## Ramakrishnan

Ramakrishnan Vaidyanathan

## 289 Reputation

5 years, 140 days

## Social Networks and Content at Maplesoft.com

With twenty years of Industrial experience and twenty years of teaching experience, I am now as retired Professor, using Maple to teach mathematics subject for students studying X to XII standards.

## More on solution plot...

Enclosed curves give more clear picture of what the solutions mean to tell.

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## Usage of Transpose...

The second  answer given by you was the most beautiful answer for me. I used it as follows.

B:=LinearAlgebra:-Transpose(A)

Somehow i feel insecure with % sign.

Thanks a lot.

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Thank you sir for the above two answers.

Ramakrishnan V

## Transpose usage...

Thank you very much for the answer is very useful.

## Transpose multiplication...

Thank you very much

## Try Axis Range - 4*pi is not a number...

I am just adding my another attempted doc for experts to help me state 'follow the rules'. Maple will not accept the pi as a number in the axis range for a plot.

I tried to enter 4*pi as one of the axis range in plot prperties. Maple says it is not a number and in my opinion it has the right to say so because of so many restrictions it has with alphabets pi as a number as well as NAN.

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## maple suggests...

Contrary assumptions means we have made impossible assumptions like less tha a parameter 10 (say X) and greater than a vale ten (or say Y, both X and Y unknown to me). So this is not a bug, a clear statement of what maple feels about our assumptions. Experts have maade rule for maple to follow and maple follows meticulously!). My attempt in a document would explain probably with less confusion and more clarity. Thanks for not throwing stones at me.

Regards;

We may see pi value will not be considered unless we evaluate its value.

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The pi value is to be evaluated for maple to recognize, else it takes as undeclared variable (value uknown not zero) and hence declares we have made contrary assumptions.

## Clear statements - Quite an useful thing...

Thank you very much. At later stage while doing dynamic analysis, i shall follow the partial derivative terms. I understood the importance of clear communication with maple. Thanks again for the useful tip.I have gone through and saw the maple error free statements for your commands. Thanks again for the useful tips.

Ramakrishnan V

## Maple is intelligent- straight forward c...

It just says 'i dont understand!! In the limits, you give only limits, no recursive calculations, first multiplying and then dividing!! We do it separately and maple comes forward to give best answers for better problems. Following is my sincere attempt to gat help and maple obliges.

Hope my document gives clearer perspective.

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## @EugeneKalentev  I have found out t...

I have found out the answer  thanks to Professor Robert Lopez.

Since l__e was redefined in the line where q__0 was defined, % is taken. The doubt was clarified by Professor Robert Lopez.

My sincere thanks go on record for him for the immediate responses from him whenever i send a mail.

Thanks to you also for the attempt. Since I have not used dimensions in the command mode, problem was not from units and dimensions. Cheers.

Ramakrishnan V

## Embedded components...

Thanks.I will try maplesim. Hope to use this later. I am learning but difficult without a book it seems. Some basic learning from class room is essential. Thanks for your response. Regards.

Ramakrishnan V

## Product line...

Maple 2015.2 is the one i use and Windows10 is in my PC. I shall fill up the product(s) data in my questionnaire hereafter. Thanks for letting me know my mistakes,however small it may look from one end. Regards.

Ramakrishnan V

## one dimensional heat equation solution...

Hope the attached equation and solution for the heat transfer problems would help understand the method.There may be unprofessional use of commands and manual mode since i am only learning

# Determine the temperature distribution in a fin of circular cross section shown in Fig. Base of the fin is maintained at 100 0C and convection occurs at the tip:

W/m2K; W/cmK:   cm;   cm;

0C;
=

=

cm: =  cm

= cm: =  cm

Discretisation:

Number of linear one dimensional elements = 2;
Total Number of nodes = 3;
Global stiffness matrix is 3 x 3
For element 1
Stiffness matrix for conduction:

=

Stiffness matrix for convection:

=

=

For element 2
Stiffness matrix for conduction:

=

Stiffness matrix for convection:

=

Stiffness matrix for convection:

=

=

=

for element 1

=

for element 2

=

=
Tip insulation does not contribute to any load vector since heat transferred is zero.
Assembling matrices give the system equation

=

# Base of the fin is at 1000C;

Multiply the first column by  to get the modified column.

We know

= Substituting and eliminating the first row and column and transferring the modified first column to RHS

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=

Solving for ,

=

=

# Determine the temperatures at the nodal interfaces for the two layered wall shown in Fig. The left face is supplied with heat flux of  W/cm2;

# and right face is maintained at  0C;

W/m2K; W/mK;  = cm2;
We shall consider two one dimensional linear elements
For element 1
cm;

[K] =  =

[F]1=  =

or element 2
cm;

[K] =  =

[F]1=  =

Assembling the matrices, we get the global system equation,

Global Stiffness matrix

=

=  =

=

We know Substituting and solving for ,

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0C

maple.

## variational methods...

I am sorry if I am not confusing. I know a reduction method in which the second order differential equations are converted to lower order and solvd. Galerkin and Ritz are the contributers. I attach with a doc detailing this methods with complete answers. Hope this will be useful. Ramakrishnna V

 Consider a cantilever beam subjected to uniformly distributed load as shown in Fig. Calculate the displacement and compare with exact solution.   The governing differential equation is  ; Boundary conditions are ; Boundary conditions These conditions enforce zero displacement and slope at fixed end; Boundary conditions These conditions enforce zero bending moment and shear force at free end; Let us find the approximate solution using single continuous trial function. Step 1: Assumption of trial solution (guess solution) The constants  are determined using boundary conditions ; The constants  are determined in terms of  using boundary conditions ; Substituting the values for  in the trial solution, we get  [Note that finding the trial function satisfying all the boundary conditions is thus cumbursome] Step 2: Find the domain residual. Substituting the trial solution in the governing equation, we get      =   The governing equation is ; Step 3: Minimise the residual.  The final solution is The following differential equation is available for a physical phenomenon.   Trial function is Boundary conditions are  Find the solution byGalerkin method   Step 1: Assumption of trial solution (guess solution)  = Step 2: Find the residue by substituting in the LHS of the governing equation The governing equation is  = : The condition for Galerkin's method Weighing function = Trial function (solution guessed)  =        =    =  = Answer considered is  =    Hence the final solution is  =

## evalf, assign - working fine - thanks...

Your suggestion is very useful and working when i call back eta later. Thanks. Ramakrishnan V

## midrich...

Thanks

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