Traction Vector

[@MegaVDubMan]

Excellent video thanks for the help!

[@ilangomahalingam7159]

Excellent!

[@shashvataghosh6313]

Sir if we use cylindrical or spherical system for the same tetrahedron you solve, will the answer remain same? What will be our approach if we want to solve this same problem in other co-ordinate system.

[@ajeet3856]

Dear Shasvata, are you suggesting how to solve this problem if the tetrahedron's planes are curved (and not flat)? I am guessing it will be difficult to begin with. In the limit of tetrahedron vanishing to a point, its curved faces will be better approximated by flat planes and then one can use our derivation. Another query could be how to solve the tetrahedron problem if the three planes (other than the inclined plane) of the tetrahedron are not perpendicular to each other - this is solvable but will be more difficult than the derivation done in this video.

[@scythe36]

Sir, why did we choose tetrahedron for our analysis? Is it possible to derive this formula using another 3D shape ??

[@ajeet3856]

a very good question. one can start with a shape having n (more than 4 ) faces too but the final result will simply relate tractions on each of the n-planes. It turns out, at a point, traction on 3 planes (at most) are independent of each other: any fourth plane's traction will be related to these three tractions. This gets nicely proved with the tetrahedron shape.

[@scythe36]

@@ajeet3856 Thank you sir

[@dokepravin2916]

yes sir enjoyed!

[@priyaluckytiwari]

Sir you explained very well. I am trying to think when we cut the body in two part using imaginary plane and show the inter force on exposed surface But not showing moment vector Why do not present moment? I can think but not ensure My thinking are Moment is not part of traction Or individual moment at a point will be zero as due dimension of point Thanks

[@ajeet3856]

Very nice question. Your thinking is right - moment per unit area would be zero at a point. This is classical elasticity picture. In more advanced elasticity theories (such as cosserat elasticity), one will have both traction (force /area) as well as moment/area distributed over the cut surface.

[@MEBMEESALACHANDU]

sir, those 3 independent planes should be perpendicular to each other?????? (sir if we consider those 3 planes not neccesarly be perpendicular then how can we assign e1,e2,e3 along those edges).

[@ajeet3856]

They do not need to be perpendicular but the derivation gets more complex if you choose such a set of three planes.

[@DidYouknow-ze8nd]

the force calculation is not meant for any single direction like in x y and z.why in tetrahedron

[@ajeet3856]

I am unable to understand your question. We are simply using Newton's 2nd law in its vector form (no specific direction is chosen). In the end, we get a vector equation relating traction vector on an inclined plane with traction on set of three perpendicular planes. ANother way is to do force balance x,y and z directions separately but that would require us to do three separate calculations.

[@nncgaming2637]

Sir why is it -e1, -e2

[@ajeet3856]

it has to be normal to the tetrahedral surface and pointing out of the tetrahedral volume. therefore you have those normals with minus signs. remember whenever you want to obtain force on a part of the body from remaining part of the body, the normal vector must be taken pointing from the part of the body (on which total force is needed) into the remaining part of the body

[@piyushraj9749]

@@ajeet3856 what is the exact tetrahedron shape? Is O is the meeting point of A,B,C or there is another view? I am unable to imagine the tetrahedron correct shape with these notations. In actual tetrahedron which point is O , A ,B,C ? PLEASE ANSWER IF YOU KNOW

[@ajeet3856]

Please listen to the video lecture starting at 24:30. It has been explained what these four points are and what are the four planes of the tetrahedron. O,A,B,C are the four vertices of the tetrahedron