Tank - 01
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Transcript of Tank - 01
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Instructor: eng \ just for help Subject: Water Tanks2000-2001 Solved Examples
1. Fundamentals
(1)
(a) Statical system for each tank
Fig. 1
For H = 1.5 m
The walls of the tank can be treated as cantilever , i.e., the total load acting in each
wall will be transmitted in vertical direction
For H = 4 m
Top horizontal beam should be used. In this case, the wall will be considered as twoway slab.
Horizontal strip at the sec.
of max. horizontal load
H
Vertical Sec.
Statical system inconjunction with
the loads
Vertical strip
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Instructor: eng \ just for help Subject: Water Tanks2000-2001 Solved Examples
Fig. 2
Fig. 3
* If the tie is removed
Fig. 4
Statical system for
vl strip (Box. Sec)
Structural systemfor vl. strip
Tie
Statical system for vertical strip
H=2mThe walls of the tank can be considered as cantilever walls
H
Statical system for vl. Strip
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Instructor: eng \ just for help Subject: Water Tanks2000-2001 Solved Examples
* H = 4 m
Top horizontal beam should be used.
Fig. 5
For large diameter tank, the part of floor with the length (L) with the wall adjacent to
it can be assumed as a retaining wall connecting to the central part.
* If the tank has sliding base, the wall and floor will be treated separately.
Fig. 6
b. Acting loads & case of loading
Fig. 1 to Fig. 4 are elevated tanks
Acting loads are: * own weight of concrete elements
* water load vl. Load on floor
lateral load on walls
* External lateral load wind pressure
Earthquake Excitation
- case of loading
1- Case of water load (No external lateral load)2- Empty tank and consider lateral load.
vl. Strip
Horizontal Strip vl. Strip in short dir.vl. Strip in long dir.
Treated as cantileverretaining wall H
L
Central part
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Instructor: eng \ just for help Subject: Water Tanks2000-2001 Solved Examples
Fig. 5 is rest on soil tank
Acting loads are : own weight of tank: Water load
: Soil reaction
there is only one case of loading under the effect of all acting loads
Fig. 6 is under ground tank
Acting load are : - Own weight of concrete
- Weight of water
- Weight of soil above the tank
- Lateral earth pressure
- Lateral hydrostatic pressure due to ground
water.- Soil reaction below the tank
- Uplift force due to ground water
Case of loading
tank is full of water and no soil around tank (Check
bearing capacity of soil)
tank is empty and consider soil around tank (Check
uplift)
(2)
- The sections that labeled by (w.s) are water side sections.
- The sections that labeled by (a.s) are air side sections.
Case 1
Case 2Empty
w.s
a.s a.s
a.s
w.s
a.s
w.s
w.sw.s
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Instructor: eng \ just for help Subject: Water Tanks2000-2001 Solved Examples
(3)sec. 1 Mu = 0.0
Nu = +150 kN
- Check of tensile stress for concrete.
Sec. 2 Mu =100 kN.m , Nu = 0.0 (water side sec.)
Water side section
o.kff
99.165.1
306.06.0f
1.65
cm50t])(
)(ft[1t
MPa1.6500*100010*66.7
btM6
ZMfwhere
ff
cm50
mm500M50tcm47.23
1000*6.67
3
1000*Mt,7.66
5.1
tct
tc
c
v
2
6
2t
tct
=======
=
cu
tc
O
t
u
f
Mf
N
MM
side//m`105use
cm2.825.64/2/sideA
cm637.5mm7.635360*0.85
1000*1.15*150
/
TA
0.8510use-
cm20taket
cm20cm610*0.6T0.6t-
kN1005.1
15
5.1
2
s
22
cr
us
cr
==
====
=
=
cu
t
f
f
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Instructor: eng \ just for help Subject: Water Tanks2000-2001 Solved Examples
Sec. 3 Mu = 100 kN.m & Tu = 10 kN
/m'105A'
/m'129or
/m'165use
cm9.75mm975826.0*460*360*75.0
10*100A
0.7516use-
/m'1011use
cm6.8mm608826.0*460*360*85.0
10*100A
0.8510use-
f
MA
0.826J
7.97c1000*30
10*100
c460
f
Mcd
mm460dmm500t&kN.m100
s
226
s
cr
226
s
cr
y
u
s
1
6
1
cu
u
1
=
===
=
===
=
=
=
==
=
===
dJ
b
M
cr
u
1.65
mm18.045]16
0.0134500[1
)(
)(ft[1t
N/mm28.3f0.6f,f
f
N/mm6134.10134.06.1f
N/mm6.1500*1000
10*66.7*6
bt
6M)(f
N/mm0134.0500*1000
1000*6.7
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mm500ttakemm384307.66503050t
kN7.61.5
10T&.7.665.1
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2
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2
c
tc
tctct
=+=+=
===
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====
Mf
N
M
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N
M
mkNM
tc
t
rt
rt
0.75m16use
0.826J
8.05c1000*30
10*97.9c460
f
Mcd
kN.m9.799.79*10eTM
m9.790.040.25-10
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100
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Me
o.kfN/mm98.165.1
28.3
crmax
1
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===
>==
m
b
er
ftc
/m'105A'
/m'165use
mm983360*0.75
1.15*10*6.7
0.826*360*0.75
10*97.9
/
T
s
236
cr
u
=
=+=
+=
syycr
us
sfdJf
MA
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Instructor: eng \ just for help Subject: Water Tanks2000-2001 Solved Examples
Sec. 4 Mu = 10 kN.m & Tu =150 kN
N/mm3.28f,1.45
mm5.9921.005
0.51200
)(
)(f1tt
N/mm005.1200*1000
10*6.7*6bt6M)(f
N/mm5.0200*1000
1000*100
A
T)(f
/f(M)f)(ff
mm200tuse
60100*6.0T0.6
mm4.591307.650
3050t
kN100
1.5
150T&kN.m7.6
5.1
10
2
rtc
c
v
2
2
6
2tc
2
c
tc
rtcctctt
=
=
+=
+=
===
===
>+=
=
==
=+=
+=
====
Mf
N
M
N
N
mm
M
M
tc
t
/m'107choose
mm175360*0.85
1.15*137.5
/
T
A
0.85,10use
kN5.21137.5-150T-TT
kN5.3710.16
10
2
150
d'-d
M
2
TT
cover-2
tm0.067
150
10
T
Me
kN150T&kN.m10M
o.kff
N/mm26.21.45
3.28f,N/mm505.1005.15.0
2
cr
1
s
cr
12
uu1
u
u
uu
tct
2
tc
2
===
=
===
=+=+=