An intermediate course for future civil, mechanical and materials engineers who have finished Physics for Engineering & Science and Calculus & Linear Algebra for STEM. You will solve 2D equilibrium problems with free-body diagrams, analyse a truss by the method of joints, compute the centroid and second moment of area of a built-up section, calculate axial stress, strain and elongation, draw shear and bending moment diagrams for a simply supported beam, check a beam for bending stress and deflection, apply F = ma to engineering problems, compute a natural frequency, and learn how factors of safety are chosen and justified. Most lessons have a short Python exercise where the code already runs and you write one key formula. The worked examples teach the mechanics and are not design methods: real structures and machines are designed to codes and standards, and their designs must be checked and sealed by a licensed professional engineer. Engineering careers also need an accredited degree and, for many roles, the FE and PE exams, which are beyond this course.

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.

Statics

## 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.

Strength of Materials

## Why it matters

Statics assumes nothing accelerates. As soon as something speeds up, slows down or turns, Newton's second law takes over: the net force equals mass times acceleration. Engineers use it to size brakes and motors, to find the force on a lift cable while it accelerates, to set conveyor speeds, and to estimate stopping distances. This lesson applies F = m·a to the kind of particle problems that appear in engineering dynamics courses.

## Kinetics of a particle

A body can be treated as a **particle** when its rotation does not matter, only its overall motion. Then:

```text
ΣF = m·a          (as vectors)
ΣFx = m·ax        ΣFy = m·ay
```

The method is the same as statics, with one change: the right-hand side is m·a instead of zero.

1. Draw a free-body diagram showing every force on the body.
2. Choose axes, ideally with one axis along the acceleration.
3. Write ΣF = m·a for each axis.
4. Solve, then use kinematics (v = u + a·t, v² = u² + 2·a·s) if you need speeds, times or distances.

## Friction

Dry friction opposes sliding. When the body slides, the kinetic friction force is

```text
f = μk · N
```

where N is the normal force and μk the coefficient of kinetic friction. Before sliding starts, static friction can resist up to μs·N, with μs usually a little larger than μk. Friction coefficients vary a lot with surfaces, contamination and wear, so treat textbook values as rough.

## Worked example 1: pulling a crate

A 50 kg crate is pulled across a floor by a rope at 30° above horizontal with a tension of 200 N. The coefficient of kinetic friction is 0.30. Find the acceleration once the crate is moving.

```text
y-axis (no vertical acceleration):
  N + 200·sin30° − 50·9.81 = 0       →  N = 490.5 − 100 = 390.5 N
x-axis:
  200·cos30° − 0.30·N = 50·a
  173.2 − 117.2 = 50·a               →  a ≈ 1.12 m/s²
```

The upward part of the rope pull reduces the normal force and therefore the friction. Pulling slightly upward is easier than pushing slightly downward, which increases N.

## Worked example 2: stopping distance

A 1,500 kg car travels at 25 m/s (90 km/h) and brakes hard on dry pavement with an effective friction coefficient of 0.70. Assume the brakes hold the tyres at the edge of grip.

```text
ΣFx = −μ·m·g = m·a      →  a = −μ·g = −0.70 · 9.81 = −6.87 m/s²
v² = u² + 2·a·s         →  0 = 25² − 2·6.87·s   →  s ≈ 45.5 m
```

The mass cancels out: a heavier car needs more braking force but has the same deceleration for the same μ. Notice that stopping distance grows with the *square* of speed: at 35 m/s the same car needs about 89 m, plus the distance covered during the driver's reaction time.

## Worked example 3: lifting a load

A crane lifts a 2,000 kg load, accelerating it upward at 0.5 m/s². The cable tension is

```text
T − m·g = m·a      →  T = m·(g + a) = 2,000 · 10.31 = 20,620 N
```

That is about 5 % more than the static weight of 19,620 N. Sudden starts and stops (snatch loading) can raise cable forces far more, which is one reason lifting equipment carries rated capacities and design factors set by standards.

## Work and energy (a shortcut)

When you only need speeds, not times, the work-energy principle avoids solving for acceleration:

```text
Σ(work done by forces) = ½·m·v₂² − ½·m·v₁²
```

For the braking car, friction does work −μ·m·g·s and the kinetic energy drops from ½·m·25² to zero, giving the same 45.5 m.

## Common mistakes

- **Putting m·a on the free-body diagram as a force.** It is the result of the forces, not one of them.
- **Assuming N = m·g.** Any force with a vertical component, or a slope, changes N.
- **Using static friction for a sliding body.** Once it slides, use μk.
- **Forgetting that a can be negative.** Choose a positive direction and stick with it.

## Recap

- ΣF = m·a: statics with the right-hand side no longer zero.
- Friction while sliding is μk·N, and N depends on all the vertical forces.
- Combine F = m·a with kinematics for times and distances, or use work-energy.
- Stopping distance grows with speed squared.

## Try it

On the Do tab, finish the acceleration line of a crate-pulling function and check the worked example.


Apply Newton's second law with friction to engineering problems such as pulling, braking and lifting.

Particle dynamics

F=ma Rectangular Coordinates | Equations of motion | (Learn to Solve any Problem)

Pulley System Dynamics: Kinetics of Particles Example (Newton’s Second Law)

## Why it matters

Anything with mass and stiffness can vibrate: a pump on its mounts, a footbridge under walkers, a car on its suspension, a building in an earthquake. Each has a **natural frequency** at which it prefers to oscillate. If something pushes it at or near that frequency, the motion can grow far larger than the push alone would cause. This is **resonance**, and avoiding it (or controlling it) is a routine part of mechanical and structural engineering.

## The spring-mass model

The simplest vibrating system is a mass m on a spring of stiffness k (N/m), with no damping. Pull the mass down and let go: the spring force −k·x pulls it back, it overshoots, and it oscillates. Newton's second law gives

```text
m·ẍ + k·x = 0
```

whose solution is a sine wave with **natural circular frequency**

```text
ωn = √(k / m)        (rad/s)
fn = ωn / (2π)       (Hz, cycles per second)
T  = 1 / fn          (s, period)
```

Stiffer springs raise the frequency; heavier masses lower it. Doubling the mass lowers fn by a factor of √2, not 2.

## Worked example: a machine on springs

A 20 kg compressor sits on four mounts with a combined stiffness of 80,000 N/m.

```text
ωn = √(80,000 / 20) = √4,000 = 63.2 rad/s
fn = 63.2 / (2π) = 10.1 Hz
```

If the compressor's motor runs at 600 rpm, its forcing frequency is 600/60 = 10 Hz, right on top of the natural frequency. The mounts would amplify the shaking instead of isolating it. Choosing softer mounts lowers fn well below the running speed.

## A shortcut from static deflection

Under its own weight the mass compresses the spring by δst = m·g/k. Substituting k = m·g/δst:

```text
fn = (1 / 2π) · √(g / δst)
```

For the compressor, δst = 20 × 9.81 / 80,000 = 2.45 mm, and √(9.81 / 0.00245) / 2π gives the same 10.1 Hz. This is handy in practice: measure how far something sags under its own weight and you can estimate its natural frequency.

## Springs in combination

- **Parallel** springs (side by side, sharing the load) add: k = k1 + k2.
- **Series** springs (one on top of another) combine like resistors in parallel: 1/k = 1/k1 + 1/k2.

Four 20,000 N/m mounts under one machine act in parallel: 80,000 N/m. Two 20,000 N/m springs stacked in series give only 10,000 N/m.

## Resonance and the frequency ratio

When a harmonic force at frequency f drives the system, the response depends on the **frequency ratio** r = f / fn:

- r much less than 1: the system follows the force almost statically.
- r near 1: resonance. Without damping the amplitude grows without limit; with damping it peaks at a large finite value.
- r well above 1 (in practice above about √2 ≈ 1.41): the mass barely moves, and less force passes to the floor. This is how vibration isolation works.

## Damping

Real systems lose energy to friction, air and material hysteresis, so free vibrations die away. Damping is described by the **damping ratio** ζ. Many structures have ζ of only a few percent, which is why resonance can still be dramatic: in 2000 the London Millennium Bridge swayed noticeably under crowds of walkers because of a lateral resonance, and it was fixed by adding dampers.

## Beams and structures

A beam is not a single spring and mass; it has many natural frequencies (modes). For a first estimate, the stiffness of a beam at a point comes from its deflection formula: a simply supported beam with a central load has k = 48·E·I / L³. Floors are often checked to keep their first natural frequency above a few hertz so that footsteps do not excite them. Real limits and methods come from design guides and codes.

## Common mistakes

- **Mixing rad/s and Hz.** ωn = 2π·fn. Most specifications and machine speeds are in Hz or rpm.
- **Using grams or kN without converting.** Use kg, N/m and m to get rad/s.
- **Adding series springs directly.** Series stiffness is always less than the softest spring.
- **Assuming damping fixes everything.** Light damping still allows large resonant motion; move the natural frequency away from the forcing frequency first.

## Recap

- ωn = √(k/m), fn = ωn/2π.
- fn = (1/2π)·√(g/δst) from the static deflection.
- Parallel springs add; series springs combine reciprocally.
- Keep forcing frequencies away from natural frequencies; isolation needs r above about 1.41.

## Try it

On the Do tab, finish a natural frequency function and check the compressor example.


Compute the natural frequency of a spring-mass system and explain resonance and isolation.

Intro to vibration

Understanding Vibration and Resonance

## Why it matters

Every calculation in this course used exact loads, exact dimensions and exact material properties. Real life has none of those. Loads are estimated, materials vary from batch to batch, workmanship is imperfect, corrosion and fatigue reduce strength over time, and the analysis itself is a simplified model. The **factor of safety** is how engineers make room for all of that uncertainty, so that a structure or machine survives conditions worse than the ones it was designed for.

## The definition

The factor of safety (FoS, or FS) compares what the part can withstand with what it is expected to carry:

```text
FoS = failure load / working load        (or  failure stress / working stress)
```

A bracket that would fail at 30 kN and carries 10 kN in service has FoS = 3. The **margin of safety** is sometimes quoted instead: MS = FoS − 1 (here 2, or 200 %).

"Failure" must be defined. For a ductile metal part it is usually **yielding**, because permanent deformation already ruins the part. For a brittle material, a fatigue-loaded part or a slender column, the governing failure might be fracture, fatigue life or buckling, each with its own capacity.

## Worked example

A steel rod with a yield strength of 250 MPa carries a working stress of 100 MPa.

```text
FoS against yield = 250 / 100 = 2.5
```

If a designer requires FoS = 2.0, the **allowable stress** is 250 / 2.0 = 125 MPa, and the minimum area for a 50 kN load is 50,000 / 125 = 400 mm².

## What drives the choice

A larger factor is justified when:

- loads are uncertain or can increase (impact, misuse, future change of use);
- material properties are variable or poorly known (castings, timber, new composites);
- the analysis is approximate or the part is hard to inspect;
- failure would be sudden (brittle fracture, buckling) rather than gradual and visible;
- the consequences are severe: injury, loss of life, or high cost.

A smaller factor is acceptable when loads and materials are well characterised, parts are tested and inspected, and weight is critical. Aircraft structures, for example, use relatively low factors (a load factor of 1.5 on limit loads is a well-known figure) but compensate with extensive testing, certified materials and regular inspection.

Textbooks often quote rough ranges, from around 1.25 to 1.5 for well-controlled conditions up to 3 or 4 or more for uncertain loads, untried materials or brittle failure. Those are guides for learning, not rules you may apply to a real product.

## Codes do this for you

In regulated work you rarely choose a single factor freely. Codes and standards set it, in one of two styles:

- **Allowable stress design (ASD):** working stresses are kept below the material strength divided by a factor of safety.
- **Load and resistance factor design (LRFD), or limit state design:** loads are multiplied by load factors greater than 1 (bigger for variable loads like live load and wind than for dead load), and the member's nominal strength is multiplied by a resistance factor less than 1. The design passes if the factored resistance is at least the factored load effect.

Building structures in the US follow ASCE 7 for loads and material codes such as AISC 360 (steel) or ACI 318 (concrete); pressure vessels follow the ASME Boiler and Pressure Vessel Code; cranes and lifting gear have their own standards. Editions change, so always check which edition applies to your project.

## Justifying a factor

Choosing and defending a factor of safety means writing down the reasons:

1. What failure mode governs (yield, buckling, fatigue, fracture)?
2. How well are the loads known, and can they be exceeded?
3. How consistent is the material, and is it certified?
4. How accurate is the analysis, and has it been checked by test?
5. What happens if it fails, and is the failure gradual or sudden?
6. What does the applicable code or standard require?

The last question usually wins. Where no code applies, a documented, reasoned choice is still essential.

## Licensure and responsibility

Engineers who design public structures and many machines must be licensed. In the United States the path is an accredited degree, the Fundamentals of Engineering (FE) exam, supervised experience, then the Principles and Practice of Engineering (PE) exam, administered through NCEES (https://ncees.org/) with licences issued by each state board. Checked on 2026-10-23. Only a licensed engineer may seal designs for public use, and nothing in this course replaces that review.

## Common mistakes

- **Treating FoS as a measure of over-design.** It covers real, unavoidable uncertainty.
- **Checking only one failure mode.** A part can have a large factor on yield and a small one on buckling or fatigue.
- **Stacking factors.** Applying a code's load factors and then an extra personal factor on top can make designs unnecessarily heavy; follow the code's method.
- **Assuming a big factor makes poor details safe.** Bad connections, corrosion and fatigue cracks can defeat any factor.

## Recap

- FoS = capacity / demand, for a defined failure mode.
- Uncertainty, failure consequences and failure type drive the size of the factor.
- Codes set factors through ASD or LRFD; follow the applicable edition.
- Real designs need review and sealing by a licensed professional engineer.


Define the factor of safety and learn how engineers choose and justify one within codes.

Factors of safety

Mechanics of Materials: Lesson 6 - Factor of Safety Explained!

Factor of Safety

Move from bodies at rest to bodies in motion: Newton's second law for particles, the natural frequency of a spring-mass system, and how engineers choose factors of safety.

Engineering Mechanics: Statics & Strength

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Dynamics

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