×
INTELLIGENT WORK FORUMS
FOR ENGINEERING PROFESSIONALS

Log In

Come Join Us!

Are you an
Engineering professional?
Join Eng-Tips Forums!
  • Talk With Other Members
  • Be Notified Of Responses
    To Your Posts
  • Keyword Search
  • One-Click Access To Your
    Favorite Forums
  • Automated Signatures
    On Your Posts
  • Best Of All, It's Free!
  • Students Click Here

*Eng-Tips's functionality depends on members receiving e-mail. By joining you are opting in to receive e-mail.

Posting Guidelines

Promoting, selling, recruiting, coursework and thesis posting is forbidden.

Students Click Here

Jobs

Peripheral thrust formula for pyramid.

Peripheral thrust formula for pyramid.

Peripheral thrust formula for pyramid.

(OP)
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.
 

RE: Peripheral thrust formula for pyramid.

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
 

RE: Peripheral thrust formula for pyramid.

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.

RE: Peripheral thrust formula for pyramid.

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.
 

RE: Peripheral thrust formula for pyramid.

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.

Red Flag This Post

Please let us know here why this post is inappropriate. Reasons such as off-topic, duplicates, flames, illegal, vulgar, or students posting their homework.

Red Flag Submitted

Thank you for helping keep Eng-Tips Forums free from inappropriate posts.
The Eng-Tips staff will check this out and take appropriate action.

Reply To This Thread

Posting in the Eng-Tips forums is a member-only feature.

Click Here to join Eng-Tips and talk with other members!


Resources