Foundation Load Calcs

[Benpointer]

Hi all, 

 

I have this from our TF co's designers, which will presumably dictate the foundation depths.  

 

Can anyone help me decipher it?  I don't absolutely need to know but as part of our self-build journey, I'd like to understand it if at all possible.  Presumably "15/5" "12.5/5" etc. means something important but what?

 

I assume the loadings are all quite light, at it's a single storey TF construction clad in a combination of timber and render/cement panels.

 

Thanks in advance for enlightening me!

 

1829 - 01 Frog Lane-LP.pdf

 

[JohnMo]

I would say you now take that drawing to your structural engineer, he then determines the foundation structure based on your ground makeup and imposed loads.

[IanR]

23 minutes ago, Benpointer said:

, I'd like to understand it if at all possible.  Presumably "15/5" "12.5/5" etc. means something important but what?

 

The drawing Key helps you here:

 

 

First figure is dead load, second is live, the units will be kN/m.

[nod]

Your ground conditions will dictate the depth of your foundation 

The weight of the building won’t make much difference 

[Alan Ambrose]

Unfactored, if I remember rightly, means no safety factor applied i.e. the actual calculated numbers. (Because, you only want one safety factor applied - if the timber frame guy applies a 2.5x safety factor and the foundation person applies 2.5x also, then you soon end up with something much stronger / heavier / more expensive than you need. ‘Dead’ is loading of construction materials, ‘live’ is assumed load from people, furniture, dancing, your grand piano etc - usually taken from NHBC assumptions. There’s also wind load, not listed there.

[SteamyTea]

20 hours ago, IanR said:

units will be kN/m.

Should that be kN/m²?

 

If I remember correctly,  N is a newton, which is mass in kg multiplied by acceleration.

So a 100 kg sitting on the ground is:

 

100 (kg) x 9.81 (m.s^-2) =981 N.

 

Spread that over an area of 0.25 m² and you get 3924, so with a bit of rounding, 4kN.m^-2.

[IanR]

1 hour ago, SteamyTea said:

Should that be kN/m²?

 

Nope, line loads are used to calculate the loads the walls put into the foundation. ie. 10kN/m = just over 1 Tonne-force per meter length of wall.

Edited to add:

From the Drawing

Edited April 12 by IanR

[SteamyTea]

21 minutes ago, IanR said:

line loads are used to calculate the loads the walls put into the foundation

Right, did not know that is a convention.

[saveasteading]

31 minutes ago, IanR said:

line loads are used to calculate the loads

Because the walls are linear.

Then the width of the footing will turn it into /m2.

 

Safety factors...different factors are applied to dead/live/wind in a realistic way. As the chance of everything being at maximum*, plus 1m of snow on the roof plus a hurricane at the same time is low. 

 

* the building weight is known. Then you allow for a reasonable amount of furnishing and people. Not grand pianos everywhere and dancing. These figures are standard and published, but are changed by purpose if it is e.g. a fitness room.

There are other safety factors built into the structure itself, for inferior materials or workmanship.

 

The other thing the SE looks at is proximity of trees and the effect on the ground.

[Gus Potter]

6 hours ago, saveasteading said:

Because the walls are linear.

Then the width of the footing will turn it into /m2.

 

Safety factors...different factors are applied to dead/live/wind in a realistic way. As the chance of everything being at maximum*, plus 1m of snow on the roof plus a hurricane at the same time is low. 

 

* the building weight is known. Then you allow for a reasonable amount of furnishing and people. Not grand pianos everywhere and dancing. These figures are standard and published, but are changed by purpose if it is e.g. a fitness room.

There are other safety factors built into the structure itself, for inferior materials or workmanship.

 

The other thing the SE looks at is proximity of trees and the effect on the ground.

This is a good point.

 

On normal low rise houses we may go to site and dig a few trial pits and find CLAY soil. 

 

We can do some tests on the soil (or just look at and mould it in our hand) and based on that information we can calculate how much load we can put on the soil before it fails, like just totally gives up the ghost big time. We don't want that to happen so we may say let's put a factor of safety of 2.0 on that. An example would be.. say we calculate that we can put thirty tonnes (about 300 kilo Newtons kN) per square metre on the ground before it totally fails. Puttinga factor of safety on that of 2.0 and that gives us a number of say 15 tonnes a square metre. 

 

But CLAY soils settle over time. The next calculation is for settlement. In the UK for normal low rise housing we often limit this to 25mm over 50 years. This gives us another load that the ground can carry without settling excessively. We pick the lower of the two numbers.. and often you'll see on you SE drawings what is referred to as say.. "this design is based on an allowable soil bearing capacity of 100 - 150 kN/m squared which is about 10-15 tonnes.

 

When we have a SAND soil ( I capitalise as the use of upper case denotes the dominant component of the soil) we carry out a similar calculation but are very keen to know about the level of the ground water and where it might be in the future as this can half the strength of a SAND soil.

 

On 11/04/2025 at 12:30, Benpointer said:

Can anyone help me decipher it?

Ben.

 

The biggest number on your drawings is 12.5 /5.0 unfactored load. Now take that as is and add the dead (the weight of things) and live load (people, snow, roof access and so on) together which give 12.5 + 5.0 = 17.5 kN per metre run of wall. Now say that sits on a simple strip concrete foundation with a bit of A142 mesh to stop it cracking too much. Also say that your soil can carry an allowable bearing pressure of 100 kN per metre squared.

 

Let's see how wide the foundation needs to be to carry that load. 17.5 kN per metre run  / 100  kN/ m squared capacity= 175mm! But that is the theory.. Incedentally if any of you have renovated old Victorian houses you may find internal walls that are just sitting on bricks laid sideways on a bed of lime mortar on the soil. So while the number above look crazy it's not actually in old money.

 

Anyway back to your build Ben. You can't reasonably dig or fit a wall onto a found that narrow. When we go to set things out for a DIY self build found.. we may be not get the marking that straight. If your builder  / you get a local hire JCB they often pitch up on site with a 600mm and 450mm wide bucket and a blade bucket for scraping the ground and tidying things up.. Thus on a house like yours I would say can we just do the narrower internal wall founds with the 450 bucket, the external walls with a 600 bucket. Make all the foundations 200 mm thick.

On 11/04/2025 at 17:36, Alan Ambrose said:

Unfactored, if I remember rightly, means no safety factor applied i.e. the actual calculated numbers. (Because, you only want one safety factor applied - if the timber frame guy applies a 2.5x safety factor and the foundation person applies 2.5x also, then you soon end up with something much stronger / heavier / more expensive than you need. ‘Dead’ is loading of construction materials, ‘live’ is assumed load from people, furniture, dancing, your grand piano etc - usually taken from NHBC assumptions. There’s also wind load, not listed there.

Alan is spot on, this happens! In the UK we still use a combination of the British Standards and the Eurocodes. These codes have different factors of safety and we combine the loads in different ways depending on which code we are designing to. By providing the unfactored loads we allow the different suppliers / package designer more scope. That drives down cost.

 

[Benpointer]

Ok thanks for all the replies, especialliy Gus's - very imformative.  

 

So... We are not engaging an SE for the foundations because neither our architect not our groundworker think it necessary; the ground is pure, solid, Kimmeridge clay (for hundreds of metres down).  We have one nearby oak tree, 15m high, with 10.2m radius root protection area, and the tree is 14m from the nearest edge of our foundations.  So I am thinking we can use the NHBC foundation tables (https://nhbc-standards.co.uk/4-foundations/4-2-building-near-trees/4-2-13-foundation-depth-tables/) which gives 1.75m depth in the corner closest to the tree, reducing to 1.0m deep once we get 20m away from the tree.

 

Hopefully the BCO will agree.

Edited April 12 by Benpointer

[SteamyTea]

6 minutes ago, Gus Potter said:

300 kilo Newtons kN

Oi, newtons.

 

7 minutes ago, Gus Potter said:

Make all the foundations 200 mm thick.

I take it that becomes 0.6 or 0.45m (wide) by 0.2m (high).

As that has a mass, (volume times density, so around 250 to 327 kg, or in Roman Catholic up to 3.2 kN) it needs to be added to the static force.

Is that right?

 

Also, how how are unusual impact loads dealt with, these can be much, much higher that normal daily loads i.e. a piano can sit for years on the floor, but when Elton John pops around, dumps himself on the stool, it can bring the house down.

May account for the extra large footware.

 

 

 

[Benpointer]

@SteamyTea  These days when Elton comes round ours for a cuppa he doesn't do so much jumping up and down on the piano.

[SteamyTea]

33 minutes ago, Benpointer said:

he doesn't do so much jumping up and down on the piano

Too much information; where is the mind bleach.

[Gus Potter]

34 minutes ago, SteamyTea said:

Oi, newtons

xxx for you!

 

36 minutes ago, SteamyTea said:

I take it that becomes 0.6 or 0.45m (wide) by 0.2m (high).

As that has a mass, (volume times density, so around 250 to 327 kg, or in Roman Catholic up to 3.2 kN) it needs to be added to the static force.

Is that right?

Yes you are right.. I can't think of a time when you have been wrong! wish I could., just kidding.

 

For low rise domestic design we try and limit the SE fee. At foundation formation level (the depth of the hole we dig) we dig a bit of soil out and put some concrete back in. But then we often make the founds wider for buildability and call it a day.

41 minutes ago, SteamyTea said:

Also, how how are unusual impact loads dealt with, these can be much, much higher that normal daily loads i.e. a piano can sit for years on the floor, but when Elton John pops around, dumps himself on the stool, it can bring the house down.

May account for the extra large footware.

Good point. We as SE's have a term called load sharing and load spread..,. similar to electrical diversity when sizing a supply or consumer unit. The piano and Eltons "charisma" gets spread about by the time it reaches the founds. While we do calculations some of these are based on a best guess and experience. The soil under the founds is variaible , best guess. If we are blessed with say 4 Eltons on a floor then that is where the safety factors come in. Anyway Elton does not stay more than a few hours which will not impact on the long term settlement.

 

[SteamyTea]

10 minutes ago, Gus Potter said:

While we do calculations some of these are based on a best guess and experience

You could try this one, is used in cold forging.

 

 

Where:

Yf = flow stress of the material (Pa)

A = cross-section area of the workpiece (m2)

r = instantaneous radius of the workpiece (m)

h = instantaneous height of the workpiece (m)

µ = coefficient of friction between the die and the workpiece

[Benpointer]

16 minutes ago, SteamyTea said:

You could try this one, is used in cold forging.

 

 

Where:

Yf = flow stress of the material (Pa)

A = cross-section area of the workpiece (m2)

r = instantaneous radius of the workpiece (m)

h = instantaneous height of the workpiece (m)

µ = coefficient of friction between the die and the workpiece

I am struggling to understand how the 'coefficient of friction between the die and the workpiece' is going to affect my foundations...  

 

I an only conclude that Yr Afin a L1/2

[SteamyTea]

2 minutes ago, Benpointer said:

I am struggling to understand how the 'coefficient of friction between the die and the workpiece' is going to affect my foundations...  

It will be caused by the backfill, so probably very low in reality.

It is more a starting point, rather than a finished formula.

I am not sure how 'soils' react over time, the N.s will have an affect, but it is probably a differential equation dN/dt of some sort, i.e. as time passes, the movement gets less.

 

One thing about foundations for lightweight buildings is that they may be limited if say, a bungalow is converted to a 2 storey place.

[Benpointer]

5 minutes ago, SteamyTea said:

...One thing about foundations for lightweight buildings is that they may be limited if say, a bungalow is converted to a 2 storey place.

Fair point but we have zero interest in ever doing that so will leave it to those who follow to worry about.  It's 170m2 house, which is plenty for us.