Example: Example: Design Design of com posite slab Check the design of a composite slab 160 mm deep, with beams 4.5 m centers, and using the deck shown below. (This design is carried out to EN1994-1-1 part 1-1). Steel thickness, t =0.86 =0.86 mm (bare thickness of steel) Steel yield strength, f yp.d ( γ A = 1.0) yp.d = 450 MPa (γ 2 Effective area of cross section, A pe = 1276 mm 4 6 4 Second moment of area of cross section. I p = 92.77 cm /m = 0.9277 ×10 mm /m Characteristic plastic sagging moment of resistance, M Pa Pa = 9.3 kN-m/m Characteristic plastic hogging moment of resistance, 7.50 kN-m/m Distance of centroidal axis above base, e = 29.6 mm Weight of the steel deck profile = 0.103 kN/m 2 2 For resistance to longitudinal shear, m = 212 N/mm 2 = 0.1409 N/mm k = 2 For partial-interaction design, τ u.Rd = 0.240 N/mm u.Rd = 3 Volume of lightweight concrete , 0.127 m per sq.m of floor 3 Weight of concrete (dry density), 14.5 kN/m Compressive strength, f cm =55 MPa, E cm = 17.5 GPa cm = 60 MPa and f ck ck =55 Loads 2 Imposed load = 5.0 kN/m 2 Partitions (imposed load) = 1.2 kN/m 2 Floor finish = 1.3 kN/m 2 Construction load (temporary) = 1.0 kN/m 9m
4.5 m
Plan
Figure 25 Design example –structure for a typical floor At construction condition
Self weight of slab is,
Composite Slabs with Profiled Steel Sheeting
2
g k1 = 0.103 +0.127×14.5=1.9445 kN/m 3 In EN 1991-1-1, for fresh concrete the weight density should be increased by 1.0 kN/m , so initial loading on the sheeting 3 g k1 = 1.9554+0.127 =2.0715 kN/m
steel mesh (bar dia 10mm) Lightweight filler
ht =160
d d
130
Figure 26 Cross section of profiled sheeting and composite slab 2 Total design load, w d = 2.0715 ×1.35 + 1.0 ×1.5 = 4.30 kN/m It is assumed that the sheet is propped at mid-span during construction. Effective length, L e = (4.5-0.1)/2 = 2.2 m Wet concrete
Imposed load
Sheeting A
50 mm
B 2200 mm
2200 mm
C 50 mm
Fig. 27 Profile sheeting during construction
2
Design sagging moment, M Ed = 4.30 ×2.2 /8 = 2.60 kN-m/m < M Pa =9.3 kNm/m 2 Design hogging moment at prop location, M Ed =0.125×4.3×2.2 =2.6 kNm/m < 7.5 kNm/m (considering both span loaded) Deflection 2 Permanent load on the sheeting = 2.0715 + 1.0 =3.0715 kN/m Maximum deflection in AB, if BC is unloaded and C is attached with beam flange 4
δ max =
wLe
185 Ea I p
=
3.0715 × 2.24 185 × 0.210 × 0.9277
= 2.0 mm , (span/1102)
2
Composite Slabs with Profiled Steel Sheeting
If C is not attached with beam flange, 4
δ max =
5wLe
384 Ea I p
=
5 × 3.0715 × 2.24 384 × 0.210 × 0.9277
= 4.81 mm , (Span/457)
These are less than the limit of span/180 and also less than 1/10 of slab thickness = 16 mm, so the ponding effect may be ignored in the design of steel sheeting.
At composite condition 9.7.2 & 9.7.5
Flexural and vertical shear All slabs are simply supported with effective length, L e = 4.40 m for moment calculation. Dead load, g k = 1.9445 (slab) + 1.3 (floor finish) = 3.2445 kN/m 2 2 Live load, q k = 5.0 (imposed) + 1.2 (partitions) = 6.2 kN/m 2 The design ultimate load, w d = 3.2445 ×1.35 + 6.2 ×1.5 = 13.68 kN/m Applied moment in composite condition 2
2
M Ed = w d × L e /8 = 13.68 ×4.4 /8 = 33.11 kN-m/m Flexural resistance of the slab:
For equilibrium, N p = N cf Force in steel: N cf = A pe × f yp.d = 1276×450 = 574.2 kN/m
hc = 100 mm
Neutral axis depth in concrete N cf 574.2 ×103 x pl = = 18.424 mm < h c = 100 mm = 0.85 b f cd 0.85 ×10 3 × (55 /1.5 )
∴ z = d p − 0.5x =130.4 − 0.5×18.424 = 121.188 mm Plastic moment resistance of the composite section M Rd = N cf × z = 574.2 ×121.188/1000 = 70.0 kN-m/m ≥ applied, M Ed = 33.11 kN-m/m
Therefore, the plastic moment resistance of the section is more than adequate, subject to check on longitudinal shear.
3
Composite Slabs with Profiled Steel Sheeting
Vertical Shear Resistance The design vertical shear for a span of 4.5 m is V Ed = 2.25(3.2445 ×1.35 + 6.2 ×1.5) = 30.78 kN /m
b =300
b0
Shear strength of concrete, 32
vmin
= 0.03 1 + (200 / d p )1 2
vmin
= 0.03 1 + (200 / 200)1 2
12
f ck
32
551 2 =0.629 N/mm
2
From Equation 27 with b 0 = 144.5 mm, b = 300 mm and d p = 130.4 mm (Fig. 20), Vv. Rd = ( b0 b ) d pvmin = (144.5 300) × 130.4× 0.734 = 39.51 kN/m > V Ed = 30.78 kN/m
which is sufficient.
Composite condition - Longitudinal shear 9.7.3
Longitudinal shear will be checked by both ‘ m-k’ and ‘ partial-interaction’ methods. Longitudinal shear by ‘m-k’ method : For longitudinal shear, it is assumed that there is no end anchorage, so that clause 9.7.3 is applicable. From clause 9.7.3(4), the design shear resistance is Vl.Rd = b.d p ( m
A pe bLs
+ k)
1
γ VS 2
Where, b = 1.0 m, d p = 130.4 mm, A pe = 1276 mm /m, L s = L/4 = 4500/4 = 1125 mm, m 2 2 = 212 N/mm , k = 0.1409 N/mm and γ VS =1.25
V l.Rd = 1000 ×130.4 × 212 ×
1 = 39.78 kN /m > V Ed = 30.78 kN/m + 0.1409 1000 ×1125 1.25 ×1000 1276
So, the slab is strong enough using this method.
Longitudinal shear resistance by ‘ partial-interaction ’ method: The design shear resistance of the profile with this concrete, τ u.Rd = 0.240 N/mm
2
For full composite or full shear connection, the required length of shear span to develop the force N cf (in absence of any anchorage system) is Lsf =
N cf
τ u , Rd b
= 574.2 /(0.240×1000) = 2.393 m > Span(4.4 m)/2 = 2.2 m
So full interaction is not achieved in a span of this length. The depth of full interaction stress block in the concrete is x pl = 18.424 mm,
4
Composite Slabs with Profiled Steel Sheeting
The degree of shear connection is given by η = N c / N cf = L x / L sf = x/ x pl N c is the force in the concrete block and x is the depth of stress block with partial interaction z = h t – 0.5η x pl – e p =160-0.5×η×18.424– 32.3 = 127.7- 9.212 η mm [Note: e p should be given by the manufacturer, if not given, assume e = e p ] M pr =1.25 × 9.3 × (1- η ) = 11.625 - 11.625η ≤ M pa = 9.3 kNm so η ≥ 0.2
The plastic resistance is 2 M Rd = N c z + M pr = 574.2 η( z/1000)+ M pr = 73.325η -5.29η + 11.625-11.625η M Rd = 11.625 + 61.7η − 5.29η
2
for η ≥ 0.2
For η ≤ 0.2 2 M Rd = N c z + M pr = 73.325η -5.29η +9.3 (here M pr = M pa ) 2 = 9.3 + 73.325η -5.29η
70 Bending moment 60 kNm
50
M Rd
40 30
M Ed
20
9.3 10 0 0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Fig. 28 Design partial-interaction diagram
The curve M Rd is plotted with curve M Ed . The curve M Rd lies above the curve M Ed at all the cross-sections, showing that there is sufficient resistance to longitudinal shear. Local effects of point load
The design point load is, Q Ed = 7.5×1.5 = 11.25 kN acting on any area 50 mm square. a p = b p = 50 mm, thickness of floor finish h f = 20 mm, d p = 130.4 mm. The small rib of the sheeting is neglected, so h c = 100 mm 5
Composite Slabs with Profiled Steel Sheeting
Punching area, cP = 2π hc + 2 ( bP + 2hf ) + 2 ( aP + 2hf + 2d P − 2hc )
=2π×100+2(50+2×20)+2(50+2×20+2×130.4-2×100)=1110 mm Steel mesh (10 mm bar at 300 mm spacing, both ways) is used on the top of the sheeting. 2 2 The area of the steel bar per unit width of slab, A sx = A sy = π(10 /4)/300 = 0.262 mm . The reinforcement ratios are, 1/2 p x = p y = 0.262/100=0.00262 , therefore, p l = (0.00262 ×0.00262) = 0.00262 < 0.02 The design shear stress for lightweight concrete is, 1 1 vl.Rd = ClRd,cη1 1 + ( 200 / d ) 2 (100 ρ l f lck ) 3 ≥ η 1vmin
where, η 1 = 0.40+0.60ρ/2200 in which ρ is the upper limit of the density for the relevant class in accordance with EN 1992-1-1, Table 11.1. The recommended value for C lRd,c is 0,15/γ c . d is the mean of the effective depths of the two layers of reinforcement, but not less than 200 mm. η 1 = 0.40+0.60 ×1450/2200 = 0.795 v l.Rd = (0.15/1.5) ×0.795×[1+1]×(100×0.00262×55) = 0.387 n/mm 1/3
2
The design punching resistance, V Rd = v l.Rd c p d = 0.387 × 1110 × (100-10) =38.66 kN > Q Ed This is more than sufficient to resist concentrated load Q Ed =11.25 kN. Serviceability of composite slab Deflection The modular ratio n can be considered as the average value of the short and long term modular ratio. n 0 = E s /E cm = 210/17.5 = 12 Note: The long-term compressive strain of these concretes under permanent loads is about three times the initial elastic strain, due to creep. In elastic analysis, this would require separate calculations for permanent and variable loads. For buildings, EN 1994 permits the simplification that all strains may be assumed to be twice their short-term value. This is done by using modular ratios n = 2n 0 . Therefore, n =2n 0 =2×(210/17.5) = 24 Neutral axis form top surface of the cracked composite section, xc =
nA pe
1 + 2bd p − 1 = 24 ×1276 1 + 2 ×1000 ×130.4 − 1 = 63.845 mm 1000 24 × 1276 b nA pe
For sagging loading, the second moment of area I cc per unit width of the cracked section is,
6
Composite Slabs with Profiled Steel Sheeting
x 2 bxc ( c ) 2 + I cc = 12n n 3
bxc
+ A p (d p − xc ) 2 + I p =
1000 × 63.8453 3 × 24
+ 1276 × (130.4 − 63.845) 2 + 0.9277 ×10 6
∴ I cc =10.2 ×106 mm4 /m For un-cracked section Position of the neutral axis bhc hc b bd d b bd d + A pe d p + s d d + hc + u d d + hc n 2 nCs 3 nCs 2 =79.65 mm xu = bhc bd (b + b ) + A pe + d u s n nC s
Where, W r = average rib width d d = steel deck depth = 60 mm, b u = Bottom flange dimension = 120 mm b s = Horizontal width of the inclined web = 25 mm Second moment of area for un-cracked composite section,
3
I u =
bhc
12n
bhc xu −
+
hc 2
n
2 2 bd d bs d d
2 2 2 dd dd + + + bs ht − xu + + bu ht − xu − + Ape ( d p − xu ) + I p 12 3 2 nCs 18 2
bu d d
∴Iu = 16.4×106 mm4 /m I avg =
I c + I u
2
= (10.2 ×10 +16.4×10 )/2 = 13.33×10 mm /m 6
6
6
4
2
The self weight of the slab is 1.9445 kN/m , so the load on the prop at B, treated as the central support of a two span beam, is F = 2×0.625×1.9445 ×2.2 =5.35 kN/m It is assumed that this is a line load acted on the slab at mid-span after removing of the prop. In service stage, the total un-factored load acting on the slab, w k = Dead load, g k + Live 2 load, q k = 3.2445 +6.2 = 9.4445 kN/m
For a simply supported slab, the mid-span deflecti on, 3
δ =
Le F
48 Es I a ve
+
5 384
4
wk Le =
3.40 + 16.5 = 19.9 mm
Hence, δ/Le = 19.9/4400 =1/221 The calculated deflection should be checked with the national annex. If it is found that the deflection is excessive, additional measure should be taken to reduce the deflection.
7
