Thursday, March 29, 2007
How Radar Works
Meteorologists use radar to track storms, hurricanes and tornadoes. You even see a form of radar at many grocery stores when the doors open automatically! Obviously, radar is an extremely useful technology.
When people use radar, they are usually trying to accomplish one of three things:
Detect the presence of an object at a distance - Usually the "something" is moving, like an airplane, but radar can also be used to detect stationary objects buried underground. In some cases, radar can identify an object as well; for example, it can identify the type of aircraft it has detected.
Detect the speed of an object - This is the reason why police use radar.
Map something - The space shuttle and orbiting satellites use something called Synthetic Aperture Radar to create detailed topographic maps of the surface of planets and moons.
All three of these activities can be accomplished using two things you may be familiar with from everyday life: echo and Doppler shift. These two concepts are easy to understand in the realm of sound because your ears hear echo and Doppler shift every day. Radar makes use of the same techniques using radio waves.
Echo and Doppler Shift
Echo is something you experience all the time. If you shout into a well or a canyon, the echo comes back a moment later. The echo occurs because some of the sound waves in your shout reflect off of a surface (either the water at the bottom of the well or the canyon wall on the far side) and travel back to your ears. The length of time between the moment you shout and the moment that you hear the echo is determined by the distance between you and the surface that creates the echo.
Doppler shift is also common. You probably experience it daily (often without realizing it). Doppler shift occurs when sound is generated by, or reflected off of, a moving object. Doppler shift in the extreme creates sonic booms (see below). Here's how to understand Doppler shift (you may also want to try this experiment in an empty parking lot). Let's say there is a car coming toward you at 60 miles per hour (mph) and its horn is blaring. You will hear the horn playing one "note" as the car approaches, but when the car passes you the sound of the horn will suddenly shift to a lower note. It's the same horn making the same sound the whole time. The change you hear is caused by Doppler shift.
Here's what happens. The speed of sound through the air in the parking lot is fixed. For simplicity of calculation, let's say it's 600 mph (the exact speed is determined by the air's pressure, temperature and humidity). Imagine that the car is standing still, it is exactly 1 mile away from you and it toots its horn for exactly one minute. The sound waves from the horn will propagate from the car toward you at a rate of 600 mph. What you will hear is a six-second delay (while the sound travels 1 mile at 600 mph) followed by exactly one minute's worth of sound.
Now let's say that the car is moving toward you at 60 mph. It starts from a mile away and toots it's horn for exactly one minute. You will still hear the six-second delay. However, the sound will only play for 54 seconds. That's because the car will be right next to you after one minute, and the sound at the end of the minute gets to you instantaneously. The car (from the driver's perspective) is still blaring its horn for one minute. Because the car is moving, however, the minute's worth of sound gets packed into 54 seconds from your perspective. The same number of sound waves are packed into a smaller amount of time. Therefore, their frequency is increased, and the horn's tone sounds higher to you. As the car passes you and moves away, the process is reversed and the sound expands to fill more time. Therefore, the tone is lower.
You can combine echo and doppler shift in the following way. Say you send out a loud sound toward a car moving toward you. Some of the sound waves will bounce off the car (an echo). Because the car is moving toward you, however, the sound waves will be compressed. Therefore, the sound of the echo will have a higher pitch than the original sound you sent. If you measure the pitch of the echo, you can determine how fast the car is going.
Understanding Radar
We have seen that the echo of a sound can be used to determine how far away something is, and we have also seen that we can use the Doppler shift of the echo to determine how fast something is going. It is therefore possible to create a "sound radar," and that is exactly what sonar is. Submarines and boats use sonar all the time. You could use the same principles with sound in the air, but sound in the air has a couple of problems:
Sound doesn't travel very far -- maybe a mile at the most.
Almost everyone can hear sounds, so a "sound radar" would definitely disturb the neighbors (you can eliminate most of this problem by using ultrasound instead of audible sound).
Because the echo of the sound would be very faint, it is likely that it would be hard to detect.
Radar therefore uses radio waves instead of sound. Radio waves travel far, are invisible to humans and are easy to detect even when they are faint.
Let's take a typical radar set designed to detect airplanes in flight. The radar set turns on its transmitter and shoots out a short, high-intensity burst of high-frequency radio waves. The burst might last a microsecond. The radar set then turns off its transmitter, turns on its receiver and listens for an echo. The radar set measures the time it takes for the echo to arrive, as well as the Doppler shift of the echo. Radio waves travel at the speed of light, roughly 1,000 feet per microsecond; so if the radar set has a good high-speed clock, it can measure the distance of the airplane very accurately. Using special signal processing equipment, the radar set can also measure the Doppler shift very accurately and determine the speed of the airplane.
In ground-based radar, there's a lot more potential interference than in air-based radar. When a police radar shoots out a pulse, it echoes off of all sorts of objects -- fences, bridges, mountains, buildings. The easiest way to remove all of this sort of clutter is to filter it out by recognizing that it is not Doppler-shifted. A police radar looks only for Doppler-shifted signals, and because the radar beam is tightly focused it hits only one car.
Police are now using a laser technique to measure the speed of cars. This technique is called lidar, and it uses light instead of radio waves. See How Radar Detectors Work for information on lidar technology.
Thursday, March 15, 2007
How Gas Turbine Engines Work
You may have never heard of gas turbine engines, but they are used in all kinds of unexpected places. For example, many of the helicopters you see, a lot of smaller power plants and even the M-1 Tank use gas turbines. In this article, we will look at gas turbine engines to see what makes them tick!
A Little Background
There are many different kinds of turbines:
You have probably heard of a steam turbine. Most power plants use coal, natural gas, oil or a nuclear reactor to create steam. The steam runs through a huge and very carefully designed multi-stage turbine to spin an output shaft that drives the plant's generator.
Hydroelectric dams use water turbines in the same way to generate power. The turbines used in a hydroelectric plant look completely different from a steam turbine because water is so much denser (and slower moving) than steam, but it is the same principle.
Wind turbines, also known as wind mills, use the wind as their motive force. A wind turbine looks nothing like a steam turbine or a water turbine because wind is slow moving and very light, but again, the principle is the same.
A gas turbine is an extension of the same concept. In a gas turbine, a pressurized gas spins the turbine. In all modern gas turbine engines, the engine produces its own pressurized gas, and it does this by burning something like propane, natural gas, kerosene or jet fuel. The heat that comes from burning the fuel expands air, and the high-speed rush of this hot air spins the turbine.
Advantages and Disadvantages
So why does the M-1 tank use a 1,500 horsepower gas turbine engine instead of a diesel engine? It turns out that there are two big advantages of the turbine over the diesel:
Gas turbine engines have a great power-to-weight ratio compared to reciprocating engines. That is, the amount of power you get out of the engine compared to the weight of the engine itself is very good.
Gas turbine engines are smaller than their reciprocating counterparts of the same power.
The main disadvantage of gas turbines is that, compared to a reciprocating engine of the same size, they are expensive. Because they spin at such high speeds and because of the high operating temperatures, designing and manufacturing gas turbines is a tough problem from both the engineering and materials standpoint. Gas turbines also tend to use more fuel when they are idling, and they prefer a constant rather than a fluctuating load. That makes gas turbines great for things like transcontinental jet aircraft and power plants, but explains why you don't have one under the hood of your car.
Sunday, March 4, 2007
Heat Treatment of Steel
On the other hand, we can make steel harder by rapid coolin. We heat it up beyond the critical temperature, and then quench it in wateror some ather liquid. The rapid temperature drop fixes the structural change in the steel which occured at the critical temperature, and makes it very hard. But a bar of this hardened steel is more liable to fracturethan normal steel. We therefore heat it again to temperature below the critical temperature , and cool it slowly. This treatment is called tempering. It helps to relieve the internal stresses, and makes the steel less brittle than before. The properties of tempered steel enable us to use it in the manufacture of tools which need a fairly hard steel. High carbon steel is harder than tempered steel, but it is much more difficult to work.
These heat treatments take places during the various shaping operations. We can obtain bars and sheets of by rolling the metal through huge rolls in a rolling-mill. The roll pressures must be much greater for cold rolling than for hot rolling, but cold rolling enables the operators to produce rolls of great accuracy and uniformity, and with a better surface finish. Other shaping operations include drawing into wire, casting in moulds, and forging.
Modes of deformation
Figure:(a) The strain diagram showing the different deformation modes corresponding to different strain ratios. (b) Equibiaxial stretching at the pole of a stretched dome. (c) Deformation in plane strain in the side-wall of a long part. (d) Uniaxial extension of the edge of an extruded hole. (e) Drawing or pure shear in the flange of a deep-drawn cup, showing a grid circle expanding in one direction and contracting in the other. (f) Uniaxial compression at the edge of a deep-drawn cup. (g) The different proportional strain paths shown in Figure plotted in an engineering strain diagram.
which β = 1. As stated above, the principal tension and principal stress in the direction, 1, will always be tensile or positive, i.e. σ1 ≥ 0. For the extreme case in which σ1 = 0 we find from Equations 2.6 and 2.14, that α = −∞ and β = −2. Therefore all possible straining paths in sheet forming processes will lie between 0A and 0E in Figure(a) and the strain ratio will be in the range −2 ≤ β ≤ 1.
Equal biaxial stretching, β = 1
The path 0A indicates equal biaxial stretching. Sheet stretched over a hemispherical punch will deform in this way at the centre of the process shown in Figure (b). The membrane strains are equal in all directions and a grid circle expands, but remains circular. As β = 1, the thickness strain is ε3 = −2ε1, so that the thickness decreases more rapidly with respect to ε1 than in any other process. Also from Equation 2.19(c), the effective strain is ε = 2ε1 and the sheet work-hardens rapidly with respect to ε1.
Plane strain, β = 0
In this process illustrated by path, 0B, in Figure (a), the sheet extends only in one direction and a circle becomes an ellipse in which the minor axis is unchanged. In long, trough-like parts, plane strain is observed in the sides as shown in Figure (c). It will be shown later that in plane strain, sheet is particularly liable to failure by splitting.
Uniaxial tension, β = −1/2
The point C in Figure (a) is the process in a tensile test and occurs in sheet when the minor stress is zero, i.e. when σ2 = 0. The sheet stretches in one direction and contracts in the other. This process will occur whenever a free edge is stretched as in the case of hole extrusion in Figure (d).
Constant thickness or drawing, β = −1
In this process, point D, membrane stresses and strains are equal and opposite and the sheet deforms without change in thickness. It is called drawing as it is observed when sheet is drawn into a converging region. The process is also called pure shear and occurs in the flange of a deep-drawn cup as shown in Figure (e). From Equation 3.1(b), the thickness strain is zero and from Equation 2.19(c) the effective strain is ε = 2/√3ε1 = 1.155ε1 and work-hardening is gradual. Splitting is unlikely and in practical forming operations, large strains are often encountered in this mode.
Uniaxial compression, β = −2
This process, indicated by the point E, is an extreme case and occurs when the major stress σ1 is zero, as in the edge of a deep-drawn cup, Figure (f). The minor stress is compressive, i.e. σ2 = −σf and the effective strain and stress are ε = −ε2 and σ = −σ2 respectively. In this process, the sheet thickens and wrinkling is likely.
Thinning and thickening
Plotting strains in this kind of diagram, Figure (a), is very useful in assessing sheet forming processes. Failure limits can be drawn also in such a space and this is described in a subsequent chapter. The position of a point in this diagram will also indicate how thickness is changing; if the point is to the right of the drawing line, i.e. if β > −1, the sheet will thin. For a point below the drawing line, i.e. β < −1, the sheet becomes thicker. The engineering strain diagram
In the sheet metal industry, the information in Figure (a) is often plotted in terms of the engineering strain. In Figure (g), the strain paths for constant true strain ratio paths have been plotted in terms of engineering strain. It is seen that many of these proportional processes do not plot as straight lines. This is a consequence of the unsuitable nature of engineering strain as a measure of deformation and in this work, true strains will be used in most instances. Engineering strain diagrams are still widely used and it is advisable to be familiar with both forms. In this work, true strain diagrams will be used unless specifically stated.
Saturday, March 3, 2007
Uniform sheet deformation processes
Tension as a measure of force in sheet forming
In sheet processes, deformation occurs as the result of forces transmitted through the sheet. The force per unit width of sheet is the product of stress and thickness and in Figure (c) is represented by, T = σt where, T , is known as the tension, traction or stress resultant. Because this is the product of the current thickness t as well as the current stress σ, it is the appropriate measure of force and will be used throughout this work in modelling processes. The term, tension, will be used even though this suffers from the disadvantage that the force is not always
a tensile force. If the tension is negative, it indicates a compressive force. This is not a serious problem as in plane stress sheet forming, almost without exception, one tension will be positive, i.e. the sheet is always pulled in one direction. It is impractical to forms sheet by pushing on the edge; the expression used by practical sheet formers is that ‘you cannot push on the end of a rope’. In the convention used here, the principal direction 1 is that in which the principal stress has the greatest (most positive) value, and the major tension T1 = σ1t will always be positive. In stretching processes, the minor tension T2 = σ2t is tensile or positive. In other processes, the minor tension could be compressive and in some cases the thickness will increase. If T2 is compressive and large in magnitude, wrinkling may be a problem. In discussing true stress in Section , it was shown that for most real materials, strain-hardening continues, although at a diminishing rate, and true stress does not reach a maximum. As tension includes thickness, which in many processes will diminish, T may reach a maximum; this limits the sheet’s ability to transmit load and is one of the reasons for considering tension in any analysis.
Strain distributions
In the study of any process, we usually determine first the strain over the part. This can be done by measuring a grid as in Figure 3.1, or by analysis of the geometric constraint exerted on the part. An example is the deep drawing process in Figure (a) and in the Introduction, Figure I.9. As the process is symmetric about the axis, we need only consider the strain at points on a line as shown in Figure (b). Plotting these strains in the principal strain space, Figure (c), gives the locus of strains for a particular stage in the process. As the process continues, this locus will expand, but not necessarily uniformly; some points may stop straining, while others go on to reach a process limit. For any process, there will be a characteristic strain pattern, as shown in Figure (c). This is sometimes known as the ‘strain signature’. Considerable information can be
obtained from such a diagram and the way it is analysed is outlined in the following section.
Strain diagram
The individual points on the strain locus in second Figure (c) can be obtained from measurements of a grid circles as shown in first Figure . (If a square grid is used, the analysis method is outlined in Appendix A.2.) If the major and minor axes are measured and the current thickness determined, the analysis is as follows.
Maximum shear stress
The three maximum shear stresses for the element are
τ1 =(σ1 − σ2)/2 ; τ2 =(σ2 − σ3)/2 ; τ3 =(σ3 − σ1)/2
From the discussion above, it might be anticipated that yielding would be dependent on the shear stresses in an element and the current value of the flow stress; i.e. that a yielding condition might be expressed as
f (τ1, τ2, τ3) = σf
Yielding in plane stress
There are a number of theories available for predicting the stresses under which a material element will deform plastically. Each theory is based on a different hypothesis about material behaviour, but in this work we shall only consider two common models and apply them to the plane stress process described by Equations ε1; ε2 = βε1; ε3 = −(1 + β)ε1. Over the years, many researchers have conducted experiments to determine how materials yield. While no single theory agrees exactly with experiment, for isotropic materials either of the models presented here are sufficiently accurate for approximate models.
With hindsight, common yielding theories can be anticipated from knowledge of the nature of plastic deformations in metals. These materials are polycrystalline and plastic flow occurs by slip on crystal lattice planes when the shear stress reaches a critical level. To a first approximation, this slip which is associated with dislocations in the lattice is insensitive to the normal stress on the slip planes. It may be anticipated then that yielding will be associated with the shear stresses on the element and is not likely to be influenced by the average stress or pressure. It is appropriate to define these terms more precisely.
Friday, March 2, 2007
General sheet processes (plane stress)
Stress and strain ratios
It is convenient to describe the deformation of an element, as in Figure (b), in terms of either the strain ratio β or the stress ratio α. For a proportional process, which is the only kind we are considering, both will be constant. The usual convention is to define the principal directions so that σ1 > σ2 and the third direction is perpendicular to the surface where σ3 = 0. The deformation mode is thus:
ε1 ; ε2 = βε1 ; ε3 = −(1 + β)ε1
σ1 ; σ2 = ασ1 ; σ3 = 0
The constant volume condition is used to obtain the third principal strain. Integrating thestrain increments in dε1 + dε2 + dε3 = 0 shows that this condition can be expressed in terms of the true or natural strains:
i.e. the sum of the natural strains is zero.
For uniaxial tension, the strain and stress ratios are β = −1/2 and α = 0.
True, natural or logarithmic strains
*the principal strain increments all increase smoothly in a constant direction, i.e. dε1 always increases positively and does not reverse; this is termed a monotonic process;
*during the uniform deformation phase of the tensile test, from the onset of yield to the maximum load and the start of diffuse necking, the ratio of the principal strains remains constant, i.e. the process is proportional; and
*the principal directions are fixed in the material, i.e. the direction 1 is always along the axis of the test-piece and a material element does not rotate with respect to the principal directions.
If, and only if, these conditions apply, we may safely use the integrated or large strains defined in Principal strain increments. For uniaxial deformation of an isotropic material, these strains are
ε1 = ln(l/l0) ; ε2 = ln(w/w0 = (−1/2)ε1 ; ε3 = ln(t/to) = (−1/2)ε1
Stress and strain ratios (isotropic material)
dε2 = dε3 = −(1/2)dε1
(In the previous chapter we considered the case in which the material was anisotropic where dε2 = Rdε3 and the R-value was not unity. We can develop a general theory for anisotropic deformation, but this is not necessary at this stage.) We may summarize the tensile test process for an isotropic material in terms of the strain increments and stresses in the following manner:
dε1 =dl/l ; dε2 = −(1/2)dε1 ; dε3 = −(1/2)dε1
and
σ1 =P/A ; σ2 = 0 ; σ3 = 0
Constant volume (incompressibility) condition
d(lwt) = d(lowoto) = 0
and we obtain
(dl . wt) + (dw . lt) + (dt . lw) = 0
or, dividing by lwt,
(dl/l )+(dw/w) +(dt/t) = 0
i.e.
dε1 + dε2 + dε3 = 0
Thus for constant volume deformation, the sum of the principal strain increments is zero.
Principal strain increments
dε1 =dl/l
i.e. the increase in length per unit current length. Similarly, across the strip and in the through-thickness direction the strain increments are
dε2 =dw/w and dε3 =dt/t
Uniaxial tension
Thursday, March 1, 2007
Tensile test
A tensile test-piece is shown in Figure 1.1. This is typical of a number of standard test-pieces having a parallel, reduced section for a length that is at least four times the width, w0. The initial thickness is t0 and the load on the specimen at any instant, P, is measured by a load cell in the testing machine. In the middle of the specimen, a gauge length l0 is monitored by an extensometer and at any instant the current gauge length is l and the extension is l = l − l0. In some tests, a transverse extensometer may also be used to measure the change in width, i.e. w= w − w0. During the test, load and extension will be recorded in a data acquisition system and a file created; this is then analysed and various material property diagrams can be created. Some of these are described below.
The load–extension diagram
Figure 1.2 shows a typical load–extension diagram for a test on a sample of drawing quality steel. The elastic extension is so small that it cannot be seen. The diagram does not represent basic material behaviour as it describes the response of the material to a particular process, namely the extension of a tensile strip of given width and thickness. Nevertheless it does give important information. One feature is the initial yielding load, Py, at which plastic deformation commences. Initial yielding is followed by a region in which the deformation in the strip is uniform and the load increases. The increase is due to strain-hardening, which is a phenomenon exhibited by most metals and alloys in the soft condition whereby the strength or hardness of the material increases with plastic deformation. During this part of the test, the cross-sectional area of the strip decreases while the length increases; a point is reached when the strain-hardening effect is just balanced by the rate of decrease in area and the load reaches a maximum Pmax .. Beyond this, deformation in the strip ceases to be uniform and a diffuse neck develops in the reduced section; non-uniform extension continues within the neck until the strip fails.
The extension at this instant is lmax ., and a tensile test property known as the total
elongation can be calculated; this is defined by
Etot. =[(lmax − l0)/l0].100%
The engineering stress–strain curve
Prior to the development of modern data processing systems, it was customary to scale the load–extension diagram by dividing load by the initial cross-sectional area, A0 = w0t0, and the extension by l0, to obtain the engineering stress–strain curve. This had the advantage that a curve was obtained which was independent of the initial dimensions of the test-piece, but it was still not a true material property curve. During the test, the cross-sectional area will diminish so that the true stress on the material will be greater than the engineering stress. The engineering stress–strain curve is still widely used and a number of properties are derived from it. Figure 1.3(a) shows the engineering stress strain curve calculated from the load, extension diagram in Figure 1.2.
Engineering stress is defined as ; σ = P/A0 (1.2)
and engineering strain as ; e = (l /l0). 100% (1.3)
In this diagram, the initial yield stress is; (σf)0 = Py/A0 (1.4)
The maximum engineering stress is called the ultimate tensile strength or the tensile
strength and is calculated as ; T S = Pmax/A0 (1.5)
As already indicated, this is not the true stress at maximum load as the cross-sectional area is no longer A0. The elongation at maximum load is called the maximum uniform elongation, Eu. If the strain scale near the origin is greatly increased, the elastic part of the curve would be seen, as shown in Figure 1.3(b). The strain at initial yield, ey, as mentioned, is very small, typically about 0.1%. The slope of the elastic part of the curve is the elastic modulus, also called Youngs modulus:
E = (σf)0 / ey (1.6)
If the strip is extended beyond the elastic limit, permanent plastic deformation takes place; pon unloading, the elastic strain will be recovered and the unloading line is parallel to the initial lastic loading line. There is a residual plastic strain when the load has been removed as shown in Figure 1.3(b).
In some materials, the transition from elastic to plastic deformation is not sharp and it is difficult to establish a precise yield stress. If this is the case, a proof stress may be quoted. This is the stress to produce a specified small plastic strain – often 0.2%, i.e. about twice the elastic straint yield. Proof stress is determined by drawing a line parallel to the elastic loading line which is offset by the specified amount, as shown in Figure 1.3(c). Certain teels are susceptible to strain ageing and will display the yield phenomena illustrated in Figure 1.4. This may be seen in some hot-dipped galvanized steels and in bake-hardenable steels used in autobody panels. Ageing has the effect of increasing the initial yielding stress to the upper yield stress σU; beyond this, yielding occurs in a discontinuous form. In the tensile test-piece, discrete bands of deformation called L¨uder’s lines will traverse the strip under a constant stress that is lower than the upper yield stress; this is known as the lower yield stress σL. At the end of this discontinuous flow, uniform deformation associated with strain-hardening takes place. The amount of discontinuous strain is called the yield point elongation (YPE). Steels that have significant yield point elongation, more than about 1%, are usually unsuitable for forming as they do not deform smoothly and visible markings, called stretcher strains can appear on the part.
***if some figures looking small, you can click on the pictures for magnify.