optimization Engineering

1. Optimization Methods
in Engineering Design
Day-2a

2. Course Objectives
• Learn basic optimization methods and how they are applied in
engineering design
• Use MATLAB to solve optimum engineering design problems
– Linear programming problems
– Nonlinear programming problems
– Mixed integer programing problems

3. Course Materials
• Arora, Introduction to Optimum Design, 3e, Elsevier,
(https://www.researchgate.net/publication/273120102_Introductio
n_to_Optimum_design)
• Parkinson, Optimization Methods for Engineering Design, Brigham
Young University
(http://apmonitor.com/me575/index.php/Main/BookChapters)
• Iqbal, Fundamental Engineering Optimization Methods, BookBoon
(https://bookboon.com/en/fundamental-engineering-optimization-
methods-ebook)

4. Engineering Design Examples
• Soda can design
• Insulated spherical tank
• Two-bar truss
• Three-bar overhead truss
• Cantilever column
• Stepped cantilever beam
• Coil spring
• Heat pump

5. Design Example: Soda Can Design
Problem: A soda can of 125 𝑚𝑙 capacity is to be designed to minimize
material costs (proportional to surface area. For esthetic reasons,
the height to diameter ratio should be at least 2.
Design variables: height (ℎ), diameter (𝑑)
Objective: 𝑚𝑖𝑛ℎ,𝑑 𝑓 =
1
2
𝜋𝑑2
+ 𝜋𝑑ℎ
Subject to: 𝑔1:
1
4
𝜋𝑑2ℎ − 125 = 0 (volume)
𝑔2: 2𝑑 − ℎ ≤ 0
Trial design:
Let 𝑑 = 4𝑐𝑚, ℎ = 8𝑐𝑚
Then 𝑓 = 125.66 𝑐𝑚2
𝑔1 𝑥 = −24.47; 𝑔2 𝑥 = 0

6. Design Example: Insulated Spherical Tank
Problem: choose the insulation thickness (𝑡) to minimize the life-cycle
costs of a spherical tank of radius 𝑅.
Life cycle costs: 𝑐2𝐴𝑡 + 𝑐3𝐺 + 𝑐4𝐺 ∗ 𝑝𝑤𝑓
Annual heat gain: 𝐺 = 365 × 24 × Δ𝑇 ×
𝐴
𝜌×𝑡
Surface area: 𝐴 = 4𝜋𝑅2 [𝑚2
]
Thermal resistivity: 𝜌 𝑚 ⋅ 𝑠𝑒𝑐 ⋅ °𝐶/𝐽
Equipment insulation cost: 𝑐2 $/𝑚3
Equipment refrigeration cost: 𝑐3 $/𝑊ℎ
Annual operating cost: 𝑐4 $/𝑊ℎ
Present worth factor: 𝑝𝑤𝑓 =
𝐴
𝑖
1 −
1
1+𝑖 𝑛
Note, there are no constraints in this problem

7. Design Example: Rectangular Beam
Problem: design a rectangular beam of minimum cross-section area to
withstand specified bending moment (𝑀) and shear load (𝑉).
Design variables: width (𝑏), depth (𝑑)
Objective: 𝑓 𝑏, 𝑑 = 𝑏𝑑
Constraints:
Bending stress: 𝜎 =
6𝑀
𝑏𝑑2 ≤ 𝜎𝑎
Shear stress: 𝜏 =
3𝑉
2𝑏𝑑
≤ 𝜏𝑎
Aspect ratio:
𝑑
𝑏
≤ 2
Problem specific data
𝑀 = 140 𝑘𝑁; 𝑉 = 24 𝑘𝑁;
𝜎𝑎 = 165 𝑀𝑃𝑎; 𝜏𝑎 = 50𝑀𝑃𝑎

8. Design Example: Hollow Cantilever Beam
Problem: design a hollow cantilever beam of length 𝐿 of square cross-
section to support a load 𝑃 with minimum mass (Ref: Arora, p. 18)
Design variables: width 𝑤 , thickness 𝑡
Objective: minimum 𝑚𝑎𝑠𝑠 = 𝜌𝑙𝐴 [𝐾𝑔]
Cross-section area: 𝐴 = 4𝑡 𝑤 − 𝑡 [𝑚𝑚2]
Moment of inertia: 𝐼 =
1
12
𝑤4 − 𝑤 − 2𝑡 4 [𝑚𝑚4]
Moment about neutral axis: 𝑄 =
1
8
𝑤3 − 𝑤 − 2𝑡 3 [𝑚𝑚3]
Bending stress: 𝜎 =
𝑃𝑙𝑤
2𝐼
[𝑁 ⋅ 𝑚𝑚−2]
Shear stress: 𝜏 =
𝑃𝑄
2𝐼𝑡
[𝑁 ⋅ 𝑚𝑚−2]
End deflection: 𝑞 =
𝑃𝐿3
3𝐸𝐼
[𝑚𝑚]

9. Design Example: Hollow Cantilever Beam
Assume the following parameter values:
𝑃 = 20,000 𝑁; 𝐿 = 2 𝑚; 𝐸 = 21 × 104
𝑁 ⋅ 𝑚𝑚−2
; 𝐺 = 8 × 104
𝑁 ⋅ 𝑚𝑚−2
;
𝜎𝑎 = 165 𝑁 ⋅ 𝑚𝑚−2; 𝜏𝑎 = 90 𝑁 ⋅ 𝑚𝑚−2; 𝑞𝑎 = 10 𝑚𝑚
The design optimization problem is stated as:
Objective: min
𝑤,𝑡
4𝑡(𝑤 − 𝑡)
Subject to:
𝑃𝑙𝑤
2𝐼
− 𝜎𝑎 ≤ 0
𝑃𝑄
2𝐼𝑡
− 𝜏𝑎 ≤ 0
𝑃𝐿3
3𝐸𝐼
− 𝑞𝑎 ≤ 0
𝑤 − 15𝑡 ≤ 0
Variable bounds:
50 ≤ 𝑤 ≤ 200, 5 ≤ 𝑡 ≤ 20

10. Design Example: Minimum Mass Tubular Column
Problem: Design a minimum mass tubular column of length 𝑙 to
support a load 𝑃 without buckling or overstressing (Ref: Arora, p. 40)
Design variables: outer and inner radii 𝑅0, 𝑅𝑖 , alternatively, (𝑅, 𝑡)
Objective: min 𝜌𝑙𝐴, where 𝐴 = 𝜋 𝑅0
2
− 𝑅𝑖
2
= 2𝜋𝑅𝑡
Moment of Inertia: 𝐼 =
𝜋
4
𝑅0
4
− 𝑅𝑖
4
= 𝜋𝑅3
𝑡
Constraints: 𝑃 < 𝑃cr, 𝜎 ≤ 𝜎𝑎,
𝑅
𝑡
≤ 𝑘
Axial stress: 𝜎 =
𝑃
𝐴
Buckling load: 𝑃cr =
𝜋2𝐸𝐼
4𝑙2
Allowable stress: 𝜎𝑎

11. Design Example: Tubular Column
Assume the following parameter values:
𝑃 = 10 𝑀𝑁; 𝐿 = 5𝑚; 𝐸 = 210𝐺𝑃𝑎; 𝜎𝑎 = 250𝑀𝑃𝑎
The resulting optimization problem is formulated as:
Objective: min
𝑅𝑜,𝑅𝑖
𝜋 𝑅𝑜
2 − 𝑅𝑖
2
Subject to:
𝑃
𝜋 𝑅𝑜
2−𝑅𝑖
2 𝜎𝑎
− 1 ≤ 0,
16𝑃𝐿2
𝜋3𝐸 𝑅𝑜
4−𝑅𝑖
4 − 1 ≤ 0,
𝑅𝑜+𝑅𝑖
2𝑘 𝑅𝑜−𝑅𝑖
− 1 ≤ 0
Alternative problem formulation:
Objective: min
𝑅,𝑡
2𝜋𝑅𝑡
Subject to:
𝑃
2𝜋𝑅𝑡𝜎𝑎
− 1 ≤ 0,
4𝑃𝐿2
𝜋3𝐸𝑅3𝑡
− 1 ≤ 0,
𝑅
𝑘𝑡
− 1 ≤ 0
Variable bounds: 50 ≤ 𝑅 ≤ 200, 5 ≤ 𝑡 ≤ 20

12. Design Example: Coil Spring
Problem: design a minimum mass spring to carry a given axial load 𝑃 without
material failure while satisfying minimum deflection and minimum surge
wave frequency requirements
Design variables: mean coil diameter 𝐷 , wire diameter 𝑑 , number of
active coils (𝑁)
Design equations:
Spring mass: 𝑚 =
1
4
𝑁 + 𝑄 𝜋2
𝐷𝑑2
𝜌
Load deflection: 𝑃 = 𝐾𝛿, where 𝐾 =
𝑑4𝐺
8𝐷3𝑁
Shear stress: 𝜏 =
8𝑘𝑃𝐷
𝜋𝑑3
Stress concentration factor: 𝑘 =
4𝐷−𝑑
4(𝐷−𝑑)
+ 0.615
𝑑
𝐷
Frequency of surge waves: 𝜔 =
𝑑
2𝜋𝑁𝐷2
𝐺
2𝜌

13. Design Example: Coil Spring
The optimization problem is formulated as:
Objective: min 𝑓 𝑁, 𝑑, 𝐷 = 𝑁 + 𝑄 𝐷𝑑2
Constraints: 𝜏 ≤ 𝜏𝑎, 𝜔 ≥ 𝜔0, 𝐷 + 𝑑 ≤ 𝐷0, 𝛿 =
𝑃
𝐾
≥ Δ,
Variable bounds:
𝑑𝑚𝑖𝑛 ≤ 𝑑 ≤ 𝑑𝑚𝑎𝑥, 𝐷𝑚𝑖𝑛 ≤ 𝐷 ≤ 𝐷𝑚𝑎𝑥, 𝑁𝑚𝑖𝑛 ≤ 𝑁 ≤ 𝑁𝑚𝑎𝑥
Assume the following parameter values:
𝑃 = 10 𝑙𝑏, Δ = 0.5 𝑖𝑛, 𝛾 = 0.285
𝑙𝑏
𝑖𝑛3 , 𝜔0 = 100 𝐻𝑧,
𝐷0 = 1.5 𝑖𝑛, 𝜏𝑎 = 80,000
𝑙𝑏
𝑖𝑛2 , 𝐺 = 1.15 × 107
𝑙𝑏
𝑖𝑛2 , 𝑄 = 2
0.01 ≤ 𝑑 ≤ 0.15, 0.5 ≤ 𝐷 ≤ 1.5, 1 ≤ 𝑁 ≤ 10

14. Design Example: Stepped Cantilever Beam
A fixed load 𝑃 = 50𝑘𝑁 is supported at the end of a stepped cantilever
beam of 𝐿 = 500𝑐𝑚. The sections are of equal length (𝑙 = 100𝑐𝑚).
The design objective is to minimize the mass of the beam by
reducing its total volume.
Design variables: 𝑏𝑖, ℎ𝑖, 𝑖 = 1, … , 5
Objective: min 𝑉 = 𝑙 𝑏1ℎ1 + 𝑏2ℎ2 + 𝑏3ℎ3 + 𝑏4ℎ4 + 𝑏5ℎ5

15. Design Example: Stepped Cantilever Beam
Design constraints:
Bending stress: 𝜎𝑏𝑖
< 𝜎𝑚𝑎𝑥 = 14000
𝑁
𝑐𝑚2 , 𝑖 = 1, … , 5
where 𝜎𝑏5
=
6𝑃𝑙
𝑏5ℎ5
2 , 𝜎𝑏4
=
12𝑃𝑙
𝑏4ℎ4
2 , 𝜎𝑏3
=
18𝑃𝑙
𝑏3ℎ3
2 , 𝜎𝑏2
=
24𝑃𝑙
𝑏2ℎ2
2 , 𝜎𝑏1
=
30𝑃𝑙
𝑏1ℎ1
2
End deflection: 𝛿 =
𝑃𝑙3
3𝐸
61
𝐼1
+
37
𝐼2
+
19
𝐼3
+
7
𝐼4
+
1
𝐼5
≤ 𝛿𝑚𝑎𝑥 = 2.7𝑐𝑚
where 𝐼𝑖 =
𝑏𝑖ℎ𝑖
3
12
, 𝑖 = 1, … , 5; 𝐸 = 2 × 107 𝑁
𝑐𝑚2
Aspect ratio:
ℎ𝑖
𝑏𝑖
≤ 𝑎𝑚𝑎𝑥 = 20, 𝑖 = 1, … , 5
Discrete variable ranges:
𝑏1 ∈ 1: 5 , ℎ1 ∈ 30: 5: 55 , 𝑏2, 𝑏3 ∈ 2.4, 2.6, 2.8, 3.1 ,
ℎ2, ℎ3 ∈ 45, 50, 55, 60 ,
Continuous variable bounds: 1 ≤ 𝑏4, 𝑏5 ≤ 5, 30 ≤ ℎ4, ℎ5 ≤ 55

16. Design Example: Symmetric Two-Bar Truss
Problem: a symmetrical two-bar truss is to be designed with hollow
tubes to support a fixed load 𝑃. The truss has height 𝐻, and span 𝐵.
Objective: minimum weight structure that will withstand 𝑃, and not
yield, buckle, or deflect excessively.
Design variables: diameter 𝑑 , thickness 𝑡
Alternately, choose: diameter 𝑑 , height 𝐻
𝑙 = (𝐵/2)2+𝐻2, 𝐴 = 𝜋𝑑𝑡
Total weight: 𝑊 = 2𝜌𝑙𝐴,
Axial stress: 𝜎 =
𝑃𝑙
2𝜋𝑑𝑡𝐻
Buckling stress: 𝜎𝑏 =
𝜋2 𝑑2+𝑡2 𝐸
8𝑙2
Deflection: 𝜀 =
𝑃𝑙3
2𝜋𝑑𝑡𝐻2𝐸

17. Design Example: Symmetric Two-Bar Truss
The optimum design problem is defined as:
Objective: min
d,t
𝑊 = 2𝜋𝑑𝑡𝜌 (𝐵/2)2+𝐻2
Subject to: 𝜎𝑏 ≤ 𝜎 ≤ 𝜎𝑎; 𝜀 ≤ 𝜀𝑚𝑎𝑥;
Simplify objective and scale the constraints:
Objective: min
d,t
𝑓 = 𝑑 × 𝑡
Subject to:
𝜎𝑏
𝜎
− 1 ≤ 0,
𝜎
𝜎𝑎
− 1 ≤ 0 ,
𝜀
𝜀𝑚𝑎𝑥
− 1 ≤ 0,
For a particular problem, let 𝑃 = 66,000 𝑙𝑏; 𝐻 = 30 𝑖𝑛; 𝐵 = 60 𝑖𝑛;
𝜌 = 0.3
𝑙𝑏
𝑖𝑛3
; 𝐸 = 30 × 106
𝑙𝑏
𝑖𝑛2
; 𝜎𝑎 = 1 × 105
𝑝𝑠𝑖; 𝜀𝑚𝑎𝑥 = 0.25 𝑖𝑛
Then 𝑊 = 79.97 𝑑 × 𝑡, 𝜎 = 14,855 𝑑 × 𝑡, 𝜎𝑏 = 20,562 𝑑2
+ 𝑡2
, 𝜀 =
0.0297
𝑑×𝑡
;

18. Design Example: Three-Bar Overhead Truss
Problem: A minimum-weight symmetric three-bar overhead truss is to
be designed to support a load applied at an angle (Ref: Arora, p. 46).
The truss has height 𝑙 and span 2𝑙; member 1 and 3 have length 2𝑙
and area 𝐴1; member 2 has length 𝑙 and area 𝐴2.
Load 𝑃 is applied at the joint at an angle 𝜃; the horizontal and vertical
components of the applied load are: 𝑃𝑢 = 𝑃 cos 𝜃 , 𝑃𝑣 = 𝑃 sin 𝜃.
Design variables:
cross-sectional areas, 𝐴1 and 𝐴2
Design objective:
minimize 𝑊 = 𝜌𝑙𝑔(2 2𝐴1 + 𝐴2)
Design constraints:
Stress, deflection, buckling, resonance

19. Design Example: Three-Bar Overhead Truss
The stresses in the three structural members are given as:
𝜎1 =
1
2
𝑃𝑢
𝐴1
+
𝑃𝑣
(𝐴1+ 2𝐴2)
; 𝜎2 =
2𝑃𝑣
(𝐴1+ 2𝐴2)
; 𝜎3 =
1
2
−
𝑃𝑢
𝐴1
+
𝑃𝑣
(𝐴1+ 2𝐴2)
The horizontal and vertical deflections of the load point are:
𝑢 =
2𝑙𝑃𝑢
𝐴1𝐸
, 𝑣 =
2𝑙𝑃𝑣
(𝐴1+ 2𝐴2)𝐸
The buckling load for the members in compression is given as:
𝜋2𝐸𝐼
𝑙𝑖
2 ,
where the moment of inertia is: 𝐼𝑖 = 𝛽𝐴𝑖
2
, where 𝛽 = constant.
The lowest resonance frequency of the structure is given as:
𝜁 =
3𝐸𝐴1
𝜌𝑙2(4𝐴1+ 2𝐴2)
, where 𝜌 is the mass density.

20. Design Example: Three-Bar Overhead Truss
The optimization problem is formulated as:
Objective: Minimize 𝑓 𝐴1, 𝐴2 = 2 2𝐴1 + 𝐴2
Design constraints:
Axial stress: 𝜎𝑖 ≤ 𝜎𝑎, 𝑖 = 1,2,3
Buckling stress: −𝜎3 ≤
𝜋2𝐸𝛽𝐴1
𝑙3
2 ≤ 𝜎𝑎
End deflections: 𝑢 ≤ ∆𝑢, 𝑣 ≤ ∆𝑣
Resonance frequency: 𝜁 ≥ 2𝜋𝜔0
2
Variable bounds: 𝐴1, 𝐴2 ≥ 𝐴𝑚𝑖𝑛

21. Design Example: Three-Bar Overhead Truss
For a particular problem, let 𝑙 = 1.0𝑚, 𝑃 = 50𝑘𝑁, 𝜃 = 30°, 𝜌 = 7850
𝑘𝑔
𝑚3 ,
𝐸 = 210𝐺𝑃𝑎, 𝜎𝑎 = 150𝑀𝑃𝑎, ∆𝑢= ∆𝑣= 0.5 𝑐𝑚, 𝜔0 = 50𝐻𝑧, 𝛽 = 1.0,
and 𝐴𝑚𝑖𝑛 = 2𝑐𝑚2
. Then, 𝑃𝑢 =
3𝑃
2
, 𝑃𝑣 =
𝑃
2
; after normalizing the
constraints, the optimal design problem is stated as:
Minimize 𝑓 𝐴1, 𝐴2 = 2 2𝐴1 + 𝐴2
Subject to: 2.357 × 10−4 0.866
𝐴1
+
0.5
𝐴1+ 2𝐴2
− 1 ≤ 0,
2.357×10−4
𝐴1+ 2𝐴2
− 1 ≤ 0, 2.357 × 10−4 −
0.866
𝐴1
+
0.5
𝐴1+ 2𝐴2
− 1 ≤ 0,
6.9077 × 104
𝐴1 − 1 ≤ 0, 3.4117 × 10−4 0.866
𝐴1
−
0.5
𝐴1+ 2𝐴2
− 1 ≤ 0,
5.8321×10−5
𝐴1
− 1 ≤ 0,
3.3672×10−5
𝐴1+ 2𝐴2
− 1 ≤ 0,
0.0012 4𝐴1+ 2𝐴2
𝐴1
− 1 ≤ 0,
𝐴1, 𝐴2 ≥ 0.01

22. Design Example: Heat Pump
A heat pump is to be designed to extract residual heat from outside air
and supply it indoor. Electric resistance heating is used as backup.
Design variables: 𝑡𝑐, 𝑡𝑒, 𝑡𝑎𝑐𝑜, 𝑡𝑎𝑒𝑜

23. Design Example: Heat Pump
The design problem is to choose the size of the compressor, condenser
and evaporator to minimize the life-cycle costs over 10 years.
The following assumptions are made:
– The average outside air temperature is 0°𝐶
– Evaporator inlet air temperature is 24°𝐶
– The approach temperature in the exchanger should be at least 4°𝐶.
– The airflow rate through the coils is 𝑚 = 5 𝑘𝑔/𝑠𝑒𝑐
– The pump operates for 4000 hours every year
– Compressor cost is $220/𝑘𝑊 of motor
– Cost of coils is $100/𝑚2
of the air-side area
– Power cost is $0.10/𝐾𝑊ℎ𝑟

24. Design Example: Heat Pump
The variables are defined as:
Condenser inlet air temperature, 𝑡𝑎𝑐𝑖 = 0°𝐶
Condenser outlet air temperature, 𝑡𝑎𝑐𝑜
Evaporator inlet air temperature, 𝑡𝑎𝑒𝑖 = 24°𝐶
Evaporator outlet air temperature, 𝑡𝑎𝑐𝑜
Temperature of approach for condenser, 𝑡𝑎𝑝𝑝𝑐 = 𝑡𝑐 − 𝑡𝑎𝑐𝑜
Temperature of approach for evaporator, 𝑡𝑎𝑝𝑝𝑒 = 𝑡𝑎𝑒𝑜 − 𝑡𝑒
Heat transferred in condenser, 𝑄𝑐 = 𝑚𝐶𝑝 𝑡𝑎𝑐𝑜 − 𝑡𝑎𝑐𝑖 = 𝑈𝐴𝑐Δ𝑇𝑐
Heat transferred in evaporator, 𝑄𝑒 = 𝑚𝐶𝑝 𝑡𝑎𝑒𝑖 − 𝑡𝑎𝑒𝑜 = 𝑈𝐴𝑒Δ𝑇𝑒
Log mean temperature difference across condenser,
Δ𝑇𝑐 =
𝑡𝑐−𝑡𝑎𝑐𝑖 − 𝑡𝑐−𝑡𝑎𝑐𝑜
ln 𝑡𝑐−𝑡𝑎𝑐𝑖 / 𝑡𝑐−𝑡𝑎𝑐𝑜
Log mean temperature difference across evaporator,
Δ𝑇𝑒 =
𝑡𝑎𝑒𝑖−𝑡𝑒 − 𝑡𝑎𝑒𝑜−𝑡𝑒
ln 𝑡𝑎𝑒𝑖−𝑡𝑒 / 𝑡𝑎𝑒𝑜−𝑡𝑒

25. Design Example: Heat Pump
Variable definitions:
Heat transfer coefficient, 𝑈 = .025
𝑘𝑊
𝑚2°𝐶
Heat supplied by the electric heater, 𝑄𝑟 = 𝑚𝐶𝑝 35 − 𝑡𝑎𝑐𝑜
Electrical energy input to the compressor, 𝑊 = 𝑄𝑐 − 𝑄𝑒
Coefficient of performance for compressor,
𝐶𝑂𝑃 =
𝑄𝑒
𝑊
= 7.24 + 0.352𝑡𝑒 − 0.096𝑡𝑐 − 0.0055𝑡𝑐𝑡𝑒
Capital cost of the heat pump, 𝐶ℎ𝑝 = 100 𝐴𝑐 + 𝐴𝑒 + 220𝑊
Operating cost of the heat pump, 𝐶𝑒ℎ𝑝 = 0.06 × 4000 × 𝑃𝑊𝐹 × 𝑊
Operating cost of the heater, 𝐶𝑒𝑟ℎ = 0.06 × 4000 × 𝑃𝑊𝐹 × 𝑄𝑟
Present worth factor, 𝑃𝑊𝐹(𝐴, 𝑖) =
𝐴
𝑖
1 −
1
1+𝑖 𝑛
Total cost, 𝐶𝑡𝑜𝑡𝑎𝑙 = 𝐶ℎ𝑝 + 𝐶𝑒ℎ𝑝 + 𝐶𝑒𝑟ℎ

26. Design Example: Heat Pump
Using three independent variables, an unconstrained optimization
problem is defined as:
Objective: min 𝐶𝑡𝑜𝑡𝑎𝑙(𝑡𝑐, 𝑡𝑒, 𝑡𝑎𝑒𝑜)
Alternatively, the constrained optimization problem is defined as:
Objective: min 𝐶𝑡𝑜𝑡𝑎𝑙(𝑡𝑐, 𝑡𝑒, 𝑡𝑎𝑐𝑜, 𝑡𝑎𝑒𝑜)
Subject to: 𝑄𝑐 − 𝑄𝑒 − 𝑊 = 0
Variable bounds:
35°𝐶 ≤ 𝑡𝑐 ≤ 70°𝐶, −20°𝐶 ≤ 𝑡𝑒 ≤ 20°𝐶,
0°𝐶 = 𝑡𝑎𝑐𝑖 ≤ 𝑡𝑎𝑐𝑜 ≤ 35°𝐶, 0°𝐶 ≤ 𝑡𝑎𝑒𝑜 ≤ 𝑡𝑎𝑒𝑖 = 24°𝐶

27. Graphical Optimization Method
• Before delving into the formal optimization methods, let’s examine
the graphical optimization method, which is applicable for problems
formulated with one or two variables
• The graphical method involves the following steps:
– Establish a (2D) grid for variable ranges considered
– Plot the constraint boundaries and establish the feasible region
– Using the grid plot contours (level curves) of the cost function
– Identify the minimum by inspection

28. Graphical Method for Problems in Two Variables
• Assume that the optimization problem is given as:
Objective: min
𝑥1,𝑥2
𝑓(𝑥1, 𝑥2)
Subject to: 𝑔𝑖 ≤ 0, ℎ𝑗 = 0
Variable bounds: 𝑥𝑖
𝐿
≤ 𝑥𝑖 ≤ 𝑥𝑖
𝑈
, 𝑖 = 1,2
• We use MATLAB to
– Define a grid of pairs of values over the ranges of variables
– Plot the constraint boundaries and locate the feasible region
– Plot the contours of the objective function to graphically locate a
minimum

29. Example: Soda Can Design
Optimization Problem:
Objective: 𝑚𝑖𝑛ℎ,𝑑 𝑓 ℎ, 𝑑 =
1
2
𝜋𝑑2
+ 𝜋𝑑ℎ
Subject to: 𝑔1:
1
4
𝜋𝑑2
ℎ = 125, 𝑔2: 2𝑑 − ℎ ≤ 0
Select variable bounds: 1 ≤ 𝑑 ≤ 8; 2 ≤ ℎ ≤ 16

30. Example: Soda Can Design
MATLAB Commands:
>> d=1:.05:8; h=2:.1:16;
>> [D,H]=meshgrid(d,h);
>> f=pi*D.*(D/2+H);
>> g1=pi/4*D.*D.*H-125;
>> g2=2*D-H;
>> figure, xlabel('D'), ylabel('H'), hold
>> contour(d,h, g1,[0 0],'r'),
>> contour(d,h, g2,[0 0],'r'), pause
>> [c,h]=contour(d,h,f);
clabel(c,h), hold

31. Design Example: Hollow Cylindrical Cantilever Beam
Problem: Design a minimum-mass cantilever beam of length 𝐿 with
circular cross-section (outer radius 𝑅𝑜, inner radius 𝑅𝑖), such that:
Bending stress: 𝜎 =
𝑃𝐿𝑅𝑜
𝐼
≤ 𝜎𝑎,
Shear stress: 𝜏 =
𝑃
3𝐼
𝑅𝑜
2 + 𝑅0𝑅𝑖 + 𝑅𝑖
2
≤ 𝜏𝑎,
where 𝐼 =
𝜋
4
(𝑅𝑜
4
− 𝑅𝑖
4
) is the cross-section moment of inertia
Design variables: outer radius 𝑅𝑜, inner radius 𝑅𝑖
The optimization problem is defined as:
Minimize 𝑓 𝑅0, 𝑅𝑖 = 𝜋𝜌𝐿(𝑅0
2
− 𝑅𝑖
2
)
Subject to:
𝜎
𝜎𝑎
− 1 ≤ 0,
𝜏
𝜏𝑎
− 1 ≤ 0; 𝑅0, 𝑅𝑖 ≤ 0.2𝑚

32. Design Example: Hollow Cylindrical Cantilever Beam
The following parameter values are assumed:
𝑃 = 10𝑘𝑁, 𝐿 = 5𝑚, 𝜎𝑎 = 250𝑀𝑃𝑎, 𝜏𝑎 = 90𝑀𝑃𝑎,
𝐸 = 210𝐺𝑃𝑎, 𝜌 = 7850 𝑘𝑔/𝑚3
After dropping the constant terms in 𝑓, the problem is stated as:
Minimize 𝑓 𝑅0, 𝑅𝑖 = 𝑅0
2
− 𝑅𝑖
2
Subject to:
𝑔1:
8×10−4𝑅𝑜
𝜋(𝑅0
4−𝑅𝑖
4
)
− 1 ≤ 0; 𝑔2:
4 𝑅𝑜
2+𝑅0𝑅𝑖+𝑅𝑖
2
27𝜋 𝑅0
4−𝑅𝑖
4 − 1 ≤ 0; 𝑅0, 𝑅𝑖 ≤ 20𝑐𝑚
The graphical solution obtained from the Matlab plot is: 𝑅𝑜 = 12𝑐𝑚,
𝑅𝑖 = 11.5𝑐𝑚, 𝑓∗
= 11.75 𝑐𝑚2

33. Graphical Solution
Hollow Cylindrical Cantilever Beam Design
0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2
0.02
0.04
0.06
0.08
0.1
0.12
0.14
0.16
0.18
0.2
0
.
0
0
1
0.001
0.001
0.001
0
.
0
0
2
0.002
0.002
0.002
0
.
0
0
3
0.003
0.003
0
.
0
0
4
0.004
0.004
0
.
0
0
5
0.005
0.005
0
.
0
0
6
0.006
0.006
0
.
0
0
7
0.007
0.007
0
.
0
0
8
0
.
0
0
8
0.008
0
.
0
0
9
0
.
0
0
9
0.009
0
.
0
1
0
.
0
1
0.01
X= 0.12
Y= 0.115
Level= 0.001175
Ro
Ri
Hollow Cylindrical Cantilever Beam Design
Optimal solution: 𝑅𝑜 = 0.12𝑚, 𝑅𝑖 = 0.115𝑚, 𝑓∗ = 0.001175
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