BME 5210 – Musculoskeletal Biomechanics

Final Exam

May 3, 1999

This is a closed book exam. Only the information provided with the exam may be used.

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True and False (1 point each)

1. The cartilage surface in a healthy joint is smoother than that of bearings typically used in mechanical devices. T F

2. The frictional resistance in a healthy joint increases as that joint is cycled for extended periods. T F

3. Synovial fluid, due to its high viscosity, causes knee friction to increase during quick motions. T F

4. Diarthrodial joints are enclosed in a fibrous capsule. T F

5. Filtered lubricant tests have shown that the macromolecule HA is the only molecule in the synovial fluid responsible for the low friction in joints. T F

6. Hydrodynamic lubrication assumes that the articulating surfaces do not actually touch. T F

7. Boosted lubrication theory assumes that the synovium secretes additional synovial fluid to lubricate the joint. T F

8. According to biphasic-poroelastic models for joint contact, fluid will be pushed out of the cartilage as the opposing articular surface slides over it. T F

9. Ateshian’s theory assumes that fluid weeps out of the cartilage and forms a lubricating layer that lower friction. T F

10. Squeeze film lubrication has been shown to be reasonable based on fluid film thickness calculations. T F

11. The knee can be accurately modeled as a commonly available mechanical bearing. T F

12. During pedaling, forces generated by the dominant leg are greater along all axes than the non-dominant leg. T F

Multiple Choice (1 point each) (#13 – 18 taken from student papers)

13. What was the main conclusion drawn in the paper describing the effects of exercise on osteoporosis?

a. Endurance exercise definitely increases BMD

b. Nutrition does not play a role in bone loss

c. Longitudinal studies are a perfect way to study the effects of exercise on bone mass

d. Exercise decreases fracture risk by decreasing bone loss and increasing bone mass

14. Osteoporosis is thought to possibly result from:

a. An increase in bone resorption due to a reduction in estrogen

b. A decrease in bone formation due to an increase in estrogen

c. A shift in bone remodeling due to increased levels of Vitamin D

d. More osteoblast activity than osteoclast

15. “Whiplash” injuries are thought to be caused by the following sequence of motions:

a. Compression - shear

b. Tension - shear

c. Extension - flexion

d. Flexion - extension

16. Automotive impact induced cervical spine injuries in children are most often due to:

a. The head impacting vehicular structures

b. The inertia of the head pulling on the spine

c. Flexion of the neck at C7

d. Correctly used child restraints

17. Fractures of the cervical spine can be classified based on:

a. The magnitude of the applied load

b. Whether failure occurs before or after impact with the vehicle interior

c. The doctor who diagnoses the condition

d. The combination of axial and bending loads placed on the spine

18. “Muscling” of the golf swing is most likely a result of:

a. Maximizing the efficiency of the golf swing

b. Compensating for weak or fatigued muscles

c. Over coiling of the body during the take-away phase of the swing

d. Gripping the club too far down the shaft

19. A torsional load on a bone results in which of the following fracture patterns:

a. b.

c. d.

Short Answer and Problem

19. Name 3 methods that can be used to obtain a geometric description of joint contact.

20. If the effective stress in the cartilage of the knee is calculated to be 5 MPa for a 150 lb (667N) person standing on straight legs (knee contact area = 100 mm²), what is the effective frictional coefficient according to Ateshian’s theory? State any assumptions you make.

21. Four biomechanical stages of fracture repair have been identified based on how the bone refractures. Explain how bone remodeling results in each of these stages of fracture repair. (12 points)

I. Bone fails through the original fracture site with a low stiffness, rubbery pattern.

II. The bone fails through the original fracture site with a high stiffness, hard tissue pattern.

III. The bone fails partially through the original fracture site and partially through the previously intact bone with a high stiffness, hard tissue pattern.

IV. The site of failure is not related to the original fracture site and occurs with a high stiffness pattern.

I. Bone remodeling begins with the formation of fibrous tissue, which is collagen based. Due to the soft-tissue nature of this structure, it will behave mechanically like a soft tissue (toe-region, low stiffness, large deformations). II. Following the formation of the fibrous tissue bridge, the damaged area is mineralized. However, this structure is woven bone and not lamellar bone. Thus, while it behaves like a hard tissue, it has reduced strength and will tend to be the site of failure. III. The woven bone at the fracture site and the adjoining bone are remodeled into osteonal bone, which involves removal of portions of the bone via osteoclastic resorption. Due to this resorption before deposition cycle, the fracture site will remain weaker than the surrounding tissue. A fracture will involve both segments of the bridging section. IV. Remodeling will have returned the bone to its pre-fracture state. There will be no difference between properties at the fracture site and the surrounding bone.

22. Which of the following long bone plating techniques will be most effective in stabilizing a diaphyseal fracture of the femur and why? Bones are viewed from the anterior side. Assume all plates have the same dimensions. (5 points)

A. B. C. D.

C - Bones should be plated to resist tensile loads – the loads that will act to open a fracture surface. The loads on the femoral head will cause a medially directed moment. That causes tension on the lateral (right) side of the bone. The plate should be placed on the lateral side to resist this force.

23. There is conflicting data on the role of the cortical shell in carrying the load seen by a lumbar vertebral body. Some researchers show that removing the cortical shell only reduces the vertebral strength by 10%. Others indicate that it may carry up to 65% of the total load. One of the differences in these studies may be the age of the cadaver subjects used in the study. Based on your knowledge of vertebral geometry and bone loss, explain how the cortical shell may play an increasingly dominant role in carrying spinal loads as we age. (10 points)

The cortical shell comprises only a small fraction of the total cross-sectional area of a vertebral body. In young individuals, where the trabecular bone is at its peak properties, the geometric stiffness of the cortical shell is low compared to the trabecular centrum. Thus, the majority of the load is carried by (and the majority of the strength is provided by) the trabecular bone. As bone is lost with age, the modulus and geometric stiffness of the trabecular bone decreases. As this occurs, the cortical shell carries a higher percentage of the load and thus provides a greater contribution to the overall strength. Therefore, as an individual ages, the cortical shell plays an increasingly important role in the load bearing functions of a vertebral body.

24. A child holds a 10 kg mass in front of himself at a distance of 25 cm from his chest. The mass of the child’s body segments and their moment arms about the lumbar spine are given below:

Body Part	Mass	Moment Arm Length
Head	4 kg	5 cm
Torso	30 kg	2 cm
Arm	2 kg	15 cm
Anterior Chest	NA	7 cm

The pair of erector spinae muscles are symmetrically distributed on either side of the spine. They act 1 cm from the center of rotation of the spine in the anteroposterior direction. What is the required force generated by the erector spinae muscles in order to allow the child to carry this weight? If the muscles have a maximum intensity of 1000 kPa and the cross-sectional area of each is 5 cm², what is the maximum possible load that can be generated by this child’s erector spinae. Assuming they are the only muscles acting, will the child be physically able to lift this weight? (10 points)

SM = 0

m_box*g*d_box + m_head*g*d_head + m_torso*g*d_torso + m_arm*g*d_arm – F_es*d_es = 0

g*(10 kg * 0.32 m + 4 kg * 0.05 m + 30 kg * 0.02 m + 2 kg * 0.15 m) – F_es * 0.01 m = 0

F_es = 4214 N

F_max = s_max*A_xs

F_max = 1000 kPa * 5 x 10^-4 m² = 500 N

2 muscles à F_max2 = 1000 N

F_max2 < F_es – child not able to lift weight