Help Finding horizontal loads when pouring a column at an angle 2

[vlad1981]

Hi
I need some help checking my line of reasoning, as it seems we can not get to a decision at work.
I am pouring a column, the column is inclined at an angle (Say 52.2 deg), and it will be poured in two lifts (No more than 13ft in vertical).
The columns are formed (Not concerned about this), and the scaffolding system is generic 10K or other tower like system.
The problem starts on the second pour, to get the horizontal loads due to the inclined concrete. This horizontal loads will be transferred to the Forming/ Struts & scaffolding system, and the bracing/ lacing/ tie down will be designed based on them.
See attached pdf.
1. First, the previously poured lift, concrete has already hardened, and it doesn't exert any pressure over the forms. Only its weight, which acts vertically, and needs to be adjusted for the inclination. (vertical distance "t" * Ɣc (150 pcf) * cos(θ)^2 * influence (2 supports))
2. Second, freshly poured concrete, the final pressure on the forms will be due to the weight (vertical) & the horizontal pressure (Fixed at a max. of 600 psf or 4ft of liquid head), all adjusted for the angle of inclination of the struts. For weight same as point 1, for pressure Pmax. (600 psf) * sin(θ)^2 * influence.
All the weights & pressures being adjusted by the inclination, I can treat it all as a beam, and find reactions, with reactions and angle of inclination, I can find Horizontal & Vertical (Taken by tower legs) loads.
Can someone confirm or correct me?

 

[1503-44]

--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."

 

[1503-44]

There.
When it is liquid, why would it not be like a uniformly increasing liquid pressure load?

With 1500 psf across the bottom at the solid concrete surface.

If the two forms shown are connected, like making a box form, the hydrostatic forces are contained within the form and then you can just consider the weight of concrete as a force at its centroid. Column pinned at the solid interface? Maybe you only need one support. Otherwise you will have to consider the flexibility and stiffness of the form to find the other reactions.

Now break R down into horizontal and vertical components and sum forces in X and Y to find loads at the pin at the liquid/solid interface.

--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."

 

[HTURKAK]

I am sorry if my respond disappoints you but i will not confirm you. Your approach neglects the top form . If the slope steeper than 30 degr, you need to use top form together with ties .

The following excerpt from ( Concrete Society Staff-Formwork_ a Guide to Good Practice )

Pls notice that , the horizontal load from fresh conc. pressure essentially cancels with the use of ties.




Use it up, wear it out;
Make it do, or do without.

NEW ENGLAND MAXIM

 

[1503-44]

I finally reached the same conclusion, but it took awhile.
Or the form walls, if they are connected together to make a box, or a tube form, etc. Such as a pipe filled with concrete.

--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."

 

[vlad1981]

@HTURKAK
No need to feel sorry, if I was looking only for answers that 100% satisfied me, I wouldn’t post in this forum.
I think you’re right, @1503-44 is too. I was neglecting the top form, right on, the forms are a box (not ties) with sufficient clips and other connecting parts to hold it together for the stated pressure.
As such, I will only have to bother about the horizontal component of the concrete weight + forms weight + construction loads. No matter whether liquid or solid, specific weight being the same (150 pcf), every time a pour a new lift, weight will add to previous lift.
Thanks ya’ll