Elasticity and Strength of Materials: Biophysics Applications

Elasticity and Strength of Materials

Introduction

This section explores the physical principles underlying the elasticity and strength of materials, with a focus on biological tissues such as bone and muscle. The concepts of stress, strain, Young's modulus, and the mechanics of fracture are discussed, along with applications to real-world scenarios such as bone fractures and car collisions.

Stretch and Compression

• Stress (S): The internal force per unit area acting on a material. Defined as: $S = \frac{F}{A}$ where $F$ is the applied force and $A$ is the cross-sectional area.

• Strain (S_l): The fractional change in length due to applied force. For longitudinal strain: $S_l = \frac{\Delta l}{l}$ where $\Delta l$ is the change in length and $l$ is the original length.

• Hooke's Law: Within the elastic limit, the ratio of stress to strain is constant: $Y = \frac{S}{S_l}$ where $Y$ is Young's modulus, a measure of material stiffness.

Young's Modulus

Young's modulus quantifies the stiffness of a material. Higher values indicate stiffer materials. The following table summarizes Young's modulus and rupture strength for various materials:

Material	Young's modulus (dyn/cm2)	Rupture strength (dyn/cm2)
Steel	200 × 1010	450 × 107
Aluminum	69 × 109	6.2 × 107
Bone	14 × 1010	100 × 107 (compression)
Tendon	27.5 × 107	1.2 × 107 (stretch)
Muscle	4.0 × 106	8.55 × 105 (stretch)

Additional info: Young's modulus is typically measured in Pascals (Pa) in SI units, where 1 Pa = 1 N/m2.

Spring

• Hooke's Law for Springs: The force required to stretch or compress a spring is proportional to the displacement: $F = -K \Delta l$ where $K$ is the spring constant.

• Potential Energy in a Spring: $E = \frac{1}{2} K \Delta l^2$

• Elastic Body Analogy: An elastic body under stress behaves like a spring with an effective spring constant: $K = \frac{Y A}{l}$

• Potential Energy in an Elastic Body: $E = \frac{1}{2} \frac{Y A}{l} (\Delta l)^2$

Bone Fracture

• To calculate the energy required to break a bone of area $A$ and length $l$:

• Assume the bone remains elastic until fracture. The breaking stress is $S_B$.

• The force to fracture the bone: $F_B = S_B A = \frac{Y A}{l} \Delta l$

• The compression at the breaking point: $\Delta l = \frac{S_B l}{Y}$

• The energy stored at fracture: $E = \frac{1}{2} \frac{Y A}{l} (\Delta l)^2 = \frac{1}{2} \frac{A l S_B^2}{Y}$

Example Calculation

• Given: Two leg bones, combined length $l = 90$ cm, area $A = 6$ cm2, $S_B = 10^8$ dyn/cm2, $Y = 14 \times 10^{10}$ dyn/cm2.

• Energy absorbed by one leg at fracture: $E = \frac{1}{2} \frac{6 \times 90 \times 10^{18}}{14 \times 10^{10}} = 19.25 \times 10^6 \text{ erg} = 192.5 \text{ J}$

• Combined energy for two legs: $385$ J.

• This is equivalent to the energy from a 70-kg person jumping from a height of 56 cm.

Impulsive Force

• A large force exerted over a short time interval during a collision is called an impulsive force.

• The exact magnitude is difficult to determine, but the average value can be calculated using momentum:

• Average impulsive force: $F_{av} = \frac{m v_f - m v_i}{\Delta t}$

Fracture Due to a Fall

• When a person falls from height $h$, the velocity on impact: $v = \sqrt{2gh}$

• Momentum on the ground: $mv = m \sqrt{2gh}$

• Average impact force: $F = \frac{m}{\Delta t} \sqrt{2gh}$

Airbags

• Airbags expand to reduce the force on passengers during a collision by increasing the stopping distance.

• Average deceleration: $a = \frac{v^2}{2s}$

• Average force: $F = m a = m \frac{v^2}{2s}$

• For a 70-kg person with a 30-cm stopping distance: $F = \frac{70 \times 10^3 v^2}{2 \times 30} = 1.17 \times 10^3 v^2$ (dyn)

• If the force is distributed over 1000 cm2, the applied force per cm2 is $4.42 \times 10^3$ dyn, just below the estimated strength of body tissue.

Summary Table: Key Equations

Concept	Equation	Description
Stress	$S = \frac{F}{A}$	Force per unit area
Strain	$S_l = \frac{\Delta l}{l}$	Fractional change in length
Young's modulus	$Y = \frac{S}{S_l}$	Stiffness of material
Hooke's Law (spring)	$F = -K \Delta l$	Force-displacement relation
Potential energy (spring)	$E = \frac{1}{2} K \Delta l^2$	Energy stored in spring
Impulsive force	$F_{av} = \frac{m v_f - m v_i}{\Delta t}$	Average force during collision
Impact velocity (fall)	$v = \sqrt{2gh}$	Velocity on impact from height $h$
Average impact force (fall)	$F = \frac{m}{\Delta t} \sqrt{2gh}$	Force during landing
Average deceleration (airbag)	$a = \frac{v^2}{2s}$	Deceleration over stopping distance $s$
Average force (airbag)	$F = m \frac{v^2}{2s}$	Force on passenger

Applications and Examples

• Bone fracture energy: Calculating the energy required to break bones helps in understanding injury mechanisms in falls and impacts.

• Airbags: The physics of airbags demonstrates how increasing stopping distance reduces force and prevents injury.

Additional info: These principles are foundational in biomechanics, safety engineering, and medical physics, illustrating the intersection of physics and biology.