maths question, need help please ..

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i need to fill a circular drain pipe with concrete and find out how heavy it will be .

the inside dimension of the pipe is 600mm in (edit ...diameter ) circumference x 250mm tall

any ideas please as my brain has gone into meltdown

cheers
 
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This will be roughly 70.65 litre of concrete if I have got this about right.

Volume = pi * radius squared * height

= 3.14 * 300^2 * 250 = 70,650,000 cubic millimetres
= 70.65 litres

Concrete roughly weighs 2.41kg per litre, so 168kg+ the weight of the pipe.
= 2.41kg * 70.65 = 168.7kg

All this from a couple of quick Google searches. Pipe volume and weight of 1 litre of concrete.

Disclaimer: this could be completely wrong!!

Edit: added workings
 
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What's the density of the concrete? You need to do density x volume. Going to convert to metres to make it less painful.

You've got to work back on the area, so if circumference is 600 mm, C = 2 x Pi x r, rearrange for r = C / 2 x Pi.

That gives you r = 0.6 / (2 x Pi) = 0.095 m

Now you need the cross sectional area - Pi x 0.095^2 = Pi x 0.6^2 = 0.028 m2

Volume = area x length = 0.028 x 0.25 = 0.007 m3.

Average density of concrete is 2400 kg per m3, so multiply that out 2400 x 0.007 = 17 kg of concrete (I think).

Bit different from the guy above!

Edit: plus weight of pipe
 
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Syla5 said:
This will be roughly 70.65 litre of concrete if I have got this about right.

Volume = pi * radius squared * height

= 3.14 * 300^2 * 250 = 70,650,000 cubic millimetres
= 70.65 litres

Concrete roughly weighs 2.41kg per litre, so 168kg+ the weight of the pipe.
= 2.41kg * 70.65 = 168.7kg

All this from a couple of quick Google searches. Pipe volume and weight of 1 litre of concrete.

Disclaimer: this could be completely wrong!!

Edit: added workings
Click to expand...

600 isn't the radius! OP isn't easy to read.
 
do you mean the pipe is 600mm diameter rather than circumference?

taking it as diameter you're talking a volume of pi*0.3 (remember radius=diameter/2)^2*0.25=0.0707m^3 and going by google's 2400kg/m^3 figure for density it'll weigh 169.6kg

taking it as circumference we convert back to diameter so d=circumference/pi=0.6/pi=0.19m, then as before we calculate the volume which ends up agreeing with chinchilla's 0.007m^3 and using google's 2400kg/m^3 again gives us 17kg

so depending on wether it's a diameter or radius is going to make a slight difference to how much it weighs!
 
so what do we all know about wind load ?

going into the concrete base is a 2.5m x 76mm aluminium post with a 600mm x 600mm sign on top , how much battering will this take ?

Ta
 
signs said:
so what do we all know about wind load ?

going into the concrete base is a 2.5m x 76mm aluminium post with a 600mm x 600mm sign on top , how much battering will this take ?

Ta
Click to expand...

Is this just a sneaky way to find out how much load your washing line can take to optimise the laundry cycle??
 
Syla5 said:
This will be roughly 70.65 litre of concrete if I have got this about right.

Volume = pi * radius squared * height

= 3.14 * 300^2 * 250 = 70,650,000 cubic millimetres
= 70.65 litres

Concrete roughly weighs 2.41kg per litre, so 168kg+ the weight of the pipe.
= 2.41kg * 70.65 = 168.7kg

All this from a couple of quick Google searches. Pipe volume and weight of 1 litre of concrete.

Disclaimer: this could be completely wrong!!

Edit: added workings
Click to expand...
seriously? a pipe 2 feet acoss and less than a foot tall will be over 160kg?!
 
Syla5 said:
This will be roughly 70.65 litre of concrete if I have got this about right.

Volume = pi * radius squared * height

= 3.14 * 300^2 * 250 = 70,650,000 cubic millimetres
= 70.65 litres

Concrete roughly weighs 2.41kg per litre, so 168kg+ the weight of the pipe.
= 2.41kg * 70.65 = 168.7kg
Click to expand...

this is about what i got it too also
 
signs said:
so what do we all know about wind load ?

going into the concrete base is a 2.5m x 76mm aluminium post with a 600mm x 600mm sign on top , how much battering will this take ?

Ta
Click to expand...

assuming a turbulent flow, maximum wind speed of 60mph (27m/s) for a good gale, density of air is in the region of 1.225kg/m^3 (google), and the plate being perpendicular to the wind direction (ie the worst possible force) giving a drag coeffecient of 1.28 (wikipedia)

then the drag force is going to be roughly:
e9582e0099f3d4ead6a6933118ed73bc99bd7cc6
(wiki)

=0.5*1.225*27^2*1.28*(0.6^2)=205N or the equivalent of the weight of a 20kg object

now the mechanics of will it tip come down to moments, or force*distance

so our wind force is creating a moment around the base and the top of the sign which is 205*2.5=514Nm

assuming the base doesn't slide along the ground and in fact tries to topple, pivoting around the edge of the cylinder then the moment it'll give is based on the radius of the cylinder and the force of mavity, in reality it's much more complex than that but this is as much detail as i can be bothered to give out for free on the internet but in this simplified version the mavity moment is mg*r=9.81*170kg*0.3=499.8Nm

now we could juggle this around, find out what wind speed would be needed to tip exactly (although we're not too far over with a 60mph gale), work out if the sign would slide along the ground or tip over, or even figure out if the aluminium pole would break before any of this happens, but that's all a pretty deep rabbit hole.
 
wolfie138 said:
seriously? a pipe 2 feet acoss and less than a foot tall will be over 160kg?!
Click to expand...
It does entirely depend on the concrete density, but consider comparing it to a strong man Atlas stone that has been squashed in to a disc, it really isn't hard to comprehend that something that doesn't seem overly large can weigh that much.
 
adolf hamster said:
assuming a turbulent flow, maximum wind speed of 60mph (27m/s) for a good gale, density of air is in the region of 1.225kg/m^3 (google), and the plate being perpendicular to the wind direction (ie the worst possible force) giving a drag coeffecient of 1.28 (wikipedia)

then the drag force is going to be roughly:
e9582e0099f3d4ead6a6933118ed73bc99bd7cc6
(wiki)

=0.5*1.225*27^2*1.28*(0.6^2)=205N or the equivalent of the weight of a 20kg object

now the mechanics of will it tip come down to moments, or force*distance

so our wind force is creating a moment around the base and the top of the sign which is 205*2.5=514Nm

assuming the base doesn't slide along the ground and in fact tries to topple, pivoting around the edge of the cylinder then the moment it'll give is based on the radius of the cylinder and the force of mavity, in reality it's much more complex than that but this is as much detail as i can be bothered to give out for free on the internet but in this simplified version the mavity moment is mg*r=9.81*170kg*0.3=499.8Nm

now we could juggle this around, find out what wind speed would be needed to tip exactly (although we're not too far over with a 60mph gale), work out if the sign would slide along the ground or tip over, or even figure out if the aluminium pole would break before any of this happens, but that's all a pretty deep rabbit hole.
Click to expand...
brilliant , thanks , i've just commissioned an engineer to take a look (before reading this ) but i really do appreciate your response .
 
signs said:
so what do we all know about wind load ?

going into the concrete base is a 2.5m x 76mm aluminium post with a 600mm x 600mm sign on top , how much battering will this take ?

Ta
Click to expand...

How much is burried and how much isn't? Where on the post is the sign pinned? Is the centre 300 down from the top or at the top?

Afaik you need to work out where the load is concentrated on the post to work out the max bending moment (would be found at the base) then compare that with the failure stress of the wood in that axis. You'd also need to work out an average force from wind pressure acting on the sign. Do you really need to do this?
 
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