This article presents a comparative analysis between analysis of flat slab using moment coefficients method (MCM) in table 3.12 of BS 8110 and equivalent frame method (EFM).
Moment coefficient method is a simplified method in which with aid of some certain coefficients, a slab can be easily analyzed. Table 3.12 of BS 8110 provides some of these coefficients that can be used to analyze a one-way slab or a flat slab in each orthogonal direction under certain conditions. You can refer to this article here on flat slab and read these conditions under the “simplified method of moment and shear coefficients”.
The equivalent frame analysis on the other hand is a method that divides flat slab into equivalent frames of columns and slabs. The moment is these constituents’ elements is then determined from elastic analysis using the stiffness of each member.
The plan of the flat slab to be analyzed is shown below. It shows a 250mm thick office slab of 18 x 18m, which is meant to bear an impose load of 4.0KN/m2 and finishes of 1.0KN/m2. The slab is supported on 300 x 300mm columns, and the concrete strength for both slabs and columns is C25/30.
Analysis using Moment Coefficients
Permanent action
Characteristic Self-weight of slab = 0.25 x 25 = 6.25KN/m²
Partition Load on Slab = 1.0KN/m²
Characteristic Permanent Load on Slab = 6.25 + 1.0 = 7.25KN/m²
Variable action
Variable load on slab = 4.0KN/m²
Width of Slab
The width of slab is 6m
Total Design Load
Ultimate Load acting on Slab = (1.35(7.25) + 1.5(4))6 = 94.74KN/m²
Applicability of table 3.12 of BS 8110
Before table 3.12 of BS 8110 can be used for evaluating the effects of loading, three conditions must be met:
- Condition 1: Bay area > 30m2
Check: Bay area = (6+6+6) x 6 = 108m2
108m2 > 30m2.
Result: Condition met
- Condition 2: qk/gk ≤ 1.25
Check: = 4.0/7.25 = 0.55
0.55 < 1.25.
Result: Condition met
- Condition 3: qk ≤ 5KN/m2
Check: qk = 4KN/m²
Result: Condition met
Since the three conditions are satisfied, table 3.12 can be used.
Loading Arrangement
A load arrangement of maximum ultimate load (of 1.35gk + 1.5qk) on all spans shall be used. 20% redistribution of support moment from linear elastic analysis to the span shall be performed while using the equivalent frame analysis in accordance with UK national annex to BS EN 1992-1-1-2004. No redistribution shall be performed when using table 3.12 as allowance for this has already made in the coefficients.
Computation of shear forces and bending moments using moments coefficients
The beam is taken to be continuous rather than simple at the end span to account its monolithic connection with the end beams
F = n x l = 94.74 x 6 = 568.44KN
At Outer Support
Moment = 0.04FL = 0.04 x 568.44 x 6 = 136.43KNm
Shear force = 0.46 x F = 0.46 x 568.44 = 261.48KN
At First Interior Support
Moment = 0.086FL = 0.086 x 568.44 x 6 = 293.31KN
Shear force = 0.6 x F = 0.6 x 568.44 = 341.06KN
Near Middle of end Span
Moment = 0.075FL = 0.075 x 568.44 x 6 = 255.80KNm
Middle interior span
Moment = 0.063FL = 0.063 x 568.44 x 6 = 214.87KNm
Analysis using Equivalent Frame Method
The analysis of this flat slab using equivalent frame has already been performed in a separate article published here. However, since ultimate load is applied on all spans, the elastic analysis results using equivalent frame cannot be used directly for design nor comparable with analysis results from moments coefficients prior to 20% redistribution of span moment.
The bending moment diagram from the equivalent frame analysis without redistribution is presented below for easy reference:
Moment Redistribution
Redistribution of support moment of span BE
Support moment = 336.66
Reduce support moment by 20% = 0.8 x 336.66 = 269.328
Calculate the shear force at B using the formular below
VBE = wl/2 + (MBE – MEB)/L
VBE= 94.74 x 6/2 + (79.506 – 269.328)/6
VBE= 252.583KN
Considering that summation of upward forces equals zero
VBE + VEB = 94.74 x 6
VEB = 568.44 – 252.583
VEB = 315.857KN
Calculate the Span moment after redistribution
Position of maximum moment (x) = 252.583/94.74 = 2.67m
Span moment = 0.5 x 252.583 x 2.67 = 257.195KNm
Redistribution of support moment of span EH
Support moment 1= 315.92KNm
Support moment 2= 315.06KNm
Reduce support moment 1 by 20% = 0.8 x 315.92 = 252.736
Reduce support moment 2 by 20% = 0.8 x 315.06 = 252.048
Calculate the shear force at E using the formular below
VEH = wl/2 + (MEH – MEB)/L
VEH = 94.74 x 6/2 + (252.048 – 252.736)/6
VEH = 284.1KN
Considering that summation of upward forces equals zero
VEH + VHE= 94.74 x 6
VHE = 568.44 – 284.1
VHE = 284.33KN
Calculate the Span moment after redistribution
Position of maximum moment (x) = 284.1/94.74 = 3m
Span moment = 0.5 x 284.1 x 3 = 173.93KNm
Redistribution of support moment of span HK
Support moment = 336.04KNm
Reduce support moment by 20% = 0.8 x 336.66 = 270.43
Calculate Shear force at H using the formular below
VHK = wl/2 + (MHK – MKH)/L
VHK = 94.74 x 6/2 + (270.43 – 81.061)/6 = 252.658KN
Considering that summation of upward forces equals zero
VHK + VKH = 94.74 x 6
VKH = 568.44 – 252.658
VKH = 315.782KN
Calculate the Span moment after redistribution
Position of maximum moment (x) = 252.658/94.74 = 2.67m
Span moment = 0.5 x 252.658 x 2.67 = 255.8KNm
The combine elastic and redistributed bending moment diagram is given below:
Click here to study a worked example where analysis results using EFM is applied in designing the flat slab
Tabular Comparison
For the sake of easy comparison, we shall tabulate the moment from coefficients method and moment from equivalent frame analysis after redistribution
Conclusion
It can be observed that the end moment due to coefficient method is highly conservative than the equivalent frame analysis method. This is due to a conservative assumption due to lack of information about the boundary condition and properties of the supporting member. On the other hand, the equivalent frame analysis is able to estimate the stiffness of the supporting column thereby having much more economical values due to availability of information. The same thing as aforementioned can also be said for the moment over internal supports
The span moments from the moment coefficient method and equivalent frame method are very close. This can be attributed to the lack of the need to determine the properties of supporting members hence almost the same information is available to both methods such that a moment coefficient for span moment results in realistic and also economical moments.
