Exploring Engineering. Chapter 14 Bioengineering. Topics to be Covered. What is bioengineering? Impact damage to the human body Fracture criterion Injury potential Gadd Severity Impact Parameter Examples. What is Bioengineering?.
Bioengineering applies engineering methods and techniques to problems in biology and medicine.
DNA Expression Arrays
Fluorescent Stained Myocyte
ΔV = [60 - 0 miles/hr](5280 ft/mile)(1 hr/3600 s) = 88 ft/s
So aavg:0-60 = (88 ft/s)/(2.6 s) = 33.8 ft/s2
and since g = 32.2 ft/s2 in the English unit system,
aavg:0-60 = 33.8/32.2 = 1.05 g, and
aavg:0-100 = 147/5.30 = 27.7 ÷ 32.2 = 0.86 g
aavg:0-100 = (0 – 147)/3.60 = - 40.7 ÷ 32.2 = -1.26 g
Note:a = v/t then F = ma = mv/t. Since the area under the V – t plot is Ds, then Ds = ½ Vt, and t = 2Ds/V and F = mV2/2Ds
An unconscious 8 weeks old infant was admitted to a hospital and found to have bilateral, subdural and retinal hemorrhages. He died the following day.
The parents explanation for the infant’s injuries was that the infant had been in a baby-rocker and that they had seen the rocker being rocked vigorously by their 14 months old daughter on two separate occasions.
A biomechanical analysis of the infant in the baby-rocker was used to estimate the maximum forces generated. These forces were then compared with those necessary to cause the subdural hemorrhage.
The Gadd Severity Index was developed with data from tests on human cadavers, and supported with real accident data. GSI = a2.5tS
where tS is the duration of the impact in seconds, and a is the deceleration in g’s. A human head can sustain values as high as 1,000 without serious injury as long as the peak value does not last for more than 10 or 15 milliseconds. For comparison, a hammer hitting a nail into wood gives a value of about 3,000, a baseball hitting a concrete wall is about 10,000, and a hammer hitting a concrete wall is about 3,600,000.
GSI is the equation of this line
A car traveling at 30. miles per hour runs into a sturdy stone wall. Assume the car is a totally rigid body that neither compresses nor crumples during the collision. The wall “gives” a distance of DS = 0.03 meters in the direction of the collision as the car is being brought to a stop. Assuming constant deceleration, calculate the deceleration.
Need: Deceleration = ____ m/s2
Know - How: First sketch the situation to clarify what is occurring at the impact. Then use speed - time graph to model the collision. The slope of speed - time graph is acceleration.
Solve: We first need to calculate the stopping time, ts , to decelerate from 30. mph (13.4 m/s) to 0. From Chapter 10, if deceleration is constant, we know that distance equals the area under the V – t graph or DS = 0.03 [m]
= ½ 13.4[m/s] ts [s] where ts is the stopping time.
Therefore ts = 0.06/13.4 = 0.0045 s.
Also the deceleration rate is the slope od the V-t graph,
V/t = (0 - 13.4)/ (0.0045 - 0) = -2980 = -3,000 m/s2
(about 306 g’s)
Need: Force stopping 75 kg driver on sudden impact in SI. Force = _____ N
Know:The driver’s weight is 75 [kg] 9.8 [m/s2] = 735 N and deceleration rate is 306 g.
How:F = ma = mga/g
Solve:F=735 306 = 2.24 105 N (or ~50,000 lbf!).
Even though the driver is belted in position, this is a very substantial force. What are the effects on human tissue of such large forces?
(made up of
Structure of the Body’s Long Bones
V = 0
Seat Belts & Crumple Zones
The purpose of a seat belt is to attach the driver to a rigid internal passenger shell while the rest of the car shortens by crumpling. A seat belt is most effective when a car significantly crumples during a collision.
The crumple zone is the area of a car designed to absorb energy upon impact. Engineers deliberately place weak spots in a car\'s structure to enable the metal work to collapse in a controlled manner to direct energy from the impact from the passenger area, and channeled to the floor, bulkhead, roof, or hood. Energy from the impact is converted into heat and sound (loud noise).
Ds= ½ V0× ts
= 0.10 m
ts = .015 sExercise 1 Solution
This gives the deceleration and thus the acting force.
(Constant) deceleration rate = V0/ts = 13.4/0.0149 [m/s][1/s] = 899 m/s2.
Force on head at contact is m × a = 5.0 × 899 [kg][m/s2] = 4490 = 4,500 N and lasting 0.015 s.
A rear facing child safety seat holds a child of mass 12 kg rigidly within the interior of a car. The area of contact between the seat and the child is 0.10 m2. The car undergoes a 30. mph collision. The car’s crumple zone causes the distance traveled by the rigid interior to be 1.0 m. Give the stress experienced by the child’s body in terms of a fraction of the breaking strength of bone assuming an infant’s bone breaks at a stress of 10. MN/m2.
Hence deceleration rate for infant = 13.4/0.15 = 90. m/s2. Force on baby = 12. × 90. [kg][ m/s2] = 1.08 × 103 N.
Stress = 1.08 × 103/0.10 = 1.08 × 104 N/m2.
As fraction of breaking stress, 1.08 × 104/ 10. × 106 = 1.08 × 10-3 Bones should hold and infant should be safe.
A car strikes a wall traveling 30. mph. The driver’s cervical spine (basically the neck) first stretches forward relative to the rest of the body by 0.01 m, and then recoils backward by 0.02 m, as shown below. Assume the spine can be modeled by a material of a modulus E = 10. GPa and a strength of 1.00 102 MPa. Will the maximum stress on the cervical spine during this “whiplash” portion of the accident exceed the strength of the spine? The length of the cervical spine is 0.15 m.
During stopping period, T2, spine stretches forward 0.01 m and on stopping at T3 backwards by 0.02 m. Elastic constants are E = 10. GPa and “transverse” strength of 1.00 × 102 MPa for this failure mode.
Maximum strain of spine = 0.02/0.15 [m][1/m] = 0.133
Stress on cervical spine = modulus × strain = 1010 × 0.133 = 1.3 × 109 N/m2.
Compare maximum stress on spine to tensile strength of spine: 1.3 × 109 N/m2 (max stress) > 108 N/m2 (strength)
Stress on cervical spine is greater than its tensile strength.