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 the stress-strain curve and identify key material properties such as yield strength and ultimate strength.

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.

ΣF=0\Sigma \vec{F} = 0

Variables

SymbolDescriptionUnit
ΣF\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.

Στ=0\Sigma \vec{\tau} = 0

Variables

SymbolDescriptionUnit
Στ\Sigma \vec{\tau}Net external torqueNmN\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.

Interactive Physics Simulation

Torque Balance & Rotational Equilibrium Simulator

Place different masses along a balanced beam. Observe CCW and CW torques, pivot reaction forces, and translational/rotational equilibrium states.

✓ Equilibrium (Στ = 0)
Left Block Parameters (CCW)
20 kg
-3.0 m
Right Block Parameters (CW)
30 kg
2.0 m
Live Torque Summation Equation
Στ=τCCWτCW=588.6 Nm588.6 Nm=0.0 Nm\Sigma\tau = \tau_{CCW} - \tau_{CW} = 588.6 \text{ N}\cdot\text{m} - 588.6 \text{ N}\cdot\text{m} = 0.0 \text{ N}\cdot\text{m}

For **Rotational Equilibrium**, counter-clockwise (CCW) torque must exactly equal clockwise (CW) torque ($\Sigma \tau = 0$).

CCW Torque (τ1 = m1·g·r1)
588.60 N·m
CW Torque (τ2 = m2·g·r2)
588.60 N·m
Fulcrum Reaction Force (FR)
588.60 N
Net Residual Torque (Στ)
0.00 N·m

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=1M0Lxdmx_{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.

Interactive Physics Simulation

Tensile Specimen & Stress-Strain Curve

Apply axial tensile force on a specimen. Observe Hooke's Law in the linear elastic region, and the horizontal curving path representing plastic flow beyond the yield threshold.

Elastic Region
Material Preset
35 kN
250 mm²
200 GPa
250 MPa
Governing Formulas
Normal Stress
σ=PA\sigma = \frac{P}{A}
Hooke's Law (Elastic Strain)
ϵ=σE(σσy)\epsilon = \frac{\sigma}{E} \quad (\sigma \le \sigma_y)
Stretching specimen and physical stress-strain graph showing elastic and plastic yielding zonesPStrain (ε)Stress (σ)Yield Threshold (σy = 250 MPa)
Normal Stress (σ)
140.0 MPa
Strain (ε)
7.000e-4
Deformation Mode
ELASTIC

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 temperatureC^\circ C

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.

Regions of the Stress-Strain Curve

    1. Proportional Limit: The highest stress where Hooke's law is valid (the curve is a straight line). The slope of this line is Young's Modulus (EE).
    1. Elastic Limit (Yield Strength): The maximum stress a material can withstand without permanent (plastic) deformation. Up to this point, if you remove the load, the material snaps back to L0L_0.
    1. Plastic Region: Beyond the yield strength, the material deforms permanently. It behaves more like putty.
    1. Ultimate Tensile Strength (UTS): The maximum stress the material can sustain before necking and eventual fracture.
    1. Fracture Point: The stress at which the material breaks.

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 (ΣF=0\Sigma \vec{F} = 0) and zero net torque (Στ=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).
  • Materials have limits (Yield Strength, Ultimate Strength) beyond which they deform permanently or break. Engineering designs must account for these limits.