Concrete Pressure on Inclined Formwork 1

[Bridge100]

Could somebody please direct me to a good concrete formwork publication or help me with the following problem? I’m designing formwork for an inclined concrete column but I can’t find anything that deals with this problem. The contractor would like to pour the entire 40 Ft tall column in one day.

Looking at the design of the underside incline bulkhead I am considering the following:

All vertical and lateral loads must be converted to force components normal to the forming face. (See attached method B)

The shear components of the vertical and lateral loads will not be resisted by the form since the form does not have shear connectors or friction with the fluid concrete (See attached method A). Therefore, the shear component of the vertical concrete load will travel through the previously cast pier stem below.

The lateral load would be based on an assumed allowable form pressure. I know that ACI 347 suggests designing for full liquid head for columns this tall; however, it has been the contractor’s experience to use penetrating rods to determine when initial concrete set has occurred.

The vertical normal component would vary from zero at the top of the form to 150pcf*40’*cos2(A) at the bottom.

The lateral normal component would be a uniform load of 150pcf*Liquid Head (Ft)*sin2(A).

Can anybody see problems with this approach?

 

[SAIL3]

Agree, BA, the formwork in question will only experience horizonal tension.
For my own enlightenment, coming back to case#2. If the four sides
were sloped say 80deg(10deg to horiz), I would expect, again, horizontal tension in the sides. Now, instead of four sides, say have 16 sides etc. approaching a circular vessel as an upward bound. Still tension in sides.
What I suspect, and not sure of is that a component of this tension reacts the press(or gravity)load locally at the sides and that the compression at the base is only a local condition
confined to the vicinity of the base.
Using this logic and coming back to the formwork in question I
think that a vertical component of this tension load reacts the
vertical component of the press. from the conc.
If one inverted the formwork(small opening at top) one would expect the formwork to lift up as the conc. was poured.

 

[BAretired]

SAIL3. It is always best to draw a sketch of the situation, particularly when you are dealing with a shape you are not familiar with.

You are describing a shape approaching a conical shell with apex down. You can think of it as a series of rings of varying diameter, one on top of another. When the shell is filled with water, each ring is in pure tension which is constant around the circumference. The magnitude of ring tension is gamma*z*R*dy where z is the depth of water, R is the ring radius at depth z and dy is the incremental thickness of the ring.

The weight of water resting on one ring is pi*R^2*z*dR where dR is the change of radius in a height of dy. The axial compression in any ring parallel to the slope is the sum of all the ring loads above it modified to account for the slope of the cone.

While it is true that the compressive stress along the slope line in a cone diminishes rapidly as you move up from the apex, it does not entirely disappear until you reach the water surface, so I do not think it is accurate to label it a local condition.

I agree that inversion of the form would tend to make it lift as the concrete is poured.

BA

 

[Bridge100]

I would like to attempt to prove my point with one last sketch (see attachment). The 10 kip block is on a frictionless surface. A column support keeps the block from sliding downhill. The normal force is a small portion of the block weight (3.42 k). The biggest component of the block weight exerts a 9.4 kip axial load on the column support, not the inclined surface. The free body diagragm proves that only 1.18 kip of the 10 kip block is applied to the inclined surface. In the case of the concrete form example, the column support would be the concrete medium down to the lower pier stem.

 

[BAretired]

Bridge100. I am not sure what point you are trying to prove. Your model reflects a situation totally different from the one at hand. You do not have a sloping column under a 10 kip block.

What you have is a triangular wedge of concrete 40' high x 14.5' wide at the top tapering down to 0' at the bottom, weighing nearly 44,000# per foot of thickness. As a fluid, there is a lateral pressure pushing outward varying from 0 at the top to 6000 psf at the bottom resulting in a horizontal force of 120,000# per foot of width. The statics of the inclined form supporting a fluid are elementary.

If concrete is poured slowly, the lower parts start to set, reducing the pressure on the forms, but in no case can any part of the concrete be considered to function as a column or strut during the pour.

BA

 

[Bridge100]

BA. The 10 kip block is used to compare to the vertical load of concrete. I still feel that I need to treat the lateral load separately. If we only considered pressure as a function of depth and unit weight then I could never reduce the pressure due to any amount of initial set - right?

Let's say that the concrete sets up in some amount of time that causes the form to see a max. pressure of 1000 psf. Would you then apply only 1000 psf normal to the bulkhead all the way up?

Why wouldn't the concrete function as a support? If you poured a plumb column, certainly the concrete at the lower end of the column is experiencing the most load, which never diminishes with concrete setup. However, the lateral form pressure at this same location will not increase.

I appreciate all of your time on this matter. I hope to put an end to this subject soon!

 

[ishvaaag]

Just a publication from APA dealing with concrete forming.

 

[BAretired]

Bridge100. With a plumb, prismatic column, the base pressure is 150*H psf where H is the height of pour at any time. It doesn't matter how fast or how slowly you pour it, it remains constant. Even after the concrete sets for 28 days, the base pressure remains the same. Slow pouring changes only the lateral pressure on the bulkheads.

With the tapered column, the central section is prismatic and the pressure on the pier is 150*H. That must not change throughout the pour and until the column has completely set. After the forms are removed, the weight of the two triangular sections will be added to the bearing pressure on the pier.

Place a sloping beam under each inclined form and parallel to it. Hinge both beams at the bottom and tie them together at intervals with horizontal ties. If the weight of each triangular part is W, each sloping beam must carry an axial load of W/sin70 = 1.06*W.

The tension in the horizontal ties depends on the rate of pour. If the entire column is poured in a short time, the ties must resist liquid pressure. The sum of all tie tensions will be 120,000#.

Pouring more slowly will reduce tie tension but the the weight W remains constant, so the axial load in each beam is still 1.06*W. The sum of tie tensions cannot be less than W*tan20.

I agree it would be nice to put an end to this topic, but we have to get it right first.


BA

 

[doka1]

I would start refer to MK Hurd guide to formwork, best book out there on form design. BA is the closest to comprehending whats going on, but does need to consider the factors in mk hurd's book on rate of placement and different types of mixes.

 

[BAretired]

doka1,

There are many aspects about formwork design about which I am ignorant. Pressures resulting from different rates of placement or different types of mixes are outside my area of experience as I have never designed formwork during fifty four years of practice.

One thing I do know is this. Fresh concrete must be supported by forms until the concrete has attained sufficient strength to act as a structural unit. Until then, it is not safe to assume that the fluid concrete will possess properties which permit the material to act in a structural way, i.e. to exhibit shear strength or bending strength, even though we suspect it has some of these capabilities.

BA

 

[Bridge100]

doka1,

Does MK Hurd's book touch on the subject of inclined forms? I do not have this book but I would purchase it.

I have talked to 2 other formwork companies about this subject. One sees my point but has still designed as BA is recommending. The other designs as I'm suggesting; no axial load is transferred to the bulkhead unless you consider some friction coefficient between the concrete and form. If there is no friction between the liquid and the inclined wall then there can be no axial load caused by the self-weight of the material above. I will include a model of this case tomorrow when I get to work. The model does not permit shear transfer between the supported material and the inclined support.

BA,

I understand your example of the inclined beam under the form which is pinned at the bottom and tied along the remaining length. The loading would add axial load to the beam and tension to the ties. I would assume that your weight W is a vertical load and then the axial load should be W*sin 70.

What if you kept your linearly distributed vertical load on your beam and then added another inclined beam under and parallel to it? Connect the 2 beams with "compression only" members at any interval. Keep the pin connection at the base of the upper inclined beam and remove the horizontal ties from it. Now pin the lower inclined beam at the top and bottom. The compression will be in the top inclined beam while the lower beam has bending and some amount of reaction.

 

[Bridge100]

I have attached the file that I mentioned in my previous post.

The model only considers the self weight of the element. The lateral form pressure would be a separate component to be added and would be resisted by the form ties.

 

[doka1]

OK BA, I respect your experience, but let me give you some insight on formwork design. Mind you, I am a superintendent for a concrete construction company, although graduated with a civil engineering degree, and received my EIT, I don't actually perform design work. According to ACI 347, the basic lateral pressure formula is p=wh. Where p is the lateral pressure in psf, and w is the unit weight of the fresh concrete in pcf, and h is the depth of the fluid or plastic concrete from the top of placement. But ACI goes on to create a basic pressure formula, p=CwCc(150+9000R/T). Where r is the rate of placement in ft per hour, and T is the temperature of concrete during placing. That is from MK Hurd's book, where is does make mention of additional considerations on uplift pressure on sloping surfaces, but does not go in to detail on the calcs. Wouldnt you just take the pressure and transfer the lateral pressure to a vector normal to the surface like you posted originally(again I havent done any design work in 10 years). From there wouldnt you just size up the required hold down and spacings?

I just reread your original post Bridge100, your contractor seems pretty competetent as he will be rodding the concrete. Will they be inside the column vibrating as the go up or will they have external vibrators?

 

[BAretired]

If there is no friction between the liquid and the inclined wall then there can be no axial load caused by the self-weight of the material above.

Not true! The ties are horizontal. They put axial load in the bulkhead because they are not normal to it.

I would assume that your weight W is a vertical load and then the axial load should be W*sin 70.

Not true! The axial load in the bulkhead must be W/sin70 because W is resolved into two components, one horizontal, the other parallel to the bulkhead.

BA

 

[Bridge100]

BA,

I agree that the ties would exert an axial load on the form - that is not the problem. A load applied normal to the surface will in turn create axial load through the horizontal ties because the ties have a positive connection to the form. However, my ties will actually be installed normal to the bulkhead since I have an inclined column. Given that, the form ties will not exert an axial load on the bulkhead.

I have a problem with your component that is parallel to the bulkhead. I continue to ask you this question - how does this parallel load add axial load to the bulkhead? Could you please address this question?

 

[psmaxtor]

BA,
I seem to disagree with your assumption that the axial component is W/sin 70. Axial component should be W*sin 70 as Bridge 100 has insisted. And if dealing with a uniform load say (x) then axial load would be (w)sin^2 70 as was discussed in bridge 100 first post.

When dealing with concrete there are two load types that need to be addressed. One is concrete load and other fresh concrete pressure)
For the concrete load acting perpendicular to the form the following is valid:
The first cosine of the angle you take into account because of the trigonometry.
With the second cosine of the angle you consider the fact that the inclined length where loads is longer than the horizontal length.

The fresh concrete pressure also acts perpendicular to the form.
This also has to be resolved into components based on geometry and length. This load in Bridge 100 case will be taken by the ties positioned perpendicular to the form face. They will only take on the perpendicular component of pressure.

Both the concrete weight and pressure components can be resolved into parallel components acting along the form face. It seems to me this is what Bridge 100 is trying to ask. If concrete behaves as a fluid and no shearing stresses are present how can this force be transferred into the form as he indicated in his sketches.

So the question remains where does this axial/parallel load go.
I agree with the assumption that the pier footing takes on this load. Imagine if the pier footing were not there and the form was held in space. The concrete would slide out the bottom since there is no resistance to this axial component.

PS

 

[BAretired]

Bridge100,

From the beginning, I have assumed you would be tying the inclined bulkheads using horizontal ties. Now you are telling me you will install them normal to the bulkhead.

The attached sketch indicates four ties near the bottom and four ties near the top. If this is not how you are designing it, then please show me where your ties are located.

BA

 

[SAIL3]

Looks like this subject still has some life left in it.
To make this simple for myself:
Assume conc. is in liguid state.
Assume two vert. sides opposite each other and two sloped sides
opposite each other.
Liquid conc. can ONLY exert pressure load normal to all sides.
This press. load causes a shear in the formwork that is normal(perpendicular)to the surface.
At the intersection of the sloped side and the vert side(corner),this shear in the sloped side is resolved into two components,one vertical and one horizontal.
The vert. component goes into the vertical side causing compression in that side.
I believe that when the conc. is in the liquid state that this
is the mechanism that results in a compression force in the formwork, varying from max at the base to zero at the top.
Ofcourse there is also a horizontal tension in all of the sides that all seem to agree on.
BA had it correct many posts back, but the subject since then has
become muddled by jumping back and forth between conc. in a liguid
form, semi-solid conc, and solid conc.

 

[BAretired]

The attached sketch shows the statics of the forces acting on the shaded triangular chunk of concrete assuming it is in the liquid state.

BA

 

[BAretired]

Sorry, I made a slight error in that last post...F_tie is the sum of the horizontal tie forces from top to bottom.

BA

 

[BAretired]

Okay, now let us consider a solid concrete wedge supported by four roller bearings as shown in the attached sketch. We assume zero friction between the concrete and the inclined bulkhead (a moot point).

Whaddya know! It's the same diagram we had before. The axial force in the inclined bulkhead is still W/sin[α].

BA