budynas_SM_ch01.pdf

(62 KB) Pobierz
budy_sm_ch01.qxd
Chapter 1
Problems 1-1 through 1-4 are for student research.
1-5
(a) Point vehicles
v
x
hour = x
42
.
1
v − v
2
Q
=
=
0
.
324
Seek stationary point maximum
dQ
d
=
0
=
42
.
1
−
2
v
∴ v
*
=
21
.
05 mph
v
0
.
324
Q *
=
42
.
1(21
.
05)
−
21
.
05 2
=
1368 cars/h Ans
.
0
.
324
(b)
v
l
2
x
l
2
v
0
.
324
l
v
− 1
Q
=
l =
2 +
x
+
v
(42
.
1)
− v
Maximize Q with l
=
10 / 5280 mi
v
Q
22.18
1221.431
22.19
1221.433
22.20
1221.435
←
22.21
1221.435
22.22
1221.434
% loss of throughput
=
1368
1221
1221 =
−
12% Ans.
(c) % increase in speed 22
.
2
−
05 =
21
.
05
5
.
5%
21
.
Modest change in optimal speed Ans.
cars
672657746.005.png 672657746.006.png 672657746.007.png
 
2
Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
1-6 This and the following problem may be the student’s first experience with a figure of merit.
•Formulate fom to reflect larger figure of merit for larger merit.
• Use a maximization optimization algorithm. When one gets into computer implementa-
tion and answers are not known, minimizing instead of maximizing is the largest error
one can make.
F V
=
F 1 sin
θ −
W
=
0
F H =−
F 1 cos
θ −
F 2 =
0
From which
F 1 =
W
/
sin
θ
F 2 =−
W cos
θ/
sin
θ
fom
= −
$
=−
¢
γ
(volume)
¢
γ
( l 1 A 1 +
l 2 A 2 )
A 1 =
S =
W
S sin
, l 2 =
l 1
cos
θ
θ
=
A 2 =
F 2
S
W cos
θ
S sin
θ
l 2
cos
fom
= −
¢
γ
W
S sin
+
l 2 W cos
θ
θ
θ
S sin
θ
1
= −
¢
Wl 2
S
+
cos 2
θ
cos
θ
sin
θ
Set leading constant to unity
θ ◦
fom
θ
*
=
54
.
736 ◦ Ans.
0
−∞
20
−
5.86
fom*
=−
2
.
828
Alternative:
d
d
30
−
4.04
1
40
−
3.22
+
cos 2
θ
=
0
45
−
3.00
θ
cos
θ
sin
θ
50
−
2.87
And solve resulting tran-
scendental for
54.736
−
2.828
θ
*.
60
−
2.886
Check second derivative to see if a maximum, minimum, or point of inflection has been
found. Or, evaluate fom on either side of
θ
*.
=−
F 1
γ
672657746.001.png
Chapter 1
3
1-7
(a) x 1 +
x 2 =
X 1 +
e 1 +
X 2 +
e 2
error
=
e
=
( x 1 +
x 2 )
−
( X 1 +
X 2 )
=
e 1 +
e 2 Ans.
(b) x 1 −
x 2 =
X 1 +
e 1 −
( X 2 +
e 2 )
e
=
( x 1 −
x 2 )
−
( X 1 −
X 2 )
=
e 1 −
e 2 Ans.
(c) x 1 x 2 =
( X 1 +
e 1 )( X 2 +
e 2 )
e
=
x 1 x 2 −
X 1 X 2 =
X 1 X 2 e 1
X 1 e 2 +
X 2 e 1 +
e 1 e 2
=
X 1 e 2 +
X 2 e 1 =
X 1 +
e 2
X 2
Ans.
1
(d)
x 2 =
X 2 + e 2 =
e 1
X 1
X 2
+
e 1 /
X 1
1
+ e 2 / X 2
1
− 1
1
1
e 2
X 2
=
e 2
X 2
e 1
X 1
e 2
X 2
=
e 1
e 2
X 2
+
1
−
and
+
−
1
+
X 1 −
e 1
e
=
x 2 −
X 1
X 2
=
X 1
X 2
X 1 −
e 2
X 2
Ans.
1-8
(a)
x 1 =
√ 5
=
2
.
236 067 977 5
X 1 =
√ 6
.
23 3-correct digits
x 2 =
=
2
.
449 487 742 78
X 2 =
√ 5
.
44 3-correct digits
√ 6
x 1 +
x 2 =
+
√ 5
=
4
.
685 557 720 28
e 1 =
x 1 −
X 1 =
√ 6
−
2
.
23
=
0.006 067 977 5
e 2 =
x 2 −
X 2 =
√ 5
−
2
.
44
=
√ 6
0.0 09 489 742 78
e
=
e 1 +
e 2 =
−
2
.
23
+
−
2
.
44
=
0.015 557 720 28
Sum
=
x 1 +
x 2 =
X 1 +
X 2 +
e
=
2
.
23
+
2
.
44
+
0
.
015 557 720 28
=
4.685 557 720 28 (Checks) Ans.
(b) X 1 =
√ 5
.
24, X 2 =
2
.
45
e 1 =
√ 6
−
2
.
24
=−
0
.
003 932 022 50
e 2 =
−
2
.
45
=−
0
.
000 510 257 22
e
=
e 1 +
e 2 =−
0
.
004 442 279 72
Sum
=
X 1 +
X 2 +
e
=
2
.
24
+
2
.
45
+
(
−
0
.
004 442 279 72)
=
4
.
685 557 720 28 Ans.
x 1
X 1 +
x 1
2
2
2
672657746.002.png
4
Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
1-9
(a)
σ =
20(6.89)
=
137.8 MPa
(b) F
=
350(4.45)
=
1558 N
=
1.558 kN
(c) M
=
1200 lbf
·
in (0.113)
=
135.6 N
·
m
(d) A
=
2
.
4(645)
=
1548 mm 2
(e) I
=
17
.
4in 4
(2
.
54) 4
=
724
.
2cm 4
(f ) A
=
3
.
6(1
.
610) 2
=
9
.
332 km 2
(g) E
=
21(1000)(6
.
89)
=
144
.
69(10 3 )MPa
=
144
.
7GPa
(h)
v =
45 mi/h (1.61)
=
72.45 km/h
(i) V
=
60 in 3
(2
.
54) 3
=
983
.
2cm 3
=
0
.
983 liter
1-10
(a) l
=
1.5 / 0.305
=
4.918 ft
=
59.02 in
(b)
σ =
600 / 6.89
=
86.96 kpsi
(c) p
=
160 / 6.89
=
23.22 psi
(d) Z
=
1
.
84(10 5 )
/
(25
.
4) 3
=
11
.
23 in 3
(e)
w =
38.1
/
175
=
0.218 lbf/in
(f)
δ =
0.05
/
25.4
=
0.00197 in
(g)
v =
6.12
/
0.0051
=
1200 ft /min
(h)
=
0.0021 in/in
(i) V
=
30
/
(0
.
254) 3
=
1831 in 3
1-11
200
15
(a)
σ =
.
3 =
13
.
1MPa
(b)
σ =
42(10 3 )
6(10 − 2 ) 2
=
70(10 6 ) N/m 2
=
70 MPa
(c) y
=
1200(800) 3 (10 − 3 ) 3
3(207)10 9 (64)10 3 (10 − 3 ) 4
=
1
.
546(10 − 2 )m
=
15
.
5mm
(d)
θ =
1100(250)(10 − 3 )
32)(25) 4 (10 − 3 ) 4 =
9
.
043(10 − 2 ) rad
=
5
.
18 ◦
79
.
3(10 9 )(
π/
1-12
600
20(6) =
(a)
σ =
5MPa
1
(b) I
=
12 8(24) 3
=
9216 mm 4
64 32 4 (10 − 1 ) 4
(c) I
=
=
5
.
147 cm 4
16(16)
(d)
τ =
=
5
.
215(10 6 ) N/m 2
=
5.215 MPa
π
(25 3 )(10 − 3 ) 3
672657746.003.png
Chapter 1
5
1-13
(a)
τ =
120(10 3 )
=
382 MPa
(
π/
4)(20 2 )
(b)
σ =
32(800)(800)(10 − 3 )
π
(32) 3 (10 − 3 ) 3
=
198
.
9(10 6 ) N/m 2
=
198
.
9MPa
(c) Z
=
32(36) (36 4
π
−
26 4 )
=
3334 mm 3
(d) k
=
(1
.
6) 4 (10 − 3 ) 4 (79
.
3)(10 9 )
=
286
.
8 N/m
8(19
.
2) 3 (10 − 3 ) 3 (32)
672657746.004.png

Plik z chomika:

Inne pliki z tego folderu:

Inne foldery tego chomika:

Zgłoś jeśli naruszono regulamin