Statics, strength of materials and an introduction to dynamics: equilibrium, trusses, section properties, stress and strain, beam diagrams, bending and deflection, F = ma, vibration and factors of safety, with Python exercises.

Engineering Mechanics: Statics & Strength

## Why it matters

Statics is the study of bodies that are not accelerating: a bridge, a shelf bracket, a crane holding a load still. Almost every structural or machine calculation starts with a statics step, because before you can ask "is this member strong enough?" you have to know what force it carries. Those forces come from one idea: if a body is in equilibrium, the forces and moments on it add up to zero.

## The equilibrium equations

For a **particle** (a point where forces meet, such as a ring where cables join), only the forces matter:

```text
ΣFx = 0
ΣFy = 0
```

For a **rigid body** (a beam, a bracket), forces can also make it rotate, so a third equation is added. The moment of a force about a point is the force times its perpendicular distance from that point (units N·m or kN·m):

```text
ΣFx = 0
ΣFy = 0
ΣM about any point = 0
```

In 2D that gives at most three independent equations per body, so you can solve for at most three unknowns. A body with more unknown support reactions than equations is **statically indeterminate** and needs strength-of-materials methods, which come later in a degree.

## The free-body diagram comes first

1. Isolate the body: draw it alone, cut free from its supports.
2. Replace each support with the reactions it can provide. A **pin** resists movement in both directions (two unknowns, Ax and Ay). A **roller** resists movement only perpendicular to the surface (one unknown). A **fixed support** resists both directions and rotation (three unknowns).
3. Draw every applied load at its real location, with its direction.
4. Add dimensions, then choose axes and a sign convention (for example, up and right positive, counter-clockwise moments positive).

If you guess a reaction's direction wrong, the algebra simply returns a negative number. That is fine; just report it in the opposite direction.

## Worked example 1: a sign on two cables

A 1,000 N sign hangs from a ring held by two cables. Cable 1 pulls up and to the left at 30° above horizontal; cable 2 pulls up and to the right at 45°. Find the tensions.

```text
ΣFx = 0:  −T1·cos30° + T2·cos45° = 0      →  T1 = T2 · (0.7071 / 0.8660) = 0.8165·T2
ΣFy = 0:   T1·sin30° + T2·sin45° − 1000 = 0
           0.8165·T2·0.5 + 0.7071·T2 = 1000  →  1.1154·T2 = 1000
T2 ≈ 896.6 N,   T1 ≈ 732.1 N
```

The steeper cable carries more of the load. Here neither tension exceeds the weight, but if both cables were nearly horizontal the tensions would grow very large: a flat cable has only a small vertical component to hold the load with.

## Worked example 2: reactions on a simply supported beam

A 6 m beam rests on a pin at A (left end) and a roller at B (right end). A 12 kN load acts 2 m from A. Ignore the beam's own weight.

Take moments about A, so the unknowns at A drop out:

```text
ΣMA = 0:  RB·6 − 12·2 = 0        →  RB = 4 kN
ΣFy = 0:  RA + RB − 12 = 0       →  RA = 8 kN
ΣFx = 0:  Ax = 0 (no horizontal loads)
```

The support nearer the load carries more of it. A quick check: take moments about B instead. RA·6 − 12·4 = 0 gives RA = 8 kN, the same answer.

## Choosing where to take moments

Pick the point where the most unknown forces act. Their moment arms are zero, so they vanish from that equation and you solve for one unknown at a time. This trick saves more algebra than any other in statics.

## Common mistakes

- **Forgetting a force.** The weight of the body itself, or a reaction at a support you forgot to draw. Check every place the body touches something.
- **Mixing units.** Keep kN with m (giving kN·m), or N with mm (giving N·mm). Do not mix kN with mm.
- **Using the wrong distance.** A moment arm is the *perpendicular* distance to the line of action of the force, not the distance to where the force is drawn.
- **Including internal forces.** Forces between parts of the same body cancel and do not appear on its free-body diagram.

## A note on real design

These are idealised textbook supports and loads. Real structures carry combinations of dead, live, wind, snow and seismic loads defined by building codes, and real connections are never perfect pins or rollers. Use these methods to learn and to sanity-check, not to size a real structure: that work is done to code and reviewed by a licensed professional engineer.

## Recap

- Equilibrium means ΣF = 0 and, for rigid bodies, ΣM = 0.
- Always start with a free-body diagram showing supports as reactions.
- Take moments about the point with the most unknowns.
- A negative answer means the force acts the other way.

## Try it

On the Do tab, finish a Python function that finds the support reactions of a simply supported beam carrying several point loads.


Draw free-body diagrams and use the equilibrium equations to solve 2D problems for cable tensions and support reactions.

Equilibrium of particles and bodies

Equilibrium of Rigid Bodies (2D - Coplanar Forces) | Mechanics Statics | (Solved examples)

Statics - Free Body Diagram

## Why it matters

A truss is a frame of straight members joined at their ends to form triangles. Roof trusses, bridge trusses, crane booms, transmission towers and bicycle frames all use the idea. Triangles are used because a triangle cannot change shape without changing the length of a side, so a truss carries load mostly by stretching or squashing its members rather than bending them. That makes trusses light and stiff, and it makes them pleasant to analyse.

## The idealisation

Textbook truss analysis assumes:

1. Members are straight and connected only at their ends, by frictionless pins.
2. Loads and reactions are applied only at the joints.
3. The members' own weight is ignored, or shared out to the joints.

Under these assumptions every member is a **two-force member**: it carries a single axial force along its length, either **tension** (pulling on the joints, the member is being stretched) or **compression** (pushing on the joints, the member is being squashed). The usual sign convention: tension positive, compression negative.

A planar truss is statically determinate when m + r = 2j, where m is the number of members, r the number of support reactions and j the number of joints.

## The method of joints

1. Find the support reactions from the equilibrium of the whole truss (previous lesson).
2. Pick a joint with **no more than two unknown member forces**.
3. Draw that joint's free-body diagram. Assume every unknown member force is tension, so it points *away* from the joint, along the member.
4. Apply ΣFx = 0 and ΣFy = 0 at the joint. Solve the two unknowns.
5. Move to the next joint that now has at most two unknowns, and repeat.

Because each joint is a particle, there is no moment equation at a joint: two equations, two unknowns.

## Worked example: a triangular roof truss

Joints: A (pin) at (0, 0), B (roller) at (4, 0) and apex C at (2, 1.5), in metres. Members AB, AC and BC. A 10 kN load acts downward at C.

Check determinacy: m = 3, r = 3, j = 3, so m + r = 6 = 2j. Good.

**Reactions.** By symmetry, Ay = By = 5 kN up and Ax = 0.

**Geometry of member AC.** It runs 2 m across and 1.5 m up, so its length is 2.5 m (a 3-4-5 triangle). The unit vector from A toward C is (0.8, 0.6).

**Joint A.** Unknowns: F_AC (along (0.8, 0.6)) and F_AB (along (1, 0)).

```text
ΣFy = 0:  5 + 0.6·F_AC = 0          →  F_AC = −8.33 kN  (compression)
ΣFx = 0:  F_AB + 0.8·F_AC = 0       →  F_AB = +6.67 kN  (tension)
```

**Joint C** (check): BC is the mirror image of AC, so F_BC = −8.33 kN. Vertical forces at C: the two rafters push up on C with 2 × 8.33 × 0.6 = 10 kN, balancing the 10 kN load. The bottom chord AB is in tension, tying the feet of the rafters together so they do not spread apart. That is exactly why the bottom chord of a roof truss is called a tie.

## Solving a joint as a 2 × 2 system

At any joint you can write the two equations as a linear system:

```text
ux1·F1 + ux2·F2 = −Px
uy1·F1 + uy2·F2 = −Py
```

where (ux, uy) are the unit vectors along the two unknown members pointing away from the joint and (Px, Py) is the sum of the known forces at the joint. Cramer's rule solves it:

```text
det = ux1·uy2 − ux2·uy1
F1  = (−Px·uy2 + Py·ux2) / det
F2  = (−Py·ux1 + Px·uy1) / det
```

A large truss is the same idea scaled up: computer programs assemble every joint's equations into one big matrix and solve them together, which is the linear algebra you met in the prerequisite course.

## Zero-force members

At an unloaded joint where only two non-collinear members meet, both carry zero force. At an unloaded joint where three members meet and two are collinear, the third carries zero force. Spotting these first shortens the work. Zero-force members are still needed: they brace other members against buckling and carry load when the loading changes.

## The method of sections

When you only need the force in one member in the middle of a long truss, cut the truss through that member (and at most two others), and apply the three rigid-body equilibrium equations to one side. Taking moments about the point where the other two cut members meet gives the force you want in one line.

## Common mistakes

- **Sign confusion.** Always assume tension (arrow pointing away from the joint). Then a negative answer simply means compression.
- **Starting at a joint with three unknowns.** You only have two equations per joint.
- **Wrong angles.** Get member directions from coordinates (Δx/L, Δy/L), not by eyeballing the sketch.
- **Forgetting buckling.** A compression member can fail by buckling long before its material yields. Truss design checks slender compression members for this, which is a design-code topic.

## Recap

- Ideal truss members are two-force members: pure tension or compression.
- Method of joints: reactions first, then joints with at most two unknowns, ΣFx = ΣFy = 0.
- Each joint is a 2 × 2 linear system.
- Method of sections: cut and take moments to get one member quickly.

## Try it

On the Do tab, finish a small Cramer's-rule solver and use it to solve joint A of the roof truss.


Solve for the tension and compression in truss members using the method of joints.

Trusses

Understanding and Analysing Trusses

Statics: Lesson 48 - Trusses, Method of Joints

## Why it matters

The shape of a cross-section matters as much as the material. A plank laid flat sags easily; the same plank on edge is many times stiffer. Steel I-beams put most of their material far from the middle for the same reason. Two section properties capture this: the **centroid** (the geometric centre of the area, where the bending axis passes) and the **second moment of area** I (often just called the moment of inertia of the section), which measures how far the area is spread from that axis. Bending stress and deflection, two lessons from now, both depend on I.

## The centroid of a composite area

For a shape built from simple pieces (rectangles, circles, triangles), the centroid is the area-weighted average of the pieces' centroids:

```text
ȳ = Σ(Ai · yi) / Σ Ai
```

where Ai is the area of piece i and yi is the distance from a reference line (say, the bottom of the section) to that piece's own centroid. Holes are included as pieces with negative area. The centroid of a rectangle is at its middle; of a triangle, one third of the height from its base.

## The second moment of area

For a rectangle of width b and height h, about a horizontal axis through its own centroid:

```text
I = b·h³ / 12
```

The h³ is why the plank on edge is so much stiffer. A 50 mm × 150 mm board on edge has I = 50 · 150³ / 12 = 14.1 × 10⁶ mm⁴. Laid flat (b = 150, h = 50) it has I = 150 · 50³ / 12 = 1.56 × 10⁶ mm⁴: nine times less, for the same amount of wood.

Other common results: a solid circle of diameter d has I = π·d⁴ / 64.

## The parallel axis theorem

The I of each piece must be about the *whole section's* centroid, not its own. The parallel axis theorem shifts it:

```text
I = I_own + A · d²
```

where d is the distance between the piece's own centroid and the section centroid. The A·d² term is usually the big one, which is why material far from the axis is so effective.

## Worked example: a T-section

A T-section has a flange 100 mm wide and 20 mm thick sitting on top of a web 20 mm wide and 100 mm tall. Measure y from the bottom of the web.

| piece | b × h (mm) | A (mm²) | y (mm) | A·y (mm³) |
|---|---|---|---|---|
| web | 20 × 100 | 2,000 | 50 | 100,000 |
| flange | 100 × 20 | 2,000 | 110 | 220,000 |
| total | | 4,000 | | 320,000 |

```text
ȳ = 320,000 / 4,000 = 80 mm from the bottom
```

Now the second moment of area about the horizontal axis through ȳ:

```text
web:    I_own = 20·100³/12 = 1,666,667    d = 80 − 50 = 30    A·d² = 2,000·900 = 1,800,000
flange: I_own = 100·20³/12 =    66,667    d = 110 − 80 = 30   A·d² = 1,800,000
I = 1,666,667 + 1,800,000 + 66,667 + 1,800,000 = 5,333,333 mm⁴ ≈ 5.33 × 10⁶ mm⁴
```

The flange's own I is tiny (it is thin), but its A·d² term is large because it sits 30 mm from the axis.

## Units

Area is mm² or m²; the first moment A·y is mm³; the second moment I is mm⁴ or m⁴. Convert carefully: 1 m⁴ = 10¹² mm⁴. Steel section tables usually list I in mm⁴ × 10⁶ or in in⁴.

## Common mistakes

- **Forgetting the parallel axis term**, or adding I_own values about different axes directly.
- **Using d from the reference line** instead of from the section centroid.
- **Swapping b and h.** h is always the dimension perpendicular to the bending axis, and it is the one that is cubed.
- **Confusing I with mass moment of inertia.** The second moment of area (mm⁴) is a geometric property used for bending; the mass moment of inertia (kg·m²) is used for rotation in dynamics.

## Recap

- Centroid: ȳ = Σ(Ai·yi) / ΣAi.
- Rectangle: I = b·h³/12 about its own centroid.
- Parallel axis: I = I_own + A·d², with d measured to the section centroid.
- Putting area far from the axis increases I much faster than adding area near it.

## Try it

On the Do tab, finish the parallel-axis step in a function that computes the centroid and I of any section made of rectangles, then check the lesson's T-section.


Find the centroid and second moment of area of a built-up section with the parallel axis theorem.

Centroids and moments of inertia

Understanding the Area Moment of Inertia

Statics: Lesson 67 - Introduction to Area Moment of Inertia

Keep bodies in equilibrium: free-body diagrams and the equilibrium equations, trusses by the method of joints, and the centroid and second moment of area of a cross-section.