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 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.