Elements and Units of Machines
4.5 ANALYSIS OF SHAFT DEFORMATIONS
Do you like calculations? If so, we envy you. The point is that most engineers like to draw all kinds of machines and to enjoy their pictures. Some of them avoid even the simplest calculation for strength and deformation, relying on their intuition and hoping for the best. What self-confidence!
Well, deformation analysis is not a luxury. Sometimes it takes quite some time, and we often tend to use our intuition and decide whether this work is a must or not. Occasionally, it is really obvious that deformations are negligible. For example, if a fly alights on a table, the latter will be bent, but we don’t take this deflection into consideration when placing a plateful of soup on it!
Our experience says that the real danger for the soup comes when the fly flies off the table. With this is mind, let’s consider a shaft of 2 m length and of about 0.5 m in diameter. The first impression is that no power on earth can deform this iron log! No power on earth? Let’s calculate the bending deformations of such a shaft.
Analysis of shaft deformations is time consuming. It is better to do this work using computer programs, particularly when the diameter of the shaft is variable. For our readers, who might not have such a program, the authors have decided to give formulas for calculation of deflection y and slope θ (shown in Figure 4.15b). These formulas are accompanied by graphs (Figure 4.16 to Figure 4.19) for estimation of slopes, which are usually important to know because they affect the alignment of mating parts and load distribution in their contact. Using these graphs, we are able to estimate very quickly the order of the slope magnitude and decide whether it is worthwhile to analyze the slopes precisely or not. In addition, the simplified estimation may serve as a reference point for the precise calculation and possibly prevent a mistake. For Figure 4.16 and Figure 4.17, the slopes are to be determined from the equation
(4.5)
FIGURE 4.16 Angular deformation (slope) of shaft cross sections (radial force applied between the bearings).
θ =KFL
FIGURE 4.17 Angular deformation (slope) of shaft cross sections (overhung radial force).
FIGURE 4.18 Angular deformation (slope) of shaft cross sections (bending moment applied between the bearings).
For Figure 4.18 and Figure 4.19,
(4.6)
where
F = radial force M = bending moment
E = modulus of elasticity of the shaft material I = moment of inertia of the shaft section L = shaft length between bearings
K = value taken from corresponding graphs
Dear reader, if you try to use all the quantities in compatible units (e.g., F in N (newton), M in N⋅mm, L in mm, E in N/mm2 (MPa), and I in mm4), the slope angle θ will be obtained in radians.
But, if the dimensions are incompatible, the result will have no physical meaning. (See Chapter 13, Subsection 13.4.2 about the choice of units.)
FIGURE 4.19 Angular deformation (slope) of shaft cross sections (overhung bending moment).
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8
0
–0.1
–0.2
K
X/L 0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 M L
X X1
θ =KML EI
The exact formulas for deformations of shafts with uniform cross section are as follows.
1. A shaft loaded by a radial force F between bearings (see draft in Figure 4.16):
where b = L − a.
2. A cantilever shaft loaded by a radial force F on the overhanging end (see draft in Figure 4.17):
3. A shaft loaded by a bending moment M between bearings (see draft in Figure 4.18):
y Fbx
4. A cantilever shaft loaded by bending moment M on the overhanging end (see draft in Figure 4.19):
You should not read at leisure this frightening amount of equations, as you would a Harry Potter Book. But if you really must check your shaft for rigidity, these equations and graphs may be helpful.
EXAMPLE 4.5
If you remember, our intention was to calculate the deformations of some large shaft. Here you are: Figure 4.20 shows the main shaft and part of the drive gear of a subway escalator, which is designed to transport passengers through a height of 65 m. (Dimensions of the shaft and the loading force vectors are given in Figure 4.20 and Figure 4.21a). Forces applied to the hauling chains are
FIGURE 4.20 Escalator main drive.
created from below (Fcl) by the weight of the chains and steps, and from above (Fcu) by the same weight of the chains and steps plus the weight of the passengers. The forces loading the shaft are as follows:
Forces Fcu= 30 · 104 N and Fcl= 10 · 104 N act in the horizontal plane and are distributed equally between the two sprockets.
Gear force Fg= 24 · 104 N acts in the plane of action of the gear.
Gear axial force Fa= 5.1 · 104 N.
The directions of the forces are shown in Figure 4.21, where the picture is turned for convenience by 45° counterclockwise.
Of interest to us is the influence of the shaft resilience on gear teeth misalignment (skewness).
For that reason, the shaft deformations should be calculated in the plane of action of the gear, and all the forces should be projected onto this plane (i.e., on the plane where force Fg acts). The angle between the forces Fcl, Fcu, and the gear action plane measures about 25°. The loading layout of the shaft in this plane is given in Figure 4.21b. Here are the forces:
F1= Fg= 24 · 104 N
F2= F3= 0.5(Fcl+ Fcu) cos 25°= 18.1 · 104 N
M = Fa·0.5dg·sin 20°= 5.1·104 · 0.5·1264·sin 20°= 1.1·107 N·mm (Here dg [the gear wheel pitch diameter] = 1264 mm.)
Let’s estimate the slope angle of the shaft section in the middle of the gear face width, i.e., at a distance of 243 mm from the left bearing. This will be done using Figure 4.16 and Figure 4.18, FIGURE 4.21 Escalator main drive shaft with a pinion.
243
(a) Fa
F2 F3
F1 Fg
350
M
(b)
243 728
1948 2070 1/2 Fcu
1/2 Fcl 1/2 Fcl
1/2 Fcu
Fcl
Fcu
Fg 45°20°
R 2070
485 1220
∅450
∅290
as well as Equation 4.5 and Equation 4.6. But these figures and formulas are intended for shafts of uniform cross section. Therefore, we have to determine the uniform diameter of the equivalent shaft, which has nearly the same rigidity as the real one. The real shaft is made up mainly of two segments: diameter d1 = 290 mm (with a total length L1 ≈ 850 mm) and diameter d2 = 450 mm (L2 ≈ 1220 mm). The longer the segment, the greater its influence on shaft rigidity. For this reason, the short segments are ignored regardless of their diameter.
(Not only the length and diameter of the segment but also its position relative to the bending moment diagram is important; the bigger the bending moment that the segment is exposed to, the greater its contribution to the resilience of the shaft. But for the approximate calculation, this factor may be omitted.)
The deformation of a segment depends on its length and the moment of inertia of its section (I).
To take account of these two factors, the moment of inertia of the equivalent shaft Ie is determined as follows:
(For the curious, the equivalent diameter corresponding to this value of Ie is about 406 mm. We don’t need it for the following analysis.)
Now, we go to Figure 4.16 and Figure 4.18. The section in question is placed at a distance of 243 mm from the left bearing, so for this section x/L = 243/2070 = 0.12. Coefficients KF1, KF2, and KF3 for forces F1, F2, and F3 correspondingly found from Figure 4.19 are as follows:
x/L = 0.12, a/L = 243/2070 = 0.12, KF1= 0.026 x/L = 0.12, a/L = 728/2070 = 0.35, KF2 = 0.057 x/L = 0.12, a/L = 1948/2070 = 0.94, KF3 = 0.006 For the bending moment M, the KM value is found from Figure 4.21:
x/L = 0.12, a/L = 0.12, KM = 0.22
Using Equation 4.5 and Equation 4.6 we calculate the following results:
I I L I L
The total slope of the shaft in the middle of the gear width,
θ = θF1 + θF2 + θF3 + θM = (9.77 + 16.15 + 1.7 + 1.83)10−5 = 2.945·10−4 (rad)
The face width of the gear measures 350 mm, and thus the error in parallelism of the teeth is about 0.1 mm. It is a large error, which results in such an unevenness of load distribution along the teeth that the maximal unit load is more than twice the average. But in this case, the tilt of the gear wheel (along with the adjoined shaft section) is largely compensated by the elastic deformation of the pinion, which tilts in the same direction (as shown in Chapter 1, Figure 1.5e).
Computer calculation of slope angle θ gives θ = 3.5 · 10−4 rad. That means that the error of the rough estimation was 19%. It is too much, but the preliminary estimate lets us know if a more precise deformation analysis should be performed. In this case, the answer is “yes.”
The torque, which loads the main shaft between sprockets, is
(Here, R [the pitch radius of the sprocket] = 706 mm.)
The diameter of the shaft having the required strength may be estimated by Equation 4.1, where the tensile strength of the shaft material Su is taken equal to 930 MPa:
If the shaft segment between the sprockets had been chosen with a diameter d2 = 250 mm (instead of 450 mm), the slope angle of the gear wheel would have grown to θ= 1.53 · 10−3 rad. This results in a gear tooth parallelism error of about 0.5 mm, which is absolutely unacceptable.
REFERENCES
1. Timoshenko, S., Vibration Problems in Engineering, D. Van Nostrand Company, Toronto, New York, London, 1955.
2. Birger, I.A., Shorr, B.F., and Ioselevitz, G.B., Calculations for Strength of Machine Elements, Mashinostroenie, Moscow, 1979 (in Russian).
3. Peterson, R.E., Stress Concentration Factors, John Wiley and Sons, New York, London, Sydney, Toronto, 1974.
4. Philimonov, G.N. and Balazky, L.T., Fretting in joints of naval parts, Sudostroenie, Leningrad, 1973 (in Russian).
T=0 5. (Fcu−F Rcl) =0 5 30 10. ( ⋅ 4−10 10 706⋅ 4) =7 06. ⋅1107N.mm
d= ÷( ) . ⋅ = ÷ 5 6 7 06 10
930 7 212 254
3 mm
67