Manufacturing Methods MAE 4500 Homework # 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 and identify all phase fields. b. Make two sketche
Manufacturing Methods
MAE 4500 Fall, 2006 1. Homework #4
A 1015 steel disc of 8.0 in. diameter and 1.0 in. in height is forged cold in a press of v=300 ipm. No lubricant is used. Calculate: (a) the starting workpiece length, if bar diameter is 5.0 i
Manufacturing Methods
MAE 4500 Fall, 2006 1. Homework #4 A 1015 steel disc of 8.0 in. diameter and 1.0 in. in height is forged cold in a press of v=300 ipm. No lubricant is used. Calculate: (a) the starting workpiece length, if bar diameter is 5.0 in
MAE 4500 Manufacturing Methods
Engine Lathe Project
Group Members Mark Dalton Kyle Collier Nathan Jones Josh Smith-Moore Michael Koch
FS06
Table of Contents 1. 2. 3. 4. 5. 6. 7. 8. Abstract Introduction Objectives Experimental Set-Up and Procedur
University of Missouri Columbia
Department of Mechanical and Aerospace Engineering
Manufacturing Methods
MAE 4500
Fall 2009
(1)
Assignment # 3
A part given in figure 1 is to be made by casting. The following
materials are considered: Al-3%Si and Al-12%Si.
MAE 4500
Manufacturing Methods
Assignment #1
REMEMBER:
Due September 15, 2009
Make engineering quality sketches
Identify all coordinates and curves
Use brief, clearly worded explanatory sentences.
Review Question 1.
a. Define allowance, tolerance, and fit
Fall 2006 1.
University of Missouri-Columbia Department of Mechanical and Aerospace Engineering Manufacturing Methods MAE 4500
HW #5-Solutions
Sheet Forming A bracket made of leaded brass (Ebrass=105103MPa) is manufactured by fine blanking, pierci
Fall 2006 1.
University of Missouri-Columbia Department of Mechanical and Aerospace Engineering Manufacturing Methods MAE 4500
HW #5
Sheet Forming A bracket made of leaded brass (Ebrass=105 103MPa) is manufactured by fine blanking, piercing, and b
Manufacturing Methods
MAE 4500
Fall Semester 2009
Solutions to Homework #3
1. A part given in figure 1 is to be made by casting. The following materials are considered: Al3%Si and Al-12.6%Si.
a) Copy the Al-Si equilibrium diagram and identify all phase fi
MAE 4500 Manufacturing Methods PLC Project Fall 2006
Doug Koch 900196 Preston Clark 905452 Travis Cardoza 911831
PLC and History Programmable Logic Controller (PLC) is a user friendly, microprocessor based, specialized computer that carries out control fu
MAE 4500 Manufacturing Methods
Assignment #1
REMEMBER: Make engineering quality sketches Identify all coordinates and curves Use brief, clearly worded explanatory sentences. Review Question 1. a. Define allowance, tolerance, and fit. b. Give examples wher
MAE 4500
Manufacturing Methods
Assignment #1
Solutions
1. a. Define allowance, tolerance, and fit.
b. Give examples where the following fits are used: RC5, LC3, LN2, FN3.
c. Make a sketch to show a typical surface roughness one would obtain from a stylus
University of Missouri Columbia
Department of Mechanical and Aerospace Engineering Manufacturing Methods MAE 4500 Fall 2009 Assignment # 3
(1)
A part given in figure 1 is to be made by casting. The following materials are considered: Al-3%Si and Al-12%Si.
Manufacturing Methods
Fall, 2009 MAE 4500
1.
Homework #4
A 1015 steel disc of 8.0 in. diameter and 1.0 in. in height is forged cold in a press of v=300 ipm. No lubricant is used. Calculate: (a) the starting workpiece length, if bar diameter is 5.0 in.,
c, we have |F(x) F(c)| = Z x a f(t) dt Z c a f(t) dt = Z x c f(t) dt
Z x c |f(t)| dt Z x c M dt. Since the constant function x M has U(P) = M |x c| for the
trivial partition P of the interval with endpoints x and c, |F(x) F(c)| M |x c|. Now given >
0, c
is denoted by (xk). Definition 49.4. A vector x R n is said to be the limit of the sequence (xk)
if for any > 0 there exists N = N() > 0 such that for any k > N we have kxk xk < . In this
case we write lim k xk = x. The limit of the sequence (xk), if it e
convergent on I. Proof. The proof is rather technical and it will omitted. 29 Series of functions.
Convergence and uniform convergence. Let be A R 1 and (fn) a sequence of functions fn : A
R 1 . Definition 29.1. It is said that the symbol X n=1 fn is a c
case when = m () July 2008 10 / 44 Riemann versus Caputo The Laplace transform D f (t) s
ef (s) m1 k=0 D k J (m) f (0 +)s m1k , m 1 < m Requires the knowledge of the (bounded)
initial values of the fractional integral J m and of its integer derivatives o
two representations of this kind corresponding to the two orientations of C. 71 Line integral of
first type Let now an elementary curve C and one of its parametric representations x(s) = (x1(s),
x2(s), x3(s) 0 s l in function of the arc length s. In order
( + 1 ) t , > 0 , > 1 ,t > 0 The fractional derivative D f is not zero for the constant
function f (t) 1 if 62 N D 1 = t (1 ) , 0 ,t > 0 Is 0 for 2 N, due to the poles of the
gamma function () July 2008 7 / 44 Caputo fractional derivative D f (t) := J m D
coordinate axis Ox1, Ox2, Ox3, respectivelly. In terms of an arbitrary parametric representation
: [a, b] C which corresponds to the same orientation, these line integrals of second type are
equal to the following Reimann integrals: Z C f dx1 = Z b a f(t
> 0 () July 2008 4 / 44 Fractional integral according to Riemann-Liouville Alternatively (leftsided integral) J b f (t) := 1 () Z b t ( t) 1 f () d , 2 R (a = 0, b = +) Riemann (a = , b
= +) Liouville Let J := J 0+ Semigroup properties J J = J + , , 0 Com
dag1, , dagp are linearly independent and f has a conditional extremum at a, then there exist p
constants 1, , p such that daf = X p i=1 idagi or f xk (a) = X p i=1 i gi xk (a) k = 1,
n Proof. The proof is rather technical and it will be skipped. Example
evaluation of the integral to the evaluation of integrals for functions of one variable. Z b1 a1
Z bn an f(x1, , xn) dx1 dx2 dxn = Z b1 a1 dx1 Z b2 a2 dx2 Z bn an f(x1, , xn) dxn
Remark 69.2. Theorem 69.1 is valid also for more complex sets A as in 2-dim
Power Series Definition 31.1. A series of functions of the form X n=0 an x n is called power
series. Clearly, any power series converges when x = 0. 58 Theorem 31.1 (The set of the
convergence of power series. Abel-Cauchy-Hadamard theorem). - The power se
real number L is called the limit of f(x) as x tends to a if for any > 0, there exists = () > 0
such that 0 < kx ak < |f(x) L| < . We write limxa f(x) = L. 107 Just like in the case of
functions of one variable, this definition is technically difficult to
x ux + f y uy = y ux + x uy. Definition 53.5. Let f = (f1, . . . , fm) be a vector valued n
variables function f : A R n R m; a an interior point of A and u a unit vector in R n (i.e. kuk
= 1). If the limit lim t0 f(a + t u) f(a) t exists, it is called th
00, y00) Ai p (x 0 x 00) 2 + (y 0 y 00) 2 < for i = 1, n. Hence Uf (P) Lf (P) = Xn i=1
(Mi mi)area(Ai) < area(A) Xn i=1 area(Ai) Hence f is Riemann-Darboux integrable on A.
Definition 61.1. A function f is called piecewise-continuous on A if there exists
idea we may extend the concept of orientation to surfaces which can be decomposed in
elementary surfaces, as follows: A surface S which can be decomposed in elementary surfaces is
said to be orientable if we can orient each elementary piece of 160 S in su
MAE 4500: Manufacturing
Methods
Powder Metallurgy
Diagram of Powder Metallurgy Process
Advantages
Suitable for making parts of high melting point
materials
Production of parts with controlled porosity
(porous bearings)
Capability of making bimetallic pr
Manufacturing Methods
MAE 4500
Sheet Metal Forming
Examples of sheet metal forming processes:
shearing
bending
drawing and stretching
spinning
Sheet Metal Properties
All sheet metal forming processes
involve some type of stretching and
bending. These t
is a bijection. Its inverse function is given by f 1 (x) = x. Now f is differentiable for x > 0 and
f 0 (x) = 2x 6= 0. Hence, by the inverse rule, f 1 (x) = x is differentiable and (f 1 ) 0 (x) = 1
2 x . 67 Example 34.5. The function g : 2 , 2 (1, 1) give
(u k1 )d k a f(u 1 )(u 2 ) (u k )k = 0 If the function is k-times differentiable at a, then the
following relations hold: k fi xj1 xj2 xjk (a) = k fi x(j1)x(j2) x(jk) (a)
Theorem 55.3 (Criterion for k-times differentiability). If the partial derivatives o
47.2 we conclude that limn Sn(x0) = 1 2 f(x + 0 ) + f(x 0 ) . If f is continuous at x0,
then limn Sn(x0) = f(x0). We have thus proved one form of the Fourier theorem on
convergence of Fourier series. Example 47.2. a) Deduce the Fourier series expansion of