كتاب تصميم الخزانات by Mebuild

Reinforced Concrete IV Second Semester Structural Engineering Department Faculty of Engineering Tanta University

General Notes and Solved Examples

Eng \ Just for help

First Edition

Instructor: eng \ just for help

2000-2001

Subject: Water Tanks Solved Examples

1. Fundamentals

Instructor: eng \ just for help

Subject: Water Tanks Solved Examples

2000-2001

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

Statical system in conjunction with the loads

Vertical Sec.

For H = 4 m Top horizontal beam should be used. In this case, the wall will be considered as two way slab.

Vertical strip

Horizontal strip at the sec. of max. horizontal load

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2000-2001

Subject: Water Tanks Solved Examples

Fig. 2

Statical system for vl strip (Box. Sec)

Tie

Fig. 3

Structural system for vl. strip

* If the tie is removed

Statical system for vertical strip

Fig. 4

H=2m The walls of the tank can be considered as cantilever walls

Statical system for vl. Strip

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Subject: Water Tanks Solved Examples

2000-2001

*H=4m Top horizontal beam should be used.

vl. Strip

Fig. 5

Treated as cantilever retaining wall

H Central part L

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

Horizontal Strip

vl. Strip in long dir.

vl. Strip in short dir.

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 -

12-

case of loading Case of water load (No external lateral load) Empty tank and consider lateral load.

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Subject: Water Tanks Solved Examples

2000-2001

•

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

Case 1

Empty

Case 2

• 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) a.s

a.s a.s

w.s w.s

a.s w.s

w.s w.s

- The sections that labeled by (w.s) are water side sections. - The sections that labeled by (a.s) are air side sections.

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Subject: Water Tanks Solved Examples

2000-2001

(3) sec. 1

Mu = 0.0 Nu = +150 kN

Water side section

Tu 15 = = 100 kN 1.5 1.5 - t = 0.6 T = 0.6 * 10 = 6 cm < 20 cm take t = 20 cm

10 → β cr = 0.85

- use

Tu 150 * 1.15 * 1000 = = 563.7 mm 2 = 5.637 cm 2 β cr f y / γ s 0.85 * 360

As =

A s /side = 5.64/2 = 2.82 cm 2 use 5

10 /m` / side

- Check of tensile stress for concrete. f t > f cto = (f ctr /η ) , η = 1.7 ft =

T 150 * 1000 = = 0.75 MPa A c 200 * 1000

f ctr =

0.6 f cu

f t < f ctr

Sec. 2

0.6 300 = 1.93 MPa 1.7 η (o.k) =

Mu =100 kN.m

, Nu = 0.0

(water side sec.)

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Subject: Water Tanks Solved Examples

2000-2001 Mu M * 1000 = 66.7, t = = 1.5 3

∴M =

6.67 * 1000 = 47.2 cm 3 ≅ 50 cm

ft > fct where f t =

M 6M = Z bt 2

66.7 * 10 6 1000 * 500

= 1.6 MPa

t v = t[1 +

fct (N ) ] = t = 50 cm f c t (M ) → η = 1.65

0.6 f cu

∴fct =

→ ft < fct

0.6 30 = 1.99 1.65 o.k =

M u = 100 kN.m & t = 500 mm → d = 460 mm d = c1

Mu f cu b

100 * 10 6 → c1 = 7.97 30 * 1000 ∴ J = 0.826 Mu → As = β cr f y dJ 460 = c1

- use

10 → β cr = 0.85

100 * 10 6 = 860 mm 2 = 8.6 cm 2 0.85 * 360 * 460 * 0.826 use 11 10 /m' - use 16 → β cr = 0.75 As =

100 * 10 6 = 975 mm 2 = 9.75 cm 2 0.75 * 360 * 460 * 0.826 use 5 16 /m' or 9 12 /m' A's = 5 10 /m' As =

t = 50 M = 500 mm

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2000-2001 Sec. 3

Subject: Water Tanks Solved Examples

Mu = 100 kN.m & Tu = 10 kN

100 10 = 66.7 kN .m & T = = 6.7 kN 1.5 1.5 t = 50 M + 30 = 50 66.7 + 30 = 438 mm take t = 500 mm f t = f c t ( N ) + f c t (M) M =

fc t (N ) =

T 6.7 * 1000 = = 0.0134 N/mm 2 A c 1000 * 500

6M 6 * 66.7 * 10 6 = = 1.6 N/mm 2 2 2 bt 1000 * 500 f t = 1.6 + 0.0134 = 1.6134 N/mm 2 f c t (M ) =

fct r

fct =

t v = t[1 +

, f c t r = 0.6 f cu = 3.28 N/mm 2 fc t (N ) 0.0134 ] = 504.18 mm 2 = 500[1 + f c t (M ) 16 ∴η ≅ 1.65

3.28 = 1.98 N/mm 2 > f t o.k 1.65 M 100 = 10 m, outside sec. (big ecc.) e= u = Tu 10 fct =

t + cov er 2 = 10 - 0.25 + 0.04 = 9.79 m M us = Tu e s = 10 * 9.79 = 97.9 kN.m

es = e -

d = c1

use

As =

M us f cu b

max

∴ 460 = c1

97.9 * 10 6 → c1 = 8.05 30 * 1000

∴ J = 0.826 = 16 m m → β cr = 0.75

Tu M us + β cr f y dJ β cr f y / γ s

97.9 * 10 6 6.7 * 10 3 * 1.15 + = 983 mm 2 0.75 * 360 * 0.826 0.75 * 360 use 5 16 /m' A's = 5 10 /m' =

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2000-2001 Sec. 4

Subject: Water Tanks Solved Examples

Mu = 10 kN.m & Tu =150 kN

10 150 = 6.7 kN.m & T = = 100 kN 1.5 1.5 t = 50 M + 30 M =

= 50 6.7 + 30 = 159.4 mm 0.6 T = 0.6 * 100 = 60mm ∴ use t = 200 mm f t = f ct ( N ) + f ct (M) > f c t r /η fct (N ) =

T 100 * 1000 = = 0.5 N/mm 2 A c 1000 * 200

6M 6 * 6.7 * 10 6 = = 1.005 N/mm 2 2 2 bt 1000 * 200  f (N )  0.5   t v = t 1 + c t = 299.5 mm  = 200 1 + f c t (M )   1.005   ∴η ≅ 1.45 , f c t r = 3.28 N/mm 2

f c t (M ) =

f t = 0.5 + 1.005 = 1.505 N/mm 2 , f c t = ∴ ft < fct

3.28 = 2.26 N/mm 2 1.45

o.k

M u = 10 kN.m & Tu = 150 kN e=

Mu 10 t = = 0.067 m < - cover Tu 150 2

Tu M u 150 10 + = + = 137.5 kN 2 d - d' 2 0.16 T2 = T - T1 = 150 - 137.5 = 12.5 kN T1 =

→ use

, β cr = 0.85

T1 137.5 * 1.15 As = = = 517 mm 2 0.85 * 360 β cr f y / γ s choose 7 A's =

10 /m'

T2 12.5 * 1.15 = = 47 mm 2 β cr f y / γ s 0.85 * 360 use 5

10/m'

As` T+As

T2 T T1

Instructor: eng \ just for help

:2000-2001

Subject: Water Tanks Solved Examples

2. Rectangular tanks with sliding Base (I)

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Subject: Water Tanks Solved Examples

:2000-2001

2. Rectangular tanks with sliding Base (I) * Consider 1.0 m Strip at lower end of wall. b

a 40 kN/m` 40 kN/m`

40 kN/m`

3.0 m

4.0 m γh = 40 kN/m

40 kN/m` d

5.0 m

* Applying 3- moment Eqn to calculate the bending moment: apply 3- moment Eqn at (a) Md (3) + 2 Ma (3+5) + Mb(5) = -6Rel Re1 = -(20.83 +4.50) 253.3 Ma =Mb= Md R1 = 45

3Ma +16Ma + 5Ma = 6 * 253.3 24 Ma = 1519.8 ∴Ma = 630 m.t

4 t/m`

100

R2 =208.3

w 3 24

100

(2) 63 +

(3)

(1)

100 60

(3)

(2)

100 60 62

N. F. D

B. M. D

Instructor: eng \ just for help

Subject: Water Tanks Solved Examples

:2000-2001

Calculation of moment by using method of moment distribution: a

d 4 t/m`

Joint

member

R. S D. F F. E. M 0B

ad 0.33 0.62 30 33

b ab 0.2 0.38 83.3 20.3

-63

ba 0.2 0.38 +83.3

bc 0.33 0.62 -30

10.15 -30.4 63

16.5 -49.6 -63

cb 0.33 0.62 +30 33

d cd 0.2 0.38 -83.3 20.3

-63

dc 0.2 0.38 +83.3

Da 0.33 0.62 -30

10.15 -30.4 +63

16.5 -49.6 -63

* Design of Section (1): t = 50 M + 30 = 50 69.1 + 30 = 445.6 take t = 500 mm A = 1000 * 500 = 5 * 10 5 = 500 mm 1000 * 500 3 = 1.042 * 1010 mm 4 I= 12

500 mm 1000 mm

100 * 10 3 = 0.2 5 * 10 5 63 * 10 6 f ct (M) = * 250 = 1.51 N/mm 2 10 1.042 * 10 f (N ) t v = t(1 + ct ) = 566.22 mm f ct ( M ) f ct (N) =

f ctr

0.6 25 = 1.81 N/mm 2 1.65 η f f ct (N) + f ct (M) = 1.71 < ctr ok

∴

M = 69.1 kN.m Mu = 94.5 kN.m

T = 100 kN Tu= 150 kN.m

, Water- side section

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:2000-2001

Subject: Water Tanks Solved Examples

Mu = 63.0cm > t / 2 Tu

t + cov er = 0.45m 2 = T .e s = 67.5 kN.m

es = e − M us

M us f cu B

d = c1

67.5 * 10 6 430 = c1 25 * 10 ∴ c1 = 8.28 j = 0.826 As =

M us Tu 67.5 * 10 6 15 * 10 3 + = + = 1130 mm 2 β cr f y jd β cr f y / γ s 0.85 * 360 * 0.826 * 430 0.85 * 360 / 1.15 = 11.3 cm 2

12 /m` use 10 Sec. Steel = 6 10 /m`

* Design of section (2) M u = 93 kN.m , Tu = 90 kN t = 30 cm e=

;

Air - side section

Mu t = 1.03 m > 2 Tu t + cover = 0.953 2 = 85.8 kN.m

es = e M us

d = C1

M us f cu B

∴ C1 = 3.93 As =

230 = C1

85.8 * 10 6 250 * 1000

j = 0.802

M us Tu 85.8 * 10 6 90 * 10 3 + = + f y jd f y / γ s 360 * 0.802 * 230 360 / 1.15 = 20.85 cm 2 use 10

16 /m`

Instructor: eng \ just for help

Subject: Water Tanks Solved Examples

:2000-2001

Design of Section (3) M u = 27 kN.m

Tu = 150 kN.m

t = 300 mm , Water - side section

A = 1000 * 300 = 3 * 10 5 mm 2 1000 * 300 3 = 2.25 * 10 9 mm 4 12 100 * 10 3 = 0.33 N/mm 2 f ct ( N ) = 5 3 * 10 18 * 10 6 * 150 = 1.2 f ct ( M ) = 2.25 * 10 9 f (N ) ) = 382.5 mm ∴η = 1.5 t v = t (1 + ct f ct ( M ) I=

f ctr /η = 2 N/mm 2 f ct ( N ) + f ct ( M ) = 1.53 <

f ctr

M t = 0.18 m > N 2 t e s = e - + cover = 0.10 m 2 M us = 15 kN.m

d = C1

M us f cu b

230 = C1 As =

15 * 10 6 25 * 1000

∴ C1 = 9.39

M us Tu + β cr f y jd β cr f y / γ s

= 8.22 cm 2 use 10 12 /m`

j = 0.826

Instructor: eng \ just for help

:2000-2001

Subject: Water Tanks Solved Examples

3. Rectangular tanks with Sliding Base (II)

Instructor: eng \ just for help

:2000-2001

Subject: Water Tanks Solved Examples

•

Prob. No. (1) [Elevated tank with Sliding Base] Take thickness of base = 20 cm

•

Loads w kN/m2 = weight of covering + weight of water + weight of walls + o.w of base Weight of covering = 4 kN/m2 Weight of water = γ h = 40 kN/m2 Weight of walls : assume thickness of wall = 30 cm = 0.3*25.0*4* (5*2+4.40*2) = 564 kN Weight of walls /m2 = 56.40/(5*5) = 2.256 t/m2 Weight of base = 0.2 *25 +0.5 = 5.5 kN/m2 wtot t/m2 = 4+ 40 +22.56 +5.5 = 72.06 kN/m2 36.03 kN/m`

5.0 r= = 1.0 5.0 α = 0.5 β = 0.5

5.0 m 36.03 kN/m`

w α = w β = 36.03 kN/m 2 5.0 m

•

Consider 1.0 m width: M u = 169 kN/m` d = C1

(Air - side section)

36.03 kN/m`

Mu f cu B

5.0 m +

169 * 10 6 25 * 1000 ∴ C1 = 1.95 C1 < C1 min

(1)

112.6 kN.m

160 = C1

take t = 300 mm ∴ C1 = 3.16 j = 0.757 As =

Mu 169 * 10 6 = = 2385 mm 2 / m = 23.85 cm 2 / m f y jd 360 * 0.757 * 260

use 10

19 /m`

in both directions bottom reinforcement

Use top reinforcement = 0.2 * 23.85 = 4.77 cm 2 /m → Choose 7

Design of beams: assume beam 40 * 80 cm O.w = 0.4 (0.8-0.30) *2.50= 0.50 t/m` wslab= 7.206 t/m2 Equivalent load for shear ws= 0.50 + 7.206 *2.50*0.5 =9.508 t/m` Qu max = 35.655 t

10 /m`

5.0 m ws = 95.08 kN/m` wb = 125.7 kN/m`

5.0

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Subject: Water Tanks Solved Examples

:2000-2001 Equivalent load for flexure (bending) wb= 0.50+7.206*2.50*0.67= 12.57 t/m` Mu max = 58.92 m.t d = C1

Mu f cu B 6 t s + b = 220 cm

B = the least of

∴ 75 = C1 As =

L2 + b = 90 cm (controls) 10 1 C.L → C.L = 250 cm 2

58.92 * 10 5 250 * 90

Mu = 26.58 cm 2 f y jd

∴ C1 = 4.63 use 7

j = 0.821 22

As '= 3

check of shear Qu 35.655 * 10 3 qu = = = 11.885 kg / cm 2 > q u c = 9.68 kg/cm 2 unsafe 40 * 75 bd use 7 φ 8/m` 4 branches q u act = q u s + q u c /2 = 15.79 > 11.885 kg/cm 2

∴ O.K.

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:2000-2001

Prob. No. (2) •

• • •

Subject: Water Tanks Solved Examples

[Resting on soil tank with sliding base]

Check of bearing capacity: Assume t = 250 mm ∑ vl load = wall load + base load + weight of water (assume t wall = 300 mm) Weight of walls = 0.3*25*3 (2*4+2*3.40) =33.30 t w t/m2= 333 / (4*4) = 20.8 kN/m2 Wall load = 20.8 kN/m2 Base load = 0.25 *25 = 6.25 t/m2 +0.5 (flooring) = 6.75 kN/m2 Weight of water = γh = 10 *3 = 30 kN/m2 Total load = 57.55 kN/m2 < 100 kN/m2 ok Loads w kN/m2= wt of walls (assume t wall = 300 mm) w kN/m`= 333 / (4*4) = 20.8 r = 1.0 α = β = 0.5 Wα = Wβ = 10.4 kN/m2

10.4 kN/m`

Consider 1.0 m width Mu max = 23.4 kN.m (Water-side section)

0.25

3.50 m

Water side 0.25

t = 50 M = 50 15.6 = 197.4 mm take t = 250 mm

10.4 kN/m` 15.6 kN/m` -

A = 1000 * 250 = 25000 mm 2 bt 2 1000 * 250 2 z= = = 10.42 * 10 6 6 6 f ct (N) = 0 M 15.6 * 10 6 = = 1.49 N/mm 2 Z 10.42 * 10 6 t v = 250 mm f ctr /η = 2.2

f ct (M) =

f ct (M) = 1.498 <

f ctr

0.25

3.50 m

+ 0.3

250 1000

Instructor: eng \ just for help

Subject: Water Tanks Solved Examples

:2000-2001

d = C1

Mu f cu B

23.4 * 10 6 200 = C1 25 * 1000 â&#x2C6;´ C1 = 6.54 j = 0.826 As = use 7

Mu 23.4 * 10 6 = = 463 mm 2 Î˛ cr f y jd 0.85 * 360 * 200 10 /m`

(Top reinforcement)

Use bottom reinforcement 5

10 /m`

Note that in this case, walls of the tank act as supporting beam with adequate capacity.

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:2000-2001

Subject: Water Tanks Solved Examples

4. Elevated Conduits

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:2000-2001

(1) (i)

0.3 m

1m 7m

Sec. Elevation x-x

30 5m

6 10 /m` (U) 9 10 /m` (L)

6 10 /m` (U) 6 10 /m` (L)

Plan

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Subject: Water Tanks Solved Examples

:2000-2001

(ii) Acting load assume ts = 30 cm - Walls: the wall will be considered as one way slab in vertical direction.

wβ

p= γ h = 1 * 17 = 17 kN/m2

wα

- Floor: w = γ ts + γ wh = 25 * 0.3 + 1 * 17 = 24.5 kN/m 2 4 5 * 0.76 r= = 1.25 4 * 0.76 r4 = 0.7 → β = 1 - α = 0.3 (Grashoff ' s coefficients) α= 4 r +1 w α = 0.7 * 24.5 = 17.15 kN/m 2 , w β = 0.3 * 24.5 = 7.4 kN/m 2

Straining Actions * Vertical strip 17.2 kN/m` 17 kN/m`

17 kN/m`

8.2 + 14.45

8.2 0.82

0.82

14.45

26.2

N. F. D

B. M. D

* Section in Floor 7.4

kN/m`

5m 18.5

18.5

B. M. D

Instructor: eng \ just for help

Subject: Water Tanks Solved Examples

:2000-2001

Design of critical sections Consider fcu = 25 kN/cm2 St. 360/520

(1) (3) (4)

Sec. (1) water side section M = 8.2 kN.m t = 50 M = 50 8.2 = 143.1 mm take t s = 300 mm

f ct = f ct (M) >

f ctr

6M 6 * 8.2 * 10 6 where, f ct ( M ) = 2 = = 0.546 N/mm 2 2 bt 1000 * 300 f ctr = 0.6 f cu = 0.6 25 = 3 N/mm 2 t v = t [1 + ∴

f ctr

f ct ( N ) ] = 300 → η = 1.45 f ct ( M )

29.76 = 2.06 N/mm 2 > f ct 1.45

(o.k)

M u = 1.5 * 8.2 = 12.3 kN.m t = 300 mm → d = 260 mm 12.3 * 10 6 → c1 = 11.7 25 * 1000 → J = 0.826 ∴ use φ max = 10mm → β cr = 0.85 d = c1

As =

Mu 12.3 * 10 6 = = 187 mm 2 = 1.87 cm 2 β cr f y dJ 360 * 260 * 0.826

choose 6 10 /m` A`s = 6 10 /m`

(2)

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Subject: Water Tanks Solved Examples

:2000-2001

Sec. (2) M = 8.2 kN.m & T = 14.45 kN (water side section) t = 50 M + 30 = 173 mm take t s = 300 mm f ct = f ct ( N ) + f ct (M) > f ct ( N ) =

f ctr

T 14.45 * 1000 = = 0.048 kN/mm 2 Ac 300 * 1000

6M 6 * 8.2 * 10 6 f ct (M) = 2 = = 0.546 kN/mm 2 2 bt 1000 * 300 ∴ f ct = 0.048 + 0.546 = 0.594 kN/mm 2 → t v = t[1 +

f ct ( N ) 0.048 ] = 300[1 + ] = 326.3 mm 0.547 f ct ( M )

∴η = 1.45 , f ctr = 3 ∴ f ct <

f ctr

o.k

kN/mm 2 ∴

f ctr

= 2.06

Instructor: eng \ just for help

Subject: Water Tanks Solved Examples

:2000-2001

M u = 1.5 * 8.2 = 12.3 kN.m & Tu = 1.5 * 14.45 = 21.7 kN M u 12.3 t = = 0.57 m > Tu 21.7 2

(big ecc.)

t + cover = 0.57 - 0.15 + 0.04 = 0.46 m 2 = Tu e s = 21.7 * 0.46 = 10 kN.m

es = e M us

M us f cu b

d = c1

∴ 260 = c1

10 * 10 6 → c1 = 13 25 * 1000 J = 0.826

use φ max = 10 mm

∴ β cr = 0.85

M us Tu + β cr f y dJ β cr f y / γ s

As =

10 * 10 6 21.7 * 1000 * 1.15 + 0.85 * 360 * 260 * 0.826 0.85 * 360 = 234 mm choose 6 10 /m` A`s = 0.2 A s use 6 10 /m` =

Sec. (3) M =26.2 kN.m t = 300 mm

& T = 14.45 kN (air side section) & d = 260 mm

M u = 1.5 * 26.2 = 39.3 kN.m & Tu = 1.5 * 14.45 = 2.17 kN M u 39.3 = = 1.81 m (big ecc.) Tu 21.7

t + cover = 1.81 - 0.15 + 0.04 = 1.7 m 2 = Tu e s = 21.7 * 1.7 = 36.9 kN.m

es = e M us

d = c1

M us f cu b

∴ 260 = c1

36.9 * 10 6 → c1 = 6.77 25 * 1000 J = 0.826

use φ max = 10 mm As =

M us Tu + f y dJ f y / γ s

36.9 * 10 6 21.7 * 1000 + = 200 mm 2 360 * 260 * 0.826 360/1.15 choose 6 10 /m` A`s = 6 10 /m` =

Instructor: eng \ just for help

Subject: Water Tanks Solved Examples

:2000-2001

Sec. (4) M = 18.5 kN.m

(water side section)

t = 50 M = 50 18.5 = 215 mm take t = 300 mm f f ct = f ct (M) > ctr

6M 6 * 18.5 * 10 6 = = 1.23 N/mm 2 2 2 bt 1000 * 300 f (N ) t v = t[1 + ct ] = t = 300 mm → η = 1.45 f ct ( M )

f ct (M) =

f ctr = 0.6 f cu = 3 N/mm 2 → ∴ f ct <

d = c1

f ctr

M us f cu b

f ctr

= 2.06 N/mm 2

o.k

∴ 26 = c1

18.5 * 10 6 * 1.5 → c1 = 7.8 25 * 1000 J = 0.826

use φ max = 10 mm As =

∴ β cr = 0.85

M us 1.5 * 18.5 * 10 6 = = 422.3 mm 2 β cr f y dJ 0.85 * 360 * 260 * 0.826

choose 6 10 /m` A`s = 6 10 /m`

Design of wall as beam o.w= γ bt = 25 *0.3 *2 = 15 kN/m` wf = 24.5 kN/m2 2m 5m

(r =1.25 , α = 0.79 , β = 0.6)

Instructor: eng \ just for help

Subject: Water Tanks Solved Examples

:2000-2001

Load for shear: w = o.w + β wf x = 1.5+0.6*24.5*2 = 44.4 kN/m` Load for moment w = o.w + α wf x = 1.5+0.79*2.45*2 = 5.37 t/m` Leff= C.L-C.L= 5 m = 1.05 ln= 1.05 *4.7 = 4.94 m → ∴t/Leff = 200/494 = 0.405 > 0.4 (deep beam) wl 2 53.7 * 5 2 wl 2 53.7 * 5 2 = = 112 kN/m , M − ve = = = 134 kN/m` 12 12 10 10 For - ve moment M u = 1.5 * 134 = 201 kN.m , d = 1900 mm

M + ve =

d = c1

Mu f cu b

,1900 = c1

201 * 10 6 → c1 = 11.6 25 * 300 J = 0.826

As =

Mu 201 * 10 6 = = 360 mm 2 f y dJ 360 * 1900 * 0.826

use 3

16 for - ve &+ ve moment /m`

* Shear capacity has to be checked according to code provisions for deep beams β=0.5 α= 1/3

Max. reaction of beam R = 1.1 wl = 1.1 *44.4 *5 = 244 kN

2 a

For beam Sec. of beam = 300 * 1000 o.w = 25 *0.3*1= 8 kN/m` Load for moment Load for shear

Mu f cu b

,95 = c1

123 * 10 6 → c1 = 7.4 25 * 300 J = 0.826

As =

Mu 123 * 10 6 = = 440 mm 2 f y dJ 360 * 950 * 0.826

use 3

w = o.w +α w x = 8+2*1/3 *24.5*2= 41 kN/m` w = o.w +β w x = 8+2*1/2 *24.5*2= 57 kN/m`

M = 41 * 2 2 / 2 = 82 kN.m M u = 1.5 M = 123 kN.m

d = c1

16 , A`s = 2

Instructor: eng \ just for help

:2000-2001

Subject: Water Tanks Solved Examples

(iii) loads on columns w

vl. Load = 2 [244 + 57 *2]=716 kN N = 716 *1.1 = 788 kN

P 1m

Wwind = 716 *5 = 5 kN/m` P= 5 * 2 = 10 kN M = 10*10 = 100 kN.m 8m

Design of sec. N = 788 kN Nu = 1.5 *788 = 1182 kN Design for N only Nu = 0.35 Ac fcu +0.67 fy Asc Nu = Ac [0.35 fcu + 0.67 fy Âľ] 1182*1000= Ac [0.35*25+0.67*360*1/100] Ac = 1059 cm2 Take column 300 *400 mm To sustain the effect of buckling, take column 300 * 1000 He= k Ho = 2.2 *8 = 17.6 m Consider unbraced column

Instructor: eng \ just for help

:2000-2001

Subject: Water Tanks Solved Examples

H e 17.6 o.k = = 17.6 < λ max 1 t λ > 10 ∴ (long column)

λ=

λ2 t

17.6 * 1000 δ= = = 155 mm 2000 2000 155 = 12.2 m.t ∴ M add = Pδ = 788 * 1000 M design = M 2 + M add , where M 2 is due to wind load = 100 + 122 = 222 kN.m N u = 1182 kN , M u = 1.5 * 222 = 333 kN.m e=

Mu 333 = = 0.3 m N u 1182

e 0.3 = = 0.3 > 0.05 t 1

o.k

use interaction diagram for f y = 3600, α = 1 , ξ = 0.8 k=

Nu 1182 * 1000 = = 0.16 f cu bt 25 * 300 * 1000

Mu e 333 * 10 6 = = = 0.04 t f cu bt 2 25 * 300 * 1000 2 ρ = 0.0 0.4 use min steel = bt /side 100 0.4 = * 300 * 1000 = 1200 mm 2 100 choose 6 16 k

1000

300

Instructor: eng \ just for help

Subject: Water Tanks Solved Examples

:2000-2001

10/m`

16/m`

3 6

10/m`

16/m`

10/m`

12 2

10/m`

12 2

16 6

Cross sec. Of column

30 6 100

10/m` 6

10/m`

Instructor: eng \ just for help

Subject: Water Tanks Solved Examples

:2000-2001

Prob. (2) 40 4m 2m 60o 4m 4m x

Sec. Elevation x-x

Plan Loads The floor and wall will be considered as way slab in the plane of the paper. Assume t = 40 cm Floor wf = γ ts + γw h = 25 *0.4 + 10*2 = 30 kN/m2 Wall ww = γ ts = 25 *0.4 = 10 kN/m2 P = γw h = 20 kN/m2 Straining Actions L

4m 2.3 m 60o 1.16

31.2 _ 25.3 kN

_ + 20

25.3 kN

31.2 31.2

31.2 + 28.8

Instructor: eng \ just for help

Subject: Water Tanks Solved Examples

:2000-2001

B. M. D

N. F. D 23.1 kN T= 23.1 kN sin 60 =20 kN 60o

60o

23.1 kN water pressure 29.2 sin 60 25.3 kN

o.w of wall + water weigh = 1*23.1 +00.5*0.76*16 =29.2 kN

Design of critical sections Sec. (1) M = 31.2 kN.m

, T = 20 kN (water side section)

t = 50 M + 30 = 50 31.2 + 30 = 309 mm take t = 400 mm f f ct = f ct ( N) + f ct (M) > ctr

f ct (N) =

T 20 * 1000 = = 0.5 N/mm 2 Ac 100 * 40

6M 6 * 312 * 10 6 f ct ( M ) = 2 = = 1.17 N/mm 2 2 bt 1000 * 400 f ct = 0.5 + 1.17 = 1.67 N/mm 2

t v = t[1 +

f ct ( N ) 0.5 ] = 570 mm , η = 1.65 ] = 400[1 + f ct ( M ) 1.17

f ctr = 0.6 f cu = 3 N/mm 2 → ∴ f ct >

f ctr

= 1.82 N/mm 2

o.k

M u = 1.5 * 31.2 = 46.8 kN.m & Tu = 1.5 * 20 = 30 kN e=

M u 46.8 t = = 1.56m > Tu 30 2

(big ecc.)

t + cover 2 = 1.56 - 0.2 + 0.04 = 1.40 m M us = Tu e s = 30 * 1.4 = 42 kN.m

es = e -

(3) (2)

(1)

Instructor: eng \ just for help

Subject: Water Tanks Solved Examples

:2000-2001

d = c1

Mu f cu b

, 360 = c1

42 * 10 6 → c1 = 8.78 25 * 1000 J = 0.826

use φ max = 10 mm As =

∴ β cr = 0.85

M us Tu + β cr f y dJ β cr f y / γ s

42 * 10 6 30 * 1000 = + = 5.74 mm 2 0.85 * 360 * 360 * 0.826 0.85 * 360 / 1.15 choose 8 10 /m` A`s = 6 10 /m`

Sec. (2) M = 288 kN.m & T= 20 kN (air side section) M u = 1.5 * 288 = 43.2 m.t & Tu = 1.5 * 20 = 30 kN e=

M u 43.2 t = = 43.2 kN.m > Tu 30 2

(big ecc.)

t + cover = 1.44 - 0.2 + 0.04 = 1.28 m 2 = Tu e s = 30 * 1.28 = 38.4 kN.m

es = e M us

d = c1

M us f cu b

, 360 = c1

38.4 * 10 6 → c1 = 9.19 25 * 1000 J = 0.826

use φ max = 10 mm As =

∴ β cr = 0.85

M us Tu + f y dJ f y / γ s

38.4 * 10 6 30 * 1000 + = 455 mm 2 360 * 360 * 0.826 360 / 1.15 choose 6 10 /m` A`s = 6 10 /m` =

Instructor: eng \ just for help

Subject: Water Tanks Solved Examples

:2000-2001

Sec. (3) M = 31.2 kN.m

, N = - 25.3 kN (water side section)

Big. Ecc. compression t = 50 M − 30 ≅ 249 mm take t = 400 mm f ct = f ct (N) + f ct (M) - 25.3 * 1000 = −0.063 kN/mm 2 1000 * 400 6M 6 * 31.2 * 10 6 = 1.17 N/mm 2 f ct ( M ) = 2 = 2 bt 1000 * 400

f ct (N) =

f ct = −0.063 + 1.17 = 1.107 N/mm 2 t v = t[1 +

f ct ( N ) 0.063 ] = 400[1 − ] = 421 mm , η = 1.6 f ct ( M ) 1.17

f ctr = 0.6 f cu = 3 N/mm 2 ∴

f ctr

= 1.86 N/mm 2 > f ct

(o.k)

M u = 1.5 * 31.2 = 46.8 kN.m N u = -1.5 * 2.53 = -38 kN ∴ Use φ max = 10 mm

∴ β cr = 0.85

∴ β cr f y = 0.85 * 360 = 306 N/mm 2 ⇒ Use strength interaction diagram for f y = 2800 kg/cm 2 ,

ξ = 0.8 , α = 1 ⇒ A s = A`s = 8

10 /m` (same as for sec.1)

* Design of wall as deep beam Water weight

* Loads in the plan of the wall o.w of wall = γ b t sin θ θ =60o = 25 * 0.3 *2.31 sin 60 = 15 kN/m` - wt of water = γw A sin θ = 10 * ½ *1.16 *2 sin 60 = 10 kN/m` - Load from floor = wf *2 = 30* 2 = 60 kN/m` ∴wt = 15 +10 +60 = 85 kN/m` 85 kN/m` 4m

A Own weight

Instructor: eng \ just for help

:2000-2001

+ ve

M max

− ve M max

Subject: Water Tanks Solved Examples

wl 2 85 * 4 2 = = = 113.3 kN/m 12 12 wl 2 85 * 4 2 = = = 136 kN/m 10 10

Check of deep beam condition Le ff = C.L − C.L = 4m = 1.05 L n = 1.05 * 3.7 = 3.89 m t L e ff

2.31 = 0.59 > 0.4 Deep beam 3.89

for - ve moment d = c1

t = 2310 mm → d = 2200 mm

Mu f cu b

1.5 * 136 * 10 6 2200 = c1 25 * 300 C1 = 13.34 → J = 0.826 As =

Mu 1.5 * 136 * 10 6 = = 312 mm 2 f y dJ 360 * 2200 * 0.826

1.3 * 312 = 406 mm 2 Smaller 11 A s min = greater * 300 * 2200 = 2017 mm 2 360 0.15/100 * 30 * 2200 = 990 mm 2 ∴ use A s = 5 16 for both negative and positive momnts • Shear capacity has to be checked according to code provisions

Instructor: eng \ just for help

:2000-2001

Subject: Water Tanks Solved Examples

(iii) Design of column Load on each column concrete 1 6.32 + 4 5.52 + 3.6 * 4[( )*2 − ( ) * 1.6] * 25 2 2 2 water 1 5.52 + 3.6 + * 4[( ) * 1.6] * 10 = 295.9 2 2 due to o.w of column → Pcolumn = 1.1 * 295.9 = 325.2 kN

Wind load P = 1 * 2 * 4 = 8 kN M = 8 * 8 = 64 kN ∴M/column = 64/2 = 32 kN.m

Design for N only Nu = 1.5 * 325.5 = 488.3 kN Nu = [0.35 fcut + 0.67 fy µ] Ac 488.3 * 1000 = [0.35 *25 + 0.67 *360 * 1/100] Ac Ac = 43700 mm2 Take sec. 300 * 500 unbraced column H λ = e , H e = k H o = 1.6 * 7 = 11.2m t 11.2 * 100 λ= = 22.4 < (λ max = 23) 50 ∴ take t = 70 → λ = 16 > 10 (long column)

λ2 t

16 2 * 700 = 89.6 mm 2000 2000 89.6 = 29 kN.m M add = P.δ = 325.5 * 1000 M design = 32 + 29 = 61 kN.m

∴δ =

∴ M u = 1.5 * 61 = 92 kN.m

6.32 5.52 1.6 0.4 ≅3.6 m 4m

Instructor: eng \ just for help

Subject: Water Tanks Solved Examples

:2000-2001

* Design for M & N Mu = 9.2 m.t Nu = 48.83 t 30 cm

Mu 92 = = 0.2m 488.3 Nu

70 cm

e 0.2 = = 0.3 > 0.05 t 0.7 use interaction diagram k=

for f y = 360 , α = 1 , ξ = 0.8

Nu 488.3 * 1000 = = 0.093 f cu bt 25 * 300 * 700

Mu e 92 * 10 6 = = = 0.025 , ρ = o t f cu bt 2 25 * 300 * 700 2 0.4 use min. steel A s = * 300 * 700 = 8400 mm 2 /side 100 use 5 16

Sec. in column

300 mm 700 mm

40 cm 3

10/m` 6

40 cm 3

10/m`

Cross-section of tank

10/m`

16 6

10/m`

Instructor: eng \ just for help

Subject: Water Tanks Solved Examples

:2000-2001

10/m` 7

10/m` (L) 10/m`(L)

10/m` (U)

10/m`(U)

10/m` 8

5 5

8 10/m` 10/m`

Plan

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