# What are Load Combinations and how to calculate them?

Last updated: February 15th, 2023

As buildings and structures must withstand the heaviest storms, accidental events and combined loading scenarios, engineers multiply loads with safety factors and combine different loads in so-called Load combinations to make sure that the structure doesn’t collapse.

We’ll show step-by-step, how load combinations work, what different types we use and how to calculate them.

Before we start: Don’t forget to check out our load combination generator! ππ

## What are load combinations?

Load combinations combine different loads like snow, wind, dead, seismic and live load to represent a “real scenario”. A real scenario is for example the resulting force for a heavy wind storm. By setting up all possible load combinations we will find the worst-case scenario for a structural member which is in many cases the biggest load.

Load combinations according to Eurocode consist mostly of 3 components:

- The characteristic load value (snow, wind, dead, seismic, live load)
- Partial factor $\gamma$
- Factor for combination value of variable loads $\Psi_{0}$

So let’s look at an example π

## Loads acting on a flat roof – Example

*Example flat roof*

First, let’s define some symbols and values for our loads.

$g_{k}$ | 1.08 kN/m2 | Characteristic value of dead load |

$q_{k}$ | 1.0 kN/m2 | Characteristic value of live load |

$s_{k}$ | 1.0 kN/m2 | Characteristic value of snow load |

$w_{k}$ | -1.0 kN/m2 | Characteristic value of wind load |

## ULS load combinations

ULS stands for ultimate limit state. Due to ULS load combinations, structural members are designed for bending, shear, buckling, etc.

If you want to save time, then check out our load combination generator which automatically creates the ULS load combinations. π₯π₯

According to Eurocode EN 1990 (6.10) the load combinations can be written as

LC1 | $\gamma_{g} * g_{k} $ |

LC2 | $\gamma_{g} * g_{k} + \gamma_{q} * q_{k}$ |

LC3 | $\gamma_{g} * g_{k} + \gamma_{q} * q_{k} + \Psi_{0.s} * \gamma_{q} * s_{k}$ |

LC4 | $\gamma_{g} * g_{k} + \Psi_{0.q} * \gamma_{q} * q_{k} + \gamma_{q} * s_{k}$ |

LC5 | $\gamma_{g} * g_{k} + \gamma_{q} * q_{k} + \Psi_{0.s} * \gamma_{q} * s_{k} + \Psi_{0.w} * \gamma_{q} * w_{k} $ |

LC6 | $\gamma_{g} * g_{k} + \Psi_{0.q} * \gamma_{q} * q_{k} + \gamma_{q} * s_{k} + \Psi_{0.w} * \gamma_{q} * w_{k} $ |

LC7 | $\gamma_{g} * g_{k} + \Psi_{0.q} * \gamma_{q} * q_k + \Psi_{0.s} * \gamma_{q} * s_{k} + \gamma_{q} * w_{k} $ |

LC8 | $\gamma_{g} * g_{k} + \gamma_{q} * s_{k} $ |

LC9 | $\gamma_{g} * g_{k} + \gamma_{q} * w_{k} $ |

LC10 | $\gamma_{g} * g_{k} + \gamma_{q} * s_{k} + \Psi_{0.w} * \gamma_{q} * w_{k} $ |

LC11 | $\gamma_{g} * g_{k} + \gamma_{q} * w_{k} + \Psi_{0.s} * \gamma_{q} * s_{k} $ |

LC12 | $\gamma_{g} * g_{k} + \gamma_{q} * q_{k} + \Psi_{0.w} * \gamma_{q} * w_{k}$ |

LC13 | $\gamma_g * g_k + \gamma_q * \Psi_{0.q} * q_k + \gamma_{q} * w_{k}$ |

LC14 | $\gamma_{g.inf} * g_k + \gamma_q * w_k$ |

Where

$\gamma_{g}$ | Partial factor for permanent loads from EN 1990 Table A1.2(B) |

$\gamma_{q}$ | Partial factor for variable loads from EN 1990 Table A1.2(B) |

$\gamma_{g.inf}$ | Partial factor for permanent loads (lower value) from EN 1990 Table A1.2(B) |

$\Psi_{0.q}$ | Factor for combination value of live load from EN 1990 Table A1.1 |

$\Psi_{0.s}$ | Factor for combination value of snow load from EN 1990 Table A1.1 |

$\Psi_{0.w}$ | Factor for combination value of wind load from EN 1990 Table A1.1 |

For the case of the flat roof, we get the following values

$\gamma_{g}$ | 1.35 (unfavourable) |

$\gamma_{q}$ | 1.5 |

$\Psi_{0.q}$ | 0 |

$\Psi_{0.s}$ | 0.7 (Sweden) |

$\Psi_{0.w}$ | 0.6 |

Which we now can put in the Load combinations.

If you are unsure of how to apply the different loads to roofs, then check out my post about loads and how to apply them (link)π

LC1 | $1.35 * 1.08 \frac{kN}{m^2} $ | $1.46 \frac{kN}{m^2}$ |

LC2 | $1.35 * 1.08 \frac{kN}{m^2} + 1.5 * 1.0 \frac{kN}{m^2}$ | $2.96 \frac{kN}{m^2} $ |

LC3 | $1.35 * 1.08 \frac{kN}{m^2} + 1.5 * 1.0 \frac{kN}{m^2} + 0.7 * 1.5 * 1.0 \frac{kN}{m^2}$ | $4.0 \frac{kN}{m^2}$ |

LC4 | $1.35 * 1.08 \frac{kN}{m^2} + 0 * 1.5 * 1.0 \frac{kN}{m^2} + 1.5 * 1.0 \frac{kN}{m^2}$ | $2.96 \frac{kN}{m^2} $ |

LC5 | $1.35 * 1.08 \frac{kN}{m^2} + 1.5 * 1.0 \frac{kN}{m^2} + 0.7 * 1.5 * 1.0 \frac{kN}{m^2} + 0.6 * 1.5 * (-1.0 \frac{kN}{m^2}) $ | $3.1 \frac{kN}{m^2}$ |

LC6 | $1.35 * 1.08 \frac{kN}{m^2} + \Psi_{0} * 1.5 * 1.0 \frac{kN}{m^2} + 1.5 * 1.0 \frac{kN}{m^2} + 0.6 * 1.5 * (-1.0 \frac{kN}{m^2}) $ | $2.1 \frac{kN}{m^2} $ |

LC7 | $1.35 * 1.08 \frac{kN}{m^2} + 0 * 1.5 * 1.0 \frac{kN}{m^2} + 0.7 * 1.5 * 1.0 \frac{kN}{m^2} + 1.5 * (-1.0 \frac{kN}{m^2}) $ | $1.0 \frac{kN}{m^2}$ |

LC8 | $1.35 * 1.08 \frac{kN}{m^2} + 1.5 * 1.0 \frac{kN}{m^2} $ | $ 2.96 \frac{kN}{m^2} $ |

LC9 | $1.35 * 1.08 \frac{kN}{m^2} + 1.5 * (-1.0 \frac{kN}{m^2}) $ | $ -0.04 \frac{kN}{m^2} $ |

LC10 | $1.35 * 1.08 \frac{kN}{m^2} + 1.5 * 1.0 \frac{kN}{m^2} + 0.6 * 1.5 * (-1.0 \frac{kN}{m^2}) $ | $ 2.06 \frac{kN}{m^2}$ |

LC11 | $1.35 * 1.08 \frac{kN}{m^2} + 1.5 * (-1.0 \frac{kN}{m^2}) + 0.7 * 1.5 * 1.0 \frac{kN}{m^2} $ | $1.0 \frac{kN}{m^2}$ |

LC12 | $1.35 * 1.08 \frac{kN}{m^2} + 1.5 * 1.0 \frac{kN}{m^2} + 0.6 * 1.5 * (-1.0 \frac{kN}{m^2})$ | $2.06 \frac{kN}{m^2}$ |

LC13 | $1.35 * 1.08 \frac{kN}{m^2} + 1.5 * 0 * 1.0 \frac{kN}{m^2} + 1.5 * (-1.0 \frac{kN}{m^2}) $ | $-0.04 \frac{kN}{m^2}$ |

LC14 | $1.0 * 1.08 \frac{kN}{m^2} + 1.5 * frac{kN}{m^2} + 1.5 * (-1.0 \frac{kN}{m^2}$ | $-0.42 \frac{kN}{m^2}$ |

Puhhh that was quite a lot of typing. Forgive me if I did a typo somewhere π

Don’t worry, you do not have to do this manually every time because luckily most FE programs do that for us.

But if we were to dimension a timber beam for bending now manually we would use the biggest value of the Load combinations which is 4.0 kN/m2 (LC3) and transform it first in a line load (kN/m)

Assuming that the beams have a spacing of 0.8m we get the following line load:

$$4.0 \frac{kN}{m^2} * 0.8 m = 3.2 \frac{kN}{m}$$

That line load can now be applied to our static system

Perfect. So now we can go ahead and dimension a beam in the next blog post.

## SLS load combinations

SLS stands for Serviceability limit state. Due to SLS load combinations structural members are designed for deflection, cracks in concrete, …

First, let’s look at the **characteristic SLS load combination**.

According to EN 1990 (6.14b) the characteristic load combinations can be written as

LC1 | $g_{k} $ |

LC2 | $g_{k} + q_{k}$ |

LC3 | $g_{k} + q_{k} + \Psi_{0.s} * s_{k}$ |

LC4 | $g_{k} + q_{k} + \Psi_{0.w} * w_{k}$ |

LC5 | $g_{k} + q_{k} + \Psi_{0.s} * s_{k} + \Psi_{0.w} * w_{k} $ |

LC6 | $g_{k} + \Psi_{0.q} * q_{k} + s_{k} + \Psi_{0.w} * w_{k} $ |

LC7 | $g_{k} + \Psi_{0.q} * q_{k} + \Psi_{0.s} * s_{k} + w_{k} $ |

LC8 | $g_{k} + s_{k} $ |

LC9 | $g_{k} + w_{k} $ |

LC10 | $g_{k} + s_{k} + \Psi_{0.w} * w_{k} $ |

LC11 | $g_{k} + w_{k} + \Psi_{0.s} * s_{k} $ |

LC12 | $g_{k} + \Psi_{0.q} * q_{k} + s_{k}$ |

LC13 | $g_{k} + \Psi_{0.q} * q_{k} + w_{k}$ |

.. and if we put in the values, we get ..

LC1 | $1.08 \frac{kN}{m^2} $ | $1.08 \frac{kN}{m^2}$ |

LC2 | $1.08 \frac{kN}{m^2} + 1.0 \frac{kN}{m^2}$ | $2.08 \frac{kN}{m^2}$ |

LC3 | $1.08 \frac{kN}{m^2} + 1.0 \frac{kN}{m^2} + 0.7 * 1.0 \frac{kN}{m^2}$ | $2.78 \frac{kN}{m^2}$ |

LC4 | $1.08 \frac{kN}{m^2} + 1.0 \frac{kN}{m^2} + 0.6 * (-1.0 \frac{kN}{m^2})$ | $1.48 \frac{kN}{m^2}$ |

LC5 | $1.08 \frac{kN}{m^2} + 1.0 \frac{kN}{m^2} + 0.7 * 1.0 \frac{kN}{m^2} + 0.6 * (-1.0 \frac{kN}{m^2}) $ | $2.18 \frac{kN}{m^2}$ |

LC6 | $1.08 \frac{kN}{m^2} + 0 * 1.0 \frac{kN}{m^2} + 1.0 \frac{kN}{m^2} + 0.6 * (-1.0 \frac{kN}{m^2}) $ | $1.48 \frac{kN}{m^2}$ |

LC7 | $1.08 \frac{kN}{m^2} + 0 * 1.0 \frac{kN}{m^2} + 0.7 * 1.0 \frac{kN}{m^2} + (-1.0 \frac{kN}{m^2}) $ | $0.78 \frac{kN}{m^2}$ |

LC8 | $1.08 \frac{kN}{m^2} + 1.0 \frac{kN}{m^2}$ | $2.08 \frac{kN}{m^2}$ |

LC9 | $1.08 \frac{kN}{m^2} + (-1.0 \frac{kN}{m^2}) $ | $0.08 \frac{kN}{m^2}$ |

LC10 | $1.08 \frac{kN}{m^2} + 1.0 \frac{kN}{m^2} + 0.6 * (-1.0 \frac{kN}{m^2}) $ | $1.48 \frac{kN}{m^2}$ |

LC11 | $1.08 \frac{kN}{m^2} + (-1.0 \frac{kN}{m^2}) + 0.7 * 1.0 \frac{kN}{m^2} $ | $0.78 \frac{kN}{m^2}$ |

LC12 | $1.08 \frac{kN}{m^2} + 0 * 1.0 \frac{kN}{m^2} + 1.0 \frac{kN}{m^2}$ | $2.08 \frac{kN}{m^2}$ |

LC13 | $1.08 \frac{kN}{m^2} + 0 * 1.0 \frac{kN}{m^2} + (-1.0 \frac{kN}{m^2})$ | $0.08 \frac{kN}{m^2}$ |

Now let’s move on to the **quasi-permanent SLS load combination**.

According to EN 1990 (6.16b) the quasi-permanent load combinations can be written as.

LC1 | $g_{k} $ |

LC2 | $g_{k} + \Psi_{2.q} * q_{k}$ |

LC3 | $g_{k} + \Psi_{2.q} * q_{k} + \Psi_{2.s} * s_{k}$ |

LC4 | $g_{k} + \Psi_{2.q} * q_{k} + \Psi_{2.w} * w_{k}$ |

LC5 | $g_{k} + \Psi_{2.q} * q_{k} + \Psi_{2.s} * s_{k} + \Psi_{2.w} * w_{k} $ |

LC6 | $g_{k} + \Psi_{2.s} * s_{k} $ |

LC7 | $g_{k} + \Psi_{2.w} * w_{k} $ |

LC8 | $g_{k} + \Psi_{2.s} * s_{k} + \Psi_{2.w} * w_{k} $ |

Where

$\Psi_{2.q}$ | 0 |

$\Psi_{2.s}$ | 0.2 (Sweden) |

$\Psi_{2.w}$ | 0 |

.. and if we put in the values, we will get ..

LC1 | $1.08 \frac{kN}{m^2} $ | $1.08 \frac{kN}{m^2} $ |

LC2 | $1.08 \frac{kN}{m^2} + 0 * 1.0 \frac{kN}{m^2}$ | $1.08 \frac{kN}{m^2} $ |

LC3 | $1.08 \frac{kN}{m^2} + 0 * 1.0 \frac{kN}{m^2} + 0.2 * 1.0 \frac{kN}{m^2}$ | $1.28 \frac{kN}{m^2} $ |

LC4 | $1.08 \frac{kN}{m^2} + 0 * 1.0 \frac{kN}{m^2} + 0 * (-1.0 \frac{kN}{m^2})$ | $1.08 \frac{kN}{m^2} $ |

LC5 | $1.08 \frac{kN}{m^2} + 0 * 1.0 \frac{kN}{m^2} + 0.2 * 1.0 \frac{kN}{m^2} + 0 * (-1.0 \frac{kN}{m^2}) $ | $1.28 \frac{kN}{m^2} $ |

LC6 | $1.08 \frac{kN}{m^2} + 0.2 * 1.0 \frac{kN}{m^2} $ | $1.28 \frac{kN}{m^2} $ |

LC7 | $1.08 \frac{kN}{m^2} + 0 * (-1.0 \frac{kN}{m^2}) $ | $1.08 \frac{kN}{m^2} $ |

LC8 | $1.08 \frac{kN}{m^2} + 0.2 * 1.0 \frac{kN}{m^2} + 0 * (-1.0 \frac{kN}{m^2}) $ | $1.28 \frac{kN}{m^2} $ |

Alright, this post got longer than I thought π

We haven’t covered the **accidental and frequent load combinations**, but once we have an example where we need them we will also explain them.

With the knowledge we have got now, we can dimension our first timber beam in the next post. Hope to see you there π

## Load Combinations FAQ

**Why are load combinations important?**

Load combinations are important because they help ensure the structural integrity and safety of a building or structure. For example, without considering load combinations, a structure may be designed to withstand only one type of load (e.g. snow) but could fail under a different type of load (e.g. wind). In addition, loads usually never act alone; instead, multiple loads act simultaneously, and load combinations consider this event.

**How does the load combination affect the structural design?**

Load combinations affect the structural design because the maximum expected loads determine the strength and safety of a structure. By considering different load combinations, you can ensure that a structure can withstand the most severe loads and remain safe for its intended use.