Design Of Flat Slab In Accordance To EC2-04: Finite Element Analysis Approach

you are required to produce a words technical report that includes the following: A plan of the geometrical model. Contour lines of bending stresses and shear stresses. Recommendation on the smallest slab thickness to use, and the reasoning behind it.

Properties of the analyzed flat slab

The flat slab that I analyzed was of the following properties; the distances between the grids are X1, X2, and Y1, Y2 are 6.00m, 6.00m, 6.75m, and 7.00m. the thickness 200mm. and the supporting columns are 300mm square. The loading are the variable loads is 3 kN/m², the perment loads are partions load 1 kN/m² and other is the self- weight. The wind load on the building is taken care by the bracing. The thickness of the slab 200mm which is to the nearest 25mm and satisfy the value of the design shear resistance in the EC2without the provision of shear reinforcement.

For the analysis of the flat slabs in accordance to EC2 can be achieved in the following methods: equivalent frame method, finite element analysis, yield line analysis, and the Grillage analogy. The following steps are used in the design of the flat slabs:

Design life – for our case the building is common structure (design life is 50 years) 

Asses of the loading on the slab. in this case the ratio of the variable loads to permanent loads should be less than 1.25

1. Determine the load combination applied.in our case we considered the variable loads and permanent loads, since the assumption made is that the wind load is taken care by the bracing.
2. Determine loading arrangements. The 1.35Q_k+1.5G_k(ULS) and 1.0 (Q_k+G_k) for SLS.
3. Assess durability requirements and determine the concrete strength. In our case the grade of the concrete used is C40/50.
4. Fire resistance period check. Here we provide suitable cover for the fire resistance period. Provided 25mm cover
5. Determine the minimum cover for durability, fire and bond requirements. Best choice 25mm cover both ways.
6. Analyze structure to obtain critical moments (M_c)and shear forces (V_c).
7. Design flexural reinforcement
8. Check defection.
9. Check punching shear capacity
10. Check spacing of the bars

Finite element analysis is the use of the ordinary or partial differential equation by a computer to solve wide range of structural problems. The FEA software in our case Midas Gen has the following features applicable that assist in the analysis of the structure, flat slab:

1. The bending moments in orthogonal directions, which considers torsion moment. It is essential to design the reinforcement to resist critical moments hence full.
2. Auto-mesh generator. Mesh creation is fast which will assist in the refinement of critical areas.
3. Auto application of load pattern. Assist in the solving for the worst-case design forces.
4. Curvature due to free shrinkage strain calculation, cracked section properties calculation and recalculated for subsequent iterations, cracked section in each direction and The analysis of ULS and SLS separately. Helps in the estimation of deflection.
5. Averaging area of the required reinforcement in a specified width 

Any model should have nodes, elements, supports, and loads. Nodes can be free points. If not free, they are at the ends and intersection of elements. Elements can be inn our case plates and beams. A plate element is either triangular or quadrilateral with nodes at corners. Meshing is referred to the use of sub-division of surface members into elements. The mesh size I used is 250mm. the supports in the model will ensure that surface will not stand on its own. It provides the result of the bend moments between them which are realistic. The supports for the flat slab are columns. The column moment assist in the generation of punching shear stress.

MIDAS GEN was used to perform the FEA. I did it using the following procedure

Selected the units work with. Distance used Meters. Loading Kilonewton (KN).

The code of preference is EC2-04  

Grid creation, used grid line

Named the grid, g

Grid spacing in x-axis 6.00, 6.00, 6.00, 6.00 and y-axis 6.75, 7.00, 7.00, 6.75. used the relative option.

Design steps for flat slabs

Created nodes and add them in place of the column as the floor plan.

Created column as an element. Assigned properties, used DB option, size 0.3m square and material of grade C40/50. Added them as at the nodes as it had been showed in the plan and the height is -0.5m.

Created virtual beams as an element. Helped in producing the surface. Assigned properties, used users option size 0.001m rounded named it null. Added to connect columns, at position z0.

Modified the interior beams by translating (copying) parallel dy-0.3 and 0.3 those that are parallel to y-axis and did so those in the x-axis. The edge virtual beam translating (copying) to be at the edge of the column in the outwards side and 0.45m in the inwards side.

 Auto meshed, by selected all the null elements, mesh type selected both triangle and quadrilateral. Side used is 0.5m. used create by element. The properties used material used C40/50and named slab, thickness of 0.2m, offset plate in ratio of 0.5 in local z. Generated plates on the surface.

Provided distinguishing colors for the plates and column by material.

Defined support at every column. 

Created cases, deadload (self-weight and partition) and variable load

Self-weight  assigned -4.8KN/m². Partition load assigned -1.0KN.m² as a pressure load. Variable load -3.0KN/m² as a pressure load.

Auto-generated load combination for EC2. ccCB1 for ULS and ccCB2 for SLS

Performed analysis by pressing F5.

Clicked results to generate graphs and contours. 

Calculation

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    [[[*]]]   PUNCHING CHECK MAXIMUM RESULT DATA BY STRESS: DOMAIN 1-[10].

===============================================================================

  -. Information of Parameters.

    Node No.  : 6248

    LCB No.   :    1

    Materials :  fck =  40000.0000 KPa.

    Thickness :    0.2000 m.

    Covering  :  dB  =   0.0260 m.

                 dT  =   0.0260 m. 

 -. Information of Checking.

    v_Ed    =  12583.8867 KPa.

    u1      =   2.1803 m.

    d       =   0.1740 m.

    u0      =1.5708e-008 m. 

 -. Basic control perimeter

    rholy   =       0.0042

    rholz   =       0.0042

    rhol    = min[ sqrt(rholy*rholz), 0.02 ] =       0.0042

    k       = min[ 1+(200/d)^0.5, 2.0 ]      =   2.0000 (d in mm)

    gamma_c =    1.500

    v_Rd,c  = max[ 0.035*k^1.5*sqrt(fck), (0.18/gamma_c)*K*(100*rhol*fck)^1/3 ]

         =    626.0990 KPa. 

 RatV    = v_Ed / v_Rd,c =  20.099  >  1.0  --->  Not Acceptable !!!

              (Need Vertical Reinforcements.)

    fywd    = 347826.0870 KPa.

    fywd_ef = min[ 250+0.25*d, fywd ]                    = 293500.0000 KPa.

    Asw/sr  = (v_Ed-0.75*vRd_c)*(u1*d) / (1.5*d*fywd_ef) =   0.060 m^2/m.  (   0.060 m^2/m.)

              (Calculating the outermost perimeter of shear reinforcement.)

    uout,ef = v_Ed*(u1*d)/(v_Rd,c*d) =  43.8218 m.

Conclusions and Recommendations

To conclude this report has shown a full finite element model as well as a finite element analysis which shown above are the steps that were made to determine the most appropriate thickness for the slab which was rounded to the nearest 25 mm. Even though the flab flat of thickness 200mm is sufficient for our case. The best for flat slab should have drop heads which will be more sufficient to take the punching stress. This will lead to further reduction of the thickness of the slab of about 150mm. 

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