### Materials?

```Young’s Modulus:
Statistical Analysis
Khalid Alamri
Mohammed Alzayer
Andres Glasener
Mat E 316
April 28th, 2014
Outline
• Basics
• Objectives
• Experiment
o Apparatus
o Procedures
• Statistical Analysis:
o List of Factors
• Blocking Factors
• Experimental Factors
• Response Variable
o Modeling Deflection
o Application of Model
• Conclusion and Suggestions
Basics
• Stress and strain are related by: σ= E ε
• σ: Applied stress (Pa) usually in MPa
ε: Strain resulted from a load (mm/mm)
E: Young’s modulus (Pa) usually in GPa
• E is the slope of the elastic regime of a
stress-strain curve.
• The steeper the slope, the stiffer the
material
Objectives
• Purpose of experiment? find E for 4
samples of rods.
• Materials? steel, glass, plastic, wood.
• How? by measuring deflection resulted
• Purpose of this study? come up with a
well fit model that describes the
deflection.
Experimental Procedures1
Dial
and span
Supports
length were
measured,
and the
apparatus was
set-up.
Hook
2. The dial on
the apparatus
was calibrated
at a weight of
zero.
Rebecca. Mat E 215L – Young’s modulus. 2012. Web.
Sample
Bucket
Experimental Procedures1
3. Max affordable weight for each rod was
calculated. Given max stresses of:
Glass: 60 MPa Plastic: 60 MPa
Steel: 200 MPa Wood: 100 MPa
4. Weight of bucket was measured. First
deflection was recorded with empty
bucket.
5. Weight was added to the bucket, in the
form of water. Deflection was recorded
for 5 different weights.
Rebecca. Mat E 215L – Young’s modulus. 2012. Web.
List of Factors
Factors
Operator
Material
Type of factor
Blocking factor
Levels
1) Khalid
2) Mohammed
3) Andres
1) Steel
Experimental variable 2) Glass
(Nominal)
3) Plastic
4) Wood
List of Factors
Factors Units Type of factor
m
Experimental
variable
(Continuous)
Levels
1)
Khalid
Steel: 0.002
Glass: 0.002
Plastic: 0.004
Wood: 0.0025
2)
Mohammed
Steel: 0.002325
Glass: 0.002
Plastic: 0.00475
Wood: 0.003075
3)
Andres
Steel: 0.0024525
Glass: 0.002025
Plastic: 0.004775
Wood: 0.00396
List of Factors
Factors
Units Type of factor
Span length (L)
m
N
Deflection (d)
m
Experimental
variable
(Continuous)
Levels
1)
0.293
2)
0.285
3)
0.295
Each level corresponds
to an operator.
59
Response
variable
N/A
Modeling Deflection
• One model for the entire data? not
possible.
• Why? not enough data. Materials behave
differently.
• What to do? model each material
separately.
and interactions. Better fit.
Modeling Deflection
Material
Equation
Other
terms
R2
Steel
-0.0074 + 0.58*r +
0.020*L + 8.99E-5*P +0.0013*[(L
+ 0.08*[(r-0.29)(P0.99
0.0023)*(P-10.13)] + 10.13)]
1.44E-6*(P-10.13)2
0.98
Glass
+0.0058*[(P
-3.93)*(L0.99
0.29)]
0.98
-0.027 + 19.51*r 0.043*L + 6.15E-4*P
Modeling Deflection
Material
Equation
Other
terms
Plastic
-0.16 + 3.72*r + 0.50*L +
3.02E-4*P + 0.020*[(r+1.76E-6
0.0045)*(P-19.86)] +
*(P0.014*[(L-0.29)*(P19.86)2
19.86)]
0.99
0.98
Wood
3.58E-4*P - 8.33E4*(Andres) - 1.24E- 6.38E-4 4*[(Andres)*(P-11.75)] + 3.64E0.97
1.47E-4*[(Khalid)*(P4*(Khalid)
11.75)]
0.96
Application of Models
• Stress and strain equations derived from
the geometry of the apparatus:
• Deflections were calculated using the
new models. They were plugged into
the strain equation. Stresses were
also calculated using raw data.
• Stress versus strain plots were obtained
using Excel.
Application of Models
Original E Predicted Original New R2
(GPa)
E (GPa)
R2
Material
Operator
Steel
Mohammed
221.116
222.751
0.9847
Khalid
576.696
590.704
0.9727 0.9963
Andres
167.675
168.483
0.9918 0.9966
Mohammed
65.96
66.122
0.9981
1
Khalid
66.004
66.56
0.9882
1
Andres
62.256
63.471
0.9837
1
Glass
0.992
Application of Models
Original E Predicted Original New R2
(GPa)
E (GPa)
R2
Material
Operator
Plastic
Mohammed
4.1277
4.3817
0.9407 0.9986
Khalid
11.876
11.934
0.9903 0.9951
Andres
3.2087
3.27
0.998
0.9987
Mohammed
20.453
20.471
0.9991
1
Khalid
32.562
33.865
0.9615
1
Andres
11.651
11.875
0.9995
1
Wood
Stress-Strain Graphs
Glass
200
180
160
140
120
100
80
60
40
20
0
52.5
Stress, σ, MPa
Stress, σ, MPa
Steel
47.5
42.5
37.5
0
32.5
0.0003
0.0002 0.0004 0.0006 0.0008
Strain, ɛ
Plastic
160
120
Stress, σ, MPa
Stress, σ, MPa
140
100
80
60
40
20
0
0
0.001
0.002
Strain, ɛ
0.0007
Strain, ɛ
Wood
-0.001
0.0005
0.003
0.004
50
45
40
35
30
25
20
15
10
5
0
0
0.005
Strain, ɛ
0.01
Conclusions
• How accurate are the models? Pretty
accurate. Better stress-strain curve fit.
• Same material, different E, why? some
samples were plastically deformed, i.e.
cold worked when they were tested.
Figure: Real steel stress-strain curves.
• Steel Example:
2
Operator
E (GPa)
Mohammed
222.751
Khalid
590.704
Andres
168.483
Callister, William. Materials science and engineering: an introduction. 2010. Print.
2
Conclusions
• Operator effect? No effect. Wood is an
exception according to JMP.
• Why? It has low degree of structural
consistency, due to its composite nature.
• JMP emphasized effect of operator
why? Operator does not really affect
results. It’s JMP way to say wood
behavior is difficult to predict.
Suggestions
1. Emphasizing the role of interactions:
– Big variations in radii and lengths in plastic
data makes its model the best. It has more
effective interactions than other materials.
– Maximize the differences between a factor’s
levels for better results in future.
2. Decreasing the magnitude of applied