Shaft Design
Sections 7-1 through 7-3 in Shigley discuss a large number of qualitative considerations
that should be taken into account when designing a shaft. These are good sections, and if you
did not read them carefully as assigned, please do so.
Shigley makes what I consider to be a particularly important comment regarding axial
loads. Axial loads are generally caused by one of three situations. When the shaft supports
helical gears.
When the shaft supports bevel gears.
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When the shaft is supported by tapered roller bearings.
The axial load must be transmitted to the housing of the machine, usually through a thrust
bearing that can provide both radial and axial forces.
Shigley recommends, and I concur, that, in general, only one thrust bearing should be
used. The reason for this recommendation is to avoid large axial stresses caused by thermal
expansion of the shaft.
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Example
Consider a low-alloy steel shaft, with length L between two thrust bearings, subjected to
an average temperature increase along its length of 200  F .
 1 

  6.6 (10)  6 
 F 
(E.1)
E  29.0 (10) 3 ksi
(E.2)
 L  L  T  6.6 (10) 6 (200)  0.00132 L
(E.3)
Suppose, as a result of the two thrust bearings, that the length of the shaft between the
bearings is not free to expand. This would introduce a compressive axial stress determined by
the following relationship.
 L   0.00132 L 
L
E
   0.00132 E   0.00132 (29.0) (10) 3  38.3 ksi
(E.4)
(E.5)
A relatively small temperature increase has induced a compressive axial stress in the
shaft well above any reasonable allowable stress.
Section 7-4, Shaft Design for Stress
This is an extremely difficult section, containing a large number of very complicated
relationships. These relationships involve torsion and axial stresses, in addition to bending
stresses in more than one plane. All these stresses include both alternating and mid-range
stresses. Von Mises stresses are determined after including concentration factors.
It is my opinion that stress considerations in shaft design are almost always performed on
a computer today. However, if you are ever required to design a shaft by hand, which I consider
extremely unlikely, you will need to understand the material in this section.
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Section 7-5, Deflection Considerations
Shigley points out that deflection analysis, even at a single location on a shaft, requires
that the complete geometry of the shaft be known. He therefore recommends that stresses be
considered first, and the shaft dimensions determined to reduce stresses to acceptable levels,
prior to conducting deflection analysis.
Deflection and slope of the shaft should be checked at all gears and bearings. In general,
it is necessary to consult with the gear and bearing manufacturers to determine allowable
deflections and slopes. However, Shigley presents the following table that can be used as a
general guideline for the maximum deflection or slope at some of these devices. In the first
table, the first five entries refer to bearing types.
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Forces that cause the shaft to deflect are frequently in more than one plane, as in the
following figure.
When this is the case, it is necessary to perform two deflection-slope analyses, one in the
vertical plane and one in the horizontal plane. The results are then added as vectors, using the
Pythagorean theorem. For deflection-slope analyses in two non-perpendicular planes, or in any
three or more planes, the results are added as vectors.
As an aside, note how complicated the stress analysis of this shaft becomes.
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Section 7-6, Critical Speeds of Shafts
This is a very important topic, and unfortunately one that Shigley treats very briefly. I
want to go into more depth, and try to give you an overall understanding of this subject.
Consider figure (a), a mass m suspended from a vertical, linear spring with a spring
constant k. When the spring is un-stretched, the mass is in the top position and must be
supported there. As the support is slowly withdrawn, the mass is lowered to the central position,
a distance Yst below the top position. The mass is in vertical static equilibrium at this position.
 FY  0  k Yst  m g
(24.1)
k
g

m Yst
(24.2)
If the mass is then pulled down a distance Y from the static-equilibrium position and
released, the mass will oscillate, with an amplitude Y about the static-equilibrium position. We
determined the natural frequency, denoted by  n , expressed in rad/s, in physics I and
Engineering Mechanics II.
n 
k

m
g
Yst
(24.3)
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Now consider figure (b), a shaft with a single concentrated mass. The mass of the shaft is
assumed to be negligible in comparison to the concentrated mass. When the shaft is undeflected, the concentrated mass is in the top position and must be supported there. As the
support is slowly withdrawn, the concentrated mass is lowered to the central position, a distance
Yst below the top position. The mass is in vertical static equilibrium at this position. We define
the spring constant of the shaft in exactly the same manner as for the spring, the force required to
deflect the concentrated mass a unit distance from its position when the shaft is un-deflected.
If the concentrated mass were then displaced from its static-equilibrium position, it would
vibrate about that static-equilibrium position with a natural frequency given by equation (24.3).
However, mechanical engineers generally refer to this natural frequency as the critical
speed, denoted by N C , and expressed in units of revolutions per minute (rpm).
 60 
g
  n  9.549
N C  
Yst
 2 
(24.4)
If Yst is measured in inch, g must be expressed in inch / s 2 .
g  32.174
N C  188
ft
s2
 386.1
inch
s2
1
Yst
(24.5)
(24.6)
Equation (24.6) is an important relationship for two reasons. (1) The value of Yst is
frequently determined for other reasons, such as the clearance with the housing. (2) For shafts
with multiple concentrated or distributed masses, equation (24.6) yields a reasonable
approximation for the critical speed (first or lowest natural frequency) when the maximum
deflection is used in place of Yst .
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Whirl
Consider a shaft with either a single concentrated mass, or a distributed mass that can be
modeled as a concentrated mass. For simplicity, the static-deflection is assumed to be very small
and both vibration and whirl take place around the shaft centerline.
If the mass were vertically displaced from its equilibrium position (approximated as the
shaft centerline), as shown in the left figure above, and released, the shaft would vibrate in a
vertical plane at its natural frequency.
If the shaft were perfectly balanced, and rotated at some frequency, the shaft and
concentrated mass would rotate around the static-deflection curve (again approximated as the
shaft centerline), with zero amplitude, at that same frequency.
However, the concentrated mass might suddenly develop an eccentricity, e.g., if a chunk
of the mass broke off. If this were to happen, the shaft would whirl, i.e., develop a displaced
curve that rotates around the static-deflection curve. This is shown, greatly exaggerated, in the
right figure above.
The center of the shaft is displaced a distance Y from the original centerline of the shaft.
The CG of the concentrated mass is no longer coincident with the center of the shaft. That CG
is now located a distance e, (eccentricity) from the center of the shaft. It whirls, as shown at a
radius that differs from Y by e. In the figure, the CG is whirling at a radius greater than Y. This
may or may not be true.
This whirl creates a strong vibration in the machine at the run frequency of the shaft.
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The centripetal force necessary to hold the mass m in its position must be equal to the
restoring force provided by the deflected shaft.
k Y  m  2 ( Y  e)
(24.7)
(k  m  2 ) Y  m  2 e
(24.8)


2


Y

e
2
2
2

k  m
 n   
(24.9)
m 2 e
Y is the amplitude of the whirl of the center of the shaft. It is positive. If   n , then e
is also positive, and the CG of the mass whirls with a greater radius than the center of the shaft.
However, if   n , then e is negative, and the CG of the mass whirls with a smaller radius
than the center of the shaft.
This is analogous to the situation in which a forced, vibrating mass with no damping is in
phase with the forcing function for   n , and out of phase for   n .
Following is an extremely important concept for mechanical engineers.
If a rotating machine suddenly develops a serious vibration problem, the frequency of the
vibration is almost always either the run frequency of the machine or the lowest natural
frequency of the shaft.
If the vibration is at the run frequency, then the cause is very likely to be an eccentricity
in the rotor, resulting in shaft whirl.
If the vibration is at the lowest natural frequency of the shaft (the critical speed) , then
the cause is very likely that some portion of the rotating shaft has developed a rub (contact) with
the housing that surrounds the shaft. As a result of this rub, the shaft will vibrate laterally at its
natural frequency, or lowest natural frequency if the shaft possesses more than one degree of
freedom. This may seem counter-intuitive, but it is identical with the principle of a violin. A
stretched string is rubbed perpendicular to its length with a bow. The string “jumps” away from
the bow and vibrates, at its first natural frequency, perpendicular to plane defined by the original
string and the bow.
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Torsional Oscillation of a Shaft
I have analyzed a number of vibration problems during my career. All but one were
vibrations that “appeared” suddenly in a rotating machine that had been functioning normally. In
every one of those cases, I was able to determine the generic cause of the vibration using the
important concept on the previous page.
However, one problem involved a horrible vibration that appeared when a large fan, used
to force air into a steam generator, was initially started. This vibration, in Richmond, Indiana,
rattled dishes in kitchen cupboards more than a mile from the power plant. Inasmuch as one of
those kitchens belonged to the wife of the mayor of Richmond, we were under some pressure to
make it go away as soon as possible.
Consider two unequal masses, resting on a horizontal, frictionless surface, connected by a
linear spring with spring constant k.
Apply equal-magnitude forces to the two masses, stretching the spring. Then release the
masses. They will oscillate back and forth, out of phase, with the same frequency. There will be
some point on the spring that does not move, in effect dividing the spring into two separate
springs with lengths shown in the above figure.
Determine the spring constants of the two springs.
 L  L2 

k 1  k  1
 L1 
(24.10)
 L  L2 

k 2  k  1
 L2 
(24.11)
The two masses oscillate with the same frequency, resulting in the following relationship.
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2 
k 1 k (L 1  L 2 ) k 2 k (L 1  L 2 )



m1
m1 L1
m2
m2 L2
m1 L1  m 2 L 2
(24.12)
(24.13)
Consider two masses, mounted on a common shaft.
The unequal mass moments of inertia are with respect to the centerline of the shaft. A
common example of this configuration is an electric motor driving a centrifugal fan. Note that
lateral vibrations of the masses will occur only at very high frequencies.
Apply equal-magnitude torques to the two masses, twisting the portion of the shaft
between the masses. Then release the masses. They will oscillate back and forth torsionally, out
of phase, with the same frequency. There will be some point on the shaft that does not rotate, in
effect dividing shaft into two separate shafts with lengths shown in the above figure.
Determine the torsional spring constants of the two shafts. The torsional spring constant
of a shaft is the torque required to cause a unit rotation of one end of the shaft with respect to the
other end.
k1 
JG
L1
(24.14)
k2 
JG
L2
(24.15)
The two masses oscillate torsionally with the same frequency, resulting in the following
relationship.
2 
k1
k
JG
JG

 2 
I 1 I 1 L1 I 2 I 2 L 2
I 1 L1  I 2 L 2
(24.16)
(24.17)
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Homework
1.
Consider a simply-supported, low-alloy steel shaft, ten feet in length and two
inches in diameter. The shaft supports a single concentrated mass at its center. The weight of
the concentrated mass is 0.6 kip. Estimate the critical speed of the shaft, expressed in rpm.
2.
Consider a simply-supported, low-alloy steel shaft, ten feet in length and two
inches in diameter. The shaft supports two equal concentrated masses, each located two feet
from its center. The weight of each concentrated mass is 0.5 kip. Estimate the critical speed
(first natural frequency) of the shaft, expressed in rpm.
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