Elastic Properties

Elastic Properties

7 min read Updated Mar 26, 2026

Pull a rubber band: it stretches a lot. Hang the same weight from a steel wire: it stretches almost imperceptibly. Both deform — the question is how much and whether they snap back.

Every solid material deforms under force. Some return to their original shape (elastic), others stay deformed (plastic), and pushing too hard breaks them entirely. This section gives you the language — stress, strain, Young’s modulus, elastic limit — to describe all of that quantitatively.

The MCAT doesn’t lean heavily on this topic, but it’s on the AAMC content list and shows up in passages about bones, tendons, materials engineering, and biomechanics. It also marks the natural boundary between solid mechanics and the fluid mechanics in the rest of this chapter.

Stress: Force per Area

Stress measures how much force is applied per unit cross-sectional area. The formula looks identical to pressure — but stress applies specifically to solids being deformed.

There are three kinds of stress:

  • Tensile stress — pulling/stretching forces (a rubber band being stretched, a tendon being loaded).
  • Compressive stress — squeezing/crushing forces (stepping on a soda can, your femur supporting your weight).
  • Shear stress — forces parallel to a surface in opposite directions (sliding the top of a deck of cards sideways while holding the bottom).

Strain: The Response to Stress

Strain measures how much the material actually deforms in response to stress. It’s a dimensionless ratio.

Young’s Modulus: Stiffness

Young’s modulus (also called the elastic modulus) measures stiffness — how much stress is needed to produce a given strain.

MaterialYoung’s Modulus (GPa)Relative stiffness
Steel~200Very stiff
Bone (cortical)~18Stiff
Tendon~1.5Moderately flexible
Rubber~0.01Very flexible

The Elastic Limit

Below the elastic limit (a.k.a. yield point), a material returns to its original shape when force is removed — like a stretched rubber band snapping back. Above the elastic limit, the material undergoes plastic (permanent) deformation — it stays bent or stretched even after you let go. Push even further and you reach the fracture point, where the material breaks.

You’ve seen this every time you’ve bent a paperclip a little (it springs back) versus bent it a lot (it stays bent) versus bent it back-and-forth too many times (it snaps).

Why This Matters for Fluids

This section sits inside the fluids chapter because it marks the boundary between solid and fluid behavior. The defining property of a fluid is that it cannot resist shear stress — apply any shearing force and the fluid simply flows in response. Solids, in contrast, can hold a fixed deformation against shear (within their elastic range).

That single distinction — solids resist shear, fluids don’t — is the reason fluid mechanics exists as its own branch of physics. Once you can’t assume things hold their shape, all the equations change. Everything in the rest of this chapter (Bernoulli, continuity, viscosity, Pascal’s law) flows (literally) from this property.

Worked Example

A bone of cross-sectional area 4×1044 \times 10^{-4} m² is loaded with a 2000 N compressive force (about 200 kg pushing down). If the bone is 0.4 m long and Young’s modulus for bone is 1.8×10101.8 \times 10^{10} Pa, how much does it shorten?

  • Stress: σ=F/A=2000/(4×104)=5×106\sigma = F/A = 2000 / (4 \times 10^{-4}) = 5 \times 10^6 Pa.
  • Strain: ε=σ/E=(5×106)/(1.8×1010)2.78×104\varepsilon = \sigma/E = (5 \times 10^6)/(1.8 \times 10^{10}) \approx 2.78 \times 10^{-4}.
  • ΔL=εL0=2.78×104×0.41.1×104\Delta L = \varepsilon \cdot L_0 = 2.78 \times 10^{-4} \times 0.4 \approx 1.1 \times 10^{-4} m = 0.11 mm.

So a 200 kg load shortens the bone by only about a tenth of a millimeter. Bone is impressively stiff — which is why your skeleton can support you without obviously deforming.

A wire with cross-sectional area 2×1062 \times 10^{-6} m² and original length 3 m stretches by 0.6 mm under a 400 N load. What is Young's modulus?
Click to reveal answer
E=1×1012E = 1 \times 10^{12} Pa = 1000 GPa. Stress = F/A=400/(2×106)=2×108F/A = 400/(2 \times 10^{-6}) = 2 \times 10^8 Pa. Strain = ΔL/L0=0.0006/3=2×104\Delta L/L_0 = 0.0006/3 = 2 \times 10^{-4}. E=σ/ε=1012E = \sigma/\varepsilon = 10^{12} Pa. Extremely stiff — consistent with a very rigid metal.
What is the fundamental difference between a solid and a fluid in terms of how they respond to stress?
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Solids can resist shear stress; fluids cannot. Apply a shear force to a solid and it deforms to a fixed shape and holds it. Apply the same force to a fluid and it flows continuously for as long as the force is applied. Both solids and fluids resist compressive stress.
A tendon stretches 2% under load (strain = 0.02). If Young's modulus for the tendon is 1.5×1091.5 \times 10^9 Pa, what is the stress?
Click to reveal answer
σ=3×107\sigma = 3 \times 10^7 Pa. σ=Eε=(1.5×109)(0.02)=3×107\sigma = E \cdot \varepsilon = (1.5 \times 10^9)(0.02) = 3 \times 10^7 Pa = 30 MPa. (For context, this is roughly the maximum stress a healthy tendon can handle before approaching its fracture point.)