Continue to Site

Eng-Tips is the largest engineering community on the Internet

Intelligent Work Forums for Engineering Professionals

  • Congratulations GregLocock on being selected by the Eng-Tips community for having the most helpful posts in the forums last week. Way to Go!

Eq'n for fix-pin beam with partial trapazoidal load? 1

Status
Not open for further replies.

gb156

Structural
Dec 12, 2001
33
I'm designing a wall that is fixed at the base and pinned at the top, with various soil heights (none of them go full height). Could someone please post the equations for shear and moment for this loading/restraint condition? I can't find it anywhere!

Thanks!
 
Replies continue below

Recommended for you

If you want (relatively) exact numbers, I would suggest using software such as STAAD.

In leau of that, check out Loading Condition 14 on pg. 2-300 in the ACI ASD Steel Manual 9th Edition. This would not be as exact as using software to model the soil conditions, but if you reduce each trapeziodal load to a point load in the appropriate location, these equations would give you the answers you're looking for. I would suggest the software approach though.

Good luck!
 
See Steel Designers' Manual 4th Edition. The trapezoidal load is equivalent to a combination of a partial height rectangular distribution and a partial height triangular distribution on a propped cantilever beam (pages 72 and 74). Moment and shear equations are given.
 
gb156,

The book "Formulas for Stress and Strain" by Roark has formulas for partial triangular and partial uniform loads for the Fix-pinned end conditions.

For a trapezoidal load you would need to superimpose the triangular and uniform loads.

I could give you the formulas but without the corresponding sketches the formulas do not mean much.

AEF
 
Look at the site below under Beams->Simple bending->Supported-fixed->Distr.load1 (or 2).
If you want the equations click the button Formulas.
prex
motori@xcalcsREMOVE.com
Online tools for structural design
 
Is there anybody out there who still remembers how to do this stuff from first principles, or don't they teach that anymore? One should hardly need a software program to solve a simple bending problem involving equating two deflections.
 
Ma = Wa/60L2 (3a2 - 15aL + 20L2)
Mx = Rb*x - W/3a2 (x-b)3

Mmax when x=b + a2/2L *square root of 1-a/5L

Where L= height of wall, a= height of soil, and b= L-a
NB. L2 & a2 denote L squared & a squared respectively


Ra = W-Rb
Rb = Wa2/20L*L*L (5L-a)

W = total load & Ra is reaction at fixed end (bottom of wall) Kieran
BEng(hons),CEng, M.I.Struct.E, M.I.E.I.
 
redhead

Right on.

Seems too simple to even post.
 
Redhead and bylar, it's been years since I've had to derive a formula for even a 'simple' load case/support condition. I could do the derivation, but, to be blunt, I don't have the time or patience at this moment (pun intended). If I had derived the formula myself I would have still posted the question just to compare my derivation with others.

To everyone else, especially Prex and Kieran, thank you very much! Your help has been greatly appreciated!



 
gb156 - You asked for a trapezoidal load distribution (I assume soil pressure plus surcharge). Kieran1 equation is for a triangular partial height load (soil pressure only with no surcharge). So if you really have a trapezoidal load distribution you still have to do superposition by adding partial height rectangular load case to Kieran1 equation.
 
Thanks for the reminder Amokhta. Shortly after the intial posting, I realized trapazoidal wasn't the load condition I was seeking, but, rather, triangular. Fortunately, Kieran knew what I was seeking and supplied the equations for a triangular load.
 
Thank you Kieran1. My comment was just that, not a challenge.
 
Status
Not open for further replies.

Part and Inventory Search

Sponsor