Mechanics of Materials 1 Part 6 pps
Mechanics of Materials 1 Part 6 pps
...
replacing
T.
(14 . 16 )
Similarly,
(14 .17 )
358
Mechanics
of
Materials
Fig.
13 . 36.
Therefore at the junction
of
web and flange
75
x
10 3
x
0.07
a*
=
72. 56
x
=
72.35
x
lo6
=
72.35 ...
0,)'
+
4~2 ~1
3 36
Mechanics
of
Materials
$13 .7
re
I
up
=ux
uq
=uy
=o
r,,
:O
1 1 1
I
1
I
I
O0
45"
90"
13 5'
18 0&q...
Mechanics of Materials 1 Part 3 ppsx
...
1
2
2
3
4
5
6
0
10 0~
6=
60 0
10 0x12 =12 00
12 00
13 20
14 40
16 80
19 56
-
72
69
69
68
66 .3
61 . 6
54.5
-
10 0
10 0
12
12
12
12
12
-
3.2
6. 2
51. 3
55 .6
59 .1
64 .1
66 .0 ...
(1)
-
16 MB- 74 .67 Mc
=
398.24
x
lo3
-
71. 67 Mc
=
313 .66
x
lo3
Mc
=
-
4.37
x
lo3
Nm
Substituting in (l),
-
16 ~,
-
3(
-
4.37
x...
Mechanics of Materials 1 Part 5 pps
... (Fig.
11 .7)
( :)
;;
25’
lo6
volume of bar
=
in
x
~
x 2 .6
=
12 . 76 x
O’
x 12 . 76
x
Then
10 x 10
1. 30
O2
30+-
=
10 9
3i3x
10 12
1. 30
30
x
313
x
10 ”
f-
x
313
x
10 ” ... deflection
6,
when applied statically then, by proportion,
Substituting in eqn.
(11 .12 )
6
W(h +6) =~WX- x6
6,
6
-
26, 6
-
26, h
=
0
6
=
6,
J (6,...
Mechanics of Materials 1 Part 7 ppsx
...
0.3.
1
2E
-
cu:
+
u:
+
u:
-
2v(u,u2
+
u1u3
+
u2u3)]
[W,
60 ,70
MN/m2.]
16 .9
Mechanics of Materia/s
442
Fig. 16 .13 . Miniature strain-gauge installation. (Welwyn.)
Fig. 16 .14 . "Full ... arrangement shown
in
Fig. 16 .14 and would appear as in Fig. 16 .15 b.
Lood
M
To
R,
M
V
1. 1
Fig.
16 .1
5.
(a) Determination of bending strains independen...
Mechanics of Materials 1 Part 8 pps
...
Introduction
11 .1. 1
The SIN curve
1
1. 1.2
PISIN curves
1 1 .1. 3
Effect
of
mean stress
1 1 .1. 4
Effect
of
stress concentration
11 .1. 5
Cumulative damage
1
1. 1 .6
Cyclic stress-strain
1 1 .1. 7 ...
443
4 46
4 46
4 46
449
45
1
453
454
455
458
460
462
462
462
465
466
467
470
47
1
472
472
473
474
474
475
475
477
48
1
4...
Mechanics of Materials 1 Part 15 pps
...
(MN/m 312 ) 4 .13 4.38 4 .65 4.84 5. 26 5. 86 6.57 6. 97
h30(AK) 0. 61 6 0 .64 2 0 .66 8 0 .68 5 0.7 21 0. 769 0. 818 0.843
da
-
x
lo6
&cycle 2. 06 4.34 4. 91 6. 02 6. 97 8.38
dN
log,, (da/dN) -5 .68 -5. 36 ...
+
C)
10 23 (1, 3000
+
C)
=
10 73 (1, 500
+
C)
10 23(8.0 064
+
C)
=
10 73 (6. 214 6
+
C)
819 0.5
+
10 23C
=
66 68.27
+
10 73C
15 22.23...
Mechanics of Materials 2010 Part 6 ppsx
... terms.
c
1, 1 ,1, 1
c
1, 1 ,1, 2
c
1, 1 ,1, 3
c
1, 1,2 ,1
c
1, 1,2,2
c
1, 1,2,3
c
1, 1,3 ,1
c
1, 1,3,2
c
1, 1,3,3
c
1, 2 ,1, 1
c
1, 2 ,1, 2
c
1, 2 ,1, 3
c
1, 2,2 ,1
c
1, 2,2,2
c
1, 2,2,3
c
1, 2,3 ,1
c
1, 2,3,2
c
1, 2,3,3
c
1, 3 ,1, 1
c
1, 3 ,1, 2
c
1, 3 ,1, 3
c
1, 3,2 ,1
c
1, 3,2,2
c
1, 3,2,3
c
1, 3,3 ,1
c
1, 3,3,...
High Cycle Fatigue: A Mechanics of Materials Perspective part 2 pps
... University Press, New York, 19 98. Hertzberg, R.W.,
Deformation and Fracture Mechanics of Engineering Materials, 3rd ed., Wiley, New York, 19 89.
x
Chapter 1
Introduction
1. 1. HISTORICAL BACKGROUND
Fatigue ... consideration in the disciplines of solid
mechanics or strength of materials. In fact, one of the earliest researchers to address the
topic was Wöhler [1] , who, in...
High Cycle Fatigue: A Mechanics of Materials Perspective part 6 pdf
... 8.
0
500
10 00
15 00
2000
10
3
10
4
10
5
10
6
10
7
10
8
Failure
Run-out
Stress amplitude (MPa)
Cycles to failure
σ
m
= –784 MPa
σ
m
= +784 MPa
σ
m
= 0 MPa
Figure 2 . 16 . S–N data on SKH 51 tool ... = 1
10
7
A
B
C
Figure 2 .11 . Schematic of S–N curves obtained at constant values of R.
Alternating stress
R = 0
R
= 0.7
R
= 1
A
B
C
N
= 10
7
N = 10
4
Mean stress...
High Cycle Fatigue: A Mechanics of Materials Perspective part 7 ppsx
... −3 01 9E +11
0 1 −5956E +14
05 −4 91 0E +12
08 −7 51 9E +11
0
10
20
30
40
50
010 20304050
59 Hz HCF data
Alternating stress (ksi)
Mean stress (ksi)
R = 1
R
= –0.33
R
= 0.8
R
= 0.5
R
= 0 .1
46 ... subset Number of tests k m Walker exponent, w
A R ≤−0 1 @ 59 Hz 24 5.83E 16 −7 17 0 16 5
C R ≤−0333 @
59HzR= 0 1
@ 370–400 Hz
20 7 .63 E 16 6 98 03 817
52 Introduct...