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Stresses in a concrete dome

  1. Sep 1, 2011 #1

    Jus

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    Hi there,

    I could really do with some help with this.

    First of all, this is NOT a homework question. It is a question that is listed in Benham & Crawford’s 'Mechanics of Engineering Materials' book. I am just reading through my books again and I'm stuck on this problem. As the book is copyright, I have included only the necessities. I wouldn't normally do this and I apologize in advance if I have done wrong but this problem has been bothering me for a week now!

    Please see the attached file.

    This may sound like a trivial question but I can't seem to see how the formulation of the first equation has been done. I understand part of the first part (force = stress x area, in which case the radial component of stress is 'sigma1 sin theta' and the thickness is 't'), but where does the '2 x pi x r x sin theta' come from? This is somehow the overall surface length of the dome but I don't understand how that has been derived. If I was looking to find the length of an arc of circle, I would use l = r x theta.

    Also, in the second part we '2 x pi x r x (r - r cos theta) x q'. Again, I know why we are multiplying by q as q is the force per unit area. But from what I have read in other books, the surface area of the dome should be '2 x pi x r x h' where 'h' is the height. This '2 x pi x r x h' may well have implications on the first part of the equation as well, but I just can't seem to see it.

    Please can someone give me some advice in this?

    Once again, I apologize as I'm sure that this is a trivial question for all you intelligent people.

    Thanks again,

    Jus
     

    Attached Files:

    • Dome.pdf
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  2. jcsd
  3. Sep 2, 2011 #2
    Good morning, Jus.

    Benham is usually good at explaining his working, more especially in his other books, but this short section is seriously abbreviated.

    He is following the membrane theory method of a few pages previous to your extract.
    The formulae he quotes in your example are the results of integration of differential elements in 3D.

    It is difficult to know where to start

    How comfortable are you with membrane theory?
     
  4. Sep 2, 2011 #3

    Jus

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    Hi Studiot,

    Thank you so much for replying.

    I've actually been going through the whole book again, page by page, refreshing my memory from my university days so I should hopefully be ok with membrane theory.

    I do understand the overall process of what we're trying to achieve, it's just a couple of bits in the equations that I can't seem to get.

    Thanks again for your help.
     
  5. Sep 4, 2011 #4
    I am not the greatest draughstman but I have now had the opportunity to draw up an explanation of Benham's formula.

    Figs 1 and 2 show some preliminaries, Figs3 and 4 develop the formula.

    In Fig1 I have shown surface of revolution about an axis of symmetry - in this case a sphere.
    Since you are happy with membrane theory I will just recap on the results pertinent to this issue.
    If we cut the surface with a series of parallel planes, perpendicular to the axis we generate a series of circles on the surface, called parallels in curvilinear coordinates.
    These parallels intersect at right angles meridians drawn from pole to pole (axis of symmetry).
    Any particular parallel (plane) can be identified by the angle alpha measured from the axis.
    The radius of any particular parallel, measured in its plane, is r0 sin[itex]\alpha[/itex], where r0 is the sphere radius.

    Finally membrane theory requires that the only stresses acting are tangential to the surface at any point. These we conveniently resolve into parallel and meridional stresses, [itex]\sigma[/itex]p and [itex]\sigma[/itex]m as shown in the expansion.

    Now what we need from all this is to realise that parallel (also called hoop) stresses are purely horizontal in Benham's dome. That is they have no vertical component so play no part in the vertical equilibrium.

    So in the vertical equilibrium we only need to consider meridional stresses and the loads.

    Moving on to Fig2

    This shows a vertical force F applied tangentially at some point on a circle ( and therefore the surface of the dome).

    The important fact here is that the vertical component of F is F sin [itex]\alpha[/itex] where alpha is the angle from the parallel to the vertical axis as before.

    Benham's vertical equilibrium equation reads

    Upward forces due to meridional stress + downward forces due to self weight = 0

    In Fig3 I am looking up under the dome at the upward forces, distributed round the ring at bearing level.
    The inset considers a 1 metre length of the circumference to yield
    Force per metre = meridional stress x 1metre x thickness = [itex]\sigma[/itex]1 x 1 x t

    The total force obviously equals the force per metre times the circumference.
    Remembering from Fig 1 that the in plane radius = dome radius times sin[itex]\theta[/itex] we obtain

    Total meridional force = 2[itex]\pi[/itex]r0sin[itex]\theta[/itex]

    But this is not the vertical force. To get this we multiply by sin[itex]\theta[/itex] as per Fig2.

    To obtain the second term consider the generation of the area of revolution.
    As the position vector, given by angle [itex]\theta[/itex] from the pole, rotates about the pole it generates a ring of circumference 2[itex]\pi[/itex]r0sin[itex]\theta[/itex]
    as before.

    If we now allow theta to increase by a differenctial amount d[itex]\theta[/itex] this describes a short arc given by r0d[itex]\theta[/itex].
    If we revolve this arc around the axis we generate a small differential strip of surface area of the dome.
    Integrating this and multiplying by the (constant) load per unit area yields the second term in Benham's formula.

    go well
     

    Attached Files:

    Last edited: Sep 4, 2011
  6. Sep 5, 2011 #5

    Jus

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    Studiot, you are a legend!

    I fully understand the process that has been adopted here. I just couldn't visualise what the equation was saying. Just to make sure I have this clear in my head for the first party of the equation, please could you have a look at my attachment and just confirm (hopefully!) that my thinking is correct.

    Thanks again.

    Kind Regards,

    Jus
     

    Attached Files:

  7. Sep 5, 2011 #6
    If F is the vertical component of [itex]\sigma[/itex]1 x Area then yes you are correct.

    Remember that [itex]\sigma[/itex]1 is not in general vertical.
     
  8. Sep 5, 2011 #7

    Jus

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    Yes F would be the vertical component of [itex]\sigma[/itex]1 x Area.

    Thanks again for all your help on this.

    Kind Regards,

    Jus
     
    Last edited: Sep 5, 2011
  9. Sep 5, 2011 #8
    Feel free to ask again if your revision throws up any more queries.

    :wink:

    go well
     
  10. Sep 5, 2011 #9

    Jus

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    Yes thank you very much for that, I might have to ask for your help again very soon Studiot.

    On a completely different topic, do you recommend an particular book to purchase on the topic of materials / soild mechanics? I have to say that this Benham one is quite tough to follow. It was on the list of recommended books for my university but I've always struggled with it a bit. This might be due to the fact that it's been a good few years since I've graduated but I was just wondering if there are any other books that explain the theories better and have easier to follow worked examples. Do you have any other recommendations? I'm really sorry to bother you about this.

    Kind Regards,

    Jus
     
  11. Sep 5, 2011 #10
    This is not a problem I'm glad to help someone with a genuine interest.

    However mechanics of materials is a very broach church so you would really need to narrow it down a bit.

    You say you are a Mech Engineer, so I assume you are interested in theory for components rather than structures such as concrete domes?

    Perhaps this is better discussed by private message (PM) ?
    If you would like to post or PM me some areas of interest eg photoelasticity, fracture mechanics FE, stress analysis etc we could take it from there.
     
  12. Sep 6, 2011 #11

    Jus

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    Hi Studiot,

    Thanks you for yur help on this. I have sent you a message with some details about what I am looking for.

    Thanks again for your help.

    Kind Regards,

    Jus
     
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