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Building and Structures: free practice, theory and problems

Before you can design a beam or a floor, you need to know which loads it must carry. In Norway we follow the Eurocodes (NS-EN 1990 and 1991). You split the loads into permanent and variable loads, turn area loads into line loads on the beams, and apply safety factors.

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Loads and load combinations
Building physics: heat and moisture
Surveying

1. Loads and load combinations

What is it about?

Concepts and formulas

Permanent load $G$ : the self-weight of the structure, floors, roof and fixed installations. Computed from thickness and unit weight, for example concrete 25 kN/m³.
Variable load $Q$ : imposed load from people and furniture (housing 2.0 kN/m², offices 3.0 kN/m²), snow and wind.
Snow load on a roof: $s = \mu\,s_k$ , where $s_k$ is the ground snow load for the site and the shape factor $\mu = 0.8$ for a flat roof.
From area load to line load: a beam carries the load from a strip as wide as the spacing $c$ :

q = p\cdot c \quad [\text{kN/m}]

Ultimate limit state (simplified, one dominant variable load): the loads are multiplied by load factors.

q_d = 1.2\,G + 1.5\,Q

Simply supported beam with a uniform load: $M_{max} = \dfrac{qL^2}{8}$ and support reaction $R = \dfrac{qL}{2}$ .

How to solve the problems

Find the area loads in kN/m² (self-weight and imposed load separately).
Multiply by the spacing to get a line load in kN/m.
Combine with the load factors.
Compute the moment and support reactions for the beam.

Example

Floor joists in a house: self-weight 0.5 kN/m², imposed load 2.0 kN/m², spacing 0.6 m and span 4.0 m.

$G = 0.5\cdot 0.6 = 0.3$ kN/m and $Q = 2.0\cdot 0.6 = 1.2$ kN/m.
$q_d = 1.2\cdot 0.3 + 1.5\cdot 1.2 = 2.16$ kN/m.
$M = 2.16\cdot 4^2/8 = 4.32$ kNm.

Common mistakes

Forgetting to multiply the area load by the spacing.
Applying the load factor twice, or forgetting it.
Mixing kN and N, or mm and m, in the moment formula.

Area load × spacing = line load. Ultimate: 1.2 G + 1.5 Q. Moment: qL²/8.

Concepts in this part

Permanent and variable loadsA building must carry its own weight and everything that comes on top. Self-weight is permanent and known quit…Load factorsIn the ultimate limit state the loads are multiplied by load factors to give a safety margin. Self-weight has …Load combinationsIt is unlikely that full snow, full wind and full imposed load occur at the same time. So one variable load is…

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2. Building physics: heat and moisture

What is it about?

A good house keeps the heat in, the moisture out and the electricity bill low. Building physics is about how heat and moisture pass through walls, roofs and windows. The Norwegian building regulations (TEK17) set requirements for how well each part must be insulated.

Concepts and formulas

Thermal resistance of a layer with thickness $d$ and thermal conductivity $\lambda$ (W/mK):

R = \frac{d}{\lambda} \quad [\text{m}^2\text{K/W}]

U-value (thermal transmittance) of the whole construction. The resistances are added, including the surface resistances inside ( $R_{si} = 0.13$ ) and outside ( $R_{se} = 0.04$ ):

U = \frac{1}{R_{si} + \sum R_i + R_{se}} \quad [\text{W/m}^2\text{K}]

Heat loss through a surface: $\Phi = U\cdot A\cdot \Delta T$ (W). Energy: power times time, $E = \Phi\cdot t$ .
TEK17 requirements (U-value): external wall ≤ 0.18, roof ≤ 0.13, windows ≤ 0.80 W/m²K.
Moisture: warm air can hold more water vapour than cold air. If air is cooled below the dew point, the vapour turns into water (condensation). That is why the vapour barrier is placed on the warm side of the insulation.

How to solve the problems

Compute $R = d/\lambda$ for each layer (thickness in metres).
Add all the resistances, including $R_{si}$ and $R_{se}$ .
$U = 1/R_{tot}$ , and the heat loss is $U A \Delta T$ .

Example

A wall has 200 mm of insulation with $\lambda = 0.036$ W/mK. Ignore the other layers.

$R = 0.200/0.036 \approx 5.56$ m²K/W.
$R_{tot} = 0.13 + 5.56 + 0.04 = 5.73$ m²K/W.
$U = 1/5.73 \approx 0.175$ W/m²K, which meets the 0.18 requirement.

Common mistakes

Using the thickness in mm instead of m.
Adding U-values instead of R-values. It is the resistances that add up.
Putting the vapour barrier on the cold side. Then moisture can condense inside the wall.

R = d/λ add up, U = 1/R_tot. Heat loss = U·A·ΔT. Vapour barrier on the warm side.

Concepts in this part

U-valueThe U-value tells how much heat leaks through one square metre of wall for each degree of difference between i…Heat loss through building partsThe heat loss through a wall, roof or window is the U-value times the area times the temperature difference. T…Moisture and dew pointWarm air can hold more water vapour than cold air. When moist indoor air cools down inside a wall, it can reac…

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3. Surveying

What is it about?

Before a road, a building or a tunnel can be built, the terrain must be surveyed and the structure set out in the right place. Surveying is about coordinates, distances, directions and heights.

Concepts and formulas

Norwegian maps use coordinates north ( $N$ ) and east ( $E$ ) in metres. The distance between two points:

s = \sqrt{\Delta N^2 + \Delta E^2}

Angles in surveying are often measured in gon: a full circle is 400 gon (versus 360°). A right angle is 100 gon.
Levelling: you read a staff at the back point (known height) and at the front point. The height difference is

\Delta h = \text{back reading} - \text{front reading}

Gradient in percent: $\dfrac{\Delta h}{\text{horizontal distance}}\cdot 100$ .
Scale 1 : $M$ : 1 cm on the map is $M$ cm in the terrain. At 1 : 1000, 1 cm is 10 m.
Area from coordinates (shoelace formula) for a polygon with corners $(x_i, y_i)$ :

A = \tfrac12\left|\sum (x_i\,y_{i+1} - x_{i+1}\,y_i)\right|

How to solve the problems

Compute the differences $\Delta N$ and $\Delta E$ between the points.
Use Pythagoras for distance, and remember that levelling gives back minus front.
Convert scale by multiplying up and then changing unit.

Example

Point A has a height of 12.345 m. The staff reading at A is 1.200 m and at B 0.800 m.

$\Delta h = 1.200 - 0.800 = 0.400$ m. Positive, so B is higher.
$H_B = 12.345 + 0.400 = 12.745$ m.

Common mistakes

Computing front minus back in levelling, which gives the wrong sign.
Mixing degrees and gon.
Forgetting to convert cm on the map to metres in the terrain.
Trusting a single measurement. Check measurements, such as a closed loop, reveal errors.

Distance = √(ΔN² + ΔE²). Levelling: back − front. 400 gon = 360°.

Concepts in this part

LevellingLevelling measures height differences with an instrument that gives a horizontal line of sight. You read a sta…Coordinates and distanceIn surveying, points are given by coordinates, often north and east in metres. The distance between two points…Area from coordinatesThe area of a plot with known corner coordinates can be calculated directly with the coordinate method. You go…

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Example problems with solutions

Here are some of the problems in building and Structures. In the app, calculation problems get new numbers every time, so you can practise until it sticks – and take a graded practice exam before the real one.

Loads and load combinations: A floor has an imposed load of 2.0 kN/m². The beams are spaced 0.6 m apart. What line load does each beam get from the imposed load?

Answer: 1.2 kN/m

$q = p\cdot c = 2.0\cdot 0.6 = 1.2$ kN/m.

Building physics: heat and moisture: What is the thermal resistance of 200 mm of insulation with

\lambda = 0.036

W/mK?

Answer: 5.556 m²K/W

$R = d/\lambda = 0.200/0.036 \approx 5.56$ m²K/W.

Surveying: Point A has the coordinates N = 100, E = 200 and point B N = 130, E = 240 (metres). How far apart are the points?

Answer: 50 m

$\Delta N = 30$ and $\Delta E = 40$ , so $s = \sqrt{30^2 + 40^2} = \sqrt{2500} = 50$ m.

Loads and load combinations: A beam has a permanent load of 3.0 kN/m and an imposed load of 2.0 kN/m. What is the design load in the ultimate limit state (1.2G + 1.5Q)?

Answer: 6.6 kN/m

$q_d = 1.2\cdot 3.0 + 1.5\cdot 2.0 = 3.6 + 3.0 = 6.6$ kN/m.

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Building and Structures: free practice, theory and problems

Contents

1. Loads and load combinations

What is it about?

Concepts and formulas

How to solve the problems

Example

Common mistakes

Concepts in this part

2. Building physics: heat and moisture

What is it about?

Concepts and formulas

How to solve the problems

Example

Common mistakes

Concepts in this part

3. Surveying

What is it about?

Concepts and formulas

How to solve the problems

Example

Common mistakes

Concepts in this part

Example problems with solutions