Example Composite Floor Slab
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Example: Composite floor slab
SX009a-EN-EU
Document Ref:
CALCULATION SHEET
Sheet
7
of
12
Title
Example: Composite floor slab
Eurocode Ref
EN 1994-1-1, EN 1993-1-3, EN 1992-1-1 & EN 1993-1-1
Made by
Jonas Gozzi
Date
March 2005
Checked by
Bernt Johansson
Date
April 2005
For full shear connection: Mpl,Rd
M pl,Rd
=
Ap
⋅
fyd
⋅ ( dp −
xpl / 2
)
= 955 ⋅ 320 ⋅ (101 − 21, 6 / 2 ) ⋅ 10 −3 = 27,5 kNm/m > 20,5 = M Ed
Longitudinal shear by partial connection method: Shear span required for full shear connection Nc
= τ u,Rd ⋅ b⋅
Lx
≤ N cf
The distance to the nearest support, Lx, required for full shear connection can be determined by
t n e m e e r g A e c n e c i L l e e t S s s e c c A e h t f o s n o i t i d n o c d n a s m r e t e h t o t t c e j b u s s i t n e m u c o d s i h t f o e s U . d e v r e s e r 9 s 0 t 0 h g 2 i , r 9 l l 0 a e n t u h g J i , r y y a p d o s c e s u i l T a n i r o t e d a e t m a s i e r h C T
Lx
=
τ u,Rd Lx
=
N cf b ⋅ τ u,Rd
=
τ u,Rk γ Vs
= =
Ap ⋅ f yd b ⋅ τ u,Rd
0, 306 1, 25
955 ⋅ 320 1000 ⋅ 0,245
= 0,245 N/mm
2
= 1247 mm
Hence, at a distance of 1247 mm from the support a full shear connection is fulfilled. Design check using the simplified partial interaction diagram: For any cross section along the span it has to be shown that the corresponding design bending moment, M Ed, does not exceed the design bending resistance, M Rd. In the figure x is the distance from the support.
EN 1994-1-1 §9.7.3 (8)
Example: Composite floor slab
CALCULATION SHEET
SX009a-EN-EU
Document Ref:
Sheet
8
of
12
Title
Example: Composite floor slab
Eurocode Ref
EN 1994-1-1, EN 1993-1-3, EN 1992-1-1 & EN 1993-1-1
Made by
Jonas Gozzi
Date
March 2005
Checked by
Bernt Johansson
Date
April 2005
M Rd, M Ed
[kNm/m] 30
M pl,Rd M Rd 20
10
M= Ed
[γ G ⋅ ( g1 + g 2 ) + γ Q ⋅ q ] ⋅ x 2
M pa t n e m e e r g A e c n e c i L l e e t S s s e c c A e h t f o s n o i t i d n o c d n a s m r e t e h t o t t c e j b u s s i t n e m u c o d s i h t f o e s U . d e v r e s e r 9 s 0 t 0 h g 2 i , r 9 l l 0 a e n t u h g J i , r y y a p d o s c e s u i l T a n i r o t e d a e t m a s i e r h C T
⋅ ( −L ) x x [m]
0 0
0.4
0.8
1.2
1.6
2
Lx M Ed
≤ M Rd for all cross sections
Vertical shear: V Ed
=
V Ed
=
[γ G ⋅ ( g1 + g 2 ) + γ Q ⋅ q ] ⋅ L 2
[1, 35 ⋅ ( 2, 6 + 1, 2) + 1, 5 ⋅ 5, 0] ⋅ 3, 6 2
= 22,7 kN/m
Design vertical shear resistance: Vv,Rd
= ⎡⎣CRd,c ⋅ k ⋅ (100 ⋅ ρ I ⋅
1 / 3
f ck )
+ k1 ⋅ σ cp ⎤⎦ ⋅ bw ⋅ d p
EN 1992-1-1 §6.2.2
with a minimum of Vv,Rd,min
C Rd,c
=
k = 1 +
= (vmin + k1 ⋅ σ cp ) ⋅ bw ⋅ d p 0,18
γ C
=
200 d p
0,18 1, 5
= 1+
= 0,12 200 101
= 2, 4
See Note in EN 1992-1-1 §6.2.2
Example: Composite floor slab
CALCULATION SHEET
ρ l
=
Asl bw ⋅ d p
Document Ref:
SX009a-EN-EU
9
Sheet
of
12
Title
Example: Composite floor slab
Eurocode Ref
EN 1994-1-1, EN 1993-1-3, EN 1992-1-1 & EN 1993-1-1
Made by
Jonas Gozzi
Date
March 2005
Checked by
Bernt Johansson
Date
April 2005
≤ 0,02
Asl is the area of the tension reinforcement in [mm], i.e Asl = Ap bw = 400 mm/m, i.e. the smallest width in [mm] of the section in the tension area.
ρ l
=
σ cp
955 400 ⋅101
=
N Ed Ac
= 0,024 > 0,02 ρ l = 0,02
= 0 , since N Ed = 0, i.e. no axial forces or prestress. See Note in EN 1992-1-1 §6.2.2
k 1 = 0,15 V v,Rd t n e m e e r g A e c n e c i L l e e t S s s e c c A e h t f o s n o i t i d n o c d n a s m r e t e h t o t t c e j b u s s i t n e m u c o d s i h t f o e s U . d e v r e s e r 9 s 0 t 0 h g 2 i , r 9 l l 0 a e n t u h g J i , r y y a p d o s c e s u i l T a n i r o t e d a e t m a s i e r h C T
= ⎡⎣ 0,12 ⋅ 2, 4 ⋅ (100 ⋅ 0,02 ⋅ 25 )1 / 3 + 0,15 ⋅ 0 ⎤⎦ ⋅ 400 ⋅ 101
V v,Rd = 42,8 kN/m
Minimum value vmin
= 0, 035 ⋅ k 3 / 2 ⋅ f ck1/2 = 0 , 035 ⋅ 2, 43 / 2 ⋅ 251 / 2 = 0,65
V v,Rd,min
= (0, 65 + 0,15 ⋅ 0) ⋅ 400 ⋅ 101= 26,3 kN/m
OK
V v,Rd = 42,8 kN/m > 22,7 kNm/m = V Ed
All design checks of the composite slab in the ultimate limit state are OK. Serviceability limit state:
Cracking of concrete: As the slab is designed as simply supported, only anti-crack reinforcement is EN 1994-1-1 needed. The cross-sectional area of the reinforcement above the ribs should be §9.8.1 (2) not less than 0,4% of the cross-sectional area of the concrete above the ribs. min As = 0, 004 ⋅ b⋅ hc = 0, 004 ⋅ 1000 ⋅ 75 = 300 mm /m 2
φ8 s160 mm will be enough for this purpose.
Example: Composite floor slab
SX009a-EN-EU
Document Ref:
CALCULATION SHEET
10
Sheet
of
12
Title
Example: Composite floor slab
Eurocode Ref
EN 1994-1-1, EN 1993-1-3, EN 1992-1-1 & EN 1993-1-1
Made by
Jonas Gozzi
Date
March 2005
Checked by
Bernt Johansson
Date
April 2005
Deflection: For the calculations of the deflections of the slab, the slab is considered to be continuous. The following approximations apply:
•
the second moment of area may be taken as the average of the values for the cracked and un-cracked section;
•
for concrete, an average value of the modular ratio, n, for both longand short-term effects may be used. n=
Ep ' E cm
=
E p E ⎞ ⎛ ⋅ ⎜ E cm + cm ⎟ 2 ⎝ 3 ⎠
1
=
210000 2 ⋅ 31000 3
≈ 10
Second moment of area for the cracked section t n e m e e r g A e c n e c i L l e e t S s s e c c A e h t f o s n o i t i d n o c d n a s m r e t e h t o t t c e j b u s s i t n e m u c o d s i h t f o e s U . d e v r e s e r 9 s 0 t 0 h g 2 i , r 9 l l 0 a e n t u h g J i , r y y a p d o s c e s u i l T a n i r o t e d a e t m a s i e r h C T
Ibc
=
b ⋅ xc3
3⋅ n
+
Ap ⋅ ( dp
− xc )2 +
∑ Ai ⋅ zi n ⋅ Ap xc = = b ∑ Ai xc
I bc
=
10 ⋅ 955
=
1000
⎛ ⋅ ⎜⎜ ⎝
1+
1000 ⋅ 35, 43 3 ⋅ 10
⎛ ⋅⎜ ⎜ ⎝
1+
Ip
⎞ − 1⎟ ⎟ ⎠
2 ⋅ b ⋅ dp n⋅ Ap
2 ⋅1000 ⋅ 101 10 ⋅ 955
⎞ − 1⎟⎟ = 35,4 mm ⎠
+ 955 ⋅ (101 − 35, 4 )2 + 33,0 ⋅ 10 4 = 5,92 ⋅10 6 mm4 /m
Second moment of area for the un-cracked section 2 b0 ⋅ hp3 b0 ⋅ hp ⎛ hc ⎞ ⎛ + ⋅ ⎜ u x− ⎟ + + ⋅ ⎜ t h− buI = n ⎝ n 12 ⋅ n 2⎠ 12 ⋅ n ⎝ Ap ⋅ ( dp − xu )2 + Ip
b ⋅ hc3
2
b⋅ xu
=
hc
2
b ⋅ hc
h ⎞ ⎛ + b0 ⋅ hp ⋅ ⎜ ht − p ⎟ + n ⋅ Ap ⋅ d p 2⎠ ⎝ b ⋅ hc + b0 ⋅ hp + n ⋅ Ap
−
ux
hp
2
2
⎞ ⎟ + ⎠
EN1994-1-1 §9.8.2 (5)
Example: Composite floor slab
Document Ref:
CALCULATION SHEET
1000 ⋅ xu
12
of
Title
Example: Composite floor slab
Eurocode Ref
EN 1994-1-1, EN 1993-1-3, EN 1992-1-1 & EN 1993-1-1
Made by
Jonas Gozzi
Date
March 2005
Checked by
Bernt Johansson
Date
April 2005
45 + 650 ⋅ 45 ⋅ ⎛⎜120 − ⎞⎟ + 10 ⋅ 955 ⋅ 101 2 2 ⎠ ⎝ = 58,3 mm 1000 ⋅ 75 + 650 ⋅ 45 + 10 ⋅ 955
1000 ⋅ 753
=
11
Sheet
752
=
I bu
SX009a-EN-EU
12 ⋅ 10
+
1000 ⋅ 75 10
2
75 ⎞ 610 ⋅ 453 ⎛ ⋅ ⎜ 58, 3 − ⎟ + + 2 ⎠ 12 ⋅10 ⎝ 2
610 ⋅ 45
45 ⋅ ⎛⎜120 − 58,3 − ⎞⎟ + 955 ⋅ (101 − 58,3)2 + 10 2 ⎠ ⎝ 33, 0 ⋅ 104 = 13, 5 ⋅106 mm 4 /m Average I b of the cracked and un-cracked section I b t n e m e e r g A e c n e c i L l e e t S s s e c c A e h t f o s n o i t i d n o c d n a s m r e t e h t o t t c e j b u s s i t n e m u c o d s i h t f o e s U . d e v r e s e r 9 s 0 t 0 h g 2 i , r 9 l l 0 a e n t u h g J i , r y y a p d o s c e i s u l T a i n r o t e d a e t m a s i e r h C T
=
I bc
+ I bu 2
=
5, 92 + 13, 5 2
⋅ 106 = 9, 7 ⋅ 106 mm 4 /m
Deflections The total deflection under the worst load case should not exceed L/250.
EN1992-1-1 §7.4.1(4)
Weight of floor finishes: 0, 0068 ⋅ g 2 ⋅ L4
=
δ c,g2
E ⋅ I b
=
0, 0068 ⋅ 1, 2 ⋅ 3600 4 210000 ⋅ 9, 7 ⋅ 10
Live load, worst case:
δ c,q
=
6
= 0,67 mm
q
0, 0099 ⋅ψ 1 ⋅ q ⋅ L4 E ⋅ I b
=
q
0, 0099 ⋅ 0, 7 ⋅ 5, 0 ⋅ 3600 210000 ⋅ 9, 7 ⋅ 10
6
4
= 2,86 mm
Removal of the props: G1'
G1′ = g1 ⋅
δ
c,G1′
=
L
2
= 2, 6 ⋅
3, 6 2
G1'
= 4,68 kN/m
0, 01146 ⋅ G1′ ⋅ L3 E ⋅ I b
G1'
=
0, 01146 ⋅ 4680 ⋅ 3600 210000 ⋅ 9, 7 ⋅ 10
6
3
= 1,23 mm
e
E
g
n
x
a
m
p
l
Document Ref:
CALCULATION SHEET
i
u r
J y
p
a
d s
s i
e
u l
T a
i
n r
o e
t
d a
e m
t
a s
e i
r h
C T
C
o
SX009a-EN-EU
Sheet
12
p
of
o
12
Example: Composite floor slab
Eurocode Ref
EN 1994-1-1, EN 1993-1-3, EN 1992-1-1 & EN 1993-1-1
Made by
Jonas Gozzi
Date
March 2005
Checked by
Bernt Johansson
Date
April 2005
= δ c,G ′ + δ c,g + δ c,q = 1, 23 + 0, 67 + 2, 86 = 4,76 mm
δ c
= 4,76 mm <
1
m
Title
δc
2
o
c
:
Total deflection:
,
y
e
L
250
=
3600 250
= 14,4 mm
OK
s
i
t
Example: Composite floor slab
Example: Composite floor slab SX009a-EN-EU
Quality Record RESOURCE TITLE
Example: Composite floor slab
Reference(s) ORIGINAL DOCUMENT Name
Company
Date
Created by
Jonas Gozzi
SBI
10/03/2005
Technical content checked by
Bernt Johansson
SBI
08/04/2005
1. UK
G W Owens
SCI
7/7/05
2. France
A Bureau
CTICM
17/8/05
3. Sweden
A Olsson
SBI
8/8/05
4. Germany
C Muller
RWTH
10/8/05
5. Spain
J Chica
Labein
12/8/05
G W Owens
SCI
06/7/06
Editorial content checked by Technical content endorsed by the following STEEL Partners:
t n e m e e r g A e c n e c i L l e e t S s s e c c A e h t f o s n o i t i d n o c d n a s m r e t e h t o t t c e j b u s s i t n e m u c o d s i h t f o e s U . d e v r e s e r 9 s 0 t 0 h g 2 i , r 9 l l 0 a e n t u h g J i , r y y a p d o s c e s u i l T a n i r o t e d a e t m a s i e r h C T
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