SIMPLY SUPPORTED ONE WAY SLAB EXAMPLE 3.3

Design a one way slab which is simply supported on two brickwalls with a distance of 3

m centre to center.

Design data:

Finishes etc. (not including self-weight) = 0.35 kN/m2

Imposed load (variable action) = 2.5 kN/m2

Concrete grade = C25/30

Steel grade, fyk = 500 N/mm2

Cover, c = 20 mm

Assume bar = 10 mm

Slab thickness, h = 125 mm

SOLUTION:

1. Calculate the loads acting on the slab.

Loadings acting on slab

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

Finishes etc. = 0.35 kN/m2

Total characteristic permanent action, gk = 3.475 kN/m2

Total characteristic variable action, qk = 2.5 kN/m2

Design load, n = 1.35 gk + 1.5 qk

= 1.35 ( 3.475 ) + 1.5 ( 2.5 )

= 8.44 kN/m2

Consider 1 m width of slab = n x 1 m

= 8.44 x 1 m

= 8.44 kN/m

Min bar size = 8 mm

Max bar size = 12

mm

SIMPLY SUPPORTED ONE WAY SLAB EXAMPLE 3.3

2. Draw the shear force and bending moment diagram.

3. Design the main reinforcement.

i) Calculate the effective depth, d

Assume bar = 10 mm

d = h – c - bar/2 = 125 – 20 – 10/2 = 100 mm

𝐾 =𝑀

𝑏𝑑2𝑓𝑐𝑘=

9.5 𝑥 106

1000 𝑥 1002 𝑥 25= 0.038 < 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=

9.5 𝑥 106

0.87 𝑥 500 𝑥 0.95 100 = 230 𝑚𝑚2/𝑚

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

12.66 kN

12.66 kN

9.5 kNm

+

+

-

Maximum shear force, Vmax = wL/2

= 8.44 (3) /2

= 12.66 kN/m width

Maximum moment, M max = wL2/8

= 8.44 (32) / 8

= 9.5 kNm/m width

8.44 kN/m

L = 3 m

SFD

BMD

SIMPLY SUPPORTED ONE WAY SLAB EXAMPLE 3.3

Check As min and As max

𝐴𝑠,𝑚𝑖𝑛 = 0.26 2.6 1000 (100)

(500)= 135.2 𝑚𝑚2/𝑚 ≥ 0.0013 1000 100 = 130𝑚𝑚2/𝑚

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

As,min (135.2 mm2/m) < As,prov (262mm2/m) < As,max (5000mm2/m) => OK!!

ii) Transverse reinforcement

Provide minimum area of reinforcement = 135.2 mm2/m

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

4. Check the slab for shear

VEd = Vmax = 12.66 kN/m width

i) Calculate VRd,c

𝑘 = 1 + 200

100= 2.41 > 2.0 𝑑 𝑖𝑛 𝑚𝑚 ∴ 𝑢𝑠𝑒 = 2.0 𝑚𝑚

𝜌𝑙 =𝐴𝑠𝑙

𝑏𝑤𝑑=

262

1000 𝑥 100= 0.0026

𝑉𝑅𝑑 ,𝑐 = 0.12𝑘 100𝜌𝑙𝑓𝑐𝑘 13𝑏𝑤𝑑 ≥ 𝑉𝑚𝑖𝑛

= 0.12 2 100 0.0026 25 13 1000 100 = 44.79 𝑘𝑁

𝑉𝑚𝑖𝑛 = 0.035 𝑘3/2𝑓𝑐𝑘1/2 𝑏𝑤 𝑑

= 0.035 2 32 25

12 1000 𝑥 100 = 49.5 𝑘𝑁

VRd,c < Vmin

Use VRd,c = Vmin = 49.5 kN

SIMPLY SUPPORTED ONE WAY SLAB EXAMPLE 3.3

ii) Compare VEd with VRd,c

VEd (12.66 kN) < VRd,c (49.5 kN)

No shear reinforcement is required.

5. Deflection check

𝜌 =𝐴𝑠,𝑟𝑒𝑞

𝑏𝑤𝑑=

230

1000 𝑥 100= 2.3 𝑥 10−3

𝜌0 = 25𝑥 10−3 = 5 𝑥 10−3

< o

𝑙

𝑑= 𝐾 11 + 1.5 𝑓𝑐𝑘

𝜌𝑜𝜌

+ 3.2 𝑓𝑐𝑘 𝜌𝑜𝜌− 1

3/2

= 1.0 11 + 1.5 25 5

2.3 + 3.2 25

5

2.3− 1

32

= 40.24

From table 7.4N, K = 1.0 (one way simply supported slab)

(i) Calculate the modification factor

310

𝜎𝑠=

500

𝑓𝑦𝑘 𝐴𝑠,𝑟𝑒𝑞

𝐴𝑠,𝑝𝑟𝑜𝑣

= 500

500 230263

= 1.14

(ii) Calculate (L/d)allowable

𝐿

𝑑 𝑎𝑙𝑙𝑜𝑤𝑎𝑏𝑙𝑒

= 𝐿

𝑑 𝑏𝑎𝑠𝑖𝑐

𝑥 𝑚𝑜𝑑𝑖𝑓𝑖𝑐𝑎𝑡𝑖𝑜𝑛 𝑓𝑎𝑐𝑡𝑜𝑟

𝐿

𝑑 𝑎𝑙𝑙𝑜𝑤𝑎𝑏𝑙𝑒

= 40.24 𝑥 1.14 = 45.87

SIMPLY SUPPORTED ONE WAY SLAB EXAMPLE 3.3

a) Calculate (L/d)actual

𝐿

𝑑 𝑎𝑐𝑡𝑢𝑎𝑙

=𝑒𝑓𝑓𝑒𝑐𝑡𝑖𝑣𝑒 𝑠𝑝𝑎𝑛 𝑙𝑒𝑛𝑔𝑡𝑕

𝑒𝑓𝑓𝑒𝑐𝑡𝑖𝑣𝑒 𝑑𝑒𝑝𝑡𝑕 =

3000

100 = 30

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

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

Therefore, slab is safe against deflection.

6. 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 = 385 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!)

SIMPLY SUPPORTED ONE WAY SLAB EXAMPLE 3.3

7. Detailing

Plan view

Cross-section

H10-300 (B)

H8

-30

0 (

B)

H10-300 (B)

H8-300 (B)