BME 5210 – Musculoskeletal Biomechanics

 

Final Exam

 

April 26, 2000

 

 

 

 

 

This is a closed book exam.  Only the information provided with the exam may be used.  Answer all questions completely, succinctly, and to the best of your ability.  For short answer questions, answers may be provided in paragraph form, in phrases, or as lists.

 

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True and False (1 point each)

 

1.     The low frictional coefficient of cartilage is due to its highly smooth surface.       T      F

 

2.     Synovial fluid acts to reduce friction in articular joints.                 T          F

 

3.     Atheshian’s theory for joint friction requires that pressurized fluid leaves the articular cartilage to provide joint lubrication.                      T          F

 

4.     Boosted lubrication involves the flow of fluid out of the articular cartilage into the joint to improve lubrication.       T          F

 

5.     A lubricating glycoprotein thought to reduce friction in synovial joints has bee isolated by a group of researchers. T          F

 

6.     Fractures that occur as a result of high velocity impacts release a large amount of energy.  This explosive release of energy results in a smooth fracture surface with minimal soft tissue damage.          T          F

 

7.     During fracture healing, mechanical loading improves fracture healing because it induces motion at the fracture site thus promoting bony union.    T          F

 

8.     In order to optimize fracture healing, the goal of fracture fixation is to maximize the rigidity of the fixation system.       T          F

 

9.     The vertebrae of the spine increase in strength as you move from the cervical through the lumbar spine.   T          F

 

Matching

 

10.  In this question (5 points), match the technique with the information provided by that technique (some answers may be used more than once).

 

Stereophotogammetry                   A, B                             A.  cartilage thickness

B.  joint surface curvature

Fuji Pressure Sensitive Film          C, D                             C.  interarticular pressure

D.  joint contact area

      Cryosections                                 A, B                             E.  real time measurement of

                                                                                                            cartilage thickess

      Magnetic Resonance Images         A, B, D, E                    F.  joint friction

Fill in the Blank

Choose (from the list below) the most appropriate words to fill in the blanks in the following statements (1 point for each statement).

            Words:                        decrease, lower, exudes, increase, reabsorbs, higher

 

11.  Fluid exudes at the leading edge of a parabolic load traveling across cartilage, and reabsorbs at the trailing edge.

 

12.  Initial friction coefficient values are lower than steady state values according to Atheshian’s theory.

 

13.  Wear rates decrease if tests are performed in synovial fluid versus saline, but increase if the normal loading is increased.

 

14.  Small oscillations of the normal loading of cartilage decrease the magnitude of the frictional coefficient.

 

15.  In squeeze film lubrication, a decrease in the film layer would be expected as time passes.

 

Multiple Choice (1 point each)

 

16.  Bone adapts to mechanical stimulus through remodeling.  Based on the natural tendency of a bent bone (under axial load) to straighten, Jansen proposed the following relationship:

            a.  Tension – osteoclastic activity; Compression – osteoblastic activity

            b.  Tension – osteoblastic activity; Shear – osteoclastic activity

            c.  Torsion – osteocytic communication; Compression – osteoclastic activity

            d.  Shear – osteoclastic activity; Compression – osteoclastic inactivity

 

17.  When comparing the resultant forces at the ankle due to running and biking, which of the following is true?

            a.  The ankle forces are approximately the same

            b.  The ankle forces are more than twice as high in running compared to biking

            c.  The ankle forces are more than twice as high in biking compared to running

            d.  It is not possible to estimate ankle forces in biking as there is no ground reaction force

 

18.  Injury tolerance has been found to be dependent on all of the following EXCEPT: 

            a.  Age

            b.  Gender

            c.  Maximum weight carried on back

            d. Maximum ambulatory speed

19.  A stress-strain plot for bone provides several pieces of information, including the energy transferred to the specimen during the test.  Which of the following statements is NOT true?

a.      The energy absorbed by bone due to its viscoelastic nature is described by the area between the loading and unloading curves.

b.     The energy to failure of a specimen is calculated as the area under the force-deformation curve to the point of failure.

c.      The elastic energy of a piece of bone is described by the area under the unloading curve.

d.     All strain energy applied to bone during a compression test is released during the unloading phase.

 

20.  The goal of the European standards for pedestrian impact is:

a.     To increase the tolerance of the human leg and knee to impacts by requiring padding increased padding in this region.

b.     To decrease the aggressiveness of regions of the vehicle most likely to strike a pedestrian.

c.     To develop an accident avoidance system to better enable drivers and pedestrians to avoid an impact.

d.     To determine the maximum velocity with which a car can strike a pedestrian without causing significant damage to the lower limb.

             

21.  The spine includes both intersegmental and multisegmental muscles.  Research in the last decade has shown that:

            a.  Intersegmental muscles provide spinal stability, while multisegmental muscles provide motion.

            b.  Multisegmental muscles provide spinal stability, while intersegmental muscles provide motion.

            c.  Multisegmental muscles provide a compressive load across the entire lumbar spine, thus maximizing stability.  No intersegmental stablization is necessary.

            d.  Intersegmental and multisegmental muscles act together to increase the stability of the spine.

 

22.  Tensile loading results in a transverse fracture (Figure I).  What type of loading results in the type of fracture seen in figure II?

 


            I.                                                                                  II.

 

 

 

            a.  Torsion

            b. Axial compression

            c.  Bending with axial compression

            d.  Shear

 

23.  The role of the facet joints of the vertebrae include:

 

a.     Transferring 80% of the compressive load between vertebrae

b.     Limiting torsional and shear motions between vertebrae

c.     Carrying 10 – 20 % of the compressive load in an upright posture

d.     Maximizing lateral bending

e.     a and b

f.      b and c

 

Short Answer and Problem

 

24.  Bony healing involves three general phases, which tend to overlap.  They are: a) stablization; b) union; and c) remodeling.  Briefly describe each of these phases.  (6 points)

 

a.     stablization, either through artificial fixation or the formation of a callus, mechanically fixes the fracture segments and minimizes motion

b.     union occurs via the formation of a bony bridge between the fracture segments

c.     remodeling is the normal process in which the bony bridge, and surrounding bone, is remodeled from woven or primary osteonal bone into lamellar or osteonal bone

 

 

 

 

 

 

 

 

 

25.  A football player breaks his leg during a game but decides to “walk-it-off.”  3 weeks later, complaining of continuing leg pain, he has it x-rayed.  The film shows a misalignment of the diaphysis at the fracture site.  Initial fracture healing has begun.  Which of the main patterns of fracture healing is most likely to have occurred and why?  (4 points)

 

Non-osteonal healing (including formation of a callus) – this occurs predominantly when a fracture is misaligned or there is an extensive fracture gap

 

 

 

 

 

 

 

 

26.  Give one benefit and one drawback for each of the following means of fracture fixation.  Do not repeat any of your choices for more than one fixation method.  (6 points)

 

Benefit                                     Drawback

a.     External fixation                Minimal incision                                 Infection

b.     Plate fixation                     Multi fragment fixation                       Bone loss under plate

c.     Intramedullary nail Immediate weight bearing                    Torsional instability

 

 

 

27.  During the day, individuals tend to lose up to 1 inch in height.  They regain this height at night as they sleep.  Explain why this happens.  (4 points)

 

Under load, the intervertebral discs lose water.  As this water is lost, the discs’ stiffness is reduced and more deformation occurs due to normal posture – height of individual is reduced.  Once the load on the spine is reduced (such as while sleeping), the water is resorbed back into the discs and the stiffness returns to the base-line level.

 

 

 

 

 

 

 

28.  Weight lifters typically use belts to increase their abdominal pressure minimize the effect on the back.  You have been asked to quantify the effect of a weight belt for the U.S. Olympic weight lifting team.  What effect is seen on the erector spinae force and vertebral compressive load given the following data?  Given a maximum isometric tension in the erector spinae of 1000 kPa and a cross-sectional area of each erector spinae muscle of 75 cm2, what is the maximum weight that could be lifted by this individual wearing the weight belt?  (15 points)

 

Parameter

Value

Upper body weight

90 kg

Weight lifted

150 kg

Normal abdominal pressure

2 kPa

Abdominal pressure with belt

15 kPa

Abdominal cross-sectional area

.075 m2

Line of action of BW

10 cm

Line of action of weights

15 cm

Line of action of abdominal pressure

8 cm

Line of action of erector spinae

2 cm

 

SM = 0           (1)

            g(90 kg * 0.1 m + 150 kg * 0.15 m) – (2000 Pa * 0.075 m2 * 0.08 m) – FES * 0.02 m = 0  (2)

            FES  = 14835 N            (1)

 

            g(90 kg * 0.1 m + 150 kg * 0.15 m) – (15000 Pa * 0.075 m2 * 0.08 m) – FES * 0.02 m = 0

            FES  = 10935 N            (1)

 

SF = 0             (1)

            g(90 kg + 180 kg) + 14835 N – (2000 Pa * 0.075 m2) – Fc = 0  (2)

            Fc = 17331 N                          (1)

 

            g(90 kg + 180 kg) + 10935 N – (15000 Pa * 0.075 m2) – Fc = 0

            Fc = 12456 N                          (1)

 

Maximum weight

            FES-Max = 1000 kPa * AES * 2 = 1000 kPa * 0.0075 m2 * 2 = 15000 N          (3)

            g(90 kg * 0.1 m + FMAX * 0.15 m) – (15000 Pa * 0.075 m2 * 0.08 m) – 15000 N * 0.02 m = 0  (1)

            FMAX  = 205 kg                        (1)