From [Column Design (Part 4)], the Euler formula has been introduced. But when the slenderness ratio KL/r is less than the transition value Cc, then column is short, and the J.B. Johnson formula should be used. If we use Euler formula for the short column, it would predict too high critical load than it really is.

The J.B. Johnson formula is as follows:

From the J.B. Johnson formula we can see that the critical load for the short column is affected by the strength of the material (Sy) in addition to its stiffness (E). But for the long column as Euler formula is used, the strength of material (Sy) is not a factor for the critical load.

Let's see how we can make the excel file to help calculate critical load for both short and long columns in the next post.

## Monday, December 5, 2011

## Saturday, October 22, 2011

### Column Design (Part 4)

From Column Design (Part 3), we compute the value of column constant (Cc) and slenderness ratio (KL/r_min) to check whether the column is long or short.

If the column is long, Euler formula will be used for calculation. The Euler formula is defined as,

Another form of this equation can be calculated form substituting r

Notice that the buckling load is dependent only on the length (L), cross section (I) and the stiffness of material (E) of the column.

See the J.B. Johnson formula in the next post.

Reference:

If the column is long, Euler formula will be used for calculation. The Euler formula is defined as,

^{2}= I/A into the above equation. Then we get,Notice that the buckling load is dependent only on the length (L), cross section (I) and the stiffness of material (E) of the column.

**The strength of the material is not involved at all.**Therefore, in a long column application, there is no benefit to use a high-strength material. A low-strength material having the same modulus of elasticity (E) would perform well.See the J.B. Johnson formula in the next post.

Reference:

## Thursday, August 18, 2011

### Column Design (Part 3)

Let's continue from the previous post ...

The criteria to select whether we should use the Euler formula or the J. B. Johnson formula to calculate for the critical load (Pcr) of the column is related to the value of the actual slenderness ratio or

The criteria to select whether we should use the Euler formula or the J. B. Johnson formula to calculate for the critical load (Pcr) of the column is related to the value of the actual slenderness ratio or

**column constant**, Cc. It is defined as
where:

E = Modulus of elasticity of the material of the column

Sy = Yield strength of the column material

The use of the above column constant (Cc) is as follows,

- Determine the
**length**and**end fixity**of the column - Define the value of the
**constant (K)**according to the type of end fixity - Compute the
**effective length (Le)**from**Le = KL** - From the cross section shape and dimensions, compute the
**radius of gyration (r)**from**r = sqrt(I/A)** - Compute the slenderness ratio from
**Slenderness ratio = Le/r_min = KL/r_min** - From the material of the column, compute the
**column constant (Cc)**as per the above formula - Check whether
**KL/r > Cc?** - If yes, the column is Long: Use the Euler formula
- If no, the column is Short: Use the J. B. Johnson formula

In the next post, let's check the formulas of critical load (Pcr) of Euler and J. B. Johnson.

## Wednesday, August 17, 2011

### Column Design (Part 2)

From Column Design (Part 1), we know that a column will tend to buckle about the axis for which the radius of gyration (r) and the moment of inertia (I) are minimum. Another important parameter for column design is the effective length (Le) of the column. The effective length is defined as

where:

L = Actual length of the column between its supports

K = Constant value dependent on the end fixity of the column as following.

**Le = KL**where:

L = Actual length of the column between its supports

K = Constant value dependent on the end fixity of the column as following.

- A pinned-end column is guided so that the end of the column cannot sway from side to side, but it can rotate with no resistance at the end.
- A fixed-end column is held against rotation at the support.

The higher constant value of K as shown as the "practical values" in the above table is recommended because in reality it is particularly difficult to achieve a true fixed-end column because of lacking of rigidity of the support.

The slenderness ratio is the ratio of the effective length of the column to the least radius of gyration.

**Slenderness ratio = Le/r_min = KL/r_min**

The slenderness ratio will be used to select the method of performing the analysis of straight, centrally loaded columns. Two methods will be presented in the next post.

- The Euler formula for long, slender columns
- The J. B. Johnson formula for short columns

__Source:__

## Sunday, April 24, 2011

### Column Design (Part 1)

A

Columns that tends to buckle are

The tendency for a column to buckle is dependent on the shape and the dimensions of its cross section and how it is supported.

If we take a look at the cross section of the column, the followings are important properties for buckling.

A column tends to buckle about the axis for which the radius of gyration and the moment of inertia are minimum.

From the above picture (thin plate h x t) , we can calculate the value of radius of gyration about x-x axis and y-y axis as shown above. From calculation, we can see that r

Let's explore more in the next post and make excel sheet to calculate.

**column**in the definition of mechanical engineering does not have to be in vertical. The*column*is a structural member that*carries an*, and that**axial compressive load***tends to fail by*rather than by crushing the material. Buckling or**elastic instability**or**buckling***elastic instability*is the the failure condition in which the shape of the column is not sufficient enough to hold it straight under a*xial compressive load*. At the point of buckling, a radical deflection of the axis of the column occurs suddenly. Then if the load is not reduced, the column will collapse. It's obviously that this kind of failure must be avoided in our machine elements design.Columns that tends to buckle are

**ideally straight**and**relatively long**and**slender**. If a compression member is so short, the normal failure analysis must be used rather than the method that we're going to discuss in this post.__How will we know when a member is long and slender?__The tendency for a column to buckle is dependent on the shape and the dimensions of its cross section and how it is supported.

If we take a look at the cross section of the column, the followings are important properties for buckling.

- The
**cross sectional area, A**. - The
**moment of inertia of the cross section, I**, with respect to the axis about which__the value of I is minimum__. - The
**least value of the radius of gyration of the cross section, r**.

*radius of gyration*is computed from**r = sqrt(I/A)**A column tends to buckle about the axis for which the radius of gyration and the moment of inertia are minimum.

From the above picture (thin plate h x t) , we can calculate the value of radius of gyration about x-x axis and y-y axis as shown above. From calculation, we can see that r

_{y-y}is less than r_{x-x}because t < h. Therefore, the expected axis of buckling is y-y. We can imagine that we press a common ruler with an axial load of sufficient magnitude to cause buckling, and we can easily imagine how it will bend. The formula of radius of gyration is the tool to predict this phenomenon.Let's explore more in the next post and make excel sheet to calculate.

## Tuesday, March 29, 2011

### Chain Drives Design (Part 3)

Let's take a look at the formulas related to chain design.

The pitch diameter of a sprocket with

The center distance, C, is the distance between the center of the driver and the driven sprockets. It's the distance between the two shafts coupled by the chain drive. In typical applications, the center distance should be in the following range:

The chain length,

The center distance for a given chain length can be computed as

Please note that the computed center distance assumes no sag in either the tight or the slack side of the chain, and thus it is the maximum center distance. Negative adjustment and adjustment for wear must be provided.

The angle of contact, Î¸, is a measure of the angular engagement of the chain on each sprocket. The arc of contact Î¸

And the angle of contact of the chain on the larger sprocket, Î¸

Source:

The pitch diameter of a sprocket with

*N*teeth for a chain with a pitch of*p*is determined by__Note:__the angle of sine function must be degree (not radian)The center distance, C, is the distance between the center of the driver and the driven sprockets. It's the distance between the two shafts coupled by the chain drive. In typical applications, the center distance should be in the following range:

The chain length,

*L*, is the total length of the chain. Because the chain is comprised of interconnected links,*the chain length must be an integral multiple of the pitch*.The chain length is expressed in"It's preferable to have and odd number of teeth on the driving sprocket and an even number of pitches (links) in the chain to avoid a special link"

**number of links, or pitches**(not in mm or inches!), can be computed asThe center distance for a given chain length can be computed as

Please note that the computed center distance assumes no sag in either the tight or the slack side of the chain, and thus it is the maximum center distance. Negative adjustment and adjustment for wear must be provided.

The angle of contact, Î¸, is a measure of the angular engagement of the chain on each sprocket. The arc of contact Î¸

_{1}is for the chain on the smaller sprocket and it should be greater than 120^{o}. It can be computed asAnd the angle of contact of the chain on the larger sprocket, Î¸

_{2}, can be computed usingSource:

## Tuesday, March 15, 2011

### Chain Drives Design (Part 2)

**GENERAL CHAIN DESIGN CONSIDERATIONS****Nominal Tensile Load**

The main consideration for all types of chain is the

*nominal tensile load*that is required to perform the basic function. The

*nominal tensile load*generally fluctuates in a regular cycle. For example, the chain tension from the nominal load in a

*chain drive*increases as the chain moves around the driven sprocket. The tension remains basically constant at a high level as the chain runs through the tight strand. Tension then decreases as the chain moves around the driver sprocket. It then remains basically constant at a low level as it runs through the slack strand. This cycle then repeats again and again.

**Shock Load**

*Shock loads*are caused by the characteristics of the power source and the driven machinery. They occur repeatedly in a regular cycle, usually one or more times in each shaft revolution. They usually must be added to the nominal tensile load. Service factors are used to account for commonly known shock loads in most chain drives and conveyors.

**Inertia Load**

As the term is used here,

*inertia loads*are different from

*shock loads*.

*Inertia loads*are the occasional loads imposed on the chain by unusual, and often unexpected, events. They may come from starting a heavily loaded conveyor or a drive with a large flywheel. Or they may be caused by a sudden momentary jam in the driven machine or conveyor. The drive or conveyor designer should calculate expected starting loads and be sure that they are never more than the yield strength of the chain.

**Centrifugal Tension**

In high-speed drives,

*centrifugal force*is generated as the chain travels around the sprockets. Centrifugal force also may be generated by the chain’s travel over a curved path between sprockets. The tensile load from centrifugal force may have to be added to the nominal tension when appropriate.

**Catenary Tension**

The weight of that portion of the chain that hangs in a catenary generates additional tensile loads in the chain. The tensile load from the catenary tension must also be added to the nominal tension when appropriate. Catenary tension is usually a minor consideration in drives, but it may be a major consideration in conveyors.

**Chordal Action**

As the chain wraps a sprocket, it effectively forms a

*regular polygon*. That causes the chain strand to rise and fall each time a joint engages a sprocket tooth. This motion is called

*chordal action*. Chordal action also causes the chain speed to increase and decrease each time a joint engages a sprocket tooth.

**Vibration**

*Chain vibration*can cause very large increases in chain tensile loading if the vibration occurs at or

near the natural frequency of the chain. The added tension from vibration can sometimes be as

large as the nominal tensile load.

Source: Standard Handbook of Chains: Chains for Power Transmission and Material Handling, Second Edition (Dekker Mechanical Engineering)

## 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:__

- Chain has a speed variation, called chordal action, which is caused by the polygonal effect of the sprockets.
- Chain needs lubrication.
- Chain wears and elongates.
- Chain is weak when subjected to loads from the side. It needs proper alignment.

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.

__Source:__

- http://www.fptgroup.com/Chain/
- Machines and Mechanisms Applied Kinematic Analysis, 3rd edition, David H. Myszka
**http://chain-guide.com***** Recommended web site for chain drives.- http://tsubakimoto.com/product/drive-chains/lube-free/lube-free/class3/10/2/2/

## 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:

You'll need the following password to unzip: mechanical-design-handbook.blogspot.com

You can find the links related to this series of post below:

- Solving System of Equations using Gauss Elimination Method (Part 1)
- Solving System of Equations using Gauss Elimination Method (Part 2)
- Solving System of Equations using Gauss Elimination Method (Part 3)
- Solving System of Equations using Gauss Elimination Method (Part 4)
- Solving System of Equations using Gauss Elimination Method (Part 5)

**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 [

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.

**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

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 [

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.

*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 x

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.

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 x

_{1}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

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 x

Multiply equation (1) with coefficient of x

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'

Where: the superscript ', ", ...

We can find from the above system of equations that, first, we can compute the value of x

Then we can calculate for x

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.

*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 x

_{1}. Equation (1) now becomes:Multiply equation (1) with coefficient of x

_{1}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 eliminationWe can find from the above system of equations that, first, we can compute the value of x

_{n}from equation (n) from:Then we can calculate for x

_{n-1}, x_{n-2}, ..., x_{2}, x_{1}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:__
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