budynas_SM_ch02.pdf

(149 KB) Pobierz
budynas_SM_ch02.qxd
Chapter 2
2-1 From Table A-20
S ut =
470 MPa (68 kpsi) , S y =
390 MPa (57 kpsi) Ans.
2-2 From Table A-20
S ut =
620 MPa (90 kpsi) , S y =
340 MPa (49
.
5 kpsi) Ans.
2-3 Comparison of yield strengths:
S ut of G10 500 HR is 620
470 =
1
.
32 times larger than SAE1020 CD Ans.
S yt of SAE1020 CD is 390
340 =
1
.
15 times larger than G10500 HR Ans.
From Table A-20, the ductilities (reduction in areas) show,
SAE1020 CD is 40
35 =
1
.
14 times larger than G10500 Ans.
The stiffness values of these materials are identical Ans.
Table A-20
Table A-5
S ut
S y
Ductility
Stiffness
MPa (kpsi)
MPa (kpsi)
R %
GPa (Mpsi)
SAE1020 CD 470(68)
390 (57)
40
207(30)
UNS10500 HR 620(90)
340(495)
35
207(30)
2-4 From Table A-21
1040 Q&T
S y =
593 (86) MPa (kpsi) at 205 ◦ C (400 ◦ F) Ans.
2-5 From Table A-21
1040 Q&T R
=
65% at 650 ◦ C (1200 ◦ F) Ans.
2-6 Using Table A-5, the specific strengths are:
UNS G10350 HR steel:
S y
W =
39
.
5(10 3 )
282 =
1
.
40(10 5 )in Ans.
0
.
2024 T4 aluminum:
S y
W =
43(10 3 )
0
.
098 =
4
.
39(10 5 )in Ans.
Ti-6Al-4V titanium:
S y
W =
140(10 3 )
0
.
16 =
8
.
75(10 5 )in Ans.
ASTM 30 gray cast iron has no yield strength. Ans.
672657644.003.png
Chapter 2
7
2-7 The specific moduli are:
UNS G10350 HR steel:
E
W =
30(10 6 )
0
.
282 =
1
.
06(10 8 )in Ans.
2024 T4 aluminum:
E
W =
10
.
3(10 6 )
098 =
1
.
05(10 8 )in Ans.
0
.
Ti-6Al-4V titanium:
E
W =
16
.
5(10 6 )
0
16 =
1
.
03(10 8 )in Ans.
.
Gray cast iron:
E
W =
14
.
5(10 6 )
0
26 =
5
.
58(10 7 )in Ans.
.
2-8
2 G (1
+ ν
)
= E
⇒ ν =
E
2 G
2 G
From Table A-5
Steel:
ν =
30
−
2(11
.
5)
=
0
.
304 Ans.
2(11
.
5)
Aluminum:
ν =
10
.
4
−
2(3
.
90)
=
0
.
333 Ans.
2(3
.
90)
Beryllium copper:
ν =
18
2(7)
2(7)
=
0
.
286 Ans.
Gray cast iron:
ν =
14
.
5
2(6)
2(6)
−
=
0
.
208 Ans.
2-9
E
U
80
70
60
50
40
Y
S u 85.5 kpsi Ans.
S y 45.5 kpsi Ans.
30
E 90 0.003 30 000 kpsi Ans.
20
R
A 0 A F
A 0
0.1987 0.1077
0.1987
(100) 45.8% Ans.
10
l
l 0
l l 0
l 0
l
l 0 1
A 0
1
A
0
0
0.002
0.1
0.004
0.2
0.006
0.3
0.008
0.4
0.010
0.5
0.012
0.6
0.014
0.7
0.016
0.8
(Lower curve)
(Upper curve)
Strain,
−
−
672657644.004.png
8
Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
2-10 To plot
σ true vs.
ε
, the following equations are applied to the data.
A 0 = π
(0
.
503) 2
4
=
0
.
1987 in 2
Eq. (2-4)
ε =
ln
l
l 0
for 0
≤
L
≤
0
.
0028 in
ε =
ln
A 0
A
for
L
>
0
.
0028 in
P
A
The results are summarized in the table below and plotted on the next page.
The last 5 points of data are used to plot log
σ true =
σ
vs log
ε
The curve fit gives
m
=
0.2306
Ans.
log
σ 0 =
5
.
1852
⇒ σ 0 =
153
.
2 kpsi
For 20% cold work, Eq. (2-10) and Eq. (2-13) give,
A
=
A 0 (1
−
W )
=
0
.
1987(1
−
0
.
2)
=
0
.
1590 in 2
ε =
ln
A =
ln 0
.
1590 =
0
.
2231
0
.
Eq. (2-14):
S y = σ 0 ε
m
=
153
.
2(0
.
2231) 0 . 2306
=
108
.
4 kpsi Ans.
Eq. (2-15), with S u =
85
.
5 kpsi from Prob. 2-9,
S u =
− W =
85
.
5
2 =
106
.
9 kpsi Ans.
1
1
−
0
.
P
L
A
ε
σ true
log
ε
log
σ true
0
0
0.198 713
0
0
1 000
0.0004
0.198 713
0.000 2
5032.388
−
3.699 01
3.701 774
2 000
0.0006
0.198 713
0.000 3
10 064.78
−
3.522 94
4.002 804
3 000
0.0010
0.198 713
0.000 5
15 097.17
−
3.301 14
4.178 895
4 000
0.0013
0.198 713
0.000 65
20 129.55
−
3.187 23
4.303 834
7 000
0.0023
0.198 713
0.001 149
35 226.72
−
2.939 55
4.546 872
8 400
0.0028
0.198 713
0.001 399
42 272.06
−
2.854 18
4.626 053
8 800
0.0036
0.198 4
0.001 575
44 354.84
−
2.802 61
4.646 941
9 200
0.0089
0.197 8
0.004 604
46 511.63
−
2.336 85
4.667 562
9 100
0.196 3
0.012 216
46 357.62
−
1.913 05
4.666 121
13 200
0.192 4
0.032 284
68 607.07
− 1.491 01
4.836 369
15 200
0.187 5
0.058 082
81 066.67
− 1.235 96
4.908 842
17 000
0.156 3
0.240 083
108 765.2
−
0.619 64
5.036 49
16 400
0.130 7
0.418 956
125 478.2
−
0.377 83
5.098 568
14 800
0.107 7
0.612 511
137 418.8
−
0.212 89
5.138 046
A 0
1987
S u
672657644.005.png
Chapter 2
9
160000
140000
120000
100000
80000
60000
40000
20000
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
true
5.2
5.1
y 0.2306 x 5.1852
5
4.9
1.6
1.4
1.2
1
0.8
0.6
0.4
0.2
0
4.8
log
2-11 Tangent modulus at
σ =
0 is
E 0 = σ
=
5000
−
0
0 =
25(10 6 ) psi
ε
0
.
2(10 − 3 )
−
At
σ =
20 kpsi
E 20
=
(26
−
19)(10 3 )
=
14
.
0(10 6 ) psi Ans.
(1
.
5
−
1)(10 − 3 )
ε
(10 − 3 )
σ
(kpsi)
60
0
0
50
0.20
5
0.44
10
40
( S y ) 0.001 ˙ 35 kpsi Ans.
0.80
16
30
1.0
19
1.5
26
20
2.0
32
2.8
40
10
3.4
46
4.0
49
0
0
1
2
3
4
5
5.0
54
(10 3 )
672657644.006.png 672657644.001.png
10
Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
2-12 Since
| ε o |=| ε i |
=
=
−
ln
h
R + N
R
+
ln
R
R + N
ln
R
+
N
R
h
R + N =
+
R
+
N
R
( R
+
N ) 2
=
R ( R
+
h )
From which,
N 2
R
+
2 RN
−
Rh
=
1 / 2
0
1
h
R
The roots are:
N
=
−
1
±
+
The
+
sign being significant,
R 1
1
1 / 2
h
R
N
=
+
−
Ans.
Substitute for N in
ε o =
ln
h
R + N
+
R + h
=
ln 1
h
R
1 / 2
Gives
ε 0 =
ln
R 1
1 / 2
+
Ans.
h
R
R
+
+
−
R
These constitute a useful pair of equations in cold-forming situations, allowing the surface
strains to be found so that cold-working strength enhancement can be estimated.
2-13 From Table A-22
AISI 1212
S y =
28
.
0 kpsi ,
σ f
=
106 kpsi, S ut =
61
.
5 kpsi
σ 0 =
110 kpsi,
m
=
0.24,
ε f
=
0
.
85
From Eq. (2-12)
ε u =
m
=
0
.
24
Eq. (2-10)
A i =
− W =
1
1
2 =
1
.
25
1
1
−
0
.
Eq. (2-13)
ε i =
ln 1
.
25
=
0
.
2231
⇒ ε i <ε u
Eq. (2-14)
S y = σ 0 ε
i
=
110(0
.
2231) 0 . 24
=
76
.
7 kpsi Ans.
Eq. (2-15)
S u =
− W =
S u
61
.
5
2 =
76
.
9 kpsi Ans.
1
1
−
0
.
2-14 For H B = 250,
Eq. (2-17)
S u =
0.495 (250)
=
124 kpsi
Ans.
=
3.41 (250)
=
853 MPa
R
R
A 0
672657644.002.png

Plik z chomika:

Inne pliki z tego folderu:

Inne foldery tego chomika:

Zgłoś jeśli naruszono regulamin