peripheral thrust formula for pyramid

huangyhg

peripheral thrust formula for pyramid.
anyone has this book "design of pyramid" by terrington john stanley.
it has some formulae for peripheral thrust. any guidance will be appreciated.
thanks to all.


i don't have the book, but here is where you can find it:
amazon.co.uk
[united kingdom] publisher: concrete publications, 1939
used, very good, usually dispatched within 1-2 business days $112.55

dgkhan,
do you have a specific query regarding pyramids, we are much more helpful and responsive to those.
this seems to be very basic first principles type engineering.
i think only those that deal with grain discrete models may provide some "modern" answer to the question. for everything else one can make simplifications, such that layers parallel to the surface are free to roll over one of such surfaces and then the weight of the outer layer, produces thrust parallel to the surface that must be contained at the base. with this simplification (that implies others, over what happens at the corner of the piramid to keep formal integrity) it is a basic trigonometry problem to state a thrust. then you decide if the total thrust is to be accepted uniform or you work with slices of the outer layer, which is probably a more severe and hence a more interesting solicitation from an engineering viewpoint. taking just a slice to the top, for a piramid of slope alfa and height h, and density dens
the horizontally pushing force at the base
push hor = h·dens/ tan (alfa)
if h=140 m, dens=2.3 tonne/m3 alfa=55o
push hor = 225.45 tonne/m2 = 22.5 kgf/cm2 = 320 psi = 0.32 ksi.
hope i have not erred in the trigonometry.

i think a correction is required as per the expounded theory
weight=dens·section·sloped length=dens·dw·dt·h/tan(alfa)
tangential thrust=weight·sin(alfa)=dens·dw·dt·h·cos(alfa)
horizontal push = tangential thrust·cos(alfa)=
                  dens·dw·dt·h·(cos(alfa))^2
horizontal pressure = horizontal push / vertical area =
                      horizontal push /(dw·dt/sin(alfa))=
                      dens·h·sin(alfa)·(cos(alfa))^2
hence for the example
2.3·140·sin(55)·(cos(55))^2=86.77 tonne/m2=8.67 kgf/cm2=
122 psi = 0.12 ksi
that sounds more reasonable; the other value seemed to high to me, so i reviewed the trigonometry.