This video introduces the strain tensor and its interpretation. Lectures created for Mechanics of Solids and Structures course at Olin College.
Пікірлер: 24
@angelizquierdo32653 жыл бұрын
Just with that excellent explanation of the displacement vector field you deserve a like and a sub. Awesome video! Thanks a lot!
@ahmetozbekler3 жыл бұрын
Great explanation. Very clear and understandable. Thank you.
@why_are_kishore2 жыл бұрын
best explanation on strain so far
@eduardoschiavon56523 жыл бұрын
Great explanation, thank you!
@giovanniferreira8023 жыл бұрын
Great video! Thank you very much! =D
@javierramon87212 жыл бұрын
Thank you this explanation was very useful!
@howtoscienceandmath2 жыл бұрын
Very Awesome!
@omaryehia3572 Жыл бұрын
You, sir are a legend!
@hamaschwa Жыл бұрын
Very helpful, thanks a lot!
@얼음소년9 ай бұрын
저는 한국인 입니다. 좋은 설명과 내용을 감사합니다
@montyd74212 жыл бұрын
Nicely explained, pls do it for 3d case as well
@eduardoschiavon56523 жыл бұрын
Mr. Storey, will you explain failure theories in this course?
@thomasvarghese40853 жыл бұрын
Tysm
@leophysics2 жыл бұрын
Sir I have doubt Under rotation d(thita)n(cap) transform like contravariant or covariant . About same axis of rotation
@gaiuspliniussecundus1455 Жыл бұрын
What about the large deformations case? Any pointers?
@rudolfzhou8842 жыл бұрын
Wonderful jobs! though it takes time to relaize for me
@MukitAmin2 жыл бұрын
Best
@jokerman92952 жыл бұрын
11:33 should that not be the partial derivative of U with respect to x?
@ThePlazmapower2 жыл бұрын
yh but I'm sure it's a mistake either way d and ∂ look the same and doesn't really matter unless you're doing a degree in Mathematics
@jokerman92952 жыл бұрын
@@ThePlazmapower No i mean it should be a partial of U with respect to x, not y. Right? Because the U vector is point right, just like he did the partial of U with respect to x in the example at 8:24.
@raiden57362 жыл бұрын
@@jokerman9295 I understand what you are asking. It is a Taylor expansion of multivariable function but you only keep linear terms (terms of power 1 in x and y) of the function U=U(x,y). You are looking for the change in U as you move along ''y'' and only ''y'' (partial U / partial y). NOTE: The expansion of U(0,dy) will produce the vector pointing AA' which point in the x direction. But you still evaluating U only from (0,0) to (0,dy), that why you use the change of ''U'' as you change ''y''. Let me know if this is clear enough, this topic are confusing and i'm glad to help. -Best Yael
@raiden57362 жыл бұрын
Just a final comment... Formally you should have this: U(0,dy)=U(0,0) + (Delta y)*(∂U/∂y )|_(0,dy) + (Delta x)*(∂U/∂x )|_(0,dy) + (higher order terms). How ever "Delta x" is just (x_A' - x_A')=(0-0)=0 so the term '' (Delta x)*(∂U/∂x )|_(0,dy) '' vanish, in addition you ignore '' higher order terms''. So the final result of your expansion is simply: U(0,dy)=U(0,0) + (Delta y)*(∂U/∂y )|_(0,dy). \m/ Best Yael.
@raiden57362 жыл бұрын
@@jokerman9295 Just a final comment... Formally you should have this: U(0,dy)=U(0,0) + (Delta y)*(∂U/∂y )|_(0,dy) + (Delta x)*(∂U/∂x )|_(0,dy) + (higher order terms). How ever "Delta x" is just (x_A' - x_A')=(0-0)=0 so the term '' (Delta x)*(∂U/∂x )|_(0,dy) '' vanish, in addition you ignore '' higher order terms''. So the final result of your expansion is simply: U(0,dy)=U(0,0) + (Delta y)*(∂U/∂y )|_(0,dy).