| Rubber bands, plastic food wrap, bandages that conform to knuckles and knees, where would we be without them? These are all fairly recent inventions, but their elastic properties were imagined far before they actually came into existence.
Around the turn of the 19th Century a mathematics genius by the name of Siméon Denis Poisson dabbled in higher level mathematics. He enjoyed working with calculus and probability theories and their applications, and his work eventually led to the discovery of his own special ratio, the “Poisson ratio.” Denoted today by the Greek letter “µ,” his discovery has a great deal to do with elasticity. In fact, much of his work evolved to become the modern study of engineering. If you’ll remember from last week’s blog, we talked about the elasticity of materials, including materials you generally wouldn’t think of as being elastic. In our steel rod example we saw that when you pull on the ends of a steel rod hard enough, you can actually stretch it and make it longer. But where does this extra length come from? According to Poisson’s ratio, as the rod lengthens, its diameter decreases proportionately. The rod’s increased length comes at the expense of its diameter. You can see this effect at work by repeatedly stretching that fat rubber band whose task it is to contain your bulging Sunday paper. The more you pull on it, the skinnier the rubber band becomes. It will eventually get to the point were its elastic properties have been so compromised it won’t even be able to hold together Monday’s paper. Over the decades that have passed since Poisson’s discovery a multitude of laboratory tests have been conducted to determine µ for a vast number of materials. These values have been duly tabulated in engineering reference books, doing away with the tedious task of conducting individualized experimentation by present day design engineers. Steel, for example, has a Poisson’s ratio of around 0.28, and this number is readily available in most strength of materials reference books. It’s pretty obvious why Poisson’s contribution is important to the world of engineering, but now let’s see how his ratio can be applied. Last week we saw that a 15-foot long, 2-inch diameter round steel rod stretches by 0.115 inches when it is pulled by a steady 60,000 pound force. Poisson’s ratio tell us that this results in an accompanying decrease in diameter, but by how much? To find out, we simply multiply the stretched length of the rod by Poisson’s ratio for steel (µ = 0.28). Plugging these numbers into an equation we see that the diameter decreases by: 0.115 inches × 0.28 = 0.032 inches This is approximately the thickness of nine sheets of paper. So if the rod was 2 inches in diameter before the 60,000 pound force was applied, its new diameter after application of the stretching force would be: 2 inches – 0.032 inches = 1.968 inches The change of .032 in the rod’s diameter may not seem like much, but in the world of machine parts it could mean the difference between parts fitting properly or becoming loose. This wraps up our short series on strengths of elastic materials. Next time we’ll move on to discuss coal power plant fundamentals, an arena in which many of the things we’ve been discussing take on real world meaning. _____________________________________________
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Posts Tagged ‘strain’
Strength of Materials – Poisson’s Ratio
Sunday, January 16th, 2011
Strength of Materials – Modulus of Elasticity
Sunday, January 9th, 2011| When you think of elastic, you most likely think about the stuff that allows you to put your underwear on and helps it stay riding around your mid section. In the absence of a belt, a band of elastic is indispensable. What you probably don’t realize is that most materials, including those you consider to be just plain hard, like wood, plastic, and metal, are also elastic to some extent. They’re certainly not as elastic as the rubber that your underwear’s waist band is made of, but they do stretch, or deform, when you pull on them, depending on the force exerted.
When engineers design a machine and specify that it is to be made out of particular materials, they have to take into account how much the materials will deform during usage. If metal parts in the machine become overly deformed due to operating forces, they may start interfering with each other. When that happens the machine will suffer with premature wear or perhaps even grind to a halt or fly apart. So how do we get a handle on this deformation factor? The first factor to consider is the stiffness of the material. In other words, when you try to pull it apart, a material will better resist the pull if it has high stiffness. In the world of engineering, this material stiffness is known as “modulus of elasticity” and is denoted by the letter E. Over countless decades, testing laboratories have been helping design engineers by determining the values of E for all sorts of materials that are commonly used to fabricate machines and structures. For example, for most steels E has been determined to be about 30,000,000 pounds per square inch (Lb/in2). So let’s see how we can put E to work for us. Suppose you have to design a very heavy machine with a lot of precision parts. In that machine you have a round steel rod that has a steady 60,000 pound force pulling on it lengthwise. The rod is 2 inches in diameter and 15 feet (or 180 inches) long. How much will the rod stretch under the force? This formula provides us with the answer: Rod Stretch = (Pulling Force × Rod Length) ÷ (Rod Area × E) Since the steel rod is round in our example, its cross sectional area is that of a circle. You might recall from your math classes that the area of a circle involves a constant called π, or 3.14. So to find the Rod Area we simply multiply π times the square of the rod’s diameter, d, then divide by 4. In this case the diameter of the steel rod is 2 inches, and the Rod Area is calculated to be: Rod Area = πd2 ÷ 4 = 3.14 × (2 inches)2 ÷ 4 = 3.14 square inches (in2) So, putting this value and the other information we were given into the equation, we get: Rod Stretch = (60,000 Lb × 180 in) ÷ (3.14 in2 × 30,000,000 Lb/in2) = 0.115 inches This stretch, or deformation, resulted in the rod’s length increasing by about the thickness of two dimes. You may think this is insignificant when speaking about a rod of 15 feet, but this deviation in size could lead to trouble if another machine part is in the way when the rod stretches. Unintended collision between machine parts is undesirable, no matter how you look at it. Next time we’ll investigate what happens to the diameter of our example rod when it stretches under the 60,000 pound load. _____________________________________________
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