Design of a 1-way spanning, simply supported slab

3.5 m

Specifications

fck = 30 N/ mm2

fyk = 500 N/mm2

Gk = 5.0 kN/ m2 (exclusive of self-weight)

Qk = 2.5 kN/ m2

Assumptions: Design life is 50 years and resistance time to fire is 1 hour.

Step 1: Estimating the slab depth

Trying out a basic span to effective depth ratio of 25.

Thus, effective depth, d =

=

3500

25 =

140

. mm.

Since high strength steel is being used and the span is less than 7 m, the correction factors can

be taken as unity (1.0). Therefore, d = 140 mm.

Step 2: Nominal cover dimension and overall slab depth (h)

Cnominal = Cminimum + Cdeviation

Cmin = max [minimum cover due to bond (Cmin b); minimum cover due to environmental

conditions (Cmin dur)]

Assuming bars of 10 mm to be used, Cmin b = 10 mm.

Assuming an exposure of class XC 1 & using C 30/ 37 concrete, Cmin dur = 15 mm.

Thus, Cmin = 15 mm.

Cdeviation = allowance in design for deviation.

Let Cdev = 10 mm.

Therefore, Cnom = 15 + 10 = 25 mm.

Overall depth of slab, h = d + Cnom +

2 = 140 + 25 + 5 = 170 mm.

Thus, h = 170 mm.

Furthermore, according to specific tables from EC 2, the depth, d, and size of nominal cover

chosen is adequate for a fire resistance of 1 hour and a design life of 50 years.

Step 3: Slab loading

Self-weight of slab = Overall depth, h x unit weight of reinforced concrete

Self-weight = 170 x 10-3 x 25 = 4.25 kN/m2.

Total dead load, Gk = Self weight of slab + Permanent actions = 4.25 + 5.0 = 9.25 kN/m2.

Total live load, Qk = 2.5 kN/m2.

For a 1-metre width (b) of slab:

Ultimate load, n = [1.35 Gk + 1.5 Qk] [b]

Thus, n = [1.35 (9.25) + 1.5 (2.5)] [1] = 16.24 kN/m

Step 4: Analysis

Maximum bending moment, M =

8 =

16.24 3.5 3.5

8 = 24.87 kNm.

Maximum shear force, V =

2 =

16.24 3.5

2 = 28.42 kN.

Step 5: Flexural Design

Moment coefficient, K =

fck =

24.87 10^6

1000 140 140 30 = 0.042 (< 0.156, thus acceptable).

Lever arm, z = 0.95 d = 0.95 x 140 = 133 mm.

As =

0.87 =

24.87 10^6

0.87 500 133 = 430 mm2/m.

Assuming bars of 10 mm to be used:

1

< or = As

Spacing = 1

As =

78.5

430 = 0.182 m = 182 mm.

Thus, chosen spacing = 175 mm.

Hence, provide H10 at 175 mm (B1) spacing C/C, with As provided = 449 mm2/m.

Step 5: Deflection check

Checking for r.

Since, r = 100

=

100 430

1000 140 = 0.31 % (> 0.13 % minimum requirement)

For r = 0.31% and fck = 30 N/mm2, the basic span to effective depth ratio = 39.

Actual span to effective depth ratio = 25.

Since basic ratio > actual ratio, the depth can be considered acceptable.

Step 6: Shear check

Vmax = 28.42 kN/m

VEd = Vmax 0.14 (ultimate load) = 28.42 0.14 (16.24) = 26.14 kN.

Determination of r1 = 100

=

100 449

1000 140 = 0.32 % (< 0.4 % - hence no

correcting factor is necessary)

Therefore, V Rd, c (shear capacity without any shear reinforcement) = 0.55 (since r1< 0.4%)

x b x d = 0.55 x 1000 x 140 = 77 kN.

Since VEd < VRd, c, no shear reinforcement is required.

Step 7: Detailing checks

a) Minimum areas of reinforcement

As, min = 0.26 (fctm/fyk) btd

As, min = 0.26 [0.30 (30^0.666)

500] x 1000 x 140 = 210 mm2/m.

Thus, since As provided is greater than the minimum, H10 @ 175 mm C/C is acceptable.

b) Secondary reinforcement (Distribution steel)

Secondary reinforcement = 20 % x As required.

Thus, secondary reinforcement = (20/100) x 430 = 86 mm2/m.

Assuming bars of 8mm to be used as distribution steel.

Spacing = 1

Area =

50.3

86 = 0.584 m = 584 mm.

Since maximum spacing can be 300 mm, the spacing for the secondary reinforcement is

taken as same.

Thus, provide H08 @ 300 mm (B2 T2) spacing C/C, with secondary reinforcement area

of 168 mm2/m.

c) Crack control

Maximum spacing of bars should be < 3h or less than 400 mm, whichever is the maximum.

3 h = 3 x 170 = 510 mm.

Since maximum spacing = 300 mm, crack control is considered to be appropriate.

d) End anchorage

For fck 30/35 N/mm2, the tension anchorage length is given as 51 of main bars.

Thus, tension anchorage length = 51 x 10 = 510 mm, which is taken as 525 mm.

e) Curtailment

Curtailment of main bars: - Curtail main bars at 50 mm from support.

At supports: - 50% of As to be anchored from face of support.

Therefore, 50% of As = 0.5 x 430 = 215 mm2/m

Assuming bars of 12 mm to be used, spacing = 113.1 / 215 = 0.526 mm = 526 mm.

The maximum spacing is thus chosen. Hence, use H12 @ 300 mm B1 T1 U-bars.