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UNIVERSITEIT VAN PRETORIA
UNIVERSITY OF PRETORIA
Departement Meganiese en Lugvaartkundige Ingenieurswese
Department of Mechanical and Aeronautical Engineering
MASJIENONTWERP MOW217
MACHINE DESIGN MOW217
April 2011
Time: 3 hours/ure
Mark
WELDING 1
MOW 227 2013
Shigley chapter 9
Department of Mechanical and Aeronautical Engineering
Welding
Department of Mechanical and Aeronautical Engineering
Types of Welded Joints
Department of Mechanical and Aeronautical Engineering
Welding Symbols
Butt
Mechanical Springs
Shigley Ch 10
Lecture 21
Mechanical Springs
Why do we need springs?
Flexibility of structure
Storing and releasing of energy
Standard spring types:
Wire springs
- Helical springs of round or square wire, made to deflect under
Compres
Springs Lecture 3
Extension Springs
Extension springs
Carry tensile loading
Max. Normal stress at A due to
Bending and Axial loading
A
KA
16 D
F KA
d3
4C12 C1 1
4C1 (C1 1)
4
d2
C1
2r1
d
Max. torsional stress at B
8 FD
KB
B
d3
KB
4C2 1
4C2 4
C2
Improved d
Springs Lecture 2
Spring Design
Spring Design for Static Service
Ch 10-7
Preferred range of spring index is
4 C 12
Recommended range of active turns is
3 N a 15
Working range of the spring is 75% of the
curve between no-load and closure because
spring
Influence of Mean Stress
Discussion has been limited to completely
reversible stress thus far.
Meaning m = 0
However, there are many instances of
dynamic loading when mean stress is nonzero.
Mean Stresses
Increasing mean stress in the tensile
directio
Class example
A 40mm diameter machined bar has to
withstand a load that fluctuates from 0 to
70kN tensile force. At a critical section there
is a radius=1mm with a corresponding
stress concentration factor of 1.9.
Calculate the safety factor against fatig
Palmgren-miner method limitations
Damage method is not history dependant,
so one must take caution when using it.
Eg:
vs
Failure occurs sooner with one on right
because damage is accumulated quicker
Palmgren-miner method limitations
Sudden high stresse
Fatigue of Materials
MOW227
Semester 22013
Introduction
Repeated or cyclic loads can result in cyclic
stresses that lead to failure.
Failure can occur at stresses well below the
yield or ultimate strength of the material.
Damage accumulates with contin
WELDING 2
MOW 227 2013
Department of Mechanical and Aeronautical Engineering
Bending
Primary Shear Stress
d
V
M
y
V
A
Nominal Throat shear Stress
x
Mc
I
d
M
2
0.707h I u
1.414M
bhd
Compare 1.414 with 1.197 from
distortion theory and 1.207 from MSS
cons
SCREWS, FASTENERS AND NON
PERMANENT JOINTS
3
MOW 227 2013
Department of Mechanical and Aeronautical Engineering
Department of Mechanical and Aeronautical Engineering
Bolted and Riveted joints in shear
Department of Mechanical and Aeronautical Engineering
Stress Concentration
Mow 227
Shigley Ch 3-13
Stress Concentration
Stresses and strains are uniformly distributed
for parts of constant cross-section with
uniformly distributed loads.
However,
applied
forces
are
often
concentrated (not uniform) where cro
Kopiereg voorbehou
Copyright reserved
UNIVERSITEIT VAN PRETORIA
UNIVERSITY OF PRETORIA
Departement Meganiese en Lugvaartkundige Ingenieurswese
Department of Mechanical and Aeronautical Engineering
Masjienontwerp MOW217
Machine Design MOW217
Semester toets
Bearings
SM Roux
February 5, 2016
Bearings
Bearings permit smooth, low friction movement between two surfaces moving relative to one another.
This movement may be translational or rotational.
Bearings may employ a sliding (frictional) or rolling action
MOW 217: Belt and Chain Problems
Specify a belt and chain design for the following problems:
Machine Type
Rotational Speed [rpm]
Shaft Diameter [mm]
Centre Distance (Belts Only)
Power [kW]
Operating time [hrs/day]
Driver
Shunt-wound DC Motor
350
50
~800
1
Question 1
The gear forces in Figure 1 shown act in planes
parallel to the yz plane. The force on gear A is
1.2kN. Consider the bearings at O and B to be simple
supports.
a. Calculate the force FC
b. Calculate the reaction forces at O and B
[FC = 3 kN, FO
MOW 217: Ball Bearing Problems
Using the NTN bearing catalogue, calculate the answer which is in braces:
(All loading is purely dynamic unless stated otherwise.)
D
[mm]
70
Fr
[kN]
3.2
Fa
[kN]
0
N
[rpm]
100
Adjusted life
Code
Remarks
25 000 hrs
[6914]
25
2
SCREWS, FASTENERS AND NON
PERMANENT JOINTS
2
MOW 227 2013
Department of Mechanical and Aeronautical Engineering
Department of Mechanical and Aeronautical Engineering
Department of Mechanical and Aeronautical Engineering
Yielding take place in the first th
SCREWS, FASTENERS AND NON
PERMANENT JOINTS 1
MOW 227 2013
Shigley Ch 8
Department of Mechanical and Aeronautical Engineering
STANDARDS AND DEFINITIONS-THREAD
/Pitch diameter
Department of Mechanical and Aeronautical Engineering
Lead = distance that a nut
SCREWS, FASTENERS AND NON
PERMANENT JOINTS
2
MOW 227 2013
Department of Mechanical and Aeronautical Engineering
Department of Mechanical and Aeronautical Engineering
Department of Mechanical and Aeronautical Engineering
Yielding take place in the first th
Lecture 7
MOW 227
1
Strain Transformation
Transformation equations between stress
transformation and strain transformation are
analogous.
x, y, z are analogous to x, y, z
xy, xz, yz are analogous to xy, xz, yz
x, y, z are called the tensorial shear st
Shaft Design
MOW227
Semester 2 2012
Shaft System
A shaft is a rotating (or stationary) member
usually having a circular cross-section with
a diameter much smaller than its length.
A shaft generally has power-transmitting
elements such as gears, pulleys,
SCREWS, FASTENERS AND NON
PERMANENT JOINTS 1
MOW 227 2013
Shigley Ch 8
Department of Mechanical and Aeronautical Engineering
STANDARDS AND DEFINITIONS-THREAD
/Pitch diameter
Department of Mechanical and Aeronautical Engineering
Lead = distance that a nut
SCREWS, FASTENERS AND NON
PERMANENT JOINTS 1
MOW 227 2013
Shigley Ch 8
Department of Mechanical and Aeronautical Engineering
STANDARDS AND DEFINITIONS-THREAD
/Pitch diameter
Department of Mechanical and Aeronautical Engineering
Lead = distance that a nut
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 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 7
7-1
(a) DE-Gerber, Eq. (7-10):
A 4 K f M a 3 K fsTa 4 (2.2)(70) 3 (1.8)(45) 338.4 N m
2
2
2
2
B 4 K f M m 3 K fsTm 4 (2.2)(55) 3 (1.8)(35) 265.5 N m
2
2
2
6
8(2)(338.4) 2(265.5) 210 10
d
1 1
6
6
210 10 338.4 700 10
3
d = 25.85 (10 ) m = 25.8
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 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