Chapter 6
61 MSS: DE: 1  3 = S y /n n= Sy n= Sy 1  3
1/2
2 2 = A  AB + B
1/2
2 2 2 = x  x y + y + 3x y
(a) MSS:
1 = 12, 2 = 6, 3 = 0 kpsi 50 n= = 4.17 Ans. 12 = (122  6(12) + 62 ) 1
79
MAE 4900, Fall 2012 Semester
Exam II, November 6, 2012
Name:.(J)q!~ .
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ID# jQOLf
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Only the textbook is allowed, cell phone use is not allowed.
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MAE 4900
Mid Term Exam, 2/29/12
lllll . . ID#
You can use the text book only. And leave one empty seat between your neighbors.
Calculators allowed, no cell phone use allowed.
Q1. A l in. diameter Lsh
MAE 4900
Mid Term Exam, 11/14/11, 1:001:50 pm
Name: Zudwldi W2C . ID#i9QC707)O .
You can use the text book onl . Leave one em seat between our nei hbors.
NO CELL PHONES allowed. All uestions are 10 t
Manufacturing Methods
MAE 4500
Assignment #4
1.
Due April 7, 2016
A bolt of the indicated geometry (Fig. 1) is to be produced for high
temperature service. A material similar to H13 is proposed, and i
Manufacturing Methods
Spring 2011
MAE 4500
Assignment #1
Due February 3rd, 2011
1. A pin and a bushing assembly (figure 1) are to be used as a part of a locating fixture for a
milling machine. The pin
MAE 4500
Manufacturing Methods
Assignment #5
Due April 28, 2016
Problem #1
A sheetmetal part used to be made by bending 1015 steel of 5mm thickness on a
sharp edge (zero radius). The part needs to be
MAE 4500
Manufacturing Methods
Assignment #3
1. The part shown below is to be made by casting. The following materials are considered:
Al 3% Si and Al 12.6% Si
a. Copy the Al Si equilibrium diagram an
1. A bolt of the indicated geometry is to be produced for high temperature
service. A material similar to H13 is proposed, and it is planned to forwardextrude the shaft and, in a separate operation, b
Chapter 6: Numerical Simulation of Dynamic Systems
6.4 The mathematical model of the mechanical system is mz bz kz Fa (t ) . Because the
system has zero initial conditions we may use a transfer functi
Manufacturing Methods
MAE 4500
Inclass Assignment 5 Solution
Sheet metal working process
1)
A sheetmetal part used to be made by bending 1015 steel of 5mm thickness on a
sharp edge (zero radius). Th
tan ufacturing,illethods
MAE 45OO
Fall2012
Assignment #i! Solutions
t.
The part shown below Gig.l) is to be made by casting. The following'materials are
considered:
Al
a.

Copy the Al
b. Make
3% Si a
Chapter 5: Standard Models for Dynamic Systems
5.1
The system consists of two firstorder ODEs and one secondorder ODE. Therefore the
.
system has order n = 4 and we require four state variables: let
Chapter 4
412
=
I
=
(1.54 ) 0.2485 in 4
64
From Table A9 by superposition of beams 6 and 7, at x = a = 15 in, with b = 24 in and
l = 39 in
Fba 2
wa
=
y
[a + b 2 l 2 ] +
(2la 2 a 3 l 3 )
6 EIl
24 EI
Chapter 6
61
Eq. (221): =
Sut 3.4
=
H B 3.4(300)
= 1020 MPa
Eq. (68): =
Se 0.5
=
Sut 0.5(1020)
= 510 MPa
Table 62:
a = 1.58, b = 0.085
b
Eq. (619): =
ka aS
=
1.58(1020) 0.085
= 0.877
ut
0.107
Eq.
/ C7.
MAE 4900 ,
Mid Term Exam, 11/16/09 kme Q37 7 ghw
._ (JQWQ
Name: EMS.) . . ID# . 06370573 .
Please leave one empty seat next to you to allow space for use of textbook. Only Text Book
is allowed d
MAE 4900
Final Exam
SP 2011
Q1. Determine the support reaction R for the cantilever shown below using the strain energy
approach and Castiglianos theorem.
1
MAE 4900
Final Exam
SP 2011
Q2. A power tra
Chapter 8
81 (a)
2.5 mm 25 mm 5 mm 2.5
Thread depth = 2.5 mm
Ans.
Width = 2.5 mm Ans. dm = 25  1.25  1.25 = 22.5 mm dr = 25  5 = 20 mm l = p = 5 mm Ans. Thread depth = 2.5 mm Ans. Width at pitc
Chapter 16
161 (a) 1 = 0, 2 = 120, a = 90, sin a = 1,
120 0
a = 5 in
Eq. (162): M f =
0.28 pa (1.5)(6) 1
sin (6  5 cos ) d
= 17.96 pa lbf in pa (1.5)(6)(5) Eq. (163): M N = 1
120 0
sin2 d
Chapter 13
131 d P = 17/8 = 2.125 in dG = N2 1120 (2.125) = 4.375 in dP = N3 544
NG = PdG = 8(4.375) = 35 teeth Ans. C = (2.125 + 4.375)/2 = 3.25 in Ans. 132 n G = 1600(15/60) = 400 rev/min Ans. p
Chapter 11
111 For the deepgroove 02series ball bearing with R = 0.90, the design life x D , in multiples of rating life, is xD = The design radial load FD is FD = 1.2(1.898) = 2.278 kN From Eq. (1
Chapter 15
151 Given: Uncrowned, throughhardened 300 Brinell core and case, Grade 1, NC = 109 rev of pinion at R = 0.999, N P = 20 teeth, NG = 60 teeth, Q v = 6, Pd = 6 teeth/in, shaft angle 90, n p
Chapter 1
Problems 11 through 14 are for student research. 15
Impending motion to left E
1 f A
1 f B G Fcr F
cr
D
C
Facc
Consider force F at G, reactions at B and D. Extend lines of action fo