DESIGN OF INTZE TANK

(COLUMN,BRACING AND FOUNDATION)

CE-4102
STRUCTURAL ANALYSIS AND DESIGN-III SESSIONAL

SUBMITED BY

MD. MOIN UDDIN

CE-4/2
ID No. -171113

DEPARTMENT OF CIVIL ENGINEERING

DHAKA UNIVERSITY OF ENGINEERING & TECHNOLOGY, GAZIPUR

 2.7m

D=13.5 m H=9
m

2.75m

8m
Materials Properties:

Yield strength of steel, f y =345MPa

Crushing strength of concrete f ' c=28MPa
Allowable tensile stress f s=.40*345=138 Mpa
Allowable compressive stress f c=0.45 f ' c=12.6 Mpa
29∗106
Modular ratio, (n= )=8
57000 √28∗145

K=0.42
J=0.86
R=.5 f cj k=2.27
 INTZE TANK DESIGN
Design data:

Diameter of the tank, D=13.5m

Height of cylindrical wall, h= 9m
Solution:
1.Design of Top Dome
1.1 Load calculation
Let, the thickness of slab = 100mm
Self-weight = 0.1*25 = 2.5 kPa
Live load = 1.5 kPa
Total load = 4.0 kPa

Height of top dome =0.2D = (0.2*13.5) = 2.7 m

From figure, radius of top dome,
r2=(r-2.7)2+(13.5/2)2
r= 9.78m ≈ 9.8m
Again from figure,
sinØ= (6.75/9.8)
or Ø=43.6°˂51.8°
So, it is a compressive dome.
1.2 Check for maximum stress
2
wr cos θ+ cosθ−1
Hoop stress at any angle θ= ( )
t 1+cos θ

Maximum hoop stress occurs at, θ=0°

wr
∗1 4∗1000∗9.8
∴Hoop stress = t = 0.1∗2 = 196000Pa=0.196 MPa
2
Allowable direct compressive stress = 0.25f’c = 7.0 MPa
Hoop stress = 0.196 MPa˂7.0 MPa
So, it’s ok.
Meridional stress at any angle θ
wr 1−cosθ wr
= t ( 2
)= ( 1 )
sin θ t 1+cosθ

Maximum meridional stress occur at the base of the dome, where θ=∅ =43.6°
wr 4.0∗1000∗9.8
T = t(1+ cos ∅ ) = 0.1∗(1+cos 43.6 °)

= 227356Pa
= 0.227 MPa˂7MPa. (Ok)
1.3 Area of reinforcement:
Since the stress are within the safe limits, nominal reinforcement@ 0.2% of
gross sectional area of concrete is provided, (minimum steel)
fy= 345*145*0.001 = 50 ksi
Ast= (0.2*1000*100)/100 = 200mm2
Use #3 (10mm) bars
1000∗78.54
Spacing, S = 200
= 392.7mm c/c ≈ 350mm c/c.

2.Design of Top Ring Beam:

The horizontal component of meridional thrust, P = T*cosθ
Total tension tending to rupture the ring beam per meter length of its
circumference,
D
Pt=P* 2
D
= T*cosθ* 2
13.5
= 0.227*cos43.6°*100*1000* 2
= 110962 N
Area of reinforcement, As= Pt/fs
 110962
As= 0.4∗345 = 804 mm2

Use #5 (16mm) bars

804
No of bars = 201.06 = 3.99 ≈ 4 Nos

To fix the size of beam,

Equivalent area of the composite section,
= A+(n-1)Ast
6
29∗10
= A+(8-1)*201.06*4 (n= )=8
57000 √ 28∗145

=A + 5630
Assuming allowable tensile stress of composite section = 1.2 MPa
110962
1.2= A +5630

A= 86838 mm2
If width of beam, b= 250mm
Depth of beam, d=348mm ≈ 350mm
Size of top ring beam (350mm*250mm)
Minimum shear reinforcement:
c/c spacing of vertical stirrups to meet the requirement of minimum shear
2.5 A sv f y
reinforcement is given by S v=
b

Using 8mm∅ 2 legged stirrup, A sv=2*0.7854*(8)2=100.5 mm 2

2.5∗100.5∗345
Sv = =346≈ 300mm
250

3. Design of cylindrical wall:

Maximum hoop tension at the base per meter height of wall
F = γh*0.5D
= 10*1000*9*0.5*13.5 = 607500N
 250mm

607500
Area of steel required, Ast= 0.4∗345 = 4402 mm2

Steel provided, Ast=2201 mm2 in two layers

Use #6 (20mm) bars
#3∅ bar@2
1000∗314
S= 2201 =142.5mm ≈ 140mm c/c

Hence 20mm ∅ 140mm c/c bar is used. 20mm ∅ 140mm c/c

bar
Thickness of cylindrical wall:
Equivalent area of concrete assuming (t)mm thickness of wall
= t*1000+(n-a)Ast
= 1000t+(8-1)*4402
= 1000t + 30814
480mm

Assuming allowable tensile stress in the composite section = 1.2 MPa

607500 Fig: Cylindrical wall
1.2= 1000t +30814

=˃ t= 475.4mm ≈480mm
∴Thickness of wall at its base is 480mm and taper it to 250mm at top.

Distribution steel:
480+250
Average thickness of wall slab, t = 2
= 365mm

% of steel provided, Ast= 0.002*1000*365= 730mm2

or, 365mm2 in each face.
Use #3 (10mm)∅ bars
1000∗78.53
Spacing, S = 365
= 215.1mm ≈ 250mm c/c
4. Design of Bottom Ring Beam
4.1 Load Calculation
This ring beam is to be provided at the junction of the cylindrical wall and the
conical dome.
w 1 per meter length consists of the following:

i)Load due to top dome =area of dome slab*meridional stress*sin∅

=100*1000*0.227 sin43.6
=15654.4 N/m
ii) Load due to top ring beam=b*(d-t wall top)*γ c
=0.25*(0.35-0.25) *25000=625 N/m
ii)Load due to cylindrical wall =t av*H*γ c
0.25+0.48
=( 2
)*9*25000=82125N/m

Considering size of beam =1200mm*900mm

iv)Self weight of bottom ring beam = b*(d-t wall bottom)*γ c
=0.9*(1.2-0.48) *25000
= 16200 N/m
Total load, W 1= 114604 N/m
The horizontal component of thrust (from conical slab) due to W 1
H 1=W 1tan β I β = tan−1 (2.75/2.75)=45 0

=114604*tan45 0

=114604 N

Hoop tension due to trust= H 1*D/2=114604*13.5/2

=773577N
Hoop tension due to the horizontal force of water pressure on ring beam
, =whd*D/2=1000*9*0.9*13.5/2

=546750N
∴ Total hoop tension for which the beam has to be designed,

H=773577+546750=1320327

4.2 Area of Steel Required,

1320327
A st= =9568 mm 2
138

7836
No. of 25 mm ∅ bar = 2 =20 nos
0.7854∗(25 )

∴20 nos 25 mm ∅ bar is provided

H 1320327
Check for maximum tensile stress = A + ( n−1 )∗A = 1200∗900+ ( 8−1 )∗9828
s

=1.116N/mm 2 (ok)
Shear reinforcement:
Using 8mm 4 legged stirrup, a s=201mm 2
2.5 A sv f y
Spacing of nominal stirrup, S=
b
2.5∗201∗345
= (900−2∗50) =216 mm≈ 200mm c/c

5. Design of Conical Dome

13.5+8
Average diameter of conical dome = 2
=10.75m

Average depth of water= (9+2.75/2) =10.375m

Weight of water above the conical dome=( π *10.75*10.375*10000)*2
=7007716N
Assuming 700mm thickness of conical slab
i)self-wt. of slab =3.89*.7*10.75* π *25000=2299038 N
ii) Load from top dome, wall and
700mm

the bottom ring beam =114604* π *13.5=4860528 N

Total load at the base of conical dome slab 12 mm ∅
bar@100mmc/c
=7007716+2299038+486052
20mm@150mm
=14167282 N c/c

Total load per meter length at the base of conical slab

14167282
,W 2 = π∗8

=563698 N/m
Meridional thrust due to above loadT 1=W 2 /cos 45 ° Fig : Conical dome

=563698/cos 45 °=797190
797190
Meridional stress = 700∗1000 =1.138 N/mm 2 (safe)

5.1 Hoop tension in conical dome:

To have a general expression,
diameter of conical dome at any height, h m above its base
Dh=8+2*(2.75/2.75) *h

=8+2h
Intensity of water pressure at this height, P=((9+2.75)-h) *10000
=(11.75-h) *10000
Weight of conical dome slab w 3=0.7*25000=175000 N/m2

Total hoop tension, T=(p*cos β +w 3tan β )* Dh /2

8+2 h
=((11.75-h)*10000cos45° +17500tan45° ¿*( 2 )

=((11.75-h) *7071+17500) *(4+h)

Now,
When H=0 T=402337N
 H=1.5 T=494878 N
H=2.75 T=547688 N
∴ Maximum hoop tension occurs at the top of conical dome slab at h=2.75m
547688
For maximum hoop tension , A s= 138 =3969 m2

A s on each face=1984.5 mm 2

1000∗314.16
Spacing of 20mm ∅ bar= 1984.5
=158≈ 150mm c/c

The conical dome also act as a slab spanning between beam B2 and B3 and
loaded with the wt. of water plus self-wt. of slab.
WL
The approx. B.M for design may be taken = 12

7007716+2299038
W=( π∗8
)=370304 N, L=2.75m

370304∗2.75
∴ Bending Moment= =84861.5 N-m
12

Effective depth of slab=700-50= 650mm

84861.5∗1000
A st= =1100mm 2
.86∗650∗138
113∗1000
Spacing of a s=113mm 2, S= 1100
=103mm≈ 100 mmc /c

Distribution steel =0.002*700*1000=1400mm 2

Steel on each face =1400/2= 700 mm 2
which is less then steel of BM consideration
∴ 12 mm ∅ bar @100 mm c/c as bottom face of the slab to act as
reinforcement for BM as well as distribution steel.
6.Design of Bottom Spherical Dome:
Let, the thickness of slab = 300mm
Radius of top dome,
r2=(r-2.6)2+(8/2)2
 r= 4.38m ≈ 4.4m
Self-weight of the dome =2r* π *2.6*0.3*25
=2*4.4* π *2.6*0.3*25000=539000 N
Volume of water above the dome
π 2 2 π 82∗(4.44−2.6)
= 4 *8 *(9+2.75)-( 3 * *4.44 *2.6- 4 *
π 2
)
3

=590.7-(107.35-30.83)=514.18 m3
=10000*514.18=5141800 N
w
Meridional Thrust= πDsinθ =
539000+5141800
π∗8∗0.90
I θ=sin−1 ( 4.444 )=64.27 °
=251146.5 N/m
251146.5
Meridional stress= 300∗1000 = 0.84 N/mm 2 (safe)

2
wr cos θ+ cosθ−1
Hoop stress at any angle θ= ( )
t 1+cos θ

Maximum hoop stress occurs at, θ=0°

wr
∗1 251146.5∗4.44
∴Hoop stress = t = 300∗1000∗2 = =1.85 MPa
2

Allowable direct compressive stress = 0.25f’c = 7.0 MPa

Hoop stress = 1.85 MPa˂7.0 MPa
So, it’s ok.
Minimum steel required=0.002*300*1000=600mm 2
113∗1000
Using 12mm ∅ bar, Spacing= 600
=188mm≈ 180 mmc /c

12mm ∅ bar @180mm c/c is used in both way

7.Design of Bottom Circular Beam:

Inward thrust from conical dome H1=T1*Cos(α)

 Where α=tan-1(2.75/2.75) =45°
W
Now, Meridional thrust from conical dome, T1= cos(α )

563698
= cos(45) =797190 N

Inward Thrust from conical dome H1=797190*Sin(α)

= 797190*Sin(45) =563698.5 N
Outward thrust from bottom spherical dome
H2=Meridional thrust*Cos(β)
(where β=Cos-1((4.44-2.6)/4.44)=65.51
= 251146.5*Cos(65.51) =104109N
Net inward thrust, P=H1-H2 =563698.5-104109 =459589.5 N
PD 459589.5∗8
Hoop compressive force in beam = 2 = 2
=1838358N

Assume the size of the beam = 800 mm*1400 mm

1838358
Hoop stress = 800∗1400 =0.73 N/mm2 <1.2 Mpa (Hence ok)

This stress is very small; its effect is neglected.

Vertical load per meter of the beam =T1*Sin(α)+T2*Sin(β)

=797190*Sin (45) + 251146.5*Sin(65.51)

=792250 N/m

Self-weight of the beam=0.8*1.4*25000=28000 N/m

Total design load =792250+28000 =820250 N/m

7.1 Moment Calculation:

Assume that the beam supported by 8 columns.
 π
Max. negative B.M at support=K1*w*R2*θ =0.066*820250*42* 4

=680300 N-m
π
Max. positive B.M at mid span = K2*w*R2*θ =0.03*820250*42* 4

=309227 N-m

Max. torsional moment at 9.5 degree from support =K3*w*R2* θ

π
=0.005*820250*42* 4 =51538 N-m

Required effective depth of the beam considering max. B.M

√
d= M
Rb
Where b=0.6m

√
Now d= 680300∗1000 =612 mm ¿1400 mm (Hence safe)
2.27∗800

Assuming 25mm dia bar use and clear cover 50 mm for bottom and top ,

Effective depth of beam=1400-50-25/2= 1337.5mm

M 680300∗1000
Area of steel for negative moment, At= fsjd = 138∗0.86∗1337.5 =4285.77 mm2

4285.77
No. of bars= 491 =8.73 =9 nos

Use 9-25mm ∅ bar.

M 309227∗1000
Reinforcement for positive B.M, At= fsjd = 138∗0.86∗1337.5 =1948 mm2

Use 25mm ∅ bar ,

1948
No. of bars= 491 =3.97 =4 nos

7.2 Shear Calculation:

π
Load on each column, w=820250*4* 4 =2576892N

S.F. at support, V=0.5*w=0.5*2576892 =1288446 N

 V 1288446
Shear stress developed,τ v =¿ bd = 800∗1337.5 =1.20 N/mm2

For M20 grade concrete allowable shear stress=1.7 N/mm2 . Hence Safe.
100∗As 100∗9∗491
p= bd
= 800∗1337.5 =0.41%

For 0.41% and M20 permissible shear stress in concrete

0.39−0.35
, Tc=0.35+ 0.25
*(.41-.40)

=0.352 Mpa

Since 1.2 N/mm2 >.352 N/mm2, Shear reinforcement will be required.

Design of shear reinforcement

Vc=Tc*bd =0.352*800*1337.5=376640 N

Vs=V-Vc =1288446-376640 =911806 N

Asv∗fsv∗d
Use 10 mm dia 4 legged stirrups, Spacing Sv= V −Vc

π
(where Asv=4* 4 *102=314 mm2)

314∗138∗1337.5
Spacing Sv= 911806
=63.56 mm =60mm c/c.

7.3 Design for Torsion

The maximum torsion occurs at the contra flexure which is situated at
an angle of 9.5 from either support
9.5 9.5
S.F. at the point of maximum torsion, V1=V-V* 22.5 =1288446-1288446* 22.5

=744436 N
T 51538∗1000
Equivalent Shear, Ve=V1+1.6* b =744436+1.6* 800
=847512 N

Equivalent nominal shear stress,

 847512
, τ ve=Ve /bd =
800∗1337.5
=0.792 N/mm2 ¿ 1.7 N/mm2

Hence Safe.