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Concentrated load on structural slab 1
- Thread starter MikeE55
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ACI does not give any guidelines for this, but lucky for us Westegaard solved this problem back in the 30's when wheel loads on bridge slabs started becoming large.
Westergaard, a professor at the University of Illinois, issued a comprehensive report in March of 1930 to the Bureau of Public Roads "Computation of Stresses in Bridge Slabs due to Wheel Loads". I believe Westergaard's report is the basis for the modern day AASTHO requirements. The AASTHO spec., unfortunately, has taken Westergaard's recommendations and converted them into engineering data that works only for truck loads and bridges. Building engineers must work with varying wheel bases, axle lengths and tire contact areas. You must go back to Westegaard's original report to find useful information. You should be able to order this report thru your Public Library.
This report has charts that give coefficients both for principal and distribution reinforcements for one way slabs subjected to concentrated loads. It has charts to handle overlapping stresses, when wheel loads are parallel or perpendicular to the span. It even takes into account the contact area of the concentrated loads.
I hope this helps!
Westergaard, a professor at the University of Illinois, issued a comprehensive report in March of 1930 to the Bureau of Public Roads "Computation of Stresses in Bridge Slabs due to Wheel Loads". I believe Westergaard's report is the basis for the modern day AASTHO requirements. The AASTHO spec., unfortunately, has taken Westergaard's recommendations and converted them into engineering data that works only for truck loads and bridges. Building engineers must work with varying wheel bases, axle lengths and tire contact areas. You must go back to Westegaard's original report to find useful information. You should be able to order this report thru your Public Library.
This report has charts that give coefficients both for principal and distribution reinforcements for one way slabs subjected to concentrated loads. It has charts to handle overlapping stresses, when wheel loads are parallel or perpendicular to the span. It even takes into account the contact area of the concentrated loads.
I hope this helps!
A rough and dirty way of approaching the problem....
Sometimes what I do is find out the actual bearing area of the concentrated load (say it is distributed uniformly over a 2ft by 2ft area). Treat the width of the loaded area as the width of a "beam" element within the slab (but the "beam" depth is the same thickness as the slab of course!)
Then you calculate the effective overhanging flange width on either side of the width of the loaded area.
The total effective width in each direction is the width of the loaded area plus the width of the effective slab overhang. This is what I use for the effective resisting width of point loads.
Sometimes what I do is find out the actual bearing area of the concentrated load (say it is distributed uniformly over a 2ft by 2ft area). Treat the width of the loaded area as the width of a "beam" element within the slab (but the "beam" depth is the same thickness as the slab of course!)
Then you calculate the effective overhanging flange width on either side of the width of the loaded area.
The total effective width in each direction is the width of the loaded area plus the width of the effective slab overhang. This is what I use for the effective resisting width of point loads.
Please note that Westegaard's method is only valid for solid, monolithic, poured in place, formed, concrete slabs. It does not apply to precast plank, poured in place slabs over composite metal deck, grating, floor plate, one way concrete joists, etc.
The amount of reinforcement, placed at right angles to the main reinforcement (often called distribution steel), is very important and is generally more than the normal requirement for temperature and shrinkage steel. This steel is what helps the concentrated load(s) spread out over the effective width.
The amount of reinforcement, placed at right angles to the main reinforcement (often called distribution steel), is very important and is generally more than the normal requirement for temperature and shrinkage steel. This steel is what helps the concentrated load(s) spread out over the effective width.
The Mathcad Library has a Mathcad file for designing a concrete slab with a column load applied. Try the following link if you have MathCad.
Australian Standard AS 3600–2001 Concrete structures (Clause 9.6) calculates the moment resisting width for a one-way slab supporting concentrated loads thus:
b_ef = b_load + 2.4*a*(1.0 – a/L_n)
where:
- ‘b_ef’ is the effective width
- ‘b_load’ is the load width
- ‘a’ is the perpendicular distance from the nearer support to the section under consideration
- ‘L_n’ is the length of clear span
As ‘b_load’ approaches 0 and ‘a’ approaches L_n/2, ‘b_ef’ approaches 60% of L_n.
b_ef = b_load + 2.4*a*(1.0 – a/L_n)
where:
- ‘b_ef’ is the effective width
- ‘b_load’ is the load width
- ‘a’ is the perpendicular distance from the nearer support to the section under consideration
- ‘L_n’ is the length of clear span
As ‘b_load’ approaches 0 and ‘a’ approaches L_n/2, ‘b_ef’ approaches 60% of L_n.
Try my quick homeopathic formula. First, "ASSUME" a slab thickness that works for all the typical ACI Code criteria using a 12 inch trial width. Then, to be conservative I would use the width of the loaded area plus the 45 degree projection down and out from both sides of the actual loaded area and compare my results. Good Luck.
AS 3600 doesn’t specifically address overlapping loads. I guess it’s up to designers to calculate the unit moment (kNm/m) for each concentrated load and sum them where the effective widths overlap.
The design of flexural reinforcement is based on a rectangular stress block with a uniform compressive stress of 0.85*f_c (where f_c is the characteristic compressive cylinder strength of concrete at 28 days) and a maximum strain in the extreme compressive fibre of 0.003.
Cross-sectional steel area is calculated from the following equation for the moment capacity (kNm/m):
phi*m_u = (phi*f_c*q*(1 – q/1.7)*d^2)/1000 [kNm/m]
where:
- phi: strength reduction factor, equal to 0.8 for slabs [dimensionless]
- m_u: ultimate moment capacity [kNm/m]
- f_c: characteristic compressive cylinder strength of concrete at 28 days [MPa]
- q = (A_st*f_sy)/(1000*d*f_c)
where:
- A_st: cross-sectional area of tension reinforcement [mm^2]
- f_sy: yield strength of reinforcing steel [MPa]
- d: effective depth of a cross-section, i.e. the distance fro extreme compressive fibre to the resultant tensile force in the reinforcing steel [mm]
The design of flexural reinforcement is based on a rectangular stress block with a uniform compressive stress of 0.85*f_c (where f_c is the characteristic compressive cylinder strength of concrete at 28 days) and a maximum strain in the extreme compressive fibre of 0.003.
Cross-sectional steel area is calculated from the following equation for the moment capacity (kNm/m):
phi*m_u = (phi*f_c*q*(1 – q/1.7)*d^2)/1000 [kNm/m]
where:
- phi: strength reduction factor, equal to 0.8 for slabs [dimensionless]
- m_u: ultimate moment capacity [kNm/m]
- f_c: characteristic compressive cylinder strength of concrete at 28 days [MPa]
- q = (A_st*f_sy)/(1000*d*f_c)
where:
- A_st: cross-sectional area of tension reinforcement [mm^2]
- f_sy: yield strength of reinforcing steel [MPa]
- d: effective depth of a cross-section, i.e. the distance fro extreme compressive fibre to the resultant tensile force in the reinforcing steel [mm]
Based upon an effective width of .6 L, for a single concentrated load, the previously suggested 45 degree distribution may be unconservative. A 30 degree distribution away from the concentrated load in each direction corresponds more closely with the .6 L effective width.
James G MacGregor's book "Reinforced Concrete" contains a section on Yield Line Theory for slabs. One part of the section addresses analysis and design for concentrated loads. I have used it from time to time for heavy concentrated loads in industrial facilities.
Good luck.
Good luck.
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