Very very fascinating and extremely well communicated. Most surprising to me are that the location of the load (x) depends on the length of the column and that the change in diameter is proportional to the original diameter. Thanks for such great content!
I got to the same result by using the value for modulus of elasticity earlier, computing both Fa and Fb as a function of the common deformation ΔL and then using equilibrium equation for the beam to solve for ΔL. Your solution was 10 times easier tho
How come published real world data for metals never agree with the E, G, Poisson's Ratio relationship? It is not even close. For example: 1045 Steel E = 200 GPa G = 80 GPa v = .29 If you were to calculated Poisson's Ratio from E and G, you would come up with .25, not .29. I have checked many materials, and found the real world G vs E is not consistent with G = E/2*(1+v)
F column B is F column A, and the opposite is true. You mistook the result, because it's not logical the have a weak force on A then B, when 80KN is applied much close to A. So, FA would be 40.93KN and FB = 39.07kn. For this, the new diameter will be, 30.008316 mm
You guys probably know by now, but for anyone else wondering, he goes through this in Lesson 13 - Shear Stress Strain Diagram. Which is the next lesson
Probably too late but since it is just a column it is equivalent to a roller, which has no moment, only a force perpendicular to the axis the roller is on
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Dr Hanson, You are the best, #1 by far. Thanks! Hope you finished the dynamic and thermodynamic courses as well.
Very very fascinating and extremely well communicated. Most surprising to me are that the location of the load (x) depends on the length of the column and that the change in diameter is proportional to the original diameter.
Thanks for such great content!
Next series : Mechanics of machinery, heat transfer lets gooo !
if youtube didn't exist i could never be an engineer. thank you mr hanson.
I got to the same result by using the value for modulus of elasticity earlier, computing both Fa and Fb as a function of the common deformation ΔL and then using equilibrium equation for the beam to solve for ΔL. Your solution was 10 times easier tho
at 8:18, wouldn't the 80(x) be negative in the equation?
You're correct, but he does get the correct X value. So he probably mentally self-corrected it and accidentally forgot to change it on the board.
How come published real world data for metals never agree with the E, G, Poisson's Ratio relationship? It is not even close. For example:
1045 Steel
E = 200 GPa
G = 80 GPa
v = .29
If you were to calculated Poisson's Ratio from E and G, you would come up with .25, not .29.
I have checked many materials, and found the real world G vs E is not consistent with G = E/2*(1+v)
Wonderful Lectures ! Thanks.
F column B is F column A, and the opposite is true. You mistook the result, because it's not logical the have a weak force on A then B, when 80KN is applied much close to A. So, FA would be 40.93KN and FB = 39.07kn. For this, the new diameter will be, 30.008316 mm
Fascinating!
Thank YOU Teacher
What video do you go over G?
Wondering the same thing, annoying af
You guys probably know by now, but for anyone else wondering, he goes through this in Lesson 13 - Shear Stress Strain Diagram. Which is the next lesson
@@samasterchief thanks for taking time 2 reply
@@lukecaetano8405No problem!
How is there not a Ma on column A? I thought only pin connections didn’t have moments...
Probably too late but since it is just a column it is equivalent to a roller, which has no moment, only a force perpendicular to the axis the roller is on
Sounds a lot like my…. Never mind doesn’t grow much, but it grows
At 11:40, dividing 39 000 by pi x 15^2 gives us 550 not 55 Mpa.
I love your videos, but is 8 ads a video necessary?
An Adblocker will solve that problem
Take It as a joke, Do we also have to remember the sound effects you make during the Poisson ratio, squeeze it Woooaaaaa, the diameter Woooooooopp.