Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Stress in Reinforced Concrete Column

  1. Apr 24, 2010 #1
    When we calculate the stresses in the section of a Reinforced Concrete column reinfored with Steel Rods, it is seen that the percentage Load carried by the Steel is much greater than the percentage area it contributes in the cross sectional area of the reinforced concrete section.

    The load taken by steel is much greater than what it should have taken as per its area (As) in comparison to the area of concrete (Ac) where total area (A) is : A = As +Ac.

    And if load applied is P so Ps/Pc should have been equal to As/Ac if we follow the general rule of load distribution as P/A.

    Anyways further on , the equations used are :

    σs/σc = Es/Ec , where σs and σc are the stress induced in steel and concrete respectively and the Es and Ec are the Modulus of Elasticity of Steel and concrete respectively.

    This means that the stress induced in steel is much larger than the stress in concrete. It means that the load taken by steel is still much greater than that taken by concrete ( in comparison to its percentage area relative to concrete area) as the value of Es is much greater than Ec and also as the area contibuted by steel rods in the section as a whole is much smaller.

    And the reason expalined for this is that : Steel having greater Modulus Of Elasticity ( so greater strength) takes a larger ratio of the applied load as compared to concrete.

    Bu then if we have a picture of load distribution over a surface as say the section being divided in virtual small square units each of a unit area say 1 cm2 then the load acting on each square would be Total Load (P)/Total Area (A), and this would be the load per unit area ie the stress σ.

    Then as per the theory of reinforced sections as mentioned with the
    σs/σc = Es/Ec equation, it suggests that the load is not distributed accrding to the above mentioned criteria ( of uniform distribution per unit area), stating that steel being stronger tkes up greater load as a virtue of its Material Strength.

    So, I could never clearly understand HOW THIS HAPPENS?
    The Steel does not have a mind of its own neither does the load so as to think and know that steel is stronger, so it( the load) has to enter in larger proortions in steel and less in concrete or neiher the steel has the ability to think and attract larger load portion onto itself bein stronger.

    You can compare this with an analogy of the load being like unifromly distributed Iron grits over the cross section of the reinforced beam, and the steel rods being like magnets which will automatically attract larger amount of Iron grits to itself on account of its larger magnetic strength as compared to concrete ( which is not magnetic at all) thereby having a larger concentration of grits on to steel just like larger stress induced in steel ( this analogy is to point out the question - doubt I have)

    So HOW DOES STEEL TAKE UP GREATER LOAD just almost INTUITIVELY ( such intution is not possible for non living objects!!!) So, HOW DOES THIS HAPPEN - WHATS THE PHYSICS BEHIND THIS UNPROPORTIONAL LOAD DISTRIBUTION.

    Please explain in detail as it will equally appply to Flitched Beams also.
     
  2. jcsd
  3. Apr 24, 2010 #2
    No it does not apply to flitched beams in the same way.

    When concrete sets it shrinks.
    If the setting concrete contains steel reinforcement it grips the reinforcement very tightly.
    Reinforcement is often surface patterned and shaped into hooks to enhance or increase this grip.

    So in order for there to be relative movement (slip) between the concrete and the steel the frictional force of this grip needs to be overcome.

    Reinforced concrete is designed on the 'no-slip' principal. Since there is no slip between the steel and concrete the strains are equal. This condition isused in calculation.

    In force terms, some (most) of the load is transferred from the concrete to the steel, by the frictional grip (also called bond) forces. These appear as shear forces and stresses.
     
  4. Apr 24, 2010 #3
    Didnt quite get it, can you please explain it - specially how exactly is force transfered through bonds that too frictional force ( nearly cohesion)

    And also the forces in steel are either tensile or compressive ( depends on the type of loading it supports) and not shear , so can you alrify the shear part too.

    A clearer picture would be better.



    Then what is the principle on which flitched beams work, ( there too the steeel pplates take up a major proportion of the Bending Stress. An explaination can clear the concept, so can you please elaborate.
     
  5. Apr 24, 2010 #4
    In all fairness you did ask about reinforced columns. For members such as beams the gripping mechanism is the only one available.

    There is another mechanism in play when axial compressive stresses are considered.
    When you apply a compressive load to anything, there will be a first point of contact. This will temporarily carry the entire load. This point will deform plastically, allowing the objects to approach more closely. This will bring further points into contact, each of which will deform until the total contact area is capable of bearing the load.

    In my sketch 1
    I have shown this as a stiff plate loading three columns A, B and C. On first contact A will deform until contact is made with B then A and B will deform until contact with C and so on.

    In my sketch 2
    I have shown only two supports, but these are made of different materials and are flush with the loading platten.
    This business of being flush is important because when they are squashed down they suffer equal deformation i.e. equal strain.
    So they are squashed down to the dashed lines.

    But it take 15 times as much force to squash down steel as it does to squash concrete by the same amount. This is because the Young's modulus for steel is 15 times that of concrete.
    We say that the modular ratio is 15.

    strain in concrete = strain in steel

    By Hookes law

    [tex]\frac{{Stres{s_{concrete}}}}{{{E_{concrete}}}} = \frac{{Stres{s_{steel}}}}{{{E_{steel}}}}[/tex]

    rearrange

    [tex]Stres{s_{steel}} = 15xStres{s_{concrete}}[/tex]

    In columns both the gripping mechanism and this mechanism operate.

    In sketch 3
    I have shown that a structural member in compression has a tendency to bulge outwards

    In sketch 4
    I have shown how this tendency is counteracted in a reinforced concrete column. Steel links are placed around the vertical reinforcement rods. The bursting force is transferred to these links by the gripping mechanism mentioned in the previous post.

    Forces in reinforcing steel can be tensile or compressive but there will usually be shear forces as well.
    This is always the case in beams and often the case in uniaxially loaded members such as columns because they also support side loads, wind loads, bursting and buckling loads etc.

    Loads are transferred by what are known as shear connectors.
    In the case of flitch beams between timber and steel these are usually heavy duty bolts in shear but they may be cramp type connectors.
     

    Attached Files:

    • rc1.jpg
      rc1.jpg
      File size:
      9.7 KB
      Views:
      377
  6. Apr 24, 2010 #5
    It is evident from the condition of the reinforced section in practicality that the concrete and steel both undergo an extension or compression by the same amount, and that there is no slip between them.

    So quite visually we infer that the Strain is same in both, and hence apply the same strain condition This was known to me earlier, so thats not the issue.

    And also I always thought that the contact force ( or grip as you say) between the concrete and steel ( through bonding) must be stronger than the active loads since there is no slip between the two.Thanks for confirming it, so i was on the right path.

    And I've already mentioned the equal strian condition used and I do unnderstand that since the Modulus Of Elasticity of steel is greater than that of concrete so being stronger requires greater force for the same strain.

    Have a look at my first post :

    But I had asked you to explain how force is transmitted through gripping ( nearly like shear - can compare it to friction between two bodies almost adhered to each other , if you meant this) I cant grasp the fact that one body can transmit a force on the other through gripping -thats what I wanted to understand bette, so can you please explain it.

    Also for the flitched beams I had always wondered that there should be some technique by which the two ( wood and steel) are joined togther, I had screws in mind and bolts as you said is not far away, thanks for the insight. unfortunately such simple practical things are not at all mentioned in any textbooks covering the topic - they directly jump to solutions leaving the basic concepts.

    So can you kindly elaborate the gripping working.

    And thanks for the diagrams.
     
  7. Apr 25, 2010 #6
    OK I am going to assume you are some sort of Mechanical Engineer since this is in the ME section.
    Hopefully this means that you have some knowledge of stress analysis (often called mechanics of meaterials or strength of materials).
    It is important to understand the difference between the external loads on a body, which is normally the province of mechanics and conducted in terms of forces, and the internal responses of that body which is the province of strength of materials and conducted in terms of stresses.
    I going to further assume you have at least an elementary knowledge of the distribution of stresses in beams. There is no point discussing reinforced concrete without this.

    The answer to your question about flitched beams is that the connectors have to be able to transfer the horizontal shear that appears at any section of a beam.
    This means that the product of their total cross sectional area and the their shear strength must be greater than the shear forces acting at that section.

    Back to reinforced concrete, consider a steel bar, partly embedded in a block of concrete.
    It will take a force, P, pulling on the bar to pull it out from the concrete.

    I have shown this in sketch 1

    The only forces acting are the pull, P and the 'frictional' grip force, F.
    Since we want standardisation we measure this in terms of a pull out contact stress, u.
    This stress (as all stresses) is a force divided by an area.

    Unlike friction in normal mechanics this force is proportional to contact area.

    The contact area is the embedded length times the bar perimeter as shown.

    The pull out bond stress, u is a measured parameter. There is no magic crystal lattice theory to derive it. Safe values are given in design codes, as a result of many years of experience (measurements).

    This is analagous to hold the bar (or your finger) in a pair of pliers and pulling.

    So how do we use this information in a structure?

    In sketch 2 I have show a very simple beam with a single reinforcement bar in the bottom.

    It should be noted that the neutral axis is not midway up the section in such a beam. I have shown it dashed.
    All the concrete above the neutral axis is available to resist the horizontal compressive stress, brought about by the beam loads.
    For simplicity we assume that all the tensile stresses developed below the neutral axis are resisted by the reinforcement.

    In sketch 3 I have taken a typical vertical section, somewhere along the beam.

    In order to perform mechanical calculations, stresses have been converted to stress resultant forces. So the compressive stress resultant shown is actually the resultant of the compressive stress over the whole area of the compression zone.

    As shown we take a moment balance about the point M, for the forces acting on our block of length [tex]\Delta[/tex]L.

    Now the net force acting within the bar, over the length [tex]\Delta[/tex]L is [tex]\Delta[/tex]T.
    This is the value of the force P trying to pull out the bar from the concrete as in sketch 1above.

    This yields two equations.

    [tex]\begin{array}{l}
    V\Delta L = q\Delta T \\
    \Delta T = u\pi d\Delta L \\
    \end{array}[/tex]

    A bit of shuffling with these leads to the leads to the requirement

    [tex]\frac{V}{{q\pi d}} \le u[/tex]

    Now u is a measured quantity and V is the vertical shear at the section, as calculated by normal mechanics of loads.

    This is exactly the condition that has to be satisfied when designing rc structures.

    In fact rather than use the stress parameter u, it is more usual to quote the
    bond length. This is the length of bar that needs to be embedded to fully resist a pull out force equal to the strength of the bar (ie the max force allowable in the bar).
    Codes quote bond lengths for different types of bar, a good rule of thumb is 20 times the diameter.
     

    Attached Files:

    • rc2.jpg
      rc2.jpg
      File size:
      16 KB
      Views:
      289
    Last edited: Apr 25, 2010
  8. Apr 27, 2010 #7
    I am a qualified Mechanical Engineer so it is not some sort of Mechanial Engineer, though I may be lacking in experience.




    Dear studiot :

    My knowledge of elementary stress analysis and distribution (of UG level ) is sound enough and its certainly beyond the need of your scrutiny, and your need to assume or be hopeful about.



    Infact the question I have posed shows itself that I am looking for something beyond what is found in textbooks used by UG level courses which as I've already said just focus on relations derived without going any deeper or giving reasons for what is being applied.


    So instead of giving synonyms to the subject name ( honestly which doesnt make sense in the relevance of the question) please re-read my question previously asked.


    I had the idea that the gripping force ( due to the interatmic bonds between set concrete and steel periphery) is actually what is trying to prevent the pull out of the bar from withtin the column ( for a vertical column loaded vertically - so no bending considered).


    But what I wanted to understand was farther than this ( and I dont think many students would even desire to go this deep in understanding the theory)

    - I want to understand and visualise that how does interaction between the circumfrential layer of atoms ( atoms along the periphey of the bar) and the atoms of concrete surrounding it ( these two pairs actually form the bond between concrete and steel) which is in the form of shear or bond strength ( the V in your formulae) affect the normal stress (tensile or compressive) in the bar.
    And how does the steel end up carrying greater proportion of the load than concrete.

    What I actually want to visualise is how does the atoms respond to the tension being created and how the shear which again opposes the movements of the atoms of steel upwards due to the tension in normal direction in the zone of the bonds formation ( ie peripheral layer of steel and gripping layer of concrete).

    I cannot very clearly mention the exact question clearly in writing ( verbally it would be much much easier to explain what exactly I want to know - its easier to convey and easier for the listner to understand verbally ) but if you or anyone is interested I can certainly elaborate and much clearly write down the exact thing.

    So if you are interested to shed light we can continue.
     
  9. Apr 27, 2010 #8
    Some sort of Mechanical Engineer is complimentary not perjorative.

    Mechanical Engineering is a wide and august field, embracing amongst other things,
    rocket engineering, production engineering, plant engineering, aerospace engineering, pipeline engineering, automotive engineering, industrial engineering etc etc.

    In order to understand reinforced concrete you need to know the following

    1) Concrete exerts a triaxial compressive stress on the reinforcement embedded in it. It is this stress that is the source of the grip. It is purely mechanical.

    2) The term bond is used by civil and structural engineers to refer to this grip. This bond is not chemical and has no relation to the interatomic forces. Concrete does not chemically interact with steel, except insofar as it passivates the steel against corrosion.

    3) The effects of the triaxial stress within the steel can be analysed by normal continuum mechanics. You can find the principal axes and axes of maximum shear in the usual way. This is not usually done in structural engineering.

    4) It is a fallacy to think that there are no shear stresses within the reinforcement.

    5) The structural action of reinforced concrete cannot be analysed by the methods of continuum mechanics.

    I have no idea what you mean by this.
     
  10. Apr 28, 2010 #9
    Studiot,
    Do excuse me if I've mistaken and misunderstood your intent behind the statement regarding ME previously. I regret my reaction sincerely.


    Regarding the reinforced structure


    Yes I can understand that the concrete and steel would not react chemically, but ultimately any force exerted between two bodies ( even of different materials - as is the case here) should have the electrostatic forces acting between their atoms ( and due to the close proximity exsisting between them) as the only source for the origin.

    This can be compared to the Force of Friction between two bodies ( of different materials) which is built up only due to the interatomic ( dont confuse it with chemical reaction -- typical chemical bond formation) forces of attraction between the two bodies due to their extremely close contact with each other ( the influence circles overlap). Is this view correct.


    Secondly,


    Yes , thats the problem I addressed earlier, I had already said that its too messy to convey my exact idea through writing, it would be much much better and very simple to have spoken it out rather than writng it. Could we have spoken it would be pretty much easier to convey, and understand what I meant exactly.


    Anyways as I had mentioned earlier I could try make a effort to explain in a more clear manner if you are interested. The task to literally put the thing ongoing in the mind in exact words seems a bit difficult.
     
  11. Apr 28, 2010 #10
    I think it would be useful to step back and summarise the different ‘levels’ of treating the action of contact between two bodies where there is no chemical bonding by chemical action or welding.

    I would like to distinguish 5 levels, 3 macroscopic and 2 microscopic. Of course the methods appurtenant to each level must not be inconsistent with each other and there is some overlap. I think it confusing to mix up examples appropriate to any particular level. Perhaps that is what has been happening here.

    1) At the top level we have analyses by the normal methods of Structural Engineering. I think we have covered these in some detail. Flitched beams and reinforced concrete fit here. The analyses are normally in terms of forces and displacements.
    2) The next level uses the methods of continuum mechanics. The models are normally in terms of stresses and strains. Internal stresses distributions in homogenous and other materials are dealt with here.
    3) The third macroscopic level is called tribology or perhaps contact mechanics. It is still much coarser than the direct interactions between individual atoms, the quantities described being the aggregate effects of many atoms.
    I have shown a sketch of the analysis at this level of contact between a rough surface and a smooth one. The contact points are known as asperities and consideration is given to the stress states at these asperities in the light of failure criteria such as those of Tresca or Von Mieses.
    4) The first microscopic level considers the forces between individual atoms and here we move out of the province of the engineer and into the province of the materials scientist or physicist.
    This question is really about how objects touch and was asked here recently in another thread. No successful resolution was offered, or exists to my knowledge. It may be that by solving the energy equations for the approach of nuclei we can make a more elaborate theory than simply saying each atom transmits a push to the next but is it any better?
    5) At the level of subatomic particles we dispense with forces altogether and consider exchange or other exotic theories of interaction.
     

    Attached Files:

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Stress in Reinforced Concrete Column
  1. Reinforced concrete (Replies: 1)

Loading...