Portble Notes of material slection course

Material slection

1. CHAPTER 4 : Past Year Questions

2. SEMESTER I SESI 2011/2012
A particular design asks us to choose a
material using
For a plot of log ρ [X axis] versus logE [Y
axis], determine the slope of the selection
line
Materials Selection BDA 20402 2

3. Solution:
Take the log of both sides of the M equation
and arrange as Y = mX + C
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4. Materials Selection BDA 20402 4

5. Materials Selection BDA 20402 5

6. Materials Selection BDA 20402 6

7. Materials Selection BDA 20402 7

8. Materials Selection BDA 20402 8
SEMESTER II SESI 2012/2013

9. Solution
Step 1: Identify the objective and constraint
Step 2: Solve the deflection (constraint)
equation for the free parameter, t.
Step 3: Substitute the value for t into the mass
(objective) equation to get an equation for t
that depends only on the material properties
and design fixed parameters.
Step 4: Separate out the materials property
information from the performance equation to
find the materials selection criterion, M.
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10. Solution
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The measure of performance, P, for this design is given in the
problem statement as the feature to be minimized.
In this case, the weight or mass, m, is to be made as small
as possible (lightweight), so that
1

11. Solution
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We know the deflection from the given equation.
Solve this for the free parameter, t, and plug into
the perfomance equation:

12. SEMESTER I SESI 2012/2013
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1
2

13. Solution
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Step1:Identifytheobjective(deflection)andconstraint
(stress)
Step2:Solvethestress(constraint)equationforthefree
parameter,t.
Step3:Substitutethevaluefortintothedeflection
(objective)equationtogetanequationfortthatdepends
onlyonthematerialpropertiesanddesignfixed
parameters.
Step4:Separateoutthematerialspropertyinformation
fromtheperformanceequationtofindthematerials
selectioncriterion,M.

14. Solution
The measure of performance, P, for this
design is given in the problem statement as
the feature to be maximized (large
deflection).
In this case, deflection, δ, is to be made as
large as possible, so that
P = δ
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15. Solution
•Constraint is not fail under loading,
which is related to σ
max
•Solve free variable (thickness, t) by
using equation 1 (σ
max )
t=(∆pa
2
/2σ
max )
1/2
•Then substitute t into equation 2
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16. Solution
P = δ = 3/8 (1-v
2
) a
2
(σ
max
3/2
/E)
So, M = ???
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M =σ
max
3/2
/E

17. SEMESTER II SESI 2014/2015
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18. Solution
(a)
(b) P = ???
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FunctionBeam/Rectangular plank
ObjectiveTo maximize the length (longest as possible)
Constraint(i)W and t are fixed
(ii)Not deflect larger than max δ
P = L

19. Solution
(c) From the equation,
P = L = (4Wt
3
δE/F)
1/3
= (4Wt
3
δ/F)
1/3
E
1/3

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M = E
1/3

20. Semester 1 2015/2016
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21. Solution
(a)
(b) P = ???
*r = free variable
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FunctionBracket
ObjectiveTo minimize mass (small mass)
Constraint(i)L is fixed
(ii)Rotational deflection, θ
P = 1/m

22. Solution
From equation : θ = 4LT/πr
4
G
So, r = (4LT/πGθ)
1/4
Next, P =1/m = 1/ ρV = 1/ρπr
2
L
Substitute equation 1 into equation 2
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1
2

23. Solution
Substitute equation 1 into equation 2:
P =1/m= 1/ρπL(4LT/πGθ)
1/2


1/m = (θ/πL
3
T)
1/2
(G
1/2
/ρ)
Then G = E/2(1-v)
Therefore : M = E
1/2
/ρ
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24. Materials Selection BDA 20402 24
2018-2019 SEM 1

25. Materials Selection BDA 20402 25
Solution

26. Materials Selection BDA 20402 26
2018-2019 SEM 1

27. Materials Selection BDA 20402 27
Solution