Forces & Equilibrium

Before you can understand how structures carry loads, you need to understand how forces work. When loads act on a building, they create internal forces within structural members — tension, compression, shear, bending, and torsion. Structural analysis determines the magnitude and distribution of these internal forces so that each member can be sized to safely resist them.

This lesson introduces the fundamental concepts of force mechanics — vector quantities, equilibrium, support reactions, free body diagrams, and the five types of internal stress. These concepts form the language of structural engineering that every construction professional should understand.

A force is a push or pull that tends to change the motion or shape of an object. Every force has three characteristics:

1. Magnitude: How strong the force is (measured in pounds, kips [1 kip = 1,000 pounds], or newtons)
2. Direction: Which way the force acts (up, down, horizontal, angled)
3. Point of application: Where the force is applied on the object

Forces are vectors — quantities with both magnitude and direction. This means forces can be added, subtracted, and resolved into components using vector mathematics.

Concurrent forces are forces whose lines of action pass through a common point. Non-concurrent forces do not share a common point and can create rotation (moments).

In building structures, forces are categorized by how they are applied:

• Concentrated load (point load): A force applied at a single point — such as a column bearing on a beam, or a heavy piece of equipment on a floor. Symbol: P (measured in pounds or kips).
• Distributed load: A force spread over a length or area — such as the weight of a floor slab on a supporting beam, or wind pressure on a wall. A uniformly distributed load (UDL) has constant intensity along its length (symbol: w, measured in pounds per linear foot, plf, or kips per linear foot, klf). A non-uniformly distributed load varies along its length (such as triangular hydrostatic pressure on a wall).
• Moment (couple): A rotational force that tends to cause turning about a point. A moment is calculated as force × distance (M = F × d), measured in foot-pounds (ft-lbs) or foot-kips (ft-k). Moments occur whenever a force is applied at a distance from a point of interest.

A structure must be in static equilibrium — it must be stationary (not moving, not rotating, not accelerating). For a structure to be in equilibrium, three conditions must be satisfied simultaneously:

1. ΣFx = 0: The sum of all horizontal forces equals zero (no net horizontal movement).
2. ΣFy = 0: The sum of all vertical forces equals zero (no net vertical movement).
3. ΣM = 0: The sum of all moments about any point equals zero (no net rotation).

These three equations are the foundation of all static structural analysis. If the applied loads (forces on the structure) and the reactions (forces from the supports) satisfy all three equations, the structure is in equilibrium.

Structural members are connected to their supports (foundations, other members) at points called supports or connections. The type of support determines what reaction forces it can provide:

Pin support (hinged):

• Restrains translation in both horizontal and vertical directions
• Allows rotation (the member can rotate freely about the pin)
• Provides two reaction forces: horizontal (Rx) and vertical (Ry)
• Example: A beam resting on a bearing pad that allows rotation but not sliding

Roller support:

• Restrains translation in one direction only (typically vertical)
• Allows movement in the other direction and allows rotation
• Provides one reaction force (perpendicular to the rolling surface)
• Example: One end of a bridge that must accommodate thermal expansion — it rests on rollers that allow horizontal movement

Fixed support:

• Restrains translation in both directions AND rotation
• Provides three reactions: horizontal force (Rx), vertical force (Ry), and moment (M)
• Example: A flagpole embedded in a concrete base; a cantilever beam welded to a column

Simple support:

• Restrains vertical translation only, allows horizontal movement and rotation
• Provides one vertical reaction force
• Example: A beam sitting on a wall by gravity alone

Understanding support types is critical for construction because the connection details must be built to match the structural engineer's assumptions. If the engineer designed a pin connection (allowing rotation) but the builder creates a rigid connection (resisting rotation), the forces in the structure will be different from what was designed — potentially causing cracking or failure.

A free body diagram (FBD) is the most fundamental tool of structural analysis. It isolates a structural element or a portion of a structure, shows all external forces and reactions acting on it, and allows the equilibrium equations to be applied.

To draw a free body diagram:

1. Isolate the element (mentally "cut" it free from its surroundings)
2. Show all applied loads (dead, live, wind, etc.)
3. Replace each support with its reaction forces (based on support type)
4. If the element was cut from a larger structure, show the internal forces at the cut faces
5. Include dimensions necessary for calculating moments

Example: A simply supported beam (pin at one end, roller at the other) carrying a uniformly distributed load:

• The FBD shows the beam with the distributed load (w, plf) along its length
• At the pin support: vertical reaction (RA) and horizontal reaction (HA)
• At the roller support: vertical reaction (RB)
• Applying ΣFx = 0: HA = 0 (no horizontal loads applied)
• Applying ΣFy = 0: RA + RB = w × L (total vertical reactions equal total load)
• Applying ΣM about A = 0: RB × L = w × L × L/2, so RB = wL/2
• Therefore RA = wL/2 (by symmetry, which is confirmed by the equation)

This simple example illustrates the fundamental process: isolate, identify forces, apply equilibrium, solve for unknowns.

When external forces act on a structural member, they create internal stresses within the material. There are five types:

1. Tension

Tension is the internal stress that occurs when a member is pulled or stretched. The fibers of the material are being pulled apart. Examples:

• The bottom chord of a truss supporting a floor
• A steel cable in a suspension bridge
• Reinforcing steel (rebar) in the bottom of a concrete beam
• A tie-down strap holding a roof to the wall in hurricane zones

Steel is an excellent tension material — it has high tensile strength and is ductile (stretches before breaking). Concrete is poor in tension (only about 10% of its compressive strength), which is why it is reinforced with steel.

2. Compression

Compression is the internal stress that occurs when a member is pushed or shortened. The fibers are being squeezed together. Examples:

• Columns supporting floor loads
• The top chord of a truss
• Bearing walls supporting the structure above
• Foundation footings pressing against the soil

Concrete and masonry are excellent compression materials. Steel is also strong in compression but can fail by buckling (sudden sideways bending) if the member is too slender — an issue discussed in Lesson 4.4.

3. Shear

Shear is the internal stress that occurs when forces act in opposite directions along parallel planes, causing one part of the material to slide relative to the adjacent part. Imagine cutting paper with scissors — the two blades push in opposite directions on adjacent planes.

Examples of shear in structures:

• Vertical shear in beams near their supports (the beam wants to "tear" where the support pushes up and the load pushes down)
• Horizontal shear in beams (the layers of a beam want to slide relative to each other during bending)
• Shear in bolts and nails connecting two members
• Punching shear in flat concrete slabs around columns

Shear failure can be sudden and dangerous — particularly in concrete beams without adequate shear reinforcement (stirrups).

4. Bending (Flexure)

Bending occurs when a member curves under load — one face is in compression while the opposite face is in tension. The internal stress varies linearly from maximum compression on one face to maximum tension on the other, with a neutral axis at the center where stress is zero.

Bending is the primary action in beams and floor slabs. When a beam bends under load:

• The top fibers are shortened (compression)
• The bottom fibers are stretched (tension)
• The neutral axis has zero stress
• The maximum bending stress occurs at the top and bottom faces

The bending stress at any point is calculated by:

σ = M × y / I

Where:

• M = bending moment at the cross-section
• y = distance from the neutral axis
• I = moment of inertia (a geometric property of the cross-section that measures its resistance to bending)

This equation explains why I-beams (W-shapes) are so efficient: they concentrate material at the top and bottom flanges (where bending stress is maximum) and minimize material at the neutral axis (where stress is zero).

5. Torsion

Torsion is the internal stress from twisting. A member in torsion experiences shear stress spiraling through its cross-section. Examples:

• A spandrel beam (an edge beam that supports an eccentric load on one side)
• A cantilevered canopy loaded off-center
• A beam carrying an eccentric concentrated load

Torsion is less common than the other four stress types in building structures, but it must be accounted for when it occurs. Closed cross-sections (tubes, box sections) resist torsion much better than open cross-sections (W-shapes, channels).