WORKSHOP - FLOOR LOADING NOTES Workshop = 2.1m x 2.1m The floor joists running through the room span 3.2m
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My workshop is in a small first floor bedroom. The room is over the lounge.
I must be careful and not overload the floor. If I overload the floor it probably would not collapse but I'm sure the joists will bend beyond their design limit and affect the ceiling below. So, I have been mindful of the weights I introduce into the bedroom.

Here are some random notes and extracts from the net that you may find useful if you are in a similar position.

According to BS6399-1, 1996 (Loading for buildings) a residential structure should be designed for a distributed load of 1.5kN/m2, and a point load of 1.4kN.

1 NEWTON is the force required to give a mass of 1KG an acceleration of 1m/s/s (1m/s2)
1NEWTON = 1KG*(M/S2)

F=MA

If the acceleration is due to gravity then

F=M*9.81m/s2

1.5KN=M*9.81
1.5*1000=M*9.81
M=1500/9.81=153KG

If the room is 2.13m*2.13m then the maximum weight allowed is

Max Weight =Area*153KG
Max Weight =2.1*2.1*153
Max Weight =4.41*153
Max Weight =674kg (uniformly distributed over the whole area!)

If the room is 2.1m*3.2m then the maximum weight allowed is

Max Weight =Area*153KG
Max Weight =2.1*3.2*153
Max Weight =6.72*153
Max Weight =1028kg (uniformly distributed over the whole area!)

mill weighs 300kg
lathe weighs 200kg
total weight = 500kg
the 500kg weight must be distributed over an area, so that the 1.5KN/m2 loading is not exceeded.
the 500kg weight must be distributed over an area, so that the 153kg/m2 loading is not exceeded.
hence
Area=500/153
Area=3.27m2
say we will build a small wooden dias, that will be used to distribute the load
if the width of the room is 2.1m then the length of the dias should be 3.27/2.1=1.56m
hence a dias of 2.1m*1.56m will distribute the load over a 3.27m2 area.

best keep the loads close to the ends of the joists that bear on the supporting wall.

over time an excessive static load will cause the joists to sag, maybe beyond the design limit.
over time the excessive static load on the room joists may cause the wall to settle and crack.

Is it safe to put a 250kg weight on a wooden domestic first floor?

According to BS6399-1, 1996 (Loading for buildings) a residential structure should be designed for a distributed load of 1.5kN/m2, and a point load of 1.4kN.

Consideration if the load is acting on a single point.
The force from that is 250*10=2500N, which is 2.5kN (error, this is an approximation).
So the floor can't take that high a load.

If we can have 1.5kN/m2, then for a load of 2.5kN, you would need it to be acting over an area of 2.5/1.5=1.67m2.
So as long as it is acting over an area of 1.67m2, you'll be fine.

NB If it has say four legs, then you need to check the point load at each leg as well.
2.5/4=0.63kN which is much less than the 1.4kN, so you'll be fine.

Civil Engineering degree (structures comes under civils)

Offices structures designed to 100psf
Parking structures designed to 50psf
UBC, IRC Table 16-A Uniform and Concentrated loads

A lot depends on how you distribute the weight.
For example, weight near to walls is obviously better since the support point is nearer.
As a rule of thumb, a weight of about 130kg per square metre should be fine.
So, 3.2m2 will take 416kg ? unverified

If the design load is 1.5kN/m2 then :

1.5kNm2 = 150kgm2. Hence 3.2m2 will take 480kg
A room 2.13mx2.13m=4.54m2. Hence 4.54m2 will take 681kg. GULP

The weight of a waterbed (a bed of 160 x 200 x 020 cm will weigh 640 kg), will put a strain on many floorboards. But because the waterbed is 35-42 sq. ft. in size, the floor load stress is 40 lbs./sq. ft. or less, and the load is of no more concern than a refrigerator or filled bathtub.[citation needed]
Many modern codes require a minimum floor load of 150 lbs./sq. ft. (mmm 150lbs??)

question:
a water bed has deminsions of 1.83 m X 2.13 m X 0.229 m.
the floor of the bedroom will tolerate an additional weight of no more than 6660 N.
find the weight of the water in the bed to determine whether it should be purchased

unverified comment:

Well, we know the density of water is about 1g/cm^3. (.998)

This bed is 183cm x 213cm x 22.9cm.

This gives a volume of 892619.1 cm^3.

Easy enough - This gives a total mass of water of 892619.1 grams.

Let's get that to Newtons (kg m/s^2)...So, first up, we convert grams to kilograms by dividing by 1000 - 892.6 kilograms.

We know that the acceleration due to gravity is 9.8 m/s^2.

F=ma.

F=892.6 kg * 9.8 m/s^2 = 8747.7 N

The room can only handle 6660N so we'd better not buy that bed!

unverified comment:

Weight = mass* gravitational acceleration with the mass equal to the volume * density.

Weight = Volume * Density * Gravity

....for volume just multiply the bed's dimensions together.

Weight = 1.83m X 2.13m X 0.229m X 1000Kg/m^3 X 9.8 m/s^2

Weight = 8748 N
STEP AWAY FROM THE BED!!!

unverified comment:
WHAT IS THE WEIGHT OF A WATERBED ?
It depends on the size and style of bed.
It can vary from 200 lbs up to 1800 lbs (90 kgs - 800 kgs).
The soft-sided divan models are lighter, size for size, than rigid-sided models.
A full depth, king size soft-sided divan bed is the equivalent weight of about 5-6 adults
How much does a waterbed weigh?
A waterbed can weigh up to 1500lbs, but because the weight is distributed over a large area,
weight is not a concern.
A waterbed weighs less per square foot than a refrigerator.
Any house built to modern building codes can handle the weight of a waterbed without a problem.

question:

am i at risk putting a filing cabinet on the first floor? (and I worry!)

The normal carrying capacity of a residential floor in a modern building is 40 pounds per square foot for the main level and until recently 30 pounds per square foot for the upper floors.
Using these figures, your file cabinet could weigh 233 pounds if placed on the main level or 175 pounds if placed on the upper level.
Having said that, there are several factors that weigh on the answer (Sorry I could not avoid the pun.) First, this is the maximum allowable weight if THE ENTIRE FLOOR were covered with file cabinets full of files. Obviously that is not the case in your situation, so you would average the weight of all the objects in the room and all the empty space together, which means that the actual weight per square foot would be far less than 40 psf (pounds per square foot).
In other words, the average floor in a room can carry a total of about 4,000 pounds.
The second important factor is that floors are much more sturdy near the bearing end point of the joists (beams) under the floor than it is in the middle of the span.
(If you can't see the beams, they normally run perpendicular to the floorboards).
If you alcove is near where the joists meet the exterior wall, it will take a lot more than 40 psf to strain the floor in that specific area.
Finally, there is the confounding effect of load bearing walls and "pseudo" load bearing walls, which are walls that carry weight by chance but are not structurally required. The net effect of all this is usually to make the floor stronger.
The fact that it is a 1900 house may increase or decrease the expected strength. If it is like most old, urban structures I have seen, the wood used for the joists is significantly larger and denser than today's modern lumber, which would argue for more strength. Also, second floor of such structures are usually just as strong as the main levels, which suggests a higher than 30 psf capacity on the upper level. On the other hand, older beams and joists were not "graded" selectively, and often contained more large knotholes and one or two individual boards may exhibit a localized weakness that could make the floor locally less strong.

So how much will your file cabinet weigh? I estimate if it were totally filed will paper files, it would weigh 622 pounds, which sounds like a lot but is within the safety factors cited above and is equal to the weight of just two professional football players.

The bottom line on all this is that there is near zero chance that your floor will be damaged by excessive loading, although there is a slightly higher chance that the floor could sag over time due to a phenomenon known as "creep" where the cells in the wood fibers deform under a load without losing much (if any) structural strength. If you are still concerned, locate the file cabinet in a portion of the floor near the bearing ends of the underlying joists, or perhaps use one of the drawers to store kleenex or other lightweight supplies. I realize my answer is not completely scientific, but I think it's the best you can get from an online advice column. If you want a more scientific answer, hire the services of a structural engineer who can use a tape measure and a stud finder to locate the exact shape and size of the hidden floor structure and calculate the maximum load in a given part of the floor...but they will charge you the cost of three file cabinets.

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