Example 1: Design the roof slab, beam and column of house given in figure 1.

Concrete compressive strength (fc′) = 3 ksi.

Steel yield strength (fy) = 40 ksi.

Load on slab:

4″ thick mud.

2″ thick brick tile.

Live Load = 40 psf

Solution: -

(1) Design of slab “S2”:

Step No 1: Sizes.

lb/la = 24.75/8 = 3.09 > 2 “one way slab”

Assume 5″ slab.

Span length for end span according to ACI 8.7 is minimum of:

(i) l = ln+ hf = 8 + (5/12) = 8.42′

(ii) c/c distance between supports = 9.0625′

Therefore l = 8.42′

Slab thickness (hf) = (l/20) × (0.4+fy/100000) [for fy < 60000 psi]

= (8.42/20) × (0.4 + 40000/100000) × 12

= 4.04″ (Minimum requirement of ACI 9.5.2.1).

Therefore take hf = 5″

d = hf – 0.75 – (3/8)/2 = 4″

Step No 2: Loading.

Service Dead Load (D.L) = 0.0625 + 0.04 + 0.02

= 0.1225 ksf

Service Live Load (L.L) = 40 psf or 0.04 ksf

Factored Load (wu) = 1.2D.L + 1.6L.L

= 1.2 × 0.1225 + 1.6 × 0.04

= 0.211 ksf

Step No 3: Analysis.

Mu = wul2/8 (l = span length of slab)

Mu = 0.211 × (8.42)2/8

= 1.87 ft-k/ft = 22.44 in-k/ft

Step No 4: Design.

Asmin = 0.002bhf (for fy 40 ksi, ACI 10.5.4)

= 0.002 × 12 × 5 = 0.12 in2

a = Asminfy/ (0.85fc′b)

= 0.12 × 40/ (0.85 × 3 × 12) = 0.156 in

ΦMn(min) = ΦAsminfy (d – a/2)

= 0.9 × 0.12 × 40 × (4 – 0.156/2)

= 16.94 in-k < Mu

Therefore,

• As = Mu/ {Φfy (d – a/2)}

Take a = 0.2d

As = 22.44/ {0.9 × 40 × (4 – (0.2 × 4)/2)}

As = 0.173 in2

• a = 0.173 x 40/ (0.85 x 3 x 12) = 0.226 in

As = 22.44/ {0.9 × 40 × (4 – (0.226/2))}

= 0.160 in2

• a = 0.160 × 40/ (0.85 × 3 × 12) = 0.209 in

As = 22.44/ {0.9 × 40 × (4 – (0.209/2))}

= 0.160 in2, O.K.

Using ½″ Φ (#4) {#13, 13 mm}, with bar area Ab = 0.20 in2

Spacing =Area of one bar (Ab)/As

= [0.20 (in2)/0.160 (in2/ft)] × 12 = 15 in

Using 3/8″ Φ (#3) {#10, 10 mm}, with bar area Ab = 0.11 in2

Spacing = Area of one bar (Ab)/As

= [0.11(in2)/0.160(in2/ft)] × 12 = 7.5″ ≈ 6″

Finally use #3 @ 6″ c/c (#10 @ 150 mm c/c).

Shrinkage steel or temperature steel (Ast):

Ast = 0.002bhf

Ast = 0.002 × 12 × 5 = 0.12 in2

Using 3/8″ Φ (#3) {#10, 10 mm}, with bar area Ab = 0.11 in2

Spacing = Area of one bar (Ab)/Asmin

= (0.11/0.12) × 12 = 11″ c/c

Finally use #3 @ 9″ c/c (#10 @ 225 mm c/c).

• Maximum spacing for main steel in one way slab according to ACI 7.6.5 is

minimum of:

i) 3hf =3 × 5 =15″

ii) 18″

Therefore 6″ spacing is O.K.

• Maximum spacing for shrinkage steel in one way slab according to ACI 7.12.2

is minimum of:

i) 5hf =5 × 5 =25″

ii) 18″

Therefore 9″ spacing is O.K.

Concrete compressive strength (fc′) = 3 ksi.

Steel yield strength (fy) = 40 ksi.

Load on slab:

4″ thick mud.

2″ thick brick tile.

Live Load = 40 psf

Solution: -

(1) Design of slab “S2”:

Step No 1: Sizes.

lb/la = 24.75/8 = 3.09 > 2 “one way slab”

Assume 5″ slab.

Span length for end span according to ACI 8.7 is minimum of:

(i) l = ln+ hf = 8 + (5/12) = 8.42′

(ii) c/c distance between supports = 9.0625′

Therefore l = 8.42′

Slab thickness (hf) = (l/20) × (0.4+fy/100000) [for fy < 60000 psi]

= (8.42/20) × (0.4 + 40000/100000) × 12

= 4.04″ (Minimum requirement of ACI 9.5.2.1).

Therefore take hf = 5″

d = hf – 0.75 – (3/8)/2 = 4″

Step No 2: Loading.

Service Dead Load (D.L) = 0.0625 + 0.04 + 0.02

= 0.1225 ksf

Service Live Load (L.L) = 40 psf or 0.04 ksf

Factored Load (wu) = 1.2D.L + 1.6L.L

= 1.2 × 0.1225 + 1.6 × 0.04

= 0.211 ksf

Step No 3: Analysis.

Mu = wul2/8 (l = span length of slab)

Mu = 0.211 × (8.42)2/8

= 1.87 ft-k/ft = 22.44 in-k/ft

Step No 4: Design.

Asmin = 0.002bhf (for fy 40 ksi, ACI 10.5.4)

= 0.002 × 12 × 5 = 0.12 in2

a = Asminfy/ (0.85fc′b)

= 0.12 × 40/ (0.85 × 3 × 12) = 0.156 in

ΦMn(min) = ΦAsminfy (d – a/2)

= 0.9 × 0.12 × 40 × (4 – 0.156/2)

= 16.94 in-k < Mu

Therefore,

• As = Mu/ {Φfy (d – a/2)}

Take a = 0.2d

As = 22.44/ {0.9 × 40 × (4 – (0.2 × 4)/2)}

As = 0.173 in2

• a = 0.173 x 40/ (0.85 x 3 x 12) = 0.226 in

As = 22.44/ {0.9 × 40 × (4 – (0.226/2))}

= 0.160 in2

• a = 0.160 × 40/ (0.85 × 3 × 12) = 0.209 in

As = 22.44/ {0.9 × 40 × (4 – (0.209/2))}

= 0.160 in2, O.K.

Using ½″ Φ (#4) {#13, 13 mm}, with bar area Ab = 0.20 in2

Spacing =Area of one bar (Ab)/As

= [0.20 (in2)/0.160 (in2/ft)] × 12 = 15 in

Using 3/8″ Φ (#3) {#10, 10 mm}, with bar area Ab = 0.11 in2

Spacing = Area of one bar (Ab)/As

= [0.11(in2)/0.160(in2/ft)] × 12 = 7.5″ ≈ 6″

Finally use #3 @ 6″ c/c (#10 @ 150 mm c/c).

Shrinkage steel or temperature steel (Ast):

Ast = 0.002bhf

Ast = 0.002 × 12 × 5 = 0.12 in2

Using 3/8″ Φ (#3) {#10, 10 mm}, with bar area Ab = 0.11 in2

Spacing = Area of one bar (Ab)/Asmin

= (0.11/0.12) × 12 = 11″ c/c

Finally use #3 @ 9″ c/c (#10 @ 225 mm c/c).

• Maximum spacing for main steel in one way slab according to ACI 7.6.5 is

minimum of:

i) 3hf =3 × 5 =15″

ii) 18″

Therefore 6″ spacing is O.K.

• Maximum spacing for shrinkage steel in one way slab according to ACI 7.12.2

is minimum of:

i) 5hf =5 × 5 =25″

ii) 18″

Therefore 9″ spacing is O.K.

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