The general case of unfavourable loadings is described in §4.2.3 for every type of slab. This paragraph focuses on imposing and determining unfavourable loadings in the special case of a symmetric group of twoway slabs.
Unfavourable loadings are always considered as of zatrikion (chess) form. This property leads to the consideration of two equivalent loading cases, one with uniform positive loads of equal magnitude and another of zatrikion form with uniform loads of equal magnitude but alternating sign.
This technique is applied in the next example. It is worth noting that the paragraph containing the exercises of this chapter deals with the analysis of a large symmetric group of twoway slabs.
Determine the moments for ultimate limit state of the slabs illustrated in the figure; covering loads g_{επ}=1.5 kN/m^{2} and live loads a) q= 2.0 kN/m^{2}, b) 5.0 kN/m^{2}, c) 10.0 kN/m^{2}.
Figure 4.6.41: Project <Β_464> Figure 4.6.41: Project <Β_464>
To determine the critical support moment, both slabs are loaded by the total load:
For the determination of maximum and minimum spam moments, one slab is loaded by the minimum design load g_{d}=g and the other is loaded by the minimum one p_{d}=1.35g+1.50q.
The two following independent loadings p_{1} and p_{2 }are considered:
g_{d}=p_{1}p_{2}και p_{d}=p_{1}+p_{2}
g=0.15x25.0+1.5=5.25 kN/m^{2}, thus, for
a) For live load q=2.0 kN/m^{2}
p_{1}=5.25+(0.35x5.25+1.5x2.0)/2= =5.25+2.42=7.67 kN/m^{2}
p_{2}=(0.35g+1.50q)/2=2.42 kN/m^{2} >
g_{d}=p_{1}p_{2}=5.25 & p_{d}=p_{1}+p_{2}=10.09 kN/m^{2}
b) For live load q=5.0 kN/m^{2}
p_{1}=5.25+(0.35x5.25+1.5x5.0)/2= =5.25+4.67=9.92 kN/m^{2},
p_{2}=(0.35g+1.50q)/2=4.67 kN/m^{2 }>
g_{d}=p_{1}p_{2}=5.25 & p_{d}=p_{1}+p_{2}=14.59 kN/m^{2}
c) For live load q=10.0 kN/m^{2}
p_{1}=5.25+(0.35x5.25+1.5x10.0)/2= =5.25+8.42=13.67 kN/m^{2}
p_{2}=(0.35g+1.50q)/2=8.42 kN/m^{2 }>
g_{d}=p_{1}p_{2}=5.25 & p_{d}=p_{1}+p_{2}=22.09 kN/m^{2}
Each slab is solved for two types of support by means of proper coefficients for aspect ratio
Figure 4.6.42 Figure 4.6.42
Figure 4.6.43 Figure 4.6.43
Table b5.2 for slab s1(one fixed edge and the rest simply supported) gives k_{x,1}=0.453, k_{y,1}=0.547, v_{x,1}=0.738, v _{y,1}=0.631, while table b5.1 for slab s1 (simply supported all four edges) gives k_{x,2}=0.675, k_{y,2}=0.325, v_{x,2}=v_{y,2}=0.610.
The most unfavourable support moment results from the maximum loading p_{d} of both slabs. Since both geometry and loading are symmetric static behaviour of the slabs is identical. Consequently the support moment s1s2 is equal to the fixity moment of each slab.
Figure 4.6.44 Figure 4.6.44

a) For live load q= 2.0 kN/m^{2}
b) For live load q= 5.0 kN/m^{2}
c) For live load q= 10.0 kN/m^{2}

The maximum moment of each span results from loading one span with p_{d} and the other span with g_{d}, as shown in the first out of the three following figures. These loadings are equal to p_{d}=p_{1}+p_{2} and g_{d}=p_{1}p_{2} respectively. Analysis is performed by means of equivalent loadings using p_{1} and p_{2}.
Both 1^{st} equivalent loadings are symmetric, thus both slabs function as fixed on one edge and simply supported on the rest. On the contrary 2^{nd} equivalent loadings are antisymmetric, thus both slabs function as simply supported on all four edges. Consequently the two individual analyses as well as their resultants will be accurate.
  1^{st} loading of upper slab
  2^{nd} loading of upper slab

Figure 4.6.45 Figure 4.6.45
(1) elastic line, (2) moment diagram
  1^{st} loading of lower slab
  2^{nd} loading of lower slab
