text stringlengths 30 4k | source stringlengths 60 201 |
|---|---|
costly to store and
costly to invert. Good practical alternatives include quasi-Newton methods
such as LBFGS, which attempt to partially invert the wave-equation Hessian.
3.5 Exercises
1. Repeat the development of section (3.1) in the frequency domain (ω)
rather than in time.
2. Derive Born series with a multiscale exp... | https://ocw.mit.edu/courses/18-325-topics-in-applied-mathematics-waves-and-imaging-fall-2015/15588369c29a866ba59b450518ba486a_MIT18_325F15_Chapter3.pdf |
δm means the linear form that takes a function m1 and
returns the operator of multiplication by m1. We may also write it as
δm1
3.5. EXERCISES
69
the identity Im1 “expecting” a trial function m1. A second derivative
with respect to m(cid:48)
1 gives
δm
δm1
∂2
∂t2
δF(m)
δm(cid:48)
1
+
δm
δm(cid:48)
1
∂2
∂t2
δF(m)
δm1
(... | https://ocw.mit.edu/courses/18-325-topics-in-applied-mathematics-waves-and-imaging-fall-2015/15588369c29a866ba59b450518ba486a_MIT18_325F15_Chapter3.pdf |
for the second-order field u2 when the respective model perturbations
are m1 + m(cid:48)
1, and take a combination of those two fields.
1 and m1 − m(cid:48)
olarization:
6. Consider the setting of section 3.2 in the case M = I. No perturbation
will be needed for this exercise (no decomposition of M into M0 +εM1).
Prove t... | https://ocw.mit.edu/courses/18-325-topics-in-applied-mathematics-waves-and-imaging-fall-2015/15588369c29a866ba59b450518ba486a_MIT18_325F15_Chapter3.pdf |
Find a similar inequality to control the time derivative of w − w0.
(c) Find an equation for w − w0 − w1. Prove that
(cid:107) − 0 − 1(cid:107)∗ ≤ (cid:107) 1(cid:107)∞
w w
(ε M ΩT )2
w
(cid:107)w(cid:107)∗
8. Consider the gradient descent method applied to the linear least-squares
problem minx (cid:107)Ax − b(cid:107)... | https://ocw.mit.edu/courses/18-325-topics-in-applied-mathematics-waves-and-imaging-fall-2015/15588369c29a866ba59b450518ba486a_MIT18_325F15_Chapter3.pdf |
, F
[m] − d(cid:105).
(3.14)
3.5. EXERCISES
71
Note: F ∗F is called the normal operator.
Solution. To compute Hessians, it is important to expand the notation
δm1δm(cid:48) A first
.
to keep track of the different variables, i.e., we compute
derivative gives
δ2J
1
δJ
δm1
= (cid:104)
δF(m)
δm1
, F(m) − d(cid:105),
where ... | https://ocw.mit.edu/courses/18-325-topics-in-applied-mathematics-waves-and-imaging-fall-2015/15588369c29a866ba59b450518ba486a_MIT18_325F15_Chapter3.pdf |
δm1
(cid:48) m1, m1(cid:105) = (cid:104)u1, u(cid:48)
(cid:48)
1(cid:105) + (cid:104)v, u0 − d(cid:105),
(3.15)
where v was defined in the solution of an earlier exercise as
v = (cid:104)
δ2F(m0)
δm1δm(cid:48)
1
m
1, m(cid:48)
1(cid:105).
11. Show that the spectral radius of the Hessian operator δ2J ,
δm2 when data
are ... | https://ocw.mit.edu/courses/18-325-topics-in-applied-mathematics-waves-and-imaging-fall-2015/15588369c29a866ba59b450518ba486a_MIT18_325F15_Chapter3.pdf |
MIT 2.852
Manufacturing Systems Analysis
Lecture 10–12
Transfer Lines – Long Lines
Stanley B. Gershwin
http://web.mit.edu/manuf-sys
Massachusetts Institute of Technology
Spring, 2010
2.852 Manufacturing Systems Analysis
1/91
Copyright (cid:13)2010 Stanley B. Gershwin.
c
Long Lines
M
1
B
1
M
2
B
2
M
3
B... | https://ocw.mit.edu/courses/2-852-manufacturing-systems-analysis-spring-2010/155e83dd472fcab7095db53005c093bb_MIT2_852S10_long_lines.pdf |
◮ Then we extend it to assembly/disassembly systems without loops.
◮ Then we look at systems with loops.
◮ Etc., etc. if there is time.
2.852 Manufacturing Systems Analysis
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Copyright c(cid:13)2010 Stanley B. Gershwin.
Decomposition — Concept
◮ Conceptually: put an observer in a buffer, and tell him that he i... | https://ocw.mit.edu/courses/2-852-manufacturing-systems-analysis-spring-2010/155e83dd472fcab7095db53005c093bb_MIT2_852S10_long_lines.pdf |
in a buffer of a long line.
◮ The two-machine lines are sometimes called building blocks.
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Copyright c(cid:13)2010 Stanley B. Gershwin.
Decomposition — Concept
◮ Consider an observer in Buffer Bi .
◮ Imagine the material flow process that the observer sees entering and
the... | https://ocw.mit.edu/courses/2-852-manufacturing-systems-analysis-spring-2010/155e83dd472fcab7095db53005c093bb_MIT2_852S10_long_lines.pdf |
. Gershwin.
Decomposition — Concept
There are 4(k − 1) unknowns for the deterministic processing time line:
ru(1), pu (1), rd (1), pd (1),
ru(2), pu (2), rd (2), pd (2),
...,
ru (k − 1), pu (k − 1), rd (k − 1), pd (k − 1)
Therefore, we need
◮ 4(k − 1) equations, and
◮ an algorithm for solving those equations.... | https://ocw.mit.edu/courses/2-852-manufacturing-systems-analysis-spring-2010/155e83dd472fcab7095db53005c093bb_MIT2_852S10_long_lines.pdf |
◮ unknowns, or
◮ functions of parameters or unknowns derived from the two-machine line
analysis.
◮ This is a set of 4(k − 1) equations.
2.852 Manufacturing Systems Analysis
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Copyright c(cid:13)2010 Stanley B. Gershwin.
Decomposition Equations
Overview
Notation convention:
◮ Items that pertain to two-mac... | https://ocw.mit.edu/courses/2-852-manufacturing-systems-analysis-spring-2010/155e83dd472fcab7095db53005c093bb_MIT2_852S10_long_lines.pdf |
Equations
Flow Rate-Idle Time
Ei = ei prob [ni −1 > 0 and ni < Ni ]
ei =
ri
ri + pi
where
Problem:
◮ This expression involves a joint probability of two buffers taking
certain values at the same time.
◮ But we only know how to evaluate two-machine, one-buffer lines, so
we only know how to calculate the probabi... | https://ocw.mit.edu/courses/2-852-manufacturing-systems-analysis-spring-2010/155e83dd472fcab7095db53005c093bb_MIT2_852S10_long_lines.pdf |
Rate-Idle Time
Then
prob [ni −1 > 0 and ni < Ni ]
= prob [NOT {ni −1 = 0 or ni = Ni }]
= 1 − prob [ni −1 = 0 or ni = Ni ]
= 1 − { prob (ni −1 = 0) + prob (ni = Ni )
− prob (ni −1 = 0 and ni = Ni )}
≈ 1 − { prob (ni −1 = 0) + prob (ni = Ni )}
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Copyright c(cid:13)2010 S... | https://ocw.mit.edu/courses/2-852-manufacturing-systems-analysis-spring-2010/155e83dd472fcab7095db53005c093bb_MIT2_852S10_long_lines.pdf |
1 −
E (i − 1)
ed (i − 1)
;
pb(i) = 1 −
E (i)
eu (i)
so (replacing ≈ with =),
Ei = ei
1 −
1 −
(cid:20)
(cid:26)
E (i − 1)
ed (i − 1)
−
1 −
(cid:27)
(cid:26)
E (i)
eu (i)
(cid:27)(cid:21)
The goal is to have E = Ei = E (i − 1) = E (i), so
E (i) = ei
1 −
1 −
(cid:20)
(cid:26)
E (i)
ed (i − 1)
... | https://ocw.mit.edu/courses/2-852-manufacturing-systems-analysis-spring-2010/155e83dd472fcab7095db53005c093bb_MIT2_852S10_long_lines.pdf |
2 equations.
2.852 Manufacturing Systems Analysis
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Copyright c(cid:13)2010 Stanley B. Gershwin.
Decomposition Equations
Resumption of Flow
M
i−4
B
i−4
M
i−3
B
i−3
M
i−2
B
i−2
M
i−1
B
i−1
M
i
B
i
M
i+1
B
i+1
M
i+2
B
i+2
M
i+3
When the observer sees Mu(i ) down, Mi may actually be down..... | https://ocw.mit.edu/courses/2-852-manufacturing-systems-analysis-spring-2010/155e83dd472fcab7095db53005c093bb_MIT2_852S10_long_lines.pdf |
−2
B
i−2
M
i−1
B
i−1
M
i
B
i
M
i+1
B
i+1
M
i+2
B
i+2
M
i+3
0
0
... or Mi −2 may be down and Bi −1 and Bi −2 may be empty, ...
M (i)
u
M (i)
d
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Copyright c(cid:13)2010 Stanley B. Gershwin.
Decomposition Equations
Resumption of Flow
M
i−4
B
i−4
M
i−3
... | https://ocw.mit.edu/courses/2-852-manufacturing-systems-analysis-spring-2010/155e83dd472fcab7095db53005c093bb_MIT2_852S10_long_lines.pdf |
0
0
0
M (i)
u
M (i)
d
... etc.
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Copyright c(cid:13)2010 Stanley B. Gershwin.
Decomposition Equations
Resumption of Flow
M
i−4
B
i−4
M
i−3
B
i−3
M
i−2
B
i−2
M
i−1
B
i−1
M
i
B
i
M
i+1
B
i+1
M
i+2
B
i+2
M
i+3
M (i−1)
u
M (i−1)
d
Similarly for the o... | https://ocw.mit.edu/courses/2-852-manufacturing-systems-analysis-spring-2010/155e83dd472fcab7095db53005c093bb_MIT2_852S10_long_lines.pdf |
M (i)
d
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Copyright c(cid:13)2010 Stanley B. Gershwin.
Decomposition Equations
Resumption of Flow
M
i−4
B
i−4
M
i−3
B
i−3
M
i−2
B
i−2
M
i−1
B
i−1
M
i
B
i
M
i+1
B
i+1
M
i+2
B
i+2
M
i+3
0
M (i−1)
u
M (i−1)
d
M
i−4
B
i−4
M
i−3
B
i−3
M
i−2
B
i... | https://ocw.mit.edu/courses/2-852-manufacturing-systems-analysis-spring-2010/155e83dd472fcab7095db53005c093bb_MIT2_852S10_long_lines.pdf |
−3
M
i−2
B
i−2
M
i−1
B
i−1
M
i
B
i
M
i+1
B
i+1
M
i+2
B
i+2
M
i+3
0
0
0
M (i)
u
M (i)
d
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Copyright c(cid:13)2010 Stanley B. Gershwin.
Decomposition Equations
Resumption of Flow
M
i−4
B
i−4
M
i−3
B
i−3
M
i−2
B
i−2
M
i−1
B
i−1
M
i
B
i
M
i+1 ... | https://ocw.mit.edu/courses/2-852-manufacturing-systems-analysis-spring-2010/155e83dd472fcab7095db53005c093bb_MIT2_852S10_long_lines.pdf |
� either real machine Mi is down,
M
M
i−4
i−4
B
B
i−4
i−4
M
M
i−3
i−3
B
B
i−3
i−3
M
M
i−2
i−2
B
B
i−2
i−2
M
M
i−1
i−1
B
B
i−1
i−1
M
M
i
i
B
B
i
i
M
M
i+1
i+1
B
B
i+1
i+1
M
M
i+2
i+2
B
B
i+2
i+2
M
M
i+3
i+3
◮ or Buffer Bi −1 is empty and the Line L(i − 1) observer sees a failure in
Mu(i − 1).
B
M B
... | https://ocw.mit.edu/courses/2-852-manufacturing-systems-analysis-spring-2010/155e83dd472fcab7095db53005c093bb_MIT2_852S10_long_lines.pdf |
30/91
Copyright c(cid:13)2010 Stanley B. Gershwin.
Decomposition Equations
Resumption of Flow
Also, for the Line L(j) observer to see Mu(j) up, Mj must be up and Bj−1 must
be non-empty. Therefore,
{αu(j, τ ) = 1} ⇐⇒ {αj (τ ) = 1} and {nj−1(τ − 1) > 0}
{αu(j, τ ) = 0} ⇐⇒ {αj (τ ) = 0} or {nj−1(τ − 1) = 0}
2.852... | https://ocw.mit.edu/courses/2-852-manufacturing-systems-analysis-spring-2010/155e83dd472fcab7095db53005c093bb_MIT2_852S10_long_lines.pdf |
prob (U|V or W ), where
U = {αi (t + 1) = 1} and {ni −1(t) > 0}
V = {αi (t) = 0}
W = {ni −1(t − 1) = 0}
Important: V and W are disjoint.
prob (V and W ) = 0.
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Copyright c(cid:13)2010 Stanley B. Gershwin.
Decomposition Equations
Resumption of Flow
prob (U|V or W ) = ... | https://ocw.mit.edu/courses/2-852-manufacturing-systems-analysis-spring-2010/155e83dd472fcab7095db53005c093bb_MIT2_852S10_long_lines.pdf |
=
prob (V and (V or W ))
prob (V or W )
=
prob (V )
prob (V or W )
so
prob (U|V or W ) = prob (U|V )prob (V |V or W )
+prob (U|W )prob (W |V or W ).
2.852 Manufacturing Systems Analysis
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Copyright c(cid:13)2010 Stanley B. Gershwin.
Decomposition Equations
Resumption of Flow
Then, if we plug U, V , a... | https://ocw.mit.edu/courses/2-852-manufacturing-systems-analysis-spring-2010/155e83dd472fcab7095db53005c093bb_MIT2_852S10_long_lines.pdf |
ni −1(t − 1) = 0 or αi (t) = 0
(cid:12)
(cid:12)
(cid:12)
(cid:12)
,
(cid:21)
= prob [ni −1(t) > 0 and αi (t + 1) = 1 | αi (t) = 0] ,
X ′ (i) = prob (V |V or W )
= prob [αi (t) = 0 | {ni −1(t − 1) = 0 or αi (t) = 0}] .
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Copyright c(cid:13)2010 Stanley B. Gershwin.
... | https://ocw.mit.edu/courses/2-852-manufacturing-systems-analysis-spring-2010/155e83dd472fcab7095db53005c093bb_MIT2_852S10_long_lines.pdf |
time t − 1, it
must have gained 1 part. For it to gain a part when αi (t) = 1, Mi must not have
been working (because it was previously starved). Therefore, Mi could not have
failed and A(i − 1) can therefore be written
A(i − 1) = prob
»
ni −1(t) > 0
˛
˛
˛
˛
ni −1(t − 1) = 0
–
2.852 Manufacturing Systems Analys... | https://ocw.mit.edu/courses/2-852-manufacturing-systems-analysis-spring-2010/155e83dd472fcab7095db53005c093bb_MIT2_852S10_long_lines.pdf |
ved.
That is, if ni −1(t) > 0, then αu (i − 1, t) = 1.
Therefore,
A(i − 1) = prob
»
αu (i − 1, t) = 1
˛
˛
˛
˛
αu (i − 1, t − 1) = 0
–
= ru (i − 1)
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Copyright c(cid:13)2010 Stanley B. Gershwin.
Decomposition Equations
Resumption of Flow
Similarl... | https://ocw.mit.edu/courses/2-852-manufacturing-systems-analysis-spring-2010/155e83dd472fcab7095db53005c093bb_MIT2_852S10_long_lines.pdf |
is down
◮ plus ru(i − 1) times the probability that Mu(i) is down because
Mu(i − 1) is down and Bi −1 is empty.
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Copyright c(cid:13)2010 Stanley B. Gershwin.
Decomposition Equations
Resumption of Flow
X (i)= the probability that Mu(i) is down because Mu(i − 1) is down ... | https://ocw.mit.edu/courses/2-852-manufacturing-systems-analysis-spring-2010/155e83dd472fcab7095db53005c093bb_MIT2_852S10_long_lines.pdf |
1(t − 1) = 0]
prob [ni −1(t − 1) = 0 or αi (t) = 0]
=
ps (i − 1)
prob [ni −1(t − 1) = 0 or αi (t) = 0]
2.852 Manufacturing Systems Analysis
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Copyright c(cid:13)2010 Stanley B. Gershwin.
Decomposition Equations
Resumption of Flow
To analyze the denominator, note
◮ {ni −1(t − 1) = 0 or αi (t) = 0} = {αu ... | https://ocw.mit.edu/courses/2-852-manufacturing-systems-analysis-spring-2010/155e83dd472fcab7095db53005c093bb_MIT2_852S10_long_lines.pdf |
c(cid:13)2010 Stanley B. Gershwin.
Decomposition Equations
Resumption of Flow
Therefore,
and
X (i) =
ps (i − 1)ru(i)
pu(i)E (i)
ru (i) = ru (i − 1)X (i) + ri (1 − X (i)), i = 2, . . . , k − 1
This is a set of k − 2 equations.
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Copyright c(cid:13)2010 Stanley B. Ger... | https://ocw.mit.edu/courses/2-852-manufacturing-systems-analysis-spring-2010/155e83dd472fcab7095db53005c093bb_MIT2_852S10_long_lines.pdf |
48/91
Copyright c(cid:13)2010 Stanley B. Gershwin.
Decomposition Equations
Algorithm
We use the conservation of flow conditions by modifying these equations.
Modified upstream equations:
ru (i) = ru (i − 1)X (i) + ri (1 − X (i));
X (i) =
ps (i − 1)ru (i)
pu (i)E (i − 1)
pu (i) = ru (i)
1
E (i − 1)
„
... | https://ocw.mit.edu/courses/2-852-manufacturing-systems-analysis-spring-2010/155e83dd472fcab7095db53005c093bb_MIT2_852S10_long_lines.pdf |
rd (i ), pd (i ) parameter,
i = 1, ..., k − 1 is less than ǫ, or
◮ etc.
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Copyright c(cid:13)2010 Stanley B. Gershwin.
Decomposition Equations
Algorithm
DDX algorithm : due to Dallery, David, and Xie (1988).
1. Guess the downstream parameters of L(1) (rd (1), pd (1)). ... | https://ocw.mit.edu/courses/2-852-manufacturing-systems-analysis-spring-2010/155e83dd472fcab7095db53005c093bb_MIT2_852S10_long_lines.pdf |
[ni −1(t − 1) = 0 and ni (t − 1) = Ni ] ≈ 0
Question: When will this work well, and when will it work badly?
2.852 Manufacturing Systems Analysis
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Copyright c(cid:13)2010 Stanley B. Gershwin.
Examples
Three-machine line
Three-machine line – production rate.
E
.8
.7
.6
.5
.4
.3
.2
.1
p2 = .05
.15... | https://ocw.mit.edu/courses/2-852-manufacturing-systems-analysis-spring-2010/155e83dd472fcab7095db53005c093bb_MIT2_852S10_long_lines.pdf |
Stanley B. Gershwin.
Examples
Long lines
20
15
10
5
l
e
v
e
L
r
e
f
f
u
B
e
g
a
r
e
v
A
0
0
50 Machines; r=0.1; p=0.01; mu=1.0; N=20.0
Distribution of
material in a line
with identical
machines and buffers.
Explain the shape.
10
20
30
40
50
Buffer Number
2.852 Manufacturing Systems Analysis
55/91 ... | https://ocw.mit.edu/courses/2-852-manufacturing-systems-analysis-spring-2010/155e83dd472fcab7095db53005c093bb_MIT2_852S10_long_lines.pdf |
Stanley B. Gershwin.
Examples
Long lines
20
15
10
5
l
e
v
e
L
r
e
f
f
u
B
e
g
a
r
e
v
A
0
0
50 Machines; r=0.1; p=0.01; mu=1.0; N=20.0 EXCEPT N(25)=2000.0
Same as Slide 55
except that Buffer 25
is now huge.
Explain the shape.
10
20
30
40
50
Buffer Number
2.852 Manufacturing Systems Analysis
5... | https://ocw.mit.edu/courses/2-852-manufacturing-systems-analysis-spring-2010/155e83dd472fcab7095db53005c093bb_MIT2_852S10_long_lines.pdf |
15; p=0.01; mu=1.0, N=50.0
20
15
10
5
l
e
v
e
L
r
e
f
f
u
B
e
g
a
r
e
v
A
0
0
Upstream same as
Slide 58; downstream
faster.
Explain the shape.
10
20
30
40
50
Buffer Number
2.852 Manufacturing Systems Analysis
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Copyright c(cid:13)2010 Stanley B. Gershwin.
Examples
Long lines
50 Machines... | https://ocw.mit.edu/courses/2-852-manufacturing-systems-analysis-spring-2010/155e83dd472fcab7095db53005c093bb_MIT2_852S10_long_lines.pdf |
; p=0.01; mu=1.0, N=15.0
20
15
10
5
l
e
v
e
L
r
e
f
f
u
B
e
g
a
r
e
v
A
0
0
Downstream same as
downstream half of
Slide 57; upstream
faster.
Explain the shape.
10
20
30
40
50
Buffer Number
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Copyright c(cid:13)2010 Stanley B. Gershwin.
Examples
Long... | https://ocw.mit.edu/courses/2-852-manufacturing-systems-analysis-spring-2010/155e83dd472fcab7095db53005c093bb_MIT2_852S10_long_lines.pdf |
; N=20.0 EXCEPT mu(10)=0.8
20
15
10
5
l
e
v
e
L
r
e
f
f
u
B
e
g
a
r
e
v
A
0
0
Operation time
bottleneck. Identical
machines and buffers,
except for M10.
Explain the shape.
10
20
30
40
50
Buffer Number
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Copyright c(cid:13)2010 Stanley B. Gershwin.
Exa... | https://ocw.mit.edu/courses/2-852-manufacturing-systems-analysis-spring-2010/155e83dd472fcab7095db53005c093bb_MIT2_852S10_long_lines.pdf |
r
e
f
f
u
B
e
g
a
r
e
v
A
0
0
Repair time
bottleneck.
Explain the shape.
10
20
30
40
50
Buffer Number
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Copyright c(cid:13)2010 Stanley B. Gershwin.
Examples
Infinitely long lines
Infinitely long lines with identical machines and buffers
ri = r
pi = p
Ni ... | https://ocw.mit.edu/courses/2-852-manufacturing-systems-analysis-spring-2010/155e83dd472fcab7095db53005c093bb_MIT2_852S10_long_lines.pdf |
ru (i ) =
E (i ) + 1
1
ei
− 2
2pu
r = 1
E + 1
e − 2
2.852 Manufacturing Systems Analysis
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Copyright c(cid:13)2010 Stanley B. Gershwin.
Examples
Infinitely long lines
In the last equation, pu is unknown and E is a function of pu. This is one
equation in one unknown.
.35
.34
.33
.32
.31
.30
.29... | https://ocw.mit.edu/courses/2-852-manufacturing-systems-analysis-spring-2010/155e83dd472fcab7095db53005c093bb_MIT2_852S10_long_lines.pdf |
l
e
v
e
L
r
e
f
f
u
B
e
g
a
r
e
v
A
n1
n2
n3
n4
n5
n6
n7
Continuous material model.
◮ Eight-machine,
seven-buffer line.
◮ For each machine,
r = .075, p = .009,
µ = 1.2.
◮ For each buffer (except
Buffer 6), N = 30.
0
0
5
10
15
20
N
6
25
30
35
40
45
50
M
1
B
1
M2
B2
M
3
B
3
M4
B4
M
5
... | https://ocw.mit.edu/courses/2-852-manufacturing-systems-analysis-spring-2010/155e83dd472fcab7095db53005c093bb_MIT2_852S10_long_lines.pdf |
2
B2
M
3
B
3
M4
B4
M
5
B
5
M6
B6
M
7
B
7
M8
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Copyright c(cid:13)2010 Stanley B. Gershwin.
Examples
Buffer allocation
Which has a higher production rate?
◮ 9-Machine line with two buffering options:
◮ 8 buffers equally sized; and
M1
B1
M
2
B
2
M
... | https://ocw.mit.edu/courses/2-852-manufacturing-systems-analysis-spring-2010/155e83dd472fcab7095db53005c093bb_MIT2_852S10_long_lines.pdf |
8 buffers
2 buffers
2 buffers
2 buffers
1000
1000
1000
1000
2000
2000
2000
2000
3000
3000
3000
3000
4000
4000
4000
4000
5000
5000
5000
5000
6000
6000
6000
6000
7000
7000
7000
7000
8000
8000
8000
8000
9000
9000
9000
9000
10000
10000
10000
10000
Total Buffer Space
◮ Continuous model; all
machines have
r = .019... | https://ocw.mit.edu/courses/2-852-manufacturing-systems-analysis-spring-2010/155e83dd472fcab7095db53005c093bb_MIT2_852S10_long_lines.pdf |
5
1111
8 buffers
8 buffers
8 buffers
8 buffers
2 buffers
2 buffers
2 buffers
◮ Is 8 buffers always
faster?
◮ Perhaps not, but
difference is not
significant in systems
with very small buffers.
10101010
100100100100
1000100010001000
10000
10000
10000
10000
Total Buffer Space
2.852 Manufacturing Systems Analysis
73/91... | https://ocw.mit.edu/courses/2-852-manufacturing-systems-analysis-spring-2010/155e83dd472fcab7095db53005c093bb_MIT2_852S10_long_lines.pdf |
/91
Copyright c(cid:13)2010 Stanley B. Gershwin.
Long Lines — Exponential Processing Time Model
The observer thinks he is in a two-machine exponential processing time line with
parameters
ru (i)δt =
probability that Mu (i) goes from down to up in (t, t + δt), for small δt;
pu (i)δt =
probability that Mu (i) go... | https://ocw.mit.edu/courses/2-852-manufacturing-systems-analysis-spring-2010/155e83dd472fcab7095db53005c093bb_MIT2_852S10_long_lines.pdf |
B. Gershwin.
Long Lines — Exponential Processing Time Model
Equations
We have 6(k − 1) unknowns, so we need 6(k − 1) equations. They are
◮ Interruption of flow , relating pu (i) to upstream events and pd (i) to
downstream events,
◮ Resumption of flow,
◮ Conservation of flow,
◮ Flow rate/idle time,
◮ Boundary con... | https://ocw.mit.edu/courses/2-852-manufacturing-systems-analysis-spring-2010/155e83dd472fcab7095db53005c093bb_MIT2_852S10_long_lines.pdf |
(cid:12)
(cid:12)
.
(cid:21)
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Copyright c(cid:13)2010 Stanley B. Gershwin.
Long Lines — Exponential Processing Time Model
Interruption of Flow
We define the events that a pseudo-machine is up or down as follows:
Mu(i) is down if
1. Mi is down, or
2. ni −1 = 0 and Mu... | https://ocw.mit.edu/courses/2-852-manufacturing-systems-analysis-spring-2010/155e83dd472fcab7095db53005c093bb_MIT2_852S10_long_lines.pdf |
line L(i − 1) is in state
(0, 0, 1) and p(i + 1; N10) is the steady state probability that line L(i + 1) is in
state (Ni +1, 1, 0).
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Copyright c(cid:13)2010 Stanley B. Gershwin.
Long Lines — Exponential Processing Time Model
Resumption of Flow
ru (i) =
ru (i − 1)
pi −1... | https://ocw.mit.edu/courses/2-852-manufacturing-systems-analysis-spring-2010/155e83dd472fcab7095db53005c093bb_MIT2_852S10_long_lines.pdf |
2, . . . , k − 1.
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Copyright c(cid:13)2010 Stanley B. Gershwin.
Long Lines — Exponential Processing Time Model
Flow Rate/Idle Time
The flow rate-idle time relationship is, approximately,
Pi = ei µi (1 − prob [ni −1 = 0] − prob [ni = Ni ]) .
which can be transformed into... | https://ocw.mit.edu/courses/2-852-manufacturing-systems-analysis-spring-2010/155e83dd472fcab7095db53005c093bb_MIT2_852S10_long_lines.pdf |
Analysis
83/91
Copyright c(cid:13)2010 Stanley B. Gershwin.
Long Lines — Exponential Processing Time Model
Boundary Conditions
Md (1) is the same as M1 and Md (k − 1) is the same as Mk . Therefore
ru(1) = r1
pu(1) = p1
µu(1) = µ1
rd (k − 1) = rk
pd (k − 1) = pk
µd (k − 1) = µk
2.852 Manufacturing Systems A... | https://ocw.mit.edu/courses/2-852-manufacturing-systems-analysis-spring-2010/155e83dd472fcab7095db53005c093bb_MIT2_852S10_long_lines.pdf |
B
2
M
3
B
3
M
4
B
4
M
5
B
5
M6
Conceptually very similar to exponential processing time model. One
difference:
◮ prob (xi −1 = 0 and xi = Ni ) = 0 exactly .
2.852 Manufacturing Systems Analysis
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Copyright c(cid:13)2010 Stanley B. Gershwin.
Long Lines — Continuous Material Model
New approximation ... | https://ocw.mit.edu/courses/2-852-manufacturing-systems-analysis-spring-2010/155e83dd472fcab7095db53005c093bb_MIT2_852S10_long_lines.pdf |
u(i) < µd (i)) or 0;
2.852 Manufacturing Systems Analysis
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Copyright c(cid:13)2010 Stanley B. Gershwin.
Long Lines — Continuous Material
New approximation
M
i−2
B
i−2
M
i−1
B
i−1
M
i
B
i
M
i+1
B
i+1
M
i+2
B
i+2
M
i+3
M (i)
u
M (i)
d
Assume that ... < µi −2 < µi −1 < µi < µi +1... | https://ocw.mit.edu/courses/2-852-manufacturing-systems-analysis-spring-2010/155e83dd472fcab7095db53005c093bb_MIT2_852S10_long_lines.pdf |
6(k − 1) equations. They are,
as before,
◮ Interruption of flow ,
◮ Resumption of flow,
◮ Conservation of flow,
◮ Flow rate/idle time,
◮ Boundary conditions.
They are the same as in the exponential processing time case except for the
Interruption of Flow equations.
2.852 Manufacturing Systems Analysis
89/91
Cop... | https://ocw.mit.edu/courses/2-852-manufacturing-systems-analysis-spring-2010/155e83dd472fcab7095db53005c093bb_MIT2_852S10_long_lines.pdf |
µd (i + 1)
P(i) − pi(0, 1, 1)µu (i) «
„
rd (i + 1), i = 1, · · · , k − 2
2.852 Manufacturing Systems Analysis
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Copyright c(cid:13)2010 Stanley B. Gershwin.
To come
◮ Assembly/Disassembly Systems
◮ Buffer Optimization
◮ Effect of Buffers on Quality
◮ Loops
◮ Real-Time Control
◮ ????
2.852 ... | https://ocw.mit.edu/courses/2-852-manufacturing-systems-analysis-spring-2010/155e83dd472fcab7095db53005c093bb_MIT2_852S10_long_lines.pdf |
MIT OpenCourseWare
http://ocw.mit.edu
18.014 Calculus with Theory
Fall 2010
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. | https://ocw.mit.edu/courses/18-014-calculus-with-theory-fall-2010/15694ffbb31f9061bbe041a05f5433aa_MIT18_014F10_ChJnotes.pdf |
Chapter 1
Introduction
This course will be organized around algorithmic issues that arise in machine learn
ing. The usual paradigm for algorithm design is to give an algorithm that succeeds on
all possible inputs, but the difficulty is that almost all of the optimization problems
that arise in modern machine learnin... | https://ocw.mit.edu/courses/18-409-algorithmic-aspects-of-machine-learning-spring-2015/15c289f8a75b18b46c7d9eb9e5fb8485_MIT18_409S15_intro.pdf |
work?
Algorithmic Aspects of Machine Learning
© 2015 by Ankur Moitra.
Note: These are unpolished, incomplete course notes.
Developed for educational use at MIT and for publication through MIT OpenCourseware.
3
4
CHAPTER 1. INTRODUCTION
This course will focus on
(a) nonnegative matrix factorization
(b) topic mod... | https://ocw.mit.edu/courses/18-409-algorithmic-aspects-of-machine-learning-spring-2015/15c289f8a75b18b46c7d9eb9e5fb8485_MIT18_409S15_intro.pdf |
3.37 (Class 3)
Review:
The inherent strength of all bonds (even van der Waals) is extremely high
Primary (1-3eV) (cid:198) 1,000,000 – 3,000,000 psi
van der Waals (0.1-0.2eV) (cid:198) 100,000 – 200,000 psi
graph of energy vs. distance
graph of force vs. distance
interatomic distance
can get bulk compressibilit... | https://ocw.mit.edu/courses/3-37-welding-and-joining-processes-fall-2002/15d1ad7edaf8a60be7ad51059c1875aa_33703.pdf |
aminant”, in this case
oxygen
Can think of surface energy as surface tension
• Surface energy
o Units of J/m
• Surface tension
o Units of N/m^2
• Can think of as either as energy or force
Cleaving a material in a vacuum, then if stick it back together (no contamination) will
bond with much of its original stre... | https://ocw.mit.edu/courses/3-37-welding-and-joining-processes-fall-2002/15d1ad7edaf8a60be7ad51059c1875aa_33703.pdf |
How much contact do I get when surfaces touch
• Asperity contact
• Squeeze it together: Can I get 100% contact? No
• Compressive yield strength (sigma-yield) of the material when two small pieces
pushed together
• 3 times sigma-yield to indent a flat object with a punch since need to push aside
material on sides ... | https://ocw.mit.edu/courses/3-37-welding-and-joining-processes-fall-2002/15d1ad7edaf8a60be7ad51059c1875aa_33703.pdf |
MATH 18.152 COURSE NOTES - CLASS MEETING # 1
18.152 Introduction to PDEs, Fall 2011
Professor: Jared Speck
Class Meeting # 1: Introduction to PDEs
1. What is a PDE?
We will be studying functions u = u(x1, x2, · · · , xn) and their partial derivatives. Here x1, x2, · · · , xn
are standard Cartesian coordinates on Rn. We... | https://ocw.mit.edu/courses/18-152-introduction-to-partial-differential-equations-fall-2011/15e3e62a513bb3bdc5de09374f03517d_MIT18_152F11_lec_01.pdf |
· , iN ∈ {1, 2, · · · , n}
for some function F.
Here N is called the order of the PDE. N is the maximum number of derivatives appearing in
the equation.
Example 1.0.1. u = u(t, x)
(1.0.3)
is a third-order nonlinear PDE.
Example 1.0.2. u = u(t, x)
(1.0.4)
is a second-order linear PDE.
−∂2u + (1 + cos u)∂3
t
xu = 0
−∂2
t... | https://ocw.mit.edu/courses/18-152-introduction-to-partial-differential-equations-fall-2011/15e3e62a513bb3bdc5de09374f03517d_MIT18_152F11_lec_01.pdf |
recipe for answering it!
In practice, good models are often the end result of confrontations between experimental data and
theory. In this course, we will discuss some important physical systems and the PDEs that are
commonly used to model them.
Now let’s assume that we have a PDE that we believe is a good model for ou... | https://ocw.mit.edu/courses/18-152-introduction-to-partial-differential-equations-fall-2011/15e3e62a513bb3bdc5de09374f03517d_MIT18_152F11_lec_01.pdf |
ı∂tu + ∂2
• ut + ux = 0, transport equation, first-order, linear, homogeneous
•
ut + uux = 0, Burger’s equation, first-order, nonlinear, homogeneous
MATH 18.152 COURSE NOTES - CLASS MEETING # 1
E = (cid:0)E1(x, y, z), E2(x, y, z), E3(x, y, z)(cid:1), B = (cid:0)B1(x, y, z), B2(x, y, z), B3(x, y, z)
3
(cid:1) are vectors... | https://ocw.mit.edu/courses/18-152-introduction-to-partial-differential-equations-fall-2011/15e3e62a513bb3bdc5de09374f03517d_MIT18_152F11_lec_01.pdf |
= u(x, y), Lu = ∂2
L(u + v) = ∂2
x(u + v) + (u + v)2∂2
xu + u2∂2
y u does NOT define a linear operator:
y (u + v) = ∂xu + u ∂y u + ∂xv + v ∂y v = Lu + Lv
2 2
2 2
2
2
Definition 4.0.3. A PDE is linear if it can be written as
(4.0.9)
Lu = f (x1, · · · , xn)
for some linear operator L and some function f of the coordinates.... | https://ocw.mit.edu/courses/18-152-introduction-to-partial-differential-equations-fall-2011/15e3e62a513bb3bdc5de09374f03517d_MIT18_152F11_lec_01.pdf |
L is
linear.
As we will see in the next proposition, inhomogeneous and homogeneous linear PDEs are closely
related.
Proposition 4.0.2 (Relationship between the inhomogeneous and homogeneous linear
PDE solutions). Let Sh be the set of all solutions to the homogeneous linear PDE
(cid:3)
(4.0.13)
Lu = 0,
and let uI be a “... | https://ocw.mit.edu/courses/18-152-introduction-to-partial-differential-equations-fall-2011/15e3e62a513bb3bdc5de09374f03517d_MIT18_152F11_lec_01.pdf |
data.”
6. Some simple PDEs that we can easily solve
6.1. Constant coefficient transport equations. Consider the first-order linear transport equa-
tion
a∂xu(x, y) + b∂yu(x, y) = 0,
(6.1.1)
where a, b ∈ R. Let’s try to solve this PDE by reasoning geometrically. Geometrically, this equation
v = 0, where ∇u = (∂xu, ∂yu) and ... | https://ocw.mit.edu/courses/18-152-introduction-to-partial-differential-equations-fall-2011/15e3e62a513bb3bdc5de09374f03517d_MIT18_152F11_lec_01.pdf |
Let’s consider the following example:
(6.2.1)
y∂xu + x∂yu = 0.
Let P denote a point P = (x, y), and let V denote the vector V = (y, x). Using vector calculus
notation, (6.2.1) can be written as ∇u(P ) · V = 0, i.e., the derivative of u at P in the direction
of V is 0. Thus, equation (6.2.1) implies that u is constant a... | https://ocw.mit.edu/courses/18-152-introduction-to-partial-differential-equations-fall-2011/15e3e62a513bb3bdc5de09374f03517d_MIT18_152F11_lec_01.pdf |
y)
.
7. Some basic analytical notions and tools
We now discuss a few ideas from analysis that will appear repeatedly throughout the course.
6
MATH 18.152 COURSE NOTES - CLASS MEETING # 1
7.1. Norms. In PDE, there are many different ways to measure the “size” of a function f. These
measures are called norms. Here is a s... | https://ocw.mit.edu/courses/18-152-introduction-to-partial-differential-equations-fall-2011/15e3e62a513bb3bdc5de09374f03517d_MIT18_152F11_lec_01.pdf |
a-
tion about f in one variable compared to another. For example, if f = f (t, x), then we use notation
such as
|+sup(x,y)∈Ω |∂2
y f (x, y)|.
(7.1.3)
(cid:107)f (cid:107)
def
C1,2 =
1
(cid:88)
a
=0
a
sup |∂t f (t, x)| +
(t,x)
∈
R
2
2
(cid:88)
a=1
a
|.
sup |∂xf (t, x)
(t,x)
∈
R2
Above, the “1” in C 1,2 refers to the t c... | https://ocw.mit.edu/courses/18-152-introduction-to-partial-differential-equations-fall-2011/15e3e62a513bb3bdc5de09374f03517d_MIT18_152F11_lec_01.pdf |
)Lp(Ω)
• Triangle inequality: (cid:107)f + g(cid:107)Lp(Ω) ≤ (cid:107)f (cid:107)Lp(Ω) + (cid:107)g(cid:107)Lp(Ω)
Similarly, (cid:107) · (cid:107)Ck(Ω) also has all the properties of a norm. All of these properties are very easy to
show except for the last one in the case of (cid:107) · (cid:107)Lp(Ω). You will study t... | https://ocw.mit.edu/courses/18-152-introduction-to-partial-differential-equations-fall-2011/15e3e62a513bb3bdc5de09374f03517d_MIT18_152F11_lec_01.pdf |
.2 (Divergence). Recall that ∇·F, the divergence of F, is the scalar-valued function
on Rn defined by
(7.2.2)
∇ ·
def
F =
n
(cid:88)
i=1
∂iF i.
We are now ready to recall the divergence theorem.
Theorem 7.1 (Divergence Theorem). Let Ω ⊂ R3 be a domain2 with a boundary that we denote
by ∂Ω. Then the following formula hol... | https://ocw.mit.edu/courses/18-152-introduction-to-partial-differential-equations-fall-2011/15e3e62a513bb3bdc5de09374f03517d_MIT18_152F11_lec_01.pdf |
.152 Introduction to Partial Differential Equations.
Fall 2011
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. | https://ocw.mit.edu/courses/18-152-introduction-to-partial-differential-equations-fall-2011/15e3e62a513bb3bdc5de09374f03517d_MIT18_152F11_lec_01.pdf |
1 2 3 4 5 6 7 8 MIT OpenCourseWare
http://ocw.mit.edu
6.890 Algorithmic Lower Bounds: Fun with Hardness Proofs
Fall 2014
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. | https://ocw.mit.edu/courses/6-890-algorithmic-lower-bounds-fun-with-hardness-proofs-fall-2014/163d244c8ab074e246c843ccaf2fa7ae_MIT6_890F14_L01.pdf |
Elasticity
(and other useful things to know)
Carol Livermore
Massachusetts Institute of Technology
* With thanks to Steve Senturia, from whose lecture notes some
of these materials are adapted.
Cite as: Carol Livermore, course materials for 6.777J / 2.372J Design and Fabrication of Microelectromechanical Devices, Spri... | https://ocw.mit.edu/courses/6-777j-design-and-fabrication-of-microelectromechanical-devices-spring-2007/1654337d7c64fdb073b2ab3b4aac7688_07lecture06split.pdf |
4 (1999): 338-347.
www.dlp.com
1 mm
Cantilever
Veeco.com
0.5 mm
Silicon
Pull-down
electrode
Anchor
Adapted from Rebeiz, Gabriel M.
Hoboken, NJ: John Wiley, 2003. I
RF MEMS: Theory, Design, and Technology.
SBN: 9780471201694.
Image by MIT OpenCourseWare.
AFM cantilevers
Courtesy of Veeco Instruments, Inc. Used
with pe... | https://ocw.mit.edu/courses/6-777j-design-and-fabrication-of-microelectromechanical-devices-spring-2007/1654337d7c64fdb073b2ab3b4aac7688_07lecture06split.pdf |
, or at the point of maximum stress)?
• How much load can I apply without breaking the structure?
Cite as: Carol Livermore, course materials for 6.777J / 2.372J Design and Fabrication of Microelectromechanical Devices, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downlo... | https://ocw.mit.edu/courses/6-777j-design-and-fabrication-of-microelectromechanical-devices-spring-2007/1654337d7c64fdb073b2ab3b4aac7688_07lecture06split.pdf |
are the loads, and where on the structure are they applied?
1 mm
Cantilever
0.5 mm
Silicon
Pull-down
electrode
Anchor
F
Adapted from Rebeiz, Gabriel M. RF MEMS: Theory, Design, and Technology.
Hoboken, NJ: John Wiley, 2003. ISBN: 9780471201694.
Image by MIT OpenCourseWare.
> Given the loads, what is going on at point (... | https://ocw.mit.edu/courses/6-777j-design-and-fabrication-of-microelectromechanical-devices-spring-2007/1654337d7c64fdb073b2ab3b4aac7688_07lecture06split.pdf |
77J/2.372J Spring 2007, Lecture 6 - 8
Outline
> Overview
> Some definitions
• Stress
• Strain
> Isotropic materials
• Constitutive equations of linear elasticity
• Plane stress
• Thin films: residual and thermal stress
> A few important things
• Storing elastic energy
• Linear elasticity in anisotropic materials
• Beh... | https://ocw.mit.edu/courses/6-777j-design-and-fabrication-of-microelectromechanical-devices-spring-2007/1654337d7c64fdb073b2ab3b4aac7688_07lecture06split.pdf |
kPa
• 1 dyne/cm2 = 0.1 Pa
> Notation: τface,direction
x
z
sz
tzy
tyz
sy
tyx
txy
Dx
Dz
tzx
txz
sx
Dy
y
Adapted from Senturia, Stephen D. Microsystem Design. Boston, MA:
Kluwer Academic Publishers, 2001. ISBN: 9780792372462.
Image by MIT OpenCourseWare.
Cite as: Carol Livermore, course materials for 6.777J / 2.372J Desig... | https://ocw.mit.edu/courses/6-777j-design-and-fabrication-of-microelectromechanical-devices-spring-2007/1654337d7c64fdb073b2ab3b4aac7688_07lecture06split.pdf |
12
Strain
> Strain is a continuum description of deformation.
> Center of mass translation and rigid rotation are NOT strains
> Strain is expressed in terms of the displacements of each point
in a differential volume, u(x) where u is the displacement and x is
the original coordinate
> Deformation is present only wh... | https://ocw.mit.edu/courses/6-777j-design-and-fabrication-of-microelectromechanical-devices-spring-2007/1654337d7c64fdb073b2ab3b4aac7688_07lecture06split.pdf |
. Downloaded on [DD Month YYYY].
C. Livermore: 6.777J/2.372J Spring 2007, Lecture 6 - 14
Shear Strains (γxy, γxz , γyz)
> Angles change
> Comes from shear stresses
> Quantified as change in angle
in radians
Δux
θ2
Δy
Δuy
Δx
θ1
xyγ
=
u
Δ
x
y
Δ
⎛
⎜⎜
⎝
+
Δ
u
y
x
Δ
⎞
=⎟⎟
⎠
⎛
⎜⎜
⎝
u
∂
x
y
∂
+
u
∂
y
x
∂
⎞
⎟⎟
⎠
Ci... | https://ocw.mit.edu/courses/6-777j-design-and-fabrication-of-microelectromechanical-devices-spring-2007/1654337d7c64fdb073b2ab3b4aac7688_07lecture06split.pdf |
(http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
C. Livermore: 6.777J/2.372J Spring 2007, Lecture 6 - 16
Outline
> Overview
> Some definitions
• Stress
• Strain
> Isotropic materials
• Constitutive equations of linear elasticity
• Plane stress
• Thin films: residual and ther... | https://ocw.mit.edu/courses/6-777j-design-and-fabrication-of-microelectromechanical-devices-spring-2007/1654337d7c64fdb073b2ab3b4aac7688_07lecture06split.pdf |
6 - 18
Linear Elasticity in Isotropic Materials
> Poisson ratio, ν
• Some things get narrower in the transverse direction when
you extend them axially.
• Some things get wider in the transverse direction when you
compress them axially.
• This is described by the Poisson ratio: the negative ratio of
transverse strai... | https://ocw.mit.edu/courses/6-777j-design-and-fabrication-of-microelectromechanical-devices-spring-2007/1654337d7c64fdb073b2ab3b4aac7688_07lecture06split.pdf |
on [DD Month YYYY].
C. Livermore: 6.777J/2.372J Spring 2007, Lecture 6 - 20
Isotropic Linear Elasticity
> For a general case of loading, the constitutive relationships
between stress and elastic strain are as follows
> 6 equations, one for each normal stress and shear stress
ε
x
=
ε
y
=
ε
z
=
1
E
1
E
1
E
[
(
σσνσ
z
−... | https://ocw.mit.edu/courses/6-777j-design-and-fabrication-of-microelectromechanical-devices-spring-2007/1654337d7c64fdb073b2ab3b4aac7688_07lecture06split.pdf |
. Downloaded on [DD Month YYYY].
C. Livermore: 6.777J/2.372J Spring 2007, Lecture 6 - 22
Plane stress
> Special case: when all stresses are confined to a single plane
Often seen in thin films on substrates (will discuss origin of
these stresses shortly)
> Zero normal stress in z direction (σz = 0)
> No constraint on... | https://ocw.mit.edu/courses/6-777j-design-and-fabrication-of-microelectromechanical-devices-spring-2007/1654337d7c64fdb073b2ab3b4aac7688_07lecture06split.pdf |
J Design and Fabrication of Microelectromechanical Devices, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
C. Livermore: 6.777J/2.372J Spring 2007, Lecture 6 - 24
Stresses on Inclined Sections
> Can resolve axial forces into normal and shea... | https://ocw.mit.edu/courses/6-777j-design-and-fabrication-of-microelectromechanical-devices-spring-2007/1654337d7c64fdb073b2ab3b4aac7688_07lecture06split.pdf |
0.5
Adapted from Figure 9.4 in Senturia, Stephen
Kluwer Academic Publishers, 2001, p. 206. ISBN: 9780792372462.
D. Microsystem Design. Boston, MA:
Image by MIT OpenCourseWare.
Failure in shear occurs at an angle of 45 degrees
Cite as: Carol Livermore, course materials for 6.777J / 2.372J Design and Fabrication of Micr... | https://ocw.mit.edu/courses/6-777j-design-and-fabrication-of-microelectromechanical-devices-spring-2007/1654337d7c64fdb073b2ab3b4aac7688_07lecture06split.pdf |
A thin film on a substrate can have residual stress
• Intrinsic stress
• Thermal stress
> Mostly well-described as a plane stress
Thin film
Plane stress region
Edge
region
Substrate
Adapted from Figure 8.5 in: Senturia, Stephen D. Microsystem Design. Boston, MA: Kluwer
Academic Publishers, 2001, p. 190. ISBN: 97807923... | https://ocw.mit.edu/courses/6-777j-design-and-fabrication-of-microelectromechanical-devices-spring-2007/1654337d7c64fdb073b2ab3b4aac7688_07lecture06split.pdf |
length changes
> This is a thermally-induced strain
> An unopposed thermal
expansion produces a strain, but
not a stress
> If you oppose the thermal
expansion, there will be a stress
> Coefficient of thermal expansion,
αT
thermal
ε
x
α
T
Δ
T
(
)
T
=Δ
⇓
)
(
T
0
+
( )
T
=
ε
x
ε
x
α
T
(
TT
−
0
)
and
V
Δ
V
=
3
α
T
(
TT... | https://ocw.mit.edu/courses/6-777j-design-and-fabrication-of-microelectromechanical-devices-spring-2007/1654337d7c64fdb073b2ab3b4aac7688_07lecture06split.pdf |
uced Residual Stress
Substrate:
Δ
T
s
,αε
−=
sT
where
=Δ
TT
d
−
T
r
Some of the final strain is
accounted for by the strain that the
film would have if it were free. The
remainder, or mismatch strain, will
be associated with a stress through
constitutive relationships.
Film:
ε
f
,
free
−=
α
fT
,
α
,
sT
ε
f
,
atta... | https://ocw.mit.edu/courses/6-777j-design-and-fabrication-of-microelectromechanical-devices-spring-2007/1654337d7c64fdb073b2ab3b4aac7688_07lecture06split.pdf |
[DD Month YYYY].
C. Livermore: 6.777J/2.372J Spring 2007, Lecture 6 - 33
Edge effects
> If a bonded thin film is in a state of plane stress due to residual
stress created when the film is formed, there are extra stresses
at the edges of these films
Shear stresses
F = 0
F = 0
Extra peel force
Adapted from Figure 8.6 i... | https://ocw.mit.edu/courses/6-777j-design-and-fabrication-of-microelectromechanical-devices-spring-2007/1654337d7c64fdb073b2ab3b4aac7688_07lecture06split.pdf |
6 - 35
Storing elastic energy
> Remember calculating potential energy in physics
x
f
U
−= ∫
=
> Deforming a material stores elastic energy
> Stress = F/A, strain = ΔL/L
example,
dxF
x
(for
U
x
i
mgh
)
ε(x,y,z)
∫
0
σ(ε)dε
=
???
> Together, they contribute 1/length3: strain energy density at a
point in space
Cite as:... | https://ocw.mit.edu/courses/6-777j-design-and-fabrication-of-microelectromechanical-devices-spring-2007/1654337d7c64fdb073b2ab3b4aac7688_07lecture06split.pdf |
].
C. Livermore: 6.777J/2.372J Spring 2007, Lecture 6 - 37
Including Shear Strains
> More generally, the energy density in a linear elastic medium is
related to the product of stress and strain (both normal and
shear)
For
axial
strains
:
For
shear
strains
~
W
σε
τγ
=
1
2
1
2
strain
=
~
W
:
a
to... | https://ocw.mit.edu/courses/6-777j-design-and-fabrication-of-microelectromechanical-devices-spring-2007/1654337d7c64fdb073b2ab3b4aac7688_07lecture06split.pdf |
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
=
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
C
11
C
12
C
13
C
14
C
12
C
22
C
23
C
24
C
13
C
23
C
33
C
34
C
14
C
24
C
34
C
44
C
15
C
25
C
35
C
45
C
16
C
26
C
36
C
46
C
C
15
16
C
C
25
26
C
C
35
36
C
C
45
46
C
C
55
56
C
C
56
66
ε
ε
ε
x
y
z
γ
γ
γ
yz
zx
xy
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
Cite as: Carol ... | https://ocw.mit.edu/courses/6-777j-design-and-fabrication-of-microelectromechanical-devices-spring-2007/1654337d7c64fdb073b2ab3b4aac7688_07lecture06split.pdf |
is denoted by Sij
> Yes, the notation is cruel
> Some texts use different symbols,
but these are quite widely used in
the literature
=σ
I
=ε
I
C
ε
J
IJ
S
σ
J
IJ
∑
J
and
∑
J
Cite as: Carol Livermore, course materials for 6.777J / 2.372J Design and Fabrication of Microelectromechanical Devices, Spring 2007. MIT
OpenCo... | https://ocw.mit.edu/courses/6-777j-design-and-fabrication-of-microelectromechanical-devices-spring-2007/1654337d7c64fdb073b2ab3b4aac7688_07lecture06split.pdf |
2.372J Design and Fabrication of Microelectromechanical Devices, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
C. Livermore: 6.777J/2.372J Spring 2007, Lecture 6 - 41
Materials with Lower Symmetry
> Examples:
• Zinc oxide – 5 elastic const... | https://ocw.mit.edu/courses/6-777j-design-and-fabrication-of-microelectromechanical-devices-spring-2007/1654337d7c64fdb073b2ab3b4aac7688_07lecture06split.pdf |
3
2
1
0
-1
-2
)
s
t
i
n
u
y
r
a
r
t
i
b
r
a
(
s
s
e
r
t
S
Loading curve
Unloading curve
Strain if unloaded to zero stress
Stress if unloaded to zero strain
0
1
2
4
Strain (arbitrary units)
3
5
6
Adapted from Figure 8.8 in: Senturia, Ste
Kluwer Academic Publishers, 2001, p. 198. ISBN: 9780792372462.
phen D. Microsystem... | https://ocw.mit.edu/courses/6-777j-design-and-fabrication-of-microelectromechanical-devices-spring-2007/1654337d7c64fdb073b2ab3b4aac7688_07lecture06split.pdf |
Image by MIT OpenCourseWare.
Cite as: Carol Livermore, course materials for 6.777J / 2.372J Design and Fabrication of Microelectromechanical Devices, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
C. Livermore: 6.777J/2.372J Spring 2007, Lec... | https://ocw.mit.edu/courses/6-777j-design-and-fabrication-of-microelectromechanical-devices-spring-2007/1654337d7c64fdb073b2ab3b4aac7688_07lecture06split.pdf |
Behavior at large strains
> Using this to find the stiffness of structures
Cite as: Carol Livermore, course materials for 6.777J / 2.372J Design and Fabrication of Microelectromechanical Devices, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]... | https://ocw.mit.edu/courses/6-777j-design-and-fabrication-of-microelectromechanical-devices-spring-2007/1654337d7c64fdb073b2ab3b4aac7688_07lecture06split.pdf |
deformation.
Stretched: tensile stress
Compressive stress
> Next time!
Cite as: Carol Livermore, course materials for 6.777J / 2.372J Design and Fabrication of Microelectromechanical Devices, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
C... | https://ocw.mit.edu/courses/6-777j-design-and-fabrication-of-microelectromechanical-devices-spring-2007/1654337d7c64fdb073b2ab3b4aac7688_07lecture06split.pdf |
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