Eng-Tips is the largest engineering community on the Internet
Intelligent Work Forums for Engineering Professionals
-
Congratulations waross on being selected by the Tek-Tips community for having the most helpful posts in the forums last week. Way to Go!
How to calculate point load on suspended slab 1
- Thread starter yahoo123
- Start date
- Thread starter
- #3
COEngineeer
Structural
a point load of a 1 lb will be 1 lb anywhere you put it. are you a bioengineer?
Never, but never question engineer's judgement
Never, but never question engineer's judgement
- Thread starter
- #6
A couple of relevent questions:
Is this a solid slab formed with plywood or is it a slab on metal deck or a precast slab or some other type? These will act differently under concentrated loads.
Are there several concentrated loads such as wheels from a fork truck? These will most likely have overlapping stresses.
Is this a solid slab formed with plywood or is it a slab on metal deck or a precast slab or some other type? These will act differently under concentrated loads.
Are there several concentrated loads such as wheels from a fork truck? These will most likely have overlapping stresses.
- Thread starter
- #9
A simple way of calulating primary bending with a single concentrated load on a one way slab is to assume it spreads out over 0.6 times the span (sometimes refered to as the effective width). With a single load at midpan, the moment equals PL/4 divided by the effective width (0.6 x L). M = 0.42 P per foot of width. If the contact area of the concentrated load is significant, you can also add its width to the effective width.
One word of caution: if you have two concentrated loads, the effective width needs to be reduced. I would suggest that the effective width be 0.3xL plus half the distance between loads.
I hope this helps!
One word of caution: if you have two concentrated loads, the effective width needs to be reduced. I would suggest that the effective width be 0.3xL plus half the distance between loads.
I hope this helps!
- Thread starter
- #11
Both AS3600 and BS8110 give the same method, as can be found here;
-
1
- #13
ThePenguin
Structural
I believe that the equation to which Jike refers can be found in "Computation of Stresses in Bridge Slabs Due to Wheel Loads." by Westergaard, published in Public Roads Vol 11 No. 1 1930. In this paper Westergaard provides an analytical elastic analysis of concentrated loads on a slab, and gives an equation that gives the effective width of slab for an applied force at the center of the span as 0.58 * Span + 2*c, where c is the diamater of the assumed circular contact area. For a truly concentrated load c=0 and you are left with 0.58 * Span, or as jike stated, 0.6 * Span. As jike also mentioned this equation is for a single concentrated load at the center, so be careful. The entire document however has lots of charts and equations for different cases.
Penguin: Thanks for the backup. Westergaard's charts are the method that I regularly use but wanted to provide a simple method without charts. Westergaard also has an easy way to determine distribution steel. I know of no way to do this with the simplified approach.
Yahoo123: Usually, slabs don't span 30 feet unless they are PT.
Yahoo123: Usually, slabs don't span 30 feet unless they are PT.
- Thread starter
- #15
yahoo,
Look at the link which apsix referenced above. The equations are in a contribution by "dbuzz". Note that if your load is near the edge, you should use an effective width equal to half the value plus the distance to the free edge.
Look at the link which apsix referenced above. The equations are in a contribution by "dbuzz". Note that if your load is near the edge, you should use an effective width equal to half the value plus the distance to the free edge.
- Status
Similar threads
- Question
- Question