Korea Advanced Institute of Science and Technology
MAE 307

Spring 2015
1
Chapter 12
Lubrication and Journal Bearings
Mechanical Design Lab. with Advanced Materials
Carbon fiber phenolic composite journal bearings
2
High thermal resistance
High wear resistance
High strength
< The stern of a ship >
< Schematic of journal be
Korea Advanced Institute of Science and Technology
MAE 307

Spring 2015
1
Chapter 20
Statistical Considerations
Mechanical Design Lab. with Advanced Materials
2
201 Random Variables
Statistics in mechanical design provides a method of dealing
with characteristics whose values are variable (random,
stochastic).
When two dic
Korea Advanced Institute of Science and Technology
MAE 307

Spring 2015
1
Chapter 8 Screws, Fasteners,
and the Design of
Nonpermanent Joints
Mechanical Design Lab. with Advanced Materials
2
Home work Problems:
82, 89, 820, 835, 864
Due Nov. 19, 2013
Mechanical Design Lab. with Advanced Materials
3
Mechanical Joining (1)
Korea Advanced Institute of Science and Technology
MAE 307

Spring 2015
1
Chapter 11 Rolling contact Bearings
Mechanical Design Lab. with Advanced Materials
2
Homework
(Use the Weibull parameters for Manufacturer 2 on p. 608 for all the
problems.):
[1] Problem (1113) with af =1.
[2] Problem (1120)
[3] Calculate the combined
Korea Advanced Institute of Science and Technology
MAE 307

Spring 2015
1
Chapter 2 Materials
Mechanical Design Lab. with Advanced Materials
2
Chapter 2 Materials
Design:
Stiffness Design : Elastic or stiffness properties of the
material are required.
Deformation or accuracy is most important.
Strength Design : Strength or
Korea Advanced Institute of Science and Technology
MAE 307

Spring 2015
1
Chapter 5
Failures Resulting from Static Loading
Mechanical Design Lab. with Advanced Materials
Chapter 5 Failures Resulting from Static Loading
Strength is a property or characteristic of a mechanical
element.
It results from the material identity, t
Korea Advanced Institute of Science and Technology
MAE 307

Spring 2015
1
Chapter 1
Mechanical Design Lab. with Advanced Materials
2
11 Design
The designers (mechanical engineer, electrical engineer, mayor,
CEO, etc) must do the following.
1. Know or understand their customers needs.
2. Define the problem they must solve to
Korea Advanced Institute of Science and Technology
MAE 307

Spring 2015
Chapter 20 Homework solution
[1] Should an automotive engineer increase the cost per car by
10000 Won in order to avoid 100 failures in a production
run of a million cars, where the failure would not involve
safety, but would entail a 100,000 Won repair?
Korea Advanced Institute of Science and Technology
MAE 307

Spring 2015
MAE341 MECHANICAL COMPONENT DESIGN FINAL, 2013
Name_ Student Number_
Only correct numerical answers will be counted!
Since this final examination is open book, writing equations without numerical results will
not be counted.
[1] The hotrolled steel bar (
Korea Advanced Institute of Science and Technology
MAE 307

Spring 2015
27 To plot 0' [we v3.6, the following equations are applied to the data.
P
otme :
A
Eq. (24)
£=lnli forOSAlS0.07mm
0
.3 =1n§ for Al > 0.07 mm
12. 2
where A0 =li 2128.7 mm2
4
The results are summarized in the table below and plotted on the next page. T
Korea Advanced Institute of Science and Technology
MAE 307

Spring 2015
Chapter 10 Homework solution
[102] A = Sutdm
Dim(ASI)/Dim(Auscu)= [Dim (S) Dim(dm)]SI /[Dim (S) Dim(dm)]uscu
= MPammm/(kpsiinm)
= MPammm /(MPa/6.891/25.4m)=6.89(25.4)m
For music wire, from Table 104:
Auscu = 201 kpsi inm,
m = 0.145; what is ASI?
ASI = 6
Korea Advanced Institute of Science and Technology
MAE 307

Spring 2015
[614]
Given: w = 60 mm, t = 10 mm, d = 12 mm, nd = 1.8. From Table A20, for AISI 1020 CD,
Sut = 440/0.814=541 MPa, and Sy = 370 MPa (R=0.99).
Eq. (68): Se=0.5(541) =270 MPa
Table 62: ka=4.51(541)0.265=0.851
Eq. (621): kb = 1 (axial loading)
Eq. (62
Korea Advanced Institute of Science and Technology
MAE 307

Spring 2015
Chapter 5 Homework solution
[Equation 5.14]
The stress vectors (force per unit area in the figure below) x , y and z on planes
that are perpendicular, respectively, to the x, y, and z axes are
x = xx i + xy j + xz k
y = yx i + yy j + yz k
z = zx i +
Korea Advanced Institute of Science and Technology
MAE 307

Spring 2015
(82)
From Table 81,
d r = d 1.226869 p
d p = d 0.649519 p
(d 1.226869 p )(d 0.649519 p ) = d 0.938194 p
d=
2
At =
4
(d 0.938194 p )2
(89)
dm = 40 6/2 = 37 mm, l = 2(6) = 12 mm
From Eq. (81) and Eq. (86)
TR =
( )
( )
10 103 (0.037 ) 0.012 + (0.10 )(0.
Korea Advanced Institute of Science and Technology
MAE 307

Spring 2015
17 From Fig. 12, cost of grinding to 0.0125 mm is 260%.
Cost of turning to 0.075 mm is 60%.
Relative cost of grinding vs. turning = 260/60 = 4.3 times Ans.
114
a = 37.5 0.025 mm
b = 50 0.075 mm
c = 75 0.1 mm
d = 163 0.25 mm
w = d a b c = 163 37.5 50 75
Korea Advanced Institute of Science and Technology
MAE 307

Spring 2015
1.3409
1.3409
628.68
628.68
Notice: In this solution, wrong value of the spring diameters (2.7 mm and 3.1 mm)
were considered on the table below. Please reconsider spring diameter 2.6 mm and
3.0 mm for the exact solution.
n
Korea Advanced Institute of Science and Technology
MAE 307

Spring 2015
84
Given F 5 kN, [ 5 mm, and dm d p/'2 25 5/2 22.5 mm, the torque required to
raise the load is found using Eqs. (81) and (86)
H 5(22.5) 5+;r(0.09)22.5 5(0.06)45
1R : + 215.85 N m Am
2 E(22.5)0.09(5) 2
The torque required to lower the load, from Eqs
Korea Advanced Institute of Science and Technology
MAE 307

Spring 2015
or SAE 1045 HR with
There is no need to obtain the yield strength and ultimate strength after the
coldwork operation because the material was fractured.
(b) For the same reason of (a), there is no need to answer.
Korea Advanced Institute of Science and Technology
MAE 307

Spring 2015
Homework #6 Solution
(a)
a= 820.8MPa,
b=b, Ne=7287912000=60879 cycle
New endurance limit:
(b)
S f 511.7 MPa
Se ' a ' Neb ' 311.05MPa
511.7
)
378.6 0.0733
b'
1000
log(
)
60879
New endurance limit:
log(
Se ' a ' Neb ' 308.4MPa
a'
378.6
848.95MPa
(60879)