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Shear Understanding

  1. Jul 13, 2012 #1
    Guys when beam like this subjected to axial load only,it will create shear forces if we take a section inclined at an angle
    and this make since ,however when we design we don't look at inclined sections ,we just
    draw axial force diagram and we notice that there is not shear diagram , so this implies no shear force in section,however there is shear force at inclined surfaces
    So wht we dont consider it in design
    I'm really confused.
  2. jcsd
  3. Jul 13, 2012 #2
    Have you seen a sample of steel tested in tension, or concrete tested in compression? In both cases, the axial load failure surface is characterised by being inclined to the member axis, illustrating that shear stresses are working as part of the mechanism of failure. In everyday design practice, this is simplified to be axial stress=P/A where the area A is perpendicular to the applied load, and the permissible stress related to test results calculated in the same way, even though the local failure is one of shear. In research or specialised design practice (such as some aspects of prestressed concrete design) it might be necessary to approach the issues more fundamentally in the way your question is indicating.
  4. Jul 16, 2012 #3
    Thank you brother for the explanation.
    I think this also related to principle axes of maximum stress (axial and shear) that we study in mechanics of materials ,and as you said in concrete design they don’t look deeply in this, they just consider axial for vertical plan and 45 degree inclined for shear, which is correct if the beam is subjected to pure axial or pure tension or compression, while in reality they are subjected to mix of them ,which implies that they are neither vertical nor 45 degree. However I think that this still acceptable and provide reasonable analysis.

    That What I think

    Am I right Sir??????????, or there is something wrong in my understanding.

    Thank you another time
  5. Jul 16, 2012 #4
    I think you are ok. Conclusion? An axial force can be considered to have shear and normal components at some section other than the section at right angles to the member axis. This can be used to find explanations of failure surfaces in non-homoegenic materials (which all are in reality), or in specialised design situations, especially where members are not uniform in their cross section. Designers of conventional everyday buildings will not need this. Designers of nuclear facilities, for example, will want to take a more considered approach.
  6. Jul 19, 2012 #5
    Thank you
  7. Jul 26, 2012 #6
    Originally Posted by pongo38

    "Also in concrete beam ,we put stirrups for vertical shear so what about longitudinal shear that come from equilibrium(no reinforcement)"
    The stirrups are not for vertical shear. What vertical stirrups do is resist the vertical component of (approximately) diagonal shear in tension. The diagonal shear in compression is resisted by the concrete. From a practical point of view, the stirrups help keep the whole reinforcement cage together during the concreting and vibration process of manufacture. So why don't we use diagonal stirrups? Partly for the practical reason just explained, but also because it has been shown that diagonal bars - which used to be much more common - are only fully effective when used in conjunction with vertical stirrups.

    Thank you very much, i read about this subject and understand it, but still something not clear, you said that failure occur from diagonal tension, so why equations use Vu in calculation ,Also, We say Vu=vs+vc, vc implies interlock in concrete and vs steel stirrup in tension which hold system togethor,so maybe i guess this will increase vc ,so this why we sum vc and vs,,Right!!!!

    The second question, we know that fail occur due to indirect shear, so if direct shear occur
    How much concrete can handle.

    Please expalin in details to end this misunderstanding
    Thank You another time
    Last edited: Jul 26, 2012
  8. Jul 26, 2012 #7
    "Also, We say Vu=vs+vc, vc implies interlock in concrete and vs steel stirrup in tension which hold system togethor". Shear in rc in complex and the statement you made is over-simplified. In the first place, there is a significant contribution to shear resistance from the longitudinal steel, called dowel action. Secondly, the compression zone in the concrete arising from bending action has a significant increase in shear resistance just due to the longitudinal compression (like picking up a row of books from a shelf by squeezing the ends - the analogy often used to demonstrate how prestressed concrete works. Agregate interlock in the failure zone can make a contribution, but the fact is that all these resistances can't add up (like vs+vc) because they appear and disappear at different stages of the failure process. If you are trying to relate all this to first explanations of shear in beams of uniform properties, as found in basic strength of materials texts, then you need to remember that reinforced concrete is anything BUT isotropic and uniform. Even the variation in vibration of the concrete in suitable pockets available in the reinforcement cage is a variation of E. To get a better understanding, you need to read the research papers on which the codes of practice are based (and I don't have them to hand right now).
    Your second question "we know that fail occur due to indirect shear, so if direct shear occur How much concrete can handle." I don't understand the question. If you are referring to unreinforced concrete, you still have a material that is not uniform and not isotropic. As with rc, it is better for design purposes to do lots of tests and draw conclusions from results that envelope all the variables. I don't think elementary mechanics of materials is applicable to these problems.
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