. American engineer and railroad journal . fci ^ ^i fc il Fig. 4. Fig. 5. Fig. 6. Fig. 7. Fig. 8. Fig. 5, = 56,520 = 9,311 lbs. per square inch to resist rg the force will be less. The simplest way to calculate the bend-ing moment is to make a diagram like Fig. 3, in which:. Fig. 3. L = the length of the rod between = the radius of the crank. The ellipse in the center, of which A, B and C are a part, isthe path of the center of the rod. Then the radius of the curvepassing through the points ABC can be obtained. Assumingthe velocity between the points A and B to equal that
. American engineer and railroad journal . fci ^ ^i fc il Fig. 4. Fig. 5. Fig. 6. Fig. 7. Fig. 8. Fig. 5, = 56,520 = 9,311 lbs. per square inch to resist rg the force will be less. The simplest way to calculate the bend-ing moment is to make a diagram like Fig. 3, in which:. Fig. 3. L = the length of the rod between = the radius of the crank. The ellipse in the center, of which A, B and C are a part, isthe path of the center of the rod. Then the radius of the curvepassing through the points ABC can be obtained. Assumingthe velocity between the points A and B to equal that of thecrank circle, the centrifugal force can easily be calculated bysubstituting the radius thus found for the radius of the centrifugal force for any portion of the rod can be found,approximately, by the following formula:Let F = centrifugal force at crank circle. X = distance from center of front end to point where the stress is = centrifugal force at = length of rod between Then Y = ■ . LAs the transverse bending In the center of the rod due to thecentrifugal force will be only equal to half that due to thevelocity and radius of the crank pin, it will generally be foundthat the modulus of section, taken at the center, for the usu
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Keywords: ., bookcentury1800, bookdecade1890, booksubjectrailroadengineering
