Figure 4.9.11: Project <B_491> Figure 4.9.11: Project <B_491>
Given: Covering load g_{e}=1.00 kN/m^{2}, live load q=5.00 kN/m^{2}, concrete quality C30/37 (E=32.8 GPa).
Question : Perform static analysis to determine bending moments, shear forces, support reaction forces and elastic deflection.
Selfweight: g_{o}=0.17m·25.0kN/m^{3}= 4.25 kN/m^{2}
Covering load: g_{e}= 1.00 kN/m^{2}
Total dead load: g= 5.25 kN/m^{2}
Total live load: q= 5.00 kN/m^{2}
The design combination in ultimate limit state (ULS) is obtained by load p=γ_{g}·g+γ _{q}·q equal to:
p=1.35g+1.50q=1.35x5.25+1.50x5.00=14.59 kN/m^{2}
Three different methods for the calculation of the twoway slab, by means of Marcus, Czerny and finite element method are presented below.
l_{x}= 4.00 m,l_{y}=5.00 m → ε=l_{y}/l_{x}=5.00/4.00=1.25
Table b5.1 for ε=1.25 gives:
k_{x}=0.709 , k_{y}=0.291, v_{x}=v_{y}=0.622 →
p_{x}=k_{x}·p·1.0m=0.709x14.59kN/m^{2}x1.0m=10.34 kN/m
p_{y}=k_{y}·p·1.0m=0.291x14.59kN/m^{2}x1.0m=4.25 kN/m
M_{x}=v_{x}·p_{x} · l_{x}^{2}/8=0.622x10.34x(4.0^{2}/8)=12.86 kNm
M_{y}=v_{y}·p_{y} · l_{y}^{2}/8=0.622x4.25x5.0^{2}/8=8.26 kNm
a_{m}=0.0689 → f_{m}=a_{m}·p·l_{x}^{4}/(E·h^{3})= 0.0689x14.59x10^{3}N/m^{2}x4.0^{4}m^{4}/(32.80x10^{9}N/m^{2}x1.0mx0.17^{3}m^{3})
→ f_{m} =1597x10^{6}m=1.597 mm
Equivalent uniform reaction forces :
υ_{xr}=0.300, υ_{yr}=0.250
Table 46, for ε=1.25 gives:
m_{x}=17.80 , m_{y}=29.90
M_{x}=p·l_{x}^{2}/m_{x}=14.59x4.00^{2}/17.80=13.11 kNm
M_{y}=p·l_{x}^{2}/m_{y}=14.59x4.0^{2}/29.90=7.81 kNm
V_{xr}=ρ_{xr}·p·l_{x}=0.39x14.59kN/m^{2}x4.00m=22.76 kN/m
V_{yr}=ρ_{yr}·p·l_{y}=0.36x14.59kN/m^{2}x4.00m=21.01 kN/m
f_{m}=a·p·l_{x}^{4}/(E·h^{3})=0.0728x14.59x10^{3}N/m^{2}x4.0^{4}m^{4}/(32.80x10^{9}N/m^{2}x0.17^{3}m ^{3})=1.687 mm
Equivalent uniform reaction forces:
Equivalent uniform reaction forces are determined in exactly the same way as in Marcus method.
In project < B_491>, the following results are obtained by activating module "SLABS" in piFES:
Figure 4.9.12: Results summary from "Slab Results" module in piFES Figure 4.9.12: Results summary from "Slab Results" module in piFES

M_{x}=12.9 kNm, M_{y}=7.6 kNm  V_{x}=22.1 kN, V_{y}=20.1 kN  

Figure 4.9.13: Analytical results from "FEM Results" of piFES Figure 4.9.13: Analytical results from "FEM Results" of piFES

Bending moment distribution M_{11}  Shear force distribution V_{11}  
Equivalent uniform reaction forces are determined in exactly the same way as in Marcus method.
M_{x} / M_{y}=12.86 / 8.26 kNm
V_{xr} / V_{yr}=29.18 / 29.18 kN
M_{x} / M_{y}=13.11 / 7.81 kNm
V_{xr} / V_{yr}=22.76 / 21.01 kN
 Finite element method gives:
M_{x}/M_{y}=12.9 / 7.6 kNm
V_{xr}/V_{yr}=22.1 / 20.1 kN


Note that the maximum bending moment of the twoway slab, i.e. M=13.00 kNm, is lower by 50% than the one of the oneway slab of identical area, i.e. M=29.18 kNm (§3.3.1, §3.3.2).

From §4.6.2.1 it is obvious that the deflection of the oneway slab (ε=l_{y}/l_{x}=?), k_{x}=1.0, ν_{x}=1.0 → a _{m}=12·c_{x}·k_{x}·ν_{x}=12x5/384=0.156, is 0.156/0.0689 = 2.26 times greater than the deflection of the twoway slab.