Load Analysis
Part of Structural Engineering
Tracing how forces travel through a structure from where loads are applied to where they reach the ground.
Why This Matters
Building a wall that is thick enough to resist the loads it carries requires knowing what those loads are. Choosing the right beam size requires calculating the bending forces inside it. Sizing a column requires knowing the compression it must sustain. All of these calculations require load analysis — systematically identifying every force acting on a structure, determining how those forces flow through the structural elements, and calculating the maximum stresses in each member.
Load analysis is the bridge between observation (I can see this structure needs to support this load) and design (here is how large each element must be to safely carry that load). Without it, you are guessing at structural adequacy. With it, you can make informed decisions, use materials efficiently, and identify which elements are critical versus which have excess capacity.
The methods here are simplified but adequate for the scale of structures a rebuilding civilization would typically encounter. Full engineering analysis requires calculus and matrix methods — the simplified approaches here give reasonable answers for ordinary structures and identify when problems are likely.
Free Body Diagrams
The fundamental tool of load analysis is the free body diagram — a sketch showing a structural element with all forces acting on it.
Steps to draw a free body diagram:
- Isolate the element (imagine cutting it free from everything connected to it)
- Draw the element as a simple shape (line for a beam, box for a column)
- Mark every external force with an arrow showing direction and magnitude
- Mark every reaction force at supports
- Check that forces balance: sum of forces in each direction = zero (for static equilibrium)
Reactions: Where a beam sits on a wall or column, the support pushes up on the beam. This upward force (the reaction) equals the portion of the beam’s load carried by that support. For a beam with equal spans on both sides, reactions are equal. For an asymmetric beam, reactions are proportional to the inverse of the distance to the load.
Example — simply supported beam with uniform load: A 12-foot beam carries 100 lb/ft uniform load (its own weight plus floor load above). Total load = 12 ft × 100 lb/ft = 1,200 lb With equal supports at each end: each reaction = 1,200 / 2 = 600 lb upward
Shear Force and Bending Moment
For beams and horizontal elements, the critical internal forces are shear and bending.
Shear force: The vertical force trying to cut the beam at any cross-section. Think of it as the force you would need to apply at that cross-section to prevent the two halves from sliding past each other vertically.
Calculating shear: Starting from one end, the shear at any point = (reaction at that end) minus (any loads between the end and that point).
For the 12-foot beam above (uniform load, 600 lb reactions):
- At the left support: shear = +600 lb
- At 6 feet (midspan): shear = 600 - (100 × 6) = 600 - 600 = 0 (maximum bending occurs where shear = 0)
- At the right support: shear = -600 lb
Bending moment: The tendency of the beam to bend at any cross-section. At any point, bending moment = sum of (force × distance from that force to the cross-section) for all forces on one side of the section.
Maximum bending moment for uniform load on simply supported beam: M_max = w × L² / 8 Where w = load per unit length (lb/ft), L = span (ft) M_max = 100 × 12² / 8 = 100 × 144 / 8 = 1,800 ft-lb
This maximum moment occurs at midspan. The beam must be designed to resist 1,800 ft-lb of bending.
Maximum bending moments for common loading cases:
| Loading case | Maximum moment |
|---|---|
| Uniform load on simply supported beam | w × L² / 8 |
| Point load P at midspan | P × L / 4 |
| Point load P at any position a from left | P × a × (L-a) / L |
| Fixed ends, uniform load | w × L² / 12 (at supports), w × L² / 24 (at midspan) |
Beam Sizing Using Bending Moment
Once you have the maximum bending moment, you can size the beam cross-section.
The bending stress formula: σ = M / Z
Where σ = bending stress (PSI), M = bending moment (in-lb), Z = section modulus (in³) Z = b × h² / 6 for a rectangular cross-section of width b and depth h
Required section modulus: Z_required = M / σ_allowable
For timber with allowable bending stress of 1,500 PSI and M_max = 1,800 ft-lb = 21,600 in-lb: Z_required = 21,600 / 1,500 = 14.4 in³
Choose a cross-section with Z > 14.4 in³:
- A 4-inch × 8-inch beam: Z = (4 × 8²) / 6 = 256/6 = 42.7 in³ → adequate, with 3× margin
- A 4-inch × 6-inch beam: Z = (4 × 6²) / 6 = 144/6 = 24 in³ → adequate, with 1.7× margin
- A 3-inch × 6-inch beam: Z = (3 × 6²) / 6 = 108/6 = 18 in³ → marginally adequate
The importance of depth: Notice that Z depends on h² — doubling the beam depth quadruples its bending resistance while using only twice the material. Always make beams deep rather than wide for bending resistance.
Column Analysis
Columns carry compressive axial loads. Two failure modes must be checked:
Compressive crushing: Stress = Load / Area. Must be below allowable compressive stress. For a 10,000 lb load on a timber column: If allowable compressive stress = 1,200 PSI: Required area = 10,000 / 1,200 = 8.3 in² → a 3-inch × 3-inch timber (9 in²) is sufficient for crushing
Buckling: Slender columns fail by buckling at loads far below the crushing load. The Euler buckling formula: P_cr = π² × E × I / (K × L)²
Where E = modulus of elasticity, I = moment of inertia, K = end condition factor, L = length
Slenderness ratio (L/r): A simpler check. If slenderness ratio exceeds about 100–120 for timber (or 150 for iron), buckling is the controlling failure mode.
r = √(I/A) for the minimum radius of gyration; for a rectangular section: r_min = h/(2√3) where h is the smaller dimension
If L/r > 100 for timber, use the reduced allowable stress from column design tables.
Lateral Load Analysis
Wind and seismic (earthquake) forces act horizontally on structures. These lateral forces must be carried to the ground through the building’s lateral force-resisting system.
Wind force on a building wall: Total wind force = pressure (PSI or lb/sq ft) × exposed wall area
This force must be carried by:
- Shear walls: walls stiff enough in their plane to carry horizontal loads
- Diagonally braced frames: an X-pattern of diagonal bracing members
- The roof and floor diaphragm: the horizontal structure that transfers wind loads to the shear walls
Checking wall overturning: A wall under horizontal wind pressure wants to overturn (tip over) about its base. Check: Overturning moment = wind pressure × wall area × height to centroid of wind load Resisting moment = weight of wall × half-width of wall
If resisting moment > overturning moment with a safety factor of 1.5, the wall is stable against overturning from wind.
Checking Your Work
Always do a final equilibrium check:
- Sum all vertical forces (loads down, reactions up): should equal zero
- Sum all horizontal forces: should equal zero
- Sum all moments about any point: should equal zero
If any sum is non-zero, you have missed a force or made an arithmetic error. Fix it before proceeding to design.
A second useful check: estimate the result by a different method. If calculating a beam reaction by moments and checking by a simpler method gives the same answer, your analysis is likely correct. Disagreement means an error somewhere.