LESSION 3 SHAFT DESIGN
鸡茸蘑菇汤Shafts are usually of circular cross ction; either solid or hollow ctions can be ud. A hollow shaft weighs considerably less than a solid shaft of comparable strength, but is somewhat more expensive. Shafts are subjected to torsion, bending, or a combination of the two; in unusual cas, other stress might also become involved. Careful location of bearings can do much to control the size of shafts, as the loading is affected by the position of mountings. After a shaft size is computed, its diameter is often modified (upward only) to fit a standard bearing. Calculations merely indicate the minimum size.
Common Shaft Sizes
Table 3-1 lists some of the common available sizes for steel (round, solid) shafting. The are nominal sizes only. A designer must accurately compute the exact size so that it will fit properly into bearings. Since any machining is costly, a minimal amount of metal should be removed from stock sizes. Any metal removed in certain locations changes the shaft diameter in various axial positions. Therefore, proper radii must be provided to minimize str
ess concentrations. Abrupt changes in diameter without sufficient radii produce so-called ‘stress rairs.’ Thus it is desirable——from the standpoint to provide large radii. However, large radii also make it difficult to mount such other components as pulleys, cams, gears, and so on becau of radius interference. Often, the bore of such other components has to be chamfered to clear radii at the point where a shaft changes diameter. In the final analysis, a compromi has to be made between ideal shaft radii and the undercutting of other components.
Metric Shaft Sizes
The diameters of shafts made compatible with metric-sized bores of mechanical components (such as antifriction bearings) are specified in millimeters. Although any shaft size can be turned to provide extremely accurate fits, Table 3-2 shows popular nominal sizes.
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Torsion
Equations for a shaft in pure torsion are listed below; the equations are for round solid and round hollow ctions only:
T= for solid shafts;
T= for hollow shafts;
where
T= torque (in-lh),
= design stress in shear (psi),
D= diameter of solid shaft (in.),
= outside diameter of hollow shaft (in.),
= inside diameter of hollow shaft (in.)
hp = horpower,
许少安N = revolutions per minute (rpm)
J = polar moment of inertia (in),
c = distance from neutral axis to outermost fiber (in.)
Example 1
Compute the diameter of a solid shaft that rotates 100 rpm and transmits 1.2 hp. The design stress for shear is to be 6000 psi and the shaft is subjected to torsion only.
Solution
T===756 in-lh
D===0.863 in. (minimum)
U -in. diameter.
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Torsional Deflection (Solid Shaft)
The amount of twist in a shaft is important. One rule of thumb is to restrict the torsional deflection to one degree in a length equal to 20 diameters. For example, if the active part of a shaft is 40 in. and the shaft diameter is 2 in. , 1 deg of torsional deflection would be permitted. In some applications, the angle of twist must be smaller than this. The following equation applies to torsional deflection:
D=
where
L= length of shaft subjected to twist (in.)
G=shear modulus of elasticity (psi)
θ＝ angle of twist (rad)
Fig.3-1 shows the angle of twist (greatly exaggerated) that appears when torque is applied to a shaft.
Example 2
A 2-in. lineshaft transmits 20 hp and rotates at 200 rpm. Two pulleys spaced 30 in. apart cau a torsional deflection. If the shear modulus of elasticity is 12,000,000 psi, find the angle of twist in degree.
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领导动员大会发言稿Solution:
T===6300 in-lh
θ＝＝＝0.0100 rad；
0.0100×57.3＝0.573 deg.
Combined Torsion and Bending (Solid Shafting)没有胎心怎么办
A shaft is often subjected to combined torsion and flexure. There are numerous ways of computing a shaft diameter under the conditions. The simplest is to compute equivalent bending and twisting moments for the shaft and then substitute the values into the regular equations for torsion and bending. Equations for equivalent moments are as follows:
T=
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where
T= equivalent twisting moment (in-lb),
M=equivalent bending moment (in-lb),
The torsion equation then becomes
D=
and the bending (or flexure) equation becomes
D=
Both equations must be solved; the larger of the two diameters is then ud for the calculated size. It is important to remember that the allowable shearing stress is ud in the torsion formula; the allowable stress in tension is ud in the flexure (or bending) formula. The following example shows how the method is applied.
