Wednesday, February 23, 2011

Chain Drives Design (Part 1)

Chain drives are used to transmit rotational motion and torque from one shaft to another, smoothly, quietly and inexpensively. Chain drives provide the flexibility of a belt drive with the positive engagement like a gear drive. Therefore, the chain drives are suitable for applications with large distances between shafts, slow speed and high torque.

Usually, chain is an economical part of power transmission machines for low speeds and large loads. However, it is also possible to use chain in high-speed conditions like automobile engine camshaft drives. This is accomplished by devising a method of operation and lubrication.

Compare to other forms of power transmission, chain drives have the following advantages:
  • Chain drives have  flexible shaft center distance, whereas gear drives are restricted. The greater the shaft center distance, the more practical the use of chain and belt, rather than gears. Chain can accommodate long shaft-center distances (less than 4 m), and is more versatile.
  • Chain drives require little adjustment, whereas belts require frequent adjustment.
  • Chain drives are less expensive than gear drives.
  • Chain drives have no slippage, as with belts, and provide a more efficient power transmission.
  • Chain drives are more effective at lower speeds than belts.
  • Chain drives have lower loads on the shaft bearings because initial tension is not required as with belts.
  • Chain drives have longer life service life and do not deteriorate with factors such as heat, oil, or age, as do belts. Sprockets are subject to less wear than gears because sprockets distribute the loading over their many teeth.
  • The sprocket diameter for a chain system may be smaller than a belt pulley, while transmitting the same torque.
Points of Notice:
  1. Chain has a speed variation, called chordal action, which is caused by the polygonal effect of the sprockets.
  2. Chain needs lubrication.
  3. Chain wears and elongates.
  4. Chain is weak when subjected to loads from the side. It needs proper alignment.
Standardization of chains under the American National Standards Institute (ANSI), the International Standardization Organization (ISO), and the Japanese Industrial Standards (JIS) allow ease of selection.

The followings are some example of chains from British and ISO standard.

BRITISH STANDARD ROLLER CHAINS
BS 228, ISO R606, DIN 8187

ISO Chain no.
The first two digits are related to the chain pitch. See below table.


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Monday, February 21, 2011

Solving System of Equations using Gauss Elimination Method (Part 6)

In this post, you'll find the video clip to show how to use the excel program to solve system of equations using Gauss Elimination method and download link.

You can find the links related to this series of post below:
 Free download excel file of equations solver with Gauss elimination method
You'll need the following password to unzip: mechanical-design-handbook.blogspot.com

Solving System of Equations using Gauss Elimination Method (Part 5)

Let's continue from [Solving System of Equations using Gauss Elimination Method (Part 4)]. Now you know how to enter data and solve a set of linear equations using our program. Now it's time to see how to setup another set of equations. Let's try to solve system of equations with 10 unknowns.

From the following screen, click "Main Menu".


Program will move to main screen with pop-up windows. Enter number of equations to be solved, for this example, enter 10 and click OK.


Warning screen will appear as follows. Please note that the program allows to keep only 1 set of equations at a time. Existing equations will be deleted we you set new equations. If you wish to solve new set of equations, click "Yes".


Program will move back to the calculation screen. You'll find matrix [A] with 10x10 dimensions, vector {B} with 10x1 dimensions and vector {X} with 10x1 dimensions. The program erased all existing data and create new table automatically.


Try enter the following values.


And click "Solve" to see the results.


You can recheck the result by multiplying matrix [A] with vector {X} using "=MMULT($B3:$K3,$L$3:$L$12)" in cell M3 of the following screen. I just copy it from excel program to another excel workbook and put those formula to confirm the computation results. You will find that the results are the same as vector {B}.


Let's watch the video clip and get a download link for this Gauss Elimination Equations Solver in the next post.

Sunday, February 20, 2011

Solving System of Equations using Gauss Elimination Method (Part 4)

After you download an excel program Gauss Elimination Method from our web site and open it, you'll find the following screen.


You have to click "Options..." and "Enable this content" to enable the VBA otherwise the VBA code will be blocked and can't run the program.

The program will show a form asking for your agreement confirmation.


After you've agreed, you'll find the main screen, then click Start.


You'll find the menu whether you need to review or recalculate the previous equations or you want to add new equations. Let's try review the earlier equations by clicking at "Existing equations" button.


The program will move to another screen where you can change values in matrix [A] or vector {B} and recalculation to solve the equations with the same number of equations. For this example, the dimension of matrix [A] is 3x3. So if you want to solve another set of linear equations with 3 unknowns then you can modify this existing table or you can either create new table which we will explain later.


In this screen, you will be able to change values only in matrix [A] and vector {B}. Other cells are locked. The program will show the values of solution vector {X} as soon as the calculation is complete.

Now, let's test the case of "division by zero" that we explained in [Solving System of Equations using Gauss Elimination Method (Part 3)] by swapping equation (1) and (3) as follows then click "Solve" button. The screen will show the same result in solution vector {X}.


Let's try to change equations and click "Solve". If there's nothing wrong with equations, you'll find solutions.


In case something wrong with equations, the program will shows the error message. The following equations have no solution and program terminated automatically.


Let's see how to setup the program to solve more equations such as 10 unknowns in the next post.

Solving System of Equations using Gauss Elimination Method (Part 3)

From post [Solving System of Equations using Gauss Elimination Method (Part 2)], we know the procedures of Gauss Elimination method to solve system of linear equations. But if we write excel VBA program based on those procedures, we may encounter with problems and cannot get the calculation result. You can find more details regarding problems and how to correct them from [Numerical methods in engineering]. We will explain one problem that may occur. It's division by zero.

For example, we want to solve a set of simultaneous equations as follows using Gauss Elimination method.



The first step of Gauss Elimination method is to divide equation (1) with a coefficient of x1 which is 5. This is fine to do that. But imagine if the equation (3) and (1) are swapped as follows, we will encounter with "division by zero" problem.



Actually we will have problem if the value in diagonal term of matrix [A] is zero or has very small value. So we can avoid this problem by swapping equations to get the highest absolute value of coefficient of x in diagonal term.

In the next post, let's have a look the excel program to solve system of equations using Gauss Elimination method.

Saturday, February 19, 2011

Solving System of Equations using Gauss Elimination Method (Part 2)

The the previous post [Solving System of Equations using Gauss Elimination Method (Part 1)], the basic information regarding Gauss Elimination Method has been shared. In this post, we will have at more details about Gauss Elimination method.

Why called "elimination"?

The general form of system of equations is like this.


The Gauss Elimination method starts from forward elimination by dividing equation (1) with coefficient of x1. Equation (1) now becomes:

Multiply equation (1) with coefficient of x1 from equation (2) then we get:



Then we subtract equation (2) with equation (1). Equation (2) becomes:


Or we can write it as


Repeat the same procedures for the remaining equations and we get the system of equations as follows:


We can see that the first term in equation (2) to (n) are eliminated for this round of calculation. For the next round, we will repeat the same procedure, only we change the coefficient to equation (2) i.e. dividing equation (2) with a'22 and multiplying it with a'32 ... As soon as we do the forward elimination until round (n-1) we will get the system of linear equations that is ready for backward substitution as follows.


Where: the superscript ', ", ... (n-1) means number of round of forward elimination
We can find from the above system of equations that, first, we can compute the value of xn from equation (n) from:


Then we can calculate for xn-1, xn-2, ..., x2, x1 by backward substitution using the following equation.

The excel program we developed to solve system of equations using Gauss Elimination method is based on the above equations with some improvements. We will discuss later in the next post about the limitation of Gauss Elimination method with improvement method.

Further reading:

Solving System of Equations using Gauss Elimination Method (Part 1)

Numerical Methods in Engineering: Theories With Matlab, Fortran, C and Pascal Programs
Sometimes engineers need to solve a system of equations to obtain solutions for their engineering design works. This kind of problems consists of several number of unknowns and equations that cannot be solved easily without good knowledge of system of equations solving method and understanding of computer programming. If we need to solve a set of linear equations with 2 - 3 unknowns, it will be easily to solve by hand calculation. There are several methods available for solving them. But when we need to solve much more unknowns e.g. 20 unknowns or 1,000 unknowns, computer software is unavoidable.

System of equations with "n" linear equations can be expressed as follows:



We can write it the matrix form as [A]{X} = {B}

The dimension of matrix [A] is (nxn), vector {X} is (nx1) and vector {B} is (nx1).
There are several methods to solve a set of simultaneous equations e.g. Cramer's rule, Gauss elimination or Gaussian Elimination method, Gauss-Jordon method, matrix inversion method, LU decomposition method, Cholesky decomposition method, etc.

Cramer's rule can easily solve a set of equations with small number of equation. The Cramer's rule uses determinant to solve for the unknowns. The unknowns can be solved from:

However, the number of calculations using Cramer's rule is approx. (n-1)(n+1)!, where n is number of equations to be solved. Therefore, if we want to solve a set of 10 linear equations, the number of calculations will be approx 360,000,000 times. But if we use Gauss Elimination method for 100 linear equations, the number of calculations is only about 700,000. Thus, practically, Gauss Elimination Method is the most favorite method for general use.

Gauss Elimination Method or Gaussian Elimination Method is widely used for solving most of engineering problems. It can be used to program in the computer easily. Gauss Elimination method can be categorized as follows:
  1. Forward elimination
  2. Back substitution
If you're interested in more details, please check the link of article sources below.

Mechanical Design Handbook has developed a VBA excel program to solve a set of simultaneous equations using original FORTRAN code from "Numerical Methods in Engineering by Dr. Pramote Dechaumphai". The program is written in VBA excel with simple user interface. It can solve up to 10 numbers of equations. Mechanical Design Handbook is going to share more details about how to use the program and download link in later posts [Solving System of Equations using Gauss Elimination Method (Part 2)].

Here are some pictures of the program that we will discuss in more details later.

Main screen


User interface


Input and Solution Screen



Source:

Tuesday, February 8, 2011

Peaucellier–Lipkin and Sarrus Straight-line Mechanism

The Peaucellier–Lipkin linkage (or Peaucellier–Lipkin cell), invented in 1864, was the first planar linkage capable of transforming rotary motion into perfect straight-line motion, and vice versa. It is named after Charles-Nicolas Peaucellier, a French army officer, and Yom Tov Lipman Lipkin, a Lithuanian Jew and son of the famed Rabbi Israel Salanter.

Until this invention, no planar method existed of producing straight motion without reference guideways, making the linkage especially important as a machine component and for manufacturing. In particular, a piston head needs to keep a good seal with the shaft in order to retain the driving (or driven) medium. The Peaucellier linkage was important in the development of the steam engine.



The mathematics of the Peaucellier–Lipkin linkage is directly related to the inversion of a circle.


There is an earlier straight-line mechanism, whose history is not well known, called "Sarrus linkage". This linkage predates the Peaucellier–Lipkin linkage by 11 years and consists of a series of hinged rectangular plates, two of which remain parallel but can be moved normally to each other. Sarrus' linkage is of a three-dimensional class sometimes known as a space crank, unlike the Peaucellier–Lipkin linkage which is a planar mechanism.


The Sarrus linkage, invented in 1853 by Pierre Frédéric Sarrus, is a mechanical linkage to convert a limited circular motion to a linear motion without reference guideways. The linkage uses two perpendicular hinged rectangular plates positioned parallel over each other. The Sarrus linkage is of a three-dimensional class sometimes known as a space crank, unlike the Peaucellier–Lipkin linkage which is a planar mechanism.

Source:
More information about Peaucellier–Lipkin linkage

    Sunday, February 6, 2011

    Hoekens Straight-line Mechanism

    The Hoekens linkage is a four-bar mechanism that converts rotational motion to approximate straight-line motion. The Hoekens linkage is a cognate linkage of the Chebyshev linkage.

    "DESIGN OF MACHINERY" by Robert L. Norton shows the link ratios that give the smallest possible structural error in either position or velocity over values of Δβ from 20° to 180°.

    The followings are some interesting examples of Hoekens straight-line mechanism from youtube.

    Walking robot


    Marble machine





    Source: