Smart questions
Smart answers
Smart people
Join Eng-Tips Forums
INTELLIGENT WORK FORUMS
FOR ENGINEERING PROFESSIONALS

Member Login




Remember Me
Forgot Password?
Join Us!

Come Join Us!

Are you an
Engineering professional?
Join Eng-Tips now!
  • Talk With Other Members
  • Be Notified Of Responses
    To Your Posts
  • Keyword Search
  • One-Click Access To Your
    Favorite Forums
  • Automated Signatures
    On Your Posts
  • Best Of All, It's Free!

Join Eng-Tips
*Eng-Tips's functionality depends on members receiving e-mail. By joining you are opting in to receive e-mail.

Posting Guidelines

Promoting, selling, recruiting, coursework and thesis posting is forbidden.
Jobs from Indeed

Link To This Forum!

Partner Button
Add Stickiness To Your Site By Linking To This Professionally Managed Technical Forum.
Just copy and paste the
code below into your site.

Bridge100 (Civil/Environmental) (OP)
25 Jan 11 10:43
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?
 
BAretired (Structural)
25 Jan 11 12:41
If you consider concrete as a liquid with unit weight of 150 pcf, the pressure is 150*H where H is the head.  The pressure acts normal to any surface.

If the rate of pour is slow enough to allow the concrete in the lower part to attain initial set before the upper part is cast, the form pressure will be reduced by a factor which depends on the rate of pour.  I don't know what that factor is, but if the contractor can determine when initial set has taken place the pressure on the sloping form should be no different than that of a vertical form, should it?

A book on formwork would be a good source of information, but in any case, a little conservatism seems warranted.

BA

Bridge100 (Civil/Environmental) (OP)
25 Jan 11 13:30
So you believe that the majority of the concrete dead load above the bulkhead is acting on the lower stem more than the bulkhead itself?
BAretired (Structural)
25 Jan 11 19:32
If the concrete remains liquid throughout the pour, the weight of concrete acting on the bulkhead is the full weight of the shaded triangle on your lower diagram.  This must be transferred to the pier by compressive stresses in the bulkhead varying from 0 at the top to maximum at the bottom.

If the concrete has partially set up in the lower part when the pour is completed, some of the load will be transferred to the pier by shear in the set concrete.  The pressure on the bulkhead at any given time will depend on the rate of concrete placement.

BA

jberg (Structural)
25 Jan 11 21:38
BA,

You are correct about the rate of placement affecting the lateral pressure on the formwork.  Attached is a link to a Dayton Superior forming design guide.  If you refer to Page 14 of the attached PDF, they list lateral pressures based on the rate of concrete placement.

Hope it helps!

JWB

BAretired (Structural)
25 Jan 11 23:25
Thanks, jberg that is a big help and not something that could be calculated from scratch.

With a height of 40' and the requirement that it must be poured within a day, it would be necessary to place concrete at a rate of 40/24 = 1.67 (say 2) feet per hour.  Assume a temperature of 500F.  

Then P = 150 + 9000*R/T = 150 + 9000*2/50 = 510 psf maximum.

But this would run into overtime.  If the whole pour had to be completed in 8 hours, R = 40/8 = 5 feet per hour.  Then, P = 150 + 9000*5/50 = 1050 psf max.

If the contractor really wants to speed things up, let us say he wants to complete the pour in 4 hours.  Then R = 40/4 = 10 feet per hour.  Now, with this high rate of pouring, we have a different formula for pressure, namely:
P = 150 + 43,400/T + 2800*R/T
  = 150 + 43400/50 + 2800*10/50 = 1578 psf max.

The rate of placement will affect the pressure, P but whatever value we assign to P, it must be applied normal to the bulknead.  It is not necessary to consider the vertical and horizontal components separately.

The cost of additional formwork pressure versus savings in time are factors for the contractor to consider.  

BA

Bridge100 (Civil/Environmental) (OP)
26 Jan 11 13:57
BA,

You stated earlier, "This must be transferred to the pier by compressive stresses in the bulkhead varying from 0 at the top to maximum at the bottom." This is the main point of my question: How can there be a compressive force in the bulkhead if the loads are applied normal to the bulkhead form face without applying a shear force?

I am breaking my loads into vertical and lateral components because I feel that the vertical load cannot be reduced regardless of the adjustment to the rate of pour, unless we consider the shear strength of the initially set concrete. I would not dare do that! Only the lateral component can be adjusted based on rate of pour.

One additional item of infomation that I should have mentioned before is that the pier is a mass concrete pour. All guidelines that I have seen recommend full liquid head design; however, the contractor has seen setup in previous jobs by using tamping rods and load cells within the form ties to guage pour rate.
 
Helpful Member!  hokie66 (Structural)
26 Jan 11 15:03
BA is correct.  Pressure by concrete is in all directions.  The pressure should be taken normal to all surfaces, and it is not necessary to resolve the pressure into components.

A separate issue is that the formwork must be able to support the gravity weight of the concrete globally between the formwork supports.

Don't confuse or combine the two loading conditions.
BAretired (Structural)
26 Jan 11 19:26
Bridge100,

Quote:

How can there be a compressive force in the bulkhead if the loads are applied normal to the bulkhead form face without applying a shear force?
I assumed that your walers would carry only horizontal forces, leaving the bulkhead to provide static equilibrium.

Your "column" is more like a trapezoidal shaped wall.  I am assuming that the trapezoidal sides will have walers and snap ties through the wall to hold the bulkheads in place.  Say that 'S' is the waler spacing and 'W' is the width of column or wall.

The sloping bulkheads will have short walers spaced at 'S' vertically (S/sin70 diagonally) and held in place horizontally by the long walers which act in tension.

The sloping bulkhead normal force per waler set is F = P*W*S/sin70 where P is the concrete pressure.  This can be resolved into two components, one horizontal, the other, parallel to the bulkhead.  So the tension carried by the two side walers is P*W*S/sin270 = 1.132P*W*S and the axial force added to the bulkhead is P*W*S*cot70/sin70 = 0.387P*W*S.

BA

Grahammachin (Structural)
27 Jan 11 8:25
I have a question regarding the pouring of the concrete itself rather than the formwork itself.

How is the contractor planning on pouring the concrete? Will he be pouring from the top so that the concrete will be hitting all the reinforcement on the way down? Or will he pouring from the bottom up through a tube which he will be pulling up as he fills the column?

I believe there is a rule of thumb about not allowing concrete to free fall more than 1.5-2m (5'-7') through a reinforced cage as this will cause segregation of the concrete.

Also is the contractor going to guarantee that the concrete at the bottom of the formwork will be sufficiently compacted so as to minimise the risk of honeycombing at the bottom of the pour?
BAretired (Structural)
27 Jan 11 10:57
Another option to avoid segregation is to pour concrete through "elephant trunks", a flexible hose made of canvas which prevents the aggregate from colliding with reinforcement.  This technique is sometimes used when pouring deep piles.

BA

Bridge100 (Civil/Environmental) (OP)
27 Jan 11 15:23
Grahammachin - The concrete will be placed with a pipe from within the form. The pier form is plenty wide to fit crew members inside to work the concrete from bottom up.

BA - I can't figure out your last 2 equations. Do you have a reference that you can provide? I am still confused as to how we have an axial load on the bulkhead. I have attached another sketch to try to help clear up my question.
BAretired (Structural)
27 Jan 11 16:27
Bridge100,

I don't have a reference offhand, although I might be able to find one.  The principal is essential for you to understand if you are designing formwork.  Let me try again and if anyone else can explain it more clearly, please join in.

Suppose the elevation on your attachment shows an aquarium built with glass walls.  The triangular area shown hatched has to be carried by the sloping wall.  Its area is H2tan20 = 0.364H2 where H is the height of aquarium.  The volume of water above the sloping wall is 0.364H2*W where W is the width of tank.  That load must be carried entirely by the sloping wall because there is nothing else available to carry it.

The axial compression at the bottom of the sloping wall must be:

0.364H2*W*γ/sin70 = 0.3870H2*W*γ where [γ0 is the unit weight of water.

If you consider any vertical section within the shaded region, suppose the depth of water at the section is z.  Then the axial load in the sloping wall at depth z is 0.3870*z2*W*γ.

Are you okay with that, so far?

BA

Bridge100 (Civil/Environmental) (OP)
27 Jan 11 17:52
BA,

One minor correction on the area equation should read: A=.5*H^2*tan20.

Besides that, I would agree with your final equation for the axial load.

Go on....
BAretired (Structural)
27 Jan 11 19:25
Bridge100,

You are correct about the factor 0.5 for the area of a triangle.  So the axial compression, C at the bottom of the inclined form is:

C = 0.5*H2*W*γtan20/cos20
  = 0.5*402*W*150*0.387 = 46,500W#

That is a very substantial compression.  It is based on the assumption that the concrete remains fluid during the entire pour and also that the side forms are capable of resisting horizontal tension only where they connect to the end forms.

If the contractor adjusts his rate of pour, the lateral pressure and hence the axial force will be reduced accordingly.

BA

doka1 (Structural)
28 Jan 11 17:29
When we form battered walls we stab coil rod into the footing to have somewhere to hold the form down.  Columns are tricky because people like to fill them up quickly, but if they take there time and rod the concrete they wont have any problems.  Rodding is the method by which you can determine how the concrete lifts are setting up.  Formulas are great for the initial design, but the contractor should follow this procedure.  Concrete sets up differently depending on the temperature and admixtures. Not sure what MK Hurd's book has in reference to battered wall formwork design, but there are adjustment factors for temperature etc..  But your typical batter is 1:12, with a coil rod at 3/4" spaced at like 3' horizontally (conservative), that would be a good place to start calculations.  The weakest part of the system is the attachment point on the forms.  
cap4000 (Civil/Environmental)
29 Jan 11 10:11
I have done a battered concrete dam holding back water which the structural analysis is similar. I used 10 foot concrete pour based on 150pcf. That 10 feet must set and then another 10 feet could get poured. I get the two force components based on 1 foot width of formwork Fy= 10.9k and Fx = 30k. The resulatant is 31.9k. with the angle between the component as 20 degrees which checks with your 70 degrees. Clearly a lot of load must be ditributed to the formwork and or wall ties. Good Luck.
SAIL3 (Structural)
29 Jan 11 15:31
Agree that press. load is normal to all surfaces. If consider the
conc. as a liquid, then by what mechanism does this axial load
end up in the sloped side, since there is no friction between the
sloped side and the conc.
Suppose one has a liquid container shaped like a square box.
Case#1...All sides vert...fill with liquid.
         No axial load in sides, just bottom perimiter load.
Case#2...slope one side, say 60deg...fill with liquid.
         sum up reactions at base...results in unbalanced moment.
         The two sides(closest to the sloped side) react this
         moment in bending. This results in a stress distribution
         of tension in portion furthest away and compression
         in portion closest to the sloped side. The sloped side
         will have hoop tension and shear normal to the face.
         
I could be all wet on this one(excuse the pun).
           
            
BAretired (Structural)
29 Jan 11 16:52

Quote:

Case#1...All sides vert...fill with liquid.
         No axial load in sides, just bottom perimiter load.
Neglecting wall dead load, axial load in walls = 0 vertically, variable horizontally, uniform water pressure on container base.  

Quote:


Case#2...slope one side, say 60deg...fill with liquid.
         sum up reactions at base...results in unbalanced moment.
         The two sides(closest to the sloped side) react this
         moment in bending. This results in a stress distribution
         of tension in portion furthest away and compression
         in portion closest to the sloped side. The sloped side
         will have hoop tension and shear normal to the face.
Pressure on base remains same as Case #1.

All four walls act like a hollow section of varyiable depth to carry the additional eccentric mass of water.  Sloping wall acts in compression along slope.  Opposite wall acts in vertical tension.  Side walls act in flexure with variable vertical stress.  All walls carry horizontal tension.

BA

BAretired (Structural)
29 Jan 11 17:03
The container in Case #2 does not perform quite the same as formwork.  The walls and base of a container are connected together to form one rigid body.  The side bulkheads of this thread are constrained to resist only horizontal tension from the inclined bulkheads.

BA

SAIL3 (Structural)
29 Jan 11 18:28
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 (Structural)
29 Jan 11 20:40
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 (Civil/Environmental) (OP)
31 Jan 11 11:34
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 (Structural)
31 Jan 11 23:07
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 (Civil/Environmental) (OP)
1 Feb 11 0:06
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 (Structural)
1 Feb 11 12:40
Just a publication from APA dealing with concrete forming.
BAretired (Structural)
1 Feb 11 12:43
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 (Structural)
5 Feb 11 20:10
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 (Structural)
5 Feb 11 21:39
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 (Civil/Environmental) (OP)
7 Feb 11 20:22
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 (Civil/Environmental) (OP)
8 Feb 11 9:02
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 (Structural)
8 Feb 11 17:59
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 (Structural)
9 Feb 11 11:37

Quote:

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.

Quote:

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 (Civil/Environmental) (OP)
9 Feb 11 13:42
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 (Structural)
9 Feb 11 14:43
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 (Structural)
9 Feb 11 15:50
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 (Structural)
9 Feb 11 16:35
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 (Structural)
9 Feb 11 21:43
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 (Structural)
9 Feb 11 21:49
Sorry, I made a slight error in that last post...Ftie is the sum of the horizontal tie forces from top to bottom.   

BA

BAretired (Structural)
9 Feb 11 22:27
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/sinα.

BA

BAretired (Structural)
9 Feb 11 23:26
Oops, I didn't attach my sketch, so here it is.

BA

Bridge100 (Civil/Environmental) (OP)
10 Feb 11 12:39
BA,

I like the sketch that you included in your last post - it added a new perspective to the solution.

From your sketch, are you saying that there is an axial load on your inclined form because of the load or because of the tie? This is the most important question. You show a tie component in the lower sketch but the upper sketch does not show any ties; it only shows reactions normal to the inclined form.

I have included some analysis of my own based on your wedged mass. The analysis on the left is similar to your sketch. I created a triangular concrete element that is 10 Ft Tall by 3.64 Ft Wide 1 Ft Thick. The summation of the reactions on the inclined form makes sense. I used 3.64 Ft to keep the 70 degree incline.

The total weight of the element is (0.150 kcf)(0.5)(10')(3.64')=2.73 kip

The summation of the inclined reactions = 2.73/cos 70 = 7.98 kip or 3.64+4.34=7.98 kip.

The second case uses a linearly distributed load placed normal to the surface. This case uses full liquid head of (0.150 kcf)(10') = 1.5 k/ft.  The summed reactions of the inclined forms are agreeable between cases 1 and 2, although the individual reactions differ greatly due to the varying locations of the load centroids.

The pressure within the form will be constant at any given depth and will act normal to any surface if we are considering a fluid. I would like to refer back to my attachment from my 8 Feb 11 9:02 posting. The pressure at the bottom left is the greatest, not uniform. I would attribute this to the initial setting of the concrete. The maximum pressure from the result view shows a maximum pressure of 120 psi. If the concrete could support 120 psi well before it actually experienced this loading, then the forms will not see this value applied because the concrete is strong enough to support it. If not, then the forms will experience 120 psi or 17,280 psf which is well beyond the 1200 psf design pressure that many forms are designed for.
 
Bridge100 (Civil/Environmental) (OP)
10 Feb 11 12:42
Bridge100 (Civil/Environmental) (OP)
10 Feb 11 12:58
I would like to make one more point. I have attached a sketch of an external frame. If the force on the bulkhead is normal only and does not apply an axial load, as I am arguing, then the inclined member of an external bulkhead support frame could actually experience an axial tension force from the overturning forces on the frame. Let's say the form ties (not shown) are installed normal to the inclined form. They do not support any self-weight of concrete - they only help to hold down the other form on the top of the inclined concrete pier stem.
BAretired (Structural)
10 Feb 11 15:17
Bridge100,
  
As shown in your latest sketch, the statics of this problem are very dependent on the method chosen to support the inclined form.  You show a framed retaining wall which develops tension in the inclined form and compression in the vertical member.  Under liquid pressure, the moment about the toe is readily calculated as are the three forces shown.

Under liquid pressure, the total force on the bulkhead is normal and does not stress the bulkhead axially.  If the inclined bulkhead could be held in position in some way, it would have only normal pressure to resist but the vertical component of that normal force would precisely equilibrate the liquid weight.  The forces acting on the bulkhead supports depend on the design selected.

If we consider modifying the liquid pressure as recommended by M. K. Hurd, the horizontal pressures are reduced but the vertical pressure remains the same.  Assume your client pours at a rate of 5 feet per hour at a temperature of 50F.

At depth z, the form supports a pressure of 150*z vertically but only 1050 psf laterally if z > 7'.  In the top 7', the pressure varies from 0 to 1050 psf, a substantial reduction from liquid pressure.  This greatly reduces the moment about the base and results in much smaller forces for the bulkhead support structure.

BA

Bridge100 (Civil/Environmental) (OP)
10 Feb 11 15:51
BA,

So if we have a 40' Column on a 20 degree incline and 1050 psf lateral load, what would you say the maximum normal load is to the form at the lowest point? I come up with 1630 psf. What do you think? I think we're almost done here!!
BAretired (Structural)
10 Feb 11 16:32
Sounds right to me, Bridge100.

BA

psmaxtor (Structural)
10 Feb 11 17:22
BA,
Appreciate your posts/sketches.  I agree that concrete axial force does not transfer thru bulkhead form. Your vector method and Bridge 100 cos^2 method produces the same results.

One last thing..Looking at the liquid Concrete sketch you provided the Normal force (N) shall read:
150h^2/(2 sin alpha).  Not (2 cos alpha).

PS
Bridge100 (Civil/Environmental) (OP)
10 Feb 11 23:11
From what I have seen, we need full liquid head applied normal to the inclined form to account for the full weight of the concrete above. If you reduce the lateral pressure to a value less than that figured from the full depth, then we would not see the full load of the concrete on the form. Does everybody agree?
BAretired (Structural)
11 Feb 11 11:16
psmaxtor,

You are correct about the value of the normal force.  Thank you for pointing that out.

Bridge100,

What would happen if, instead of an inclined bulkhead, you used a stepped bulkhead with risers of 12" and treads of 12tan20 = 3.5"?  Would you not get 1050# pushing against each riser and the full head of concrete bearing on each tread.

Is the assumption of zero friction between concrete and inclined form valid, or is it possible that the bulkhead feels axial load resulting from friction.

I just don't feel comfortable saying that freshly placed concrete can be expected to take substantial shear stresses before it has reached its final set and has had time enough to make strength gains.

BA

Bridge100 (Civil/Environmental) (OP)
11 Feb 11 12:26
BA,

I agree that if you used a stepped bulkhead as you describe that the concrete pressure would bear normal to the riser and tread with the lateral and vertical forces. You would have axial load on the bulkhead from these forces.

I share your same concern about the zero friction value between the concrete and the inclined form. However, if one did assume some amount of friction, then the friction force would be the result of the friction coefficient times the normal force to the form. The friction coefficient should be small, if any. I have seen an example of a grain bin design with sloped walls where the grain pressure acts normal to all walls of the bin and the friction force is due to the coefficient of friction between the grain and the wall material.

I am currently considering the following for the pour:
1) Smaller, plumb piers will be poured first with the same concrete mix. This will give us a chance to study the concrete strength.
2) The crew will rod the concrete as the pour advances.
3) We will install load cells in line will several of the side form ties to verify the expected pressure.
4) We can take early concrete cylinder breaks to check concrete strength. If strength is lacking, we will have quite the pour and use a construction joint half way up.
BAretired (Structural)
11 Feb 11 13:49
B100,

I found the following link on the coefficient of friction between wood and concrete, but I imagine they are talking about hardened  concrete on non-oiled wood.

http://www.engineershandbook.com/Tables/frictioncoefficients.htm

As you can see, the coefficient for static friction is listed as 0.62 but I'm not too sure whether this applies to your situation.

This is an interesting question, one that I have never really thought about prior to this thread.

BA

BAretired (Structural)
12 Feb 11 2:17
It might be useful to perform a test on recently placed concrete on a horizontal form to determine the coefficient of friction between the two materials.  The test should be relatively easy to perform.

BA

BAretired (Structural)
13 Feb 11 11:52
After further thought, I don't think friction is critical.  The attached sketch shows a bulkhead with a hinge at the bottom and a horizontal reaction as shown at height 'a' above the hinge.  The choice was arbitrary.  The bulkhead could be supported in a variety of other ways.  

Reactions are shown.  Axial force, shear and bending moment can be calculated at any point along the bulkhead.

Adding a multitude of horizontal ties instead of one changes the bulkhead to a continuous beam instead of a simple beam with cantilever.

BA

Reply To This Thread

Posting in the Eng-Tips forums is a member-only feature.

Click Here to join Eng-Tips and talk with other members!

Close Box

Join Eng-Tips® Today!

Join your peers on the Internet's largest technical engineering professional community.
It's easy to join and it's free.

Here's Why Members Love Eng-Tips Forums:

Register now while it's still free!

Already a member? Close this window and log in.

Join Us             Close