throughout its entire range. This is evident by noting that it crosses the yield line in
Fig. 627. The Gerber and modied Goodman criteria do not guard against yielding,
requiring a separate check for
the coordinate system of Fig. 88, we note
x =
6F
dr n t p
yz =
y = 0
z =
x y = 0
4F
dr2
16T
dr3
zx = 0
then use Eq. (514) of Sec. 55.
The screw-thread form is complicated from an analysis viewpoint.
The pressure p generated at the interface of the interference t, from Eq. (356)
converted into terms of diameters, is given by
p=
(739)
2
d d 2 + di2
d do + d 2
+ o +
i
2
E o do d 2
E i d 2 di2
or, i
1
From Table A17 the next fraction size bolt is L = 2 4 in. From Eq. (813), the thread
length is L T = 2(0.625) + 0.25 = 1.50 in. Thus the length of the unthreaded portion
in the grip is ld = 2.25 1.5
With = 30 , this becomes
k=
0.5774 Ed
(1.155t + D d)(D + d)
ln
(1.155t + D + d)(D d)
(820)
Equation (820), or (819), must be solved separately for each frustum in the
joint. Then individual stiffnesse
For a F = 16 kN load nd
(a) The resultant load on each bolt
(b) The maximum shear stress in each bolt
(c) The maximum bearing stress
(d) The critical bending stress in the bar
Solution
(a) Point O, th
The nominal stresses at the angle in the weldment, and , are
=
Fs
F sin (cos + sin )
F
=
= (sin cos + sin2 )
A
hl
hl
(d)
=
Fn
F cos (cos + sin )
F
=
= (cos2 + sin cos )
A
hl
hl
(e)
The von Mises stre
The angle subtended by the end deection of a cantilever, when viewed from the
built-in ends, is y/l rad. From Table A91,
e =
Fl 2
Fl 2
64Ml
y
=
=
=
l
3E I
3E(d 4 /64)
3d 4 E
(1046)
For a straight tors
F = 10 kip
12 in
6 in
6 in
d/5 R.
Problem 611
d
R1
1.5 d
d/10 R.
R2 d
1 in
612
A bar of steel has the minimum properties Se = 276 MPa, Sy = 413 MPa, and Sut = 551 MPa.
The bar is subjected to a steady
DE-Gerber
1
8A
=
n
d 3 Se
8n A
d=
Se
where
1+ 1+
2B Se
ASut
2B Se
ASut
1+ 1+
2 1/2
2 1/2
(79)
1/3
(710)
A=
4(K f Ma ) 2 + 3(K f s Ta ) 2
B=
4(K f Mm ) 2 + 3(K f s Tm ) 2
DE-ASME Elliptic
1
16
=
4
n
material tested. This means that the strength of parts could, if desired, be tested part
by part during manufacture.
For steels, the relationship between the minimum ultimate strength and the Brinell
315
Stresses in Rotating Rings
Many rotating elements, such as ywheels and blowers, can be simplied to a rotating
ring to determine the stresses. When this is done it is found that the same tangential
loading are symmetric relative to the midspan. However, we will use the given boundary conditions of the problem and verify that the slope is zero at the midspan. Integrating
Eq. (1) gives
EIy =
M dx
h = depth of section
co = distance from neutral axis to outer ber
ci = distance from neutral axis to inner ber
rn = radius of neutral axis
rc = radius of centroidal axis
e = distance from centroidal a
B
Figure 511
(a)
Load lines for Example 51.
Sy
(b)
B
A
Sy
Sy
A
(c)
Sy
DE
MSS
Load lines
(d )
the only plane stress case given in which the two theories agree, thus giving the same
factor of safety.
The next step is to estimate the bending stress at point B. The bending moment
is
M B = R1 x =
225(6.8)
225F
250 =
250 = 695.5 N m
550
550
Just to the left of B the section modulus is I /c = d 3 /32 =
To determine the surface fatigue strength of mating materials, Buckingham designed
a simple machine for testing a pair of contacting rolling surfaces in connection with his
investigation of the wear o
Combining these stresses in accordance with the distortion energy failure theory,
the von Mises stresses for rotating round, solid shafts, neglecting axial loads, are given
by
a
m
=
2
(a
=
2
(m
+
2
3a
The diametral clearance between the inside of the spring coil and the pin at load is
= D d D p = 0.494 0.072 0.400 = 0.022 in
Answer
(c) Fatigue:
Ma = (Mmax Mmin )/2 = (5 1)/2 = 2 lbf in
Mm = (Mmax +
and I2 = 0,
E=
in Btu. The temperature rise
1 I
2 2
f
2 9336 o
(1658)
T due to a single stop is
T =
E
WC
(1659)
Tmax has to be high enough to transfer E Btu in t1 seconds. For steady state, rearrange
where is the nominal angular velocity, given by
=
2 + 1
2
(1663)
Equation (1661) can be factored to give
E2 E1 =
I
(2 1 )(2 + 1 )
2
Since 2 1 = Cs and 2 + 1 = 2, we have
E 2 E 1 = C s I 2
(1664)
Equat
The constants K and b have their ranges of validity. If N P > 109 , report that N P = 109
and t > N P L p /(720V ) without placing condence in numerical values beyond the
validity interval. See the st
types of elements available for finite-element analysis for structural problems. Not
all elements support all degrees of freedom. For example, the 3-D truss element supports only three translational d
The sample standard deviation, dened as the square root of the sample variance, is
1
N 1
sx =
N
i=1
(xi x)2
(207)
Equation (207) is not well suited for use in a calculator. For such purposes, use the
Table A6
Properties of StructuralSteel Angles
w = weight per foot, lbf/ft
m = mass per meter, kg/m
A = area, in2 (cm2)
I = second moment of area, in4 (cm4)
k = radius of gyration, in (cm)
y = centroid
18
Mechanical Engineering Design
the design factor in terms of a stress and a relevant strength. Thus Eq. (11) can be
rewritten as
nd =
S
loss-of-function strength
=
allowable stress
(or )
(13)
The s
155
Design of a Straight-Bevel Gear Mesh
A useful decision set for straight-bevel gear design is
Function
Design factor
Tooth system
Tooth count
Pitch and face width
Quality number
Gear material, core