Smart questions
Smart answers
Smart people
INTELLIGENT WORK FORUMS
FOR ENGINEERING PROFESSIONALS

Member Login




Remember Me
Forgot Password?
Join Us!

Come Join Us!

Are you an
Engineering professional?
Join Eng-Tips now!
  • 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!

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

Donate Today!

Do you enjoy these
technical forums?
Donate Today! Click Here

Posting Guidelines

Promoting, selling, recruiting, coursework and thesis posting is forbidden.
Jobs from Indeed

Link To This Forum!

Partner Button
Add Stickiness To Your Site By Linking To This Professionally Managed Technical Forum.
Just copy and paste the
code below into your site.

Pete2006 (Structural) (OP)
17 Apr 07 8:01
Existing 1/4" thick steel checker plate floor is under scrutiny.  I'm having difficulty finding a method to determine the maximum allowable concentrated load.  A typical plate is rectangular and simply supported on four sides, is removable so not fastened to supports.  The load is gravitational so is perpendicular to the plane of the plate.  The plates vary in size but an example would be 3 feet by 4 feet.  Any thoughts?
Helpful Member!  ChipB (Structural)
17 Apr 07 8:21
To truly calculate it, you could do a FEA.

Personally, I'd use the "Floor Plate Bending Capacity" tables in the Steel manual, and calculate the equivalent point load from the moment the uniform load would create.
Page 2-145 in the 9th Ed.
Pages 3-154 & 155 in the 13th Ed.

Hope it helps
brane23 (Structural)
17 Apr 07 10:48
I agree with ChipB... Also, in Table 3-18a, you have the moment of intertia per one inch strip. You can calculate your maximum allowable deflection... I'm sure "what it looks like" is more of a concern than the actual yielding anyway.
Bagman2524 (Structural)
17 Apr 07 12:48
Roark's formulas for stress and strain. 6th ed. Table 26 Case 1b.

a=4 b =3.  b/a = 1.33   Beta = 0.74  Alpha = 0.0093

Max def = alpha x W x b x b / ( E x t x t x t)

Max stress = (3W /(2 x pi x t x t)) x ((1 + poison ratio) x
ln(2b/pi x r'o) + Beta))

r'o = sqrt (1.6 ro x ro + t x t).
ro = radius of load contact
miecz (Structural)
17 Apr 07 13:14
Bagman

My 5th Edition Table 25 shows:

alpha=.1478 at a/b=1.2
alpha=.1621 at a/b=1.4
JStephen (Mechanical)
17 Apr 07 13:17
I think to get Roark's formula to work, you'll need dimensions in inches (to match PSI).

If I remember correctly, don't the building codes define what a "concentrated load" is?  It's not a point load, it's something like 2' square (ie, space a person occupies) and that will make a sizable difference in your case.  And that would put you closer to formula 1c rather than 1b in Roark's book.
Pete2006 (Structural) (OP)
18 Apr 07 7:56
My calcs show that the method of finding an equivalent concentrated load from the distributed load tables in the steel manual is very conservative in comparison to the result given by Roark's formula from Table 26 case 1b.  The resulting max point load per Roark is about twice that of the first method.  The plate's support on four edges apparently makes the difference.  The first method assumes support on only two edges.  Thank you all for your help!
Pete2006 (Structural) (OP)
18 Apr 07 8:04
Side note:  Table 3-18a in Steel Manual gives moment of inertia per one foot of width, not one inch - a minor correction to the contribution by brane23.
JStephen (Mechanical)
18 Apr 07 8:09
ASCE 7-05, 4.3- Concentrated load is over a 2.5' square area.
Pete2006 (Structural) (OP)
18 Apr 07 8:33
I have ASCE 7-02 which reads "Unless otherwise specified, the indicated concentration shall be assumed to be uniformly distributed over an area 2.5 ft square and shall be located so as to produce the maximum load effects in the structural members."  My particular situation falls under "otherwise specified".  It is, however, good to know for other cases.
miecz (Structural)
18 Apr 07 10:11
The alpha figures I posted earlier were, in fact, for Table 26, case 1b.  Interpolating, I get alpha=.158, which gives 0.46" deflection under a 1 kip load.
prex (Structural)
19 Apr 07 3:05
In the first site below there are calculation sheets for rectangular plates under concentrated loads.

prex
http://www.xcalcs.com : Online tools for structural design
http://www.megamag.it : Magnetic brakes for fun rides
http://www.levitans.com : Air bearing pads

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!

Close Box

Join Eng-Tips® Today!

Join your peers on the Internet's largest technical engineering professional community.
It's easy to join and it's free.

Here's Why Members Love Eng-Tips Forums:

Register now while it's still free!

Already a member? Close this window and log in.

Join Us             Close