Conditional statements and loops

Write a program that plots the deflection of the beam y as a function of x. First create a 100 element vector for x, then by using the LOOP and CONDITIONAL STATEMENTS calculate valute of y for the corresponding value of x.

[James Bowler]

Been trying to use conditional statements and loop commands to do this question. Any help is appreciated.

[Ravi Narasimhan]

I don't know if you are required to use loops and/or conditionals. If not, you might consider adapting the following approach.

Suppose assuming B is not equal to 0. Your equations are more complicated but this should illustrate

You can define an anonymous function in Matlab:

y = @(A,B,C,D,x) (A*x.^2/B).*(C*x.^3-D*x) % Note .* and .^ vs plain *s

y = function_handle with value:
    @(A,B,C,D,x)(A*x.^2/B).*(C*x.^3-D*x)

Initialize some values

A = 1; B = 2; C = 3; D = 4;
x = linspace(0,10,100); %  equally spaced pts between 0 and 10

Pass the arguments and linearly spaced array of x points to the anonymous function and get the result in one shot. No loop needed.

result = y(A,B,C,D,x);
plot(x,result)

You could create anonymous functions for each of the two regions and pass each a linearly spaced array to get your answers.

[James Bowler]

Thanks for the answer. Yes it is required to use conditional statements and loops

[Ravi Narasimhan]

Then, you can

1) Define your constants

2) Define an x array of some appropriate length in terms of L. e.g. x = linspace(0,L,50);

3) Define an empty y array. y= [];

4) Loop over the elements of the x array

• If the current element is less than L/2,
• evaluate your first equation and append the result to y: y = [y, evaluated function value]
• else
• If the current element is greater than L/2,
• evaluate your second equation and append the result to y

5) End the loop

This is not optimal or efficient but it satisifies the requirements.