BME 5210 – Musculoskeletal Biomechanics

 

Exam 1

 

February 14, 2000

 

 

 

 

This is a closed book exam.  Only the information provided with the exam may be used.  Answer each question completely but concisely.  You may use lists, phrases, or sentences. 

 

By signing below you indicate that your have been informed about and are in compliance with the policy on academic dishonesty as it is applied to this class and this exam.

 

 

 

RESULTS

 

            Mean:              40.4

            SD:                  15.7

            Possible:          85.0

 

            Minimum points required to obtain listed grade:

                        A:        48.2

                        B:        32.55

                        C:        20.81

 

            This exam is worth 20% of your final grade.

1.     (30 points) A parade organizer is being sued by some of the “waving marchers” for repetitive motion injuries.  You are serving as a consultant on the case, and have been asked to estimate the forces and moments at the elbow.  You are provided with the following information for one of the plaintiffs:

 

            Mass of hand and forearm:     2.5 kg

            Length of forearm/hand:          0.5 m

Moment of intertia about elbow:          0.0127 N-m-sec2

            Description of waving movement:

                        Rotation of the forearm at the elbow about the axis of the humerus (arm)

                        All motion in the coronal plane

                        Assume pure curvilinear motion

 

 

 

 

 

 

 

 

 

 

 


            Kinematic data:

 

Time (sec)

X1

Y1

X2

Y2

10

0.05 m

0.02 m

0.225 m

0.14 m

10.1

0.045 m

0.04 m

0.20 m

0.185 m

10.2

0.04 m

0.055 m

0.15 m

0.237 m

10.3

0.03 m

0.063 m

0.10 m

0.264 m

10.4

0.02 m

0.069 m

0.05 m

0.279 m

 

Calculate the following:

a)     The segment angle at the above 5 time points (4 points)

b)    The angular velocity and acceleration of the forearm/hand complex, given the above data (6 points)

c)     The force and moment at the elbow at t = 10.2 seconds (10 points)

 

State all assumptions that you make.

 

NOTE:  Remember that radians are the equivalent of a unitless measure of angle.

Answer to Question 1

 

a.     f = arctan [(yj - yi)/(xj - xi)]  (2)

 


t = 10 s                  34.5 deg/0.601 rad

t = 10.1 s               43.1 deg/0.752 rad

t = 10.2 s               58.9 deg/1.03 rad

t = 10.3 s               70.8 deg/1.24 rad

t = 10.4 s               81.9 deg/1.42 rad

 

 

 

b.     wI= (fI+1 – fI-1)/2Dt  (3)

 

t = 10.1 s               122 deg/s or 2.13 rad/s

t = 10.2 s               138 deg/s or 2.42 rad/s

t = 10.3 s               115 deg/s or 2.01 rad/s

 

aI = (wI+1 – wI-1)/2Dt  (2)

 

t = 10.2 s               -34.9 deg/sec2  or -0.0608 rad/sec2                  (+9)

 

c.     SM = Ia  (2)

 

      mg*(0.5L*sinf) + M = Ia                       (3)

            2.5 kg*9.8 m/s2*0.5*0.5 m * cos(58.9) + M = 0.0127 N-m-sec^2* (-0.608 rad/sec2)

            M = -3.17 N-m  (clockwise)               (2)

 

      SF = ma  (2)

 

Choice 1 (simple):

      a = 0 (pure curvilinear motion) (1)

      Rx = 0

      mg + Ry = 0    (2)

                  2.5 kg * 9.8 m/s2 = Ry            Ry = 24.5 N     (2)

 

Choice 2 (accurate dynamics)

      SFx = m(-rCGw2) (1)

      SFy = mrCGa

 

      SFx = Rx = 2.5 kg*(-0.25 m * (2.42 rad/s)2)

                  Rx = -3.66 N

 

      SFy = Ry – mg = mra

                  Ry = 2.5 kg * 0.25 m * 0.6 rad/s2 + 2.5 kg * 9.81 m/s2

                  Ry = 24.88 N

2.     (10 points) Explain what is happening to the foot with respect to the leg during normal walking, as is described by the following plot.  Use anatomically correct terms for planes and motions.  Based on the accompanying leg motion figure, sketch an equivalent plot for foot segment angle (general patterns only).

 

Variation in ankle joint angle with time within the sagittal plane during normal gait.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 


The 1st graph portrays the change in ankle angle with time.

 

Foot motion occurs in the sagittal plane (1)

At initial contact, foot is slightly plantarflexed (1)

As gait moves through mid-stance and terminal stance, foot increases the degree of dorsiflexion (1)

As swing begins, foot plantarflexes , but plantarflexion decreases during mid-swing in order to clear the toe (2)

 

 

Foot Angle w/r/t ground

 

 

(5)

 

 

 

 

 

 

 

3.  (15 points) Two patients present to your diagnostic laboratory for assessment of bone loss and fracture risk.  Using experimental techniques, you are able to make the following measurements in the spine:  (NOTE: Reduced to 10 points)

 

Measurement

Patient A

Patient B

Bone Density

1.5

1.5

Mean Trabecular Plate Separation

Vertical Axis:  200 µm

Horizontal Axis:  250 µm

Vertical Axis:  500 µm

Horizontal Axis:  250 µm

Mean Trabecular Plate Thickness

Vertical Axis:  150 µm

Horizontal Axis:  200 µm

Vertical Axis:  250 µm

Horizontal Axis:  250 µm

 

Describe the progression of the bone loss seen in the two patients and compare your assessment of their fracture risk.  Why would you expect this difference in fracture risk?  Other than bone properties, what other risk factors would you assess to further enhance your estimate of fracture risk?

 

(6 points based on first 2 sections)

 

Patient A:  Thinning of trabeculae with maintenance of trabecular spacing; greater spacing between vertical beams than horizontal struts

Patient B:  Loss of horizontal trabeculae with maintenance of trabecular thickness.  Space between vertical beams remains normal.

 

(2) Patient A has less risk of fracture than Patient B.  While their densities are the same, Patient B has experienced a greater loss of the 3D connectivity in the trabecular structure.  As the horizontal connecting struts are lost, the vertical beams are more likely to fail by buckling.

 

Other risk factors: (4)

      Low body weight/low soft tissue thickness

      Tendancy to fall (cognitive impairment/balance impairment)

      Increased height

      Increased fall severity

 

3.     (5 points) It is known that when a trabeculae is completely perforated during bone remodeling it can not be replaced/reconnected.  What basic rule of bone physiology underlies this phenomenon?  (Provide both the name of the rule and a brief description).  What experiment would you design to assess the validity of this rule?

 

Wolff’s Law – Bone remodels in response to the load that it sees  (3)

 

Experiment (2) – an experiment based on increasing or decreasing loading in the bones

4.     (20 points) A nursing aid is being assessed for low back pain in your clinic.  One of her main tasks is bed making.  She reports that she often leans over the bed to tuck sheets in on the far side.  You want to analyze the forces and moments at the L5 lumbar disc seen in her technique compared to the “proper” technique.  See the attached figure for clarification.

 

Information:

                  Weight of head and trunk:                   340 N

                  Weight of arms:                                    60 N

                  Moment arms

                              Bending over – arms c-of-g:    0.67 m

                              Bending over – body c-of-g:   0.31 m

                              Upright – arms c-of-g: 0.18 m

                              Upright – body c-of-g:            0.17 m

                              Erector spinae:             0.05 m

                  Angle of inclination of L5 disc

                              Bending over:                          70 deg

                              Upright:                                   30 deg

 

a.     How large is the bending moment on the L5 disc when the patient is making the bed while bending over?  While standing upright?

b.     How large must the erector spinae muscle force be to counteract the bending moment for both cases?

c.     How large will the compressive and shear force on the L5 disc be in the two cases.  Assume that the line of application of the erector spinae muscle force is perpendicular to the disc, independent of its degree of inclination.

 

a.  SM = 0  (1)

      Bending over:              340 N * 0.31 m + 60 N * 0.67 m + M = 0  (2)

                                          M = - 145.6 N-m (clockwise)

      Upright:                      340 N * 0.17 m + 60 N * 0.18 m +M = 0  (2)

                                          M = - 68.6 N-m (clockwise)

 

b.     FES = M/0.05 m

Bending over:              FES = - 2912 N  (1)

Upright:                      FES = - 1372 N  (1)

 

d.     SF = 0   (1)                 Arrange axes in line with disc  (2)

 

Bending over:              FES + 340 N * cos(70) + 60 N * cos(70) + RC = 0   (2)

                                    RC = 3049 N  (1)

                                    340 N * sin(70) + 60 N *sin(70) + RS = 0 (2)

                                                RS = 375 N (1)

Upright:                      FES + 340 N * cos(30) + 60 N * cos(30) + RC = 0 (1)

                                    RC = 1718 N (1)

                                    340 N * sin(30) + 60 N * sin(30) + RS = 0 (1)

                                                RS = 200 N (1)

5.     (5 points) You are designing an impact test to assess the properties of the bones of the leg for crash simulations.  Due to the low availability of cadaveric tissue, you decide to use your first cadaver to estimate the input variables for your study.  You insert load cells into the tibia and femur, slowly apply a deformation up to 2 cm at a strain rate of 0.01 sec-1 and measure the force in the bones.  Based on this information, you determine that you can deform the bones by the same 2 cm in your impact test at 10 sec –1.  However, when you conduct your test, the bones fracture and the failure loads measured by the load cells are higher than was seen in your pre-test.  Why did this occur?

 

 

Bone is viscoelastic (2).  At the higher deformation rate, the applied deformation would induce a higher load in the bone.  At higher deformation rates, bone is stiffer and will fail at higher loads, but lower deformations. (3)

 

 

 

 

 

 

 

 

 

 

6.     (5 points)Optimization techniques in kinetic and kinematic analysis often make use of the theory of maximum muscle force.  Explain why an optimization technique would be needed.  Describe the main idea behind the theory of maximum muscle force.

 

Optimization techniques are used for indeterminant static and dynamic problems when there are more unknowns than equations – typically when the goal is to solve for forces in multiple muscles. (2)

 

Maximum muscle force theory comes from the idea that each muscle force is capable of exerting a maximum force/unit area.  Based on the physiological cross-sectional area of the muscle and this maximum stress level, the maximum force can be calculated.  In optimization theories, this can be modeled as each model acting only until its maximum force is reached.  The first muscle is recruited, based on the maximum product of cross-sectional area and moment arm length, and other muscles are recruited in descending order of A*L once each previous muscle has reached its maximum force. (3)