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Algebra for Athletes.com |
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The weight machine problem in the last section showed how to determine the amount of force on a part. If someone were designing the machine, the next step would be to determine how large the part would need to be to support the force. To keep a machine from breaking, its parts have to be large enough to support the forces. The larger the parts are, the more force they will be able to support. But if the parts are too big, the machine will be larger and more expensive than it needs to be. To find the perfect size of the parts, engineers use a number of formulas from the field of material mechanics. If you know how much force will be on a part, the required size of the part will depend on its material. For example, most metal parts are stronger than plastic parts of the same size. Some metals are stronger than others. A type of metal or plastic can be described by how much stress it can safely handle. To find the perfect size for a part, engineers use formulas that determine the required size based on how much stress the material can handle. Stress is measured in units of force per unit area. A common unit of stress is pounds per square inch or psi. Engineers are often interested in a type of stress known as axial stress. The stress is called "axial" because the stress occurs along the axis of a bar or rod. The concept of axial stress is shown in the figures below. In the figure on the left, two forces are pushing on the rod or putting it in compression. In the figure on the right, two forces are pulling it apart or putting it in tension.
There are two things that determine how much force the rod can support: 1. The cross-sectional area XE "cross-sectional area" , A, of the part; and 2. The amount of stress, S, the material can safely handle. The cross-section of a part is the surface you would get if you sliced it perpendicular to its axis. The cross-sectional area is the area of this surface. The cross-sections of two rods are shown in Figure 3.8.
The formula that relates force, area, and axial stress is:
F= force F= SA A = cross-sectional area The formula holds true for both compression and tension. Example 3.4: The rod shown in the figure is expected to carry an axial load, F, of 5,000 lb. The metal of the rod can handle a stress of 20,000 lb./in2 (psi). What should the cross-sectional area, A, of the rod be? Solution: To determine the required area, we solve Equation 3-2 for A.
SA = F Dividing both sides by S
The rod needs a cross sectional area of 0.25 square inches to safely support a 5,000 lb. load.
In the last example, we found the area we needed for a square rod. To make a rod like this, you would need to know more than just the required area of the cross-section. You would need to know the dimensions or width of the rod. The problem stated that the rod was square in cross-section. If the rod has a width of length w, the rod will have an area of:
Since it represents the area of a square, the term w2 is said "w squared." The "2" in the term is known as the exponent. The problem required that the area equal 0.25 square inches. To find the value of w, we need to solve the following equation for w:
None of the properties we've learned so far have shown how to solve this kind of problem. To find the value of w, we need to take the square root of 0.25. This is done by finding a number which, when multiplied by itself, equals 0.25. Taking the square root of both sides of the equation,
We need to say that w equals plus or minus the square root of 0.25 because both two negative and two positive numbers will multiply to a positive number. For example: 3 x 3 = 9 and (-3) x (-3) = 9 When we take the square root of a number, we have to say it may be either positive or negative. Therefore:
Since we know that the width of the rod can only be positive, we can say that the rod needs to be 0.5 in. in width. 3.2.1 Area The concept of area is an important one used in lots of different fields. Some equations for areas of other shapes are shown in Figure 3.12.
The area of a rectangle is the height times the width. A square is a rectangle whose height equals its width. The area of a square is therefore the square of the side. The area of a triangle is one-half the height times the width. 3.2.2 Circles To find the area of a circle, you need to use a special equation. To use the equation, you have to understand the concept of
If you cut a piece of string so that it wrapped perfectly around a circle and then stretched it out as shown in the figure below, the string would have a length of times the diameter of the circle. The length of the string equals the perimeter or circumference of the circle. The circumference, c, of a circle is equal to
where c = the circumference of the circle (3-3) d = the diameter of the circle r = the radius of the circle The relationship holds true for any size of circle.
The number
Example 3.5: If the rod in Example 3.4 had a circular cross-section, what would the required radius of the bar be? Solution: In Example 3.4 the required area was 0.25 in.2. The required radius of the bar would ber, where:
Again, since we know the radius is a positive number, we can ignore the negative value. Therefore:
The required radius of a bar with circular cross section would be 0.28 in.
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whereS = stress (3-2)
Multiplying both sides by A,
or pi. Pi (pronounced "pie") is the Greek letter for "p." In math it represents the ratio of the circumference of a circle to its diameter.
d or 2
r. In mathematical terms: 
is what is known as an irrational number because its value neither terminates like the number "1.25" nor repeats like the number "0.333. . .". It is approximately equal to 3.1415. The exact value of
is described in the chapter on teaching infinite series The area of a circle is given by the equation: 
(3-4) 
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