UPDATED AS PER LIMIT STATE METHOD

1. DharamsinhDesai University
Name : Jayswal.Shashank.M
Roll Number : MS004
Subject : Seminar – 2
Branch : M-tech (Structural
engineering)

2. RECTANGULAR TANK
LIMIT STATE METHOD

3. Advantage Of Limit State Design
• Limit state method is widely used at present in comparison to working stress
method with the
following advantages:
(i) Materials are treated according to their properties.
(ii) Loads are treated according to their nature.
(iii) Structures generally fail when they reach their limit state, not their
elastic state.
• However, when structures reach to their limit state, the cracking width in the
structure may be significantly higher comparative to a structure designed by
working stress method at the same
stage. IS: 3370

4. • i.e. the Indian Standard specifications for construction of liquid retaining
structures did not adopt limit state design method for long. However,
IS:3370 has adopted the limit state design method after considering checks
over the cracking width.
• It has been recently adopted in the new version of IS 3370-2021concrete
structures for storage of liquids – code of practice, while going through IS
3370 – 2021 it can be found that three methods of design are available.
(I) Limit State Design Method – Crack width check.
(II) Limit State Design Method – Deemed to satisfy.

5. Introduction
• Storage reservoirs and overhead tanks are used to store water, liquid
petroleum, petroleum products and similar liquids.
• In general there are three kinds of water tanks:
i) Tanks resting on ground.
ii) Underground tanks.
iii) Elevated tanks.
• The tanks resting on ground like clear water reservoirs, settling tanks,
aeration tanks etc. are supported on the ground directly. The walls of
these tanks are subjected to pressure and the base is subjected to
weight of liquid and upward soil pressure. The tanks may be covered on
top.

6. • From design point of view, the tanks may be classified as per their
shape as following
i) Rectangular tanks
ii) Circular tanks
iii) Over Head Service Reservoir (OHSR)
iv) Intze Tank i.e. OHSR for large capacity.
• Rectangular tanks are provided for smaller to moderate capacity. For
small capacities, circular tanks prove uneconomical as the formwork
for circular tanks is very costly.
• The rectangular tanks should be preferably square in plan from point
of view of economy. It is desirable that longer side should not be
greater than twice the smaller side.

7. Limit State Design Method
 Limit state design method, though semi-empirical approach, has been found
to be the best for the design of reinforced concrete structures over the elastic
theory of design where the level of stresses in concrete and steel are limited
so that stress-deformations are taken to be linear.
 There are two limit states:
i) Limit state of collapse.
ii) Limit state of serviceability which includes
a) Deflection
b) Cracking

8.  The structure is first designed under Limit State of Collapse and then
checked under serviceability. Because of its superiority to other two methods
, IS 456:2000 has been thoroughly updated in its fourth revision in 2000
taking into consideration the rapid development in the field of concrete
technology and incorporating important aspects like durability etc.
 This standard has put greater emphasis to limit state method of design by
presenting it in a full section.
 It is important to point out here that a structure designed through limit state
method when fails, the failure will be in plastic stage and not in elastic stage.
 Therefore, the cracking and cracking width can be significant at the failure
stage.

9. • A rectangular water tank is to be designed to store
2500m3 water. The tank is to be made just above the
ground level and the safe bearing capacity of the soil is 75
kN/m2.
Design data & main dimensions of the tank :
Capacity of tank = Q = 2500m3
Safe bearing capacity of soil =75 kN/m2
Free Board=0.15m

10. • The allowable stresses are :
M – 30 grade concrete and Fe-415 grade tor steel, for calculations
relating to resistance to cracking (IS:3370-2021)
σct =1.5 N/mm2
σcb =2 N/mm2
σst=150 N/mm2
• For strength calculations the stresses in concrete and steel are same
as that recommended in IS:456 2000.
σcc= 8.0 N/mm2
m = 9.34
σcbc= 10.0 N/mm2
Q = 1.3
J = 0.9
Assuming a clear height of the tank as 5.25 m,
Clear base=2500/5.25-0.15= 490m2

11. • Assuming the clear dimensions of the tank as followings:
• Height = 5.4 m , length = 24.6 m , width = 19.6 m
• Let the roof slab is supported by columns spaced 5 m apart in both
directions. Let the center to center distance between the walls and the
head of water be :
Lx = 24.6+0.4 = 25 m , Ly = 19.6+0.4 = 20.0 m , H = 5.4-0.15 =
5.25 m
Assuming the thickness of the wall at the base as 400 mm.

12. Column
Let the size of the column be assumed as 440 mm2
The capacity of the tank is ( gross volume ) = 24.6 × 19.6 ×5.25 =
2531.3 m2
Net Volume = Gross volume – volume of 12 columns
= 2531.3 - ( 12×5.25×0.44×0.44 )
= 2518.3 m3

13. 1. Design of Roof Slab :
• The roof slab is designed as a flat slab with columns spaced at 5 m
apart. Let the thickness of the slab be 240 mm for self-weight purpose
and the loads on the slab are
Live load = 1.5 kN/m2
Surface finish load = 2.0 kN/m2
Self weight = 0.24 ×25 = 6.0 kN/m2
Total load = w = 9.5 kN/m2
Centre to centre of panel L = 5 m
Clear panel L0 = 5 - 0.44 = 4.56 m
Total design load on the panel is = W = w*L*L0 = 9.5 × 5 ×
4.5=216.6Kn
Total factored design load on the panel is = 1.5 × 216.6 kN =
325 kN
• The sum of the magnitudes of the positive and negative bending
moments in a panel is
M0 = 1/8 W * L0 = 325 × 4.56 / 8 = 185.25 kNm

14. • Using table 4.72 , c1 = 0.308 ,c2 = 0.702 ,c3 = 0.497 & cl = 0.65
• The magnitude of the bending moments (in kN.m) in a panel are :
• Mn1= c1 M0= 0.308 × 18525 × 103 = 57
• Mp3= c3 M0 = 92.02
• Mn2= c2 M0= 130
• Mni= ci M0 = 120.4
Where
1 : the inside face of the outer wall
2 : the face at the inner column of the outer panel
3 : the middle point of the outer panel
i : the interior panel
• The bending moments on the panel are distributed between the
column and the middle strips as per table 4.6. The bending moments
in the column strip which are indicated by a subscript c are:
• Mnc1 = Mn1 = 57 kN.m
• Mpc3 = 0.6 Mp3 = 55.2 kN.m
• Mnc2 = 0.75 Mn2 = 97.50 kN.m
• Mnci= 0.75Mni = 90.3 kN.m

15. • ( 100 % of the negative bending moment in the end support, 75% of
the negative BM in the interior supports and 60% of the negative BM
are assigned to the column strip ).
• In case of an end wall support 75% can be assigned to the column
strip sectional. The effective depth of the section is calculated from
the maximum negative bending moment and is;
• M = 0.138 fck . b .d2
0.0975 ×109 = 0.138 ×30 ×1000 × d2
d = 154 mm provide 200 mm
( Roof slab has been designed as a cracked section as it is not in
contact with the water )

16. Check for Shear Stress :
• The allowable shear stress based on diagonal tension is
τc=0.25(fck)1/2=0.25(30)1/2=1.3693
• The critical shear plane is the peripheral plane which is at a distance 0.5 d from the face
of the column. The length of the plane is
b0 = 4 ( a+d ) = 4 ( 0.44+0.20) = 2.56 m
The shear force on the plane is , V = w ( L2 – (a+d)2 )
= 9.5 ( 52 – 0.642 )
=233.6 kN
The nominal shear stress on the plane is,
τv=v/b0.d
=0.456 Mpa < τc (1.3693)
• Thus shear stress is within the allowable limits; therefore the depth is adequate against
shear stress.
• The overall depth of the slab can be ; t = d + 0.03 = 0.23 m
• Design of reinforcement in the column strip using.
Mu =0.87Fy.Ast.(d-0.42*0.48d)=0.7Fy.Ast.d
The area of reinforcement, Ast=Mu/0.7fy.d

17. • The required areas of the reinforcement at different sections are ;
Ast1= Mnc1/0.7.Fy.d=981mm2
Ast2=Mnc2/0.7.Fy.d=1678mm2
Ast3=Mnc3/0.7.Fy.d=950mm2
Asti=Mnci/0.7.Fy.d=1554mm2
The minimum area of steel required is 0.12% ,
Astm = 0.12*b* t / 100
= 0.12 ×2500×200/100
= 600 mm2
There is direct tension in the slab as it supports the vertical wall.
The tension force is =1/4(γH2/2)= 10*5.252 /8 =34.5Kn/m
Direct tension per 2.5 m panel is, T = 34.5×2.5 = 86.25 kN
Factored Tension = F.T. = 1.5 ×86.25 = 130 kN
The area of tensile reinforcement for direct tension is,
Ast =F.T./0.87Fy= 1.5*130*103/0.87*415= 540mm2

18. • At any given section the total area of the reinforcement is equal to the
sum of the area needed for bending and direct tension. The direct
tensile stress caused in the concrete is,
• σt=T/Ac
• =86250/2500*230
• =0.15 Mpa which is very small.
• Design of reinforcement in the middle strip
The positive bending moment on the middle strip in the exterior
panel is
Mmp3=0.35 × 0.25 × M0 = 0.35×0.25×185.25 = 16.2 kNm
The area of the reinforcement is ,
Ast= Mu/0.7Fy.d
=16.2*106/0.7*415*200
=280mm2

19. 2. Design Of Column (M30 Grade concrete )
• The columns are spaced at 5 m and are subjected to axial force only. The
load from the roof slab on each column is,
Ps = wl2 = 9.5 × 25 = 237.50 kN
Let self wt. = 32.5 kN
Total load P = 270 kN
Factored Load = 1.5 × 270 = 405 kN
Assume only 0.8% of reinforcement in the column, the capacity of the
column is then given by:
Pu=0.4σck.Ac+0.67fyAsc
405 ×103=0.4 × 30 × (1-0.008) a2+ 0.67 × 415 × 0.008 a2
a2=28665
square column of 200 mm size.
Area of steel provided = 1206 mm2
• Provide 350 mm size of column with 6 bars of 16 mm dia in each
column. Also provide 6 mm ties at 200 mm spacings
The slenderness factor for the column is , g = Le / a = 3.9 / 0.35 =
11.14

20. • i) Check for crack width :
• Minimum Reinforcement :
• ρcrit=Critical steel ratio, that is, the minimum ratio, of steel area to the
gross area of the whole concrete section, required to distribute the
cracking.
• Fct= direct tensile strength of the immature concrete.
• Fy= characteristic strength of the reinforcement.
• ρcrit= 1206/(350×350) = 0.00984
• Fct/ Fy = 1.3/415 = 0.003
• ρcrit > Fct/ Fy
• Maximum spacing of crack (Smax)
• Smax =Fct/Fb*φ/2ρ , where
Fct/Fb= ratio of the tensile strength of the concrete to the average
bond strength between concrete and steel which can
be taken as 2/3 in this case.
φ= size of each reinforcing bar , and
ρ= steel ratio based on the gross concrete section
Smax=542

21. • Width of fully developed crack, Wmax
Wmax=Smax*α/2*T
• α=1*10-50C-1 (Coefficient of thermal expansion of concrete)
T= 40 degree celsius
Wmax=542 × 10-5× 40/2 = 0.1084 mm < 0.2 mm ( permissible)

22. 3. Design Of Vertical Wall :
M = 0.138 fck . b .d2
91.16 ×106 = 0.138 ×30 ×1000 × d2
d = 148 mm provide 200 mm
Providing a clear cover of about 30 mm, the overall thickness of the
wall needed is about 230mm .
However, the smaller value is selected.
Let t = 230 mm and d= 200 mm
Area of Steel Required :
Ast= Mu/0.7Fy.d = 91.16*106/0.7*415*200 = 1570mm2
provide 16 mm dia bars @ 125 mm c/c . (1610 mm2)

23. • i) check for crack width :
Minimum Reinforcement :
• ρcrit = Critical steel ratio , that is , the minimum ratio, of steel area to
the gross area of the whole concrete section, required to distribute the
cracking
• Fct = direct tensile strength of the immature concrete.
• Fy =characteristic strength of the reinforcement
• ρcrit = 1610/200×1000 = 0.00805
• Fct/ Fy = 1.3/415 = 0.003
• ρcrit > Fct/ Fy
• Maximum spacing of crack (Smax)
• Smax =Fct/Fb*φ/2ρ , where
Fct/Fb= ratio of the tensile strength of the concrete to the average
bond strength between concrete and steel which can be taken as 2/3 in
this case.
φ= size of each reinforcing bar , and
ρ= steel ratio based on the gross concrete section

24. • Smax =2/3*16/2(0.00805) =662mm2
Width of fully developed crack, Wmax
Wmax=Smax*α/2*T
• α=1*10-50 C-1 (Coefficient of thermal expansion of concrete)
T= 40 degree celsius
• Wmax = 662 × 10-5× 40/2 = 0.1324 mm < 0.2 mm ( permissible)

25. • Reinforcement At Mid-Height :
• M = 0.138 fck . b .d2
• 43.140 ×106 = 0.138 ×30 ×1000 × d2
• d = 102 mm provide 150 mm
• Providing a clear cover of about 30 mm, the overall thickness of the
wall needed is about 180mm .
• However, the smaller value is selected.
• Let t = 180 mm and d = 150 mm
• Area of Steel Required :
Ast = Mu/0.7Fy.d=825mm2
Provide 12 mm dia bars @ 125 mm c/c .( 904 mm2 )

26. • i) Check for crack width :
• Minimum Reinforcement :
• ρcrit= Critical steel ratio, that is, the minimum ratio, of steel area to the
gross area of the whole concrete section, required to distribute the cracking.
Fct =direct tensile strength of the immature concrete.
Fy = characteristic strength of the reinforcement.
ρcrit = 904/180×1000 = 0.0050
Fct/ Fy = 1.3/415 = 0.003
ρcrit > Fct/ Fy
Maximum spacing of crack (Smax)
• Smax =Fct/Fb*φ/2ρ , where
Fct/Fb= ratio of the tensile strength of the concrete to the average bond
strength between concrete and steel which can be taken as 2/3 in this case.
φ= size of each reinforcing bar , and
ρ= steel ratio based on the gross concrete section
• Smax =2/3*12/2(0.005)=800

27. • Width of fully developed crack,Wmax
• Wmax=Smax*α/2*T
• α=1*10-50C-1 (Coefficient of thermal expansion of concrete)
T= 40 degree celsius
• Wmax = 800 × 10-5× 40/2 = 0.16 mm < 0.2 mm ( permissible)

28. Design of the horizontal reinforcement :
• The positive bending moment on the horizontal fibres of the wall is
zero; therefore, only nominal reinforcement be provided. However,
there is negative bending moment at the corners of the wall. Its value
at about the mid height of the wall is
Mxa = -28940 Nm/m
• There is an axial tension coming from the hydrostatic force acting on
the walls normal to this wall. This force at mid height is of H/2 width of
the wall. Therefore, the tension force can be taken approximately as
• T =γ*H2/2 = 10*(5.25)2/2 =13.8Kn/m
• The tensile stress due to the combined action on the vertical plane of
the wall is
σt =M/Z+T/A =6M/t2+ T/t
The thickness of the wall at mid height is 190 mm so
σt = 4.87 Mpa>2Mpa

29. • The axial tension contribution is negligible but the tension caused by
bending moment exceeds the allowable value of 2.0 MPa. Therefore,
a fillet be provided at the corners so as to reduce the tension.
• Provide 150 mm thick haunch at the corners of the wall. The overall
thickness at the joint including fillet is
190+150 = 340 mm and the
Bending stress is σt = 6M/t2 =1.5Mpa < 2Mpa
The effective depth of the section at this point can be taken as
340-100=240 mm.
Mxa = -28940 Nm/m
Factored Moment = 1.5 ×28940 = 43410 Nm/m
• Thickness Required :
• M = 0.138 fck . b .d2
• 43410 ×103= 0.138 ×30 ×1000 × d2
• d = 102 mm provide 150 mm

30. • Area of Steel Required :
Ast =Mu/0.7Fy.d
= 43410*103/0.7*415*150
=997mm2
Provide 16 mm dia bars @ 150 mm c/c .( 1340 mm2 )
• i) Check For Crack Width :
• Minimum Reinforcement :
• ρcrit = Critical steel ratio , that is , the minimum ratio, of steel area to the gross
area of the whole concrete section, required to distribute the cracking
• Fct = direct tensile strength of the immature concrete.
• Fy =characteristic strength of the reinforcement
• ρcrit = 1340/150×1000 = 0.00893
• Fct/ Fy = 1.3/415 = 0.003
• ρcrit > Fct/ Fy
• Maximum spacing of crack (Smax)
• Smax =Fct/Fb*φ/2ρ , where
Fct/Fb= ratio of the tensile strength of the concrete to the average bond
strength between concrete and steel which can be taken as 2/3 in this case.
φ= size of each reinforcing bar , and
ρ= steel ratio based on the gross concrete section

31. • Smax =2/3*16/2(0.00893)=597
Width of fully developed crack, Wmax
Wmax=Smax*α/2*T
• α=1*10-50C-1 (Coefficient of thermal expansion of concrete)
T= 40 degree celsius
• Wmax = 597 × 10-5× 40/2 = 0.1194 mm < 0.2 mm ( permissible)

32. 4. Design Of Base Slab
• The bottom slab is resisting on soil and it supports the water and the
columns. The weight of the water is directly transferred to the soil.
Therefore, the bearing capacity has to be checked.
• The load from the column is transferred through the bottom slab. The
slab need to be designed for tank empty condition as a two way slab
subjected to net pressure from the soil. First the bearing capacity of
the soil is computed
• Load from the roof =9.5Kn/m2
• Load from the columns =1.5Kn/m2
• Load from the water = 52.5Kn/m2
• Load from the bottom slab = 6Kn/m2
• Total Load = 69.5Kn/m2
• The bearing pressure is 69.5 kN/m2 as against the safe bearing
capacity of 75 kN/m2 . Safe

33. • STRUCTURAL DESIGN OF BOTTOM SLAB
• The net bearing pressure on the soil in the tank empty condition is
that due to the roof and the column load and it is
• w = 9.5 + 1.5 = 11 kN/m2
• The bottom slab is designed as a flat slab subjected to the soil
pressure and supported by the columns spaced at 5 m apart. Assume
a widening of the column.
• The widened size of the column base be same as the column head
and it is 440mm square.
• Efective span of the slab L = 5 m
• Clear Span = L0 = L – a = 4.56 m
• Total design load on the panel is = W = w* L* L0 = 11 × 5 × 4.56 =
250.8 kN
• The sum of the magnitudes of the positive and negative bending
moments in a panel is
• M0 = 1/8 W . L0 = 250.8 × 4.56 / 8 = 143 kNm
• The effective height of the wall be Lw=5.4 m
• The relative stiffness of the wall is , Kw =Iw/Lw=5*0.223/12*5.4 =8*10-4

34. • Where the average thickness of the wall is taken as 220 mm. The relative
stiffness of the panel slab by assuming the thickness 220 mm is
Ks=Is/L=0.223/12=8.9*10-4
The ratio of the relative stiffness of the wall and the roof slab panel is ;
αc =Kw/Ks =8/8.9 =0.9
the ratio of live load to the dead load is ,Wl/Wd =1.5/8 = 0.1875
• The designed bending moments in the flat slab are computed using the
moment coefficients given in table 4.6 and 4.7.
• α =1+1/αc = 2.111
• Using table 4.72 , c1 = 0.308 ,c2 = 0.702 ,c3 = 0.497 & cl = 0.65
• The magnitude of the bending moments ( in kNm ) in a panel are :
• Mn1= c1 M0= 0.308 × 143 = 45 , Mp3= c3 M0 = 72 ,
• Mn2= c2 M0= 100.4 , Mni= ci M0 = 93
Where 1 : the inside face of the outer wall
2 : the face at the inner column of the outer panel
3 : the middle point of the outer panel
i : the interior panel

35. • The bending moments on the panel are distributed between the
column and the middle strips as per table 4.6. The bending moments
in the column strip which are indicated by a subscript c are:
• Mnc1 =Mn1= 45 kNm
• Mpc3 =0.6Mp3= 43.2 kNm
• Mnc2 =0.75Mni= 75.3 kNm
• Mnci= 0.75Mni = 69.75 kNm
• ( 100 % of the negative bending moment in the end support, 75% of
the negative BM in the interior supports and 60% of the negative BM
are assigned to the column strip ). In case of an end wall support 75%
can be assigned to the column strip sectional. The effective depth of
the section is calculated from the maximum negative bending
moment, and is ;
• M = 0.138 fck . b .d2
• 113.25 ×106 = 0.138 ×30 ×1000 × d2
• d = 165 mm provide 200 mm

36. • Check for Shear Stress :
• The allowable shear stress based on diagonal tension is
• τc= 0.25(Fck)1/2= 0.25√30 = 1.3693
• The critical shear plane is the peripheral plane which is at a distance
0.5 d from the face of the column. The length of the plane is
• b0 = 4 ( a+d ) = 4 ( 0.44+0.20) = 2.56 m
• the shear force on the plane is , V = w ( L2 – (a+d)2 ) = 9.5 ( 52 – 0.642
) = 233.6 kN
• The nominal shear stress on the plane is, τv = V/b0.d
=0.2336/2.56*0.2 =0.456Mpa < τc(1.3693)
• Thus shear stress is within the allowable limits, therefore the depth is
adequate against shear stress.
• The overall depth of the slab can be ;
• t = d + 0.03 = 0.23 m

37. • Design Of Reinforcement In The Column Strip
Using Mu= 0.87.fy.Ast(d-0.42*0.48d) =0.7fy Ast d
The area of reinforcement , Ast = Mu/0.7 fy.d
The required areas of the reinforcement at different sections are ;
Ast1= Mnc1/0.7.Fy.d =1140mm2
Ast2=Mnc2/0.7.Fy.d =1950mm2
Ast3=Mnc3/0.7.Fy.d =1104mm2
Asti=Mnci/0.7.Fy.d =1805mm2
The minimum area of steel required is 0.20% ,
Astm= 0.20 b t / 100 = 0.27 ×2500×230/100 = 1552 mm2
Ast= 16mm#-125mm c/c (Ast provided-1608.49 mm2)

38. Design of Reinforcement In The Middle Strip
• The positive bending moment on the middle strip in the exterior panel
is
• Mmp3= 0.35 × 0.25 × M0 = 0.35×0.25×215 = 0.01881 MNm
• The area of the reinforcement is ,
• Mu/0.7.Fy.d = 0.01881*109/0.7*415*200 =323mm2

39. Reference :
1.P.c varghese
( Advance reinforce concrete design)
2. Research paper
(See discussions, stats, and author profiles for this publication
at: https://www.researchgate.net/publication/315710218)
3. H.J shah ( working stress design ) vol.2
4. Indian standard code:
(I) IS 456:2000 :Plain and Reinforced Concrete,
Code of Practice [CED 2: Cement and Concrete]
(II) IS 3370 : 2021 Concrete structures for retaining aqueous
liquids – Code of practice