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 yielding. A von Mises maximum stress is calculated 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. Remember the
origin of the tensile-stress area At , wh
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, in the case where both members are of the same material,
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 1.50 = 0.75 in. The threaded length in the grip is lt = l
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 stiffnesses are assembled to obtain km using Eq. (818).
If the me
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, the centroid of the bolt group in Fig. 828, is found by s
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 stress at angle is
= ( 2 + 3 2 )1/2 =
F
[(cos2 + sin cos )
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 torsion end spring, end corrections such as Eq. (1046) must
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 torsional stress of 103 MPa and an alternating bending
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
d 3
2
K f Ma
Se
+3
2
K f s Ta
Se
+4
2
K f Mm
Sy
+3
K f
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
hardness number for 200 HB 450 is found to be
Su =
0.49
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 and
radial stresses exist as in the theory for thick-w
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 =
w
wl 3
x x 4 + C1 x + C2
12
24
(2)
The boundary condi
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 axis to neutral axis
M = bending moment; positive M decr
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.
56
Coulomb-Mohr Theory for Ductile Materials
Not all ma
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 = 323 /32 = 3.217 (103 )mm3 .
The reversing bending stre
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 of gear teeth. Buckingham and, later, Talbourdet gathere
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 )1/2
+
2
3m )1/2
2
=
32K f Ma
d 3
2
=
32K f Mm
d 3
16K
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 + Mmin )/2 = (5 + 1)/2 = 3 lbf in
r=
a = K i
m =
2
Ma
=
M
Figure 927
Shear stress (psi)
4000
Plot for Ex. 97.
Combined
3000
2000
Thermal
1000
Load induced
x (in)
0.4
0.2
0.2
0.4
1000
2000
avg = P/(2bl) = 1000 psi. Equation (1) produced a maximum of 1922 psi, almost
double the average.
Although design considerat
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
Eq. (1656) as
Tmin T
= exp(t1 )
Tmax T
where = h CR A/(
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)
Equation (1664) can be used to obtain an appropriate ywheel
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 statement about N P and t near the conclusion of Ex. 174.
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 degrees of freedom at each node. Connecting elements
wit
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
alternative form
N
i=1
sx =
2
N
xi2
N
N
xi
i=1
i=1
=
N
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 = centroidal distance, in (cm)
Z = section modulus, in3, (cm3)
Si
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 stress and strength terms in Eq. (13) must be of the sam
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 and case hardness
Pinion material, core and case hardn