This article presents a thorough worked example on the analysis of flat slab using equivalent frame method. At the end of the article, the analysis of the equivalent frame is validated with analysis results from Staad Pro programme.
The figure below shows a two-storey building with flat slab at each level. The first-floor slab shall be analyzed using equivalent frame method.
The plan of the first-floor slab is as 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 is for slabs and columns is C25/30.
Equivalent Frame
The slab can be divided into equivalent frame in each orthogonal direction as shown below: Each orthogonal direction consists of 4 frames, two internal frames and two edge/external frames. The internal slab strips consist of half the width of two adjacent slab on both sides from the gridlines, which makes the slab in the equivalent frame to be 6m. The edge strips only consist of half the width of the slab to only one side which makes their width to be 3m.
Since the slab is square and symmetrical, only one internal frame and external frame need to be analyzed and the analysis results can be replicated for the remaining internal and external frames respectively.
However, in this worked example only the internal strip along gridline 3 shall be analyzed and the result replicated in the other orthogonal spans. The elevation of the equivalent frame along gridline 3 is as shown.
Load Analysis
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
Load Arrangement
A load arrangement of maximum ultimate load (of 1.35gk + 1.5qk) on all spans with allowance 20% redistribution of support moment shall be used in accordance with UK national annex to BS EN 1992-1-1-2004.
Width of Slab
The width of equivalent frame slab is 6m
Total Design Load
Ultimate Load acting on Slab = (1.35(7.25) + 1.5(4))6 = 94.74KN/m²
Computation of Support Moment using Moment Distribution Method
End moments
Since the slabs are of equal spans and a single load case is applied on all spans, then the fixed end moment shall be the same all through
$
_{M_{BE\,\,}}\,\,=\,\,\frac{-w\,\,*\,\,\begin{array}{l}
L^2\\
\end{array}}{12}
$ = $\frac{-94.7\,\,*\,\,\begin{array}{l}
6^2\\
\end{array}}{12}
$ = -284.1KNm
$
_{K_{BE\,\,}}$ = $
_{K_{EH\,\,}}$ = $
_{K_{HK\,\,}}$ = -284.1KNm
$
_{M_{EB\,\,}}\,\,=\,\,\frac{w\,\,*\,\,\begin{array}{l}
L^2\\
\end{array}}{12}
$ = $\frac{94.7\,\,*\,\,\begin{array}{l}
6^2\\
\end{array}}{12}
$ = 284.1KNm
$
_{K_{EB\,\,}}$ = $
_{K_{HE\,\,}}$ = $
_{K_{KH\,\,}}$ = 284.1KNm
Second Moment Area
Slab Second moment area = bh³/12 = (6 x 0.25³)/12 = 0.00781
Column Second Moment Area = bh³/12 = (0.3 x 0.3³)/12 = 0.000675
STIFFNESS
SLAB STIFFNESS
$
_{K_{slab\,\,}}$ = 4EI/L = 4 x E x 0.00781/6 = 0.0052E
$
COLUMN STIFFNESS
$
_{K_{column\,\,}}$ = 4EI/L = 4 x E x 0.000675/6 = 0.0009E
$
DISTRIBUTION FACTOR
$
DF_{BA}\,\,=\,\,DF_{BC}\,\,=\,\,\frac{K_{BA}}{K_{BA}\,\,+\,\,K_{BC}\,\,+\,\,K_{BE}}\,\,=\,\,\frac{0.0009E}{0.0009E\,\,+\,\,0.0009E\,\,+\,\,0.0052E\,\,}\,\,=\,\,0.1284
$
$
DF_{BE}\,\,=\,\,\frac{K_{BE}}{K_{AB}\,\,+\,\,K_{BE}\,\,+\,\,K_{BC}}\,\,=\,\,\frac{0.0052E}{0.0009E\,\,+\,\,0.0052E\,\,+\,\,0.0009E\,\,}\,\,=\,\,0.7432
$
$
DF_{ED}\,\,=\,\,DF_{EF}\,\,=\,\,\,\,\frac{K_{DE}}{K_{ED}\,\,+\,\,K_{EF}\,\,+\,\,K_{EB}\,\,+K_{EH}\,\,}\,\,=\,\,\frac{0.0009E}{0.0009E\,\,+\,\,0.0009E\,\,+\,\,0.0052E\,\,+0.0052E\,\,\,\,}\,\,=\,\,0.07367
$
$
DF_{EB}\,\,=\,\,DF_{EH}\,\,=\,\,\,\,\frac{K_{EB}}{K_{ED}\,\,+\,\,K_{EF}\,\,+\,\,K_{EB}\,\,+K_{EH}\,\,}\,\,=\,\,\frac{0.0052E}{0.0009E\,\,+\,\,0.0009E\,\,+\,\,0.0052E\,\,+0.0052E\,\,\,\,}\,\,=\,\,0.4263
$
$
DF_{HG}\,\,=\,\,DF_{HI}\,\,=\,\,\,\,\frac{K_{HG}}{K_{HG}\,\,+\,\,K_{HI}\,\,+\,\,K_{HE}\,\,+K_{HK}\,\,}\,\,=\,\,\frac{0.0009E}{0.0009E\,\,+\,\,0.0009E\,\,+\,\,0.0052E\,\,+0.0052E\,\,\,\,}\,\,=\,\,0.07367
$
$
DF_{HE}\,\,=\,\,DF_{HK}\,\,=\,\,\,\,\frac{K_{HE}}{K_{HG}\,\,+\,\,K_{HI}\,\,+\,\,K_{HE}\,\,+K_{HK}\,\,}\,\,=\,\,\frac{0.0052E}{0.0009E\,\,+\,\,0.0009E\,\,+\,\,0.0052E\,\,+0.0052E\,\,\,\,}\,\,=\,\,0.4263
$
$
DF_{KJ}\,\,=\,\,DF_{KL}\,\,=\,\,\frac{K_{KJ}}{K_{KJ}\,\,+\,\,K_{KL}\,\,+\,\,K_{KH}}\,\,=\,\,\frac{0.0009E}{0.0009E\,\,+\,\,0.0009E\,\,+\,\,0.0052E\,\,}\,\,=\,\,0.1284
$
$
DF_{KH}\,\,=\,\,\frac{K_{KH}}{K_{KJ}\,\,+\,\,K_{KL}\,\,+\,\,K_{KH}}\,\,=\,\,\frac{0.0052E}{0.0009E\,\,+\,\,0.0009E\,\,+\,\,0.0052E\,\,}\,\,=\,\,0.7432
$
To make the moment distribution easier to construct and less clumsy to read and interpret, columns meeting at a point shall be combined. This will also necessitate combining the distribution factors of columns meeting at a point.
DF of Column at B = 0.1284 x 2 = 0.25684
DF of column at E = 0.07367 x 2 = 0.14734
DF of column at H = 0.07367 x 2 = 0.14734
DF of Column at K = 0.1284 x 2 = 0.25684
Moment Distribution Table
Discretizing the Slab into component Slabs to obtain Reactions at each Support
Since the support moments are obtained through moment distribution, we can now obtain the reactions acting on individual span of the slab. The discretized slab span is shown below:
From SLAB BE
Taking Moment about B (clockwise direction +)
–REBx L + MEB + MBE + 94.7 x L² x 0.5= 0
REB = (336.66 – 79.506 + 94.7 x 6² x 0.5)/6
REB = 326.958KN
Summation of vertical forces must be equal to zero
REB + RBE = 94.7 x 6
RBE = 568.2 – 326.958
RBE = 241.242KN
From SLAB EH
Taking Moment about E (clockwise direction +)
–RHE x L + MHE + MEH + 94.7 x L² x 0.5= 0
RHE = (316.207 – 315.06 + 94.7 x 6² x 0.5)/6
RHE = 284.291KN
Summation of vertical forces must be equal to zero
REH + RHE = 94.7 x 6
REH = 568.2 – 284.291
REH = 241.242KN
From SLAB HK
Taking Moment about H (clockwise direction +)
–RKH x L + MHK + MKH + 94.7 x L² x 0.5= 0
RHE = (-338.04 – 81.0612 + 94.7 x 6² x 0.5)/6
RHE = 241.27KN
Summation of vertical forces must be equal to zero
RHK + RKH = 94.7 x 6
REH = 568.2 – 241.27
REH = 326.93KN
Resultant reactions
To get the resultant reaction, all the reactions at a common support are summed together
RB = 241.242KN
RE = (REB + REH) = (326.958 + 241.242) = 610.868KN
RH = (RHE + RHK) = (284.291 + 326.93) = 611.22KN
RK = 241.27KN
Since all the reactions are known, we can now compute the internal shear at desired section and then draw the shear force diagram.
Computation of Shaar Force at each joint and plotting of the Shear Force Diagram
For Span BE
Shear force at 0m = 241.242KN
Shear force at 3m = 241.242 x 94.7 x 3 = -42.858
Shear Force at 6m = 241.142 – 94.7 x 6 = -326.96KN
For Span EH
Shear Force at 6m = -326.96 + 610.868 = 283.909KN
Shear Force at 9m = 283.909 – 94.7 x 3 = -0.1906
Shear force at 12m = -0.1906 – 94.7 x 6 = -284.29KN
For Span HK
Shear Force at 12m = -284.29 + 611.22 = 326.93KN
Shear Force at 15m = 326.93 – 94.7 x 3 = 42.8298KN
Shear force at 18m = 42.8298 – 94.7 x 6 = -241.27
These values are plotted to get a shear force diagram as shown below:
Computation of Slab’s Span moment and plotting of the Bending Moment Diagram
The maximum span moment occurs at the point of zero shear in the span. To get the maximum span moment at each span, the position of the point of zero shear has to be calculated. Then we find the area under the shear diagram at that point.
Span BE
Take the Point of zero shear to be x as shown below
Then from static, the total vertical forces on the slab have to be equal to zero;
241.24 – 94.1x = 0
X = 241.24/94.7
X = 2.547
Span moment at 2.547 shall be equals to area under the shear force diagram
Area of shear force diagram = 0.5 x 241.24 x 2.547 = 307.219KNm
Preceding support moments = 79.506KNm
Span moment = 307.219 – 79.506 = 227.767KNm
Span EH
Take the Point of zero shear to be x as shown below
Then from static, the total vertical forces on the slab have to be equal to zero;
283.909 – 94.1x = 0
X = 283.909/94.7
X = 2.998
Span moment at 8.998 shall be equals to area under the shear force diagram
Area of shear force diagram = 0.5 x 283.909 x 2.998 = 425.58KNm
Preceding support moments = 315.06KNm
Span moment = 425.58 – 315.06 = 110.515KNm
Span HK
Take the Point of zero shear to be x as shown below:
Then from static, the total vertical forces on the slab have to be equal to zero;
326.93 – 94.1x = 0
X = 326.93/94.7
X = 3.45
Span moment at 15.45 shall be equals to area under the shear force diagram
Area of shear force diagram = 0.5 x 326.93 x 3.45 = 563.954KNm
Preceding support moments = 338.04KNm
Span moment = 563.954 – 338.04 = 226.285KNm
Since the span moments for the three spans have been calculated, then we plot the bending moment diagram thus:
Staad Pro Model
The equivalent frame is modeled in Staad Pro programme. The slab at the first level is modeled together with adjoining columns at the top and bottom. The slab is modelled as a line element of section geometry of 6 x 0.25m while the column is assigned a section geometry of 0.3m x 0.3m. The material is chosen as concrete so that the program can compute the second moment area and determine deformation properties like elastic modulus. A fixed supports is assigned to all far ends of the columns while the beam is assigned a uniform load of 94.74KNm. The analytical model of the equivalent frame is shown below.
Below is the bending moment and Shear force diagram from Staad Pro
The diagram from Staad Pro BMD features moments of the adjoining columns and the results of the slab support moment are not visible. The table below provides a summary of the necessary result from the Staad Pro as well as that from hand computation for easy comparison.
Click here to study a worked Example where the analysis is results are used in designing the flat slab
