"Ut tensio, sic vis."
— Robert Hooke, 1678 | As the extension, so the force.

M5L4 assumed that everything was perfectly rigid — forces acted, torques balanced, and structures stayed exactly the same shape. Real materials do not work that way. Every bridge cable stretches, every bone compresses, every rubber band deforms. The question is not whether a material deforms under load, but by how much, and whether it snaps or springs back. This lesson builds the quantitative framework for those answers: stress (force per unit area), strain (fractional deformation), and the three elastic moduli that connect them. Hooke's law lives at the heart of it — but here it applies not just to springs but to every solid material in every engineering structure ever built.

Three Ways a Material Can Deform

Each type of deformation has its own stress, its own strain, and its own modulus. All three follow the same linear relationship in the elastic regime.

Tensile / Compressive

Y = stress / strain

Stress = F/A (Pa)
Strain = ΔL/L₀ (dimensionless)
Y = Young's modulus

Shear

S = stress / strain

Stress = F_∥/A (Pa)
Strain = Δx/L (dimensionless)
S = shear modulus

Bulk (Volume)

B = stress / strain

Stress = Δp (Pa)
Strain = −ΔV/V₀ (dimensionless)
B = bulk modulus

All three moduli have units of pascals (Pa = N/m²). Steel: Y ≈ 200 GPa • S ≈ 82 GPa • B ≈ 160 GPa

Course Competencies and Learning Objectives

A ★ indicates that this page contains content related to that LO.

CC5.5 Apply stress, strain, and elastic moduli to analyze deformation of materials

★ LO5.5.1 Define stress and strain for tensile, compressive, shear, and bulk deformation

★ LO5.5.2 State Hooke's law for elastic materials and identify the conditions under which it applies

★ LO5.5.3 Apply Young's modulus to calculate tensile or compressive deformation: ΔL = (F/A)(L₀/Y)

★ LO5.5.4 Apply the shear modulus to calculate shear deformation: Δx = (F_∥/A)(L/S)

★ LO5.5.5 Apply the bulk modulus to calculate volume change under uniform pressure: ΔV = −(Δp/B)V₀

★ LO5.5.6 Interpret the stress-strain curve: elastic limit, yield point, plastic deformation, and fracture

Required Reading

Chapter 12.3 covers all three moduli. Chapter 12.4 extends the discussion to practical failure conditions and engineering safety factors. Read both sections carefully — the modulus formulas all look similar, and the differences in how stress and strain are defined for each case are where students most often make errors.

12.3 Stress, Strain, and Elastic Modulus 12.4 Elasticity and Plasticity

Optional Reading

Explore More

Physics Classroom: Elastic Properties Khan Academy: Hooke's Law PhET: Hooke's Law Simulation Engineering Toolbox: Young's Modulus Table

Media

Video 1 establishes stress, strain, and Young's modulus; Video 2 covers shear and bulk moduli; Video 3 interprets the stress-strain curve and connects elasticity to engineering failure analysis.

Video 1: Stress, Strain, and Young's Modulus

Why “How Much Force?” Is the Wrong Question

Whether a wire breaks depends not on the force alone but on the force per unit cross-sectional area. And whether it stretches much depends not on the absolute extension but on the extension relative to the original length. Stress and strain normalize both quantities to make them material properties rather than geometry properties.

Stress = F/A: why a thin wire breaks before a thick one under the same force
Strain = ΔL/L₀: fractional change, dimensionless, independent of original length
Young's modulus Y = stress/strain: the material's resistance to being stretched or compressed
Working formula: ΔL = FL₀/(AY)
Comparing Y values: steel (~200 GPa) vs bone (~15 GPa) vs rubber (~0.01 GPa) — what these numbers mean in practice

Time: 10:00

Video 2: Shear Modulus and Bulk Modulus

Two More Ways a Material Can Deform

Shear deformation happens when forces slide parallel layers of a material past each other. Bulk compression happens when uniform pressure acts on all surfaces simultaneously, changing volume without changing shape.

Shear stress = F_∥/A; shear strain = Δx/L (the shear angle φ ≈ tanφ for small deformations)
Shear modulus S = shear stress / shear strain; working formula: Δx = F_∥L/(AS)
Examples: glue joints, earthquake waves in rock, bolts in shear, book pages sliding
Bulk modulus B = −Δp / (ΔV/V₀); negative sign because compression (positive Δp) reduces volume
Why liquids and gases have bulk moduli but no shear moduli — they cannot resist shear
Comparing B values: water (2.2 GPa) vs steel (160 GPa) vs air (~0.000142 GPa)

Time: 9:30

Video 3: The Stress-Strain Curve and Engineering Failure

From Hooke's Law to Fracture

A material under increasing stress does not simply stretch proportionally forever. The stress-strain curve tells the complete story — from perfect elasticity through yield and plastic deformation to failure.

The elastic region: Hooke's law holds, slope = Young's modulus, material returns to original shape on unloading
The proportional limit: where the curve first deviates from linear
The elastic limit: the maximum stress for which the material still returns to its original shape
The yield point: where plastic (permanent) deformation begins; engineering yield strength
The ultimate tensile strength: the maximum stress the material can sustain
Fracture point: where the material fails completely
Ductile vs brittle materials: very different curves, very different failure modes
Engineering safety factors: why structures are designed to operate well below the yield point

Time: 11:00

Deeper Look: Where Elasticity Comes From and How It Connects to Everything Else

The Atomic Origin of Hooke's Law

Why F = kx Emerges from Interatomic Potential Wells

Hooke's law is not a fundamental law of nature — it is an approximation that works whenever atoms are displaced only slightly from their equilibrium positions. Here is why.

The potential energy between two atoms looks like a Lennard-Jones well: steep repulsion at short range (nuclei fighting each other), attractive at medium range (bonding), and zero at large separation (atoms apart). Near the equilibrium separation r₀, the potential U(r) is approximately parabolic:

                        U(r) ≈ U(r0) + ½U''(r0)(r − r0)²
                    

The force is F = −dU/dr = −U''(r₀)(r − r₀), which is exactly F = −kΔr with k = U''(r₀) — the curvature of the potential well at its minimum.

Key consequences:

Any smooth potential well looks parabolic near its minimum — so Hooke's law holds for any material as long as displacements are small. This is why it is so universal.
The modulus (Y, S, or B) is proportional to k, which is the curvature of the interatomic potential. Stiffer bonds (deeper wells, sharper curvature) give higher moduli.
When displacements are large, higher-order terms matter and the material becomes nonlinear — this is where yield and fracture begin.

Connection to Simple Harmonic Motion

The same parabolic potential that gives Hooke's law for solids also gives simple harmonic motion for springs and pendulums (small angle). The SHM you study in the oscillations module is the same physics as elasticity, applied to different systems. F = −kx is the universal signature of a parabolic potential well.

The Relationship Between Y, S, and B

Three Moduli, One Material — Are They Independent?

For an isotropic material (properties the same in all directions), Y, S, and B are not independent. They are related through a single additional parameter called Poisson's ratio ν.

Poisson's ratio (ν) measures the lateral contraction that accompanies axial stretching. If you pull a rod along its length and it stretches by ΔL/L, its diameter contracts by ν times that fraction:

                        ν = −(lateral strain) / (axial strain) = −(Δd/d) / (ΔL/L)
                    

For most metals, ν ≈ 0.25–0.35. For rubber, ν ≈ 0.5 (nearly incompressible). A hypothetical material with ν = 0.5 exactly would have infinite bulk modulus — it would not change volume at all under any load.

The three moduli are related by:

                        S = Y / [2(1 + ν)]    •    B = Y / [3(1 − 2ν)]
                    

So given any two of {Y, S, B, ν}, you can find the other two. A table of moduli is really a table of two independent numbers per material.

Why S < Y Always

From S = Y/[2(1+ν)] and ν > 0, we get S < Y/2 < Y. Shear moduli are always less than Young's moduli for the same material. This means materials are always easier to shear than to stretch — which is why earthquake shear waves travel slower than compressional waves through the same rock.

Elasticity Meets Equilibrium: Internal Forces in Structures

Closing the Loop on Module 5

Static equilibrium (M5L4) told you the forces in a structure. Elasticity (this lesson) tells you what those forces do to the material. The two ideas always work together in real engineering.

Example: A steel cable supporting a load.

From M5L4 equilibrium, you find the cable tension T. From this lesson, you find how much the cable stretches:

                        ΔL = TL0 / (AY)
                    

The safety question has two parts:

Will it break? — Compare tensile stress T/A to the material's ultimate tensile strength. Use a safety factor of 3–10 depending on application.
Will it deform too much? — Calculate ΔL and check that it stays within geometric tolerances, even if it will not break.

Thermal stress: Temperature changes cause materials to expand or contract. If constrained from expanding, a thermal stress develops:

                        thermal stress = Y α ΔT
                    

where α is the linear thermal expansion coefficient. For steel (α = 12 × 10⁻⁶ /°C, Y = 200 GPa), a 10°C temperature change produces a stress of 24 MPa — substantial in long bridge spans, which is why expansion joints exist.

The Tacoma Narrows Bridge

The 1940 Tacoma Narrows Bridge collapse was not a simple material failure — the steel was not overstressed in tension. It was a resonance failure: wind created oscillating aerodynamic forces that matched the bridge's natural frequency, causing ever-larger oscillations until the elastic limit was exceeded and the deck tore apart. The lesson: equilibrium and elasticity together are necessary but not sufficient for safety — dynamics matters too.

Practice and Apply — Conceptual

Matching the Moduli

Match Each Term to Its Correct Definition

Drag each definition on the right to the matching term on the left.

Tensile stress

Tensile strain

Young's modulus

Shear modulus

Bulk modulus

Elastic limit

Maximum stress a material can sustain and still return to its original shape when unloaded

Ratio of pressure change to fractional volume change; describes resistance to uniform compression

Perpendicular force per unit cross-sectional area; units of Pa

Ratio of tensile stress to tensile strain; slope of the linear region of the stress-strain curve

Fractional change in length: ΔL/L₀; dimensionless

Ratio of shear stress to shear strain; describes resistance to parallel-layer sliding

Which Modulus Applies?

Sort Each Scenario into the Correct Modulus Category

Each scenario describes a type of deformation. Identify which elastic modulus (Young's, shear, or bulk) is the appropriate one to use.

Scenarios to Sort

A steel column in a building supports the weight of the floors above it
A submarine hull is compressed uniformly as it dives deeper
A rubber eraser is pushed sideways on a table while the bottom is held fixed
A cable car wire stretches as passengers board
A bolt holding two metal plates together is pulled in opposite directions along its axis
A cube of water is subjected to increased pressure from all sides in a hydraulic press
A bone is compressed along its length by body weight during standing
Two layers of adhesive slide past each other when the joint is loaded in the plane of the glue

Young's Modulus (Y)

Shear Modulus (S)

Bulk Modulus (B)

Common Misconceptions

Think It Through — Then Flip

Each card states a common student belief about stress, strain, or elasticity. Decide whether it is correct or incorrect before flipping.

"A thicker wire is stronger because it can withstand more force before breaking."

Partially Correct — But Strength Is a Stress, Not a Force

A thicker wire does survive a larger force before breaking — but that is because it has more cross-sectional area, not because the material itself is stronger. The material strength is the breaking stress (force per area), which is the same for both wires if they are made of the same material. Ultimate tensile strength is a material property measured in Pa, not N.

"Elastic means a material returns to its shape. So elastic materials are stretchy, like rubber."

Incorrect — Steel Is Elastic; Rubber Is Merely Compliant

In physics, elastic means returning to the original shape after the load is removed. Steel is highly elastic — stretch a steel cable within its elastic limit and it snaps back perfectly. Rubber is elastic too, but also very compliant (low Y). A material can be elastic and stiff (steel), elastic and compliant (rubber), or inelastic (clay, dough) — these are two independent properties.

"If a steel rod and an aluminum rod of the same dimensions carry the same load, they have the same stress."

Correct — Stress Depends Only on Force and Area

Stress = F/A. If F and A are the same for both rods, the stress is identical regardless of the material. However, the strain differs: ΔL/L = stress/Y, and Y_steel (200 GPa) ≫ Y_aluminum (70 GPa), so the aluminum rod stretches about 2.9× as much. Same stress, very different deformation. This is a key distinction for material selection in design.

"Liquids cannot be compressed because their bulk modulus is effectively infinite."

Incorrect — Liquids Compress, Just Very Little

Water's bulk modulus is 2.2 GPa — large but finite. A pressure increase of 1 atm (~100 kPa) compresses water by ΔV/V ≈ 4.5 × 10⁻⁵ (0.0045%). That is tiny, so the “incompressible” approximation is excellent for most engineering problems. But in deep-sea contexts, submarine hulls, or hydraulic shock systems, the compressibility of fluids is engineered deliberately.

Practice and Apply — Computational

Quick Reference: Elastic Moduli

                        Tensile/compressive: ΔL = FL0/(AY)

                        F = normal force, A = cross-section, Y = Young's modulus

                        Shear: Δx = F∥L/(AS)

                        F∥ = parallel force, L = layer thickness, S = shear modulus

                        Bulk: ΔV = −(Δp/B)V0

                        Δp = pressure change, B = bulk modulus

                        Key values (GPa):

                        Steel: Y=200, S=82, B=160  |  Bone: Y=15  |  Water: B=2.2

Problem 1 — Stretching a Steel Cable

A steel cable 20 m long with a circular cross-section of diameter 1.5 cm supports a load of 8000 N. (a) Find the tensile stress in the cable. (b) Find the strain. (c) Find the elongation. (d) Check whether the cable operates within the elastic limit (ultimate tensile strength of steel ≈ 400 MPa; use a safety factor of 4).

Cross-sectional area: A = π(d/2)² = π(0.0075)² = 1.767 × 10⁻⁴ m²

(a) Tensile stress:

σ = F/A = 8000 / (1.767 × 10⁻⁴) = 45.3 MPa

(b) Strain:

ε = σ/Y = 45.3 × 10⁶ / (200 × 10⁹) = 2.27 × 10⁻⁴ (dimensionless)

(c) Elongation:

ΔL = ε × L₀ = 2.27 × 10⁻⁴ × 20 = 4.53 mm

(d) Safety check:

Allowable stress = UTS / safety factor = 400 MPa / 4 = 100 MPa

Actual stress = 45.3 MPa < 100 MPa ✓ The cable operates safely within the elastic regime.

Note the 4.53 mm elongation over 20 m — about 0.023%. This seems tiny but matters in precision applications like suspension bridge design, where cable elongation must be accounted for in deck geometry.

Problem 2 — Bone Under Compression

The femur (thigh bone) has a roughly cylindrical cross-section with an effective diameter of 2.5 cm and a length of 40 cm. Estimate the compressive stress and strain in the femur of a 75 kg person standing on one leg. (Y_bone = 1.5 × 10¹⁰ Pa; compressive strength of bone ≈ 170 MPa)

Force: The femur supports the full body weight on one leg: F = mg = (75)(9.8) = 735 N. (In reality, muscle forces increase this substantially — this is a lower bound.)

Area: A = π(0.0125)² = 4.91 × 10⁻⁴ m²

Compressive stress:

σ = F/A = 735 / (4.91 × 10⁻⁴) = 1.50 MPa

Compressive strain:

ε = σ/Y = 1.50 × 10⁶ / (1.5 × 10¹⁰) = 1.0 × 10⁻⁴

Compression amount:

ΔL = ε × 0.40 m = 0.040 mm (40 micrometers)

Safety margin: 1.50 MPa vs. 170 MPa compressive strength → safety factor of ~113. Bone has enormous strength reserve for normal standing, consistent with its design to handle impact forces many times body weight.

During running or jumping, ground reaction forces can reach 5–8 times body weight, reducing the safety factor to 15–25. Still comfortable — but a stress fracture can occur when repetitive loading exceeds the bone's fatigue limit without adequate recovery time.

Problem 3 — Shear Deformation in a Rubber Mount

A rubber vibration isolator consists of a 5.0 cm × 5.0 cm square rubber pad, 2.0 cm thick. A horizontal force of 150 N is applied to the top face while the bottom is bonded to a rigid plate. (a) Find the shear stress. (b) Find the shear deformation Δx. (c) Find the shear angle φ. (S_rubber = 0.60 MPa)

(a) Shear stress:

A = (0.05)(0.05) = 2.5 × 10⁻³ m²

τ = F_∥/A = 150 / (2.5 × 10⁻³) = 60,000 Pa = 60 kPa

(b) Shear deformation:

Δx = F_∥L / (AS) = (150)(0.020) / [(2.5 × 10⁻³)(0.60 × 10⁶)]

Δx = 3.0 / 1500 = 0.0020 m = 2.0 mm

(c) Shear angle:

tanφ = Δx/L = 0.0020/0.020 = 0.10 ⇒ φ = arctan(0.10) ≈ 5.7°

This is why rubber mounts are used in machinery and vehicle suspensions: large shear deformation (here 10% of the pad thickness) allows significant vibration absorption without generating large restoring forces. The low shear modulus of rubber is the engineering feature, not a deficiency.

Problem 4 — Bulk Compression of Water

A hydraulic press subjects 2.0 liters of water to a pressure of 50 atm (5.065 MPa) above atmospheric. (a) Find the fractional volume change ΔV/V₀. (b) Find the actual volume change in mL. (c) What pressure would be needed to compress the water by 1.0%? (B_water = 2.2 × 10⁹ Pa)

(a) Fractional volume change:

ΔV/V₀ = −Δp/B = −5.065 × 10⁶ / (2.2 × 10⁹) = −2.30 × 10⁻³

The negative sign confirms compression (volume decreases). Fractional decrease = 0.230%.

(b) Actual volume change:

|ΔV| = 2.30 × 10⁻³ × 2000 mL = 4.6 mL

At 50 atm overpressure, the water compresses by only 4.6 mL out of 2000 mL — confirming the “practically incompressible” approximation for most purposes.

(c) Pressure for 1.0% compression:

Δp = B × (ΔV/V₀) = (2.2 × 10⁹)(0.010) = 22 MPa ≈ 217 atm

Ocean pressure increases by ~1 atm per 10 m depth. To compress seawater by 1%, you would need to dive to ~2170 m. Deep-ocean water is in fact measurably denser than surface water due to bulk compression, which affects ocean circulation models.

Problem 5 — Equilibrium + Elasticity: Cable Extension on a Beam

A horizontal steel beam (length 3.0 m, mass 12 kg) is attached to a wall by a hinge at its left end. A steel cable (diameter 0.80 cm, length 2.0 m, Y = 200 GPa) connects the right end of the beam to the wall at a point directly above the hinge, making a 40° angle with the beam. A 60 kg sign hangs from the midpoint of the beam. (a) Use equilibrium to find the cable tension. (b) Find the elongation of the cable under this tension.

This problem combines M5L4 and M5L5 directly.

(a) Equilibrium — pivot at the hinge:

Forces: beam weight W_b = (12)(9.8) = 117.6 N at 1.5 m; sign weight W_s = (60)(9.8) = 588 N at 1.5 m; cable tension T at 40° at 3.0 m.

Στ = 0: T sin40° × 3.0 − W_b × 1.5 − W_s × 1.5 = 0

T(0.643)(3.0) = (117.6 + 588)(1.5) = 705.6 × 1.5 = 1058.4

1.929 T = 1058.4 ⇒ T = 548.7 N

(b) Cable elongation:

A = π(0.004)² = 5.027 × 10⁻⁵ m²

ΔL = TL₀/(AY) = (548.7)(2.0) / [(5.027 × 10⁻⁵)(200 × 10⁹)]

ΔL = 1097.4 / (1.005 × 10⁷) = 1.09 × 10⁻⁴ m ≈ 0.109 mm

Stress check: σ = T/A = 548.7/(5.027 × 10⁻⁵) = 10.9 MPa — well below the 400 MPa UTS of steel.

The 0.109 mm elongation causes the right end of the beam to drop very slightly — the cable geometry changes negligibly. In this problem, the equilibrium and elasticity phases are fully decoupled: solve equilibrium first, then use the result as input to the elasticity calculation.

Problem 6 — Choosing a Material: Engineering Design

A tension rod in a machine must: (1) support a tensile load of 25,000 N, (2) have length 1.0 m, (3) elongate by no more than 0.50 mm under load, and (4) have a safety factor of at least 5 against breaking. Three candidate materials are given below. Determine which material(s) meet all four requirements, and for the one with the lowest mass, find the required diameter.

Material	Y (GPa)	UTS (MPa)	Density (kg/m³)
Steel	200	400	7850
Aluminum alloy	70	270	2700
Titanium alloy	110	900	4430

Constraint 3 (elongation ≤ 0.50 mm) gives minimum area A_min for each material:

ΔL = FL₀/(AY) ⇒ A ≥ FL₀/(Y ΔL_max)

Steel: A ≥ (25000)(1.0)/[(200×10⁹)(5×10⁻⁴)] = 25000/(10⁸) = 2.50×10⁻⁴ m²

Aluminum: A ≥ 25000/[(70×10⁹)(5×10⁻⁴)] = 25000/(3.5×10⁷) = 7.14×10⁻⁴ m²

Titanium: A ≥ 25000/[(110×10⁹)(5×10⁻⁴)] = 25000/(5.5×10⁷) = 4.55×10⁻⁴ m²

Constraint 4 (safety factor ≥ 5) gives maximum stress σ_max = UTS/5 and therefore minimum area:

Steel: A ≥ F/(σ_max) = 25000/(400×10⁶/5) = 25000/80×10⁶ = 3.13×10⁻⁴ m² ← governs

Aluminum: A ≥ 25000/(54×10⁶) = 4.63×10⁻⁴ m² ← elongation governs (7.14×10⁻⁴)

Titanium: A ≥ 25000/(180×10⁶) = 1.39×10⁻⁴ m²; elongation governs (4.55×10⁻⁴)

All three materials meet both constraints. Governing minimum areas: Steel 3.13×10⁻⁴ m², Aluminum 7.14×10⁻⁴ m², Titanium 4.55×10⁻⁴ m².

Mass comparison (mass = ρAL):

Steel: 7850 × 3.13×10⁻⁴ × 1.0 = 2.46 kg

Aluminum: 2700 × 7.14×10⁻⁴ × 1.0 = 1.93 kg

Titanium: 4430 × 4.55×10⁻⁴ × 1.0 = 2.02 kg

Aluminum is lightest. Required diameter: A = π(d/2)² = 7.14×10⁻⁴ ⇒ d = 2√(A/π) = 2√(7.14×10⁻⁴/π) = 0.030 m = 30.1 mm

This is how real material selection works in mechanical engineering: multiple constraints are applied simultaneously, each one demanding a minimum cross-section, and the governing constraint (the one requiring the largest area) determines the actual design. Here aluminum wins on mass despite needing a much larger diameter, because its density advantage outweighs the area penalty from its lower stiffness.

Module 5 Complete — What You Have Built

The Full Arc of Module 5

Each lesson added one layer. Together they form a complete framework for analyzing anything that rotates, balances, or deforms.

M5L1 — Rotational Kinematics: How to describe rotation (angles, angular velocity, angular acceleration) as a direct analogy to linear kinematics.
M5L2 — Torque and Rotational Dynamics: What causes rotation to change (τ = Iα) and how mass distribution determines resistance to rotation.
M5L3 — Rolling Motion and Angular Momentum: How translation and rotation combine in rolling, and why angular momentum is conserved when no external torque acts.
M5L4 — Static Equilibrium: The two conditions (ΣF = 0 and Στ = 0) that must hold simultaneously for a structure to remain still, and the pivot-point strategy for solving them efficiently.
M5L5 — Elasticity (this lesson): What the forces from M5L4 equilibrium analysis do to real materials — stress, strain, and the three moduli that quantify deformation.

Module 5 Self-Check

Before moving to Module 6, make sure you can answer all of these from memory:

A steel rod (Y = 200 GPa) of diameter 1.0 cm and length 2.0 m is under 50 kN tension. What is the stress, the strain, and the elongation?
Explain in one sentence why the shear modulus of any material is always less than its Young's modulus.
A rubber pad (S = 0.5 MPa) measures 4 cm × 4 cm × 3 cm thick. A 200 N horizontal force is applied to the top face. Find Δx.
On a stress-strain curve, label: proportional limit, elastic limit, yield point, ultimate tensile strength, fracture point. Identify which region represents elastic behavior.
A beam problem gives you a cable tension of 3200 N (from M5L4 equilibrium). The cable is aluminum (Y = 70 GPa), 1.5 m long, 6 mm diameter. How much does it elongate?

Module 6 moves into oscillations — the spring constant you have been using throughout M5 will now appear in its dynamic context, driving simple harmonic motion. Everything connects.