Drain Design Calculation
Drain Design Calculation
Drain Design Calculation
Dimensions
Height of wall,Hw = 600 mm
Thickness of wall, Tw = 125 mm
BS 5400-4 There is no axial load on walls, therefore it is treated as cantilever slab &
1990 designed accordance with Clause 5.4
Cl 5.6.1 Consider 1000mm length section, b = 1000 mm
Assume, Cover = 45 mm
Bar dia meter = 10 mm
Effective Depth, d = 125 -10 /2 - 45
= 75 mm
BS 5400-4
= 2
1990 0.15x 25 x 1000 x 75
Equ. 2 = 21.09 kNm > 3.270
Hence section is Singly Reinforcement
5.057
= 0.5 1 + 1
2
6
z = 0.5x 75 1+ 1 -5.057 x 3.270 x10^
25 x 1000 x 75 2
z = 72.7 mm
0.95d = 0.95 x 75
= 71.3 < 72.7
Therefore z = 71.3 mm
2
1990 Provide T 10 @ 150 , , = 524 mm T 10 @
Cl 5.8.9 To prevent shinkage & temperature cracking 150mm spacing is provided 150 mm
BS 5400-4
1990
Equation 25
Where,
Mq = 1.340 kNm
Mg = 0.640 kNm
Cover for the reinforcemen = 45 mm
Bar Spacing = 150 mm
Bar diameter = 10 mm
2 2 1/2
a cr = 150 + 45 + 10/2 - 10 /2
2
= 85.1 mm
= 200 / ( 26 /2 )
= 15.38
2 + 2
=
x =- 15.38 x 524 +
0.5
15.38 x 524 2 + 2x 1000x 15.38 x 524 x 75
15.38 x 524
1000 15.38 x 524
x = 27.63 mm 15.38 x 524
z = d-x/3
= 75 - 27.63 /3
= 65.79 mm
Srevice Moment = Mq + Mg
= 1.340 + 0.640
= 1.980 kNm
Servivce Stress, =
Service
= 1.980 x 10^6 stress,
65.79 x 524 = 57.48
2 2
= 57.48 N/mm N/mm
= 2.8779E-05 -
1- 1.340 x 103
3
0.640 x 10
BS 5400-4
1990 = 2.8779E-05 - -0.0034526
Equation 25 = 0.0035 >
= 2.8779E-05
BS 5400-4 Compress stress ditribution due to axial load for the SLS
1990 Axial Load on member = 0 kN
Table 2 Therefore uniform compressive distribution, 0 < 0.38
Hence OK
For Reinforcements
Tensile stress ditribution for the SLS
2
Sevice tensile stress of concrete = 57.48 N/mm
BS 5400-4
= 2
1990 0.15x 25 x 1000 x 75
Equ. 2 = 21.09 kNm > 3.29
Hence section is Singly Reinforcement
5.057
= 0.5 1 + 1
2
z = 72.7 mm
0.95d = 0.95 x 75
= 71.3 < 72.7
Therefore z = 71.3 mm
Required reinforcement area = 0.87
= 3.290 x 10^6
0.87 x 460 x 71.3
2
= 115 mm
BS 5400-4
1990
Equation 25
Where,
Mq = 1.330 kNm
Mg = 0.670 kNm
Cover for the reinforcemen = 45 mm
Bar Spacing = 150 mm
Bar diameter = 10 mm
2 2 1/2
a cr = 150 + 45 + 10/2 - 10 /2
2
= 85.1 mm
= 200 / ( 26 /2 )
= 15.38
2 + 2
=
x =- 15.38 x 524 +
15.38 x 524 2 + 2x 1000x 15.38 x 524 x 75 0.5
15.38 x 5241000 15.38 x 524
x = mm x 524
27.63 15.38 15.38 x 524
z = d-x/3
= 75 - 27.63 /3
= 65.79 mm
Srevice Moment = Mq + Mg
= 1.330 + 0.670
= 2.000 kNm
Servivce Stress, =
Service
= 2.000 x 10^6 stress,
65.79 x 524 = 58.06
2
= 58.06 N/mm N/mm2
= 2.907E-05 -
1- 1.330 x 103
0.670 x 103
Bearing Pressure
2
Bearing Pressure under the base = 10.0 kN/m
2
Asummed bearin pressure = 100 kN/m > 10.0 kN/m2
Hence, Bearing pressure is OK
BS 5400-4 Stress Limitations
1990 For concrete
Cl 4.1.2 Compress stress ditribution due to Bending for the SLS
= Mser /Z
Section Modulus of section, Z = 1/6 x bd2
2
= 1/6 x 1000 x 75
3
= 937,500.00 mm
For Reinforcements
Tensile stress ditribution for the SLS
2
Sevice tensile stress of concrete = 58.06 N/mm