Mechanical Design Handbook

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 .

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 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: Machine Elements in Mechanical Design (4th Edition) http://en.wikipedia.org/wiki/Buckling

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 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

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 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.

A column in the definition of mechanical engineering does not have to be in vertical. The column is a structural member that carries an axial compressive load , and that tends to fail by elastic instability or buckling rather than by crushing the material. Buckling or 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 buckl

Let's take a look at the formulas related to chain design. 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 . "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" The chain length is expressed in number of links, or pitches (not in mm or inches!) , can be computed as The center distance for a given chain length can be computed as Please no

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 i

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 ver

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: 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

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 automaticall

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 modif

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 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 swa

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 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) w