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FLOOR CALCULUS
Composite floors with profiled steel sheet-concrete cross-sections
These composite floors can be classified in 2 categories taking into account the loading stages
and the loading carry out mode. In the first category we found floors simply supported du
ring execution and prefabricated floors, where the entire load is taken by the mixed cross-
section steel-concrete.in the second category we found monolith floors unsupported where
the profiled steel sheet is in fact the formwork and takes out the dead load and the concrete
floor load.
The calculus is made in two steps:
1. Profiled steel sheet calculus;
2. Composite section calculus.
To ensure a good collaboration between the profiled steel sheet and the concrete floor at the
contact interface we must consider the floor as a composite element, the profiled steel sheet
having a reinforcement role.
Composite floor calculus
The thickness of the composite floor () will be at least 100 [mm] and the thickness of the
profiled steel sheet it will be at least 0, 7 [mm]. The concrete floor thickness above the
profiled steel sheet () will be at least 50 [mm].
European standards recommended that the dimensions of ribs of the profiled steel sheet and
of the concrete floor will be according to the following conditions:
By taking into account the general configuration of the girders systems it is recommended
that the profiled steel sheet ribs to be disposed normally on the beams. The negative moments
in the supports are taken by the reinforcement on the superior part of the floor (the minimum
reinforcement percentage is 0, 2%).
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The superior surface on the profiled steel sheet which will be in contact with the concrete
must be proper cleaned from paint, rust, greases or from other impurities before putting in
place concrete.
Profiled steel sheet calculus
The self-weight loads of the profiled steel sheet and from the concrete will be taken only by
the profiled steel sheet during execution and can represent a fraction of 20%...50% from the
total load in the service period. The profiled steel sheet has a simply supported beam as static
scheme.
We impose ][21,12
2200*3,0
100*81,87
*
%30 3
1
11 cm
fk
MWk
y
nec
=>T.C. 60/200/0, 75/600-1300
a=6[mm]
b=200[mm]
t=0, 75[mm]
B=600[mm]
l=1300[mm]
][3,12
]/[10
3
2
cmW
cmdaNg
ef
t
yfk
W
M*1
1 Where 1k =0, 2....0, 5 based on the loads from this
stage
][3,1
]/[258
1*2500*06,0*2
085,003,0125,0*1*806,0*25001***2
*1**
*
2
ml
mdaN
Aaceb
lnnhq bribsbkb
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]/[8,3058,28322
]/[8,283258*1,1*
2
2
mdaNqqq
mdaNqnq
bt
k
b
d
b
Where : tq -is the self weight on2m of profiled steel sheet;
-we impose tq = ];/[2220*1,1* 2mdaNqn nt
-q is the load on profiled steel sheet on linear meter;
q=q*1, 35=412, 83 ]/[ 2mdaN ;
]/[21,878
3,1*83,4128
* 2
22
1, mdaNlq
MEd
We impose ][21,122200*3,0
100*81,87
*%30
3
1
11 cm
fk
MWk
y
nec
=>T.C. 60/200/0, 75/600-1300
a=6[mm]
b=200[mm]
t=0, 75[mm]
B=600[mm]
l=1300[mm]
][3,12]/[10
3
2
cmWcmdaNg
ef
t
Loads evaluation from current floor
-cold floor
No. Name of the layer d Loads Loads Loads
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(m) ]/[ 3mdaN
)/( 2mdaN
qk
n
)/( 2mdaN
qd
1. False ceiling gypsum
board
0,02 500 10
1,35
12
2. Profiled steel sheet - - 10 11
3. Reinforced concrete plate 283,8
4. Mosaic cast plates 100 130
Permanent loads from cold floor 378 - 436
Loads from partitioning walls 50 1,2 60
Total permanent loads on floor 428 1,16 496
Imposed loads 300 1,3 390Total 728 - 886
-terrace floor
No. Name of the layer d Loads Loads Loads
(m) ]/[ 3mdaN
)/( 2mdaN
qk
n
)/( 2mdaN
qd
1. False ceiling gypsum
board
0,02 500 10 12
2. Profiled steel sheet - - 22 29,7
3. Reinforced concrete
plate
0,15 2500 258 283,8
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4. Equalization level
M100
0,03 2100 36
1,35
46,8
5. Barrier against
vapors
- - 3,5 4,55
6. Sand 3 3,9
7. Thermo insulation
B.C.A.
114 136,8
8. Slope granulated
concrete
65 71,5
9. M100 cement mortar 36 46,8
10. Hydro insulation 17,5 22,75
11. Precast concrete
plates(20x20x3)
0,15 2500 100 130
Total permanent loads on
floor
663 1,35 780,34
Technological load 200 1,4 280
Snow load 250 -
Total 1113 - 1328,7
Floor design
The floor design implies the pre-dimensioning of the thickness of the floor, the stability of the
reinforcement cross-section (profiled steel sheet), the computation of joints and drawings of
execution details.
The ultimate limit state computation it will be done in the plastic domain, considering that the
steel and concrete reach the effort limit states. It is recommended that the thickness of the
compressed zone shouldnt be more than 0, 5 ah , where ah is calculus height of the
composite floor.
Floor calculus
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Considering that in exploitation limit state, the effort diagram in the compressed area is a
rectangular form, the size of the unit load is cR .In the profiled steel sheet the effort in center
weight is ,Rp where R is steel resistance
Static calculation
The static calculus will be made considering the redistribution of the plastic domain bending
moments.
CLM - longitudinal marginal frame;
CLC - longitudinal current frame;
CTCtransversal current frame;
Gsbeam;
1cM - bending moment in the first field;
1rM - bending moment in the first support;
ccM - bending moment in the current field;
rcM - bending moment in the current support;
2max
22
222
1
222
1
lg0625,0
lg0625,016
lg
lg0151,014
)lg46,0(14
lg0192,011
)lg46,0(11
sqMMM
sqs
qMM
sqsqqlongM
sqsqqlongM
ccrc
ccrc
r
c
Current floor calculus
]m*daN[58,9316
3,1*88616
slgqMM
22
ccrc
Given data:
M max=93, 58[daN*m]
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b=1[m]
steel37OL]cm
daN[2200R
]cm
daN[125R
]cm[62,11
20
100*075,0*)6*2316(
b
100*t*)a*2ec(A
]cm[13]m[13,0h
2
2c
2
t
p
p
]cm[58I
]cm[3,12W
4
3
=> profiled steel sheet
Verification of the maximum unit effort ( p )
]cm[28,13,12
586
W
Ihy 2t
Determination of the computation height of the composite floor
]cm[72,1128,113yhhtpa
- coefficient computation
]cm
daN[97,551)0023216,0*5,01(*72,11*62,11
100*6,64100*58,932,525
)*5,01(*h*A
MM
]cm
daN[3,5253,12
100*6,64
W
M
0023216,00023189,0*211211
]m*daN[6,64M
MMwhere
MMM
4,00023189,0125*72,11*100
6,64100*58,93
R*h*b
M
2
ap
12
1p
21
1
1
max2
12
2
c
2
a
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Designing the composite floor taking into consideration bending moment action in supports
M=93, 58[daN*m]
]cm[100b
]cm
daN[1253R
]cm[5,115,113ahh
]cm[13h
2c
pa
p
The reinforcement is made of PC 52 steel bars (R=3000 ]cm
daN[ 2 )
00713,000710,0*211211
4,000710,0125*5,11*100
100*58,93
R*h*b
M2
c
2
a
P-reinforcement percentage
%2,0p%03,0100*3000
125
*00713,0100*R
R
*p mina
c
We impose %2,0pmin
ml/85]cm[1,25,11*100*100
2,0h*b*
100
2,0A 2aa With ]cm[51,2A
2
eff
Roof terrace computation
Verification of the roof terrace in the maximum bending moment field
]daN[06,11216
3,1*9,1060
16
l*qM
22
q
Given data:
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steel37OL]cm
daN[2200R
]cm
daN[125R
]cm[13h
]cm[100b
]m*daN[06,112M
2
2c
p
q
]cm[58I
]cm[3,12W
4
3
=> profiled steel sheet
]cm
daN[2200R]cm
daN[9,560
]cm
daN[9,560)0033094,0*5,01(*72,11*62,11
100*6,64100*05,1092,525
)*5,01(*h*A
MM
]cm
daN[3,5253,12
100*6,64
W
M
0033094,00033039,0*211211
4,00033039,0125*72,11*100
100*)6,6406,112(
R*h*b
MM
22p
2
ap
121p
21
1
2
c
2
a
12
Designing the terrace roof floor taking into consideration bending moment action in supports
M=112, 06[dan*m]
0081647,00081313,0*211211
4,00081313,0125*5,11*100
100*06,112
R*h*b
M2
c
2
a
P-reinforcement percentage
%2,0p%034,0100*3000
125*0081647,0100*
R
R*p min
a
c
We impose %2,0pmin
ml/85]cm[1,25,11*100*100
2,0h*b*
100
2,0A 2aa With ]cm[51,2A
2
eff
Verification
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Loads:
]cm[44,1f]cm[22,0f
]cm[44,1250
360
250
Lf
]cm[22,03680*10*1,2
)100*6,3(*01,0*6,1014
*384
5
I*E
l*q
*kf
]cm/daN[1300R]cm/daN[25,11603680*26
460*39,2209
I*b
S*T
]cm/daN[2200R]cm/daN[12,705282
45,1988
W
M
]daN[39.22092
6,3*44,1227
2
l*qT
]m*dan[45,19888
6,3*44,1227
8
l*qM
]m/dan[44,12271,1*2,683,1*)35,1*30016,1*428(q*nl*)n*gn*g(q
]m/daN[6,10142,683,1*)300428(ql*)gg(q
aeff
a
6
44
T
n
2
f
2
max
22
max
c
max
22c
max
n
gsgs
n
uech
n
p
c
n
gsgs
n
u
n
p
n
Shear connectors design
Characteristics:
-diameter d=19[mm]
-head diameter de=30[mm]
-height h=100[mm]
25,1
]mm/N[450f
]mm/N[380f
2
ug
2
yg
Strength requirement
)P,Pmin(P 2Rd1RdRd
For shear force we have:
]KN[7,8110*25,1
1
*4
19*
*450*8,0
1
*4
d*
*f*8,0P 3
22
ug1Rd
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For compression we have:
]KN[3,73)7,73P,7,81Pmin(P
]KN[3,7310*25,1
1*30500*25*19*450*1*029P
]mm/N[30500E
]mm/N[25f
43,519100
dhfor1
1*E*f*d*f**029P
2Rd1rdRd
32
2Rd
2
em
ck
c
emck
2
ug2Rd
Computation of number of shear connectors for e structural element
connectors5287,513,73
727,3802
P
VN
Rd
1
Temporary climatic loads
Snow load(according STAS CR 1-1-3/2005)
Snow acts upon structures through exterior systems of forces of staic nature, distributed
on exposed building elements.
The building is in the Iasi area, for which, for a medium interval of recurrence IMR = 50
years, the characteristics value of the snow load on soil is :
= 2.5 KN/.
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The characteristic value of the snow determined load upon the exposed surface of the
considered building element is determined using the relation :
Where:
Facturated intensity of the load given by snow;
:iThe shape coefficient for the snow loading on the roof;
k,0S The characteristic value of the snow loading at the ground level, in the considered
building emplacement [KN/m2];
eC : The exposure coefficient of the construction emplacement;
t
C : Thermal coefficient;
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The exposure coefficient, of the building placement is a function of the exposure
condition of the construction, it`s values being recommanded in the following table:
= 2.5 KN/;
= 1.0;
= 1.0, because the roof has a hydroinsulation;
= 1.0, because the roof is a terrace;
Results: Sk =1 *Ce *Ct *S0,k= 1.0*1.0*1.0*2.5 = 2.5KN/ m2
Wind load(according to NP 082-2004)
The wind effects upon the buildings and structures depends on the properties of the
wind , the shape , dimensions of orientation of the building upon the wind direction , the
dynamic properties of the structures , the emplacement of the structure in the natural
environment and the neighbour constructions .
The wind load action is evaluated by the pressure of the wind , or by the forces of
given by the wind on the construction on structures.
The pressure or the forces given by the wind acts normally on the exposed surface. In
some cases there must be supplementary considered the horizontal friction forces, tangential.
The wind pressure at the level z above the surface of the ground, on the rigid
surfaces of the building is given by the relation:
W(z) = qref * ce(z) * cp , where :
qref = is the reference pressure of the wind, defined on Chapter 6 of NP 082-2004;
ce(z) = the exposure factor at the height z above the ground , defined on Chapter 6 of NP
082-2004;
Exposure time
Complete 0.8
Partial 1.0
Reduced 1.2
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cp= aerodynamic coefficient of pressure in conformity with Chapter 12 from the norms.
Given the location of building, Iasi, the value of the reference pressure of the wind will be
equal to 0, 7 kPa=0,5KN/m 2 .
Taking into consideration the height of the building-30, 77[m], and the placement of theconstruction is Iasi.
Given the dimensions of the building the length 25,5[m], width 12,0[m], and height 30,77
[m] the value of C p will be taken from the tables.
Calculation of wind action on longitudinal frame Because of the wind pressure action is
uniform distributed to a height of 10[m] and then increase linear with the height we calculate
the wind pressure first at a height of 10 [m] and then from floor to floor until the last floor.
a). Wind pressure at height z=10[m]
0926,13054,0*5779,3)10(c*)10(c)10(c
c*)10(c*q)10(w
rge
peref1
Where:
)z(cg =gust factor
)z(c r = roughness factor
5779,3]3682,0*2[5,31)]10(I*2[g1)10(cg
Where:
g=3, 5peak factory
I (z) =turbulence intensity
3054,0)10(ln*24,0)z
10(ln*)z(k)10(c
36828,0)10ln(*5,2
12,2
z
10ln*5,2
I(10)
222
0
0
2
rr
0
Where:
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9815,2]2830,0*2[5,31)]20(I*2[g1)20(cg
Where:
g=3, 5peak factory
I (z) =turbulence intensity
5169,0)20(ln*24,0)z
20(ln*)z(k)20(c
2830,0)20ln(*5,2
12,2
z
20ln*5,2
I(20)
222
0
0
2
rr
0
Where:
)z(k 0r =0, 24- type of terrain factor
0z =1, 0-for high degree of terrain roughness
For each zone the value of pC will be as follows:
;3,0C
;8,0C
;8,0C
;0,1C
pE
pD
pB
pA
Taking into consideration the values already presented, we obtain:
2E
2
D
2
B
2
A
m/KN3236,0)3,0(*5411,1*7,0w
m/KN8630,0)8,0(*5411,1*7,0w
m/KN8630,0)8,0(*5411,1*7,0w
m/KN0788,1)1(*5411,1*7,0w
c).wind pressure at height z=30[m]
8290,16663,0*7451,2)30(c*)30(c)30(c
c*)30(c*q)30(w
rge
peref1
Where :
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)z(cg =gust factor
)z(c r = roughness factor
7451,2]2493,0*2[5,31)]30(I*2[g1)30(cg
Where:
g=3, 5peak factory
I (z) =turbulence intensity
6663,0)30(ln*24,0)z
30(ln*)z(k)30(c
2493,0)30ln(*5,2
12,2
z
30ln*5,2
I(30)
222
0
0
2
rr
0
Where:
)z(k 0r =0, 24- type of terrain factor
0z =1, 0-for high degree of terrain roughness
For each zone the value of pC will be as follows:
;3,0C;8,0C
;8,0C
;0,1C
pE
pD
pB
pA
Taking into consideration the values already presented, we obtain:
2
E
2
D
2
B
2
A
m/KN3841,0)3,0(*8290,1*7,0w
m/KN0242,1)8,0(*8290,1*7,0w
m/KN0242,1)8,0(*8290,1*7,0w
m/KN2803,1)1(*8290,1*7,0w
d).wind pressure at height z=30,77[m]
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8477,16762,0*7325,2)77,30(c*)77,30(c)77,30(c
c*)77,30(c*q)77,30(w
rge
peref1
Where :
)z(cg =gust factor
)z(c r = roughness factor
7325,2]2475,0*2[5,31)]77,30(I*2[g1)77,30(cg
Where:
g=3, 5peak factory
I (z) =turbulence intensity
6762,0)77,30(ln*24,0)z
77,30(ln*)z(k)77,30(c
2475,0)77,30ln(*5,2
12,2
z
77,30ln*5,2
I(30,77)
222
0
0
2
rr
0
Where:
)z(k 0r =0, 24- type of terrain factor
0z =1, 0-for high degree of terrain roughness
For each zone the value of pC will be as follows:
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;3,0C
;8,0C
;8,0C
;0,1C
pE
pD
pB
pA
Taking into consideration the values already presented, we obtain:
2
E
2
D
2
B
2
A
m/KN3879,0)3,0(*8477,1*7,0w
m/KN0347,1)8,0(*8477,1*7,0w
m/KN0347,1)8,0(*8477,1*7,0w
m/KN2934,1)1(*8477,1*7,0w
Xxxxxxxxxxxxxxxxxxxxxxx
Calculation of wind action on transversal frame
Xxxxxx
Because of the wind pressure action is uniform distributed to a height of 10[m] and then
increases linear with the height we calculate the wind pressure first at a height of 10[m] and
then at 20[m], 30[m], 30, 77[m].
a). wind pressure at height z=10[m]
0927,13054,0*5779,3)10(c*)10(c)10(c
c*)10(c*q)10(w
rge
peref2
Where :
)z(cg =gust factor
)z(c r = roughness factor
5779,3]3682,0*2[5,31)]10(I*2[g1)10(cg
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Where:
g=3, 5peak factory
I (z) =turbulence intensity
3054,0)10(ln*24,0)z
10(ln*)z(k)10(c
3683,0)10ln(*5,2
12,2
z
10ln*5,2
I(10)
222
0
0
2
rr
0
Where:
)z(k 0r =0, 24- type of terrain factor
0z =1, 0-for high degree of terrain roughness
h>2b30, 77[m]>24, 00[m] => case c)
e=min (b, 2h) =12, 0[m] =>A=e/5=2, 4[m]
=> B * =d-A=25,5-2,4=23,1 [m]
For each zone the value of pC will be as follows:
;3,0C
;8,0C
;5,0C
;8,0C
;0,1C
pE
pD
pC
pB
pA
Taking into consideration the values already presented, we obtain:
2E
2D
2
C
2B
2A
m/KN2295,0)3,0(*0927,1*7,0w
m/KN6119,0)8,0(*0927,1*7,0w
m/KN3824,0)5,0(*0927,1*7,0w
m/KN6119,0)8,0(*0927,1*7,0w
m/KN7649,0)1(*0927,1*7,0w
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b). wind pressure at height z=20[m]
5411,15169,0*9815,2)20(c*)20(c)20(c
c*)20(c*q)20(w
rge
peref2
Where :
)z(cg =gust factor
)z(c r = roughness factor
9815,2]2830,0*2[5,31)]20(I*2[g1)20(cg
Where:
g=3, 5peak factory
I (z) =turbulence intensity
5169,0)20(ln*24,0)z
20(ln*)z(k)20(c
2830,0)20ln(*5,2
12,2
z
20ln*5,2
I(20)
222
0
0
2
rr
0
Where:
)z(k 0r =0, 24- type of terrain factor
0z =1, 0-for high degree of terrain roughness
For each zone the value of pC will be as follows:
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;3,0C
;8,0C
;5,0C
;8,0C
;0,1C
pE
pD
pC
pB
pA
Taking into consideration the values already presented, we obtain:
2
E
2
D
2
C
2
B
2
A
m/KN3236,0)3,0(*5411,1*7,0w
m/KN8630,0)8,0(*5411,1*7,0w
m/KN539,0)5,0(*5411,1*7,0w
m/KN8630,0)8,0(*5411,1*7,0w
m/KN0788,1)0,1(*5411,1*7,0w
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