9.6 Example Illustrating the Tabular Method of Solution 561Table 9.1 Component 3 (Tank)
Type of tank
Load acting on
the tank,s 4
(kgf) R 3 cost ($)Self-weight of
the component
(kgf)s 3 =s 4 +
self-weight
(kgf)(a) Cylindrical RCC tank 100,000 5,000 45,000 145,000
(b) Spherical RCC tank 100,000 8,000 30,000 130,000
(c) Rectangular RCC tank 100,000 6,000 25,000 125,000
(d) Cylindrical steel tank 100,000 9,000 15,000 115,000
(e) Spherical steel tank 100,000 15,000 5,000 105,000
(f) Rectangular steel tank 100,000 12,000 10,000 110,000
(g) Cylindrical RCC tank with
hemispherical RCC dome 100,000 10,000 15,000 115,000
Table 9.2 Component 2 (Columns)
s 2 =s 3 +
Type of columns s 3 (kgf) R 2 cost ($) Self-weight (kgf) self-weight (kgf)
(a) RCC columns 150,000 6,000 70,000 220,000
130,000 5,000 50,000 180,000
110,000 4,000 40,000 150,000
100,000 3,000 40,000 140,000
(b) Concrete columns 150,000 8,000 60,000 210,000
130,000 6,000 50,000 180,000
110,000 4,000 30,000 140,000
100,000 3,000 15,000 115,000
(c) Steel columns 150,000 15,000 30,000 180,000
130,000 10,000 20,000 150,000
110,000 9,000 15,000 125,000
100,000 8,000 10,000 110,000
foundation, columns, and the tank used in the system, respectively. Thus the vari-
ablex 1 can take three discrete values, each corresponding to a particular type of
foundation (among mat, concrete pile, and steel pile types). Similarly the variable
x 2 is assumed to take three discrete values, each correspondingto one of the columns
(out of RCC columns, concrete columns, and steel columns). Finally, the variablex 3
can take seven discrete values, each corresponding to a particular type of tank listed
in Table 9.1.
Since the input load, that is, the weight of water, is known to be 100,000 kgf, s 4
is fixed and the problem can be considered as an initial value problem. We assume
that the theories of structural analysis and design in the various materials provide the
design equations
si=ti(xi, si+ 1 )