Category: Solving non linear equations

Solving non-linear equations include the following methods:

1- bisecting method.

2-False position method.

3-Fixed point iteration.

Maximum deflection distance for a simply supported beam Under triangular load-Numerically.

We want to get the maximum defection distance for a triangular load acting in a simply supported beam of length = 6.0 m. The maximum intensity of the load is 20 kNn/m. Our task is to find the x-max distance using the Numerical method. We will solve this by using the Newton-Raphson method for roof findings.

The three equations for the change of slope & slope and deflection values.

This is a list of three items; the first item is the change of curvature equation and the diagram shape for a supported beam under triangular load, where P is the maximum value of the triangular load, x is the distance for which we want to calculate the change of curvature.

The second item is the slope of curvature equation, and the diagram shape for a supported beam under a triangular load, where P is the maximum value of the triangular load, and x is the distance for which we want to calculate the slope of curvature.

The third item is the slope of the curvature equation, and the diagram shape for a supported beam under a triangular load, where P is the maximum value of the triangular load, and x is the distance for which we want to calculate the deflection value.

Maximum deflection distance value by equations.

The next slide image shows the slope and deflection equations of a simply supported beam under a uniformly Triangularly distributed load.

The values of slope and deflection for a beam under triangular load from mechanics of materials.

Estimating the maximum deflection distance from the left support x max is required. The point of the maximum deflection is the point for which the slope of curvature is zero. Usually, we use the Newton-Raphson method we want to get the point where Y value =0.

The function value f(x) was on the numerator side. Refer to the relation, where X0 is the starting point, f(x0) is the function value at x0, and f'(x0) is the function’s slope value at point X0.

The revised form of the Newton-Raphson equation is based on the slope equation.

To use the Newton-Raphson equation, we are going to deal with the y’ curve since it is changing from a +ve value to a negative value. A root point for zero slopes will give us the point of maximum deflection. Instead of dealing with the y curve as in the case of the original form of the Newton-Raphson equation, we are now dealing with the y’ function.

Modification of Newton -raphson equation for structural analysis

The steps for estimating x max using the analytical solution.

The modified equation for the root point will be as,  for the first point after setting an initial point of, which will be given a value of 2.00m, less than the L/2.

Our first trial value of x0 is =2.00m.

We need f'(2) and f”(2) to apply it in the adjusted equation. We have the expression for F'(x) for the slope value of the curve due to the triangular load also y” the change of curvature of the beam. We plug in with x0=2.0m and get the two values of f'(2) and f”(2), as shown in the next slide. We need to use 1/EI=1/8*10^4.

Our first x0 with a starting value of 2.0m will lead us to a modified value of 3.30m, as the maximum deflection distance. along with y’ function.

The first trial for maximum deflection distance.

Our second trial value of x0.

Then we plug in with x0 as a second trial=3.30m, we get the second value for the maximum deflection distance of 3.1169 m. Our third x0 with a starting value of 3.1169 m will lead us to a modified value of 3.1159778m, as the maximum deflection distance.

The Second iteration for maximum deflection distance value.

Then we plug in with x0 as a second trial=3.1159m, we get the third value for the maximum deflection distance of 3.115977 m.
The following table shows the result of x0 when we start with an initial point of 2.0m for four iterations.

Table for the value of x0 for starting value of 2.00m

This is a graph of the slope. We start with a point with x=2m, which will lead us to another point with x=3.30 m. Please refer to the previous table.

This is the slope graph and the value at x=2m.

In case our first trial value of x0 is =3.00m.

If somebody wishes to choose a starting point as=3.00m. x0 with a starting value of 3.0m will lead us to a modified value of 3.1166m, as the maximum deflection distance.

Then, we plug in with x0 as a second trial = 3.1166 m, and we get the second value for the maximum deflection distance of 3.115977 m.
The following table shows the result of x0 when we start with an initial point of 3.0m for four iterations.
If somebody wishes to choose a starting point as=4.0. x0 with a starting value of 4.00m will lead us to a modified value of 3.09m, as the maximum deflection distance.

Then, we plug in with x0 as a second trial = 3.09 m, and we get the second value for the maximum deflection distance of 3.1160 m.

The iteration table for x0=3.00m and 4.00m

The next slide contains a table for slope values according to x values. In the iterations, we used x=3m and x=4m, and both led to x=3.1159m based on the Newton-phone method.

This is a slope curve and the value of slope at x=3m and x=4m

This is the previous post-structural analysis numerically by the Newton-Raphson method.

This is a good external reference. Holistic numerical method.

This is a useful link for a numerical analysis calculator.




  • 8- Structural analysis numerically by Newton-Raphson method.

    8- Structural analysis numerically by Newton-Raphson method.

    How to use the Newton-Raphson method for structural analysis?

    The Newton-Raphson method can be used to make a structural analysis of the beam to enable us to find maximum deflection for a simply supported beam acted upon by a uniformly distributed load. Two approaches will be used the first approach is by using the Newton-Raphson method.

    The second approach will be done by using the modified Newton -Raphson Method.

    Revisit the Newton-Raphson method-The first approach.

    First, we have the y-axis which is located at the left support point, Then we select point A which is apart from the Y-axis by a distance X0, and we make a slope at that point A, shown on the graph of f(x), that slope will intersect with the x-axis at a certain point which will have a horizontal distance of X1 from the y- axis.

    Then we will measure or we will calculate the corresponding value of Y, or, f(x1) for that point.

    We repeat the process from point B we get another point B’, the second step is we are going to make a slope at point B’, and then the slope will hit again the X-axis at a certain point for that new point we call it point C.

    We are Looking for the point for zero y value. We estimate the Y value and so on we are taking several steps that’s why we are approaching the root or the Y value.

    An approach to the structural analysis for a beam under a uniform load and link to numerical analysis.

    The point of maximum deflection for a beam under uniform loading.

    As a direct application for this method, for the uniformly distributed load, for example, after making a structural analysis we have the curve of deflection and we are looking to have the point of maximum deflection, which is X maximum distance from the left support.

    Here for that curvature for the deflection, we don’t have an intersection for the deflection curve with the X-axis except for x=0 or X =L, that’s why we cannot directly use the Newton-Raphson method by that expression of X1= x0-(f(x0)/f'(x0) to get the distance to the point of maximum deflection.

    We use the graph of the slope of the beam to apply the Newton-Raphson method.

    We have the slope at the maximum point of deflection will be =0, or the curve of slope intersects the X-axis. That’s why, while checking the slope, y’ curve,  where we find that the point of zero or the roots point has the same distance which is Xmax from the left support.

    Thus comes the idea of utilizing y’ curve as a major curve to make structural analysis for a beam to get the distance of maximum deflection.

    In the next slide, we will see a graph for the expression of Newton-Raphson x1=x0-f(x0)/f'(x0), that equation deals with Y as an f(x) when the Y curve intersects with the x-axis.

    But In our case, the Y’ curve is the one that intersects with the x-axis. Now our roots point is at another curve which is y’, the slope curve for the uniformly distributed load on a simply supported beam which is a function of X’.

    Accordingly, the terms of the Newton-Raphson method will be modified, as we are going to see in the next slide. Based on the Newton-Raphson Method, x1=x0-f(x0)/f'(x0), while we were dealing with Y as a function of X.

    An approach to the structural analysis for a beam under a uniform load and link to numerical analysis.

    Now our root point is at another curve which is y’ that’s why we are going to replace the original expression, for instance, f(x0) to be replaced with f ‘(x0) and f'(x0) value by f”(x0).

    The double integration method expression or elastic curve is y”=M/EI, while y’ is the slope and it can be calculated by the integration of M/EI. for the expression of y, which is the deflection value, is given by double integration of M / EI.

    Let us see how we can proceed by using the y’- slope curve.

    First, select the X0 distance arbitrarily chosen, this is your starting point,  the slope from the left-hand side is negative so we will estimate the f'(x0.).

    This time we draw a curve. We draw a slope at that point, then we hit y’ curve by another new point, which is apart x1. The same procedure we have done earlier,  for that x1,  we are going to estimate y’ of x1 point for the f”(x0), we will proceed to the curve of y” and we get the corresponding f”(x0).

    How to use the Modified Newton-Raphson expression to get the zero point-slope?

    The new expression for the structural analysis of a beam-Numerically- by using the Modified Newton Raphson method.

    As a second approach using the modified Newton-Raphson method for the structural analysis, the expression of the equation will be modified as:

    Replace f(x) by f'(*x), and f'(x) by f”(x). How we are going to get f”‘(xi)? By recalling the expression that y” which is a curvature variation is =M/EI, when you are going to make differentiation for d2y/dx2, accordingly, you will make differentiation for M/EI, so for d3y/dx3= (dM/dx)/  y/EI, expression. We are familiar with the dM / dX, which is the shear diagram that’s why the d3y/dx3 = Q/Ei, this is our new df3y/dx3.

    How to use the Modified Newton-Raphson expression to get the zero point-slope?

    This is the PDF data used in the illustration of the post.

    The next post will include a solved problem for the application of this method, the next post link is the Maximum deflection distance by the Newton-Raphson method.

    This is a useful link for a numerical analysis calculator.

  • 7A-Solved problem-8 by Modified Newton-Raphson method

    7A-Solved problem-8 by Modified Newton-Raphson method

    Solved problem-8 by Modified Newton-Raphson method.

    For the given Solved problem-8 by the Modified Newton-Raphson method. The function f(x)=X^3-5*X^2+7x-3 with the first choice of x0=0.

    Again I make a graph by plugging different values and getting the corresponding values of f(x), I have started from x=0 to x=3.50.

    From the graph, as we can see from the next slide image, the roots are three roots x1=3 & x2=1 and x3=1 as shown in the Excel sheet for the Solved problem-8 by Modified Newton-Raphson method. the x=0 will give a negative value of -3.

    1-We start to use the modified Newton-Raphson method, consider the initial point x0 equals 0, and we start to find the expressions for f(x),f'(x) , f’^2(x), and f”(x) .
    2- Substitute at x=0 and get the values for f (0), f'(0) & f’^2(0), and f”(0). The value of f(0)=-3. The value of f'(0)=7.

    The value of f”(0)=-10. Substitute in the equation of the modified Newton-Raphson method, we get the value of x1, which will be =1.105. Please refer to the next slide image for a detailed estimation of the various parameters.

    The detailed calculation of how to get the value of x1 for solved problem #8.

    We plug in with the value of x=1.105263 and get the corresponding values of f(1.105263),f'(1.105263), and”(1.105263).

    The detailed calculation of how to get the value of x2 for the solved problem #8 from an esmated x1 value.

    4- Substitute in the modified Newton-Raphson method. We get a new point with x2=1.00. and again continue to estimate f(1.00) and f'(1.00) and then apply them in the equation to get a new point that will be point x3.

    The detailed calculation of how to get the value of x2 for solved problem #8.

    5-Continue the process until x converges to 1.00. The Excel sheet for the various values of x, based on the modified Newton-Raphson method from x0=0 till x4=1.00 is shown.  

    An excel sheet to show the values of xi till we will get the f(1.00)=0.

    This table shows how many iterations and the corresponding f(x),f'(x), and f”(x) for each case. 

       

    Solved problem-8 by Modified Newton-Raphson method using x0=4

    Now if we consider the starting point as x0=4.00, and proceed to get the x value for f(x)& f'(x) , f’^2(x) and f”(x) for x0=4.00. We have f(4)=9,f'(4)=15 and f”(4)=14. The value of x1 is found to be=2.6363.      

    Starting as the second choice by letting x0=4.00 and get the x1 value.

    2- Substitute at x1=2.6363 and get the values for f (2.6363), f'(2.6363)  & f”(2.6363). the values are shown in the next slide image.

    Substitute by the value of x1=2.6363 in the solved problem #8.

    3-Plug in with previous calculation to the Modified Newton Raphson equation and d out the value of X2 Substitute at x2=2.8202 and get the values for f (2.6363), f'(2.8202)  & f”^2(2.8202)  and f”(2.8202) and get the value of x3, it will be=2.9617.

    Get the value of x2 by the modified newton-raphson method.

    This is the Excel sheet for the calculation based on the Modified Newton Raphson Method starting from x0=4 till x5=3.00.

    Excel table for the different values of x for solved problem #8.

    Solved problem-8 by Newton Raphson method with x0=0.

    This is a comparison by the Newton Raphson method without modification for the same problem-8 but based on the Newton formula. Starting as before with x0=0 and using the equation of xi=x0-(F(x0)/f'(x0), we can get the x1 value 0.4285, then substitute and get the x2 value=0.6857.

    Resolve problem #8 by the Newton method.

    This is an Excel sheet for the points that are obtained by using the Newton-Raphson method for the Solved problem-8

    The initial point is (0) the table lists all the values of the function till point x4. That point will be equal to 0.95578.

    Excel table for the different values of x for solved problem #8.

    If we start with x0=4.00 to get the other root but based on the Newton-Raphson method. The Excel sheet with more details shows the different values of x.

    The steps to estimate x1 and x2 from starting x0=4.

    This is an Excel sheet for the points obtained while considering the initial point x0=4.00 ending with point x6 with a value of 3.

    Excel table for the different values of x for the solved problem #8 for x0=4.00.

    This is a comparison between the Newton Raphson and Modified Newton Raphson methods for the same solved problem-8 using an Excel sheet.

    Excel table for the different values of x for solved problem #8.

    This is the PDF file used for the illustration of this post.

    The next post is Structural analysis numerically by the Newton-Raphson method.

    This is a useful link for a numerical analysis calculator.

  • 7- An Easy Guide to Modified Newton-Raphson method.

    7- An Easy Guide to Modified Newton-Raphson method.

    Modified Newton-Raphson method.

    Introduction to the bracketing method. The bracketing method is a  Numerical method, that represents two values of a function having opposite signs, the root will be in -between.

    The modified Newton-Raphson method is another method for root finding.  A simple modification to the previous method of Newton-Raphson was introduced.

    I have introduced the Modified Newton Raphson method by applying the new equation for two solved problems. the first solved problem is solved problem 7. I have used the Modified Newton Raphson method to solve. I have introduced an Excel table to show the iterations starting from x0=0.50. that was a part of the video.

    Modified Newton Raphon formula.

    This formula of the Modified Newton-Raphson method is shown in the next slide image.

    Modified Newton-Raphson method

    Solved problem#7 by using the modified Newton-Raphson Method.

    We start by solving problem #7. For the modified Newton Raphson method, I used an Excel sheet to determine the root value of the given f(x)=e^x-3*x^2. I make a table between x and the corresponding value of f(x), by selecting several values of x, starting from 0 to a value of 1.10.

    Solved problem #7 by the modified Newton- Raphson method.

    After drawing the function by Excel, the point of f(x)=0 was detected to be at x=0.91 or the root value is at x= 0.91, please refer to the Excel sheet. at an initial point of 0.50, it is required to estimate the root point, the steps are as follows:

    1- Estimate f(xi),f'(xi) , f’^2(xi) and f”(xi) at the starting point of xi=0.50 for an initial i=0.

    2- Substitute at xi=0.50 and get the values for f (0.50), f'(0.50) & f’^2(0.50) and f”(0.50) and get the value of x1, it will be=0.7117.

    How to determine x1 value or the solved problem #7 by using the modified Newton- Raphson method?

     3- Substitute at x1=0.7117 and get the values for f (0.7117), f'(0.7117)  & f’^2(0.7117)  and f”(0.71112), now we consider i=1, we want to get the x value at I when=2. The value of x2 will be=0.876.

    4-Continue the process till x converges to 0.91 as shown in the next table. the next slide image shows the values of x starting from x0=0.50 to x1=0.711699, then x2=0.87601, then x3=0.909275, x4=0.91000, and then x5=0.91000. Check the f(x)=0.

    An excel sheet to show the values of xi till we will get the f(0.91)=0.

    This is the PDF file used for the illustration of this post.

    The next post is 7A- Solved problem-8 by Modified Newton-Raphson method

    This is a useful link for a numerical analysis calculator.

  • 6a- Two Solved problems for Newton-Raphson method.

    6a- Two Solved problems for Newton-Raphson method.

    Solved problems for Newton-Raphson method.

    The first problem of the two Solved problems for the Newton-Raphson method

    Another solved problem for the Newton-Raphson method for root extraction: find the roots of x^3-3x-5=0.
    1- starting by the initial point x0=3.00.


    First, we try to find that root value by plotting the curve by plugging different values of x for example starting from 0 to 3.0 and negative values from x= -0.50 to -4.00.

    We are getting the corresponding values of y accordingly. To find roots, we find that the function value between one positive and one negative review lies between x=2 and x =2.50.
    1-So, our first iteration is selecting x0 value =3.

    The first solved problem for Newton Raphson method

    2-Get f(x0=3) and f'(x0=3) f (x) is =x^3-3x-5=0 and f'(x) = 3*x^2-3.
    At x0=3 then f(3) = 3^3-3*3-5 =13 and f'(x0=3.00)=3*(3)^2-3=24.00.
    3- X1 =3-(13/24)=2.45833.

    The first solved problem solution is continued.

    For the second iteration
    4—Get the values of f(2.45833) and f'(2.45833). The Excel sheet shows a new value x2=2.2943 and the corresponding f(x) for each iteration, shown in the next slide.

    We keep making iterations till we get x4=2.279; the error value is small.

    The fourth iteration for the second solved Problem.

    The second Problem of the two Solved problems for the Newton-Raphson method.

    The second video includes an illustration of the second solved problem Number #6. This is the second solved problem of the two Solved problems for the Newton-Raphson method which is problem number 6: Use the Newton method for root extraction to find the roots of this function.

    Problem number 6: Use the Newton method for root extraction to find the roots of this function.

    This is the first derivative of the function.

    Solution: He did not give us a starting point, so it is good to make a casing for the root range by substituting
    1-We make the table, we put different values of X, and we find the corresponding values of our function. 2-We’ll find that when we plug X=0 then (0*e^0)-2, we get -2

    The fourth iteration for the second solved Problem.

    3-For the next point where x=0.25, we get (-1.678994).
    4-For x=0.50 we get -1.175639.

    5- For x=0.750, we get. (-0.41225).
    6- For x=1, we get +0.7182818.
    Between x=0 and x=1.0, forx=0 gives a minus, and for x=1 gives a positive value.

    Check the points where the value of y is positive.

    So, our root should exist between 0 and 1. You have the choice either to start point x=0 or X =1. We put X=0 f(0)=(0)*e^0-2=-2.00, for the slope value, check the next relation, it will be=1.

    x1=x0-f(x0)/f'(x0)=0-(-2/1)=+2.00.

    For f(2), we have the value of 12.778 and f'(2)=22.16716. So X2 will be =+2-f(x1=2)/f'(x1=2).X2 will be =+2-(12.778/22.16716)=1.4325. After substitution, we get the value of the function and the derivative value. We use the expression of X3=x2- f(x2)/f'(x2), we get x3=1.034936.

    The value of the function for x=2

    It is better to calculate using an Excel sheet. This sheet includes a starting point with a selected value followed by a column that represents the function value f(x), another new column for the value of f'(x), a column for the numerator, which I f(x), and the denominator, which is ‘(x), and then a column for the numerator/ denominator.

    We have two starting points: x0=0 and x0=1. This is a function value f'(x) =4.34575, and we go on until we find that f(x) is coming closer to (0) at x value=0.8526.

    An excel for x0=0.

    When X =0.853003, the f(x) is zero if our starting point x0=1.

    if the starting point is x0=1, the values of y for x


    The next post is about the Modified-Newton-Raphson method, for which the Newton-Raphson method is modified.

    This is a useful link for a numerical analysis calculator.


  • 6- Newton-Raphson method-Easy approach

    6- Newton-Raphson method-Easy approach

    Newton-Raphson method.

    Newton Raphson’s method is another method for root finding. The Newton-Raphson expression of root-finding utilizes the linear approximation which we have discussed.

    From the next slide image. L(xb)=f(xa)+f'(xa)*(xb-xa) as a is the starting point and the xb is the ending point Now if we consider that L(xb) approximately equals (xb) and create a little modification can be done on the previous equation by letting (xb-xa) on the left side, and then rewrite the equation.

    We could say x final=x intial+(1/ slope at the ix initial)+(1/ slope at x intial)*(f(x fina)l- f(x initial).
    If we are looking for xb where the root =0, or saying f(xb)=0.

    The formula will be xb=xa+(1/f'(xa)(0-f(xa)=xa-(1/f'(xa)(f(xa).
    The formula can be used to get the distance x for the root point b for which we are looking.

    Introduction to Newton -Raphson method.

    This will create another form of the equation as (xb-xa)=(1/f'(xa))*(f(xb)-f(xa). The next step is to find the value of xb which is=xa+(1/f'(xa))*(f(xb)-f(xa).

    Suppose we have a curve and that curve we are looking for the root of that curve at a certain point. We want to find what is the x value of that root point. So we are saying that if we have x1 point, we go up and then we make a tangent at the curve at that point.

    So we get another point which we call x 2 and we get a relation between x2 the new point and the old point. This relation will be x2=x1– f(x1) /f'(x1) as shown in the next slide image to continue this process till a point where we have f(x) close to or =0. This is the Newton-Raphson method.

    Newton-raphson method-sketch

    Solved problem using the Newton-Raphson method.

    First, the equation of Newton-Raphson is written. followed by a solved example #4 Example number 4. Use the Newton method to find the roots of the √29.

    Newton-Raphson equation for root finding

    The solution will be made through the next steps.
    We put x= √ 29 or it could be expressed at X^2= to 29 then let  X^2-29 =0.
    1-We readjust the formula a for the function and we equate it to 0.
    2-We put   x0=5 as starting point after that get f(5) = 5^2-29=-4.
    The negative sign will change the relationship as we will see later.
    3-Estimate the f'(x0=5) =(2*x0)-0=2*5.0=10.00.

    4- Estimate X1 value by using the Newton-Raphson method, X1=5-(-4/10.00)=5.40.
    5-This is for the first iteration. We started from X0 we got X1 then again we substituted by this new value which we are getting which is = 5.40. For the second iteration.
    6- We put   x1=5.40 as obtained from the previous iteration, f(5.40) = (5.40)^2-29=+0.16.
    7-Estimate the f'(x1=5.40) =(2*x1)-0=2*5.40=10.80.

    Solved problem for Newton-Raphson

    8- Estimate X2 value by using the Newton-Raphson method, X2=5.40-(+0.16/10.80)=5.3852.
    9-x2=5.3852, check (5.385)^2-29=-0.001775 not zero, so proceed to get the next point x3.
    For the third iteration.
    10- We put   x2=5.3852 as obtained from the previous iteration, f(5.3852) = (5.3852)^2-29=0.00022
    11-Estimate the f'(x1=5.3852) =(2*x2)-0=2*5.3852=10.7703.

    12- Estimate X3 value by using the Newton-Raphson method, X2=5.3852-(-0.00022/10.7703)=5.38516.
    13-x3=5.38516, check (5.38516)^2-29=4.2E-10 close to zero.

    Third iteration for the solved problem,

    This is the Excel sheet showing the iterations of the different f(x) values and the first derivative values.

    Excel table for the Newton-Rraphson solved problem.

    This is the pdf file used for the illustration of this post and the next post.
    The next post is about the solved problems for the Newton-Raphson method.

    This is a useful link for a numerical analysis calculator.