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تاريخ التسجيل : 21/09/2008
 |  موضوع: shear stress & strain with examples  2008-09-26, 7:37 pm | |
| Shear Stress & Strain SHEAR STRESS In additional to Axial (or normal) Stress and Strain (discussed in topic 2.1and 2.2), we may also have what is known as Shear Stress and Shear Strain.In Diagram 1 we have shown a metal rod which is solidly attached to the floor. We then exert a force, F, acting at angle theta with respect to the horizontal, on the rod. The component of the Force perpendicular to the surface area will produce an Axial Stress on the rod given by Force perpendicular to an area divided by the area, or:
The component of the Force parallel to the area will also effect the rod by producing a Shear Stress, defined as Force parallel to an area divided by the area, or:  where the Greek letter, Tau, is used to represent Shear Stress. The units of both Axial and Shear Stress will normally be lb/in2 or N/m2.Shear Strain:
Just as an axial stress results in an axial strain, which is the change in the length divided by the original length of the member, so does shear stress produce a shear strain. Both Axial Strain and Shear Strain are shown in Diagram 2. The shear stress produces a displacement of the rod as indicated in the right drawing in Diagram 2. The edge of the rod is displaced a horizontal distance,  from its initial position. This displacement (or horizontal deformation) divided by the length of the rod L is equal to the Shear Strain. Examining the small triangle made by , L and the side of the rod, we see that the Shear Strain, /L, is also equal to the tangent of the angle gamma, andsince the amount of displacement is quite small the tangent of the angle is approximately equal to the angle itself. Or we may write:
Shear Strain = As with Axial Stress and Strain, Shear Stress and Strain are proportional in the elastic region of the material. This relationship may be expressed as G = Shear Stress/Shear Strain, where G is a property of the material and is called the Modulus of Rigidity (or at times, the Shear Modulus) and has units of lb/in2. The Modulus of Rigidity for Steel is approximately 12 x 106 lb/in2. If a graph is made of Shear Stress versus Shear Strain, it will normally exhibit the same characteristics as the graph of Axial Stress versus Axial Strain. There is an Elastic Region in which the Stress is directly proportional to the Strain. The point at which the Elastic Region ends is called the elastic limit, or the proportional limit. In actuality, these two points are not quite the same. The Elastic Limit is the point at which permanent deformation occurs, that is, after the elastic limit, if the force is taken off the sample, it will not return to its original size and shape, permanent deformation has occurred. The Proportional Limit is the point at which the deformation is no longer directly proportional to the applied force (Hooke's Law no longer holds). Although these two points are slightly different, we will treat them as the same in this course. There is a Plastic Region, where a small increase in the Shear Stress results in a larger increase in Shear Strain, and finally there is a Failure Point where the sample fails in shear.To summarize our shear stress/strain/Hooke's Law relationships up to this point, we have:While we will not go in any great depth, at this point, with respect to Shear Stress and Strain, we will look at several relative easy examples. Please Select:
Example 1
Two metal plates, as shown in Diagram 1, are bolted together with two 3/4" inch diameter steel bolts. The plates are loaded in tension with a force of 20,000 lb., as shown. What is the shearing stress that develops in the steel bolts? If the Modulus of Rigidity for Steel in 12 x 106 lb/in2, what shear strain develops in the steel bolts?
As we examine the structure we see the area of the bolt where the two plates surfaces come together are in shear. That is, if we examine one bolt, the top of the bolt experiences a force to the left while the bottom of the bolt experiences and an equal force to the right. (See Diagram 2.) The surface area between the top and bottom interface is in shear. We assume the bolts carry the load equally, and so each bolt "carries" 10,000 lb. From this we can calculate the shear stress in a straight forward manner from one of our Shear Stress/Strain Relationships:
Stress = Force parallel to area/ area
Stress = F / (p * r2) = 10,000 lb/ (3.14 * .375"2) = 22, 650 lb/in2.
In similar manner the Shear Strain can be found from the appropriate form of Hooke's Law:
G = (Shear Stress) / (Shear Strain), or (Shear Strain) = (Shear Stress)/ G = (22, 650 lb/in2)/(12 x 106 lb/in2) = .00189
[size=12]Example 2
In our second example we have a inner shaft (which perhaps drives a piece of machinery) and an outer driving wheel connected by spokes to an inner ring which is connected by means of a shear key to the inner shaft. That is, the wheel, spokes, and inner ring are one structure which is connected to the inner shaft by a shear key. (See Diagram 3.) When force is applied to the wheel (through a driving belt perhaps), the wheel begins to rotate, and through the shear key causes the inner shaft to rotate. We are interested in determining the shear stress on the shear key. We will also determine the compressive stress (or bearing stress) acting on the shear key. [The purpose of a shear key is to protect machinery connected to a shaft or gear. If the driving forces and/or torque become too large the shear key will "shear off " (fail in shear) and thus disconnect the driving force/torque before it can damage the connected machinery.]
To determine the shear stress we first need to determine the force trying to shear the key. We do this by realizing, after a little thought, that the driving force is really producing a torque about the center of the shaft, and that the torque produced by the driving force(s) must equal the torque produced by the force (of the inner ring) acting on the shear key. In our problem the two 500 lb. driving forces are acting a distance of 2 ft from the center of the 1 ft diameter shaft. Calculating torque about the center of the shaft we have:
500 lb.(2 ft) + 500 lb.(2 ft) = 2000 ft-lb. This must also be the torque produced by the force acting on the upper half of the shear key (shown in Diagram 4). So we may write: 500 lb.(2 ft) + 500 lb.(2 ft) = 2000 ft-lb. = F (.5 ft), Solving for F = 4000 lb. This is the force acing on the top half of the shear key. There is a equal force in the opposite direction acting on the bottom half of the shear key. These two forces place the horizontal cross section of the key in shear. The key is 1/2 inch wide, by 3/4 inch high, by 1 inch deep as shown in Diagram 4. We calculate the shear stress by:
Shear Stress = Force parallel to area / area = 4000 lb./ (1/2" * 1") = 8000 lb/in2.
In addition to the shear stress on the horizontal cross sectional area, the forces acting on the key also place the key itself into compression. The compressive stress (also called the bearing stress) on the top half of the shear key will be given by:
Compression (Bearing) Stress = Force normal to the area / area = 4000 lb. / (3/8" * 1") = 10, 700 lb/in2. There is an equal compressive stress on the bottom of the shear key. [/size] | |
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