Prove that the warping function should satisfy the laplace...

Name: prove that the warping function should satisfy the laplace equation for torsion of non circular shaft 52306
Uploaded: 2024-07-08T12:23:03-08:00
Duration: 6 min 57 s
Channel: Adi S
Description: prove that the warping function should satisfy the laplace equation for torsion of non circular shaft 52306

Transcript

00:02 From the given question given that dou square theta divided by dou t square is equal to 4 dou square theta divided by dou x square this goes to 1 it can be solved by variable separate method separatable separable method then theta of x comma t is equal to x of x into t of t then substitute in one then x of x into dou square t of t divided by dou t square is equal to 4 t of t into dou square x divided by dou x square then 1 divided by t of t into dou square t of t whole divided by dou t t square is equal to 4 divided by x of x into dou square x divided by dou x square.

01:14 Here it depends only on t.

01:22 Here it depends only on x.

01:26 X depends only on x common these two are common value must be a constant 1 divided by t of t into dou square t of t divided by dou t square is equal to 4 divided by x of x into dou square x divided by dou x square is equal to a then dou square x divided by dou x square minus a x divided by 4 is equal to 0 dou square t divided by dou t square minus a t is equal to 0 then we can see we can set a is equal to minus omega square then the above equation can be written as us dou square x divided by dou x square plus omega square divided by 4 into x is equal to 0 this goes to 2 then dou square t divided by dou t square plus omega square t is equal to 0 this this goes to 3.

02:59 Then the solution.

03:03 The solution of 2 and 3 can be written as.

03:11 Written as x of x is equal to a cos of omega divided by 2 of x plus b sin of omega x divided by 2.

03:32 This goes to 4.

03:33 Then t of t is equal to c cos of omega t plus d sine of omega t.

03:44 This goes to 5...