Lifting force concrete foundation 700mm

[scorp1on]

It's the form factor that I'm unsure about in this case. And then I assumed the same zone to calculate the entire roof. Feels like there should be some computer software for this.

It becomes very unfavorable for that reason, all the simplifications, etc. It's no wonder it turns out like this if you're calculating 1.5 kN/m2 over the entire 30 sqm vertically upwards.

 

[HEA260]

It becomes very unfavorable precisely for that reason, all the simplifications etc. It's not surprising that it ends up that way if you calculate 1.5 kN/m2 across the entire 30 sqm vertically upwards.

If the internal wind pressure is chosen to be the most unfavorable (wind suction -0.3) and we choose the lowest zone (zone I), we get with q(z)=0.49kN/m2 that wind suction = 0.49*0.5= 0.245 kN/m2.

Load combination 6.10b = 0.245 kN/m2 * 30m2 = 11 kN or 1100kg

Remove the weight of the carport, say 600 kg

So there is 500kg left, distributed over 8 pillars = 62 kg / piece

In other words, some type of wind brace or plate under the concrete is needed to handle the current load case, if I haven't calculated incorrectly.

 

[witten]

Fun that you're trying on your own, but there are some methodological errors. You shouldn't calculate a lift force on the roof, it's the pressure + suction on the walls you should calculate. Then you use force equations on the wind braces to get horizontal and vertical components.

When the self-weight is favorable, you must reduce it in 6.10b.

 

[HEA260]

It's great that you're trying on your own, but there are some methodological errors. You shouldn't calculate a lifting force on the roof; it's the pressure + suction on the walls you should calculate. Then use force equations on the wind braces to determine the horizontal and vertical components.

When the self-weight is favorable, you must reduce it in 6.10b.

So even if walls are missing, you assume they exist and then apply it as a lifting force? (Since the air now enters and affects the roof upwards?)

I love that the self-weight is added in 6.10b, thank you for that!

 

[scorp1on]

It's great that you're trying on your own, but there's a bit of a methodological error. You shouldn't calculate a lift force on the roof; it's pressure + suction on the walls you should calculate. Then use force equations on the wind bracings to extract horizontal and vertical components.

When the self-weight is favorable, you must reduce it in 6.10b.

Why wouldn't you calculate pressure/suction on the roof?

 

[witten]

So even if walls are missing, one assumes they exist and then adds it as an uplift force? (Since the air is now entering and affecting the roof upwards?)

I love that the self-weight is added in 6.10b, thanks for that!

If you're not going to have any walls, you hardly need to do a wind load calculation. There's no wind catch to speak of.

 

[HEA260]

If you don't need any walls, you hardly need to do a wind load calculation. There's no wind capture to speak of.

Not even a wind suction/pressure against the roof?

 

[scorp1on]

If you are not going to have any walls, you hardly need to do a wind load calculation. There's no wind resistance to speak of.

Depends on how the roof is designed.

 

[HEA260]

Depends on how the roof is designed.

Mono-pitched roof 4°

 

[scorp1on]

Shed roof 4°

Yes, it is my understanding that it will be a wind load that must be considered. Pillars and foundation constructions will be affected, compared to awnings.

 

[HEA260]

Then it is my understanding that it becomes a wind load that should be taken into account. Columns and foundation constructions will be affected, similar to a canopy roof.

Thank you!
Is it possible to somehow convert the lift force using tie rods against a horizontal beam so that part of the vertical load is absorbed by a horizontal beam? In other words, some type of truss construction installed in the wall. Or am I completely off track?

 

[scorp1on]

Can you attach a sketch of what you're thinking?

 

[HEA260]

Can you attach a sketch of your idea?

F is the lift force from the wind. The horizontal beam is fixed with moment-stiff attachment to the column at points A and C. Also fixed to approximately 45-degree beams at point B.

Can the vertical force be converted to a moment/tensile force in the horizontal beam in this way?

Apologies for the poor sketch; I hope the idea still comes across.

 

[witten]

F attacks from the side as I wrote in #18, read that post carefully again. There is also a force like the one you show but it is relatively uninteresting for the plinths (unless the self-weight is extremely light on everything). Additionally, as I also wrote, it seems to also act downward because the wind also blows under the roof. Sometimes, for large roofs, you may need to add a horizontal force for friction against the roof surface.

In this case, you need to calculate the wind load from the posts and beams. It's probably according to the book, but the wind load is likely quite small compared to if the walls are solid.

 

[HEA260]

F attacks from the side as I wrote in #18, read that post carefully again. There is also a force like the one you show but it's relatively unimportant for the footings (unless it's extremely light self-weight on everything). Also, as I wrote, it seems to act downward because it also blows under the roof. Sometimes for large roofs, you might need to add a horizontal force for friction against the roof surface.

In this case, you need to calculate the wind load from posts and beams. It is probably correct by the book but the wind load is likely quite small compared to if the walls are solid.

I read your post carefully, but Eurocode 1-4 chapter 7-4 (I think, for canopies). Gives a very large lifting force, i.e., vertically. (Cp around -0.7 if I remember correctly, I only have my phone with me).

I really want to understand this completely by the book to learn. Really grateful that you/you all are willing to help.