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Exercise 4.9.2

Given : Covering load ge=1.5 kN/m2, live load q=2.0 kN/m2, concrete quality C40/50 (E=35.2 GPa).

Question : Perform static analysis to calculate bending moments, shear forces, reaction forces, deflections and equivalent uniform loadings of slabs transferred onto beams.

Figure 4.9.2-1: Project <B_49-2>
Figure 4.9.2-1: Project <B_49-2>

Solution:

 Self-weight g o =0.15m25.0kN/m3= 3.75 kN/m2 Covering load g e = 1.50 kN/m2 Total dead load g1=g2= 5.25 kN/m2 Total live load q1=q2= 2.00 kN/m2 Design load p1=p2=γgx5.25+γqx2.0= =1.35x5.25+1.50x2.0= 10.09 kN/m2

Marcus method:

Coordinate system of the slabs is considered rotated by 90ο.

Figure 4.9.2-2
Figure 4.9.2-2

s1 : ε=ly'/lx'=1.20. From table b5.2: kx=0.453, ky=0.547,

vx=0.738, vy=0.631, am=0.0523, ρxryr=0.439, ρyerm=0.761, υxr=0.220, υyr=0.246, υyerm=0.427

px=kx p=0.453x10.09=4.57, py=ky p=0.547x10.09=5.52 kN/m

Mxx·px · lx2/8=0.738x4.57x5.02/8=10.54 kNm

Myy·py · ly2/14.22=0.631x5.52x6.02/14.22=8.82 kNm

Myerm=-py·ly2/8=-5.52x6.02/8=-24.84 kNm

fm=0.0523x10.09x103N/m2x5.04m4/(35.20x109N/m2x1.0mx0.153m 3)= =2776x10-6m=2.78 mm

Vxr=Vyrxr·p·lx=0.439x10.09x5.0=22.15 kN

Vyermyerm·p·lx=0.761x10.09x5.0=38.39 kN

pxrxr·p·lx=0.220x10.09x5.0=11.10 kN/m

pyryr·p·lx=0.246x10.09x5.0=12.41 kN/m

pyermyerm·p·lx=0.427x10.09x5.0=21.54 kN/m

Figure 4.9.2-3
Figure 4.9.2-3

s2 : ε=ly/lx=4.0/5.0=0.80. From table b5.2: kx=0.141, ky=0.859, vx=0.817, vy=0.742, am=0.0180, ρxryr=0.293, ρyerm=0.507, υxr=0.146, υyr=0.207, υyerm=0.359 a

px=kx·p=0.141x10.09=1.42 kN/m, py=ky · p=0.859x10.09=8.67 kN/m Mxx·px·lx2/8=0.817x1.42x5.02/8=3.63 kNm

Myy·py · ly2/14.22=0.742x8.67x4.02/14.22=7.24 kNm

Myerm=-py·ly2/8=-8.67x4.02/8=-17.34 kNm

fm=0.0180x10.09x103N/m2x5.04m4/(35.20x109N/m2x1.0mx0.153m 3)=955x10-6m=0.96 mm

Vxr=Vyrxr·p·lx=0.293x10.09x5.0=14.78 kN

Vyermyerm·p·lx=0.507x10.09x5.0=25.58 kN

pxrxr·p·lx=0.146x10.09x5.0=7.37 kN/m

pyryr·p·lx=0.207x10.09x5.0=10.44 kN/m

pyermyerm·p·lx=0.359x10.09x5.0=18.11 kN/m

Support s1-s2: M1-2=-(24.84+17.34)/2=-21.09 kNm

Results are transferred to the initial coordinate system producing the following stress resultants

Mx,1-2=-21.09, Mx,1=8.82, My,1=10.54, Mx,2=7.24, My,2=3.63 [kNm]

Vxr,1=22.15, Vxerm,1=38.39, Vyr,1=22.15, Vxerm,2=25.58, Vxr,2=14.78, Vyr,2=14.78 [kN]

equivalent uniform support reactions:

pxr,1=12.41, pxerm,1=21.54, pyr,1=11.10, pxerm,2=18.11, pxr,2=10.44, pyr,2=7.37 [kN/m]

b1 and b3: p=12.41, b5: p=11.10, b6: p=21.54+18.11=39.65, b2 and b4: p=7.37, b7: p=10.44 [kN/m]

Finite element method:

In project < B_49-2> the following results are obtained by activating module "SLABS" in pi-FES:

 Figure 4.9.2-4: Results summary from "Slab Results" module in pi-FES Figure 4.9.2-4: Results summary from "Slab Results" module in pi-FES Mx,1-2=-20.0 kNm Mx,1=8.5 kNm, My,1=10.2 kNm Mx,2=7.1 kNm, My,2=2.7 kNm Vxr,1=16.8 kN, Vyr,1=16.7 kN Vyerm,1=28.9 kN Vxr,2=13.1 kN, Vyr,2=10.2 kN Vyerm,2=27.0 kN y1=-2.91 mm y2=-0.85 mm

 Figure 4.9.2-5: Analytical results from ‘FEM Results’ Figure 4.9.2-5: Analytical results from ‘FEM Results’ Bending moment distribution M11 Shear force distribution V11 Deflections

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