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Exercise 4.9.5

Given : Covering load gεπ=1.5 kN/m2, live load q=5.0 kN/m2, concrete quality C40/50.

Question : Perform static analysis of the slabs illustrated in the figure (shears forces, bending moments, deflections) using the finite element method with (a) global loading and (b) unfavourable loadings.


Figure 4.9.5-1: project <B_464>
Figure 4.9.5-1: project <B_464>

Solution:

Slabs are identical to those of the example of §4.6.4.

g=0.15x25.0+1.5=5.25 kN/m2

q=5.0 kN/m2

In ULS, the maximum design load is equal to:

pd=1.35g+1.50q=1.35x5.25+1.50x5.0=

=14.59 kN/m2

Software requires only characteristic loads of the structure to be inserted while the rest are obtained through internal calculations.

Due to both structure and loading being symmetric, results will also be symmetrical for both slabs.


RESULTS FOR GLOBAL LOADING 1.35g+1.50q


Figure 4.9.5-2: Bending moment diagrams
Figure 4.9.5-2: Bending moment diagrams

Figure 4.9.5-3: Shear force diagrams
Figure 4.9.5-3: Shear force diagrams

Mx=13.9 kNm, My=12.4 kNm, Myerm=-34.6 kNm

Vxr=23.6 kN, Vyr=24.1 kN, Vyerm=45.5 kN


Figure 4.9.5-4: Deflection diagrams
Figure 4.9.5-4: Deflection diagrams

Figure 4.9.5-5: Deflection diagrams and contours
Figure 4.9.5-5: Deflection diagrams and contours

ymax=4.03 mm



RESULTS FOR UNFAVOURABLE LOADINGS


Figure 4.9.5-6: Bending moment diagrams
Figure 4.9.5-6: Bending moment diagrams

Figure 4.9.5-7: Shear force diagrams
Figure 4.9.5-7: Shear force diagrams

Mx=15.5 kNm, My=12.4 kNm, Myerm=-34.6 kNm

Vxr=24.8 kN, Vyr=24.5 kN, Vyerm=45.5 kN


Figure 4.9.5-8: Deflection diagrams
Figure 4.9.5-8: Deflection diagrams

Figure 4.9.5-9: Deflection diagrams and contours
Figure 4.9.5-9: Deflection diagrams and contours

ymax=4.39 mm


Notes:

  • The resulting span moments for unfavourable loadings are higher than those for global loading, only in the span along direction x.

  • Support moment remains the same, because unfavourable and global loadings are identical.

 


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