Design OF Ground water Tank


(حبيب الملايين) #1

Part 2 : CHAPTER13

[CENTER]Design OF Ground water Tank

[/center]

Design of ground water tank:

Dimension 5.631.8 m

13.1 Case 1: tank is full of water without lateral earth pressure
gH = 1.8*1=3t/m
Mmax at foot wall =1.8 = -0.972t.m/m
Nmax (its own weight) = 0.251.82.5 = 1.125 ton/m
Assuming t= 25 cm
Rwall at the base = 0. 51.81.81 = 1.62 t/m[/left]

[LEFT]13.1.1 Loads acting on the length (L)
N1 (weight of the wall /m) = 1.125 t/m
N2 (weight of the floor /m) = 0. 52.51 = 1.25 t/m
N3 (weight of the water /m) = 0. 81.81 = 1.44 t/m
N = 3.815 t/m

M@ GG of floor = My
= 1.9721.1250.3 = 1.1655 ton/m

e = = = 0.31 >
C = - e = 0.5 – 0.31 = 0.19
3*C = 0.57
3C * = N , y= 0
Ftot. soil = =
= 13.39 < 20 t / m2 ok

13.1.2 design of bending & shear

M= 0.972 t.m
d = 25–0.8-1.4-5 = 17.8cm

= 350 kg / cm2, = 4200 kg / cm2

= 0.004 < min = 0.0018
usemin = 0.0018
As = 0.001825200 = 8.1 cm2
use 1 Ф 14 mm @ 15 cm
Vu 3.815 t/ m’
Ф Vc = 0.850.53bd
== 27 ton > Vu = 3.815 ton zone (1)
tryФ 14 mm
S = = = 20.53 cm use Ф 14 mm @15 cm
13.2 Case 2 tank is empty with the lateral earth pressure

Φ= 30°, γsoil =1.8t/m3.[/left]

[LEFT]γKH = 1.81.8 = 1.08 t/ m’
4. Mmax. ( at foot wall) = 1/2 1.081.8= 0.5832 t.m/m’
N1= wt. of wall/m’ = 1.125t/m’
N2 = wt. of floor/m’ (L=1) = 1.25 t/m
N3 (weight of the water /m) = 0 (Empty tank)
N = 2.375 t/m’
M@GG of floor slab = My
Mwall – N1(0.5- 0.125)
= 1.08 – 1.25(0.5-0.125)
= 0.61125 t.m/m’

check for stress on floor
A = 1 * 1 = 1 m2
Iy = = m4
q =
q = 3.67 + 2.375
qmax = 6.04 t/ m2
qmin = 1.29 t/ m2 ok (safe)

13.2.1 . DESIGN FOR SOLID SLABS
1) Evaluate Slab thickness
Assume that the beam webs are 30cm wide
Ln = 5.6- .3 = 5.3m[/left]
The smallest slab thickness is given by: -

[LEFT] = = 12.86cm
The largest slab thickness is given by: -
cm
Slab thickness will be taken as 15m.
2) Determine the total factored load per unit area of the slab
- own weight of slab = 0.15 2.5 = 0.38 t/m2
- Live load = 0.2t/m2
- Wu = 1.4 (.38 + 0.2) + 1.7FONT=Times New Roman = 1.152 t/m2[/font]
3) Evaluate load distribution in both direction :-
Rectangularity ratio (r) = = 1.86
Using Grashoff coefficientsα = 0.923 , β= 0.077
- Load in the short direction Ws = 0.93FONT=Times New Roman =1.071 t/m2[/font]

For a strip 1m wide, WL = 1.07 t/m
- Loadin the long direction Ws =0.077 (1.152) = 0.089 t/m2
For a strip 1m wide, WL = 0.089t/m
4) Select representative 1m wide design strips to span in the short and long direction.
5) Evaluate Shear and Bending moment at critical section locations :-
Maximum shear force in the short direction Vu,max = 1.071 = 1.6065 ton
Maximum shear force in the long direction Vu,max = 0.089 = .25 ton
Maximum moment in the short direction Mu,max = 1.071 = 1.21 t.m
Maximum moment in the long direction Mu,max = 0.089 = 35 t.m
6) Check slab thickness for beam shear :-
Short direction: -
Average effective depth d =15-2-1.2 = 11.8 cm
AssumingΦ12mm bars in both directions.
Φ Vc = 0.850.53 (100) [11.8 / 1000] = 9.945ton
i.e. , slab thickness is adequate in terms of resisting beam shear in the shortdirection .
Long direction: -
Vu,max< Φ Vc
i.e. , slab thickness is adequate in terms of resisting beam shear in the shorter direction
7) Design the reinforcement :-
The Reinforcement Short direction
= 0.0023
Area of flexural reinforcement As = 0.0033 100 11.8 = 3.894 cm2/m
UseΦ 12mm @ 20 cm at the bottom side of the slab [/left]

For main reinforcement, spacing between bars is not to exceed the smaller of

3(14= ( 42 cm and 45 cm, which is already satisfied
The ReinforcementLong direction

[LEFT] = 0.00067
Area of flexural reinforcement As = 0.0033 100 11.8 = 3.894 cm2/m
UseΦ 12mm @ 20 cm at the bottom side of the slab

[CENTER]

Figure 13.1: Reinforcement Details for Solid Slab[/center]

[CENTER]

[/center]

[LEFT]

13.2.2 Design of beam B1[/left]

[RIGHT]

[LEFT]h min = l / 16 = 560 / 16 = 35cm
Try 25cm 40cm cross section
Factored own weight of beam = 1.4(.25)(.40)(2.5) = .35 t/m
Maximum factored moment due to own weight of beam = .35 = 1.372 t.m
Maximum factored shear due to own weight of beam = .35 = .98 ton
Loads from the slab = Wu (S / 2 ) = 1.152(3/2) = 1.728t/m
Equivalent load due to shear = 1.728(1-.27) =1.26t/m
Equivalent load due to moment = 1.728(1-.096) = 1.56t/m
Maximum factored shear = 1.26 = 3.53 ton
Maximum factored moment =1.56 = 6.12 t.m

Total factored shear = .98 +3.53 =4.51 ton
Total factored moment = 1.372+6.12 =7.50 ton.m
Assuming that f 8mm stirrups and f12mm reinforcing bars are used,
d = 40 – 4 – .8 -.6 = 34.6 cm[/left]
= 0.007

As = 0.00725 34.6 = 6.04 cm2 , use 4f 14 mm bars.
Check section for shear :

f Vc = 0.850.53 (25) [34.6 / 1000] = 7.29 ton
use f 8mm @ 20cm as minimum shear reinforcement .

[CENTER]

Figure 13.2 : Beam (1)

[/center]

[RIGHT]13.2.3 Design of beam B2:

[RIGHT]

[LEFT]h min = l / 16 = 300 / 16 = 18.75 cm
Try 25cm 40cm cross section
Factored own weight of beam = 1.4(.25)(.40)(2.5) = .35 t/m
Maximum factored moment due to own weight of beam = .35 = .4t.m
Maximum factored shear due to own weight of beam = .35 = .525 ton
Loads from the slab = Wu (S / 2 ) = 1.152(3/2) = 1.728t/m
Equivalent load due to shear = 1.728/2 =.864t/m
Equivalent load due to moment = 1.728 (2/3) = 1.152t/m
Maximum factored shear =.864 = 1.296ton
Maximum factored moment =1.152 = 1.296 t.m

Total factored shear =.864+1.296 =2.16 ton
Total factored moment =1.152+1.296 = 2.448 ton.m
Assuming that f 8mm stirrups and f12mm reinforcing bars are used,
d = 40 – 4 – .8 -.6 = 34.6 cm[/left]
= 0.0022

As = 0.0033 25 34.6 = 2.86 cm2 , use 4f 12 mm bars.
Check section for shear :

f Vc = 0.850.53 (25) [34.6 / 1000] = 6.16 ton
use f 8mm @ 20cm as minimum shear reinforcement .

[CENTER]

Figure 13.3 : Beam (2)

[/center]

13.2.4 Design of tank floor :
Design for moment :
Mu = -0.6 t.m
d = 50 – 5 -1.6 -0.7 = 42.7 cm
b = 1 m
= 0.0001
Use min = 0.0033
As = 0.003350100 = 16.5 cm2
Ues 1 Φ 16 @ 10 cm

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