MECH4301 2007 Lecture # 9 Conflicting
Constraints 1/23
Advanced Methods in Materials Selection
•Conflicting Constraints
•Lecture 9 & Tutorial 4
• Conflicting Objectives
•Lecture 10 & Tutorial 5
MECH4301 2007 Lecture # 9 Conflicting
Constraints 2/23
Tutorial 4: E7.2 and E7.3
Due Oct 1
Lecture 9: Chapters 9
& 10
MECH4301 2007 Lecture # 9 Conflicting
Constraints 3/23
Outline Lecture 9
•Conflicting constraints:
• “most restrictive constraint wins”
• Case study 1: Stiff / safe / light column
• Case study 2: Safe (no yield - no fracture)/ light air tank for a truck
MECH4301 2007 Lecture # 9 Conflicting
Constraints 4/23
Multiple Constraints and Objectives • Design with multiple constraints
• Design with multiple objectives
One Objective:the performance metric
Rank by performance
metric
One Constraint
Many Constraints O
ne
Co
nst
rain
t
Man
y C
on
stra
ints
Mu
ltip
le O
bje
ctiv
es:
seve
ral p
erfo
rman
ce
met
rics
Trade-off and value function
method
Rank by most restrictive
performance metric
Function
Combinationof
methods
One Objective:one performance
metric
Rank by performance
metric
One Constraint
Many Constraints O
ne
Co
nst
rain
t
Man
y C
on
stra
ints
Mu
ltip
le O
bje
ctiv
es:
seve
ral p
erfo
rman
ce
met
rics
Penalty function method
Rank by most restrictive
performance metric
Function
Combinationof
methods
Minimise mass
Carry force Fwithout yielding,
given length
Tie rod
Simplest case:
Design with one objective, meeting a single constraint
Or several non-conflicting constraints, such as melting point, corrosion resistance, etc.
MECH4301 2007 Lecture # 9 Conflicting
Constraints 5/23
One Objective:one performance metric
Rank by performance
metric
One Constraint
Conflicting Constraints O
ne
Co
nst
rain
t
Co
nfl
icti
ng
Co
nst
rain
ts
Co
nfl
icti
ng
Ob
ject
ives
:co
nfl
icti
ng
per
form
ance
met
rics
Penalty function method
Rank by most restrictive
performance metric
Function
Combinationof
methods
One notch up in complexity: Single objective / Conflicting Constraints
Most designs are over-constrained: “Should not deflect more than something, must not fail by yielding, by fatigue, by fast-fracture …” more constraints than free variables
Minimise mass
No yield &Given
deflectionNo corrosionTmax > 100C.
The most restrictive constraint determines the performance metric (mass)
Tie rodLecture 10
MECH4301 2007 Lecture # 9 Conflicting
Constraints 6/23
Q7.1. Materials for a stiff, light tie-rod Constraint # 1
Minimise mass m: m = A L (2)
Objective
• Length L is specified• Must not stretch more than
Constraints
• Material choice• Section area A
Free variables
Equation for constraint on A: = L = L/E = LF/AE (1)
Strong tie of length L and minimum mass
L
FF
Area A
Tie-rod Function
m = massA = areaL = length = densityE= elastic modulus = elastic deflection
EL
Fm
21
Performance metric m1
Eliminate A in (2) using (1):
Chose materials with largest M1 =
E
MECH4301 2007 Lecture # 9 Conflicting
Constraints 7/23
Q7.1. Materials for a strong, light tie-rod Constraint # 2
Minimise mass m: m = A L (2)
Objective (Goal)
• Length L is specified• Must not fail under load F
Constraints
• Material choice• Section area A
Free variables
Equation for constraint on A: F/A < y (1)
Strong tie of length L and minimum mass
L
FF
Area A
Tie-rod Function
m = massA = areaL = length = density = yield strengthy
y
FLm
2
Performance metric m2
Eliminate A in (2) using (1):
Chose materials with largest M2 =
y
MECH4301 2007 Lecture # 9 Conflicting
Constraints 8/23
Q7.1: Conflicting Constraints: Strong /Stiff / Light Tie Rod
Ran
k by
the
mor
e re
stric
tive
of t
he t
wo,
mea
ning
…?
Eva
luat
e co
mp
etin
g c
on
stra
ints
an
d p
erfo
rman
ce m
etri
cs:
Max
. de
flect
ion
Mus
t no
t yi
eld
yσσ
%1.0 %;1EL
= d
efle
ctio
n
y =
yie
ld s
tren
gth
E =
ela
stic
mod
ulus
Competingperformance metrics
y
FLm
2
EL
Fm
21
Requires stiffer material
Stiffness constraint
Strength constraint
MECH4301 2007 Lecture # 9 Conflicting
Constraints 9/23
density E TS m1 (E) m2 (TS)Mg m-3 GPa MPa grams grams
Epoxy/Aramid Fibre, 1.38 80 1400 34.5 9.9Glass/Epoxy U-D Composite 1.95 45 1100 86.7 17.7Low alloy steel 7.9 217 2445 72.8 32.3Epoxy Carbon Fiber (SMC) 1.7 150 345 22.7 49.3Al 6061-T6 2.73 74 320 73.8 85.3Titanium 4.45 117 1138 76.1 39.1
Analytical solution in three steps: Rank by the more restrictive of the constraints
y
FLm
2
EL
Fm
211. Calculate m1 and m2 for given L and F
3. Find the smallest of the larger ones
The most restrictive constraint requires a larger mass and thus becomes the controlling or active constraint.
2. Find the largest of every pair of m’s
MECH4301 2007 Lecture # 9 Conflicting
Constraints 10/23
0 0.02 0.04 0.06length (m )
0
0.02
0.04
0.06
0.08
mas
s (k
g)
m E = m
= 2700 Mg/m 3 E = 69 GPaF = 10000 Nt = 300 MPaL = 20 mm; /L = 0.1%; /L = 1%
m
m E
/L = 0.1%
m E
/L = 1%
Graphical version of the analytical solution (for Aluminium)
mass
length
EL
Fm
21
y
FLm
2
y constraint active (heavier)
E constraint active (heavier) (long rod stretches too much)
/L= 1%: Strength constraint always active
Solution for /L= 1%
Less demanding E constraint => thinner rod
MECH4301 2007 Lecture # 9 Conflicting
Constraints 11/23
Graphical solution using indices and bubble charts.
More general/powerful.Allows for a visual while physically based selection.Involves all available materials.Incorporates geometrical constraints through
coupling factors.
Pros to the graphical analytical solution : it makes explicit the dependence on L and L/.Cons: it is specific to the material considered (requires a dedicated graph per material)
MECH4301 2007 Lecture # 9 Conflicting
Constraints 12/23
MECH4301 2007 Lecture # 9 Conflicting
Constraints 13/23
Graphical solution using Indices and Bubble charts
2
1
1
12221 M
FLM
LF
FLE
LF
mmy
21 ML
M
M2 =
y
M1 =
E
y
FLm
2
EL
Fm
21
make m1 = m2
Solve for M1
L
MLogM log)2()1log(Straight line, slope = 1 y-intcpt = L/
factorcouplingL
This is what we
know
MECH4301 2007 Lecture # 9 Conflicting
Constraints 14/23
MECH4301 2007 Lecture # 9 Conflicting
Constraints 15/23
Materials for High-Performance Con-Rods
MECH4301 2007 Lecture # 9 Conflicting
Constraints 16/23
MECH4301 2007 Lecture # 9 Conflicting
Constraints 17/23
MECH4301 2007 Lecture # 9 Conflicting
Constraints 18/23
Density / Fatigue strength at 10^7 cycles10 100 1000
Den
sity
/ (
31.6
2 *
Youn
g's
mod
ulus
0
.5 )
10
100
1000
Wrought magnesium alloys
Titanium alloys
Cast magnesium alloys
Age-hardening wrought Al-alloys
Cast iron, gray
High F/L2
LowF/L2
MECH4301 2007 Lecture # 9 Conflicting
Constraints 19/23
Density / Fatigue strength at 10^7 cycles1 10 100 1000 10000
Den
sity
/ (
31.6
2 *
Youn
g's
mod
ulus
0
.5 )
10
100
1000
Beryllium, grade I-250, hot isostatically pressed
Beryllium-aluminum alloy, beralcast 310, cast
Titanium, alpha-beta alloy, Ti-6Al-4V, cast
Magnesium, AZ61, wrought
Aluminum, 6061, wrought, T6
Magnesium, EA55RS, wrought
MECH4301 2007 Lecture # 9 Conflicting
Constraints 20/23
MECH4301 2007 Lecture # 9 Conflicting
Constraints 21/23
Yield strength (elastic limit) / Density *1e31 10 100 1000
Young's
modulu
s /
Densi
ty *
1e6
100
1000
10000
100000
Select with box on line of y-intercept = 100
Polyester SMC (Low Density)
Phenolic/E-Glass Fiber, Woven Fabric Composite, Quasi-isotropic Laminate
Beech (Fagus grandifolia) (l)
SiC/SiC Fiber, 35-45Vf - Woven Laminate
selection includes several fibre reinforced composites more complaint than in the other case. Timbers are included.
Cyanate Ester/HM Carbon Fiber, UD Composite, 0° Lamina
Phenolic/E-Glass Fiber, Woven Fabric Composite, Quasi-isotropic Laminate
E 7.1 tie rod Graphical solution (/L = 1% L/ = 100) Use level 3, exclude ceramics
y
2M
EM1
Simultaneously Maximise M1 and M2
m1 = m2
m1 < m2
m2 < m1
Coupling line for L/ = 100
MECH4301 2007 Lecture # 9 Conflicting
Constraints 22/23Yield strength (elastic limit) / Density *1e3
1 10 100 1000
You
ng's
mod
ulus
/ D
ensi
ty *
1e6
100
1000
10000
100000
Balsa (l) (ld)
Spruce (Picea abies) (l)
SiC/SiC Fiber, 35-45Vf - Woven LaminateBeryllium, grade I-250, hot isostatically pressed
Cyanate Ester/HM Carbon Fiber, UD Composite, 0° Lamina
Epoxy/HS Carbon Fiber, UD Composite, 0° Lamina
SiC/SiC Fiber, 35-45Vf - Quasi-isotropic Laminate Beryllium,
selection includes strong, light and very stiff composites like CFRP. Metals are generally excluded but for Be.
Select with box on line of y-intercept = 1000
E7.1 Tie Rod Graphical solution ( /L = 0.1% L/=1000) Use level 3, exclude ceramics
E
M1
y
2M
Coupling line for L/ = 1000
Coupling line for L/ = 100
Active Constrain
t? 3-D view
MECH4301 2007 Lecture # 9 Conflicting
Constraints 23/23
z = m ax(log10(1/x),log10(1/y))3-D view of the interacting constraints
E
M1
y
2M
m1 = m2
m1 > m2
m2 > m1
m2m1
lighter
•m1 = m2 on the coupling line. •The closer to the bottom corner, the lighter the component. •Away from the coupling line, one of the constraints is active (larger m)
Locus of coupling line depends on coupling factor
MECH4301 2007 Lecture # 9 Conflicting
Constraints 24/23
Yield strength (elastic limit) / Density *1e31 10 100 1000
Young's
modulu
s /
Densi
ty *
1e6
100
1000
10000
100000
Select with box on line of y-intercept = 100
Polyester SMC (Low Density)
Phenolic/E-Glass Fiber, Woven Fabric Composite, Quasi-isotropic Laminate
Beech (Fagus grandifolia) (l)
SiC/SiC Fiber, 35-45Vf - Woven Laminate
selection includes several fibre reinforced composites more complaint than in the other case. Timbers are included.
Cyanate Ester/HM Carbon Fiber, UD Composite, 0° Lamina
Phenolic/E-Glass Fiber, Woven Fabric Composite, Quasi-isotropic Laminate
Graphical solution (deflection = 1% L/=100)
y
2M
EM1
Coupling line for L/ = 100
m1 = m2
m2 < m1
m1 < m2
lighter
MECH4301 2007 Lecture # 9 Conflicting
Constraints 25/23
Case Study # 2: Quite Similar to E7.2, Air cylinder for a truckDesign goal: lighter, safe air cylinders for trucks
Compressed air tank
MECH4301 2007 Lecture # 9 Conflicting
Constraints 26/23
Case study: Air cylinder for truck
Free variables
Function Pressure vessel
Objective Minimise mass Constraints Dimensions L, R, pressure p, given
Safety: must not fail by yielding Safety: must not fail by fast fracture
Must not corrode in water or oil Working temperature -50 to +1000C
Wall thickness, t; choice of material
t
L
2R
Den
sity
Y
ield
str
engt
h y
Fra
ctur
e to
ughn
ess
K1
c
Pre
ssur
e p
Conflicting constraints lead to competing
performance metrics
MECH4301 2007 Lecture # 9 Conflicting
Constraints 27/23
Air cylinder for truck
t
L
2R
Den
sity
Y
ield
str
engt
h y
Fra
ctur
e to
ughn
ess
K1c
Pre
ssur
e p
)tR4LtR2(m 2
Ob
ject
ive:
m
ass
Vol of material in cylinder wall
f
2 SpLR2mEliminate t
f
2 SpLR2
m*m
Str
ess
in c
ylin
der
wal
l
St
Rp f
2
Failure stress
Safety factor
Asp
ect
rat
io,
)LR2
1(LtR2
May be either y or f transpose
What is the free variable?
MECH4301 2007 Lecture # 9 Conflicting
Constraints 28/23
Air cylinder : graphical solution using CES charts
CE
S S
tag
e 1;
ap
ply
sim
ple
(n
on
co
nfl
icti
ng
) co
nst
rain
ts:
wor
king
tem
p up
to
1000 C
, re
sist
org
anic
sol
vent
s et
c.R
ank
by t
he m
ore
rest
rictiv
e of
the
tw
o
y
m
*1
aKm
c /
*1
2
CE
S S
tag
e 2:
eva
luat
e co
nfl
icti
ng
per
form
ance
met
rics
:
Mus
t no
t yi
eld:
Mus
t no
t fr
actu
re
yf 1
a
Kc
f 1
2
S =
saf
ety
fact
or
a =
cra
ck le
ng
th
y =
yie
ld s
tren
gth
K1c
= F
ract
ure
to
ug
hn
ess
Competingperformance metrics for minimum mass
MECH4301 2007 Lecture # 9 Conflicting
Constraints 29/23
Air cylinder - Simple (non- conflicting) constraints
CES Stage 1: •Impose constraints on corrosion in organic solvents•Impose constraint on maximum working temperature
Max service temp= 373 K (1000C)
Organic SolventsGood Very Good
Max
imum
Ser
vice
Tem
pera
ture
(K)
1000
Air cylinder
Low Carbon Steel
Wrought Al 1080-0
Wrought Al 2014, T4
Malleable cast iron
LA steel, AISI 4140 (normalised)
Epoxy - Glass Fibre
Epoxy - carbon
Corrosion resistance in organic solvents
K service .max T
Corrosion resistance
Select above this line
MECH4301 2007 Lecture # 9 Conflicting
Constraints 30/23
Density / Yield strength (elastic limit) / 1e30.01 0.1 1
Densi
ty/
1000/F
ract
ure
toughness
*((
3.1
4*0.0
01)^
.5)
0.01
0.1
1
10
Titanium alloysNon age-hardening wrought Al-alloys
Copper
Wrought magnesium alloys
CFRP, epoxy matrix (isotropic)
Low alloy steelMedium carbon steel
Age-hardening wrought Al-alloysPolyamides (Nylons, PA) 2
1 and
1
1plot 21 M
m M
m * *
Air cylinder - Conflicting constraints
y
m*
1
aK IC
m
*2
*m*m 21
*m*m 12 *2
* 1 mm
CES Stage 2: Find most restrictive constraint using Material Indices chart
Results so far:• Epoxy/carbon fibre composites• Epoxy/glass fibre composites• Low alloy steels• Titanium alloys• Wrought aluminium alloy• Wrought austenitic stainless steels• Wrought precipitation hardened stainless steels
Lighter this way
IC
KmE
*2
:2.7
MECH4301 2007 Lecture # 9 Conflicting
Constraints 31/23
Summary
• Real designs are over-constrained and many have multiple objectives
• Method of maximum restrictiveness copes with conflicting multiple constraints
• Analytical method useful but depends on the particular conditions set and lacks the visual power of the graphical method
End of Lecture 9
• Graphical method produces a more general solution
• Next lecture will solve air cylinder problem again for two conflicting objectives: e.g., weight and cost.
MECH4301 2007 Lecture # 9 Conflicting
Constraints 32/23
There is a typographical error in
textbook, Exercise E7.2, p. 581, 8 lines from
the bottom
a / Ic
K
a Ic
KIt reads:
It should read:
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