Tank Design

MINOR PROJECT FOR WATER TANK DESIGN

Design constants:-

For M-20 concrete σ_cbc = 7N/mm² and m=13. Taking σ_st =115N/mm²,

We have

K=m σ_cbc / m σ_cbc +σ_st

=13x7/13x7+115

=0.442

J=1-k/3

=0.853

R=0.5x σ_cbc x j x k.

=0.5 x 7 x 0.853 x 0.442.

(2) Moment and shear.

L=12.6m, B=3m, H=3.5m.

L/H=12.6/3

=4.2m>2

B/H=3/3.5

=1m

S.NO.	Line CD	Line AB	Line EF
X/H	mx my	mx my	mx my
1/4	+0.026 0.013	-0.006 -0.028	+0.0002 0.008
1/2	+0.044 0.020	-0.009 -0.046	+0.007 0.014
3/4	+0.041 0.016	-0.009 -0.044	+0.013 0.013

Horizontal Moment M_H = my x WH³ .

Vertical Moment V_v = M_x x WH³

Maximum value of M_{y =} -0.046.

Hence maximum moment in horizontal direction occurs at the mid point of edge AB.

Maximum value of m_x =+0.044. Hence maximum moment on vertical occur at the mid point of edge CD.

Specific weight W=9800N/m³

Maximum moment M_H at edge AB = -0.046x9800 x (3.5)² x 3

= 16566 N-m

Maximum moment M_H at edge EF = 0.014 x 9800 x (3.5)² x3

= 5042.1 Nm.

Maximum moment M_H at edge CD (mid point) = 0.020 x 9800 x (3.5)² x 3

= 7203 Nm.

Maximum moment M_v at edge CD (mid span) = 0.044 x 9800 x (3.5)² x 3

= 15846.6 Nm.

Maximum moment M_v at edge EF (at lower quarter point) = 0.013 x 9800 x (3.5) ² x3.

=4681.95 Nm.

Shear force at mid point of fixed side edge AB of long wall = .036 x WH³

= 0.36 x 9800 x (3.5)²

= 43218 N

Shear force at mid point of fixed side edge AB of long wall = 0.258 x WH³

= 0.258 x 9800 x (3.5)²

=30972.29 N

(3) Design of thickness of tank walls: - The thickness of wall is governed by moment M_H and S.F. at mid point of AB.

M_H =16566 N-m.

S.F =43218 N-m.

Let the thickness of the wall be 150 mm.

The criterion for the safe design is: [σ_cbt ^, / σ_cbt ] +[ σ_ct ^‘/σ_ct ] <=1.

Where σ_cbt ^, = calculation bending tensile stress in concrete

= [6566 x 1000 x 6] / [1000 x 350 x 350]

= 0.81 N/mm²

σ_cbt = permissible B.M. tensile stress in concrete = 1.7 N/mm²

σ_ct’ = calculate direct tensile stress in concrete

=43218 / 1000 x 150

= 0.288 N/m²

σ_ct = permissible direct tensile stress in concrete = 1.2 N/m²

(1.7/.8) + (0.288/1.2) <=1

Hence safe

Thus keep total thickness T = 350 mm, and an effective thickness = 320 mm.

Depth of N.A. = 320 x 0.442mm

= 53 mm.

(4) Reinforcement in horizontal direction

Eccentricity of tensile force = 16566/ 43218.

= 0.38 m

=380 mm.

Eccentricity of tensile force; measured from the centre of the steel will be

C.G of the compressive zone

= 380 – (141.44/3)

= 335 mm

43218(335 +272.85) = Ast x 115 x 272.85

Ast = 837.2 mm²

Spacing of 10 mm. ¢ bars = 1000 x 78.5 / 837.22

= 93.76 mm.

= (95 approximate)

Hence provide these @ 95 mm. c/c on the water face.

Maximum sagging moment in horizontal direction occurs at mid point of CD and its value is 7203 N-m.

Tension at this point = safe at the mid point of vertical edge AB of short wall =30972.29N

Eccentricity of tension = 7203.2 /30972.29

=0.232 mm.

=232 mm.

Distance of line of action tension from water force = 232+175

= 407 mm.

Taking moment about the centre of compression zone

Ast x 115 x 272.85 = 30972.29 [407 – (14144/3)]

Ast = 355.2 mm²

Ast = (0.3 x 350 x 1000)/100

= 1050 mm²

Spacing of 8 mm Ф bars = (1000 x 50.2)/1050

= 47.80

Hence provide 8 mm Ф bars @ 50 mm c/c.

The horizontal B.M. at the line EF of the short wall is 5042.1 N-m only but the direct tension is 43218 N, Hence provide the some reinforced i.e. 8 mm Ф bars @ 50 mm c/c.

At the outer face.

Maximum shear stress = 43218 /1000 x 0.853 x 320

= 0.15 N/mm²

Since this is less than г_bd =.24 N/mm² for 0.3% reinforcement for M-20 concrete. It is safe.

(5) Reinforcement in vertical direction

The B.M. M_v in vertical direction combines with the direct compression due to weight of wall and weight of roof.

LET the thickness of the roof slab be 10 c.m. = 0.1mm

Assuming unit weight of concrete = 24000N/mm²

Dead load transferred to walls = [3.5/2+.35] x 1 x 0.1 x 25000

=5250 N/m

Weight of wall, up to its mid height =3.5/2 x 35 x 25000 x 1

=15312 N/m

Therefore thrust in wall at its mid height = 5250+15312

=20562 N/m

B.M. in vertical direction at mid point of CD = 15846.6 N/m (sagging)

Therefore eccentricity of thrust in vertical direction in long wall = 15846.6/20562

=770 mm

Its distance from water face = 770-175

=595mm

LET the cover to the centre of the steel be 20mm

Effective depth=350-20

=330mm

Depth of neutral axis = 330 x 0.442 mm

=145.86mm

Taking moment about the C.G. of compression area

=A_st x 115x 281.38 = 20562[715 + (145.46 /3)]

A_st =485.23 mm²

Minimum area of steel @ 0.3%

= 0.3/100(1000 x 350)

=1050 mm²

Hence provided 8 mm Ф bars @ 50 mm c/c at the outside faces, in the vertical direction.

The B.M. M_v in the vertical direction in short wall is 4681.95 N-m (sagging)_, against 15846.6 N-m in the long wall. Hence provided minimum area of steel @ 0.3% = 485.23 mm²

Therefore we provide 8 mm Ф @ 50 mm c/c