Point Load on Slabs 2

[engrjenjen]

I will be designing a one-way slab with 3m by 7m spans. The slab will carry point loads due to the reaction from a machine with 0.6m by 2.5m area. How will I distribute these loads over the span? I am confused whether the weight of the machine should be distributed over the entire span or should I convert first the weight (KN) over the area (m^2) covered by the machine. The two will have a big difference when it comes to the conversion of loads.

 

[Trenno]

So what I gather you have a continuous one way system with multiple 3m spans (7m transverse width).

Therefore am I correct in assuming the machine has a load 'length' of 600mm (2.5m transverse width)?

I would just take a 2.5m wide design strip and design the slab with the machine weight as a point load acting central to the 600mm.

Alternatively, you can look into a method of finding a 'contributing width' that will flexurally resist point loads, as detailed in AS3600 (Aus).

Reference material found below.

 

[engrjenjen]

Thank you. This will help a lot.

I noticed that this is based on Aus. Standards, if I use this method will there be any conflicts when I use different code? Please clarify if the value 2.4 on the formula a constant or varies depending on the code I used? Our code is based on American Standards. Thank you.

 

[Trenno]

Well then I guess you will have to assume that constant of 2.4 will only work with metric units.

Not sure where that formula came from (probably ripped it off the British code...), but you should have a quick google and see if you can find it in other reference material.

 

[engrjenjen]

I think the formula is more appropriate to use than to assume a one-meter strip and then distribute there the longitudinal bars.

Thank you @Trenno..

 

[zakmuh]

Hi,

Based on BS EN code, you can take plant live load of 7.5 kN/m2 onto the slab and design it. Taking the steel area provided, take machine point load from its leg that acting on the slab, do a punching shear check.

Cheers

 

[rapt]

I have always considered this width to be too large and normally use the about half of it!

RE use in other codes, that is up to you as an engineer to justify! Don't ask us.

 

[bowlingdanish]

Out of interest - what is the structural mechanism that formula is based on? Would having a larger amount of distribution steel increase that width? Large out of plane shear stiffness increase it? I don't see how this width would be based purely on geometry.

 

[Retrograde]

what is the structural mechanism that formula is based on

It is based on test data.

 

[RPMG]

2.4 is unitless. It may be interpreted as meter/meter, inch/inch, or ft/ft.


The formula is based concrete tests (Westergaard's?), and an ideal shell/plane (zero thickness) element. This is why it is applicable only to concrete slabs of uniform thickness, and thickness/reinforcement is not a factor in the formula.


There is no American standard. There is an AASHTO equation, which doesn't apply. There is Westergaard's formula from 1930: 2*(load width) + 0.58L, which is comparable to the code above: load width + 0.6L, if you assume the load is at midspan. I believe the discrepancy is that Westergaard assumed a thick bridge slab, where load may disperse through the slab thickness, and AU does not assume a slab thickness.


That's strength design. If the slab fails/cracks locally, more dishing of load will occur, and a greater slab width will be engaged.

 

[engrjenjen]

Hi RPMG,

I stand corrected, what I mean on the American Standard is that our code was adopted from international codes and references, and that includes some of American code counterparts. Please refer to photo below (taken from our structural code).

 

[engrjenjen]

Hi BowlingDanish,

I don't think area of reinforcement will affect the width, it's not a factor on the given formula.