Equilibrium, stability and elasticity

Most of engineering is about things that should not move: buildings, bridges, cranes on their outriggers, ladders against walls, machine frames on their foundations. Lesson 8 introduced torque and the rotational form of Newton's second law. Setting the accelerations to zero gives the conditions for a rigid body to stay at rest, and this lesson applies them to the questions engineers ask first: where is the center of gravity, what loads does each support carry, how much load will tip a crane, and when will a ladder slip? It then drops the assumption that bodies are perfectly rigid. Real materials stretch, compress and shear under load, and the ideas of stress, strain and elastic modulus describe how much, leading to the deformation of rods and cables and to the stresses caused by temperature change. ENGG-130 develops equilibrium analysis fully; this lesson lays the physical foundations.

The conditions for equilibrium

A rigid body is in static equilibrium when it has no linear or angular acceleration. From \(\sum\mathbf{F} = m\mathbf{a}\) and \(\sum\boldsymbol{\tau} = I\boldsymbol{\alpha}\) (Lesson 8), this requires

For forces in a plane, these give three scalar equations: \(\sum F_x = 0\), \(\sum F_y = 0\) and \(\sum\tau = 0\) about any point. Because the net force is zero, the net torque is the same about every point, so choose the point that eliminates the most unknowns, usually where an unknown force acts. Three independent equations can determine at most three unknown support forces; structures with more supports than that are statically indeterminate and need the deformation analysis of later courses.

The method follows the free-body procedure of Lesson 3: isolate the body, draw every external force including support reactions (with assumed directions), write the three equations and solve. A negative answer simply means the force acts opposite to the direction assumed.

Center of gravity

The weight of a body acts as if concentrated at its center of gravity. In a uniform gravitational field this is the same as the center of mass (Lesson 7), found for a system of parts by weighting each part's position by its weight: \(x_{cg} = \sum W_ix_i/\sum W_i\). For a uniform body it lies at the geometric center. Knowing it is essential for lifting (a load hangs with its center of gravity directly below the hook), for vehicle stability and for distributing loads on supports.

Stability and tipping

A body resting on a base is stable as long as the line of action of its total weight, together with any applied loads, passes inside the base. When the combined loads move the resultant to the edge of the base, the support force there becomes the only one, and any further load tips the body over that edge. Stability is improved by a wide base and a low center of gravity, which is why cranes use outriggers and counterweights and why tall cabinets are anchored to walls.

Friction and equilibrium: the ladder

Friction often keeps a structure in equilibrium, and the question becomes how close it is to slipping. The leaning ladder is the classic example, and a real hazard: falls from ladders are one of the commonest causes of serious injury at work.

Stress and strain

No real body is perfectly rigid. A rod pulled by a force F stretches; how much depends on its size and material. To separate the two, engineers use stress, force per unit area, and strain, the fractional change in length:

Stress is measured in pascals (N/m²), usually megapascals (MPa, equal to N/mm²); strain is dimensionless. For most materials under modest loads, stress is proportional to strain (Hooke's law for materials), with the elastic modulus, or Young's modulus, E as the constant:

Typical values of E are 200 GPa for steel, 70 GPa for aluminum, 25 to 35 GPa for concrete, 10 to 14 GPa for timber along the grain and 1 to 3 GPa for many plastics. The rod behaves as a spring of stiffness k = AE/L (Lesson 4). Proportionality holds up to the yield strength (about 250 to 350 MPa for structural steel), beyond which the material deforms permanently; designs keep working stresses well below yield.

Shear, bulk and thermal stress

Forces can also shear a body, sliding one face relative to the opposite face, as in a bolt holding two plates or an earthquake bearing. Shear stress τ = F/A produces a shear strain (angle) γ, with \(\tau = G\gamma\), where G is the shear modulus (about 80 GPa for steel). Pressure applied equally on all sides changes volume, described by the bulk modulus \(B = -\Delta p/(\Delta V/V)\): about 2.2 GPa for water, which is why water is often treated as incompressible, and why the pressure surge from suddenly closing a valve in a long pipeline (water hammer) can be so large.

Temperature changes also cause strain. A bar heated by ΔT expands by \(\Delta L = \alpha L\,\Delta T\), with a coefficient of thermal expansion α of about 12 × 10⁻⁶ per kelvin for steel and 10 to 12 × 10⁻⁶ for concrete. If the bar is free, this causes no stress. If it is prevented from expanding, the restraint must push back with exactly the force needed to compress it by that amount, giving a thermal stress: