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(17.1)
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(17.2)
Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright 2004 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website.
SHAFTS 17.4
POWER TRANSMISSION
FIGURE 17.1 Simply supported shafts with force Fi and couple Mi applied.
where is the absolute value of the allowable slope at the bearing. These equations are an ideal task for the computer, and once programmed interactively, are convenient to use. Example 1. A shaft is to carry two spur gears between bearings and has loadings as depicted in Fig. 17.2. The bearing at A will be cylindrical roller. The spatial centerline slope is limited to 0.001 rad. Estimate the diameter of the uniform shaft which limits the slope at A with a design factor of 1.5.
FIGURE 17.2 A shaft carries two spur gears between bearings A and B. The gear loads and reactions are shown.
Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright 2004 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website.
SHAFTS 17.5
SHAFTS
Solution. Equation (17.1) is used. d= = 32n [F1b1(b2 1 3 E
2 2 )]H + [F2b2(b2 2 1/2 1/4 2 2 )] V
32(1.5) [300(6)(62 162)]2 + [1000(12)(122 162)]2 3 30(10)6 16(0.001)
1/2 1/4
= 1.964 in Transverse bending due to forces and couples applied to a shaft produces slopes and displacements that the designer needs to control. Bending stresses account for most or all of such distortions. The effects of transverse shear forces will be addressed in Sec. 17.3. Most bending moment diagrams for shafts are piecewise linear. By integrating once by the trapezoidal rule and a second time using Simpson s rule, one can obtain deflections and slopes that are exact, can be developed in tabular form, and are easily programmed for the digital computer. For bending moment diagrams that are piecewise polynomial, the degree of approximation can be made as close as desired by increasing the number of station points of interest. See [17.1], pp. 103 105, and [17.2]. The method is best understood by studying the tabular form used, as in Table 17.1. The first column consists of station numbers, which correspond to cross sections along the shaft at which transverse deflection and slope will be evaluated. The minimum number of stations consists of those cross sections where M/EI changes in magnitude or slope, namely discontinuities in M (point couples), in E (change of material), and in I (diameter change, such as a shoulder). Optional stations include other locations of interest, including shaft ends. For integration purposes, midstation locations are chosen so that the second integration by Simpson s rule can be exact. The moment column M is dual-entry, displaying the moment as one approaches the station from the left and as one approaches from the right.The distance from the origin to a station x is single-entry. The diameter d column is dual-entry, with the entries differing at a shoulder. The modulus E column is also dual-entry. Usually the shaft is of a single material, and the column need not be filled beyond the first entry. The first integration column is single-entry and is completed by applying the trapezoidal rule to the M/EI column. The second integration column is also single-entry, using the midstation first integration data for the Simpson s rule integration.
TABLE 17.1 Form for Tabulation Method for Shaft Transverse Deflection Due to Bending Moment Moment Dist. Dia. Modulus M x d E 1 M EI
M dx EI
M dx dx Defl. EI y
Slope dy dx
Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright 2004 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website.
SHAFTS 17.6
POWER TRANSMISSION
The deflection entry y is formed from the prediction equation y=
M dx dx + C1x + C2 EI
(17.3)
The slope dy/dx column is formed from the prediction equation dy = dx
M dx + C1 EI
(17.4)
where the constants C1 and C2 are found from
xa 0 xa 0
M/(EI) dx dx
xb 0
M/(EI) dx dx
C1 = xb C2 =
xa xb
xa 0 xa 0
(17.5)
M/(EI) dx dx xa xa xb
xb 0
M/(EI) dx dx
(17.6)
where xa and xb are bearing locations. This procedure can be repeated for the orthogonal plane if needed, a Pythagorean combination of slope, or deflections, giving the spatial values. This is a good time to plot the end view of the deflected shaft centerline locus in order to see the spatial lay of the loaded shaft. Given the bending moment diagram and the shaft geometry, the deflection and slope can be found at the station points. If, in examining the deflection column, any entry is too large (in absolute magnitude), find a new diameter dnew from dnew = dold nyold yall
(17.7)
where yall is the allowable deflection and n is the design factor. If any slope is too large in absolute magnitude, find the new diameter from dnew = dold n(dy/dx)old (slope)all
(17.8)
where (slope)all is the allowable slope. As a result of these calculations, find the largest dnew/dold ratio and multiply all diameters by this ratio. The tight constraint will be at its limit, and all others will be loose. Don t be concerned about end journal size, as its influence on deflection is negligible. Example 2. A shaft with two loads of 600 and 1000 lbf in the same plane 2 inches (in) inboard of the bearings and 16 in apart is depicted in Fig. 17.3. The loads are from 8-pitch spur gears, and the bearings are cylindrical roller. Establish a geometry of a shaft which will meet distortion constraints, using a design factor of 1.5. Solution. The designer begins with identification of a uniform-diameter shaft which will meet the likely constraints of bearing slope. Using Eq. (17.2), expecting the right bearing slope to be controlling, d= 32(1.5) 600(2)(162 22) + 1000(14)(162 142) 3 30(10)616(0.001)
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