Biomechanics

Closed book with one 8.5x11 inch sheet of notes (and Midterm #1 notes) allowed.

Total 100 pts

1) PRINT YOUR NAME on this test sheet and any sheet you detach.

2) TRY TO WRITE ALL YOUR ANSWERS ON THESE SHEETS

· Generally, ample space has been provided on the front of your exam sheets. If you need additional space to work out a problem, please use the back of these sheets.

· If you still need more space or if you strongly prefer, you can write your answers in a blue test booklet. Information entered in a blue test booklet will also be graded.

· If you remove the staple, be certain to write your name on each test sheet in case they are somehow separated/misplaced.

· If you use a blue test booklet, be certain to write your name on each booklet in case they are somehow separated from your original exam sheets.

3) You do not need to re-write information given on these test sheets. Just present your solution in a clear and organized manner.

5) SHOW YOUR WORK! Even some multiple choice or short answer questions can receive partial credit for showing your work. No credit will be given for analysis problems without showing appropriate equations/work.

6) Uphold academic integrity professional ethics (i.e. do not cheat!). Read and sign the academic integrity statement below! Giving or receiving aid from another student during the exam will result in a zero score for the exam and will be reported to the Associate Dean of Engineering.

I agree that I will not will not give or receive aid during this examination, and that I will not violate any rules of the examination regarding use of notes or other resources. I also agree that I will take an active part in seeing to it that I and others uphold the spirit and letter of academic integrity, which is integral to the engineering professional code of ethics!

Signed

Section 1 – Short Answer Questions (Total 50 pts)

Note that some short answer questions may have more answers than are required. If you provide additional correct answers, you will be given additional points. However, if you provide additional incorrect answers, you will be given additional deductions. It is not recommended that you supply additional answers, unless you are certain those answers are correct.

1. The following are true/false statements. If the statement is true, place a T in the blank to the left of the statement, if the statement is false, place an F in the blank. (1 point each)

a. T Chondrocytes play an integral role in the growth of long bones.

b. F Lamellar bone is always primary bone.

c. F An osteon is a cell living inside the bone tissue.

d. F An osteoblast is a bone cell removes existing bone matrix to make way for new bone.

e. F Woven bone is generated quickly and has excellent strength properties.

fe. T Ligaments and tendons can adapt their properties to match mechanical stimulus.

g. T Extracellular matrix of ligaments and tendons is made by cells call fibroblasts.

h. F The collagen arrangement in cartilage is completely random.

i. T Proteoglycans are responsible for much of the compressive strength of cartilage.

j. T Proteoglycans are made up of collections of glycosaminoglycan molecules.

k. T The periosteum is a highly vasularized tissue that enwraps the bone.

2. Match the bone feature at the left with the correct description/function at the right, by placing the correct letter in the blank next to each structure (1 point each).

B Volkmann’s canal

F canlicuili

H Haversian system

E Haversian canal

G epiphysis

A growth plate

D medullary canal

C diaphysis

A. a layer of proliferating cartilage that traverses the entire bone

B. predominantly horizontal canal that holds blood vessels and nerves

C. the shaft of a longbone

D. hollow region inside the long bone shaft

E. predominantly vertical canal that holds blood vessels and nerves

F. tiny canals extending to/from osteocytes

G. end of the bone beyond the growth plate

H. concentric circular layers of bone with a neurovascular canal at the center

3. Circle the roman numeral of any correct answer for each multiple choice question. (4 pts each)

a. The properties of cortical bone tissue depend on:

i. The orientation of the bone sample (Correct)

ii. Variations in cell density of the bone tissue

iii. The level of mineralization of the bone tissue (Correct)

iv. The porosity of the bone tissue (Correct)

b. The properties of cancellous bone tissue always depend on…

i. The orientation of the bone sample

ii. Trabecular architecture of the bone sample (Correct)

iii. The level of mineralization of the bone tissue (Correct)

iv. The porosity of the bone tissue (Correct)

c. Mechanical properties of cartilage are contributed to by…

i. Charge-charge repulsion in proteoglycan (Correct)

ii. Water bond by charged GAG chains (Correct)

iii. Restricted flow of water through the solid matrix (Correct)

iv. Compressive properties of collagen

4. Name at least four (4) levels of structural organization in a tendon. Make sure your four levels are distinct (not redundant) and ordered from smallest to largest! (4 pts):

tropocollagen

microfibril

subfibril

fibril

filament

fascicle

tendon

E

2

E

3

h

1

E

1

E

2

E

3

h

1

E

1

5. Name three (3) constituents of ligaments from most abundant to least abundant! (3 pts):

Water

Collagen

Elastin

Proteoglycan

6. To what physical phenomenon/quantity is the real part of the complex modulus (i.e. the storage modulus) related? (1 pts):

Elastic energy stored in the tissue

7. To what physical phenomenon/quantity is the imaginary part of the complex modulus (i.e. the loss modulus) related? (1 pts):

energy dissipated /lost per loading cycle in the material/tissue

8. Name at least 3 mechanical behavior phenomena that are characteristics of viscoelasticity: (3 pts)

Example: stress history dependent response

Strain history dependent response

Creep

Stress Relaxation

Strain rate dependent response

Hysteresis

9. In your own words, describe the mechanism that accounts for the non-linear behavior of ligaments and tendons in the early portion of a load-elongation test (3 pts):

Collagen fibers are generally lax and wavy prior to loading

Early properties are dominated by elastin properties

As collagen fibers are recruited, the stiffness becomes progressively larger

When all collagen fibers are recruited, the stiffness is essentially linear

10. Name the four primary functions of bones. (4 pts):

structure for shape and locomotion

protection of vital organs

mineral reservior

blood cell production (hematopoesis)

Section 2 – Exercises (Total 50 pts)

11. Bone fracture mechanics (16)

The image at the right shows a femur with circular cross section. A small length of the whole bone (as shown) was cut for fatigue compression testing. The geometry of the bone is as shown, with an inside radius of 9 mm, and a cortical thickness of 6 mm. The compressive force profile for the fatigue test is shown in the graph below. For this bone specimen, Kc = 2.2 MN/m3/2, C = 2.5x10-6 m(MN/m3/2)-2.5, and m=2.5.

The specimen failed at 2516 cycles.

What was the initial crack length (what was the size of the initial defect)?

max = F/A = 8000 N/[(152 – 92) mm2] = 17.68 MPa

min = F/A = 200 N/[(152 – 92) mm2] = 0.442 MPa

 = 17.24 MPa

Fast fracture: Kmax = KC => af = (KCmax2  = [(2.2 MN/m3/2)/(17.68 MN/m2)] 2/ = 4.93 mm

Nf = 2516 = (0.00493-1/4 - a0-1/4 m-1/4)/[(-.25)(2.5x10-6 m(MN/m3/2)-2.5) (17.24 MN/m2) 2.5 (1.25)]

a0 = 0.05 mm

12. Cancellous Bone Anisotropy (mean intercept lengths) (16 pts)

The anisotropy of a histological sample of bone tissue is examined. The sample is shown below with test lines at zero degrees (TL1-TL4).

a) Using only the one labeled test line shown (TL3=3mm long), calculate the mean intercept length at zero degrees (horizontal direction. (3 pts)

TL3 TL3

Answer: 18 intercepts => MIL = 3 mm / 18 intercepts = 0.1667 mm

b) The mean intercept lengths for a sample of cancellous bone were determined for 5 directions as follows:  = 0° => MIL = 0.432 mm,  = 22.5° => MIL = 0.543 mm

 = 45° => MIL = 0.612 mm,  = 67.5° => MIL = 0.554 mm,  = 90° => MIL = 0.441 mm

With which of the directions is the cancellous bone tissue most highly aligned? (3 pts)

Answer:  = 45° => MIL = 0.612 mm is the highest MIL indicating best alignment

c) Using the MILs for  = 0°,  = 45°, and  = 90°, in part (b), calculate the components of the material anisotropy matrix: (10 pts)

[ M ] = [ M11 M12 ]

[ M12 M22 ] mm-2

1/L2 = M11 cos2 + M22 sin2 + 2 M12 sin cos

 = 0° => L = 0.432 mm, M11 = 1/L2 = 5.59 mm-2

 = 90° => L = 0.441 mm, M22- = 1/L2 = 5.14 mm-2

 = 45° => L = 0.612 mm, M12 = 2.695 mm-2

13. Lumped Parameter Viscoelastic Model of a soft tissue (18 pts)

The viscoelastic model at the right is appropriate for some soft tissues.

The model is subjected to a sudden (step) stress, 0 at time t = 0, and the stress is held indefinitely.

a) Is this a fluid-dominated or solid-dominated viscoelastic models? (2 pts)

b) Determine the strain in each component of the model at time t = 0+. (4 pts)

Strain in the spring with constant E1: must equal spring 2: =0 / (E1 + E2)

Strain in the damper with constant 1: damper cannot move instantly = 0

Strain in the spring with constant E2: must equal spring 1: =0 / (E1 + E2)

Strain in the spring with constant E3: unconstrained =0 / E3

b) Determine the stress in each component of the model at time t = 0+. (4 pts)

Stress in the spring with constant E1: = 0 E1/ (E1 + E2)

Stress in the damper with constant 1: must equal (E1)= 0 E1/ (E1 + E2)

Stress in the spring with constant E2: = 0 E2/ (E1 + E2)

Stress in the damper with constant E3: must equal total stress = 0

c) Determine the strain in each component of the model at time t → ∞. (4 pts)

Strain in the spring with constant E1: carries no stress: =

Strain in the damper with constant 1: same as E2 =0 / E2

Strain in the spring with constant E2: carries all stress: =0 / E2

Strain in the spring with constant E3: unconstrained =0 / E3

d) Determine the force in each component of the model at time t → ∞. (4 pts)

Stress in the spring with constant E1: (E1)= 

Stress in the damper with constant 1: (E1)= 

Stress in the spring with constant E2: equals total stress = 0

Stress in the damper with constant E3: must equal total stress = 0

Fmax=200 N

Force (N)

Fmax=8000 N

t =6 mm

H

Force (N)

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ri=9 mm

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