MCE311 Homework # 4 Due on Wed Nov 30th, 2011 Question 1 (5.16): A function has the form 3 sin 10 5 cos 40 and is sampled at a rate of 30 samples per second. What false alias frequency(ies) would you expect in the discrete data? Solution: 3 sin 2 5 5 cos
MCE311; Homework # 5; Due on Monday. Nov 21st, 2011. Question 1 A sliding psychrometer gives a dry-bulb reading of 30 C and wet-bulb reading of 15 C. Find the relative humidity and the humidity ratio?
Solution:
Question 2 If a sliding psychrometer gives a
MCE311 Homework # 3 Due on Wed. Oct 26th, 2011. Question 1: A strip of cast iron is 8in. long and has a cross section of 1.0 by 1/16 in. It is subject to an axial load of 1000lb. Two strain gauges are attached to the surface, one in the axial direction an
MCE311 Homework # 2 Solution Question 1 (3.1 textbook): An amplifier produces an output of 5 V when the input is 5V. What is the gain G and the decibel gain GdB? Solution: The gain is: 5 10 5. 10 The gain in dB is: 20 120 _ Question 2 (3.2 textbook): An a
MCE311 Homework # 1 Due on Monday, Feb 28th, 2011. Question 1 (6.1) A certain length measurement is made with the following results: Reading: 1 2 3 4 5 6 7 8 9 10 X(cm): 49.3 50.1 49.0 49.2 49.3 50.5 49.9 49.2 49.7 50.2 a) Arrange the data into bins with
39.4 A plastic extrusion process produces round extrudate with a mean diameter = 28.6 mm. The process is in statistical control and the output is normally distributed with standard deviation = 0.53 mm. Determine the process capability. Solution: Process c
27.9 A U-groove weld is used to butt weld 2 pieces of 7.0-mm-thick titanium plate. The U-groove is prepared using a milling cutter so the radius of the groove is 3.0 mm. During welding, the penetration of the weld causes an additional 1.5 mm of material t
24.4 For the following application, identify one or more nontraditional machining processes that might be used, and present arguments to support your selection. Assume that either the part geometry or the work material (or both) preclude the use of conven
20.16 A slab milling operation is performed on the top surface of a steel rectangular workpiece 30 cm long by 6.25 cm wide. The helical milling cutter, which has a 7.5 cm diameter and ten teeth, is set up to overhang the width of the part on both sides. C
19.14 What is the relationship between the coefficient of friction and the friction angle in the orthogonal cutting model? Answer. The relationship is that the coefficient of friction is the tangent of the friction angle ( = tan ). 19.16 How is the power
17.1 A 42.0-mm-thick plate made of low carbon steel is to be reduced to 34.0 mm in one pass in a rolling operation. As the thickness is reduced, the plate widens by 4%. The yield strength of the steel plate is 174 MPa and the tensile strength is 290 MPa.
14.10 Given a large volume of metallic powders, all of which are perfectly spherical and having the same exact diameter, what is the maximum possible packing factor that the powders can take? Solution: The maximum packing factor is achieved when the spher
9.8 During pouring into a sand mold, the molten metal can be poured into the downsprue at a constant flow rate during the time it takes to fill the mold. At the end of pouring the sprue is filled and there is negligible metal in the pouring cup. The downs
Chapter 17
17-1
Given: F-1 Polyamide, b = 6 in, d = 2 in with n = 1750 rev/min, H nom = 2 hp, C = 9(12) =
108 in, velocity ratio = 0.5, K s = 1.25, n d = 1
V = d n / 12 = (2)(1750) / 12 = 916.3 ft/min
Eq. (17-1):
D = d / vel ratio = 2 / 0.5 = 4 in
42
Dd
Chapter 15
15-1
Given: Uncrowned, through-hardened 300 Brinell core and case, Grade 1, N C =
109 rev of pinion at R = 0.999, N P = 20 teeth, N G = 60 teeth, Q v = 6, P d = 6
teeth/in, shaft angle = 90, n p = 900 rev/min, J P = 0.249 and J G = 0.216 (Fig.
Chapter 14
d
14-1
N
22
3.667 in
P
6
Table 14-2:
Y = 0.331
dn (3.667)(1200)
Eq. (13-34): V
1152 ft/min
12
12
1200 1152
Eq. (14-4b): K v
1.96
1200
H
15
Eq. (13-35) : W t 33 000
33 000
429.7 lbf
V
1152
K vW t P 1.96(429.7)(6)
7633 psi 7.63 kpsi Ans.
Chapter 13
d P 17 / 8 2.125 in
N
1120
dG 2 d P
2.125 4.375 in
N3
544
13-1
NG PdG 8 4.375 35 teeth
Ans.
C 2.125 4.375 / 2 3.25 in
Ans.
_
nG 1600 15 / 60 400 rev/min
p m 3 mm Ans.
13-2
Ans.
C 3 15 60 2 112.5 mm Ans.
_
NG 16 4 64 teeth
13-3
Ans.
dG NG m 64
Chapter 12
12-1
Given: d max = 25 mm, b min = 25.03 mm, l/d = 1/2, W = 1.2 kN, = 55 mPas, and N =
1100 rev/min.
b d max 25.03 25
0.015 mm
cmin min
2
2
r 25/2 = 12.5 mm
r/c = 12.5/0.015 = 833.3
N = 1100/60 = 18.33 rev/s
P = W/ (ld) = 1200/ [12.5(25)] = 3.
Chapter 11
11-1
For the deep-groove 02-series ball bearing with R = 0.90, the design life x D , in multiples
of rating life, is
L
60D nD 60 25000 350
xD D
525 Ans.
LR
L10
106
The design radial load is
FD 1.2 2.5 3.0 kN
1/3
Eq. (11-6):
525
C10 3.0
1/1.4
Chapter 10
10-1
From Eqs. (10-4) and (10-5)
KW K B
4C 1 0.615 4C 2
4C 4
C
4C 3
Plot 100(K W K B )/ K W vs. C for 4 C 12 obtaining
We see the maximum and minimum occur at C = 4 and 12 respectively where
Maximum = 1.36 % Ans.,
and Minimum = 0.743 % Ans.
_
Chapter 9
Figure for Probs.
9-1 to 9-4
9-1
Given, b = 50 mm, d = 50 mm, h = 5 mm, allow = 140 MPa.
F = 0.707 hl allow = 0.707(5)[2(50)](140)(103) = 49.5 kN Ans.
_
9-2
Given, b = 2 in, d = 2 in, h = 5/16 in, allow = 25 kpsi.
F = 0.707 hl allow = 0.707(5/16
Chapter 8
Note to the Instructor for Probs. 8-41 to 8-44. These problems, as well as many others in this
chapter are best implemented using a spreadsheet.
8-1
(a) Thread depth= 2.5 mm Ans.
Width = 2.5 mm Ans.
d m = 25 - 1.25 - 1.25 = 22.5 mm
d r = 25 - 5
Chapter 6
Eq. (2-21):
Eq. (6-8):
Table 6-2:
Eq. (6-19):
Sut 3.4 H B 3.4(300) 1020 MPa
Se 0.5Sut 0.5(1020) 510 MPa
a 1.58, b 0.085
b
ka aSut 1.58(1020) 0.085 0.877
Eq. (6-20):
6-1
kb 1.24d 0.107 1.24(10) 0.107 0.969
Se ka kb Se (0.877)(0.969)(510) 433 MPa
Chapter 4
For a torsion bar, k T = T/ = Fl/, and so = Fl/k T . For a cantilever, k l = F/ , = F/k l . For
the assembly, k = F/y, or, y = F/k = l +
Thus
F Fl 2 F
y
k
kT kl
Solving for k
kk
1
k 2
2l T
Ans.
l
1 kl l kT
kT kl
_
4-1
For a torsion bar, k T =
Chapter 2
2-1
From Tables A-20, A-21, A-22, and A-24c,
(a) UNS G10200 HR: S ut = 380 (55) MPa (kpsi), S yt = 210 (30) Mpa (kpsi) Ans.
(b) SAE 1050 CD: S ut = 690 (100) MPa (kpsi), S yt = 580 (84) Mpa (kpsi) Ans.
(c) AISI 1141 Q&T at 540C (1000F): S ut = 8
Chapter 1
Problems 1-1 through 1-6 are for student research. No standard solutions are provided.
From Fig. 1-2, cost of grinding to 0.0005 in is 270%. Cost of turning to 0.003 in is
60%.
Relative cost of grinding vs. turning = 270/60 = 4.5 times
Ans.
_
1-
MCE240 FLUID MECHANICS
HOMEWORK # 5 - Solutions
Assignment Date:
Due Date:
Monday, 05 April 2010
Sunday, 18 April 2010 (by 11:00 am)
Problem 5.60:
1
Problem 5.69:
2
Problem 5.80:
3
Problem 5.84:
4
Problem 5.100:
5
Problem 6.43:
6
Problem 6.47:
7
Problem 6
ENGR2860U FLUID MECHANICS
HOMEWORK # 4 - SOLUTIONS
Assignment Date:
Due Date:
Sunday, 14 March 2010
Sunday, 28 March 2010 (by 11:00 am)
Problem 5.7:
Problem 5.14:
1
Problem 5.19:
2
Problem 5.24:
3
Problem 5.51:
4
Problem 5.55:
5