of a textbook are you using sir(name, author and edition)?/ which book would you recommend that i must use to get more practice for this module?
Engineering Mechanics 14th ed by Hibbler
Nice
thankkkk youuuuuuuuuuu xxxxxxx
You saved me
Real G
guys i have a question.If the problem tell us that we have to calcutate the inertia about the x axis, we find the cetroids of each shape and in the paralel axis theorem we take d to be the distance betwen the cetroid of the shape and the x axis?if anyone can responde thank you becuase i am pretty confused.
No….on those problems all shapes are referenced from the x axis, not the centroid. I don’t know why authors use those problems.
please jeff, where is mohr circle kkkkriyng
It's in the solids playlist
professor if i may ask,which book were u using for this online series
@@britrossi4876 thanks man👍👍
💖💖💯
loveya
69
NICE *
Correction:you used the diameter of the circle 2, instead of the radius which is 4. thank you
That is exactly what I thought ..
The commercials in the middle of the video are killing me
please buy a mic
Nice