teams scenario 1 11-12teams.tsaweb.org/sites/teams.tsaweb.org/files/2012 scenarios 11-12.pdf ·...

Introduction The human heart consists of four valves: tricuspid, pulmonic, mitral, and aortic. These valves regulate the unidirectional blood fl ow between the heart chambers based on pressure. A typical human heart valve opens and closes around 40 million times per year and 2 billion times over an average lifetime. Many conditions might cause a heart valve to fail, including degenerative valve calcifi cation, rheumatic fever, endocarditis, and congenital birth defects. Recent studies show that doctors use nearly 250,000 artifi cial heart valves in valve replacement surgeries each year. An artifi cial heart valve consists of an opening and a mechanism that opens and closes this opening in response to the pressure gradient between two chambers of the heart. There are three general types of valves: mechanical, bioprosthetic, and polymer valves. Materials engineers, tissue engineers, and mechanical engineers work in teams with many types of scientists, including doctors and biologists, to design, build, and test artifi cial heart valves.

Valves that are made from purely synthetic materials theoretically have an unlimited lifetime and typically do not need to be replaced for up to 25 years, depending on

the type of valve. Valves made of biological materials have shorter lifetimes, typically lasting around 15 years, so this type of valve is recommended for patients over 65.

In general, artifi cial valves are very safe. However, clot formation is one major problem with artifi cial heart valves made of synthetic materials. Clots can occur through tissue factor exposure, platelet activation, or contact activation by foreign materials of the heart valve. Tissue factor exposure is caused by cells rupturing on the valve and releasing tissue factor, which causes platelets to gather and clot. Platelet activation occurs when proteins deposit on the valve surface and cause platelets to clot there. Contact with foreign materials causes clotting due to high stresses induced by blood fl ow across the valve surface. To mitigate clotting, all patients who have mechanical or polymer heart valves are given anticoagulants. Because anticoagulants damage red blood cells and require patients to undergo monthly blood tests to monitor their condition, biological valves are preferred in older and weaker patients. Another major problem with any implant is the possibility of infection. Common practices, such as sterilizing of implants before surgery, help to reduce this risk.

TEAMS Competition 2012Scenario # 1Development and Design of Artifi cial Heart Valves

2TEAMS Competition 2012 11/12 Level © Technology Student Association (TSA)

Moreover, materials that are used to make a valve can play a role in fi ghting infection; thus, coatings of titanium dioxide, silver nanoparticles, and antibiotics are used to minimize the incidence of infection.

prevent a rejection. Currently, kangaroo and sheep valves are being explored as options.

Other prosthetic valves are made in a manner similar to the bileafl et and trileafl et valves, with a metal frame (often titanium) that has tissue leafl ets sewn into it. The tissue for these leafl ets is harvested from pig, cow, horse, or human pericardial tissue. When using human pericardial tissue, the tissue can come from the patient to avoid any chance of rejection. These valves have a much shorter life span than a mechanical valve; however, biologically based valves do not have the clotting risk that mechanical valves are prone to causing.

Assumptions and Givens

An artifi cial heart valve must have the following attributes: minimal leakage of blood from the valve, minimal pressure gradient across the valve, minimal interaction with the tissue, low risk of failure, nonclot forming, self-repairing, and rapid dynamic response.

A Newtonian fl uid is a fl uid with a constant stress–strain relationship. In a Newtonian fl uid, the shear stress is directly proportional to the velocity gradient of fl uid fl ow, and the constant of proportionality is viscosity. The presence of a Newtonian fl uid simplifi es the equations that can be used to calculate velocity and stress. Although blood is not technically a Newtonian fl uid, assume that it is for the purposes of the calculations here.

Background Mechanical and Polymer Valves

There are three main types of mechanical valves: cage ball valves, bileafl et and trileafl et valves, and tilting disk valves. Most mechanical valves are composed of titanium graphite, pyrolytic carbon, and polyester or some other biocompatible polymer. Titanium acts as the housing for a valve and is used for its strength and biocompatibility. Leafl ets and disks in their respective valves often are composed of graphite coated with pyrolytic carbon and are produced by depositing hydrocarbons on a heated graphite substrate. Tungsten is sometimes added to pyrolytic carbon so that a valve can be easily seen following implantation. The material structure of pyrolytic carbon helps it resist cracking, making it ductile. The processing method that pyrolytic carbon goes through can introduce microscopic cracks that must be detected. The sewing cuff for attaching the valve to the tissue of the artery is usually constructed of double velour polyester, and the ball in the ball-cage model is typically made of silicone or a biocompatible polymer such as Tefl on, Dacron, and ePTFE.

Bioprosthetic Valves

Some biological valves are harvested from pigs or human tissue donors and transplanted directly into a human patient. A valve that comes from another species is a xenograft; a valve from another human is an allograft. The traditional pig valve lasts about 15 years before needing replacement; in younger patients, this life span can be even lower. Xenografts always have the chance of rejection by the immune system, and sometimes a patient will have to take medication to

Figure 1-1: A cage ball mechanical heart valve.

Occluder ball

Suture ring

Restraining cage

Figure 1-2: A tilting disk mechanical heart valve.

Flange

Suture ring

Outlet strut

Inlet strut

Occluder disc

Figure 1-3: A bileafl et mechanical heart valve.

Suture ring

Leaflets

Hinges

Figure 1-4: A schematic of a bioprosthetic heart valve.


Additional Assumptions and Givens The effective orifi ce area (EOA) is a common measurement of the quality of a mechanical, polymer, and bioprosthetic heart valve and can be calculated in the

following way:

Equation 1-1

EOA � Q ________

51.6 �___

�p

The higher the EOA, the better the valve and the less likely it is to cause complications such as clot formation.

• Pa � Pascals � F __ A � N/m 2 where F is the applied force and A

is the area

1. Your team has been asked to design

a valve for a patient whose mitral

valve has failed. Given that 1) the

patient is a 45-year-old male with no

other major health problems, 2) the

bioprosthetic has a Q of 0.0018 m3/s

and a �p of 0.8 Pa, and 3) the

mechanical valve has a Q of 0.0014

and a �p of 1.2 Pa, which valve should

your team choose and why?

a. mechanical valve; The effective

orifi ce area (EOA) is better than

that of the bio prosthetic valve.

b. mechanical valve; It is a more

durable valve and likely to last

longer than the bioprosthetic valve.

c. bioprosthetic valve; The effective

orifi ce area (EOA) is better than

that of the mechanical valve.

d. bioprosthetic valve; It is a more

durable valve and likely to last

longer than the mechanical valve.

e. The answer cannot be determined

from the information given.

Questions

Notes

Additional Notes


2. What is one consequence of introducing

a foreign object into the body in the

form of an artifi cial heart valve?

a. microtears in the surrounding

tissue

b. tumor growth due to contact with

the valve surface

c. clot formation through contact

with the valve surface

d. degradation of the valve due to the

body’s immune response

e. growth of tissue, encapsulating the

valve components

3. Given the data set for compression

loading of pyrolytic carbon, determine

the elastic modulus of the material.

Compressive Loading of Pyrolytic Carbon

Str

es

s (

Pa

)

0.85

0.8

0.75

0.7

0.65

0.62E-11 2.2E-11 2.4E-11 2.6E-11 2.8E-11 3E-11 3.2E-11 3.4E-11

Strain (m/m)

a. 5 MPa d. 26 GPa

b. 10 GPa c. 30 GPa

a. 26 MPa

QuestionsAdditional Background The elastic modulus, or Young’s modulus, of a material is the

ratio of the stress to the strain: E � � __ � � F/A ____ dL/L .

The elastic modulus describes the tendency of a material to deform elastically when a force is applied to it. Elastic deformation involves a material changing shape when a force is applied but returning to its original shape after the force is removed (think of pulling and releasing a rubber band).

Notes

Additional Notes


Additional Assumptions and Givens

The valve fl ow rate in m3/s can be calculated by considering the valve to be a pipe that suddenly decreases in radius, causing the fl ow to speed up. The equation for calculating the fl ow rate is:

4. Calculate percentage shortening of

titanium in a prosthetic valve due to

the force of 1 � 10�3 N exerted by

the blood on titanium leafl ets. Titanium

has an elastic modulus of 120 GPa and

the valve has a diameter of 20 mm.

b. 2.7 � 10�13% d. 2.7 � 10�6%

c. 2.7 � 10�11% e. 3.3 � 10�10%

d. 2.7 � 10�9%

5. Given the modulus of elasticity of

titanium above, calculate the load on a

5-mm long titanium rod with a radius

of 0.1 mm that undergoes a length

change of 0.0001 mm when loaded in

compression on its ends.

a. 75 mN d. 75 kN

b. 1.8 N e. 35 MN

c. 18 kN

6. Given that the pressure drop across

a polyurethane valve with an inner

diameter of 15 mm is 1 Pa when

the diameter of the aortic artery is

20 mm, calculate the blood fl ow rate.

a. 0.0000053 m3/s

b. 0.0000098 m3/s

c. 0.00031 m3/s

d. 0.00097 m3/s

e. 0.053 m3/s

Questions

Additional Notes

Q � A2 �___________

2(P1 � P2) ___________

� ( 1 � ( A2 __ A1

) 2 )

At body temperature, the density of blood is equal to 1,025 kg/m 3 .

Notes

Equation 1-2



• The Reynolds number of a fl ow is a nondimensional velocity that is defi ned as the ratio of the dynamic pressure to the

shearing stress of a fl uid: �u2

_____

�u/d�

where � is the density of

the fl uid, � is the dynamic viscosity of the fl uid, d� is the

hydraulic diameter of the pipe/artery, and u is the velocity of the fl uid calculated using the actual cross section of the artery/pipe. This simplifi es to the following equation:

7. Which of the following is not a

method of preventing infection with

bio-implants?

a. titanium dioxide coating

b. silver nanoparticle coating

c. carbon nanotube coating

d. antibiotic coating

e. presurgery sterilization

8. Calculate the Reynolds number of

a bioprosthetic aortic valve that has

a hydraulic diameter of 15 mm, and

determine if the fl ow is turbulent or

laminar. The blood’s fl ow rate after the

valve is replaced is determined to be

0.020 m3/s.

a. 1,600; laminar

b. 16,000; laminar

c. 16,000; turbulent

d. 64,000; laminar

e. 64,000; turbulent

9. What is the velocity of blood after it

passes through an 18 mm diameter

bioprosthetic valve in a 20 mm

diameter artery if the pressure drop

across the valve is 1 Pa?

a. 0.002 m/s d. 0.5 m/s

b. 0.0075 m/s e 2.0 m/s

c. 0.075 m/s

Questions

Re � �ud

� ____ �

Additional Notes

Notes

• The Reynolds number over 2,000 is turbulent. • The viscosity of blood at body temperature is 2.7 � 10�3 Pas.

Equation 1-3



• The shear force across a tilting disk valve is A 3�Q

____ �r3 .

• The shear force across a bileafl et valve is A 8�u ______

�_____

4A/� .

10. Calculate the shear force across

a disk valve and a bileafl et valve.

Assuming that they are composed of

the same material and have the same

radius, which valve would be less likely

to cause blood clots?

A � cross- sectional area of the valve,

artery diameter � 18 mm, valve

diameter � 15 mm,

�p � 1 Pa, and u � 0.05 m/s

a. tilting disk valve

b. bileafl et valve

c. Neither valve will cause blood

clots.

d. Both valves are equally likely to

cause blood clots.



Questions

Additional Notes

Notes

Introduction

The hip joint is a ball and socket joint. The spherical head of the femur (ball) fi ts snugly into the socket of the pelvic girdle, called the acetabulum. The surfaces of both the femur and the acetabulum have cartilage tissue at the joint interface to allow for smooth movement of the joint. When the hip joint is diseased or damaged, doctors may perform hip replacement surgery depending on the severity of the impairment.

Osteoarthritis is one of the top reasons for hip replacement surgeries in the United States. Each year, doctors perform between 200,000 and 300,000 of these surgeries, replacing a damaged joint with a prosthetic. Given the number of replacement procedures performed, the design of a hip replacement device is a prominent challenge in biomedical engineering. Chemical, biomedical, and mechanical engineers play important roles in understanding the mechanics of the joint and studying the change in joint function

in injury and disease. Materials engineers and biomedical engineers design and develop joint replacement devices that are used in cases of osteoarthritis or injury-induced damage.

Background

The structure of the hip joint is shown in Figure 2-1. The range of motion is determined by the angle between the shaft of the femur and the femoral head. Normally, this angle varies between 90° and 135°. This angle can decrease as a person ages, affecting the load-bearing capability of the joint. The hip joint has three degrees of freedom: extension, rotation, and torsion. These degrees of freedom allow people to perform a wide range of actions, such as walking, running, and twisting.

FemoralHead

Femur

Figure 2-1: Detailed view of the hip joint.

TEAMS Competition 2012Scenario 2:Biomechanics Joint Replacement Hip Prosthesis


The ideal biomaterial for a hip joint is one that can with-stand the forces, provide range of motion, and not interact with biological materials present in the body. In the early 1920s, surgeons experimented with various materials to develop hip prostheses. One of the chosen materials was glass, which proved to be too fragile. In the 1940s, implants were coated, which also proved unreliable, and patients continued to have pain. Finally, in the late 1960s, materials such as cobalt chrome alloy or titanium alloy were used for hip prosthesis.

FemoralHead

Femur

N � Wb

Wleg

y

x

Figure 2-2: Forces acting on a hip joint.

Figure 2-3: X-ray showing a hip replacement device.Source: http://en.wikipedia.org/wiki/Hip_replacement; Public Domain

The hip joint supports the body and experiences a wide variety of forces during walking, running, and stretching. The forces experienced by the hip joint are rather large and are one reason why this joint is susceptible to pain. Force balance in the hip joint involves two forces acting on the body: the weight of the body acting downward and a normal force acting upward. A force balance is shown in Figure 2-2.


Titanium is a light metal that does not corrode easily and does not react with biological materials; it is lighter and stronger than stainless steel. For these reasons, it is the preferred biomaterial for designing a hip joint. The material properties of the titanium alloy used are:

• Young’s modulus � 116 GPa• Shear modulus � 44 GPa• Bulk modulus � 110 GPa• Poisson ratio � 0.32

Choice of material is critical when designing a hip replace-ment device. From an engineering basis, the idea of a replacement is simple: A shaft and a ball with a hinged joint that provides the same range of motion and withstands forces similar to a hip joint should work in a simple pros-thetic device. However, challenges in implementation begin with identifying the type of material suitable for a device.


11. Determine normal stress acting on

the cross section of the titanium tube.

a. normal stress � 0.3 MPa

b. normal stress � 179.58 MPa

c. normal stress � 18.32 MPa

d. normal stress � 65.36 MPa

e. normal stress � 217.86 MPa

12. If the tube is stretched by 2.5 cm,

what is the strain experienced? What

is the fi nal diameter of the tube?

a. strain � 0.05;

fi nal diameter � 2.89 cm

b. strain � 2.5;


c. strain � 0.95;


d. strain � 45.5;


e. strain � 0.05;


13. Calculate the amount of torque and

torsion generated in a prosthesis,

when a person walks with a torque

angle of 70°.

a. torque � 179 N;

torsion � 372 N/m

b. torque � 179 N;

torsion � 12.6 � 103 N/m

c. torque � 86.2 N;

torsion � 6,070 N/m

d. torque � 86.2 N;

torsion � 179.3 N/m

e. torque � 191.1 N;

torsion � 13,457 N/m

QuestionsAdditional Assumptions and Givens

• The cross-sectional diameter of the tube is 2.84 cm.• The length of the prosthesis is 48 cm.• Force acting on the end of the tube is 0.3 times the weight

of the body.• The person receiving the implant weighs 65 kg.• Axial strain is the strain along the length of a rod and radial

strain is the strain across the cross section of a rod. Radial Strain � Poisson’s Ratio � Axial Strain.

• The angle between the femoral head and the shaft is 70°.• Poisson ratio is the ratio of radial strain to axial strain.• Given the angle, torque is calculated as T � lFsin�, where l is

the length of the femur and F is the force applied.• Torsion is defi ned by torque divided by the radius of the shaft.

Additional Notes


14. To design an implant that will mimic

the functions of the hip joint, which of

these characteristics must the implant

possess? (Choose all that apply.)

(I) a ball to replace the femoral head

(II) a socket to mimic the acetabulum

(III) a rigid joint

(IV) a liner inside the ball and socket

(V) a rough femoral head

a. I only d. I, III, IV, and V

b. I and II e. I, II, III, IV, and V

c. I, II, and IV

15. Which of these biomaterials would

your team choose for designing a hip

implant, if the ultimate tensile strength

of bone is 109 MPa? (There can be

more than one choice.)

(I) titanium

(II) hard steel

(III) rubber

(IV) polypropylene

(V) porcelain

a. I only d. I, III, IV, and V

b. I and II e. I, II, III, IV, and V

c. I, II, and IV


• The ultimate tensile strength is the maximum stress that the material can resist before fracture.

• The ultimate tensile strength of some biomaterials commonly used are: • porcelain � 55 MPa • hard steel � 827 MPa • rubber � 2.1 MPa• titanium � 1,040 MPa• polypropylene � 0.91 MPa

• Young’s modulus of bone � 1.72 � 104 MPa

Additional Notes

Notes



• The ultimate tensile strength of bone is 162 MPa.• The diameter of the bone is 2.84 cm.• The area moment of inertia for a solid cylindrical beam is

Ia � �a4

___ 4 , where a is the radius of the beam.

• Bending moment is defi ned as MB � �FI ____ 2 , where F is the force and l is the length.

• The formula relating bending moment, radius of curvature,

and moment of inertia is MB � �YIa ____ R , where R is the radius

of curvature, Ia is the area moment of inertia, Y is the Young’s modulus, and MB is the bending moment.

• For small angles, shear strain is given by tan(�).• Length of the titanium beam � 48 cm• Diameter of the beam � 2.84 cm• UTS of titanium � 1,040 MPa• Shear modulus is defi ned as shear stress over shear strain:

S � sheer stress _________ sheer strain �

F __ A ___

�x ___ L � FL ____

A�x

Additional Notes

16. What is the stress needed to stretch

the femur to a strain of 1%? What

is the stress needed to stretch the

titanium alloy tube to the same extent?

a. femur � 1.72 � 104 MPa;

titanium � 116 GPa

b. femur � 0.96 � 102 MPa;

titanium � 58 GPa

c. femur � 1.72 � 102 MPa;

titanium � 1.16 GPa

d. femur � 1.72 � 106 MPa;

titanium � 116 � 102 GPa

e. femur � 0.96 � 104 MPa;

titanium � 58 � 102 GPa

17. What is the maximum tensile stress

that a bone can withstand before

fracture?

a. 100 MPa d. 300 MPa

b. 162 MPa e. 1,040 MPa

c. 200 MPa

18. Calculate the load per unit length of

the titanium rod just before fracture.

Assume that the load is uniformly

distributed.

a. 106.2 MPa/m

b. 1,040 MPa/m

c. 1,066.66 MPa/m

d. 2,166.66 MPa/m

e. 3,150 MPa/m

Questions

Notes


Additional Notes

19. A force of 150 N is applied on the

midpoint of a titanium beam. Calculate

the bending moment and radius of

curvature at equilibrium.

a. bending moment � 72 N m;

radius of curvature � 205 m

b. bending moment � 36 N m;

radius of curvature � 1,020 m

c. bending moment � 150 N m;

radius of curvature � 9 � 10�4 m

d. bending moment � 36 N m;

radius of curvature � 102.9 m

e. bending moment � 72 N m;

radius of curvature � 102.9 m

20. If the shear angle is 10°, calculate the

shear strain and shear stress of the

titanium alloy.

a. shear strain � 10;

shear stress � 44 GPa

b. shear strain � 0.176;

shear stress � 7.76 GPa

c. shear strain � 0.2;


d. shear strain � 0.176;


e. shear strain � 0.1;

shear stress � 4.4 GPa

Questions

Notes

Introduction

The knee joint supports the weight of the body and transmits forces to the ground; it also allows for a large range of motion between the tibia and the femur. The surface where the femur and tibia meet is covered in articular cartilage and acts as a well-lubricated connection, allowing for smooth movement of the joint. The knee joint may lose its functionality in response to diseases, such as osteoarthritis, or because of injury. In such cases, a joint replacement is a widely accepted option. The basic idea of joint replacement surgery is to replace the diseased or damaged articular surface with a synthetic material. According to the American Academy of Orthopedic Surgeons, more than 500,000 knee replacement surgeries are conducted every year. Chemical engineers, biomedical engineers, mechanical engineers, and materials engineers work together to design effective and long-lasting joint replacements.

and the fi bula is the calf bone. When people refer to the knee joint, they primarily refer to the tibiofemoral joint, which is the articular joint between two of the longest and strongest bones in the body—the tibia and the femur. This joint is classifi ed as a double condyloid joint, or a modifi ed hinge joint, which contains both a hinge joint and a pivot joint. A hinge joint is a joint that only operates backward and forward, like a door opening and closing. A pivot joint allows rotation; the tibiofemoral joint, a pivot joint, has the ability for rotation, in addition to its hinge motion. The joint surface is covered by a layer of smooth tissue called cartilage, which allows for bones of the joint to slide on one another.

TEAMS Competition 2012Scenario 3:Biomechanics of Knee Prosthesis

Figure 3-1: Healthy knee joint showing the structure of the joint and the articular cartilage surface. Source: http://en.wikipedia.org/wiki/File:Knee_diagram.svg; Public Domain

Quadriceps

tendon

Patella (normally

in center of knee)

Medial collateral

ligament

Meniscus

Patellar tendon

(Ligament)

Quadriceps

muscles

Femur

Articular

cartilage

Lateral condyle

Posterior cruciate

ligament

Anterior cruciate

ligament

Lateral collateral

ligament

Fibula

Tibia

Background

The anatomy of a healthy knee joint is shown in Figure 3-1. The knee joint is a compound joint consisting of three smaller joints: tibiofemoral joint, patellofemoral joint, and the tibiofi bular joint, where the tibia is the shin bone, the femur is the thigh bone,

The knee joint is susceptible to injury because of the large mechanical demands on it and the reliance on soft articular cartilage for support. In osteoarthritis the cartilage is worn, resulting in friction and pain at the joint surface. Figure 3-2 illustrates the anatomy of a joint with arthritic damage.


The knee joint supports the body and experiences a wide variety of forces during walking, running, and stretching. The movements at the knee joint are fl exion (bending toward the body), extension (straightening away from the body), and internal and external rotation (turning in, toward, and away from the midline of the body, respectively). The normal range of movement in the knee joint is approximately 130°–145° in fl exion and 1°–2° in extension.

The knee joint is very complex. However, for design purposes, the fi rst implants were built as hinge joints. Modern designs mimic the movement of the knee joint more closely by including rolling and gliding motions as the knee bends. Now, knee implants come in more than 250 designs and include gender-specifi c implants. Implants are designed using metal and plastic components, with the metal always rubbing against the plastic to provide smooth movement. Components of implants can include metal parts for the lower end of the femur, the top part of the tibia, and the back surface of the knee cap (patella). Figure 3-3 shows the image of a knee implant used to replace a worn joint.

Choosing a material for designing knee implants depends on several factors. The material must be biocompatible and not elicit a rejection response from the immune system. Longevity of the implant is important; the prosthesis must be able to retain strength and shape for a long time. The material also must mimic the load-bearing patterns observed in a knee joint—weight bearing ability, fl exibility, and frictionless smooth movement. Titanium or cobalt-

chromium-based alloys are often chosen for metal parts, and high molecular weight polyethylene is used for plastic parts.


Titanium is a light metal that does not corrode easily and does not react with biological materials. Also, it is lighter and stronger than stainless steel. Therefore, it is the preferred biomaterial for designing a knee joint. The material properties of titanium alloy are:

• Young’s modulus � 116 GPa • Shear modulus � 44 GPa • Bulk modulus � 110 GPa • Poisson ratio � 0.32 • Yield strength � 940 MPa

The properties of human cortical bone are:

• Young’s modulus � 17.4 GPa • Shear modulus � 3.51 GPa • Poisson ratio � 0.39

Articular cartilage is an important part of a knee joint. It is a soft tissue with strength, so it is important to note its creep—the tendency of solids to permanently deform when subjected to stress and strength.

• Strength � 15 MPa • Creep modulus � 80 MPa

Figure 3-2: Anatomy of a knee with osteoarthritis where the meniscus is torn, the articular cartilage is worn, and bone spurs have formed. This condition results in chronic knee pain.

Figure 3-3: Total knee replacement hardware.Source: http://en.wikipedia.org/wiki/File:Knieprothese.png; CC BY 3.0; Graichen

Worn ArticularCartilage

Bone Spurs

Torn Meniscus


21. Determine peak normal stress acting

on the end of a patient’s femur during

normal walking.

a. 30 Pa d. 76 KPa

b. 3.3 KPa e. 302 KPa

c. 31 KPa

22. If the patient in Question 21 has a

knee replacement with a titanium

component of the same knee diameter,

would the prosthetic be able to

withstand the applied normal stress?

a. No, the normal stress is less than

the yield strength.

b. Yes, the normal stress is less than

the yield strength.

c. No, the normal stress is greater

than the yield strength.

d. Yes, the normal stress is greater

than the yield strength.

e. No, the normal stress is applied

perpendicular to the rod surface.

23. What is the minimum diameter that a

titanium rod in a prosthetic could have

and still function during the patient’s

normal use of his or her knee?

a. 0.4 mm d. 8 mm

b. 0.8 mm e. 10 mm

c. 1 mm

Questions

Notes

Additional Notes


24. Calculate the amount of torque

generated in a prosthesis, if the angle

at the knee joint is 17°.

a. 23 N m d. 105 N m

b. 48 N m e. 360 N m

c. 58 N m

25. What is an ideal interaction surface

for a knee joint replacement?

a. metal-metal d. metal-rubber

b. metal-plastic e. plastic-rubber

c. plastic-plastic

26. The patient is an avid golfer with a

prosthetic knee that must hold up

to forces experienced during a golf

game. Assuming an engineering safety

factor of 1.5 and the patient’s weight

of 80 kg, will the polyethylene-titanium

prosthetic knee work for this patient?

a. No, the polyethylene part does not

meet specs.

b. No, the titanium part does not

meet specs.

c. No, both the titanium and the poly-

ethylene parts do not meet specs.

d. Yes, both the titanium and

polyethylene parts meet specs.



Questions

Notes


• During normal walking, the peak weight on the end of a patient’s femur is 45 kg.

• The cross section diameter of the femur is 4.31 cm. • Poisson ratio is the ratio of radial strain to axial strain. • Given the angle, torque is calculated as T � rF sin( � ), where r

is the length of the femur and F is the force applied.

Additional Notes


27. What is peak strain experienced by

the titanium and plastic components

of the patient’s prosthetic knee?

a. titanium 3.9%, polyethylene 2.3%

b. titanium 0.97 � 10�4%,

polyethylene 2.3 � 10�2%

c. titanium 9.7%, polyethylene 9.3%

d. titanium 3.9 � 10�3%,


e. titanium 0.39%,


28. When a patient is squatting, the creep

modulus of the articular cartilage is

80 MPa. If the force of the patient’s

weight on the articular surface is

300 N, what is the strain on the

cartilage if it is 3 cm in diameter?

a. 1.3 � 10�7% d. 0.3%

b. 5 � 10�7% e. 0.5%

c. 0.13%

29. The presence of a knee cap increases

the moment arm of the joint. If

M � 300 N, calculate the compressive

force acting on the knee cap.

a. 20.5 N d. 95.7 N

b. 30 N e. 205.2 N

c. 86.7 N

Notes

Questions Additional Assumptions and Givens

• Young’s modulus of tibia bone � 20.1 GPa • The chart below shows the compressive forces experienced

by a knee during a golf swing. The y-axis is given in multiples of body weight.

• The titanium portions of the prosthetic have a diameter of 4 cm.

• The polyethylene portions of the prosthetic have a diameter of 8 cm.

• The yield strength of polyethylene is 27 MPa; the Young’s modulus of polyethylene is 1.2 GPa.

Additional Notes

Compressive Knee Force during a Golf Swing

Co

mp

ressiv

e F

orc

e in

xB

od

y W

eig

ht

2.5

2

1.5

1

0.5

5

4.5

4

3.5

3

00 1 2 3 4 5

Time (s)

Figure 3-4: Chart of compressive knee forces during a golf swing.


30. For an applied force of 300 N,

calculate the stiffness of a piece of

cartilage that elongates by 0.1 cm in

the direction of applied force.

a. 3 N/m d. 30,000 N/m

b. 300 N/m e. 300,000 N/m

c. 3,000 N/m

Questions

Notes

Additional Notes

F � cos � � cos �

___________

cos � M,

where

• � � 45°, � � 5°, and � � arctan sin � � sin �

___________

cos � � cos �

Additional Background

Hyaline cartilage covers the articular surface and allows for smooth movement of the knee joint. Over time, creep and wear cause damage to the cartilage; only fi brous cartilage can be used for replacement purposes, and it does not perform as well as hyaline cartilage. Loss of joint function and pain occurs.


• Creep modulus is equal to the constant applied stress divided by the total strain at a specifi c moment in time.

• M is the muscle force acting on the knee joint. • The compressive force on the knee cap is

Equation 3-1

Introduction

Drug design is the process of fi nding new medications based on an understanding of biochemical signaling, a series of chemical reactions responsible for the transmission of information from the external environment to the inside of a cell. Pharmacologic drugs are usually small synthetic molecules that target a specifi c biochemical signaling pathway. Often in diseases, a particular signaling pathway is overactivated, which means that the end result of the signaling is no longer regulated in the body. For example, in diabetes, the body may not produce insulin properly (type 1), or it may not use the produced insulin properly (type 2). In cases such as these, using synthetic insulin allows patients to metabolize carbohydrates effectively and keep blood sugar under control. Given the vast number of diseases facing humans, effective drug design continues to be a medical challenge. Chemical engineers, biomedical engineers, and computer engineers work together to develop strategies for effective drug design.

molecule to an active form. Proteins inside a cell that sense this activated form of the receptor activate more reactions inside the cell. Figure 4-1 shows an example of a biochemical reaction. Like other reactions, a biochemical reaction is characterized by reaction rates and parameters. Not all reactions inside a cell are binding reactions; many biochemical reactions are characterized by catalysts called enzymes that speed up the rate of the reaction. The end result of a signaling pathway is that the cell makes a decision for the pathway. Outcomes include gene expression or cell proliferation. In many diseases, the decision-making ability or the regulation of the signaling can be damaged. Designing a drug that will restore the functionality of the process is important for the treatment to be effective. Drugs can be designed in two main ways: ligand-based design and structure-based design. In ligand-based drug design, a small molecule acts as a ligand for a receptor in a signaling pathway, turning on or off a signaling pathway. In some diseases, such as diabetes, it is important to turn on the metabolic pathways that are required for metabolizing carbohydrates. On the other hand, for diseases such as cancer, it is important to turn off the cell proliferation pathway to control the rate of cell division. With structure-based drug design, computer programs analyze the structure of a target protein and develop synthetic molecules that mimic the function of the protein of interest. Figure 4-2 highlights the industrial fl owchart for both of these design strategies. Here, the pharmacophore is the ligand.

Assumptions and Givens For a binding chemical reaction, given as

Ligand � Receptor k1

k2

Ligand-receptor complex

L � R k1

k2

L � R

TEAMS Competition 2012Scenario 4: Drug Development

Background

The common theme in signaling pathways is that a molecule called a ligand, which is present in the external cellular environment, binds to a receptor found on the surface of a cell. This binding event is a chemical reaction and often changes the confi guration of the receptor


pharmacophoreidentification

pharmacophoremodification

fit for thereceptorNo

No

No

Yes

Yes

Yes

potential drugpotential drug

active siteidentification

ligand fragmentsgrowing

completegrowing

fit for thereceptor

change fragment

Figure 4-2: Logical steps in drug design. The left panel is using a ligand-based drug design and the right panel is the structure-based determination of drug design.

Figure 4-1: A signaling network showing some major signaling pathways.Source: http://en.wikipedia.org/wiki/File:Signal_transduction_v1.png; CC BY-SA 3.0; Roadnottaken

the rate of the reaction is given as

Equation 4-1

� d[L]

___ dt ��

d[R] ____ dt �

d[LR] ____ dt � k1[L][R] � k2[LR]

where the forward rate is given by the fi rst term and the backward rate is given by the second term.

When the reaction is an enzyme-catalyzed reaction, for example,

Substrate Enzyme, kcat Product

S Enzyme, kcat P

the rate of reaction is given by

Equation 4-2

� d[S]

____ dt �

d[P] ____ dt �

kcat[S][E] _______ Km � [S]

where k cat and K m are enzyme properties.


31. Calculate the rate of forward reaction

for a ligand receptor binding reaction.

a. 5 � 10�2 n/M � s d. 12.5 nM/s

b. 1.25 nM/s e. 50 nM

c. 5 nM/s

32. What is the rate of backward reaction?

a. 6.2 � 10�2 nM/s d. 12.5 nM/s

b. 6.2 nM/s e. 50 nM

c. 5 nM/s

33. What is the rate of total reaction at

equilibrium?

a. 0 d. 25 nM/s

b. 6.2 � 10�2 nM/s e. 12.4 nM/s

c. 12.5 nM/s

Questions

Notes


• Initial ligand concentration � 5 nM • Initial receptor concentration � 50 nM • Initial concentration of ligand receptor complex � 6.2 nM • k 1 � 5 � 10 �2 nM �1 s �1 and k 2 � 1 � 10 �2 s �1

• Dissociation constant, K d , is defi ned as the ratio k2 __

k1 . The

smaller the dissociation constant, the tighter the binding of the ligand to the receptor.

Additional Notes



• Substrate concentration [S] � 10 � M • k cat � 5 s �1 , K m � 1 � M, and [E ] � 0.5 � M • Rate of an enzymatic reaction

Equation 4-3

rate � � d[S]

____ dt �

d[P] ____ dt �

kcat[S][E] _______ Km � [S]

34. The goal for your team is to design a

drug that turns off a signaling pathway

by binding tightly to the receptor.

There are two compounds that may

be suitable. Compound A has a Kd of

4 nM and compound B has a Kd of

1 �M. Which drug would be more

suited to meet the goal?

a. A d. A or B

b. B e. neither

c. A and B

35. Which of the following is not true of

an enzyme?

a. An enzyme functions as a catalyst.

b. An enzyme is a protein.

c. An enzyme is consumed during the

course of the reaction.

d. Enzyme kinetics are nonlinear.

e. Enzymes are regulated by signaling.

36. Calculate the rate of an enzyme-

catalyzed reaction.

a. 2.3 �M/s d. 50 �M/s

b. 2.5 �M/s e. 250 �M/s

c. 11 �M/s

Notes

Questions

Additional Notes



• Sometimes, two ligands are required to activate the receptor. In this case, the rate of the reaction is given by

Equation 4-4

L1 � L2 � R kon

koff

L1L2R

reaction rate � � d[L1] ____

dt � � d[L2] ____

dt

� � d[R]

____ dt �

d[L1L2R] ______ dt

� �koff [L1L2R] � kon[L1] [L2] [L3]

37. What will happen to the reaction rate

if the concentration of the enzyme is

doubled?

a. Nothing, the enzyme does not

participate in the reaction.

b. The reaction rate will be doubled.

c. The reaction rate will be halved.

d. The reaction rate will be squared.

e. The reaction rate will grow

exponentially.

38. Calculate the reaction rate for the

enzyme reaction in Question 36 if the

substrate concentration is halved.

a. 0.2 �M/s d. 2.1 �M/s

b. 0.5 �M/s e. 5.0 �M/s

c. 1.0 �M/s

39. For a given receptor, two ligands need

to bind simultaneously to activate the

receptor. Calculate the rate of the

reaction in this case.

a. 0.563 nM/s d. 563 nM/s

b. 5.63 nM/s e. 5,063 nM/s

c. 56.3 nM/s

Notes

Questions

Additional Notes

• k 1 � 1.41 nM �2 s� 1 , k 2 � 0.5 s �1 • [L 1 ] � 5 nM, [L 2 ] � 10 nM, [R] � 8 nM, [L 1 L 2 R] � 2 nM



• Dosage � 1. 5 mg/kg • 1 lb � 0.454 kg 40. Drug dosage is determined by body

weight. Calculate the dosage of a drug

required by a patient whose weight is

140 lbs.

a. 0.21 mg d. 140 mg

b. 9.5 mg e. 210 mg

c. 95 mg

Notes

Questions

Additional Notes

Introduction

With life spans increasing due to better medical treatment, effective dental implants are important for a good quality of life. Engineers play signifi cant roles in developing dental implants, from choosing implant materials to designing and manufacturing implants. Materials engineers create alloys that can withstand the harsh environment of a mouth and the intense biting forces of which teeth are capable.

jaw and then becomes fused with the bone (called osseointegration), as shown in Figure 5-1. An artifi cial replacement tooth is placed on top of the implant. An implant can serve as a base for one or more replacement teeth. The most common materials used for dental implants are titanium alloy (a metal) and zirconium dioxide (a ceramic). Zirconium dioxide is more aesthetically pleasing and less likely to cause an allergic reaction than titanium alloy, but it is more expensive than titanium alloy.

The task of your engineering team is to develop a dental implant. Your team will need to consider the forces acting on the implant and make sure that the material chosen can withstand these forces. Other considerations include effects of temperature changes, hardness, reactivity with food, and resistance to corrosion.

Background

Two important quantities that engineers use to analyze forces on an object are stress and strain. Stress is defi ned as the force divided by the cross-sectional area over which it acts. Stress ( � ) is a measure of the internal forces on an object and is defi ned by the following equation.

TEAMS Competition 2012Scenario 5: Engineering and Biomechanics of Dental and Facial Implants

A dental implant can replace a missing tooth or a group of teeth. A patient can eat and speak more effectively when an artifi cial replacement is provided for a lost tooth. A missing tooth contributes to bone loss in the jaw, so one lost tooth can lead to additional problems with teeth and jaws if it is not replaced. A dental implant is a post that is placed into the bone of the

Figure 5-1: Dental X-ray showing implant in second spot from the right. Source: http://en.wikipedia.org/wiki/Dental_implant; CC BY-SA 3.0; Drosenbach

Equation 5-1

� � F __ A


Equation 5-2

� � �L ___ L

where �L is the change in length of an object due to a stress, and L is the original length of the object

F F

L�L

Figure 5-3: Sketch showing the small deformation as a result of an applied force.

The cross-sectional area on which the force, F , acts is denoted by A in Figure 5-2. By normalizing the force with the cross-sectional area, the stress over any section of an object can be computed as long as the composition of the object is uniform. The force can have a component that is parallel and perpendicular to the cross-sectional area. Stress due to the perpendicular component of the force is also known as the pressure.

A pair of forces can be either compressive or tensile. If the forces are pulling away from each other, then the object is in tension, as shown in Figure 5-2. Compression forces are in the opposite direction, pushing toward each other. Some materials, such as ceramics, are much stronger in compression than in tension.

The forces on an object cause it to change shape, although these changes are often too small to notice. When you sit on a mattress, it deforms or changes shape. If you sit on a tree stump, you are pushing the top of it closer to the bottom with your weight. The amount that the tree stump shortens is related to strain. If a force is not too large, then an object will return to its original shape when the force is removed. If you stretch a rubber band, it will usually return to its original shape. However, if you stretch the rubber band excessively, it will break. The stress that causes the rubber band to break is called the ultimate strength . Another important value of stress is called the yield strength , which is the maximum amount of stress a material can withstand and still return to its original shape. If the object goes back to its original shape when the force is removed, then the force causes an elastic deformation . If the object does not go back to its original shape, then the object undergoes a plastic deformation . Strain is often denoted by the symbol ε and is defi ned by the following equation.

Figure 5-2: Internal strain is caused by an external force. Source: http://en.wikipedia.org/wiki/Mechanical_stress; CC BY-SA 3.0; David Richfi eld

A graph of stress as a function of strain can give useful information about the behavior of the material when a force is applied. One type of stress–strain diagram is shown in Figure 5-4. Many materials have a linear relationship between stress and strain for small values of stress. The slope of the line related to this relationship is called Young’s modulus (E ) and it is a property that represents the material’s stiffness. In the linear region,

Equation 5-3

E � stress _____ strain � � __

�

Figure 5-4: Stress vs. Strain curve for structural steel. Reference numbers are: * 1 - Ultimate Strength * 2 - Yield Strength(elastic limit) * 3 - Rupture * 4 - Strain hardening region * 5 - Necking region * A: Engineering Stress. Source: http://en.wikipedia.org/wiki/Ultimate_tensile_strength; CC: BY-SA 3.0; David Richfi eld

1

23

3

4 5

A

B

Stress

Strain

Location 2 in Figure 5-4 is called the elastic limit . If the applied stress is greater than this, then the material will not return to its original shape. The stress at the elastic limit is called the yield strength (point 2 in Figure 5-4) and the ultimate tensile strength is shown at point 1 in Figure 5-4. Since plastic deformation is not acceptable in a dental implant, the stress on the implant must remain below the yield strength.


• Force on teeth due to biting � 1,200 N • Contact area of the teeth � 30 mm 2 • 1 Pa � 1 N/m 2 • 1 kPa � 1 � 10 3 Pa , 1 MPa � 1 � 10 6 Pa , 1 GPa � 1 � 10 9 Pa



• Stress on dental implant, � � 25 MPa • Length of implant � 10 mm • Young’s modulus, E � 110 � 10 9 N/m 2

41. Assuming the conditions given, what is

the stress on the teeth due to biting?

a. 7.5 MPa d. 40 MPa

b. 16 MPa e. 120 MPa

c. 30 MPa

42. The force on the implant from a

person biting down on it will cause

the implant to compress. What will be

the change in the size of the implant

due to this force?

a. 0.0002 mm d. 0.2 mm

b. 0.002 mm e. 2 mm

c. 0.02 mm

43. Can a titanium alloy implant 4 mm in

diameter hold up to a 1,000 N force

exerted by the teeth?

a. Yes, the stress due to the force is

less than the yield strength.

b. Yes, the stress due to the force is less

than the ultimate tensile strength.

c. No, the stress due to the force is

beyond elastic region.

d. No, the stress due to the force is in

the plastic deformation region.

e. No, the stress due to the force is

beyond the fracture point.

Questions

Notes

Additional Notes


44. Why would a titanium alloy implant

be preferred over a stainless steel

implant according to the information

in Table 5-1?

a. Because titanium alloy has a

greater yield strength

b. Because titanium alloy has a lower

Young’s modulus

c. Because alloys have more favorable

material properties

d. Because the ultimate tensile

strength of stainless steel is not

large enough to withstand biting

forces

e. Because stainless steel has a high

Young’s modulus

Questions Additional Assumptions and Givens

Notes

Additional Notes

Material Young’s Modulus

Yield Strength

Ultimate Tensile Strength

Coeffi cient of thermal expansion (K�1) at 20°C

Aluminum 68 MPa 95 MPa 110 MPa 23.1 � 10�6

Stainless steel

200 MPa 500 MPa 860 MPa

Titanium alloy

110 MPa 730 MPa 900 MPa 8.6 � 10�6

Table 5-1: Properties of some materials used in dental implants.

• Stress on dental implant due to normal biting forces, � � 25 MPa• Diameter of implant � 4 mm2



One of the newer materials used for dental implants is zirconium, often in the form of zirconium dioxide. Unlike titanium alloy, zirconium is a ceramic and its color is similar to that of natural teeth. Zirconium has some advantages over titanium. It is believed to be stronger and less likely to cause allergic reactions than titanium dioxide. On the other hand, it is a much more expensive material. The processing of zirconium has a signifi cant effect on its properties. Your team has been asked to examine zirconium dioxide to assess its suitability for dental implants.

Additional Assumptions and Givens • Tooth dimensions are approximately 5 mm length, 5 mm

width, and 10 mm depth • In a laboratory, elastic deformation of a zirconium dioxide

implant was observed up to a maximum force of 1,750 N pushing down vertically on the tooth.

• Elastic deformation as a result of the above force � 0.0007 mm • Zirconium dioxide is a ceramic, which fractures before it yields.

Additional Background Several implants were mislabeled before testing. Your team needs to determine the kind of materials used for the implants.


• Dimensions of the implants are approximately 5 mm length, 5 mm width, and 10 mm depth

• Force � 3,750 N

45. Compute Young’s modulus for

zirconium dioxide.

a. 70 MPa d. 70 GPa

b. 100 MPa e. 100 GPa

c. 700 MPa

Questions

Notes

46. A patient’s implant is bent out of

shape as a result of a force of 3,750 N.

What material is the implant?

a. aluminum

b. natural tooth

c. stainless steel

d. titanium alloy

e. zirconium dioxide

Additional Notes

Notes


Additional Background A change in temperature can cause a deformation in an implant and, therefore, contribute to a strain similar to that in Equation 5-2. A coeffi cient of thermal expansion ( � ) gives a measure of the amount of deformation due to a change in temperature. This coeffi cient is a function of temperature, but over a relatively small temperature range, it can be considered constant.

The expression for the change in volume of an object due to a change in temperature is given by the following equation.

47. A dentist has not decided whether

to use a zirconium or titanium alloy

implant. What is the largest possible

change in volume caused by the

patient drinking hot tea?

a. 0.001 mm3 d. 0.054 mm3

b. 0.017 mm3 e. 0.15 mm3

c. 0.036 mm3

48. A dental implant should have which of

the following properties?

a. Low E, low �, and low yield stress

b. High E, low �, and high yield stress

c. Low E, high �, and low yield stress

d. High E, high �, and low yield stress

e. High E, high �, and high yield stress

Notes

Questions

Equation 5-4

�V ___ V = � � �T


• Coeffi cient of thermal expansion of zirconium, � � 5.7 � 10� 6 K �1 of titanium alloy, � � 8.6 � 10 �6 K �1

• Assume these coeffi cients of thermal expansion are valid for the temperature range in this problem.

• Body temperature � 37°C • Temperature of hot tea � 87°C • Implant is a cylinder of diameter 4 mm and length 10 mm

Additional Notes


49. What is the magnitude of the shear

stress in Figure 5-7?

Notes

Questions

a. 0 Pa d. 7.5 MPa

b. 7.5 kPa e. 30 MPa

c. 1.9 MPa


Normal stress is perpendicular to the tooth surface. However, internal forces can also be parallel to the cross-sectional area of interest, causing shear stresses.

Engineers use free body diagrams to examine the forces on an object. A free body diagram of a simply supported beam is shown in Figure 5-5. This sketch shows the applied force F as well as the forces on the beam from supports at either end. From this diagram, you can compute the following because the sum of all the forces on the beam must equal zero.

Equation 5-5

F � Support A Support B

Figure 5-6: Photo of bridge used to replace a lost tooth. The middle tooth is the one that is missing and the teeth on either side support the bridge. Source: http://en.wikipedia.org/wiki/Dental_bridge; CC: BY-SA 3.0; Russ Wagoner

Support A Support B

F

Figure 5-5: Free body diagram of a simply supported beam.

Figure 5-6 shows a dental bridge, which can be used to replace a missing tooth. The end tooth on the implant fi ts over actual tooth and the middle part replaces a missing tooth. Model the bridge as a simply supported beam to simplify the calculations. In actual practice, the forces at the supports include other components and not just the vertical component.

Shear

Support B

F

Figure 5-7: Free body diagram of a section of bridge showing the forces acting on it.

Hint: Suppose you slice the bridge

at the location where you want to

compute the internal shear stress

(between the leftmost and middle

teeth). The unknown shear is

expressed as a force on this section of

the bridge, as shown in the free body

diagram in Figure 5-7.

• Cross-sectional area of the dental bridge is approximately square with dimensions 4 mm � 4mm.



Engineers can improve the design of current implants by examining the causes of failure. One common cause of failure is not osseointegrating properly. Statistical analysis helps engineers discover the conditions under which failure is most likely to occur. Here is some initial data that will help focus engineers’ efforts in searching for problems.


Total Titanium Implants

Total Stainless Steel Implants

Titanium Implants Failed

Stainless Steel Implants Failed

Offi ce #1 48 80 5 1

Offi ce #2 21 138 4 1

Table 5-2: Rate of failure of titanium and stainless steel dental implants. Data was taken from two different dental offi ces over a period of 2 years.

Questions

50. What is the ratio of the percentage of

total failures of titanium alloy implants

to the percentage of total failures of

stainless steel implants?

a. 1:1 d. 14:1

b. 8:1 e. 19:1

c. 13:1

Notes

Additional Notes

Introduction

In order to help keep people healthy, doctors need to be able to take accurate measurements related to how internal body organs function. This data must be properly interpreted and accessible to health care professionals. Engineers play vital roles in developing the tools used to take accurate measurements of important information such as blood pressure and temperature. Electrical and mechanical engineers design the tools that doctors use to help keep people healthy, including thermometers and sphygmomanometers (blood pressure meters). Chemical engineers are important members of the team that develop medicines, while biomedical engineers use their understanding of the mechanics of the human body and human body systems to develop medical devices, such as stents to regulate blood fl ow.

Thermoregulation is the body’s ability to keep its internal temperature constant. A temperature that is too high results in a fever or hyperthermia; if the temperature is too low, it results in hypothermia. A fever is a sign that the body is responding to an external agent, such as a bacteria or virus, by increasing its core temperature. Prolonged fevers

can be indicative of a disease. There are cases of hypo- and hyperthermia in which the body temperature changes beyond control. Exposure to extreme cold leads to hypothermia, and heat stroke is one type of hyperthermia.

Circulatory System

TEAMS Competition 2012Scenario 6:Bioheat Transfer and Fluid Dynamicsin the Human Body

Figure 6-1: The human circulatory system showing the paths along which blood travels.Source: http://en.wikipedia.org/wiki/File:Circulatory_System_en.svg; Public Domain


Background

Blood fl owing near the skin transfers heat to the environment to help maintain a constant body temperature. Three ways to transport heat are conduction, convection, and radiation. Conduction transfers heat on a molecular level. If you cook hamburgers in a frying pan, the transfer of heat from the pan directly to the hamburger is conduction. Convection transfers heat using a moving fl uid, so if you put your hand over the pan, you would feel heat as a result of convection. Radiation uses electromagnetic waves to transfer heat, so it does not require any contact between two substances. The sun heating the earth’s surface is an example of radiation heat transfer (i.e., moving air does not carry the sun’s heat through space to the earth). In addition

to blood circulation, perspiration helps maintain body temperature because energy is required to evaporate sweat. Generally, convection and radiation are more signifi cant heat transfer methods than conduction, which is a slow process.

An amount of heat can be measured in terms of BTUs in English units or Joules (J) in SI units. Heat transfer is usually computed as a rate of heat gained or lost per unit time and is measured in Watts, where 1 W � 1 J/s. The total amount of heat that an object emits is the sum of conduction, convection, and radiation. The rate of radiation that heat transfer from a surface is computed as follows:

Equation 6-1

qrad � � ( T 4 � Tamb ) Awhere

• qrad � rate of heat transfer due to radiation

• T � temperature of the body in Kelvin (K)• Tamb � temperature of the surroundings in Kelvin (K)

• A � surface area of the body

• � � 5.6703 x 10�8 W/(m2K4), which is called the Stefan-Boltzmann constant

The temperature in the radiation equation must be in units of Kelvin. The Kelvin scale is similar to the Celsius scale, except that absolute zero temperature is defi ned as 0 K. Celsius can be converted to Kelvin as follows:

Equation 6-2

T(°C) � 273 � T(K)

The rate of convective heat transfer from the body into ambient temperature is computed as follows:

Equation 6-3

qconv � h A (T � Tamb)

where

• qconv � rate of heat transfer due to convection (W)• A � surface area of the body (m2)• h � convective heat transfer coeffi cient of the process

(W/(m2K) or W/(m2°C)),• T � temperature of the body• Tamb � temperature of the environment

The circulatory system is responsible for transporting blood and delivering oxygen and nutrients in the body. Blood vessels are designed to help the body maintain a relatively constant fl ow rate. Because eating, digesting, and any kind of physical activity produces heat, the body must be able to release that heat to the environment in order to maintain a constant body temperature. Blood fl ow through the circulatory system is driven by the heart, and blood pressure measurements help doctors determine the health of the heart. If blood pressure is too high, it can mean that the heart is pumping blood too rigorously; if blood pressure is too low, it can mean that the heart is weak.

The human body can adjust blood fl ow according to its perception of heat or cold. It can increase or decrease the diameter of the arteries or veins to adjust the amount of blood fl ow near the skin and the amount of heat released to the environment.

The mission of your engineering team is to help doctors accurately measure and analyze temperature and blood pressure values.



The energy in food is generally measured in terms of Calories (with a capital C).


• Internal body temperature is 37°C• Assume skin temperature is about 30°C• Body surface area is 2 m2 (average for an adult)• Cold outside temperature is 0°C• Warm outside temperature is 33°C• h (convective heat transfer coeffi cient) � 10 W/(m2K)• 1 calorie � 4.2 Joules• 1 Calorie � 1,000 calories

51. Which person in the following

examples is the most comfortable,

temperature-wise?

a. person is sitting, room is 30oC

b. person is running on a treadmill,

room is 30oC

c. person is reading, room is 20oC

d. person is eating, room is 15oC

e. person is sleeping, room is 40oC

Questions

NotesAdditional Notes


52. Calculate the rate at which the human

body radiates heat in a cold room.

a. 0.1 W d. 326 W

b. 0.2 W e. 420 W

c. 30 W

53. Calculate heat fl ow around the human

body standing outside on a hot day.

a. �60 W into body

b. 60 W away from the body

c. 80 W away from the body

d. 80 W into body

e. 100 W away from the body

54. Determine the heat fl ow around the

human body outside on a cold day.

a. 140 W d. 740 W

b. 300 W e. 1,000 W

c. 600 W

55. Calculate the calories that will be lost

in one hour on a cold day

due to heat exchange with the

surroundings. Note that simplifying

assumptions are being made about

how the human body uses energy.

a. 143 C d. 800 C

b. 514 C e. 826 C

c. 740 C

Additional Notes


• On a cold day, the radiative heat transfer between aperson and his or her surroundings is 410 W and the conductive heat transfer is 516 W.

• On a hot day, the radiative heat transfer between a person and his or her surroundings is 43 W and the conductive heat transfer is 75 W.

Notes



The rate at which the human body burns energy is called the metabolic rate. It can be expressed in terms of a person’s surface area or weight.


• Express the metabolic rate in terms of surface area, in units of kJ/(m2hr).

• A person is walking at approximately 2 mph on fl at ground. • Assume that he or she must dissipate 200 W as excess heat. • This person’s body must dissipate about 80% of its energy as

excess heat.


Assume that a blood vessel is shaped like a tube. The cross-sectional area, A, is

Equation 6-4

A � � ( d __ 2 ) 2where

• � � 3.14• d � diameter of the blood vessel

Then the volumetric fl ow rate is related to the fl uid velocity and cross-sectional area as follows:

Equation 6-5

v = V __ A

where

• v � blood velocity• V � blood volumetric fl ow rate• A � cross-sectional area through which the blood fl ows

For this calculation, assume that blood fl ow out of the heart is at a constant speed and uniform across the cross-sectional area, and that the cross-sectional area of the location where the blood leaves the heart remains constant.


• Average volume of blood that the heart pumps (volumetricfl ow rate) � 5 liters/second

• Cross-sectional area of exit from the heart is a circle with aconstant radius of 1 cm

• 1 liter � 1,000 cm3

56. What is this person’s metabolic rate?

a. 6 kJ/(m2hr)

b. 7.5 kJ/(min � m2)

c. 125 kJ/(m2hr)

d. 200 kJ/(m2hr)

e. 250 kJ/(m2hr)

Notes

Questions

57. Calculate the exit blood velocity from

a pumping heart.

a. 1.6 cm/s d. 188 cm/s

b. 27 cm/s e 260 cm/s

c. 83 cm/s



Density is the amount of mass in a unit volume of a substance. A manometer is an instrument used to measure pressure. Manometers consist of a liquid inside a u-shaped tube with one end open to the atmosphere and the other end exposed to an unknown pressure to be measured. Since the gases at either end of the tube are at two different pressures, the liquid is displaced at height H, as shown in Figure 6-2, to balance the two different forces. In the case of Figure 6-2, the pressure on the right side (Pa ) is greater than the pressure on the left side (Pb). The difference between these two pressures is proportional to the height and the density of the liquid. Water and mercury are the most common liquids used in a manometer. Because mercury has a much greater density than water, it can measure a greater pressure difference with a shorter tube. Pressure is often measured in units of millimeters of mercury or inches of water.


• Pa � P0 � density of the liquid times gravitationalacceleration times H, where Pa and P0 are pressuresas defi ned in Figure 6-2, and H is the difference in the liquid height between the two sides of themanometer.

• Density of mercury �13.5 g/cm3

• Density of water � 1 g/cm3

58. If blood pressure is measured in a

mercury manometer, what would

happen if your team used water

instead of mercury?

(Side note: Mercury manometers are

not commonly used today to measure

blood pressure.)

a. The manometer would not work

with water.

b. There would be no change in

the manometer if the mercury is

replaced with water.

c. The pressure difference would not

be proportional to the

liquid height.

d. A manometer that is 7.5� times

thicker would be needed.

e. A manometer that is 13 times

bigger would be needed.

Notes

Questions

Figure 6-2: The difference in fl uid height in a liquid-column manometer is proportional to the pressure difference.Source: http://en.wikipedia.org/wiki/File:Utube.PNG; CC BY-SA 3.0; Ruben Castelnuovo (Ub)



In the United States, blood pressure is most commonly measured in mm Hg (which means millimeters of mercury), but in some countries, it is measured in Pascals (Pa). A Pascal is defi ned to be 1 N/m2.


• Pa � 101 kPa (atmospheric pressure)• Density of mercury � 13.5 g/cm3

• Maximum blood pressure � 27 kPa• Minimum blood pressure measurement � 7 kPa• Gravitational acceleration � 9.8 m/s2

• 1 in. � 2.5 cm

59. Suppose your team is designing a

mercury-based blood pressure

measuring device. The device should

measure a maximum blood pressure

of 27 kPa and a minimum blood

pressure of 7 kPa. What is the

maximum H, in inches, your team

would need to consider for your

design?

a. 22 in. d. 94 in.

b. 28 in. e. 710 in.

c. 71 in.

Notes

Questions

Additional Notes



Engineers can model whole or parts of complex systems in a laboratory or with a computer simulation. Dimensional analysis is one tool that engineers use to help create a model in the laboratory. Dimensional analysis defi nes a quantity that has no units and describes the situation being observed.

Dimensional analysis can be used to defi ne a measurement called the Reynolds number. The Reynolds number, denoted as Re, is defi ned as

Equation 6-6

Re � UD ___ v

where

• U � average velocity• D � diameter of pipe• v � viscosity

Viscosity is a measurement of how easily a liquid or gas will fl ow.

An exact duplicate of the real life fl ow is not needed to model the behavior of a real life liquid or gas in a laboratory experiment or a computer simulation. An experiment can be a more convenient size or use a more convenient liquid or gas. According to dimensional analysis, if the Reynolds number of the real life fl ow and the experiment are the same, then the experiment will still properly imitate the real life fl ow.


• Air in human windpipe:• D � 3 cm• Volumetric fl ow rate � 1 liter/sec• The equation for volumetric fl ow rate is v � V __

A .• v (viscosity) of air � 16 � 10�6 m2/s

• In the experiment, using oil for liquid:• D � 10 cm• v (viscosity) of oil � 1.5 � 10�5 m2/s

Hint: Start by calculating the Reynolds number of the human windpipe using the information given. To ensure that the fl ow through the pipe in the laboratory is similar to fl ow through the windpipe, design the experiment so that the Reynolds number is the same in both.

60. What should the velocity of the liquid

in the experiment be so that your

team can model the fl ow through a

human windpipe? (For simplicity,

assume that your team can use a

vertical rigid pipe to model the

windpipe.)

a. 7.1 cm/s d. 2,644 cm/s

b. 40 cm/s e. 3,141 cm/s

c. 141 cm/s

Notes

Questions

Introduction

When researchers develop pharmaceutical drugs in the lab, the fi rst set of testing is carried out in animal models. For a medication to reach the market, there are many steps to ensure that the medicine is safe and any side effects are well documented. One important step in this process is a clinical trial. Clinical trials are research studies in which doctors and patients work together to fi nd ways to evaluate or improve a potential medication. Clinical data management pertains to the management, validation, and verifi cation of the data acquired during the trial. Designing effi cient and meaningful clinical data management systems is essential for clinical trials. Chemical engineers, bioengineers, and computer software engineers play a big role in designing data management systems in clinical trials. From document management systems to statistical analysis of data, industrial engineers manage and interpret the data of clinical trials. Software engineers design the programs to store and analyze the data, and electrical engineers develop the interface between the medical equipment that takes the data and the computer system that stores it.

Background

A Clinical Data Management System (CDMS) is a set of computer software tools used to store and manage patient data generated when a clinical trial is conducted. Depending on the application, a CDMS can be a part of many other clinical trial tools or it can be used as a stand-alone component. The goal of a CDMS is to organize and manage the data before the statistical analysis is done.

Different steps in the design of a clinical trial infl uence the design of the CDMS. Understanding how a clinical trial is designed is important for developing CDMS software. The critical steps in a clinical data management process are: • Development of the case report form • Database development and validation • Data entry, query, and correction • Data quality assurance • Data lock, archive, and transfer

One of the biggest challenges in creating a CDMS is ensuring that the data collected is relevant to the scientifi c objectives of the clinical trial. If the data collected does not satisfy this requirement, the conclusions drawn from the analysis may not be accurate and may be misleading. To avoid these problems, implementation of good data management practices (GDMP) is necessary. GDMP is a set of rules or standards that ensures that the data collected during a clinical trial meets the quality and integrity requirements set forth in the clinical trial. Figure 7-1 shows a fl owchart with the important steps in clinical data management.

Statisticians and programmers work together closely during the development of a CDMS. Many engineering decisions made during the development of a CDMS are logical, in terms of ensuring that the software developed is bug-free, can be implemented across various platforms, and is user friendly. Also, the database must honor patient privacy and follow ethical guidelines set forth by the FDA (Food and Drug Administration).

TEAMS Competition 2012 Scenario # 7 Clinical Data Management


developing a CDMS that accurately captures the required data during the clinical trial.

The median is the value of the data point that separates the smaller half of the data from the larger half. If the number of entries is even, then the median is the average of the two middle values.

Figure 7-1: Managing clinical data. There are many steps involved in obtaining and managing clinical data. Courtesy: National Institutes of Health. Source: http://cit.nih.gov/Science/CollaborativeResearch/CMAP/CIMS.htm; Public Domain .


Your engineering team has been given the task of developing a database that will manage the data collected during a clinical trial. The trial, which will evaluate a potential tool for advanced diabetes, has 45 patients, ages 40–65. In this cohort, or group of participants, 25 people have signed up as patients with diabetes and 20 people have signed up as the control group. Usually, patients need to visit the hospital every two weeks to have their blood sugar and protein levels tested. The goal of this treatment for patients is to effectively manage their diabetes while extending the time between hospital visits. Your team will focus on

Equation 7-1

mean � sum of all entries __________________ total number of entries


Questions:

61. What is the average age of the

women in the control group?

a. 40 d. 60

b. 50 e. 65

c. 52

62. What is the median age of the

control group?

a. 40 d. 53.5

b. 50 e. 65

c. 52

63. What is the mean age of the

patient group?

a. 40 d. 52.64

b. 42 e. 65

c. 52

64. What is the minimum number

of patient entries the database

entry must allow?

a. 5 d. 45

b. 20 e. 90

c. 25

Notes


Patient number Age Gender

1 41 M 2 40 F 3 45 F 4 65 M 5 61 F 6 63 F 7 54 M 8 56 M 9 55 M

10 58 M

11 48 F

12 49 M

13 53 F

14 61 F

15 60 F

16 43 M

17 42 F

18 46 M

19 47 F20 54 M

Table 7-1: Age of participants in control group.

Patient number Age Gender

1 41 M 2 40 F 3 45 F 4 65 M 5 61 F 6 63 F 7 54 M 8 56 M 9 55 M

10 58 M

11 48 F

12 49 M

13 53 F

14 61 F

15 60 F

16 43 M

17 42 F

18 46 M

19 47 F20 54 M

21 64 F22 63 F23 41 F24 42 M25 65 M

Table 7-2: Age of patients in patient group.


65. Which of the following fi gures

represents the correct code for a

patient’s glucose level after treatment?

a. create table GLUCOSEA (glucose_

after number (2,0) not null) ;

b. create table GLUCOSEB (glucose_

before number (2,0) not null) ;

c. create table GLUCOSEA (glucose_


d. create table GLUCOSEA (glucose_


e. create table GLUCOSEB (glucose_

before number (2,0) not null) ;

66. During data entry, an error is indicated

when an attempt is made to enter the

50 th patient’s age, which is 101. Which

of the following statements is true,

according to the AGE table code?

a. The AGE table code will return

an error if data for more than

45 patients is entered.

b. The AGE table code will return

an error if a patient over 65 is

entered.

c. The AGE table code will return an

error if a number with more than

two digits is entered.

d. The AGE table code does not exist.


by the information given.

Questions: Additional Background

Structured Query Language, or SQL, is a programming language. SQL statements are used to create database tables, update data on a database, and retrieve data from a database.

The information in data tables and SQL code come from a data dictionary, which describes application tables in a database.


TABLE NAME ATTRIBUTE NAME TYPE

AGE patient_age number (2)

GLUCOSEA glucose_after number (5)

GLUCOSEB glucose_before number (5)

Figure 7-2 Subset of data dictionary table .

• The command “create table” prompts the database system to create a new table.

• The name of the table precedes the fi rst parenthesis.• The statement “not null” means an error will appear if the

value is left empty.• The “number (2, 0)” means that a number can have up to two

digits with no digit after the decimal place.

create table AGE (patient_age number (2,0) not null) ;

Figure 7-3 Code for AGE table.

Additional Notes


67. What important fi elds must be

captured during data entry into the

database?

a. age, gender, glucose levels before

treatment, and frequency of

hospital visits

b. age, gender, and glucose levels after

treatment

c. age, control or patient, gender,

glucose levels after treatment, eye

color, and hair color

d. age, control or patient, gender,

glucose levels before treatment,

glucose levels after treatment,

frequency of hospital visits, dosage,

and side effects

e. age, control or patient, gender,

glucose levels before treatment,

glucose levels after treatment,

frequency of hospital visits, ethnicity,

and dosage

68. Which of the following is NOT true

for implementing GDMP?

a. The CDMS must be designed to

collect the data necessary and

relevant to the objectives of the

study.

b. The CDMS must collect all possible

data, even if some data may not be

relevant to the scientifi c question

being addressed.

c. The CDMS must be user-friendly.

d. The CDMS must be able to verify

the integrity and validity of the data.

e. The CDMS must allow for data

import and export.

Questions:Additional Notes


69. Calculate the mean glucose levels in

patients before treatment.

a. 15.652 d. 198

b. 150 e. 203

c. 156.52

70. Calculate the mean glucose levels in

patients after treatment.

a. 98 d. 167.5

b. 113.52 e. 203

c. 150

Questions:

Additional Notes

Notes

Patient Number

Age Before After

1 41 150 98

2 40 140 85

3 45 144 87

4 65 134 85

5 61 156 98

6 63 178 112 7 54 198 130 8 56 203 143 9 55 240 156

10 58 130 103

11 48 125 94

12 49 167 115

13 53 198 149

14 61 145 116

15 60 167 130

16 43 154 118

17 42 135 90

18 46 147 119

19 47 139 10820 54 128 98

21 64 190 15922 63 168 13623 41 80 78

24 42 143 11325 65 154 118

Table 7-3: Glucose levels of test populations before and after treatment.


Introduction

Clinical trials are important steps in bringing a new drug to the market. In clinical trials, research is conducted according to ethical guidelines on a select population of patients with each patient’s informed consent. Chemical engineers, biochemical engineers, and computer engineers play an important role in the development of an effective clinical drug trial.

Drug development occurs in a number of stages. First, a compound of interest, such as barium sulfate (BaSO4), must be isolated in the laboratory; then its biochemical properties must be characterized. After testing the compound in cells and in animal models, the next step is to determine the drug or treatment’s 1) effi cacy (effectiveness) in humans and 2) benefi cial effects. Clinical trials are research studies carefully designed to address these issues.

TEAMS Competition 2012Scenario 8:Clinical Trial Design

Background

In drug development, a clinical trial focuses on scientifi c questions regarding a drug’s effectiveness. During the planning stage of a clinical trial, researchers defi ne scientifi c questions that help them determine appropriate hypotheses to evaluate. After determining appropriate questions, researchers then identify how to answer the questions; they must ensure that the clinical trial will provide valid and unbiased data.

To best address the questions, the Food and Drug Administration (FDA), which oversees the introduction of new drugs into the market, suggests that researchers include a protocol design and plan (including objective or objectives of the study, design details, statistical methods, information on patient selection criteria, dosing schedules, and other medical-specifi c information) for the clinical trial.

Figure 8-1: Steps in the development of a drug. Courtesy: FDA.Source: http://www.fda.gov/ScienceResearch/SpecialTopics/CriticalPathInitiative/CriticalPathOpportunitiesReports/ucm077262.htm; Public Domain


It is not only important for researchers to determine how to choose an appropriate design; they also must determine statistical methods they can use to analyze data they collect. To provide an unbiased, systematic analysis of the data being collected, an appropriate design must be selected for the study.

The different phases in a clinical trial are:

• Phase I trials : During the fi rst phase of a clinical trial, the treatment is given to a small group of people. The typical range of people is 20–80.

• Phase II trials : During the second phase of a clinical trial, the treatment is given to a larger group of people. The range is typically 100–300 people.

• Phase III trials : During the third phase of a clinical trial, the treatment is given to an even larger group of people (typically about 1,000–3,000 people).

• Phase IV trials : During the fourth phase of a clinical trial, researchers continue to track the drug for any unforeseen development. (Phase IV occurs after the drug has been approved by the FDA and after long-term usage.)


A large part of clinical trials includes performing statistical analyses on the data collected, determining if the trial is working as expected based on these analyses, and cataloging any adverse effects.

The following defi nitions are important in performing statistical analyses.

Equation 8-1

mean � � �x

___ n where n is the number of patients, x is the number of individual data points, and �x is the sum of all the data points.

Equation 8-2

standard deviation � � �_________

�(x � �)2

_________ n

Equation 8-3

standard error of the mean � � ___ �__

n



• The median is the middle value in an ordered list of numbers. If a data set has an even number of elements, then the median is the average of the two middle numbers.

Patient Age

1 41

2 25

3 56

4 37

5 43

6 23

7 27

8 28

9 30

Table 8-1: Age of participants in the clinical trial.

71. Given the information in Table 8-1,

what is the mean and median age of

the participants in the clinical trial?

a. mean is 3.44; median is 30

b. mean is 34.44; median is 30

c. mean is 310; median is 30

d. mean is 34.44; median is 270

e. mean is 310; median is 270

72. What is the standard deviation of

the age distribution from the data in

Table 8-1?

a. 0.107 d. 10.7

b. 1 e. 30

c. 9

Questions

Notes

Additional Notes



Range and interquartile range are two measures of spread of a data set. The range is the difference between the largest and smallest values of a data set. The interquartile range, or IQR, is the difference between the third quartile ( Q 3 ) and fi rst quartile ( Q 1 ) of a data set. Q 1 is the median of the lower half of the data and Q 3 is the median of the upper half of that data set.


Patient Before After

1 105 90

2 115 102

3 125 95

4 135 115

5 90 85

6 128 120

7 149 130

8 200 175

9 134 112

Table 8-2: Blood glucose levels of participants before and after 15 days of treatment.

Additional Notes

73. Using the data in Table 8-2, compare

the mean and standard deviation of

the participants’ blood glucose levels

before and after treatment.

a. After 15 days of treatment, both the

participants’ mean blood glucose level

and standard deviation decreased.

b. After 15 days of treatment, the

participants’ mean blood glucose

level decreased and the standard

deviation increased.

c. After 15 days of treatment, the

participants’ mean blood glucose

level increased and the standard

deviation decreased.

d. After 15 days of treatment, both the

participants’ mean blood glucose level

and standard deviation increased.

e. There is not enough data provided

for calculations.

74. Calculate the range of the glucose

levels before treatment.

a. 20 d. 128

b. 90 e. 200

c. 110

Questions

Notes


75. Calculate the IQR of the glucose

levels before treatment to the IQR

of the glucose levels after treatment.

Which data set has a greater measure

of spread according to the IQRs?

a. IQR before treatment: 20;

IQR after treatment: 25; after

treatment

b. IQR before treatment: 20;

IQR after treatment: 25;

before treatment

c. IQR before treatment: 110;


before treatment

d. IQR before treatment: 110;


after treatment


with the information given.

76. In Phase 1 assume one patient had

an adverse reaction to the treatment.

The research team is not sure if

the reaction was due to the trial

treatment or a preexisting condition.

Ethically, what should the next step be

in the clinical trial?

a. Make no change to the trial.

b. Make a note of the adverse event

and continue the trial.

c. Stop the trial until it can be

established that the treatment did

not cause the adverse event.

d. Exclude patients with preexisting

conditions for the next trial.

e. Abandon the trial and not prepare

any reports.

QuestionsAdditional Notes


77. The probability that a drug will be

effective is 0.7 and the probability that

it will have no effect is 0.14. What

is the probability that the drug will

result in an adverse event?

a. 0.14 d. 0.84

b. 0.16 e. 0.86

c. 0.7

78. Calculate the sample size required for

a trial if the standard deviation is 15

and the margin of error is 5.

a. 5 d. 36

b. 15 e. 225

c. 25

79. Will the sample size increase or

decrease if the margin of error is

decreased to 2?

a. There will be no change.

b. The sample size will increase.

c. The sample size will decrease.

d. The sample size will fi rst increase

and then decrease.

e. The sample size will fi rst decrease

and then increase.

Questions

Notes


The results of a clinical trial are uncertain. In a clinical trial, a drug might be effective, have no effect, or cause an adverse effect. Probability measures how likely it is for an event to occur.


• Sum of all probabilities for an event � 1 • Sample size is given by

Equation 8-4

n � 4�2

____ d 2 where n � sample size, � � standard deviation, and d � margin of error.

Additional Notes


Notes

80. Which of the following factors must

be considered when developing

a computer model for a clinical

trial that tests for a drug aimed at

decreasing blood glucose levels?

Select all that apply. (Note: This is not

an exhaustive list.)

(I) clinical effect of a drug

(II) dose-concentration-effect

relationships for the drug

(III) replaced animal population

a. I d. I, II, III

b. I, II e. II, III

c. I, III

Questions

Additional Notes


Animal testing continues to be a highly debated and controversial topic. Clinical trial research must comply with the Animal Welfare Act, which requires that animals be provided humane care and treatment. Among other requirements, the United States Code and the Animal Welfare Act also require research laboratories to:

• Ensure that animal pain and distress are minimized for animal care, treatment, and practices in experimental procedures.

• Consider alternatives to any procedure that will likely cause an experimental animal pain or distress.

One possible alternative to animal testing is computer modeling, where the effects of a drug are modeled using a computer simulation.

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