Elasticity, Stress, and Longitudinal Strain Reading material

• Elasticity is the property of a body that makes it return to it's original shape after a force is removed.
• A force, if large enough, can distort a body beyond its elastic limit, causing the original shape to not be restored after the force is removed, and a still larger force will rupture the body.

Longitudinal Stretch and Compression

• The force applied to a body is transmitted to every part of the body, and will pull it apart.
• This force is resisted by a cohesive force that holds the material together, until the applied force exceeds the cohesive force, and the material breaks.
• If the force is reversed, the body is compressed and its length reduced.
• Initially, the body's compression is elastic, but a large force will produce permanent deformation followed by breakage.
• Stress (S) is the internal force per unit area acting on the material, defined as S = F/A where F is the applied force and A is the area on which the force is applied.
• Longitudinal Strain (St) is the fractional change in length Al/l, caused by the force applied to a bar which causes it to elongate by an amount Al, defined as St = Al/l where l is the length of the bar and Al is the change in length caused by the applied force.
• Robert Hooke observed in 1676 that while the body remains elastic, the ratio of stress to strain is constant.
• The ratio of stress to strain: S/St = Y where Y is defined as Young's modulus.
• Young's modulus has been measured for many materials, and the breaking or rupture strength of these materials has also been measured.

Spring Properties

• A spring’s elastic properties are analogous to the the properties of a material.
• The force F required to stretch (or compress) the spring is directly proportional to the amount of stretch: F = KAl
• K is the spring constant.
• A stretched (or compressed) spring contains potential energy.
• The energy E stored in the spring is given by E = (1/2)K(Al)².
• An elastic body under stress is analogous to a spring with a spring constant YA/l, where Y is Young's modulus, A is the area, and l is the length.
• The amount of energy stored in a stretched or compressed body is E = (1/2) (YA/l) (Al)².

Bone Fracture: Energy Considerations

• Knowing the maximum energy that parts of the body can safely absorb estimates the possibility of injury under various circumstances.
• The amount of energy required to break a bone of area A and length l can be calculated by assuming the bone remains elastic until fracture, designating the breaking stress of the bone as SB.
• The corresponding force FB that will fracture the bone is FB = SBA = (YA/l)Al.
• The compression Al at the breaking point is Al = SBl/Y.
• The energy stored in the compressed bone at the point of fracture is E = (1/2) (YA/l) (Al)²
• Substituting for Al = SBl/Y, we obtain: E= (1/2) (A l S²(B)) / Y.
• Example: the fracture of two leg bones that have a combined length of about 90 cm and an average area of about 6 cm², with a breaking stress SB of 10⁹ dyn/cm² and a Young's modulus for the bone of 14 × 10¹⁰ dyn/cm².
• The total energy absorbed by the bones of one leg at the point of compressive fracture is, from Eq. 5.13, E = (1/2) ((6 × 90 × 10¹⁸) / (14 × 10¹⁰)) = 19.25 × 10⁸ erg = 192.5 J.
• This combined energy in the two legs is twice this value, or 385 J.
• The amount of energy in the impact of a 70-kg person jumping from a height of 56 cm (1.8 ft), given by the product mgh.
• It is possible to jump safely from a height greater than 56 cm if, on landing, the joints of the body bend and the energy of the fall is redistributed to reduce the chance of fracture.

Impulsive Forces

• In a sudden collision, a large force is exerted for a short period of time on the colliding object, starting at zero, increasing to some maximum value, and then decreasing to zero again.
• The time interval t2 - t1 = At is the duration of the collision.
• This sort duration force is called an impulsive force, making it difficult to determine the exact magnitude of the force during the collision.
• The average impulsive force (Fav) can be calculated from the relationship between force and momentum by using: Fav = (mvf - mvi)/At where mvi is the initial momentum of the object and mvf is the final momentum after the collision.

Fracture Due to a Fall: Impulsive Force Considerations

• Similar calculations to the injurious effects of collisions from energy considerations, can be performed using the concept of impulsive force.
• The magnitude of the force that causes the damage is computed from the average impulsive force (Fav) where Fav = (mvf - mvi)/At, however the duration of the collision At is difficult to determine precisely since it depends on the type of collision.
• Hard colliding objects create collisions where collision time is short, lasting a few milliseconds.
• If one of the objects is a soft and yields during the collision, the duration of the collision is lengthened, and as a result the impulsive force is reduced.
• When a person falls from a height h, his/her velocity on impact with the ground, neglecting air friction, is v = √2gh.
• The momentum on impact is mv = m√2gh = W√(2h/g) where W is weight.
• After the impact the body is at rest and momentum is zero (mvf = 0).
• The change in momentum is mvi - mvf = W√(2h/g).
• The average impact force, from Eq. 5.14, is F= (W/At) √(2h/g) = (m/At) √2gh.
• If the impact surface is hard, such as concrete, and if the person falls with their joints rigidly locked, the collision time is estimated to be about 10-2 sec (collision time is longer if the person bends his/her knees or falls on a soft surface).
• The force per unit area that may cause a bone fracture is 10⁹ dyn/cm².
• If the person falls flat on their heels, the area of impact may be about 2 cm².
• The force FB that will cause fracture is FB = 2 cm² × 10⁹ dyn/cm² = 2 × 10⁹ dyn = 4.3 × 10³ lb.
• The height h of fall that will produce such an impulsive force is h= (1/2g) ((FAt)/m)²
• A man of 70 kg has a fracturing average impact force from a jump (assuming At = 10-2 sec): h= ((1/2)980) (((2x10⁹) x 10⁻²)/70)) = 41.6 cm = 1.37 ft.

Airbags: Inflating Collision Protection Devices

• The impact force can be calculated from the distance the center of mass of the body travels during the collision under the action of the impulsive force.
• In automobiles, an inflatable bag is located in the dashboard, this bag expands suddenly in a collision and cushions the impact of the passenger, stopping the forward motion must be stopped in about 30 cm of motion if contact with the hard surfaces of the car is to be avoided.
• The average deceleration is a = v²/2s where v is the initial velocity of the automobile (and the passenger) and s is the distance over which the deceleration occurs.
• The average force that produces the deceleration is F=ma = mv²/2s where m is the mass of the passenger.

Whiplash Injury

• Neck bones are rather delicate and can be fractured by even a moderate force.
• Fortunately, the neck muscles are relatively strong and capable of absorbing a considerable amount of energy, but sudden impacts, as in a rear-end collision, where the body is accelerated in the forward direction by the back of the seat, will cause the unsupported neck to be suddenly yanked back at full speed.
• The muscles do not respond fast enough and all the energy is absorbed by the neck bones, causing the well-known whiplash injury.

Falling from Great Height

• Reports of people jumping out of airplanes without parachutes and surviving due to landing on soft snow have been made.
• In these cases the body made about a 1-m-deep depression in the surface of the snow on impact.
• If the decelerating impact force acts over a distance of about 1 m, the average value of this force remains below the magnitude for serious injury even at the terminal falling velocity of 62.5 m/sec (140 mph).

Osteoarthritis and Exercise

• Osteoarthritis is a joint disease characterized by a degenerative wearing out of the components of the joint among them the synovial membrane and cartilage tissue, causing the joint to lose flexibility and strength accompanied by pain and stiffness, and eventually, the underlying bone may also start eroding.
• Osteoarthritis is a major cause of disability at an older age (knees are the most commonly affected joint).
• About 60% of men and 75% of women are to some extent affected by this condition after the age of 65.
• Joint injury is most strongly correlated with subsequent development of osteoarthritis.
• There appears to be little risk associated with recreational running 20 to 40 km a week.
• Injured joints are more likely to be subsequently subject to wear and tear because the kinetic friction (µk) of an intact joint is about 0.003, but the coefficient of friction for un-lubricated bones is a hundred times higher.
• A joint injury usually compromises to some extent the lubricating ability of the joint leading to increased frictional wear and osteoarthritis.
• Osteoarthritis seems to progress at about the same rate in both regular runners and non-runners, indicating that the joints possess some ability to self-repair (these conclusions remain tentative and are subject to further study).