## Continuous oneway solid slab

This calculation is intended to show in detail the provisions of designing a slab to Eurocode 2 using essentially the same slab as used in Example 3.2.

A 175 mm thick continuous slab is required to support screed, finishes, an office variable action of 2.5 kN / m2 and demountable partitions (@ 2 kN / m). The slab is supported on 200 mm wide load-bearing block walls at 6000 mm centres. /ck = 30, /yk = 500 and the design life is 50 years. A fire resistance of 1 hour is required.

Figure 3.3

Continuous solid slab

3.3.1 Actions

Permanent

As Section 3.2 Variable

As Section 3.2

3.3.2 Cover

Nominal cover, cnom

As Section 3.2

* A free unsupported edge is required to use 'longitudinal and transverse reinforcement' <9.3.1.4(1)> generally using U-bars with legs at least 2h long. For slabs 150 mm deep or greater, SMDSC [21] standard detail recommends U-bars lapping 500 mm with bottom steel and extending 0.1/ top into span.

As Section 3.2, BS EN 1990 Exp. (6.10b) governs <Fig. 2.5>

n = 1.25 x 5.9 + 1.5 x 3.3 = 12.3 kN / m2 <BS EN 1990 Exp. (6.10b)>

3.3.4 Analysis

Clear span, lk a, = min[h / 2; t / 2] = min[175 / 2; 200 / 2] a2 = min[h / 2; t / 2] = min[175 / 2; 200 / 2]

leff

Bending moment

ist internal support MEd = 0.086 x 12.3 x 5.9752 Internal spans and supports

Shear

End support 1st interior support

VEd = 0.40 x 12.3 x 5.975 VEd = 0.60 x 12.3 x 5.975

<5.1.1(7), Concise Table 152> <5.1.1(7), Concise Table 152>

<5.1.1(7), Concise Table 152> <5.1.1(7), Concise Table 152>

= MEd / bd2fck = 37.8 x 106 / 1000 x 1442 x 30 = 0.061

3.3.5 Flexural design: span

End span (and ist internal support) Effective depth, d d = h - cnom - (/)/2

= 175 - 25 - 12 / 2 = 144 mm Relative flexural stress , K K

.'. by inspection, section is under-reinforced (i.e. no compression reinforcement required) Lever arm, z z = (d / 2) [1 + (1 - 3.53K)05] < 0.95d

= (144 / 2) [1 + (1 - 3.53 x 0.061)05] = 0.945d = 136 mm Area of steel, As

<Appendix A1>

Internal spans and supports Lever arm z

Area of steel, As

— 27.7 x 106 / (500 / 1.15 x 137) — 465 mm2 / m

3.3.6 Deflection: end span

Check end span-to-effective-depth ratio

N = basic effective depth to span ratio: p = 0.44%

p0 = fck05 x 10-3 = 0.55% .•. use Exp. (7.16a) <7.4.2(2)>

N = 11 + 1.5fck05 Po / p + 3.2fck05 (po / P- 1)15 <Exp. (7.16a)>

= 11 + 1.5 x 3005x 0.55 / 0.44 + 3.2 x 3005 (0.55 / 0.44 - 1)15 = 11.0 + 10.3 + 2.2 = 23.5

K = structural system factor

= 1.3 (end span of continuous slab) flanged section factor

factor for long spans associated with brittle partitions = 1.0 (span < 7.0 m) F3 = 310 / where a = (fyk / yY) (4., / Asp,^) (SLS loads / ULS loads) (1 / 8)

= fd x (Asre, / Afe^ x (gk+ ¥2 qk) / (YG0k + YQqk) (1 / 8 = (500 /1.15) x (639 /645) x [(5.9 + 03 x 33) /123] x 1.08s = 434.8 x 0.99 x 0.56 x 1.08 = 260 MPa F3 = 310 / 260 = 1.19

Note: A6,pr„v / Aare, < 1.50 Allowable l / d = N x K x F1 x F2 x F3 = 23.5 x 1.3 x 1.0 x 1.19 = 36.4

Max. span = 36.4 x 144 = 5675 mm, i.e. < 5795 mm a No good

Max. span = 42.5 x 144 = 6120 mm, i.e. > 5795 mm OK

<Exp. (7.17) & BS EN 1990 A1.2.2> <Concise Table 15.14>

§ The use of Concise Table 15.2 implies certain amounts of redistribution which are defined in Concise table 15.14.

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