@@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.
@@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
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.
@@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).
Just with that excellent explanation of the displacement vector field you deserve a like and a sub. Awesome video! Thanks a lot!
Great explanation. Very clear and understandable. Thank you.
Great explanation, thank you!
best explanation on strain so far
Great video! Thank you very much! =D
Thank you this explanation was very useful!
Very Awesome!
You, sir are a legend!
Very helpful, thanks a lot!
Nicely explained, pls do it for 3d case as well
Tysm
Mr. Storey, will you explain failure theories in this course?
저는 한국인 입니다.
좋은 설명과 내용을 감사합니다
What about the large deformations case? Any pointers?
Wonderful jobs! though it takes time to relaize for me
Sir I have doubt Under rotation d(thita)n(cap) transform like contravariant or covariant . About same axis of rotation
Best
11:33 should that not be the partial derivative of U with respect to x?
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
@@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.
@@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
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.
@@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).