Tank Dimensions

Total area for the four floors = 4 (16 14 14) (13 14 13) = 7040 ft2 Number of residents, N 7040 8/1000 = 56 Daily water requirement 56 5 = 280 ft3 Required volume of UG water tank, V = 2 280 = 560 ft3

Depth of tank, H V1/3 = (560)1/3 = 8.24

Freeboard, F.B. = 0.50 Total depth, D = H + F.B. = 8.74

Assuming L/B = 1.5, Tank area A = LB 560/8.24 = 1.5B B B = 6.73, L = 10.10

Loads and Material Properties

Loads: FF = 10 psf, LL = 20 psf

Materials: fc = 3 ksi, ft,ult = 5fc = 5(3/1000) = 0.274 ksi, ft,all = ft,ult/2 = 0.137 ksi, fs = 20 ksi fc = 0.45fc = 1.35 ksi, k = 0.378, j = 0.874, R = 0.223 ksi, Rt = ft,all/6 = 0.023 ksi

For soil, Angle = 30 Ka = (1 sin )/(1 + sin ) = 0.333

Design Conditions

The UG water tank has three basic components; i.e., top slab, sidewalls and base slab The top slab will be designed as normal simply supported slab based on the self-weight and superimposed loads The design of sidewalls and the base slab will be based on assuming

(i) Tank full of water but no soil outside, (ii) No water inside tank but soil pressure from outside

The other more critical condition of no water inside but saturated soil outside is avoided here because it might cause instability of the tank itself. Alternately, a provision must be made that the tank cannot be evacuated when the soil is fully saturated.

Design of Top Slab

This is designed as a simply supported slab with clear spans

Sa = 6.73, Sb = 10.10, m = 6.73/10.10 = 0.67

Required slab thickness considering deflection = 2 (6.73 + 10.10)/15 = 2.24

Assuming minimum slab thickness t = 4, wTotal = 4 150/12 + 10 + 20 = 80 psf = 0.08 ksf

Ca = 0.072 Ma(+) = 0.072 0.08 (6.73)2 = 0.261 k/

and Cb = 0.014 Mb(+) = 0.014 0.08 (10.10)2 = 0.114 k/ d(req) = (M(max)/Rb) = (0.261/0.223) = 1.08 d = 3 or 2.5, OK

Asa(+) = Ma(+)/fsjd = 0.261 12/(20 0.874 3) = 0.06 in2/

Asb(+) = Mb(+)/fsjd = 0.114 12/(20 0.874 2.5) = 0.03 in2/ As(Temp) = 0.03 t = 0.03 4 = 0.12 in2/, S(max) = 2t = 8 Use #3 @ 8c/c in both directions

Design of Sidewalls

Since both L/H and B/H are within 0.5 and 2.0, both the sidewalls have both slab and cantilever action.

Cantilever action is within the bottom H/4 or 1 m height (whichever is greater).

H/4 = 8.24/4 = 2.06, while 1 m = 3.28 2.06

Slab Action:

Case (i) Tank full of water but no soil outside

In this case, pmax = w (Hh) = 0.0625 (8.243.28) = 0.310 ksf

Case (ii) No water inside tank but soil pressure from outside

In this case, pmax = Ka s (Dh) = 0.333 0.120 (8.743.28) = 0.219 ksf The following bending moments (k/) are obtained by analyzing the structure in GRASP

Cantilever Action:

Case (i) Tank full of water but no soil outside

In this case, pmax = wH= 0.0625 8.24 = 0.515 ksf The maximum bending moment Mmax = pmax h2/6 = 0.515 3.282/6 = 0.924 k/

Case (ii) No water inside tank but soil pressure from outside

In this case, pmax = Ka sD = 0.333 0.120 8.74 = 0.350 ksf The maximum bending moment Mmax = pmax h2/6 = 0.350 3.282/6 = 0.627 k/ t(req) = (M(max)/Rtb) = (2.049/0.023) = 9.47 t = 9.5; i.e., d = 8''

As = M/fsjd = M 12/(20 0.874 8) = M/11.65 in2/

As(Temp) = 0.03 t = 0.03 9.5 = 0.29 in2/, S(max) = 2t = 19 Since M(max) = 2.049 k/ As(max) = 2.049/11.65 = 0.18 in2/ As(Temp) governs in all cases; i.e., Use #3 @ 4.5 c/c on all surfaces in all directions

(A): # 3 @ 4.5 c/c, on both faces and in both directions of the slab

(B): # 3 @ 4.5 c/c, on both faces and in both directions of the slab

Design of Base Slab

Like the top slab, the base slab is also designed as a simply supported slab with clear spans

However, the slab has to carry end moments (0.924 k/ and 0.627 k/) from the cantilever action of the sidewalls; which results in reduction of the midspan maximum moments.

Sa = 6.73, Sb = 10.10, m = 6.73/10.10 = 0.67, t(req) = 2 (6.73 + 10.10)/15 = 2.24

However since the slab carries significant load from water, the assumed slab thickness t = 7

Case (i) Tank full of water but no soil outside

wTotal = ct + FF + wH = 150 7/12 + 10 + 62.5 8.24 = 612.66 psf = 0.613 ksf Assuming negative moment of 0.924 k/

Ca = 0.072 Ma(+) = 0.072 0.613 (6.73)2 0.924 = 1.074 k/

and Cb = 0.014 Mb(+) = 0.014 0.613 (10.10)2 0.924 = 0.050 k/

Case (ii) No water inside tank but soil pressure from outside

In this case, the total weight of top slab and sidewalls

= {(10.10 + 19/12) (6.73 + 19/12) 4/12 + 2 (10.10 + 6.73 + 19/12) 8.74 9.5/12} 0.15 = 43.08 k Net upward pressure from soil, wTotal = 43.08/{(10.10 + 19/12) (6.73 + 19/12)} = 0.444 ksf Assuming negative moment of 0.627 k/

Ma(+) = 0.072 0.444 (6.73)2 0.627 = 0.820 k/

Mb(+) = 0.014 0.444 (10.10)2 0.627 = 0.006 k/

t(req) = (M(max)/Rtb) = (1.074/0.023) = 6.86 t = 7, OK t = 7 d = 5.5As = M/fsjd = M 12/(20 0.874 5.5) = M/8.01 in2/

As(Temp) = 0.03 t = 0.03 7 = 0.21 in2/, S(max) = 2t = 14 Since M(max) = 1.074 k/ As(max) = 1.074/8.01 = 0.13 in2/ As(Temp) governs in all cases; i.e., since the sidewalls require #3 rods @ 4.5 c/c

Use #3 @ 4.5 c/c on all surfaces in all directions

Total area for the four floors = 4 (16 14 14) (13 14 13) = 7040 ft2 Number of residents, N 7040 8/1000 = 56 Daily water requirement 56 5 = 280 ft3 Required volume of UG water tank, V = 2 280 = 560 ft3

Depth of tank, H V1/3 = (560)1/3 = 8.24

Freeboard, F.B. = 0.50 Total depth, D = H + F.B. = 8.74

Assuming L/B = 1.5, Tank area A = LB 560/8.24 = 1.5B B B = 6.73, L = 10.10

Loads and Material Properties

Loads: FF = 10 psf, LL = 20 psf

Materials: fc = 3 ksi, ft,ult = 5fc = 5(3/1000) = 0.274 ksi, ft,all = ft,ult/2 = 0.137 ksi, fs = 20 ksi fc = 0.45fc = 1.35 ksi, k = 0.378, j = 0.874, R = 0.223 ksi, Rt = ft,all/6 = 0.023 ksi

For soil, Angle = 30 Ka = (1 sin )/(1 + sin ) = 0.333

Design Conditions

The UG water tank has three basic components; i.e., top slab, sidewalls and base slab The top slab will be designed as normal simply supported slab based on the self-weight and superimposed loads The design of sidewalls and the base slab will be based on assuming

(i) Tank full of water but no soil outside, (ii) No water inside tank but soil pressure from outside

The other more critical condition of no water inside but saturated soil outside is avoided here because it might cause instability of the tank itself. Alternately, a provision must be made that the tank cannot be evacuated when the soil is fully saturated.

Design of Top Slab

This is designed as a simply supported slab with clear spans

Sa = 6.73, Sb = 10.10, m = 6.73/10.10 = 0.67

Required slab thickness considering deflection = 2 (6.73 + 10.10)/15 = 2.24

Assuming minimum slab thickness t = 4, wTotal = 4 150/12 + 10 + 20 = 80 psf = 0.08 ksf

Ca = 0.072 Ma(+) = 0.072 0.08 (6.73)2 = 0.261 k/

and Cb = 0.014 Mb(+) = 0.014 0.08 (10.10)2 = 0.114 k/ d(req) = (M(max)/Rb) = (0.261/0.223) = 1.08 d = 3 or 2.5, OK

Asa(+) = Ma(+)/fsjd = 0.261 12/(20 0.874 3) = 0.06 in2/

Asb(+) = Mb(+)/fsjd = 0.114 12/(20 0.874 2.5) = 0.03 in2/ As(Temp) = 0.03 t = 0.03 4 = 0.12 in2/, S(max) = 2t = 8 Use #3 @ 8c/c in both directions

Design of Sidewalls

Since both L/H and B/H are within 0.5 and 2.0, both the sidewalls have both slab and cantilever action.

Cantilever action is within the bottom H/4 or 1 m height (whichever is greater).

H/4 = 8.24/4 = 2.06, while 1 m = 3.28 2.06

Slab Action:

Case (i) Tank full of water but no soil outside

In this case, pmax = w (Hh) = 0.0625 (8.243.28) = 0.310 ksf

Case (ii) No water inside tank but soil pressure from outside

In this case, pmax = Ka s (Dh) = 0.333 0.120 (8.743.28) = 0.219 ksf The following bending moments (k/) are obtained by analyzing the structure in GRASP

Cantilever Action:

Case (i) Tank full of water but no soil outside

In this case, pmax = wH= 0.0625 8.24 = 0.515 ksf The maximum bending moment Mmax = pmax h2/6 = 0.515 3.282/6 = 0.924 k/

Case (ii) No water inside tank but soil pressure from outside

In this case, pmax = Ka sD = 0.333 0.120 8.74 = 0.350 ksf The maximum bending moment Mmax = pmax h2/6 = 0.350 3.282/6 = 0.627 k/ t(req) = (M(max)/Rtb) = (2.049/0.023) = 9.47 t = 9.5; i.e., d = 8''

As = M/fsjd = M 12/(20 0.874 8) = M/11.65 in2/

As(Temp) = 0.03 t = 0.03 9.5 = 0.29 in2/, S(max) = 2t = 19 Since M(max) = 2.049 k/ As(max) = 2.049/11.65 = 0.18 in2/ As(Temp) governs in all cases; i.e., Use #3 @ 4.5 c/c on all surfaces in all directions

(A): # 3 @ 4.5 c/c, on both faces and in both directions of the slab

(B): # 3 @ 4.5 c/c, on both faces and in both directions of the slab

Design of Base Slab

Like the top slab, the base slab is also designed as a simply supported slab with clear spans

However, the slab has to carry end moments (0.924 k/ and 0.627 k/) from the cantilever action of the sidewalls; which results in reduction of the midspan maximum moments.

Sa = 6.73, Sb = 10.10, m = 6.73/10.10 = 0.67, t(req) = 2 (6.73 + 10.10)/15 = 2.24

However since the slab carries significant load from water, the assumed slab thickness t = 7

Case (i) Tank full of water but no soil outside

wTotal = ct + FF + wH = 150 7/12 + 10 + 62.5 8.24 = 612.66 psf = 0.613 ksf Assuming negative moment of 0.924 k/

Ca = 0.072 Ma(+) = 0.072 0.613 (6.73)2 0.924 = 1.074 k/

and Cb = 0.014 Mb(+) = 0.014 0.613 (10.10)2 0.924 = 0.050 k/

Case (ii) No water inside tank but soil pressure from outside

In this case, the total weight of top slab and sidewalls

= {(10.10 + 19/12) (6.73 + 19/12) 4/12 + 2 (10.10 + 6.73 + 19/12) 8.74 9.5/12} 0.15 = 43.08 k Net upward pressure from soil, wTotal = 43.08/{(10.10 + 19/12) (6.73 + 19/12)} = 0.444 ksf Assuming negative moment of 0.627 k/

Ma(+) = 0.072 0.444 (6.73)2 0.627 = 0.820 k/

Mb(+) = 0.014 0.444 (10.10)2 0.627 = 0.006 k/

t(req) = (M(max)/Rtb) = (1.074/0.023) = 6.86 t = 7, OK t = 7 d = 5.5As = M/fsjd = M 12/(20 0.874 5.5) = M/8.01 in2/

As(Temp) = 0.03 t = 0.03 7 = 0.21 in2/, S(max) = 2t = 14 Since M(max) = 1.074 k/ As(max) = 1.074/8.01 = 0.13 in2/ As(Temp) governs in all cases; i.e., since the sidewalls require #3 rods @ 4.5 c/c

Use #3 @ 4.5 c/c on all surfaces in all directions

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