Coefcient, coefcient of linear thermal expansion, end-condition for
springs, thread angle
Bearing angle, coefcient
Eccentricity ratio, engineering (normal) strain
Normal distribution with a mean of 0 and a sta
Bending Properties of Fillet Welds*
Location of G
Unit Second Moment of Area
(6b + d )
2d 2 y + (b + 2d ) 2
In calculating the area for Eq. (851), the designer should, of course, use the
combination of rivet or bolt holes that gives the smallest area.
Figure 823e illustrates a failure by crushing of the rivet or plate. Calculation of
this stress, which is usual
Source: Joseph E. Shigley,
Chap. 24 in Joseph E.
Shigley, Charles R. Mischke,
and Thomas H. Brown, Jr.
Torsional Fatigue Strength under
Extensive tests by Smith23 provide some very interesting results on pulsating torsional
fatigue. Smiths rst result, based on 72 tests, shows that the existence of a torsional
In addition to Eq. (636), the stress ratio
and the amplitude ratio
are also dened and used in connection with uctuating stresses.
Equations (636) utilize symbols a and m as the stress components at the location under scrutiny
From Fig. A159, using D/d = 38/32 = 1.1875, r/d = 3/32 = 0.093 75, we read
the graph to nd K t = 1.65.
(a) From Fig. 620, for Sut = 690 MPa and r = 3 mm, q = 0.84. Thus, from Eq. (632)
K f = 1 + q(K t 1) = 1 + 0.84(1.65 1) = 1
Marin12 identied factors that quantied the effects of surface condition, size, loading,
temperature, and miscellaneous items. The question of whether to adjust the endurance
limit by subtractive corrections or multiplicative corrections was resolved by an
Stress Concentration and Notch Sensitivity (Sec. 610)
The actual part may have a geometric stress concentration by which the fatigue behavior depends on the static stress concentration factor and the component materials sensitivity to fatigue damage.
Mechanical Engineering Design
These are the weld-bead lengths required by weld metal strength. The attachment shear
stress allowable in the base metal, from Table 94, is
all = 0.4Sy = 0.4(36) = 14.4 kpsi
The shear stress in the base metal adjacent to
not necessary to attempt to distribute the loading very precisely. The net force and/or
moment can be applied to a single node, provided the element supports the dof associated with the force and/or moment at the node. However, the analyst should not be
the elimination of the restriction on speeds; the teeth make it possible to run at nearly
any speed, slow or fast. Disadvantages are the rst cost of the belt, the necessity of
grooving the sprockets, and the attendant dynamic uctuations caused at the belt
SPUR GEAR BENDING
BASED ON ANSIAGMA 2001-D04
V = dn
1 [or Eq. (a), Sec. 14 10]; p. 739
W t = 33 000
Eq. (14 15)
= W tKoKvKs
Eq. (14 30); p. 739
Pd Km KB
Eq. (14 40); p. 744
Fig. 14 6; p. 733
Eq. (14 27);
The next step is to estimate the size factor kb . From Table 131, the sum of the addendum and dedendum is
= 0.281 in
The tooth thickness t in Fig. 141b is given in Sec. 141 [Eq. (b)] as t = (4lx)1/2
when x = 3Y/(2P) from Eq.
The dynamic equivalent loads PA and PB are
PA = 0.4Fr A + K A Fa A = 0.4(2170) + 1.5(2522) = 4651 N
PB = Fr B = 2654 N
From Fig. 1116 for 1050 rev/min at 55 C, f T = 1.31. From Fig. 1117, f v = 1.01.
For use in Eq. (1116), a3l = f T f v = 1.31(1.01) = 1.3
The following discussion assumes the teeth to be perfectly formed, perfectly smooth,
and absolutely rigid. Such an assumption is, of course, unrealistic, because the application of forces will cause deections.
Mating gear teeth acting
It is instructive to examine the question of the units of the parameter A of Eq. (1014). Show that
for U.S. customary units the units for Auscu are kpsi inm and for SI units are MPa mmm for ASI.
which make the dimensions of both Auscu and ASI differen
Setting Sa = Sm and solving the quadratic in Sa gives
Setting Sa = /2, Sut = Se /0.5 gives
1 + 4(0.5)2 = 1.66Se
and k f = /Se = 1.66. Since a Gerber locus runs in and among fatigue data and
Goodman does not
Steel tubes with a Youngs modulus of 207 GPa have the specications:
25 0.050 mm
49.98 0.010 mm
50 0.010 mm
75 0.10 mm
These are shrink-tted together. Find the nominal shrink-t pressure and the von Mises stress in
(50) = 88.4 MPa
The yield strength is 240 MPa, and catastrophic failure occurs at 88.4/240 = 0.37, or
at 37 percent of yield. The factor of safety in this circumstance is K I c /K I =
28.3/16 = 1.77 and not 240/50 = 4.8.
Standards and Codes
A standard is a set of specications for parts, materials, or processes intended to
achieve uniformity, efciency, and a specied quality. One of the important purposes
of a standard is to place a limit on the number of items in the sp
from rst principles, textbooks, or handbooks relating the known and unknown
parameters; experimentally or numerically based charts; specic computational tools
as discussed in Sec. 14; etc.
State all assumptions and decisions. Real design problems general
This list is not complete. The reader is urged to explore the various sources of
information on a regular basis and keep records of the knowledge gained.
The Design Engineers Professional Responsibilities
In general, the design engineer is required to
Design is a communication-intensive activity in which both words and pictures are
used, and written and oral forms are employed. Engineers have to communicate effectively and work with people of many disciplines. These are important skills, and an
Identification of need
The phases in design,
acknowledging the many
feedbacks and iterations.
Definition of problem
Analysis and optimization
adverse circumstance or a set of random circumstances that
This is a list of common symbols used in machine design and in this book. Specialized
use in a subject-matter area often attracts fore and post subscripts and superscripts.
To make the table brief enough to be useful the symbol kernels are listed. See
Mechanical design is a complex undertaking, requiring many skills. Extensive relationships need to be subdivided into a series of simple tasks. The complexity of the subject
requires a sequence in which ideas are introduced and iterated.
We rst address th