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Exam 2 Solution - Bio-Solid Mechanics | EGM 4592, Exams of Engineering

University of Florida (UF)Engineering

Prof. Scott Arthur Banks

Material Type: Exam; Professor: Banks; Class: BIO-SOLID MECHANICS; Subject: ENGINEERING SCIENCE; University: University of Florida; Term: Spring 2007;

Typology: Exams

Pre 2010

Uploaded on 03/10/2009

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EGM4592 Biosolid Mechanics Exam 2 March 22, 2007

Take home exam: Open book, notes, calculator, computer, MATLAB. No consultation with other students. Students

submitting unusually similar answers will receive no credit. (60 points possible)

You are welcome to put computations, programs, graphs on extra sheets and staple them to this exam.

The exam is due at the beginning of class on TUESDAY, March 27.

1. Uniaxial tension tests of unstimulated myocardarium (papillary muscle) yielded the following data table and graphs:

You are considering deriving your stress/strain constitutive equation using an exponential form of a strain energy density

function: )(

1

2Ec

ecW = where c1 and c2 are constant coefficients.

a.) 10 pts - Determine the expression for stress (S) in terms of strain (E)

)(

21

2Ec

ecc

E

W

S=

∂

=

b.) 10 pts - Estimate values for the coefficients c1 and c2 from the data given.

The slope of the logarithmic plot is equal to c2. Two (x,y) pairs are given, from which you can determine c2=5.

At E=0, S=c1c2. Looking at the logarithm graph, the y-intercept is ln(c1c2)=2.303  c1c2=10  c1 = 2.

0

200

400

600

800

1000

1200

1400

1600

0 0.2 0.4 0.6 0.8 1 1.2

Input Strai n E

Measured Stress S (kPa)

0.000, 2.303

1.000, 7.303

0.00

1.00

2.00

3.00

4.00

5.00

6.00

7.00

8.00

0 0.2 0.4 0.6 0. 8 1 1.2

Input Strai n E

ln(S)

Input Strain

E

Measured

Stress S

(kPa)

ln(S)

0 10.00 2.30

0.05 12.84 2.55

0.1 16.49 2.80

0.15 21.17 3.05

0.2 27.18 3.30

0.25 34.90 3.55

0.3 44.82 3.80

0.35 57.55 4.05

0.4 73.89 4.30

0.45 94.88 4.55

0.5 121.82 4.80

0.55 156.43 5.05

0.6 200.86 5.30

0.65 257.90 5.55

0.7 331.15 5.80

0.75 425.21 6.05

0.8 545.98 6.30

0.85 701.05 6.55

0.9 900.17 6.80

0.95 1155.84 7.05

1 1484.13 7.30

Discover Exams of Engineering University of Florida (UF)

Partial preview of the text

Download Exam 2 Solution - Bio-Solid Mechanics | EGM 4592 and more Exams Engineering in PDF only on Docsity!

EGM4592 Biosolid Mechanics Exam 2 March 22, 2007

Take home exam: Open book, notes, calculator, computer, MATLAB. No consultation with other students. Students

submitting unusually similar answers will receive no credit. (60 points possible)

You are welcome to put computations, programs, graphs on extra sheets and staple them to this exam.

The exam is due at the beginning of class on TUESDAY, March 27.

1. Uniaxial tension tests of unstimulated myocardarium (papillary muscle) yielded the following data table and graphs:

You are considering deriving your stress/strain constitutive equation using an exponential form of a strain energy density

function:

( ) 1

W = c ec^2 E where c

1 and c 2 are constant coefficients.

a.) 10 pts - Determine the expression for stress (S) in terms of strain (E)

cc ec^2 E

E

W

S =

b.) 10 pts - Estimate values for the coefficients c 1 and c 2 from the data given.

The slope of the logarithmic plot is equal to c 2. Two (x,y) pairs are given, from which you can determine c 2 =5.

At E=0, S=c 1 c 2. Looking at the logarithm graph, the y-intercept is ln(c 1 c 2 )=2.303 c 1 c 2 =10 c 1 = 2.

Input Strain E

Measured Stress S (kPa)

0.000, 2.

1.000, 7.

0 0.2 0.4 0.6 0.8 1 1. Input Strain E

ln(S)

Input Strain

E

Measured

Stress S

(kPa)

ln(S)

c.) 5 pts - Does this model seem a reasonable fit to the data?

It is a perfect fit to the data.

d.) 5 pts - Do the data seem realistic?

Data are incredibly, perhaps unrealistically, clean and noise free.

Data show a stiffness that is uncharacteristically high for muscle.

There is a nonzero stress at zero strain.

2. Cyclic compression testing of a sample of aorta (vessel opened, spread out flat and compressed in a direction normal to

the vessel wall) gives the following test results:

Unloading Loading

Input Strain Enn

Measured Stress Snn (kPa)

Measured Stress Snn (kPa) 0 0.37 0. 0.05 0.84 0. 0.1 1.19 1. 0.15 1.41 2. 0.2 2.27 3. 0.25 2.74 4. 0.3 3.51 5. 0.35 4.75 6. 0.4 5.41 8. 0.45 6.47 9. 0.5 7.82 11. 0.55 8.99 12. 0.6 10.55 14. 0.65 11.83 16. 0.7 13.21 19. 0.75 14.72 21. 0.8 16.69 23. 0.85 18.48 26. 0.9 20.36 28. 0.95 22.06 31. 1 24.36 34.

Input Strain Enn

Measured Stress Snn (kPa)

Loading Unloading

Your job is to develop constitutive equation(s) for this biosolid under these conditions.

a.) 5 pts - What simplifying assumptions are required to allow you to utilize a pseudo-elasticity approach?

From HW#3:” Pseudoelasticity – a dramatic simplification of the quasi-linear viscoelasticity model that applies only when the material can be preconditioned to obtain a repeatable stress/strain response with constant strain rates. Pseudoelasticity assigns one material model for the loading response and a different material model for the unloading response. This is particularly useful for biosolids because they often exhibit (stress/strain) hysteresis behavior that is relatively insensitive to strain rate.”

Assumptions: Preconditioned and constant strain rate.

A_loading=E_inv*S_loading % Compute estimates of constant coefficient parameters

A_unloading=E_inv*S_unloading % Compute estimates of constant coefficient parameters

% Now that we have estimates of the constant coefficients, we can determine

% the predicted values of the stresses for the same combination of input

% strains, and plot those results with the 'data' plot.

S_loading_predicted=2A_loading(1)Enn+3A_loading(2)Enn.^2; % Compute stress values based on model

S_unloading_predicted=2A_unloading(1)Enn+3A_unloading(2)Enn.^2;

clf % Clear any plots

plot(Enn,S_loading,'ro','MarkerSize',2) % Plot one set of data with unfilled red circles of size 2

xlabel('Input Strain') % X-axis label

ylabel('Measured Stress') % Y-axis label

hold % Hold these plot settings for additional plots

plot(Enn,S_unloading,'bo','MarkerSize',2) % Plot other set of data with unfilled blue circles of size 2

plot(Enn,S_loading_predicted,'-r') % Plot synthesized data with a solid red line

plot(Enn,S_unloading_predicted,'-b') % Plot synthesized data with a solid blue line

legend('Data Loading','Data Unloading','Model Loading','Model Unloading','location','southeast') % Add legend

And my plot looks like this:

I get values for A_unloading=[3.3783, 5.8369]

and A_loading=[5.6329, 7.6629]. The actual

values for the equations (which I corrupted with

Gaussian distributed random noise) were

A_unloading=[3, 6] and A_loading=[5, 8].

If you used EXCEL’s ‘Show Trendline’ function in the

plotting tool for a polynomial fit, you get the graph to the

right. This gives pretty much the same results (much more

easily), except a constant offset (y-intercept) term is

introduced.

So why bother with MATLAB? What happens when life

does not conform to something built-in to EXCEL – now

you at least have some idea of how to set up a problem in

MATLAB.

Remember what R^2 means: R^2 can be interpreted as the

percentage of data explained by your model. If R^2 = 0.

then your model only explains 50% of the variance in the

data. Obviously, R^2 = 0.9999 is a pretty good data fit!

(^00) 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

5

10

15

20

25

30

35

Input Strain

Measured Stress Data Loading Data Unloading Model Loading Model Unloading

y = 18.353x^2 + 5.6282x + 0. R^2 = 0.

y = 23.28x^2 + 10.676x + 0. R^2 = 0.

0

5

10

15

20

25

30

35

40

0 0.2 0.4 0.6 0.8 1 Input Strain Enn

Measured Stress Snn (kPa)

Loading Unloading

Poly. (Unloading) Poly. (Loading)

Exam 2 Solution - Bio-Solid Mechanics | EGM 4592, Exams of Engineering

Related documents

Partial preview of the text

Download Exam 2 Solution - Bio-Solid Mechanics | EGM 4592 and more Exams Engineering in PDF only on Docsity!

EGM4592 Biosolid Mechanics Exam 2 March 22, 2007

Take home exam: Open book, notes, calculator, computer, MATLAB. No consultation with other students. Students

submitting unusually similar answers will receive no credit. (60 points possible)

You are welcome to put computations, programs, graphs on extra sheets and staple them to this exam.

The exam is due at the beginning of class on TUESDAY, March 27.

1. Uniaxial tension tests of unstimulated myocardarium (papillary muscle) yielded the following data table and graphs:

You are considering deriving your stress/strain constitutive equation using an exponential form of a strain energy density

function:

W = c ec^2 E where c

1 and c 2 are constant coefficients.

a.) 10 pts - Determine the expression for stress (S) in terms of strain (E)

cc ec^2 E

E

W

S =

b.) 10 pts - Estimate values for the coefficients c 1 and c 2 from the data given.

The slope of the logarithmic plot is equal to c 2. Two (x,y) pairs are given, from which you can determine c 2 =5.

At E=0, S=c 1 c 2. Looking at the logarithm graph, the y-intercept is ln(c 1 c 2 )=2.303 c 1 c 2 =10 c 1 = 2.

Input Strain

E

Measured

Stress S

(kPa)

ln(S)

c.) 5 pts - Does this model seem a reasonable fit to the data?

It is a perfect fit to the data.

d.) 5 pts - Do the data seem realistic?

Data are incredibly, perhaps unrealistically, clean and noise free.

Data show a stiffness that is uncharacteristically high for muscle.

There is a nonzero stress at zero strain.

2. Cyclic compression testing of a sample of aorta (vessel opened, spread out flat and compressed in a direction normal to

the vessel wall) gives the following test results:

Your job is to develop constitutive equation(s) for this biosolid under these conditions.

a.) 5 pts - What simplifying assumptions are required to allow you to utilize a pseudo-elasticity approach?

Assumptions: Preconditioned and constant strain rate.

A_loading=E_inv*S_loading % Compute estimates of constant coefficient parameters

A_unloading=E_inv*S_unloading % Compute estimates of constant coefficient parameters

% Now that we have estimates of the constant coefficients, we can determine

% the predicted values of the stresses for the same combination of input

% strains, and plot those results with the 'data' plot.

S_loading_predicted=2A_loading(1)Enn+3A_loading(2)Enn.^2; % Compute stress values based on model

S_unloading_predicted=2A_unloading(1)Enn+3A_unloading(2)Enn.^2;

clf % Clear any plots

plot(Enn,S_loading,'ro','MarkerSize',2) % Plot one set of data with unfilled red circles of size 2

xlabel('Input Strain') % X-axis label

ylabel('Measured Stress') % Y-axis label

hold % Hold these plot settings for additional plots

plot(Enn,S_unloading,'bo','MarkerSize',2) % Plot other set of data with unfilled blue circles of size 2

plot(Enn,S_loading_predicted,'-r') % Plot synthesized data with a solid red line

plot(Enn,S_unloading_predicted,'-b') % Plot synthesized data with a solid blue line

legend('Data Loading','Data Unloading','Model Loading','Model Unloading','location','southeast') % Add legend

And my plot looks like this:

I get values for A_unloading=[3.3783, 5.8369]

and A_loading=[5.6329, 7.6629]. The actual

values for the equations (which I corrupted with

Gaussian distributed random noise) were

A_unloading=[3, 6] and A_loading=[5, 8].

If you used EXCEL’s ‘Show Trendline’ function in the

plotting tool for a polynomial fit, you get the graph to the

right. This gives pretty much the same results (much more

easily), except a constant offset (y-intercept) term is

introduced.

So why bother with MATLAB? What happens when life

does not conform to something built-in to EXCEL – now

you at least have some idea of how to set up a problem in

MATLAB.

Remember what R^2 means: R^2 can be interpreted as the

percentage of data explained by your model. If R^2 = 0.

then your model only explains 50% of the variance in the

data. Obviously, R^2 = 0.9999 is a pretty good data fit!