# How to convert kN/m² into point loads?

I have a concrete steel ceiling. The calculations show the ceiling is designed for 1,5 kN/m² in addition to it's own weight and traffic loads including snow. So I assume I can hang 1,5 kN/m² from that ceiling, provided I use suitable anchors.

However, when I hang any load from that ceiling, I apply discrete point loads. I probably cannot simply say: Well the room is 2x4m, so I can hang 2m x 4m x 1,5kN/m² = 12kN in the middle of the room. Which is probably wrong.

So how do I arrive at the point loads I can apply?

• This is not an answer, but i prefer to hang the anchor on the steel and not in the concrete matrix for two reason if it fails it would't fail suddenly and the second reason (i'm not sure) i think the load acts as a distributed load over a line.
– user14407
Nov 2, 2018 at 11:49
• well, divide by the number of points and then make sure that the area of each point load is sufficient for the load applied or the load is applied over sufficient area... Nov 2, 2018 at 12:11
• SolarMike, if I take your suggestion 1 could (in this case) hang 12kN or 1,2 tons in the middle of the room I strongly doubt that is correct.
– mart
Nov 2, 2018 at 12:33
• @mart did you account for the area as I suggested? Nov 4, 2018 at 6:44
• So you want to hang 1223.7 tons from the ceiling as a point load? Not going to happen. In order to calculate that you need to take shear forces of the type of concrete into account and also punching stress. That means sigma=force/ area. Point loads are only really useful for calculations but in practice you need to use moments of inertia after doing moments of area calc first. Dec 8, 2018 at 10:49