Transcript IS:1893-2002 - Indian Institute of Technology Kanpur

Solution 1
Date of assignment: 18 January 2006
Date of solution: 20 January 2006
E-course on Seismic Design of Tanks
1
Part II: True / False

Identify the following statements as True or False
1.1)Hydrodynamic pressure varies linearly with liquid height.
False
Hydrodynamic pressure has curvilinear variation along the
height. In fact, Hydrostatic pressure varies linearly with
height.
1.2)Impulsive mass is less than convective mass for short
tanks.
False
In short tanks, convective liquid or liquid undergoing
sloshing motion is more. In tall tanks, impulsive liquid mass
is more.
 Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006
Solution 1 / slide 2
Part II: True / False
1.3)Net hydrodynamic force on the container wall is zero.
False
Rather, net hydrostatic force on wall is zero
1.4)Hydrodynamic pressure on base causes overturning
moment on tank.
True
Refer slide no. 46 of Lecture 1
1.5)Convective mass acts at lower height than impulsive
mass.
False
Liquid in upper portion, undergoes convective or sloshing
motion, hence, convective mass is always located at
higher height than impulsive mass. Note the graphs for hi
and hc in figure 2b and 3b of guidelines
 Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006
Solution 1 / slide 3
Part II: True / False
1.6)With the inclusion of base pressure effect, overturning
moment on tank reduces.
False
Hydrodynamic pressure on the base produces overturning
moment in the same direction as that due to
hydrodynamic pressure on wall. Hence, with the inclusion
of base pressure effect, overturning moment increases.
This can be seen from the fact that hi* is always greater
than hi and hc* is always greater than hc.
1.7)Impulsive and convective masses depend only on height
of liquid.
False
Impulsive and convective masses depend on aspect ratio
(h/D or h/L) rather than only on height, h.
 Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006
Solution 1 / slide 4
Part III: Solutions
1.1)A circular tank has internal diameter of 12 m and water
height of 5 m. Obtain mi, mc, Kc, hi, hc, hi* and hc*.
Solution:
Total volume of water = /4 x 122 x 5 = 565.5 m3
 Total water mass, m = 565.5 x 1.0 = 565.5 t
D = 12 m, h = 5 m
 h/D = 5/12 = 0.417
For this value of h/D, from Figures 2a and 2b of the Guideline,
values of various parameters can be read. One can also use
formulae given in Table C-1 of the Guideline. Here, these formulae
will be used to obtain various parameters.
 Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006
Solution 1 / slide 5
Part III: Solutions
D

tanh 0.866 
mi
h


D
m
0.866
h
h

tanh 3.68 
mc
D

 0.23
h
m
D
12 

tanh 0.866 
5 


12
0.866
5
= 0.466
5

tanh 3.68 
12 

 0.23
5
12
= 0.503
Since, h/D is less than 0.75, hi/h = 0.375
h

cosh 3.68   1.0
hc
D

 1
h
h
h

3.68 sinh 3.68 
D
D

5

cosh 3.68   1.0
12 

 1
5
5

3.68 sinh 3.68 
12
12 

= 0.579
 Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006
Solution 1 / slide 6
Part III: Solutions
hi *

h
0.866
D
h
D

2 tanh 0.866 
h

-
0.125
h

cosh 3.68   2.01
hc *
D

 1
h
h
h

3.68 sinh 3.68 
D
D

K c  0.836


0.866
12
5
12 

2 tanh 0.866 
5

= 0.947
- 0.125
5

cosh 3.68   2.01
12 

 1
5
5

3.68 sinh 3.68 
12
12 

= 0.878
mg
h

tanh2  3.68 
h
D

5

K ch / mg  0.836 tanh2  3.68 
12 

 0.694
 Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006
Solution 1 / slide 7
Part III: Solutions
Thus, we get,
mi/m = 0.466
mi = 0.466 x 565.5 = 263.5 t
mc/m = 0.503
mc = 0.503 x 565.5 = 284.5 t
hi/h = 0.375
hi = 0.375 x 5
= 1.88 m
hc/h = 0.579
hc = 0.579 x 5
= 2.90 m
hi*/h = 0.947
hi* = 0.947 x 5
= 4.735 m
hc* = 0.878 x 5
= 4.39 m
hc*/h = 0.878
Kch/mg = 0.694
Kc = 0.694mg/h = 0.694 x 565.5 x 9.81/5.0 = 770 kN/m
 Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006
Solution 1 / slide 8
Part III: Solutions
1.2) A rectangular tank has internal plan dimension of
10 m x 16 m. Water height is 8 m. Obtain mi, mc, hi, hc,
hi* and hc* in both the directions.
Solution:
Y
16 m
10 m
X
Plan view
Total volume of water = 10 x16 x 8 = 1280 m3
Total mass of water, m = 1280 x 1.0 = 1280 t
 Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006
Solution 1 / slide 9
Part III: Solutions
First, consider base motion in X-direction:
L = 16 m, h = 8 m
 h/L = 8/16 = 0.5
From Figure 3a, 3b of guidelines, for h/L = 0.5:
mi/m = 0.54
mi = 0.54 x 1280 = 691.2 t
mc/m = 0.48
mi = 0.48 x 1280 = 614.4 t
hi/h = 0.375
hi = 0.375 x 8
= 3.0 m
hc/h = 0.57
hc = 0.57 x 8
= 4.6 m
hi*/h = 0.80
hi* = 0.8 x 8
= 6.4 m
hc* = 0.87 x 8
= 7.0 m
hc*/h = 0.87
 Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006
Solution 1 / slide 10
Part III: Solutions
Now, consider base motion in Y-direction:
L = 10 m, h = 8 m
 h/L = 8/10 = 0.8
From Figure 3a, 3b of guidelines, for h/L = 0.8:
mi/m = 0.72
mi = 0.72 x 1280 = 921.6 t
mc/m = 0.32
mi = 0.32 x 1280 = 409.6 t
hi/h = 0.40
hi = 0.40 x 8
= 3.2 m
hc/h = 0.65
hc = 0.65 x 8
= 5.2 m
hi*/h = 0.58
hi* = 0.58 x 8
= 4.64 m
hc* = 0.73 x 8
= 5.84 m
hc*/h = 0.73
 Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006
Solution 1 / slide 11
Part III: Solutions
Thus, in x-direction, wherein, h/L = 0.5, impulsive and convective
masses are almost same. Whereas, in Y-direction, for which h/L = 0.8,
more liquid contributes to impulsive mass.
Also note the difference in hi and hi* values ( or hc and hc*) for two
cases. For shallow tanks, inclusion of base pressure significantly changes
the height .
 Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks / January 2006
Solution 1 / slide 12