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Resolving forces for wall formwork?

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structures2020

Structural
Jun 7, 2020
3
BM
Hi there,

My boss would like me to design single sided formwork for a retaining wall that we will be constructing. However, I haven't done design work for a few years so I want to make sure I am going about this the right way.

I was told the wet concrete pressure to design for. I converted this pressure into a moment at the base.

I used this moment and converted it into a couple to determine the horizontal force at the top of the wall.

I resolved the top force into its diagonal component (which is quite high).

I resolved the force also into its vertical component for uplift (not sure if this is normally a consideration?)

I was pretty much thinking that :

- I'll take the horizontal force as the sliding force at the base and size stakes to prevent sliding/lateral movements (83.5kN)
- I'll take the diagonal force to size the raked members (although that load is extremely high, which is why I am wondering if I am doing this correctly).
- I'll take the uplift force to size the anchors to keep the formwork system in place to prevent uplift.

Please let me know if I should be doing this a different way or if I should be considering any other factors.
 
 https://files.engineering.com/getfile.aspx?folder=bbf985ae-d5e0-421b-8ac7-c13827749b01&file=engtips_2.pdf
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The moment seems high. I get 129kNm.
The rest seems OK.
 
Thanks for the responses so far.

That's interesting about the large difference with the moments. Maybe I did the calculation completely wrong.

I was looking at this guidance by ACI regarding wet concrete weight pressures.


See attached calcs I did to determine the bending moment.

I considered a pour rate of 6'/hour, which had a max pressure of 48.4 kPa.

**also I realise the bending moment is slightly lower in the calcs I uploaded in this post compared to the original upload. I must've made a calculation error when calculating the bending moment used in the original.
 
My calculation below. M=F*(Arm). Apply load/shape factors as required.

M1 = 48.4*1.5*(1.5/2) = 54.45 - Rectangular pressure block
M2 = 48.4*2*0.5*(2/3+1.5) = 104.87 - Triangular pressure block
M3 = 10*0.433*3.5*(3.5/2) = 20.21 - Uniform surcharge pressure
Sum M = 54.45+104.87+20.21 = 179.53 - Unfactored moment about the base

Note:
1. Make sure the initial pressure is correct. It seems low.
2. The pour rate is 6' per hour, and you have 9' wall, so it takes 1.5 hours to complete the pour. Does ACI address (how it handles) varying pressure during a lengthy pour, as some of the concrete has already hardened? It potentially can reduce the design moment quite a lot.


 
strucbells,

This is to calculate overturning stability of the structure (M[sub]OT[/sub] about toe/base), then estimate the resistance force in the supporting bracket (M[sub]R[/sub]=M[sub]OT[/sub]).

Or, you may look the other way - a vertical beam simply supported on top (roller) and bottom (pin), for the given applied load, how you find the reaction at each support? Then, the roller is set on an inclined plane, how you get/adjust the force?
 
Your original calcs showed a concrete height of 3.2 m, rather than 3.5 m, which makes a big difference.

Your calcs look OK, but for a wall of that height I'd just use the full hydrostatic pressure over the full height, and not add in any surcharge load. With a load factor of 1.35 that gives a moment at the base of 241 kNm.

strucbells said:
Then solve for the reaction at the top of the wall and convert that into its diagonal component based on the angle
.

That's exactly what he did. He calculated the moment of the applied load about the base, then divided by the height to the prop to get the horizontal force on the prop.

As a general comment, when using design procedures specified in imperial units, I'd recommend using the same units in the calculations, then converting the end result to the units you want. It doesn't make any difference in this case, but often there are hidden units in code design coefficients.

Doug Jenkins
Interactive Design Services
 
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