Design of One way Slab
Q1: Design a one way slab having an effective span of 4 m. It
supports a live load of 170 kg/m2. Take fc’ = 18N/mm2 and fy = 280
n/mm2.
Given Data:
Effective span = 4 m
Live load = 170 kg/m2
fc’ = 18 N/mm2
fy = 280 N/mm2
Solution:
According to the thumb rule, the thickness of slab for a simply
supported beam is span/25.
Thickness of slab = h = span/25 = 4/25 * 1000 = 160mm
Effective depth = d = h - 27mm = 133 mm
Self load per sq.m = w1 = 1*1*(0.16)*2400 = 384 kg/m2
Live load = w2 = 170 kg/m2
Factored load = w = 1.2WD + 1.6WL = 1.2(384) + 1.6(170) = 732.8
kg/m2
= 732.8 *(9.81/1000) = 7.188 KN/m2
Factored Bending Moment = MU = W*L2/8 = (7.188)(4)2/8 =
14.37KNm
Effective depth = dmin = √MU/0.205fc’b = √14.37 * 106/0.205*18*1000
Breadth = b = 1000 mm because we are designing the beam for a strip
of 1 meter.
dmin = 62.40 mm
But the depth from the thumb rule was 133 mm so we will adopt that.
Direct Method:
w = 0.85 (fc’/fy) = 0.85 (18/280) = 0.0546
R = MU/bd2 = 14.37 * 106/1000*(133)2 = 0.812 N/mm2
ρ = w[1-√1 - 2.614R/fc’] = 0.0546[1 - √1 - 2.614(0.812/18)] = 0.00332
AS = ρbd = (0.00332)(1000)(133) = 441.79 ≈ 442 mm2
, Trial Method:
ASmin = 1.4/fy (b*d) = (1.4/280)(1000*133) = 665 mm2
ASmax = 0.375(0.85)2(18/280)(1000*133) = 2316.51 mm2
Trial 01
Assume a = d/6 = 133/6 = 22.167 mm
AS = MU/0.9fy(d - a/2) = 14.37 * 106/0.9(280)(133 - 22.167/2) =
467.72 mm2
Trial 02
a = ASfy/0.85fc’b = (467.72)(280)/0.85(18)(1000) = 8.55 mm
AS = MU/0.9fy(d - a/2) = 14.37 * 106/0.9(280)(133 - 8.55/2) = 443
mm2
NOTE:
As the Area of steel (AS) is less than the minimum steel, so we
will consider the minimum amount of steel in steel work i.e., AS Total =
665 mm2
Spacing:
Main Bars
Provide 12 mm ⌀ bars as main bars = (Area of one bar/As Total)*1000 in
case of mm
Spacing = (113/665) * 1000 = 169.92 ≈ 170mm
So, Provide 12 mm ⌀ @ 170 mm c/c
Checking the spacing of main bars,
3h = 3(160) = 480mm, this is the maximum amount of spacing that
can be provided. The calculated value is less than maximum value, so
spacing @ 170 mm c/c is OK.
Distribution Bars
Area of steel for distribution bars: It is 0.2% of gross area of cross
section.
AS = (0.2/100)[(133)(1000)] = 266 mm2
Spacing of 6mm diameter bars = Area of one bar/AS Total *1000
= (28.27/266)*1000
= 106 mm (OK)
Q1: Design a one way slab having an effective span of 4 m. It
supports a live load of 170 kg/m2. Take fc’ = 18N/mm2 and fy = 280
n/mm2.
Given Data:
Effective span = 4 m
Live load = 170 kg/m2
fc’ = 18 N/mm2
fy = 280 N/mm2
Solution:
According to the thumb rule, the thickness of slab for a simply
supported beam is span/25.
Thickness of slab = h = span/25 = 4/25 * 1000 = 160mm
Effective depth = d = h - 27mm = 133 mm
Self load per sq.m = w1 = 1*1*(0.16)*2400 = 384 kg/m2
Live load = w2 = 170 kg/m2
Factored load = w = 1.2WD + 1.6WL = 1.2(384) + 1.6(170) = 732.8
kg/m2
= 732.8 *(9.81/1000) = 7.188 KN/m2
Factored Bending Moment = MU = W*L2/8 = (7.188)(4)2/8 =
14.37KNm
Effective depth = dmin = √MU/0.205fc’b = √14.37 * 106/0.205*18*1000
Breadth = b = 1000 mm because we are designing the beam for a strip
of 1 meter.
dmin = 62.40 mm
But the depth from the thumb rule was 133 mm so we will adopt that.
Direct Method:
w = 0.85 (fc’/fy) = 0.85 (18/280) = 0.0546
R = MU/bd2 = 14.37 * 106/1000*(133)2 = 0.812 N/mm2
ρ = w[1-√1 - 2.614R/fc’] = 0.0546[1 - √1 - 2.614(0.812/18)] = 0.00332
AS = ρbd = (0.00332)(1000)(133) = 441.79 ≈ 442 mm2
, Trial Method:
ASmin = 1.4/fy (b*d) = (1.4/280)(1000*133) = 665 mm2
ASmax = 0.375(0.85)2(18/280)(1000*133) = 2316.51 mm2
Trial 01
Assume a = d/6 = 133/6 = 22.167 mm
AS = MU/0.9fy(d - a/2) = 14.37 * 106/0.9(280)(133 - 22.167/2) =
467.72 mm2
Trial 02
a = ASfy/0.85fc’b = (467.72)(280)/0.85(18)(1000) = 8.55 mm
AS = MU/0.9fy(d - a/2) = 14.37 * 106/0.9(280)(133 - 8.55/2) = 443
mm2
NOTE:
As the Area of steel (AS) is less than the minimum steel, so we
will consider the minimum amount of steel in steel work i.e., AS Total =
665 mm2
Spacing:
Main Bars
Provide 12 mm ⌀ bars as main bars = (Area of one bar/As Total)*1000 in
case of mm
Spacing = (113/665) * 1000 = 169.92 ≈ 170mm
So, Provide 12 mm ⌀ @ 170 mm c/c
Checking the spacing of main bars,
3h = 3(160) = 480mm, this is the maximum amount of spacing that
can be provided. The calculated value is less than maximum value, so
spacing @ 170 mm c/c is OK.
Distribution Bars
Area of steel for distribution bars: It is 0.2% of gross area of cross
section.
AS = (0.2/100)[(133)(1000)] = 266 mm2
Spacing of 6mm diameter bars = Area of one bar/AS Total *1000
= (28.27/266)*1000
= 106 mm (OK)
