Elasticity Objective

Elasticity Objective Explain elastic and plastic bodies Define stress , strain, Hooke’s Law, Young’s modulus,Hysteresis loop and solve problems, clinical application. Explain the principle of Hysterisis .

Elasticity • The property by which a body return to its original size and shape when the forces that deformed it are removed Stress • A measure of the strength of the agent that is causing a deformation

Stress 

force area of surfaceon which force acts



F

(SI unit is Pa)

A

3

Strain • The fractional deformation resulting from a stress • The ratio of the change in some dimension of a body to the original dimension in which the change occurred

change in dimension strain  (no unit) original dimension

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Robert Hooke

Hooke’s law • In terms of stress and strain (stress  strain) • We then define a constant , called the modulus of elasticity Modulus of elasticity = constant =

stress strain A large modulus means that a large stress is required to produce a given strain

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Hooke’s law

stress strain

Elastic limit • The smallest stress that will produce a permanent distortion in the body • When a stress in excess of this limit is applied, the body will not return to its origin state after the stress is removed

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B

C

A Proportional limit

Thawmas Young

• TT

Young’s modulus (tensile modulus Y) • Length elasticity of a material

stretching force F Tensile stress   cross - sectional area A elongates an amount l Tensile strain   origional length l F stress Fl A Young' s modulus  Y    strain l Al l SI unit-Pascal(Pa) 16

Hysteresis loop

The lack of coincidence of the curves for increasing and decreasing stress is known as elastic hysteresis. The area of the hysteresis loop is proportional to the energy(per unit volume) dissipated within the elastic material.

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Medical applications of elasticity - Sutures made from horse hair, catgut, nylon or silk are all materials with elastic properties which increases when they are wet. - The natural elasticity of the skin, ligaments and tendons.

- The elastic property of the lung and chest wall.

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1. Stress = 8×107 Nm-2 d = 0.1 cm r = 0.05 cm = 5×10-4 m Elastic limit = 8×107 Nm-2 F=? F stress  A F  8x10 7 x π r 2  8x10 7 x π (5x10 4 ) 2  62.83N

l  5m, Δl  ? Y  0.7x10 11 Pa

stress Y strain stress strain  Y Δl 8x10 7  l 0.7x10 11 8x10 7 x5 -3 Δl   5.71x10 m 11 0.7x10

(2) For a relaxed biceps muscle, F=25N l  5cm  5x10 -2 m

l  0.2m 2 -4 2 A  50cm  50x10 m Y? Fl Y A l 25x0.2 4   2x10 Pa  20000Pa -4 -2 50x10 x5x10

For the biceps muscle under maximum tension, F  500N, Y  ? Fl Y Al 500x0.2  50x10  4 x5x10  2

 400000Pa

Extra(1) 4

A  1.5cm  1.5x10 m , m  10kg l  100m, Δl  0.0065m, Y  ? 2

2

F  mg  10x10  100N

Fl Y A Δl 100x100  1.5x10 -4 x 0.0065

 1.02 x10 Pa 10

(2) l  3.1cm  3.1x10 m, -2

d  0.22mm  0.22x10 m, -3 2 r  0.11x10 m, A   r -3

Ysteel  2x10 11 Pa, Δl  0.1mm  0.1x10 -3 m, Fl F? Y  Al YAl F l

2x10 11 x π ( 0.11x10 3 ) 2 x 0.1x10 -3 F 3.1x10 -2  24.5N

(3) m  550kg, l  3m, A  0.2cm 2  0.2x10 -4 m 2 Δl  0.4cm  0.4x10 -2 m stress  ?, strain  ? Y  ? F  mg  550x10  5500N stress 

F 5500 8   2 . 75 x10 Pa 4 A 0.2 x10

Δl 0.4x10 2 strain    1.33x10 3 l 3 stress 2.75x10 8 11 Y   2.07x10 Pa 3 strain 1.33x10