Structural Statics

Jump to Examples

Every building, bridge and shelf on the wall answers the same question: where do the forces go? The weight of the roof, the people on the floor, the snow and the wind all have to travel through the structure, from element to element, until they reach the ground. Statics is the part of mechanics that studies bodies that do not move – and a building that does not move is exactly what we want.

The basic idea is surprisingly simple: if something stands still, all forces on it are in balance. Everything else – beams, trusses, columns, standards and software – is a careful, systematic way of using that one idea, and of making sure that the materials are strong and stiff enough to carry the forces that result from it.

A Bit of History

For thousands of years, people built without calculations. The Egyptian pyramids, Roman aqueducts and Gothic cathedrals were designed with experience, geometric rules of thumb and, sometimes, collapse and repair. The Roman architect Vitruvius wrote that a building must have firmitas, utilitas, venustas – strength, usefulness and beauty – but he did not explain how to calculate the first one. The theory began much earlier, in the 3rd century BC, when Archimedes described the law of the lever and the centre of gravity – the idea of the moment that is still used in every calculation today.

The modern science started in the 16th and 17th centuries. In 1586 Simon Stevin showed how forces can be added with a triangle (or parallelogram) of forces. In 1638 Galileo Galilei studied the strength of beams – the first scientific attempt, although his result was not fully correct. Robert Hooke (1678) discovered that materials stretch in proportion to the force, and Isaac Newton (1687) gave the laws of motion – the first of which, applied to a body at rest, is statics. In the 18th century Leonhard Euler and the Bernoulli family developed the beam theory and Euler's formula for the buckling of columns, both still used today (see Example 4).

In 1826 Claude-Louis Navier published his lectures on the strength of materials, which are often seen as the start of modern structural analysis: for the first time, engineers could calculate stresses and deflections in a systematic way. Railways and iron bridges needed fast, reliable methods, and the 19th century produced many of them. Karl Culmann (1866) and Luigi Cremona developed graphic statics – solving structures by drawing forces to scale – and engineers such as James Clerk Maxwell, Otto Mohr and Alberto Castigliano gave methods for statically indeterminate structures. The Eiffel Tower (1889) is a famous example of a structure designed in that era.

The 20th century brought new materials and new methods. Reinforced concrete spread at the turn of the century, and in 1930 Hardy Cross published the moment distribution method, which let engineers solve continuous frames by hand, step by step – for decades, with a slide rule. Then came the computer. In the 1950s engineers in the aircraft industry developed the matrix stiffness method (Turner, Clough, Martin and Topp, 1956), and in 1960 Ray Clough named it the finite element method (FEM). Programs such as NASTRAN (late 1960s, NASA) and SAP from Berkeley (around 1970) brought it into practice, and with personal computers in the 1980s it moved to every design office.

In our part of Europe, earthquakes have shaped structural design. The 1895 Ljubljana earthquake led to the rebuilding of much of the city, and after the destructive 1963 Skopje earthquake, Yugoslavia introduced seismic building regulations. Today, earthquake design in Slovenia follows Eurocode 8 and, since 2024, a new seismic hazard map from 2021 with higher values in most of the country. The Eurocodes themselves were developed from the 1970s onwards, published as European standards in the 2000s, and replaced the national design codes around 2010. A second generation of Eurocodes is now being introduced: CEN made the texts available in March 2026, and the first generation is to be withdrawn by March 2028.

Today, structures are usually designed as 3D computer models that are shared between architects and engineers using BIM and the open IFC format. The software does the heavy arithmetic in seconds, but the principles have not changed since Archimedes and Navier – and the engineer still has to understand them, because a computer will calculate a wrong model just as quickly as a correct one.

A short timeline:

Some resources for history:

• History of structural engineering (Wikipedia)
• Structural analysis (Wikipedia)
• Graphic statics (Wikipedia)
• History of the finite element method (Wikipedia)
• Eurocodes – the official EU website (JRC)

The Basics of Statics

The table below goes from the simplest ideas to the ones used in every real calculation. The units are those used in practice: forces in kilonewtons (kN), lengths in metres (m), stresses in megapascals (MPa = N/mm²).

The building blocks:

Loads and safety – how the Eurocodes approach a calculation:

Slovenia and Germany: the same Eurocodes, different national annexes

The Eurocodes are the same in all EU countries, but each country chooses its own nationally determined parameters – values that depend on climate, geology, tradition and the level of safety the country wants. They are published in national annexes: in Slovenia as SIST EN … /A101 (for example SIST EN 1991-1-3:2004/A101 for snow), in Germany as DIN EN … /NA. In Slovenia, the Eurocodes and their national annexes are required by the Pravilnik o mehanski odpornosti in stabilnosti objektov (Official Gazette RS 101/2005); in Germany, they are introduced through the building authorities' list of technical rules (MVV TB).

The national annexes are paid documents. The values below come from publicly available sources – the Slovenian Environment Agency (ARSO), the Slovenian Chamber of Engineers (IZS), published Slovenian design reports and German technical literature – so treat them as an overview and always check the current SIST or DIN text before using them.

Selected differences (overview, check the current national annexes):

The lesson for a designer working across borders: the method is the same everywhere, but the numbers are not. Software usually contains the national annexes of many countries, and choosing the right one is one of the first settings in every new project.

How a structural calculation is made in practice:

A few observations. First, a good engineer first estimates the answer, and only then calculates it. If the computer says a 5 m beam needs a moment capacity of 2000 kNm, something is wrong with the input. Second, the load path matters more than the numbers: a structure in which the forces cannot find a clear way to the ground will cause problems no matter how carefully it is calculated. Third, serviceability often governs – especially for steel and timber, the deflection limit, not the strength, often decides the size (Example 1).

Some resources for learning statics:

• Eurocodes (JRC) – background documents and worked examples, free
• GOV.SI – the new seismic hazard map of Slovenia (in Slovenian)
• Comparison of the 2021 seismic hazard map of Slovenia with European and neighbouring maps (Ujma)
• IZS – Slovenian Chamber of Engineers, publishes a handbook on design with the Eurocodes
• Free body diagram (Wikipedia)
• Euler–Bernoulli beam theory (Wikipedia)
• MIT OpenCourseWare – search for structural mechanics
• My Structural Analysis resources – free and open-source programs
• My Programming & Development resources – Python, JupyterLab and more

Examples: Calculating by Hand and with a Computer

The examples below go from a single beam to a small computer program. They are simplified on purpose: real designs need more checks (for example lateral-torsional buckling, connections, fire and earthquakes) and the values from the national annex. I checked all the numbers with a short Python script; the results are rounded.

Example 1 – A simply supported steel floor beam

A steel beam spans L = 5 m between two walls (a pin at one end, a roller at the other). From the floor area it carries, it gets a permanent load of g = 3.0 kN/m and an imposed load of q = 2.0 kN/m. Steel grade S235 (yield strength f_y = 235 MPa). We assume that the floor holds the top of the beam sideways, so it cannot buckle sideways.

1. Design load (ULS):

   q_Ed = 1.35 × 3.0 + 1.5 × 2.0 = 4.05 + 3.00 = 7.05 kN/m

2. Reactions – the load is symmetric, so each support takes half:

   R = q_Ed × L / 2 = 7.05 × 5 / 2 = 17.6 kN

   This is also the largest shear force, at the supports.
3. Maximum bending moment, in the middle of the span – a formula every engineer knows by heart:

   M_Ed = q_Ed × L² / 8 = 7.05 × 25 / 8 = 22.0 kNm

4. Required section modulus (plastic design, partial factor γ_M0 = 1.0):

   W_pl,required = M_Ed / f_y = 22.0 × 10⁶ Nmm / 235 MPa = 93,750 mm³ = 93.8 cm³

   From a table of steel profiles: an IPE 140 has W_pl = 88.3 cm³ – not enough (106 % utilised). An IPE 160 has W_pl = 123.9 cm³, so its moment resistance is 123.9 × 235 / 1000 = 29.1 kNm – utilisation 76 %. Strength is fine.
5. Deflection (SLS) – now with the loads without safety factors, g + q = 5.0 kN/m. For a simply supported beam with a uniform load:

   δ = 5 × q × L⁴ / (384 × E × I)
   IPE 160 (I = 869 cm⁴):  δ = 5 × 5.0 × 5000⁴ / (384 × 210,000 × 8,693,000) = 22.3 mm

   For floors in general, the Slovenian national annex to EN 1990 limits the total deflection to L/250 = 5000 / 250 = 20 mm. The IPE 160 is strong enough, but it bends too much.
6. Choose a larger profile: an IPE 180 (I = 1317 cm⁴) gives δ = 14.7 mm < 20 mm, with a strength utilisation of 56 %. The second check, for the imposed load alone (q = 2.0 kN/m), gives 14.7 × 2 / 5 = 5.9 mm, well below L/300 = 16.7 mm. This is the beam we choose.

This small example shows something that surprises many people: the beam was not chosen by its strength, but by its stiffness. In real design we would also add the self-weight of the beam (about 0.19 kN/m for an IPE 180), check the shear and the supports, and use the deflection limits agreed for the project.

Example 2 – A simple roof truss, joint by joint

A triangular truss spans 6 m and is 2 m high. A point load of F = 10 kN acts at the top. The two sloping members are the rafters, the horizontal member at the bottom is the tie. In a truss, members carry only axial forces – tension or compression – if the loads act at the joints.

1. Reactions: the truss and the load are symmetric, so each support takes F / 2 = 5 kN upwards.
2. The angle of the rafters: tan α = 2 / 3, so α = 33.7°, sin α = 0.555 and cos α = 0.832.
3. Cut out the left support joint (a free body diagram) and use the equilibrium equations. Vertically, only the reaction and the rafter act:

   ΣF_y = 0:   5 + N_rafter × sin α = 0   →   N_rafter = −5 / 0.555 = −9.0 kN  (compression)

   Horizontally, the tie must balance the horizontal part of the rafter force:

   ΣF_x = 0:   N_tie + N_rafter × cos α = 0   →   N_tie = 9.0 × 0.832 = +7.5 kN  (tension)

4. Check with a quick formula: for this truss, the tie force is F × L / (4 × h) = 10 × 6 / 8 = 7.5 kN. ✔

This is the method of joints: go from joint to joint, and at each one use ΣF_x = 0 and ΣF_y = 0. Notice what happens if the truss is made flatter: with h = 1 m, the tie force doubles to 15 kN. A low roof pushes its walls apart much harder than a steep one – which is exactly why the tie is there.

Example 3 – Reading a bending moment diagram

Before the calculation of a column, a short exercise in "seeing" internal forces. For the beam in Example 1, the bending moment is zero at the supports and grows towards the middle in the shape of a parabola, reaching 22.0 kNm. The shear force is largest at the supports (±17.6 kN) and zero in the middle. Two rules help to draw such diagrams without any calculation:

• Where the shear force is zero, the bending moment is at its maximum. That is why the largest moment is in the middle of a symmetric beam.
• A uniform load gives a straight-line shear diagram and a curved (parabolic) moment diagram; a point load gives a jump in the shear diagram and a kink in the moment diagram.

In a cantilever – a balcony, for example – the picture is reversed: the moment is largest at the wall and zero at the free end. Its value is q × L² / 2, four times more than for a simply supported beam of the same length. That is why balconies need strong, well-anchored connections, and why the reinforcement in a concrete balcony slab is at the top, where the tension is.

Example 4 – Buckling of a timber column

A timber post of grade C24, 10 × 10 cm, 3 m high, is pinned at both ends. How much load can it carry before it buckles?

1. Second moment of area of a square section:

   I = b⁴ / 12 = 100⁴ / 12 = 8,333,333 mm⁴

2. Euler's critical force, with the mean modulus of elasticity of C24 (E = 11,000 MPa):

   N_cr = π² × E × I / L² = 9.87 × 11,000 × 8,333,333 / 3000² = 100,500 N ≈ 100 kN

3. What it means: about 100 kN is the theoretical limit for a perfectly straight post. Real columns are never perfectly straight, and timber is not perfectly uniform, so the design rules in Eurocode 5 (and Eurocode 3 for steel) reduce this value with a buckling factor. For the same post at 6 m height, the critical force drops to a quarter – about 25 kN.

Example 5 – A continuous beam with the stiffness method, in Python

Now we let the same IPE beam continue over a middle support: two spans of 5 m each, three supports, the same design load of 7.05 kN/m. This beam is statically indeterminate – four unknown reactions, but only three equations – so equilibrium alone is not enough. Software solves such problems with the stiffness method, the basis of the finite element method:

1. Split the structure into elements (here: two beam elements) connected at nodes.
2. For each element, write a small stiffness matrix – how much force is needed to move or rotate its ends.
3. Assemble all element matrices into one big matrix for the whole structure, and the loads into a load vector.
4. Apply the supports (remove the movements that are blocked).
5. Solve the system of equations K × u = F for the unknown displacements u.
6. From the displacements, calculate the reactions and internal forces.

The program below does exactly these six steps. Every structural program – from free tools to the commercial ones used in design offices – does the same thing, only with thousands of elements and many more checks.

import numpy as np

# Continuous beam: two spans of 5 m, three supports, uniform load
spans = [5.0, 5.0]       # span lengths in m
q = 7.05                 # design load in kN/m (from Example 1)
EI = 1.0                 # stiffness; for forces in a uniform beam its value does not matter

n_nodes = len(spans) + 1
K = np.zeros((2 * n_nodes, 2 * n_nodes))   # global stiffness matrix
F = np.zeros(2 * n_nodes)                  # global load vector
# each node has 2 unknowns: vertical displacement (v) and rotation (theta)

for e, L in enumerate(spans):
    # stiffness matrix of one beam element (Euler-Bernoulli beam theory)
    k = EI / L**3 * np.array([
        [ 12,    6*L,   -12,    6*L  ],
        [ 6*L,   4*L*L, -6*L,   2*L*L],
        [-12,   -6*L,    12,   -6*L  ],
        [ 6*L,   2*L*L, -6*L,   4*L*L],
    ])
    # uniform load turned into equivalent forces and moments at the two ends
    f = q * np.array([-L/2, -L*L/12, -L/2, L*L/12])
    dofs = [2*e, 2*e + 1, 2*e + 2, 2*e + 3]
    K[np.ix_(dofs, dofs)] += k             # assembly: add the element into the big matrix
    F[dofs] += f

# boundary conditions: no vertical movement at the supports, rotations are free
fixed = [2*i for i in range(n_nodes)]
free = [d for d in range(2 * n_nodes) if d not in fixed]

u = np.zeros(2 * n_nodes)
u[free] = np.linalg.solve(K[np.ix_(free, free)], F[free])   # solve K * u = F

# support reactions: what the supports must push back
R = K @ u - F
for i in range(n_nodes):
    print(f"support {i + 1}: reaction = {R[2*i]:7.2f} kN")

# bending moment over the middle support, from the first element
L = spans[0]
k1 = EI / L**3 * np.array([[12, 6*L, -12, 6*L], [6*L, 4*L*L, -6*L, 2*L*L],
                           [-12, -6*L, 12, -6*L], [6*L, 2*L*L, -6*L, 4*L*L]])
f1 = q * np.array([-L/2, -L*L/12, -L/2, L*L/12])
end_forces = k1 @ u[0:4] - f1
print(f"moment over the middle support = {abs(end_forces[3]):.2f} kNm (tension at the top)")
print(f"check with the textbook formula q*L^2/8 = {q * L**2 / 8:.2f} kNm")

Output:

support 1: reaction =   13.22 kN
support 2: reaction =   44.06 kN
support 3: reaction =   13.22 kN
moment over the middle support = 22.03 kNm (tension at the top)
check with the textbook formula q*L^2/8 = 22.03 kNm

The results match the textbook formulas for a two-span beam: 3/8 × q × L at the end supports, 10/8 × q × L at the middle, and a moment of q × L² / 8 over the middle support. Two interesting things happen compared with Example 1. First, the middle support carries more than it would if the two spans were separate beams (44.1 kN instead of 2 × 17.6 = 35.3 kN). Second, the largest moment is now over the support, with tension at the top – so a concrete beam needs reinforcement at the top there, and a steel beam may need a check of its bottom flange in compression.

If you have read How AI Works, the code will look familiar: matrices, vectors and one big system of equations. Structural analysis was one of the first fields to use computers for large matrix calculations – many years before AI.

Checking with free software

To check the examples with a program, try the free 2D frame program Ftool (draw the beam, add supports and loads, and read the diagrams), or the Python library PyNite if you prefer code. For full finite element models, FreeCAD with its FEM workbench and the CalculiX solver is a good open-source start. More programs are listed in my Structural Analysis resources. In design offices, commercial programs are common – in Slovenia, for example, Dlubal RFEM, SCIA Engineer and Tower – but the principles behind them are the same as in Example 5.

Keeping It Safe

• Learning is not designing. These examples are simplified. Anything that people will live in, walk on or stand under must be designed and signed by a licensed structural engineer.
• Use the right standards. Always use the current Eurocodes together with the national annex of the country where the building stands – and keep in mind the change to the second generation.
• Check the input, not only the output. Most errors in computer calculations come from wrong loads, wrong units or wrong supports in the model.
• Make a hand estimate. A quick q × L² / 8 on paper catches surprisingly many mistakes in a large model.
• Do not forget the ground. A perfect building on poor soil still fails – geotechnics (Eurocode 7) belongs to every design.
• Think about earthquakes early. In Slovenia, with design ground accelerations up to 0.325 g (0.275 g in Ljubljana), the seismic design often decides the layout of walls and columns, so it cannot be added at the end.

A Few More Thoughts

Statics is one of the oldest branches of engineering, and one of the most honest: gravity does not negotiate. The three equilibrium equations that Stevin and Newton would recognise are still at the heart of every calculation, whether it is done on paper, with a slide rule or with software that models a whole building in 3D.

Computers have changed how fast we calculate, not what we calculate. Like in my notes on AI, the tool is powerful, but the responsibility stays with the person who uses it. A program will happily solve a wrong model, and only an engineer who understands where the forces go will notice. That is why it is still worth knowing how to calculate a simple beam by hand – it is the quickest way to know whether to trust the computer.