Formula sheet · Structional

Weight from mass

Units and maths

\force{W} = m\,g

$\force{W}$: Weight: the force gravity pulls withN (÷ 1000 for kN)
$m$: Masskg
$g$: Gravity, 9.81 m/s² (about 10)m/s²

1 kN is about 100 kg. In US units weights are already given in lb, so there is nothing to convert.

Example

m = 25\,\mathrm{kg}

\force{W} = 25 \times 9.81 = 245.3\,\mathrm{N} \approx 0.25\,\mathrm{kN}

Pythagoras

Units and maths

c^2 = a^2 + b^2

$c$: The long side (opposite the right angle)m
$a,\ b$: The two short sidesm

3, 4, 5 is the builder's check for a square corner.

Sine, cosine, tangent

Units and maths

\begin{aligned} \sin\theta &= \tfrac{\text{opposite}}{\text{hypotenuse}} \\[4pt] \cos\theta &= \tfrac{\text{adjacent}}{\text{hypotenuse}} \\[4pt] \tan\theta &= \tfrac{\text{opposite}}{\text{adjacent}} \end{aligned}

$\theta$: The angle you know or want°

Set your calculator to degrees.

Fall from a slope in %

Units and maths

\Delta h = \len{L}\,\dfrac{s}{100}

$\Delta h$: How much lower the far end ism
$\len{L}$: Length along the slope (horizontal run, near enough)m
$s$: Slope in percent%

2 % is 20 mm for every metre.

Volume of a slab

Units and maths

V = a \times b \times t

$V$: Volumem³
$a,\ b$: Length and widthm
$t$: Thickness (in m or ft, not mm or in)m

Order a few percent extra for waste.

Split a force into two parts

Forces as arrows

\begin{gathered}\force{F_x} = \force{F}\cos\theta \\[4pt] \force{F_y} = \force{F}\sin\theta\end{gathered}

$\force{F}$: The forcekN
$\theta$: Angle from the horizontal°
$\force{F_x},\ \force{F_y}$: Horizontal and vertical partskN

Example

\force{F} = 10\kN,\ \theta = 30^\circ

\force{F_x} = 8.66\kN,\ \force{F_y} = 5\kN

Add forces into one (resultant)

Forces as arrows

\len{R} = \sqrt{\left(\textstyle\sum \force{F_x}\right)^2 + \left(\textstyle\sum \force{F_y}\right)^2}

$\len{R}$: The one force that does the same jobkN
$\textstyle\sum \force{F_x}$: All the horizontal parts added (right +)kN
$\textstyle\sum \force{F_y}$: All the vertical parts added (up +)kN

Direction: tan θ = ΣF_y / ΣF_x.

Example

3\kN \rightarrow,\ 4\kN \uparrow

\len{R} = \sqrt{3^2 + 4^2} = 5\kN

Moment of a force

Moments and balance

\mom{M} = \force{F} \times \len{d}

$\mom{M}$: Moment: how hard it turnskN·m
$\force{F}$: The forcekN
$\len{d}$: Lever arm: distance at right angles to the force's linem

Example

\force{F} = 2.5\kN,\ \len{d} = 1.2\m

\mom{M} = 3\kNm

Equilibrium (it stands still)

Moments and balance

\begin{gathered}\textstyle\sum \force{F_x} = 0 \\[4pt] \sum \force{F_y} = 0 \\[4pt] \sum \mom{M} = 0\end{gathered}

$\textstyle\sum$: Add up everything acting on the body–
$\force{F_x},\ \force{F_y}$: Horizontal and vertical forceskN
$\mom{M}$: Moments about any point you choosekN·m

Take moments about a support to get rid of its unknown force.

Balance on a pivot

Moments and balance

\force{F_1}\,\len{d_1} = \force{F_2}\,\len{d_2}

$\force{F_1},\ \force{F_2}$: Loads on each sidekN
$\len{d_1},\ \len{d_2}$: Their distances from the pivotm

Example

6\kN \text{ at } 2\m,\ 4\kN \text{ on the other side}

\len{d_2} = \dfrac{6 \times 2}{4} = 3\m

Support reactions, one point load

Moments and balance

\begin{gathered}\force{R_A} = \dfrac{\force{P}\,\len{b}}{\len{L}} \\[4pt] \force{R_B} = \dfrac{\force{P}\,\len{a}}{\len{L}}\end{gathered}

$\force{R_A},\ \force{R_B}$: Push up at the left and right supportskN
$\force{P}$: The point loadkN
$\len{a},\ \len{b}$: Distances from the load to A and to Bm
$\len{L}$: Span between the supports (a + b)m

The nearer support takes more. R_A + R_B = P.

Example

\force{P} = 12\kN,\ \len{a} = 2\m,\ \len{L} = 6\m

\force{R_A} = 8\kN,\ \force{R_B} = 4\kN

Reactions each support gives

Moments and balance

\begin{gathered}\text{pin} = 2 \\[4pt] \text{roller} = 1 \\[4pt] \text{fixed} = 3\end{gathered}

$\text{pin}$: Stops up/down and sideways, lets it turn–
$\text{roller}$: Stops one direction only–
$\text{fixed}$: Stops both directions and turning–

A beam needs 3 reactions to stay put; more than 3 makes it statically indeterminate.

Spread load as one arrow

Beams and loads

\force{W} = \force{w}\,\len{L} \quad \text{acting at } \len{L}/2

$\force{W}$: The single force that replaces the spread loadkN
$\force{w}$: Load per metre (or per foot)kN/m
$\len{L}$: Length the load coversm

Example

\force{w} = 4.5\,\mathrm{kN/m},\ \len{L} = 5\m

\force{W} = 22.5\kN \text{ at } 2.5\m

Largest bending moment, point load

Beams and loads

\begin{gathered}\mom{M_{max}} = \dfrac{\force{P}\,\len{a}\,\len{b}}{\len{L}} \\[4pt] \text{at midspan: } \dfrac{\force{P}\,\len{L}}{4}\end{gathered}

$\mom{M_{max}}$: Largest bending moment, under the loadkN·m
$\force{P}$: The point loadkN
$\len{a},\ \len{b}$: Distances from the load to each supportm
$\len{L}$: Spanm

Simply supported beam (pin + roller).

Example

\force{P} = 20\kN \text{ at midspan},\ \len{L} = 6\m

\mom{M_{max}} = 30\kNm

Largest bending moment, uniform load

Beams and loads

\begin{gathered}\mom{M_{max}} = \dfrac{\force{w}\,\len{L}^2}{8} \\[4pt] \force{R_A} = \force{R_B} = \dfrac{\force{w}\,\len{L}}{2}\end{gathered}

$\mom{M_{max}}$: Largest bending moment, at midspankN·m
$\force{w}$: Load per length, over the whole spankN/m
$\len{L}$: Spanm

Simply supported beam. Doubling the span makes the moment four times bigger.

Example

\force{w} = 5\,\mathrm{kN/m},\ \len{L} = 8\m

\mom{M_{max}} = 40\kNm

Moment at the wall of a cantilever

Beams and loads

\mom{M} = \force{P}\,\len{L} + \dfrac{\force{w}\,\len{L}^2}{2}

$\mom{M}$: Moment at the wall (the largest; stretches the top)kN·m
$\force{P}$: Load at the tipkN
$\force{w}$: Load per length along itkN/m
$\len{L}$: How far it sticks outm

That's why balconies need their steel at the top.

How much a beam sags

Beams and loads

\begin{gathered}\len{\delta} = \dfrac{\force{P}\,\len{L}^3}{48\,E\,I} \\[4pt] \len{\delta} = \dfrac{5\,\force{w}\,\len{L}^4}{384\,E\,I}\end{gathered}

$\len{\delta}$: Sag at midspanmm
$\force{P}$: Point load at midspan (first formula)kN
$\force{w}$: Uniform load per length (second formula)kN/m
$E$: Stiffness of the material (Young's modulus)MPa (N/mm²)
$I$: Second moment of area of the sectionmm⁴

Simply supported. Keep units consistent (e.g. N and mm). Span cubed: a longer beam sags a lot more.

Stress

Stress and sections

\sigma = \dfrac{\force{N}}{A}

$\sigma$: Stress: force on each square millimetre (inch)MPa (N/mm²)
$\force{N}$: Pull or push along the memberN
$A$: Cross-section areamm²

1 N/mm² = 1 MPa.

Stretch (strain) and stiffness

Stress and sections

\begin{gathered}\varepsilon = \dfrac{\Delta \len{L}}{\len{L}} \\[4pt] \sigma = E\,\varepsilon\end{gathered}

$\varepsilon$: Strain: stretch per unit length (no units)–
$\Delta \len{L}$: How much it got longermm
$E$: Young's modulus: steel ≈ 210 GPa (29,000 ksi)MPa (N/mm²)

Holds while the material springs back (elastic).

Rectangle: I and W

Stress and sections

\begin{gathered}I = \dfrac{b\,h^3}{12} \\[4pt] W = \dfrac{b\,h^2}{6}\end{gathered}

$I$: Second moment of area: resistance to bending (stiffness)mm⁴
$W$: Section modulus (US: S): resistance to bending (strength)mm³
$b$: Widthmm
$h$: Depth (in the direction of bending)mm

Depth counts most: twice as deep is 8 times stiffer.

Example

b = 100\,\mathrm{mm},\ h = 300\,\mathrm{mm}

I = 225 \times 10^6\,\mathrm{mm^4}

Bending stress

Stress and sections

\sigma = \dfrac{\mom{M}}{W} = \dfrac{\mom{M}\,y}{I}

$\sigma$: Stress at the outer edge (top or bottom)MPa (N/mm²)
$\mom{M}$: Bending momentN·mm
$W$: Section modulusmm³
$y$: Distance from the middle (neutral axis) to the edgemm

Pressure under a footing

Stress and sections

q = \dfrac{\force{N}}{B \times \len{L}}

$q$: Pressure on the groundkPa (kN/m²)
$\force{N}$: Load from the columnkN
$B,\ \len{L}$: Footing width and lengthm

Keep q below what the soil can take.

Example

\force{N} = 900\kN,\ 2\m \times 2\m

q = 225\,\mathrm{kPa}

Buckling load (Euler)

Columns

\force{P_{cr}} = \dfrac{\pi^2\,E\,I}{(K\,\len{L})^2}

$\force{P_{cr}}$: Load at which a slender column bows outkN
$E\,I$: Bending stiffness of the columnkN·m²
$\len{L}$: Column heightm
$K$: End factor: pinned 1, flagpole 2, fixed + pin 0.7, fixed both 0.5–

Twice as tall carries a quarter. Short, stocky columns crush instead.

Example

EI = 1000\,\mathrm{kN{\cdot}m^2},\ \len{L} = 3\m,\ K = 1

\force{P_{cr}} = 1096.6\kN

Pull in a rope with a weight in the middle

Trusses and ropes

\force{T} = \dfrac{\force{W}}{2\sin\theta}

$\force{T}$: Pull in each half of the ropekN
$\force{W}$: The weight hanging in the middlekN
$\theta$: Rope angle above the horizontal°

At 30° each half pulls the whole weight. Flatter means much more.

Example

\force{W} = 20\kN,\ \theta = 45^\circ

\force{T} = 14.14\kN

Forces at a truss joint

Trusses and ropes

\begin{gathered}\textstyle\sum \force{F_x} = 0 \\[4pt] \sum \force{F_y} = 0 \\[4pt] \force{N} = \dfrac{\force{H}}{\cos 45^\circ} = \sqrt{2}\,\force{H}\end{gathered}

$\force{N}$: Force in a 45° diagonal (pulled +, pushed −)kN
$\force{H}$: Sideways push the diagonal has to carrykN

Members only pull or push. Pushed members must not buckle.

Moment that cracks plain concrete

Reinforced concrete

\mom{M_{cr}} = f_{ct}\,\dfrac{b\,h^2}{6}

$\mom{M_{cr}}$: Moment at which the stretched face crackskN·m
$f_{ct}$: Tensile strength of concrete (about 3 MPa: small!)MPa (N/mm²)
$b,\ h$: Width and depthm

Example

f_{ct} = 3000\,\mathrm{kPa},\ 0.3\m \times 0.5\m

\mom{M_{cr}} = 37.5\kNm

What the steel bars can hold

Reinforced concrete

\begin{gathered}\mom{M_R} = A_s\,f_y\,z \\[4pt] z \approx 0.9\,d \\[4pt] A_s = n\,\dfrac{\pi\,\phi^2}{4}\end{gathered}

$\mom{M_R}$: Moment the cracked section can holdkN·m
$A_s$: Area of the steel barsmm²
$f_y$: Yield strength of the steel (B500: 500 MPa)MPa (N/mm²)
$z,\ d$: Lever arm, and depth to the barsmm
$n,\ \phi$: Number of bars and their diametermm

No safety factors here: real design adds them.

Example

3 \text{ bars } \phi 16,\ f_y = 500\,\mathrm{MPa},\ d = 450\,\mathrm{mm}

A_s = 603\,\mathrm{mm^2},\ \mom{M_R} = 122.1\kNm

Growth with heat

Heat and expansion

\Delta \len{L} = \alpha\,\Delta T\,\len{L}

$\Delta \len{L}$: How much longer it getsmm
$\alpha$: Expansion coefficient: steel 12 × 10⁻⁶ /°C (6.5 × 10⁻⁶ /°F)1/°C
$\Delta T$: Temperature change°C
$\len{L}$: Lengthm

That's why a bridge sits on a pin at one end and a roller at the other.

Example

\len{L} = 50\m,\ \Delta T = 40^\circ\mathrm{C}

\Delta \len{L} = 24\,\mathrm{mm}

Push when growth is blocked

Heat and expansion

\force{N} = E\,A\,\alpha\,\Delta T

$\force{N}$: Push in the member when both ends are heldkN
$E$: Young's modulusMPa (N/mm²)
$A$: Cross-section areamm²
$\alpha\,\Delta T$: The stretch it wanted (strain)–

Example

E = 200\,\mathrm{GPa},\ A = 20{,}000\,\mathrm{mm^2},\ \Delta T = 30^\circ\mathrm{C}

\force{N} = 1440\kN

Heat lost through a wall

Heat and expansion

\Phi = U\,A\,\Delta T

$\Phi$: Heat flowing outW
$U$: U-value: how easily heat gets through (lower is better)W/(m²·K)
$A$: Wall aream²
$\Delta T$: Inside minus outside temperatureK (= °C difference)

Ohm's law

Electricity

U = I\,R

$U$: Voltage (US books write V)V
$I$: CurrentA
$R$: ResistanceΩ

Electric power and current

Electricity

\begin{gathered}P = U\,I \\[4pt] I = \dfrac{P}{U}\end{gathered}

$P$: PowerW
$U$: Voltage: 230 V in Europe, 120 V in the USV
$I$: CurrentA

A 2300 W heater on 230 V draws 10 A.

Energy on the bill

Electricity

E = P\,t

$E$: Energy usedkWh
$P$: PowerkW
$t$: Time switched onh

Water pressure from height

Water and pipes

p = \rho\,g\,h

$p$: PressurePa (10 m of water ≈ 1 bar)
$\rho$: Density of water: 1000 kg/m³kg/m³
$g$: Gravity, 9.81 m/s²m/s²
$h$: Height of the water above that pointm

Why the top floor has less pressure than the basement.

How much water a pipe carries

Water and pipes

\begin{gathered}Q = A\,v \\[4pt] A = \dfrac{\pi\,d^2}{4}\end{gathered}

$Q$: Flow ratem³/s (× 1000 for L/s)
$A$: Inside area of the pipem²
$v$: Speed of the waterm/s
$d$: Inside diameterm

Same flow in a smaller pipe means faster (and noisier) water.