Equilibrium and Elasticity

Learning Objectives

  • Define and apply the conditions for static equilibrium.
  • Calculate the center of mass and understand its relation to the center of gravity.
  • Define stress, strain, and understand their relationship through Hooke's Law.
  • Analyze material deformation including tension, compression, shear, and bulk stress.
  • Interpret proportional, elastic, yield, plastic, ultimate, and fracture landmarks without conflating them.
  • Calculate free thermal strain and fully restrained thermal stress for the stated linear-elastic model.

For a structure like a bridge or a building to serve its purpose, it must remain stationary and maintain its shape under various loads. This requires the principles of static equilibrium and an understanding of how materials deform (elasticity).

Static Equilibrium

Static Equilibrium Concepts

An object is in static equilibrium if it is completely at rest in our chosen frame of reference. This means it has no linear acceleration and no angular acceleration.

For a rigid body (an object whose size and shape do not change under load), two conditions must be met simultaneously for it to be in equilibrium.

Translational Equilibrium Condition

The vector sum of all external forces must be zero.

ΣF⃗=0\Sigma \vec{F} = 0

Variables

SymbolDescriptionUnit
ΣF⃗\Sigma \vec{F}Net external forceN

Translational Equilibrium in 2D

In 2D (xy-plane), the translational equilibrium condition breaks down into two scalar equations: ΣFx=0\Sigma F_x = 0 and ΣFy=0\Sigma F_y = 0.

Rotational Equilibrium Condition

The vector sum of all external torques about any axis must be zero.

Στ⃗=0\Sigma \vec{\tau} = 0

Variables

SymbolDescriptionUnit
Στ⃗\Sigma \vec{\tau}Net external torqueN⋅mN\cdot m

Choosing an Axis of Rotation

When applying the torque equation (Στ=0\Sigma \tau = 0), you are free to choose the axis of rotation anywhere you like. A strategic choice of axis (e.g., placing it exactly where an unknown force acts) will eliminate that unknown force from your torque equation, simplifying the math significantly.

Interactive Simulation

Use this torque balance model to see how lever arms and loads must offset each other for rotational equilibrium.

Torque Balance & Rotational Equilibrium Simulator

Concept and model scope

Build a complete free-body diagram for a beam supported at its center. The beam is held horizontal so gravity remains vertical; the purple indicator shows the rotational tendency. Static rotational equilibrium occurs only when the signed net torque is zero.

Model scope: Use the displayed units and idealizations, then verify the governing balance or compatibility equation before interpreting the result.

✓ Equilibrium
Left Load — CCW
Mass 1 (m₁)

Mass 1 (m₁)

Mass 1 (m₁) is an adjustable engineering parameter. The displayed result and geometry should update directly from this value.

Range: 5–50 kg. Step: 5 kg.

20 kg
Position 1 (x₁)

Position 1 (x₁)

Position 1 (x₁) is an adjustable engineering parameter. The displayed result and geometry should update directly from this value.

Range: -4.0–-0.5 m. Step: 0.1 m.

-3.0 m
Right Load — CW
Mass 2 (m₂)

Mass 2 (m₂)

Mass 2 (m₂) is an adjustable engineering parameter. The displayed result and geometry should update directly from this value.

Range: 5–50 kg. Step: 5 kg.

30 kg
Position 2 (x₂)

Position 2 (x₂)

Position 2 (x₂) is an adjustable engineering parameter. The displayed result and geometry should update directly from this value.

Range: 0.5–4.0 m. Step: 0.1 m.

2.0 m
Moment Balance about O
ΣτO=588.6 N⋅m−588.6 N⋅m=0.0 N⋅m\Sigma\tau_O = 588.6\,\mathrm{N\cdot m} - 588.6\,\mathrm{N\cdot m} = 0.0\,\mathrm{N\cdot m}

Rotational equilibrium: ∑τO=0\sum \tau_O = 0. The uniform beam weight passes through the pivot, so it contributes no moment about OO, but it is still included in the vertical reaction.

CCW Torque
588.60 N·m
CW Torque
588.60 N·m
Vertical Reaction R_y
588.60 N
Net Torque
0.00 N·m
Static Equilibrium Requires a Complete FBD

Force equilibrium and moment equilibrium must be applied to the same isolated body with all external forces and lever arms shown.

Beam free-body diagram with two support reactions, a point load, and self-weight at its center

Center of Gravity (CG)

Center of Gravity (CG) Concepts

The center of gravity is the point at which the entire weight of an object can be considered to act for the purpose of calculating torques due to gravity.

For a uniform object in a uniform gravitational field (like near the Earth's surface), the center of gravity coincides perfectly with the geometric center of mass (CM).

Center of Mass Location (xcmx_{cm})

The geometric center of a mass distribution.

Center of Mass for Discrete Masses

Calculates center of mass position for a system of discrete particles.

xcm=ΣmixiΣmix_{cm} = \frac{\Sigma m_i x_i}{\Sigma m_i}

Variables

SymbolDescriptionUnit
xcmx_{cm}Center of mass positionm
mim_iMass of particle ikg
xix_iPosition of particle im

Center of Mass for a Continuous Body

Calculates center of mass position for a continuous uniform body.

xcm=1M∫0Lx dmx_{cm} = \frac{1}{M} \int_0^L x \, dm

Variables

SymbolDescriptionUnit
xcmx_{cm}Center of mass positionm
MMTotal masskg
LLLength of the bodym
xxPosition along the lengthm
dmdmInfinitesimal mass elementkg

Elasticity and Deformation

Elasticity and Deformation Concepts

In reality, no object is perfectly "rigid." When forces are applied, all materials deform to some extent. Understanding this deformation is the bridge between basic physics and "Mechanics of Materials," a core engineering subject.

Stress (σ\sigma)

Stress characterizes the intensity of the internal forces acting within a deformable body. It is the applied force per unit cross-sectional area. The SI unit is the Pascal (Pa), where 1 Pa=1 N/m21 \text{ Pa} = 1 \text{ N/m}^2.

Stress Equation

Calculates stress from applied force and cross-sectional area.

σ=FA\sigma = \frac{F}{A}

Variables

SymbolDescriptionUnit
σ\sigmaStressPa
FFPerpendicular force appliedN
AACross-sectional aream2m^2

Strain (ϵ\epsilon)

Strain is the measure of the relative deformation (change in shape or size) of an object in response to stress. It is a dimensionless ratio.

Strain Equation

Calculates strain from change in length relative to original length.

ϵ=ΔLL0\epsilon = \frac{\Delta L}{L_0}

Variables

SymbolDescriptionUnit
ϵ\epsilonStraindimensionless
ΔL\Delta LChange in lengthm
L0L_0Original lengthm

Hooke's Law for Continua

Hooke's Law for Continua Concepts

For small deformations, most solid materials exhibit elastic behavior: they return to their original shape when the stress is removed, and the strain is directly proportional to the stress. This is Hooke's Law applied to continuous media.

Hooke's Law

General relationship between stress and strain within the elastic limit.

Stress=Elastic Modulus×Strain\text{Stress} = \text{Elastic Modulus} \times \text{Strain}

Variables

SymbolDescriptionUnit
Stress\text{Stress}Applied stressPa
Elastic Modulus\text{Elastic Modulus}Material property indicating stiffnessPa
Strain\text{Strain}Resulting straindimensionless

Interactive Simulation

Adjust force, area, and elastic modulus to connect stress, strain, and stiffness before moving into mechanics of materials.

Idealized Bilinear Stress-Strain Simulator

Concept and model scope

This is an explicit monotonic-loading teaching model, not a material database. It is linear elastic to the modeled yield point, then follows a bilinear strain-hardening branch with tangent modulus Eₜ. Unloading, cyclic response, ultimate strength, necking, and fracture are not represented.

Model scope: Use the displayed units and idealizations, then verify the governing balance or compatibility equation before interpreting the result.

Bilinear strain-hardening model
Axial force

Axial force

Axial force is an adjustable engineering parameter. The displayed result and geometry should update directly from this value.

Range: 1–180 kN. Step: 1 kN.

30 kN
Area

Area

Area is an adjustable engineering parameter. The displayed result and geometry should update directly from this value.

Range: 100–1000 mm². Step: 25 mm².

200 mm²
Young's modulus E

Young's modulus E

Young's modulus E is an adjustable engineering parameter. The displayed result and geometry should update directly from this value.

Range: 40–220 GPa. Step: 5 GPa.

200 GPa
Modeled yield stress σᵧ

Modeled yield stress σᵧ

Modeled yield stress σᵧ is an adjustable engineering parameter. The displayed result and geometry should update directly from this value.

Range: 100–500 MPa. Step: 10 MPa.

250 MPa
Post-yield tangent ratio Eₜ/E

Post-yield tangent ratio Eₜ/E

Post-yield tangent ratio Eₜ/E is an adjustable engineering parameter. The displayed result and geometry should update directly from this value.

Range: 0.005–0.080. Step: 0.005.

0.020

Constitutive definition

For σ ≤ σᵧ: ε = σ/E.

For σ > σᵧ: ε = εᵧ + (σ − σᵧ)/Eₜ.

In this deliberately simplified monotonic-loading law, the proportional limit and modeled yield point coincide. Real materials may not show that coincidence. Unloading and all ultimate/fracture behavior require a different constitutive model or material-specific data.

yieldslope Epost-yield slope Eₜstrain εstress σ (MPa)uniform axial specimenspecimen length follows ε; display capped at 25%no unloading, necking, ultimate, or fracture branch
Normal stress σ
150.0 MPa
Strain ε
7.500e-4
Yield strain εᵧ
1.250e-3
Model state
LINEAR ELASTIC
Young's modulus E
200.00 GPa
Post-yield tangent Eₜ
4.00 GPa
Explicit Bilinear Constitutive Model

The teaching model is linear elastic to its modeled yield point and then follows a lower tangent modulus. It does not represent ultimate strength, necking, or fracture.

Idealized bilinear stress-strain curve with linear elastic and strain-hardening branches

Scope of the Interactive Constitutive Model

The interactive stress-strain curve is an explicitly idealized bilinear strain-hardening law. It is not a sourced material curve for steel, aluminum, titanium, copper, or any other specific alloy. In this idealization the proportional limit and modeled yield point coincide; real material data can distinguish these quantities and may exhibit nonlinear transition behavior.

Types of Elastic Moduli

The specific "Elastic Modulus" depends on the type of stress being applied. Common types include Young's Modulus, Shear Modulus, and Bulk Modulus.

Young's Modulus (EE)

Measures resistance to tension (stretching) or compression (squeezing) along one axis. This is crucial for designing columns and cables.

Young's Modulus Stress-Strain Equation

The fundamental relationship between stress and strain for tension or compression.

σ=Eϵ\sigma = E \epsilon

Variables

SymbolDescriptionUnit
σ\sigmaStressPa
EEYoung's ModulusPa
ϵ\epsilonStraindimensionless

Young's Modulus Force-Area Equation

Relates applied force, area, and length changes for tension or compression.

FA=E(ΔLL0)\frac{F}{A} = E \left(\frac{\Delta L}{L_0}\right)

Variables

SymbolDescriptionUnit
FFForce appliedN
AACross-sectional aream2m^2
EEYoung's ModulusPa
ΔL\Delta LChange in lengthm
L0L_0Original lengthm

Shear Modulus (GG)

Measures resistance to shear forces (forces acting parallel to a surface, trying to slide layers past one another), where γ\gamma is the shear strain angle.

Shear Modulus Equation

Hooke's Law applied to shear deformation.

τshear=Gγ\tau_{shear} = G \gamma

Variables

SymbolDescriptionUnit
τshear\tau_{shear}Shear stressPa
GGShear ModulusPa
γ\gammaShear strainrad

Bulk Modulus (BB)

Measures resistance to uniform compression from all sides (like an object submerged deep in the ocean). It relates pressure (PP) to volume strain (ΔV/V0\Delta V/V_0). The negative sign in the equation indicates that increased pressure causes a decrease in volume.

Bulk Modulus Equation

Relates pressure change to volume strain.

ΔP=−B(ΔVV0)\Delta P = -B \left(\frac{\Delta V}{V_0}\right)

Variables

SymbolDescriptionUnit
ΔP\Delta PChange in pressurePa
BBBulk ModulusPa
ΔV\Delta VChange in volumem3m^3
V0V_0Original volumem3m^3

Thermal Stress

Thermal Stress Concepts

If a structural member is constrained so that it cannot expand or contract when subjected to a temperature change (ΔT\Delta T), large internal stresses develop. The thermal strain is ϵthermal=αΔT\epsilon_{thermal} = \alpha \Delta T. Because the member is constrained, the opposing stress developed is defined by the thermal stress equation.

Thermal Stress Equation

Calculates stress caused by constrained thermal expansion or contraction.

σthermal=EαΔT\sigma_{thermal} = E \alpha \Delta T

Variables

SymbolDescriptionUnit
σthermal\sigma_{thermal}Thermal stressPa
EEYoung's ModulusPa
α\alphaCoefficient of linear expansion1/∘C1/^\circ C
ΔT\Delta TChange in temperature∘C^\circ C
Elastic Deformation Modes

Young's, shear, and bulk moduli relate different conjugate stress-strain measures; thermal strain becomes stress only when expansion or contraction is restrained.

Normal, shear, volumetric, and restrained thermal deformation modes

The Stress-Strain Curve

The Stress-Strain Curve Concepts

If you steadily increase the tensile stress on a material (like a steel rod) and plot the resulting strain, you get a characteristic curve.

Landmarks on a Real Tensile Stress-Strain Curve

  • Proportional limit: highest stress over which stress and strain remain proportional within the measurement/model definition.
  • Elastic limit: highest stress from which the material can unload without permanent strain; it is conceptually distinct from the proportional limit.
  • Yield: onset of appreciable plastic deformation, defined by a material-specific convention when a sharp yield point is absent.
  • Plastic region: deformation includes a permanent component after unloading.
  • Ultimate tensile strength: maximum engineering stress reached in a tensile test before the engineering-stress curve decreases during necking.
  • Fracture: specimen separation. It must not be inferred from a model that does not include a fracture criterion.

Yield Strength Importance

Engineering designs almost always require materials to stay well below their Yield Strength, ensuring they remain in the elastic region under typical operating loads.

Shear and Bulk Moduli

Deformation Beyond Tension

While Young's Modulus (EE) handles simple stretching and compression, complex structures experience other types of stress.

  • Shear Stress and Strain: Forces acting parallel to a surface cause layers of the material to slide past one another. The Shear Modulus (GG) relates shear stress (τ=Fparallel/A\tau = F_{parallel}/A) to shear strain (γ=Δx/h\gamma = \Delta x / h). This is critical in analyzing bolts, rivets, and torsion in drive shafts.
  • Bulk Stress and Strain: Forces acting uniformly from all directions (like hydrostatic pressure underwater) cause volume changes. The Bulk Modulus (BB) relates the change in pressure (ΔP\Delta P) to the fractional change in volume (ΔV/V0\Delta V / V_0).

Shear Modulus Equation

Hooke's Law applied to shear deformation.

τ=Gγ\tau = G \gamma

Variables

SymbolDescriptionUnit
τ\tauShear stressPa
GGShear ModulusPa
γ\gammaShear strainrad
Key Takeaways
  • Static Equilibrium requires both zero net force (ΣF⃗=0\Sigma \vec{F} = 0) and zero net torque (Στ⃗=0\Sigma \vec{\tau} = 0).
  • The Center of Gravity is the point where the total weight acts. It coincides with the center of mass for uniform fields.
  • Stress (σ=F/A\sigma = F/A) measures the intensity of internal forces. Strain (ϵ=ΔL/L0\epsilon = \Delta L/L_0) measures the resulting deformation.
  • Hooke's Law states that stress is proportional to strain in the elastic region, governed by an elastic modulus like Young's Modulus (EE).
  • The proportional limit, elastic limit, yield behavior, ultimate strength, and fracture are distinct material-response concepts; the interactive bilinear model represents only the elastic and post-yield branches stated in its constitutive definition.