CONTINUOUS ONE WAY SLAB EXAMPLE 3.4

Figure below shows a first floor plan of an office building. It is estimated that the 125 mm thick

slab will carry 4.0 kN/m2 variable action and 1.0 kN/m2 load from finishes & suspended ceiling.

This building is exposed to XC1 exposure class. Using concrete class C30/37 and high yield steel,

prepare a complete design and detailing for this slab.

Design data:

For all slab panels, Ly/Lx = 8/3 = 2.67 > 2.0

one / two way continuous slab.

Variable action, qk = 4.0 kN/m2

Loads from finishes & suspended ceiling = 1.0 kN/m2

fck = 30 N/mm2

fyk = 500 N/mm2

h = 125 mm

con

tin

uo

us

4 @ 3 m

8 m

con

tin

uo

us

dis

con

tin

uo

us

CONTINUOUS ONE WAY SLAB EXAMPLE 3.4

SOLUTION:

1. Calculate the design load acting on the slab.

Self-weight of slab = 25 x slab thickness = 3.125 kN/m2

Finishes & ceiling = = 1.0 kN/m2

Total charac. permanent action, gk = 4.125 kN/m2

Total charac. variable action, qk = 4.0 kN/m2

Design load, n = 1.35 gk + 1.5 qk

= 1.35 ( 4.125 ) + 1.5 (4)

= 11.57 kN/m2

Consider 1 m width of slab, w = 11.57 x 1 m =11.57 kN/m

2. Design the main reinforcement.

i) Nominal cover

Minimum cover (bond), Cmin,b = bar = 10 mm

Minimum cover (durability), Cmin,dur = 15 mm

Minimum value = 10 mm

Cmin = maximum value = 15 mm

Cnom = Cmin + Cdev

= 15 + 10

= 25 mm

CONTINUOUS ONE WAY SLAB EXAMPLE 3.4

ii) Shear force and bending moment diagram

a) Bay area = ( 8 x 3 ) x 4 = 96 m2 > 30 m2 OK!

b) qk / gk = 4 / 4.125 = 0.97 < 1.25 OK!

c) qk = 4 kN/m2 < 5 kN/m2 OK!

4 @ 3 m

8 m

1 m

CONTINUOUS ONE WAY SLAB EXAMPLE 3.4

F = wL

= 11.57 x 3 m = 34.71 kN

0.46F

0.6F

0.5F 0.5F

0.5F 0.5F

0.075FL

0.086FL

0.063FL

0.063FL

0.063FL

+ + +

- - -

+ + +

- -

3 m 3 m 3m

w = 11.57 kN/m

0.04FL

3 m

+

-

0.6F

0.4F

0.086FL

0.086FL

+

-

(1.35 Gk + 1.5 Qk)

CONTINUOUS ONE WAY SLAB EXAMPLE 3.4

iii) Effective depth, d

Assume bar = 10 mm

d = h – c - bar/2 = 125 – 25 – 10/2 = 95 mm

a) At outer support

M = 0.04FL = 0.04 (34.71)(3) = 4.17 kNm/m

𝐾 =𝑀

𝑏𝑑2𝑓𝑐𝑘=

4.17 𝑥 106

1000 𝑥 952 𝑥 30= 0.015 < 0.167

Compression reinforcement is not required.

𝑧 = 𝑑 0.5 + 0.25 −𝐾

1.134 = 0.99𝑑 > 0.95𝑑

𝐴𝑠,𝑟𝑒𝑞 = 𝑀

0.87𝑓𝑦𝑘 𝑧=

4.17 𝑥 106

0.87 𝑥 500 𝑥 0.95 95 = 106 𝑚𝑚2/𝑚

Provide: H10 – 300 (As,prov = 262 mm2/m) > 𝐴𝑠,𝑚𝑖𝑛 143

b) At middle of end span

M = 0.075FL = 0.075 (34.71)(3) = 7.81 kNm/m

𝐾 =𝑀

𝑏𝑑2𝑓𝑐𝑘=

7.81 𝑥 106

1000 𝑥 952 𝑥 30= 0.03 < 0.167

Compression reinforcement is not required.

𝑧 = 𝑑 0.5 + 0.25 −𝐾

1.134 = 0.97𝑑 > 0.95𝑑

𝐴𝑠,𝑟𝑒𝑞 = 𝑀

0.87𝑓𝑦𝑘 𝑧=

7.81 𝑥 106

0.87 𝑥 500 𝑥 0.95 95 = 199 𝑚𝑚2/𝑚

Provide: H10 – 300 (As,prov = 262 mm2/m) > 𝐴𝑠,𝑚𝑖𝑛 143 OK! – bottom

Slab : 8 mm – 12 mm

CONTINUOUS ONE WAY SLAB EXAMPLE 3.4

c) At first interior support and near middle of end span

M = 0.086FL = 0.086 (37.81)(3) = 9.75 kNm/m

𝐾 =𝑀

𝑏𝑑2𝑓𝑐𝑘=

9.75 𝑥 106

1000 𝑥 952 𝑥 30= 0.04 < 0.167

Compression reinforcement is not required.

𝑧 = 𝑑 0.5 + 0.25 −𝐾

1.134 = 0.97𝑑 > 0.95𝑑

𝐴𝑠,𝑟𝑒𝑞 = 𝑀

0.87𝑓𝑦𝑘 𝑧=

9.75 𝑥 106

0.87 𝑥 500 𝑥 0.95 95 = 248 𝑚𝑚2/𝑚

Provide: H10 – 300 (As,prov = 262 mm2/m) > 𝐴𝑠,𝑚𝑖𝑛 143 OK!

d) At middle interior spans and interior supports

M = 0.063FL = 0.063 (37.81)(3) = 7.14 kNm/m

𝐾 =𝑀

𝑏𝑑2𝑓𝑐𝑘=

7.14 𝑥 106

1000 𝑥 952 𝑥 30= 0.03 < 0.167

Compression reinforcement is not required.

𝑧 = 𝑑 0.5 + 0.25 −𝐾

1.134 = 0.97𝑑 > 0.95𝑑

𝐴𝑠,𝑟𝑒𝑞 = 𝑀

0.87𝑓𝑦𝑘 𝑧=

7.14 𝑥 106

0.87 𝑥 500 𝑥 0.95 95 = 182 𝑚𝑚2/𝑚

Provide: H10 – 300 (As,prov = 262 mm2/m) > 𝐴𝑠,𝑚𝑖𝑛 143 OK!

CONTINUOUS ONE WAY SLAB EXAMPLE 3.4

Calculate As min and As max

𝐴𝑠,𝑚𝑖𝑛 = 0.26 (2.9) 1000 (95)

(500)= 143 𝑚𝑚2/𝑚 ≥ 0.0013 1000 95 = 123.5𝑚𝑚2/𝑚

𝐴𝑠,𝑚𝑎𝑥 = 0.04 𝐴𝑐 = 0.04 1000 125 = 5000 𝑚𝑚2/𝑚

Asmin < Asprov < As max => Ok!!

i) Transverse reinforcement

Provide minimum = 143 mm2/m

Provide: H8-300 (As,prov = 168 mm2/m)

1. Check the slab for shear

VEd = Vmax = 0.6F = 0.6 (37.81) = 22.69 kN

i) Calculate VRd,c

𝑘 = 1 + 200

95= 1.45 < 2.0 𝑑 𝑖𝑛 𝑚𝑚 ∴ 𝑂𝐾!

𝜌𝑙 =𝐴𝑠𝑙

𝑏𝑤𝑑=

262

1000 𝑥 95= 0.0028

𝑉𝑅𝑑 ,𝑐 = 0.12𝑘 100𝜌𝑙𝑓𝑐𝑘 13𝑏𝑤𝑑 ≥ 𝑉𝑚𝑖𝑛

= 0.12 1.45 100 0.0028 30 13 1000 95 = 33.6 𝑘𝑁

𝑉𝑚𝑖𝑛 = 0.035 𝑘3/2𝑓𝑐𝑘1/2 𝑏𝑤 𝑑

= 0.035 1.45 32 30

12 1000 𝑥 95 = 31.8 𝑘𝑁

VRd,c > Vmin

Use VRd,c = 33.6 kN

CONTINUOUS ONE WAY SLAB EXAMPLE 3.4

ii) Compare VEd with VRd,c

VEd (22.69 kN) < VRd,c (33.6 kN)

No shear reinforcement is required.

2. Deflection check

Check only at mid span with maximum moment

𝜌 =𝐴𝑠,𝑟𝑒𝑞

𝑏𝑤𝑑=

248

1000 𝑥 95= 2.61 𝑥 10−3

𝜌0 = 30𝑥 10−3 = 5.48 𝑥 10−3

< o

𝑙

𝑑= 𝐾 11 + 1.5 𝑓𝑐𝑘

𝜌𝑜𝜌

+ 3.2 𝑓𝑐𝑘 𝜌𝑜𝜌− 1

3/2

= 1.3 11 + 1.5 30 5.48

2.61 + 3.2 30

5.48

2.61− 1

32

= 63

From table 7.4N, K = 1.3 (one way continuous slab)

(i) Calculate the modification factor

310

𝜎𝑠=

500

𝑓𝑦𝑘 𝐴𝑠,𝑟𝑒𝑞

𝐴𝑠,𝑝𝑟𝑜𝑣

= 500

500 248262

= 1.06

(ii) Calculate (L/d)allowable

𝐿

𝑑 𝑎𝑙𝑙𝑜𝑤𝑎𝑏𝑙𝑒

= 𝐿

𝑑 𝑏𝑎𝑠𝑖𝑐

𝑥 𝑚𝑜𝑑𝑖𝑓𝑖𝑐𝑎𝑡𝑖𝑜𝑛 𝑓𝑎𝑐𝑡𝑜𝑟

CONTINUOUS ONE WAY SLAB EXAMPLE 3.4

𝐿

𝑑 𝑎𝑙𝑙𝑜𝑤𝑎𝑏𝑙𝑒

= 63 𝑥 1.06 = 66.78

a) Calculate (L/d)actual

𝐿

𝑑 𝑎𝑐𝑡𝑢𝑎𝑙

=𝑒𝑓𝑓𝑒𝑐𝑡𝑖𝑣𝑒 𝑠𝑝𝑎𝑛 𝑙𝑒𝑛𝑔𝑡𝑕

𝑒𝑓𝑓𝑒𝑐𝑡𝑖𝑣𝑒 𝑑𝑒𝑝𝑡𝑕 =

3000

95 = 31.6

b) Compare with (L/d)actual with (L/d)allowable

(L/d)actual < (L/d)allowable

Therefore, slab is safe against deflection.

3. Crack check

i) h = 125 mm < 200 mm OK ! (Section 7.3.3 EC2)

specific measures to control cracking is not necessary.

ii) Maximum bar spacing, smax,slabs (Section 9.3 EC2)

a) For main reinforcement:

Smax, slabs = 3h 400 mm = 375 mm

Actual bar spacing = 300 mm < Smax, slabs OK !

b) For transverse reinforcement:

Smax, slabs = 3.5h 450 mm = 437.5 mm

Actual bar spacing = 300 mm < Smax, slabs OK !

i) (L/d)actual ≤ (L/d)allowable – Beam is safe against deflection (OK!)

ii) (L/d)actual > (L/d)allowable - Beam is not safe against deflection (Fail!)

CONTINUOUS ONE WAY SLAB EXAMPLE 3.4

CONTINUOUS ONE WAY SLAB EXAMPLE 3.4

3. Draw the detailing.

Plan view

Cross-section

H10-300 (T)

H10-300 (B)

H10-300 (T)

H10-300 (B)

H10-300 (T)

T10-300 (B)

H8

-30

0 (

B)

H8

-30

0 (

T)

H8

-30

0 (

T) H

8-3

00

(B

)

H8

-30

0 (

T)

H10-300 (T)

H10-300 (B)

H10-300 (T) H10-300 (T)

H8-300 (B)

CONTINUOUS ONE WAY SLAB APPENDIX A