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 tells you the force in a member. Strength of materials asks what that force does to the material inside it. A 50 kN pull is nothing for a thick steel bar and catastrophic for a thin wire. To compare members of different sizes, engineers divide by area to get **stress**, and divide stretch by length to get **strain**. These two quantities, and the link between them, underpin every structural and machine calculation that follows.

## Stress

**Normal stress** is the internal force per unit area acting perpendicular to a cut through the member:

```text
σ = F / A
```

Tension (pulling) is positive and compression negative. The SI unit is the pascal, 1 Pa = 1 N/m², which is tiny, so engineers use megapascals: **1 MPa = 1 N/mm² = 10⁶ Pa**. Working in newtons and millimetres gives MPa directly, which is why many engineers prefer N and mm for stress problems.

**Shear stress** τ = V / A acts parallel to the cut, as in a bolt that a lap joint tries to slice. This lesson focuses on axial (normal) stress.

## Strain

**Normal strain** is the change in length divided by the original length:

```text
ε = δ / L
```

It has no units (mm/mm). Strains in structural metals are small: steel at a typical working stress stretches by well under a tenth of a percent. Engineers sometimes write strain in microstrain (με, 10⁻⁶).

## Hooke's law and Young's modulus

For most materials, at small strains, stress is proportional to strain:

```text
σ = E · ε
```

E is **Young's modulus** (the modulus of elasticity), a measure of stiffness. Rough typical values:

| material | E (GPa) |
|---|---|
| structural steel | 200 |
| aluminium alloys | 70 |
| concrete | 25–35 |
| softwood timber, along the grain | about 8–14 |
| rubber | 0.01–0.1 |

Combining the three equations gives the elongation of an axially loaded bar:

```text
δ = F · L / (A · E)
```

This only holds in the **elastic** range, below the yield point, where unloading returns the bar to its original length. Beyond it the material deforms permanently, which is covered in Materials Science Fundamentals.

## Worked example: a steel tie rod

A steel rod of diameter 20 mm and length 2.0 m carries a tensile force of 50 kN. E = 200 GPa = 200,000 MPa.

```text
A = π · d² / 4 = π · 20² / 4 = 314.2 mm²
σ = F / A = 50,000 N / 314.2 mm² = 159.2 MPa
ε = σ / E = 159.2 / 200,000 = 7.96 × 10⁻⁴   (about 0.08 %)
δ = ε · L = 7.96 × 10⁻⁴ · 2000 mm = 1.59 mm
```

A 2 m steel rod stretches less than 2 mm under five tonnes of pull. If the rod were aluminium (E = 70 GPa) with the same diameter, the stress would be the same 159.2 MPa (stress does not depend on the material), but it would stretch 200/70 times as much: about 4.5 mm.

## Poisson's effect

A bar that is stretched also gets slightly thinner. The ratio of sideways strain to lengthwise strain is **Poisson's ratio** ν, about 0.3 for steel and aluminium:

```text
ε_lateral = −ν · ε_axial
```

For the rod above, its diameter shrinks by about 0.3 × 7.96 × 10⁻⁴ × 20 mm ≈ 0.005 mm.

## Thermal strain

Temperature changes also cause strain: ε_thermal = α · ΔT, where α is the coefficient of thermal expansion (about 12 × 10⁻⁶ per °C for steel). If a member is prevented from expanding, that strain turns into stress: a fully restrained steel bar heated by 30 °C develops σ = E·α·ΔT = 200,000 × 12 × 10⁻⁶ × 30 = 72 MPa of compression. This is why bridges have expansion joints.

## Common mistakes

- **Mixing GPa and MPa.** 200 GPa = 200,000 MPa. Dividing 159 MPa by 200 (GPa) gives a strain a thousand times too large.
- **Using diameter instead of radius** (or forgetting the /4) in A = π·d²/4.
- **Applying δ = FL/AE beyond yield.** It is only valid in the linear elastic range.
- **Thinking stress depends on the material.** For a statically determinate member, σ = F/A depends only on load and area; the material decides whether that stress is safe and how much the member stretches.

## Recap

- σ = F/A (MPa = N/mm²), ε = δ/L (dimensionless).
- In the elastic range σ = E·ε, so δ = F·L/(A·E).
- Steel is about three times stiffer than aluminium.
- Restrained thermal expansion creates stress: σ = E·α·ΔT.

## Try it

On the Do tab, finish an elongation function and check the steel tie rod from this lesson.


Compute axial stress, strain and elongation, and use Young's modulus to link them.

Stress and strain

An Introduction to Stress and Strain

Understanding Young's Modulus

## Why it matters

A beam carries load by bending, and the internal forces change along its length. To check a beam you need to know where the internal **bending moment** is largest (that is where bending stress peaks) and where the **shear force** is largest (that is where shear and bearing problems appear). Shear force and bending moment diagrams, often called V and M diagrams, show both at a glance. Drawing them is a core skill in every structural and mechanical engineering course.

## Internal forces in a beam

Cut a beam at a distance x from the left end and look at the left-hand piece. To keep it in equilibrium, the cut face must carry:

- a **shear force V**, acting across the beam, and
- a **bending moment M**, trying to rotate the piece.

A common sign convention (used here): V is the sum of the upward forces to the left of the cut, and M is positive when it makes the beam sag (concave up, compression on top, tension on the bottom). Some textbooks use different conventions; always state yours.

For a beam with point loads, you get V and M at any x from the left-hand piece alone:

```text
V(x) = (upward reactions left of x) − (downward loads left of x)
M(x) = Σ (each force left of x) × (its distance to the cut)
```

## Worked example 1: one point load

A simply supported beam spans 6 m with a 12 kN load at 2 m from the left support. From the first lesson, RA = 8 kN and RB = 4 kN.

```text
0 < x < 2:   V = +8 kN            M = 8·x            (M = 16 kN·m at x = 2)
2 < x < 6:   V = 8 − 12 = −4 kN   M = 8·x − 12·(x − 2)
                                    at x = 6:  48 − 48 = 0
```

The V diagram is a step: +8 kN, dropping by 12 kN at the load, to −4 kN, then back to zero at the right support. The M diagram is a triangle: zero at both supports, peaking at **16 kN·m under the load**.

## The relationships that make sketching fast

For distributed load w (kN/m, downward positive), shear V and moment M:

```text
dV/dx = −w        dM/dx = V
```

So:

1. With no distributed load, V is constant and M is a straight line.
2. A point load makes V jump by the size of the load.
3. Under a uniform load, V is a sloping straight line and M is a parabola.
4. **M is at its maximum where V crosses zero.** This is the most useful rule: find where V = 0 and you have found the peak moment.
5. The change in M between two points equals the area of the V diagram between them. In example 1, the area of the V diagram from 0 to 2 m is 8 × 2 = 16 kN·m, matching M at x = 2.

At a simple support with no applied moment, M = 0.

## Worked example 2: a uniform load

The same 6 m beam carries a uniform load w = 5 kN/m along its whole length. The total load is 30 kN, so by symmetry RA = RB = 15 kN.

```text
V(x) = 15 − 5·x                  zero at x = 3 m (mid-span)
M(x) = 15·x − 5·x²/2
M(3) = 45 − 22.5 = 22.5 kN·m
```

That matches the standard result for a uniformly loaded simply supported beam:

```text
M_max = w·L² / 8 = 5 · 6² / 8 = 22.5 kN·m
```

And for a single central point load P: M_max = P·L/4.

## Drawing the diagrams step by step

1. Find the reactions.
2. Mark the key points: supports, point loads, starts and ends of distributed loads.
3. Walk from left to right, building V: jump at each point load, slope under distributed loads.
4. Build M from the area under V, starting from zero at a simple support.
5. Check that M returns to zero at the right-hand simple support. If it does not, there is an arithmetic error.
6. Label the maximum values and where they occur.

## Common mistakes

- **Wrong reactions.** Every diagram after that is wrong. Check ΣFy = 0 and moments about a second point.
- **Missing the jump at a point load** in the V diagram.
- **Looking for M_max at the load every time.** With several loads or distributed loads, find where V crosses zero.
- **Using w·L²/8 for other cases.** It applies only to a uniform load over a simply supported span. Cantilevers and continuous beams have different results.

## Recap

- V(x) and M(x) come from equilibrium of the piece on one side of a cut.
- dM/dx = V: M peaks where V crosses zero.
- Point loads make V jump; uniform loads make V slope and M curve.
- Standard results: P·L/4 for a central point load, w·L²/8 for a full uniform load.

## Try it

On the Do tab, finish the bending moment function for a simply supported beam with point loads and find the peak moment numerically.


Draw shear force and bending moment diagrams for a simply supported beam and find the peak moment.

Shear and moment diagrams

Understanding Shear Force and Bending Moment Diagrams

How to Draw Shear Force and Moment Diagrams | Mechanics Statics | (Step by step solved examples)

## Why it matters

The moment diagram tells you how hard a beam is being bent. Two questions follow. **Strength:** is the bending stress below what the material can safely take? **Serviceability:** does the beam sag so much that floors bounce, plaster cracks or doors jam? A beam can pass the first check and still fail the second, and for long spans deflection often governs. This lesson connects the moment from the last lesson and the section property I from lesson 3 to both checks.

## The flexure formula

When a beam sags, its top fibres shorten and its bottom fibres lengthen. Somewhere between them is the **neutral axis**, which passes through the centroid of the section and has zero bending stress. Stress grows linearly with distance from that axis:

```text
σ = M · y / I
```

The largest stress is at the fibre farthest from the neutral axis, at distance c:

```text
σ_max = M · c / I = M / S,     where S = I / c is the section modulus
```

For a rectangle, S = b·h²/6. Steel section tables list S directly because it is what designers use.

## Worked example: a timber beam

A rectangular timber beam, 50 mm wide and 150 mm deep, spans 3.0 m between simple supports and carries a 5.0 kN point load at mid-span. Take E = 10,000 MPa (a typical figure for softwood along the grain). Ignore the beam's self-weight.

**Bending moment.**

```text
M = P·L / 4 = 5,000 N × 3,000 mm / 4 = 3.75 × 10⁶ N·mm   (3.75 kN·m)
```

**Section properties.**

```text
I = b·h³/12 = 50 × 150³ / 12 = 14.06 × 10⁶ mm⁴
c = h / 2 = 75 mm
```

**Bending stress.**

```text
σ_max = M·c / I = 3.75 × 10⁶ × 75 / 14.06 × 10⁶ = 20.0 MPa
```

The top fibre is at 20 MPa compression and the bottom at 20 MPa tension.

## Deflection

Integrating the curvature M/(E·I) twice gives the deflected shape. You do not need to repeat that derivation for common cases; engineers use standard results:

| case | maximum deflection |
|---|---|
| simply supported, central point load P | P·L³ / (48·E·I) |
| simply supported, uniform load w | 5·w·L⁴ / (384·E·I) |
| cantilever, point load P at the tip | P·L³ / (3·E·I) |

For the timber beam:

```text
δ = 5,000 × 3,000³ / (48 × 10,000 × 14.06 × 10⁶) = 20.0 mm
```

Deflection limits are usually written as a fraction of the span. As an illustration only, L/360 is a limit often quoted for floors under live load; here L/360 = 8.3 mm, so this beam would sag well over twice that much. The fix is more depth: because I grows with h³, a 225 mm deep beam of the same width has I = 47.5 × 10⁶ mm⁴ and deflects only 5.9 mm, while its stress drops to 8.9 MPa. (A 200 mm deep beam deflects 8.4 mm, still just over the limit.)

The real limits, load combinations and allowable stresses come from the applicable building code and material standard (for timber, values also depend on species, grade, moisture and load duration), so treat these numbers as practice, not design values.

## Why deflection depends on L³ or L⁴

Doubling the span of a uniformly loaded beam multiplies its deflection by 2⁴ = 16. That is why long spans are expensive: you need much deeper sections, and why deflection, not strength, often controls the design of long beams.

## Strength check versus stiffness check

- The **stress** check depends on M, c and I (or S), and is compared with an allowable or design strength of the material.
- The **deflection** check depends on E and I, and is compared with a limit like L/240 or L/360 from the code.
- Changing to a stronger material of the same E (a higher grade of steel, say) helps strength but does nothing for deflection.

## Common mistakes

- **Mixing kN·m with mm.** Convert M to N·mm (× 10⁶) when I is in mm⁴, so σ comes out in MPa.
- **Using the wrong formula** for the support and load case. P·L³/48EI is only for a central point load on a simple span.
- **Forgetting self-weight.** Real beams carry their own weight as a uniform load; this example left it out for clarity.
- **Ignoring other checks.** Real design also checks shear, bearing at supports, lateral-torsional buckling of slender beams, and vibration. These are covered in design courses and codes.

## Recap

- σ = M·y/I, maximum at the extreme fibre: σ_max = M·c/I = M/S.
- Deflection uses standard formulas in E·I; it grows with L³ or L⁴.
- Check both strength and deflection; deeper sections help both.
- Any real beam is sized to the governing code and checked by a licensed engineer.

## Try it

On the Do tab, finish the deflection line of a small beam checker and use it on this lesson's timber beam.


Check a beam for bending stress with the flexure formula and for deflection with standard formulas.

Bending stress and deflection

Understanding Stresses in Beams

Understanding the Deflection of Beams

Relate loads to what happens inside a member: axial stress and strain, shear force and bending moment diagrams, and bending stress and deflection checks for beams.