Suppose, you put a point load on a plate that is completely supported. The pressure under this plate is Force divided by the area of the plate. If I make this base very large, then pressure will be negligible. There has to be a limit to the size of this base right, after which it becomes useless to increase the area. Also, how does thickness of that plate come into play. Like at the top layer of the plate , the point load does engage the area completely , as we move vertically downwards does the area increases. The closest to reference I got was the force transfer under wheel surfaces in bridge design. How is this assumption made that the contact area increases with 45 deg. angle.

  • $\begingroup$ Welcome to Engineering! Pressure under the plate isn't exactly force over area. It might be if you consider the plate to be infinitely stiff. But if the plate is allowed to deform, then there'll be a concentration of load (and therefore increased pressure) in the immediate area under the point load, and lower pressures elsewhere. Only the "average pressure" is equal to force over area. $\endgroup$ – Wasabi Dec 27 '18 at 21:41
  • $\begingroup$ Than you for the response. I understand that. I wanted to know how is this accounted for. We normally assume a 45 degree cone angle. I wanted to know the source for assuming 45 degree. $\endgroup$ – Uttarayana Dec 27 '18 at 21:53
  • $\begingroup$ Ah, gotcha. Well, you don't always consider 45 degrees. Many load distribution mechanisms in reinforced concrete assume a 2/3 slope, for example. 45 degrees might be an assumption for asphalt (or maybe just an assumption, period). It really depends on the stiffness ($EI$) of the plate. $\endgroup$ – Wasabi Dec 27 '18 at 21:58
  • $\begingroup$ Such stress distributions are discussed in Johnson's Contact Mechanics. $\endgroup$ – Chemomechanics Dec 27 '18 at 23:03

It depends on how the plate is supported. For example, if it is supported on two edges like a one-way slab or four edges, and these edges are simply supported or can take moment, or the plate supported by a concrete pad. AISC has some equations for calculating base plates that are based on assumption roughly that base plate distributes the load by acting as a two-way Cantilever beam.

Westrgaard equation assigns an equivalent radius to concentrated load on a very small radius circle:

$$r_\text{equivalent} = \sqrt{1.6r_0 +t^2} - 0.675t $$

where $r_0$ is the initial radius, and $t$ is the plate thickness.

The only concern for a very large concentrated load on a very small area on a steel plate is punching shear, otherwise given enough thickness a point load will penetrate into the plate and quickly crush a rather large area of the plate and then can be dealt with as a larger radius of bearing.

Steel is very tough and resilient and methods like the angle of projection are not a good model.

Much research has been done in the area of ballistics, armor, etc. but I guess it is not easily accessible.

  • $\begingroup$ Wow. Thanks for a detailed explanation. Just a follow-up question. Let us take the same example of design of base plate. We take the point load on the base plate, divide it by area of the base plate to get stress. We take the same stress as a load on the base plate and assume it to be a double cantilevered beam and design it as such. I am interested to know if I make the base plate really large, then I would be getting a huge cantilever moment which is not possible. So there has to be limit to the size of the plate for this cantilever action to be true. So is there any info on that. $\endgroup$ – Uttarayana Jan 3 '19 at 18:46

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