- Design of a Granite/Stone Beam -

1 2

the beam of the structure shown in 1.jpg and 2.jpg has to be designed, without any tensile reinforcement.. the structure is essentially made of stone..
My doubts are:

1. Will the load of the slab be distributed to the surrounding beams as per fig.7 of IS 456 - 2000 ?

2. knowing the total load of the slabs situated on top of the beam, is it possible to determine the depth of the beam section required to resist the load, using only the Bending equation?? (the values of compressive strength of the stone material is known)

3. let us assume that the load of the slabs is 500 kN (which is found by finding the total volume and multiplying by its unit weight), we know that this 500 kN is acting over the top surface area of the beam.. how do I convert this load into a UDL acting over the span of the beam ???

P.S. I'd also like to get some suggestions on the design of this beam element using only stone.. it is known that only the load transfer is only by compression

Replies

  • Kaustubh Katdare
    Kaustubh Katdare
    Are our Civil Engineers reading this? ๐Ÿ˜€
  • J. Guru Prasanna
    J. Guru Prasanna
    The_Big_K
    Are our Civil Engineers reading this? ๐Ÿ˜€
    guess not.. but I hope they'll be doing it in a while
  • Alvord12
    Alvord12
    Are you assuming that the 500 KN acts as a point load on the center of the beam? Also, I'm not too sure about using the IS 456:2000 for construction involving stone/granite. The maximum bending moment formula was found with research on concrete, can't say that the same can be applied to stone/granite.

    These are just my thoughts, feel free to correct me if you think differently! ๐Ÿ˜€

    Edit: Yup, just confirmed this with my friend, you cannot use IS 456:2000. You'll need to use IS 3316:1974 for Structural Granite. You might also want to check #-Link-Snipped-#.
  • J. Guru Prasanna
    J. Guru Prasanna
    Thanks for your reply..
    500kN is not a point load acting on the beam.. the calculated value of UDL on the beam (entire span) is approximately 14.5 kN per sq.meters..
    I feel we can use the bending equation M=fZ (M-moment, f-stress, z-section modulus) for any beam since we use the same equation for concrete as well as for steel... but i'm not sure with my answer
    and could you please upload IS 3316 here ? (I'm unable to find a pdf on the net)
    thanks in advance
  • Alvord12
    Alvord12
    Yeah, I asked my professor about the bending equation case, he says it is possible to use that equation. The equation is like; Stress/y = M/I = E/R; So, assuming that y = d/2 where d is the depth of the beam, you can calculate the the depth if you know the permissible stress for Granite/Stone.

    Else you could use a UTM to calculate the value of 'E' for the type of stone you will be using and derive the value of the radius of curvature for the same.

    As for the IS code, I'll check back with my friend if he has one. ๐Ÿ˜€
  • J. Guru Prasanna
    J. Guru Prasanna
    Thank you
  • J. Guru Prasanna
    J. Guru Prasanna
    Alvord12
    Are you assuming that the 500 KN acts as a point load on the center of the beam? Also, I'm not too sure about using the IS 456:2000 for construction involving stone/granite. The maximum bending moment formula was found with research on concrete, can't say that the same can be applied to stone/granite.

    These are just my thoughts, feel free to correct me if you think differently! ๐Ÿ˜€

    Edit: Yup, just confirmed this with my friend, you cannot use IS 456:2000. You'll need to use IS 3316:1974 for Structural Granite. You might also want to check #-Link-Snipped-#.
    I got the IS 3316.. #-Link-Snipped-#
    but nothing related to the load distribution has been mentioned there..
    is there anyother code that i should use for getting the load distribution or can I just assume the same as in IS 456 for concrete ??
  • J. Guru Prasanna
    J. Guru Prasanna
    It has been concluded that "the granite beam cannot be designed by any approximate method".. the loads acting on the structure are to be calculated and the required depth has to be obtained by a trial and error method alone.. or an experienced engineer could give a proper span to depth ratio which can be used to find out the depth..
  • s p gupta
    s p gupta
    Sir hw can I get the latest is code for my personal information about civil engineering

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