## Monday, December 24, 2007

### SCREW FASTENER THEORY & APPLICATIONS - Unbrako's guide

#### JOINT DESIGN AND FASTENER SELECTION.

Joint Length
The longer the joint length, the greater the total elongation will occur in the bolt to produce the desired clamp load or preload. In design, if the joint length is increased, the potential loss of preload is decreased.

Joint Material
If the joint material is relatively stiff compared to the bolt material, it will compress less and therefore provide a less sensitive joint, less sensitive to loss of preload as a result of brinelling, relaxation and even loosening.

Considering the material in which the threads will be tapped or the nut used, there must be sufficient engagement length to carry the load. Ideally, the length of thread engagement should be sufficient to break the fastener in tension. When a nut is used, the wall thickness of the nut as well as its length must be considered.
An estimate, a calculation or joint evaluation will be required to determine the tension loads to which the bolt and joint will be exposed. The size bolt and the number necessary to carry the load expected, along with the safety factor, must also be selected. The safety factor selected will have to take into consideration
the consequence of failure as well as the additional holes and fasteners. Safety factors, therefore, have to be determined by the designer.

#### SHEAR APPLICATIONS

Shear Strength of Material
Not all applications apply a tensile load to the fastener. In many cases, the load is perpendicular to the fastener in shear. Shear loading may be single, double or multiple loading. There is a relationship between the tensile strength of a material and its shear strength. For alloy steel, the shear strength is 60% of its tensile strength. Corrosion resistant steels (e.g. 300-Series stainless steels) have a lower tensile/shear relationship and it is usually 50-55%

Single/Double Shear
Single shear strength is exactly one-half the double shear value. Shear strength listed in pounds per square inch (psi) is the shear load in pounds divided by the cross sectional area in square inches.

#### OTHER DESIGN CONSIDERATIONS

Application Temperature
For elevated temperature, standard alloy steels are useful to about 550°F–600°F. However, if plating is used, the maximum temperature may be less (eg. cadmium should not be used over 450°F. Austenitic stainless steels (300 Series) may be useful to 800°F. They can maintain strength above 800°F but will begin to oxidize on the surface.

Corrosion Environment
A plating may be selected for mild atmospheres or salts. If plating is unsatisfactory, a corrosion resistant fastener may be specified. The proper selection will be based upon the severity of the corrosive environment.

#### FATIGUE STRENGTH

S/N Curve
Most comparative fatigue testing and specification fatigue test requirements are plotted on an S/N curve. In this curve, the test stress is shown on the ordinate (y-axis) and the number of cycles is shown on the abscissa (x-axis) in a lograthmic scale. On this type curve, the high load to low load ratio must be shown. This is usually R =.1, which means the low load in all tests will be 10% of the high load.

Increasing the R to .2, .3 or higher will change the curve shape. At some point in this curve, the number of cycles will reach 10 million cycles. This is considered the endurance limit or the stress at which infinite life might be expected.

Modified Goodman/ Haigh Soderberg Curve
The S/N curve and the information it supplies will not provide the information needed to determine how an individual fastener will perform in an actual application. In application, the preload should be higher than any of the preloads on the S/N curve. Therefore, for application information, the modified Goodman Diagram and/or the Haigh Soderberg Curve are more useful. These curves will show what fatigue performance can be expected when the parts are properly preloaded.

Elongation
The modulus for steel of 30,000,000 (thirty million) psi means that a fastener will elongate .001 in/in of length for every 30,000 psi in applied stress. Therefore, if 90,000 psi is the desired preload, the bolt must be stretched .003 inches for every inch of length in the joint. This method of preloading is very accurate but it requires that the ends of the bolts be properly prepared and also that all measurements be very carefully made. In addition, direct measurements are only possible where both ends of the fastener are available for measurement after installation. Other methods of measuring lengths changes are ultrasonic, strain gages and turn of the nut.

Torque
By far, the most popular method of preloading is by torque. Fastener manufacturers usually have recommended seating torques for each size and material fastener. The only requirement is the proper size torque wrench, a conscientious operator and the
proper torque requirement.

Strain
Since stress/strain is a constant relationship for any given material, we can use that relationship just as the elongation change measurements were used previously.

Now, however, the strain can be detected from strain gages applied directly to the outside surface of the bolt or by having a hole drilled in the center of the bolt and the strain gage installed internally. The output from these gages need instrumentation to convert the gage electrical measurement method. It is, however, an expensive method and not always practical.

Turn of the Nut
The nut turn method also utilizes change in bolt length. In theory, one bolt revolution (360° rotation) should increase the bolt length by the thread pitch. There are at least two variables, however, which influence this relationship. First, until a snug joint is obtained, no bolt elongation can be measured. The snugging produces a large variation in preload. Second, joint compression is also taking place so the relative stiffnesses of the joint and bolt influences the load obtained.

Read more details at Unbrako's web site

## Monday, December 17, 2007

### Elements of the Design Process

All design activities must do the following:
1) Know the “customers’ needs.”
2) Define the essential problems that must be solved to satisfy the needs.
3) Conceptualise the solution through synthesis, which involves the task of satisfying several different functional requirements using a set of inputs such as product design parameters within given constraints.
4) Analyse the proposed solution to establish its optimum conditions and parameter settings.
5) Check the resulting design solution to see if it meets the original customer needs.

Design proceeds from abstract and qualitative ideas to quantitative descriptions. It is an iterative process by nature: new information is generated with each step, and it is necessary to evaluate the results in terms of the preceding step. Thus, design involves a continuous interplay between the requirements the designer wants to achieve and how the designer wants to achieve these requirements.

Designers often find that a clear description of the design requirements is a difficult task. Therefore, some designers deliberately leave them implicit rather than explicit. Then they spend a great deal of time trying to improve and iterate the design, which is time consuming at best. To be efficient and generate the design that meets the perceived needs, the designer must specifically state the users’ requirements before the synthesis of solution concepts can begin.

Solution alternatives are generated after the requirements are established. Many problems in mechanical engineering can be solved by applying practical knowledge of engineering, manufacturing, and economics. Other problems require far more imaginative ideas and inventions for their solution. The word “creativity” has been used to describe the human activity that results in ingenious or unpredictable or unforeseen results (e.g., new products, processes, and systems).

In this context, creative solutions are discovered or derived by inspiration and/or perspiration, without ever defining specifically what one sets out to create. This creative “spark” or “revelation” may occur, since our brain is a huge information storage and processing device that can store data and synthesize solutions through the use of associative memory, pattern recognition, digestion and recombination of diverse facts, and permutations of events. Design will always benefit when “inspiration” or “creativity,” and/or “imagination” plays a role, but this process must be augmented by amplifying human capability systematically through fundamental understanding of cognitive behaviour and by the development of scientific foundations for design methods.

## Saturday, December 8, 2007

### Cam design

Classes of Cams

may, in general, be divided into two classes: uniform motion cams and accelerated motion cams. The uniform motion cam moves the follower at the same rate of speed from the beginning to the end of the stroke; but as the movement is started from zero to the full speed of the uniform motion and stops in the same abrupt way, there is a distinct shock at the beginning and end of the stroke, if the movement is at all rapid. In machinery working at a high rate of speed, therefore, it is important that cams are so constructed that sudden shocks are avoided when starting the motion or when reversing the direction of motion of the follower.Cams

The uniformly accelerated motion cam is suitable for moderate speeds, but it has the disadvantage of sudden changes in acceleration at the beginning, middle and end of the stroke. A cycloidal motion curve cam produces no abrupt changes in acceleration and is often used in high-speed machinery because it results in low noise, vibration and wear. The cycloidal motion displacement curve is so called because it can be generated from a cycloid which is the locus of a point of a circle rolling on a straight line.

Cam Follower Systems

The three most used cam and follower systems are radial and offset translating roller follower, Figs. 1a and 1b; and the swinging roller follower, Fig. 1c. When the cam rotates, it imparts a translating motion to the roller followers in Figs. 1a and 1b and a swinging motion to the roller follower in Fig. 1c. The motionof the follower is, of course, dependent on the shape of the cam; and the following section on displacement diagrams explains how a favorable motion is obtained so that the cam can rotate at high speed without shock.

The arrangements in Figs. 1a, 1b, and 1c show open-track cams. In Figs. 2a and 2b the roller is forced to move in a closed track. Open-track cams build smaller than closed-track cams but, in general, springs are necessary to keep the roller in contact with the cam at all times. Closed-track cams do not require a spring and have the advantage of positive drive throughout the rise and return cycle. The positive drive is sometimes required as in the case where a broken spring would cause serious damage to a machine.

Pressure Angle and Radius of Curvature

The pressure angle at any point on the profile of a cam may be defined as the angle between the direction where the follower wants to go at that point and where the cam wants to push it. It is the angle between the tangent to the path of follower motion and the line perpendicular to the tangent of the cam profile at the point of cam-roller contact.

The size of the pressure angle is important because:
1. Increasing the pressure angle increases the side thrust and this increases the forces exerted on cam and follower.
2. Reducing the pressure angle increases the cam size and often this is not desirable because:
• The size of the cam determines, to a certain extent, the size of the machine.
• Larger cams require more precise cutting points in manufacturing and, therefore, an increase in cost.
• Larger cams have higher circumferential speed and small deviations from the theoretical path of the follower cause additional acceleration, the size of which increases with the square of the cam size.
• Larger cams mean more revolving weight and in high-speed machines this leads to increased vibrations in the machine.
• The inertia of a large cam may interfere with quick starting and stopping.

The maximum pressure angle αm should, in general, be kept at or below 30 degrees for translating-type followers and at or below 45 degrees for swinging-type followers.

These values are on the conservative side and in many cases may be increased considerably, but beyond these limits trouble could develop and an analysis is necessary.

The minimum radius of curvature of a cam should be kept as large as possible
1. to prevent undercutting of the convex portion of the cam
2. to prevent too high surface stresses. Figs. 3(a), (b) and (c) illustrate how undercutting occurs.
Undercutting cannot occur at the concave portion of the cam profile (working surface), but caution should be exerted in not making the radius of curvature equal to the radius of the roller follower. This condition would occur if there is a cusp on the displacement diagram which, of course, should always be avoided. To enable milling or grinding of concave portions of a cam profile, the radius of curvature of concave portions of the cam, Rc = ρmin + rf, must be larger than the radius of the cutter to be used.

Cam Forces, Contact Stresses, and Materials

After a cam and follower configuration has been determined, the forces acting on the cam may be calculated or otherwise determined. Next, the stresses at the cam surface are calculated and suitable materials to withstand the stress are selected. If the calculated maximum stress is too great, it will be necessary to change the cam design.

Such changes may include:
1. increasing the cam size to decrease pressure angle and increase the radius of curvature
2. changing to an offset or swinging follower to reduce the pressure angle
3. reducing the cam rotation speed to reduce inertia forces
4. increasing the cam rise angle, β, during which the rise,h, occurs
5. increasing the thickness of the cam, provided that deflections of the follower are small enough to maintain uniform loading across the width of the cam
6. using a more suitable cam curve or modifying the cam curve at critical points
Although parabolic motion seems to be the best with respect to minimizing the calculated maximum acceleration and, therefore, also the maximum acceleration forces, nevertheless, in the case of high speed cams, cycloidal motion yields the lower maximum acceleration forces. Thus, it can be shown that owing to the sudden change in acceleration (called jerk or pulse) in the case of parabolic motion, the actual forces acting on the cam are doubled and sometimes even tripled at high speed, whereas with cycloidal motion, owing to the gradually changing acceleration, the actual dynamic forces are only slightly higher than the theoretical. Therefore, the calculated force due to acceleration should be multiplied by at least a factor of 2 for parabolic and 1.05 for cycloidal motion to provide an allowance for the load-increasing effects of elasticity and backlash.

The main factors influencing cam forces are:
1. displacement and cam speed (forces due to acceleration)
2. dynamic forces due to backlash and flexibility
3. linkage dimensions which affect weight and weight distribution
4. pressure angle and friction forces
5. spring forces
The main factors influencing stresses in cams are: 1) radius of curvature for cam and roller; and 2) materials.

Cam Materials: In considering materials for cams it is difficult to select any single material as being the best for every application. Often the choice is based on custom or the machinability of the material rather than its strength. However, the failure of a cam or roller is commonly due to fatigue, so that an important factor to be considered is the limiting wear load, which depends on the surface endurance limits of the materials used and the relative hardnesses of the mating surfaces.
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### Flywheels

A flywheel is a mechanical device with a significant moment of inertia used as a storage device for rotational energy. Flywheels resist changes in their rotational speed, which helps steady the rotation of the shaft when a fluctuating torque is exerted on it by its power source such as that caused by a piston-based (reciprocating) engine, or when an intermittent load, such as the motion of a piston pump, is placed on it.

Flywheels can be used to produce very high power pulses for experiments, where drawing the power from the public network would produce unacceptable spikes. A small motor can accelerate the flywheel between the pulses.

Flywheels may be classified as balance wheels or as flywheel pulleys. The object of all flywheels is to equalize the energy exerted and the work done and thereby prevent excessive or sudden changes of speed. The permissible speed variation is an important factor in all flywheel designs. The allowable speed change varies considerably for different classes of machinery; for instance, it is about 1 or 2 per cent in steam engines, while in punching and shearing machinery a speed variation of 20 per cent may be allowed.

The function of a balance wheel is to absorb and equalize energy in case the resistance to motion, or driving power, varies throughout the cycle. Therefore, the rim section is generally quite heavy and is designed with reference to the energy that must be stored in it to prevent excessive speed variations and, with reference to the strength necessary to withstand safely the stresses resulting from the required speed. The rims of most balance wheels are either square or nearly square in section, but flywheel pulleys are commonly made wide to accommodate a belt and relatively thin in a radial direction, although this is not an invariable rule.

Flywheels, in general, may either be formed of a solid or one-piece section, or they may be of sectional construction. Flywheels in diameters up to about eight feet are usually cast solid, the hubs sometimes being divided to relieve cooling stresses.

Flywheels ranging from, say, eight feet to fifteen feet in diameter, are commonly cast in half sections, and the larger sizes in several sections, the number of which may equal the number of arms in the wheel. Sectional flywheels may be divided into two general classes. One class includes cast wheels which are formed of sections principally because a solid casting would be too large to transport readily. The second class includes wheels of sectional construction which, by reason of the materials used and the special arrangement of the sections, enables much higher peripheral speeds to be obtained safely than would be possible with ordinary sectional wheels of the type not designed especially for high speeds. Various designs have been built to withstand the extreme stresses encountered in some classes of service. The rims in some designs are laminated, being partly or entirely formed of numerous segmentshaped steel plates. Another type of flywheel, which is superior to an ordinary sectional wheel, has a solid cast-iron rim connected to the hub by disk-shaped steel plates instead of cast spokes.

Steel wheels may be divided into three distinct types, including
1. those having the center and rim built up entirely of steel plates
2. those having a cast-iron center and steel rim
3. those having a cast-steel center and rim formed of steel plates.
Wheels having wire-wound rims have been used to a limited extent when extremely high speeds have been necessary.

When the rim is formed of sections held together by joints it is very important to design these joints properly. The ordinary bolted and flanged rim joints located between the arms average about 20 per cent of the strength of a solid rim and about 25 per cent is the maximum strength obtainable for a joint of this kind.

However, by placing the joints at the ends of the arms instead of between them, an efficiency of 50 per cent of the strength of the rim may be obtained, because the joint is not subjected to the outward bending stresses between the arms but is directly supported by the arm, the end of which is secured to the rim just beneath the joint. When the rim sections of heavy balance wheels are held together by steel links shrunk into place, an efficiency of 60 per cent may be obtained; and by using a rim of box or I-section, a link type of joint connection may have an efficiency of 100 percent.

Source:

### Rotary motion

In practice most machines involve rotary motion as well as linear motion. This could be such examples as electric motors, gears, pulleys and internal combustion engines. Therefore if we need to calculate how fast a machine will reach full speed (in other words calculate the acceleration of all its components) then we must consider rotary acceleration, and the associated torques, as well as linear acceleration. Fortunately Newton’s second law of motion applies equally well to rotary motion provided we use the correct version of the formula.

Suppose we are looking at the acceleration of a solid disc mounted on a shaft and rotated by a pull cord wrapped around its rim. We cannot use the conventional form of Newton’s second law, F = ma, because although there is a linear force being applied in the form of the tension in the cord, the acceleration is definitely not in a line. We therefore cannot identify an acceleration a in units of m/s2. Furthermore some of the mass of the disc is close to the axle and not moving from one point to another, only turning on the spot, while some of the mass is moving very quickly at the rim. Now in the section on kinematics we saw how the equations of uniform motion could be adapted for rotary motion by substituting the equivalent rotary quantity instead of the linear term. We found that linear distance travelled s could be replaced by angle of rotation q, while linear acceleration a could be replaced by the angular acceleration a, where a = a/r. It is also quite clear that if we are looking at rotary motion then we should be using the torque t = Fr instead of force. All that remains is to find the quantity that will be the equivalent of mass in rotary motion.

The mass of an object is really the resistance that the object offers to being moved in a straight line even when there is no friction. It can also be called the inertia of the object; an object with a large mass will accelerate much slower than one with a low mass under the action of an identical force. We are therefore looking for the resistance that an object offers to being rotated in the absence of friction. This must be a combination of mass and shape because a flywheel where most of the mass is concentrated in the rim, supported by slender spokes, will offer much more resistance than a uniform wheel of the same mass. The quantity we are seeking is something known as the mass moment of inertia or simply as the moment of inertia, I, which has units of
kilograms times metres squared (kg.m2). For a uniform disc, which is the most common shape found in engineering objects such as pulleys, this is given by the formula

I = mr2/2

We are now in a position to write Newton’s second law in a form suitable for rotary motion

t = Ia

With this equation we can solve problems involving real engineering components that utilize rotary motion and then go on to consider combinations of linear and rotary components. One of the most common rotary motion devices is the flywheel. This is a massive wheel which is mounted on a shaft to deliberately provide a great deal of resistance to angular acceleration. The purpose of the device is to ensure a smooth running speed, especially for something that is being driven by a series of pulses such as those coming from an internal combustion engine.

This general method can be applied to much more complicated shapes such as pulleys which may be regarded as a series of uniform discs on a common axis. Because all the components are all on the same axis, the individual moments of inertia may all be added to give the value for the
single shape. Sometimes the moment of inertia of a rotating object is given directly by the manufacturer but there is also another standard way in which they can quote the value for designers. This is in terms of something called the radius of gyration, k. With this the moment of inertia of a body is found by multiplying the mass of the body by the square of the radius of gyration such that

I = mk2

Example: An electric motor is being used to accelerate a flywheel from rest to a speed of 3000 rev/min in 40 s. The flywheel is a uniform disc of mass 96 kg and radius 0.75 m. Calculate the angular acceleration required and hence find the output torque which must be produced by the motor.

The first part of this problem is an application of kinematics to rotary motion where we know the time, the initial velocity and the final velocity, and we need to find the acceleration. The equation to use is therefore

w2 = w1 + at

The start velocity is zero because the flywheel begins from rest. The final angular velocity is
3000 rev/min = 3000 x 2p/60 = 314.2 rad/s

Therefore
314.2 = 0 + 40a

Now that the angular acceleration is known it is possible to calculate the torque or couple which is necessary to produce it, once the moment of inertia is calculated. The flywheel is a uniform disc so the moment of inertia is given by I = mr2/2 = 96 x 0.752/2 = 27 kg.m2

Therefore the driving torque is given by
t = Ia = 27 x 7.854 = 212 N.m

## Tuesday, December 4, 2007

### Newton’s second law of motion

Let us go back to the legend of Newton and the apple. From the work on statics we will find that the apple stays on the tree as long as the apple stalk is strong enough to support the weight of the apple. As the apple grows there will come a point when the weight is too great and so the stalk will break and the apple falls. The quantity that is being added to the apple as it grows is mass. Sometimes this is confused with weight but there have now been many examples of fruits and seeds being grown inside orbiting spacecraft where every object is weightless and would float if not anchored down.

Mass is the amount of matter in a body, measured in kilograms (kg).

The reason that the apple hangs downwards on the tree, and eventually falls downwards, is that there is a force of attraction between the earth and any object that is close to it. This is the gravitational force and is directed towards the centre of the earth. We experience this as a vertical, downward force. The apple is therefore pulled downwards by the effect of gravity, commonly known as its weight. Any apple that started to grow upwards on its stalk would quickly bend the stalk over and hang down due to this gravitational effect. As the mass of the apple increases due to growth, so does the weight until the point is reached where the weight is just greater than the largest force that can be tolerated by the stalk. As a result the stalk breaks.

Now let us look at the motion of the apple once it has parted company with the tree. It still has mass, of course, as this is a fundamental property to do with its size and the amount of matter it contains. Therefore it must also have a weight, since this is the force due to gravity acting on the apple’s mass. Hence there is a downwards force on the apple which makes it start to move downwards and get faster and faster. In other words, it accelerates. What Newton showed was that the force and mass for any body were linked to this acceleration for any kind of motion, not just falling under gravity. What Newton stated as his Second Law of Motion may be written as follows.

A body’s acceleration a is proportional to the resultant force F on the body and inversely proportional to the mass m of the body.

This seems quite complicated in words but is very simple mathematically in the form that we shall use.

F = ma

## Sunday, December 2, 2007

### Newton's law of motion

When Newton first published these laws back in the seventeenth century he caused a great deal of controversy. Even the top scientists and mathematicians of the day found difficulty in understanding what he was getting at and hardly anyone could follow his reasoning. Today we have little difficulty with the topic because we are familiar with concepts such as gravity and acceleration from watching astronauts floating around in space or satellites orbiting the earth. We can even experience them for ourselves directly on the roller coaster rides at amusement parks.

Newton lived in the second half of the seventeenth century and was born into a well-to-do family in Lincolnshire. He proved to be a genius at a very early age and was appointed to a Professorship at Cambridge University at the remarkably young age of 21. He spent his time investigating such things as astronomy, optics and heat, but the thing for which he is best remembered is his work on gravity and the laws of motion. For this he not only had to carry out an experimental study of forces but also he had to invent the subject of calculus.

According to legend Newton came up with the concept of gravity by studying a falling apple. The legend goes something like this. While Newton was working in Cambridge, England became afflicted by the Great Plague, particularly in London. Fearing that this would soon spread to other cities in the hot summer weather, Newton decided to head back to the family’s country estate where he soon found that there was little to do. As a result he spent a great deal of time sitting under a Bramley apple tree in the garden. He started to ponder why the apples were hanging downwards and why they eventually all fell to earth, speeding up as they fell. When one of the apples suddenly fell off and hit him on the head he had a flash of inspiration and realized that it was all to do with forces.

There is some truth in the legend and certainly by the time it was safe for him to return to Cambridge he had formulated the basis of his masterwork. After many years of subsequent work Newton was eventually able to publish his ideas in a finished form. Basically he
stated that:
• Bodies will stay at rest or in steady motion unless acted upon by a resultant force.
• A resultant force will make a body accelerate in the direction of the force.
• Every action due to a force produces an equal and opposite reaction.
These are known as Newton’s first, second and third laws of motion respectively.

## Monday, October 22, 2007

### Column design and analysis

In a design situation, the expected load on a column would be known, along with the length required by the application. The designer would then specify the following:
1. The manner of attaching the ends to the structure that affects the end fixity.
2. The general shape of the column cross section (for example, round, square, rectangular, and hollow tube).
3. The material for the column.
4. The design factor, considering the application.
5. The final dimensions for the column.
It may be desirable to propose and analyze several different designs to approach an optimum for the application, so software such as this facilitates the process.
It is assumed that the designer for any given trial specifies items 1 through 4. For some simple shapes, such as the solid round or square section, the final dimensions are computed from the appropriate formula: the Euler formula, or the J. B. Johnson formula. If an algebraic solution is not possible, iteration can be done.
In a design situation, the unknown cross-sectional dimensions make computing the radius of gyration and therefore the slenderness ratio, KL/r, impossible. Without the slenderness ratio, we cannot determine whether the column is long (Euler) or short (Johnson). Thus, the proper formula to use is not known.

We overcome this difficulty by making an assumption that the column is either long or short and proceeding with the corresponding formula. Then, after the dimensions are determined for the cross section, the actual value of KL/r will be computed and compared with Cc. This will show whether or not the correct formula has been used. If so, the computed answer is correct. If not, the alternate formula must be used and the computation repeated to determine new dimensions.
Design factor
Under typical industrial conditions, the design factor of N = 3 is recommended. If the application is very smooth, a value as low as N = 2 may be justified. Under conditions of shock or impact, N = 4 or higher should be used, and careful testing is advised.
Column analysis

The procedure for analyzing straight, centrally loaded columns:
1. For the given column, compute its actual slenderness ratio.
2. Compute the value of Cc.
3. Compare Cc with KL/r. Because Cc represents the value of the slenderness ratio that separates a long column from a short one, the result of the comparison indicates which type of analysis should be used.
4. If the actual KL/r is greater than Cc the column is long. Use Euler's equation:
The equation gives the critical load, Pcr, at which the column would begin to buckle.
An alternative form of the Euler formula is often desirable. Note that:
But, from the definition of the radius of gyration, r,
Then
This form of the Euler equation aids in a design problem in which the objective is to specify a size and a shape of a column cross section to carry a certain load.

Notice that the buckling load is dependent only on the geometry (length and cross section) of the column and the stiffness of the material represented by the modulus of elasticity. The strength of the material is not involved at all. For these reasons, it is often of no benefit to specify a high-strength material in a long column application. A lower-strength material having the same stiffness, E, would perform as well.

5. If KL/r is less than Cc, the column is short. Use the J. B. Johnson formula:

Use of the Euler formula in this range would predict a critical load greater than it really is. The J. B. Johnson formula is written as follows:

The critical load for a short column is affected by the strength of the material in addition to its stiffness, E. As shown in the preceding section, strength is not a factor for a long column when the Euler formula is used.

An eccentric load is one that is applied away from the centroidal axis of the cross section of the column, as shown in the graphic help entitled “Eccentric column”. Such a load exerts bending in addition to the column action that results in the deflected shape shown in the figure. The maximum stress in the deflected column occurs in the outermost fibers of the cross section at the midlength of the column where the maximum deflection, ymax occurs. Let's denote the stress at this point as sL/2. Then, for any applied load, P,

Note that this stress is not directly proportional to the load. When evaluating the secant in this formula, note that its argument in the parentheses is in radians. Also, because most calculators do not have the secant function, recall that the secant is equal to 1/cosine.

For design purposes, we would like to specify a design factor, N, that can be applied to the failure load similar to that defined for straight, centrally loaded columns. However, in this case, failure is predicted when the maximum stress in the column exceeds the yield strength of the material. Let's now define a new term, Py, to be the load applied to the eccentrically loaded column when the maximum stress is equal to the yield strength. The equation then becomes

Now, if we define the allowable load to be

or

this equation becomes

Required

This equation cannot be solved for either A or Pa, so an iterative solution is required.
Another critical factor may be the amount of deflection of the axis of the column due to the eccentric load:

### Stresses and Deformations in Beams

Reactions are the forces and/or couples acting at the supports and holding the beam and holding the beam in place. In some cases the user should enter a distributed load to account for the weight of the beam.

The shear V effective on a section is the algebraic sum of all forces acting parallel to and on one side of the section,

The bending moment is the algebraic sum of the moments due to applied loads and other applied moments to one side of the section of interest. Using value V bending moment can be calculated
where
x = position on the beam measured along its length
M0 = constant of integration evaluated from the boundary conditions.
A bending moment that bends a beam convex downward (tensile stress on bottom fiber) is considered positive, while convex upward (compressive on bottom fiber) is negative.

Moment and shear diagram constructed by plotting to scale the particular entity as the ordinate for each section of the beam. Such diagrams show in continuous form the variation among the length of the beam.

Magnitude of the direct stresses (tension and compressive) can be calculated from the direct stress formula
where
F = tensile/compressive force
A = cross sectional shape area

At the point of maximum bending stress, the flexure formula gives the stress

where
M = magnitude of the bending moment in the section
I = moment of inertia of the cross section with respect to its neutral axis
c = distance from the neutral axis to the outermost fiber of the beam cross section
A beam carrying loads transverse to its axis will experience shearing force. The resulting shearing stress can be computed from
where
I = rectangular moment of inertia of the cross section of the beam
t = thickness of the section at the place where the shearing stress is to be computed
Q = First moment of the area to the outside of the axis of interest with respect to the overall centroidal axis of the cross section of interest
V =shearing force

When a torque is applied to a member, it tends to deform by twisting, causing a rotation of one part of the member relative to another. The value of the maximum torsion shear stress can be computed

where
T = moment due to torque
c = distance from the neutral axis to the outermost fiber of the beam cross section
J = polar moment of inertia

When the beam is subjected to bending, the fibers on one side elongate, while the fibers on the other side shorten. These changes in length cause the beam to deflect. All points on the beam expect those directly over the support fall below their original position. The fundamental equation from which the elastic curve of a bent beam can be developed and the deflection of any beam obtained is,

If the loads are applied in vertical and horizontal plane or the forces are applied at angle to planes, in this case it is necessary to use a principle of superposition. Therefore result deflection at specific locations is computed from
where
fh = deflection in horizontal plane
fv = deflection in vertical plane

Using value of bending moment, slope can be calculated

where C0 = constant of integration evaluated from the boundary conditions.

The shortening/prolongation due to a direct axial tensile/compressive load is computed from

where
= total deformation of the member carrying the axial load
L = original total length of the member
E = modulus of elasticity of the material
A = cross-sectional area of the member

The following formula computes the relative elongation due to a direct axial tensile (compressive) load

When a shaft is subjected to a torque, it undergoes a twisting in which one cross-section is rotated relative to the other cross section in the shaft.

The relative twisting angle is defined

The angle of twist is computed from

where
L = length of the shaft over which the angle of twist is being computed
G = modulus of elasticity of the shaft material in shear

Note:
All applied loads, forces and moments are considered to pass through the center of gravity of the cross-section.

The mass of the beam is not considered. If the consideration of the beam's mass is necessary, then represent it with a distributed load.

Type of beam support

There are two very common beam arrangements. A beam that is supported at one fixed end is commonly referred to as a cantilever. The other is simply supported at two points along the length of the beam. Both of these arrangements are statically defined and are referred to as “determinate.”
The beam that is supported at one end can experience shearing force, longitudinal reaction, twisting, and bending moments.
The beam supported at two points can result in shear forces and reactions due to axial and shear forces.

Distance to considered cross-section

Distance to the desired cross-section, where the calculation of the force factors and deformations is needed, is defined with “Distance to considered cross-section” variable.

Type of cross-sectional shape

For standard structural shapes, click on the table icon and a menu of standard shapes will appear and the software will automatically input the geometric values into the input page. According to the selected cross-section type the following geometrical data must be entered:

Circle:
D diameter mm

Rectangle:
b width mm
h height mm

Shapes:
w depth mm
b flange width mm
t web thickness mm
t1 flange thickness mm
A area mm^2
IX-X moment of inertia mm^4
IY-Y moment of inertia mm^4
SX-X section modulus mm^3
SY-Y section modulus mm^3

Hollow round tube:
d inner diameter mm
D outer diameter mm

Modulus of elasticity in tension

For the part of the stress-strain diagram that is straight, stress is proportional to strain, and the value of E is the constant of proportionality. That is,

This is the slope of the straight-line portion of the diagram. The modulus of elasticity indicates the stiffness of the material, or the resistance to deformation.

Modulus of elasticity in shear

The modulus of elasticity in shear (G) is the ratio of shearing stress to shearing strain. This property indicates a material’s stiffness under shear loading, that is, the resistance to shear deformation. Generally this material property is available from published data. It is commonly provided in this software by clicking on the table icon and selecting the appropriate material.

There is a simple relationship between E, G and Poisson’s ratio:

where
E = modulus of elasticity in tension
= Poisson’s ratio

This equation is valid within the elastic range of the material.

The shearing forces input list “Forces and distributed loads” has the following parameters: distance from the beam’s beginning to the application point “Distance”; length of a distributed load “Length”, (should be 0 for a concentrated force); force/load value; angle of projection’s inclination of the force’s action line on the OYZ plane to Z-axis “a ”. The “Value” sign defines the direction of the force’s action. Force projections on coordinate axis are found using the method of superposition.
Note: Distributed load has unit “lb/in” or "N/mm".

Bending moments

The list of bending moments has the following parameters: distance from the beam’s beginning to the application point “Distance”; value of the moment “Value” and angle of inclination of the moment action plane to Z-axis. The “Value” sign defines the direction of the moment’s action. Moment projections on OXY and OXZ planes are found using the method of superposition.

Twisting moments

“Twisting moments” list is used for twisting moments input. The list has the following parameters:
distance from the beam’s beginning to the application point, “Distance”, in,
value of the moment, “Value”, lb.in. or N.mm The sign defines the direction of application. The default is shown in the corresponding graphic help.