Fire, durability and bond required
Minimum cover in regard to durability, Cmin, durability = 15 mm
Minimum cover in regard to bond, Cbond = 10 mm
From AS 3600: 2018. Table 5.5.2 (A)
Minimum axis distance required for resistance of FRP 120/120/120, as = 35 mm
Minimum cover in regard to fire
Minimum, fire = a – 29 mm
Allowance for deviation of design,
Nominal cover
= 29 + 10 = 39 mm
Loadings
Slab self-weight = 0.2 * 25 = 5.0 kN/m2
Dead load (self-weight excluded) = 1.50 kN/m2
Tiling and partitions load = 1.0 kN/m2
Total dead load, gk = 7.50 kN/m2
Imposed load, qk = 4.00 kN/m2
Clause 2.5.2.2 AS 3600: 2018
Ed = [1.2G, 1.5Q]
Permanent and imposed load
nd = 1.2 * 7.50 + 1.5 * 4 = 15.0 kN/m2
Analyzing
If we consider a slab of 1 m
w = 15.0 kN/m2
L L L
0.04FL 0.086FL 0.086FL 0.04FL
0.075FL 0.063FL 0.075FL
0.46F 0.5F 0.6F
0.6F 0.5F 0.46F
Main reinforcement
Calculation of effective depth
dx = 200 – 34 – 0.5*12 = 160 mm
Maximum and minimum reinforcement area
As, minimum = 0.256(fctm/fyk)bd = *
= 0.0013*1000 *160
= 208 mm2/m
As, maximum = 0.04Ac = 0.04 *1000 *200 = 8000 mm2/m
Secondary bar: Provide H12 – 500 mm c/c (220 mm2/m)
At 1st interior support
FL = 15 *5.6 = 84 kNm/m
Moment, M = 0.086F*L = 0.086 * 84 * 5.6 = 40.45 kNm/m
K =
Hence, compression reinforcement would not be required
Z =
As =
Provide H12 – 180 top (611 mm2/m)
At middle interior span
FL = 15 *5.6 = 84 kNm/m
Moment, M = 0.063F*L = 0.063 * 84 * 5.6 = 29.64 kNm/m
K =
Hence, compression reinforcement would not be required
Z =
As =
Provide H12 – 240 bottom (458 mm2/m)
Near middle interior span
FL = 15 *5.6 = 84 kNm/m
Moment, M = 0.075F*L = 0.075 * 84 * 5.6 = 35.28 kNm/m
K =
Hence, compression reinforcement would not be required
Z =
As =
Provide H12 – 200 bottom ( 550 mm2/m)
At outer support
FL = 15 *5.6 = 84 kNm/m
Moment, M = 0.04F*L = 0.04 * 84 * 5.6 = 18.82 kNm/m
K =
Hence, compression reinforcement would not be required
Z =
As =
Provide H12 – 380 bottom ( 289 mm2/m)
SHEAR
Maximum shear force design
VED = 0.6F = 0.6 * 84 = 50.4 kN
Designing shear resistance
VRd,c =
K = 1 + (200/160)1/2 = 2.12
VRd,c = *0.00424 * 25)1/3] * 1000 * 160 = 84.35 kN
Vmin = [0.035k3/2* fck1/2]*bd = 0.035 * 23/2*251/2*1000*160 = 79.2 kN
VRd,c = 84.35 kN > VED = 50.4 kN
Hence, OK
DEFLECTION
Required percentage of tension reinforcement
w = = 0.728 n mm
Span/250 [Clause 2.3.2, Table 2.3.2 AS 3600: 2018]
= 5600/250 = 22.4 mm
0.78 mm < 22.4 mm
Deflection is satisfactory
Hence, OK
CRACKING
h = 200 mm = 200 mm
Main bar
Smaximum, slab = 3h = 400
Maximum bar spacing = 300 Smaximum, slab, OK
Secondary bar
Smaximum, slab = 3.5h = 450
Maximum bar spacing = 250 OK
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PART 2
Fire, durability and bond required
Minimum cover in regard to durability, Cmin, durability = 15 mm
Minimum cover in regard to bond, Cbond = 10 mm
From AS 3600: 2018. Table 5.5.2 (A
Minimum axis distance required for resistance of FRP 120/120/120, a = 30 mm
Minimum cover in regard to fire
Assume the L-mesh used has a diameter of 10 mm
Minimum, fire = a – 25 mm
Allowance for deviation of design,
Nominal cover
= 25 + 10 = 35 mm
Loadings
Slab self-weight = 0.2 * 25 = 5.0 kN/m2
Dead load (self-weight excluded) = 1.50 kN/m2
Tiling and partitions load = 1.0 kN/m2
Total dead load, gk = 7.50 kN/m2
Imposed load, qk = 4.00 kN/m2
Clause 2.5.2.2 AS 3600: 2018
Ed = [1.2G, 1.5Q]
Permanent and imposed load
nd = 1.2 * 7.50 + 1.5 * 4 = 15.0 kN/m2
N = 15.0 *5.6 = 84 kN/m
Moment, W = = 329.28 kNm
Area of slab panels = L * W = 8.2 * 5.6 = 45.92 m2
Compressive strength = C25
Take modulus of rapture (MOR) = 33
Cracking moment, Mcr = = 0.01833 kNm of the slab width
Moment required for safety factor
Take safety factor of 2
Moment = 329.28 kNm * 2 = 658.56 kNm
Using one layer of 6 numbers of bar
Using the table below to determine moment capacity
Moment = 7.13 kip-ft at a spacing of 12 inches for 6 numbers of bar
Moment = 7.13 kip-ft = 9.67 kNm
Spacing of 12 inches = 0.305 m
Spacing = 0.305 * (9.67/ 658.56) = 0.00448 m
Therefore,
Use one layer of six bars with a spacing of 4.5 mm on Centre
SHEAR
Maximum shear force design
VED = 0.6F = 0.6 * 84 = 50.4 kN
Designing shear resistance
VRd,c =
K = 1 + (200/160)1/2 = 2.12
VRd,c = *0.00424 * 25)1/3] * 1000 * 160 = 84.35 kN
Vmin = [0.035k3/2* fck1/2]*bd = 0.035 * 23/2*251/2*1000*160 = 79.2 kN
VRd,c = 84.35 kN > VED = 50.4 kN
Hence, OK
