a guide to designing reinforced concrete water...

Oct-15

Dr. A.Helba CIV 416 E 1

A Guide to Designing Reinforced Concrete Water Tanks

Helba AlaaDr.

Examples of tank Sections Resisting Tension

Tank Sections Resisting Tension and Moments

General Design Requirements for Tank Elements

Analysis and Design of R.C. Sections under T&M

With tension on water side. (Uncracked Sections)

Appendix: Solved Examples

Lecture 2 & 3

Oct-15


Example of Cylindrical Wall of Water Tank in Hal direction

Wall SEC. PLAN Wall SEC. ELEV.

D H r Water Pressure

Sections Resisting Tension

T T

r

R

T = r R

1 m

t

R

Ring tension T in Cylindrical Wall

t h

Oct-15


Water Pressure on Walls and Floor of an elevated Tank

rests on columns

Sec. Elev.

Sections Resisting Tension and Moments

Final B.M.D. on Walls and Floor

Critical Sections in Tank Structural Elements

(walls & Floors)

Oct-15


Sections 1 & 4 Tension due to Moment is on Air-Side (Cracked Sections)

Open Tank B . M . D

1

4

2 3

Sections 2 & 3 Tension due to Moment is on Water-Side (UnCracked Sections)

1

4

2

3

B.M.D. on Walls and Floor

Oct-15


General Design Requirements for Tank Elements (Uncracked Sections)

• Consider Condition I (Tension on Water side)

1- For serviceability requirement (no cracks at

the liquid side ) ,Design a section To satisfy:

ctrct

ff

ct ct(N) ct(M)

T Mf = f +f +

A Z

cuctr

η coeff. 1 from code

table(4-16)

0.6 ff =

2- To Control the crack width (wk)

• USE - according to ECP Code 203 tables 4-13 , 4-14 and 4-15 – the appropriate :

- concrete cover (from table 4-13)

- type/diam./stress of steel.

( from tables 4-14 and 4-15)

Oct-15


Control of crack width

ECP Code 203 Recommendations

Table 4-11 Code

Control of crack width ECP Code 203 Recommendations

Oct-15


Control of crack width ECP Code 203

Recommendations

Control of crack width

ECP Code 203 Recommendations

Oct-15




Oct-15


Analysis and Design of R.C. Sections

- Under T only - Under M only - Under T & M

Controlling Tensile Strength of Concrete according to Egyptian Code

• Water Tanks are classified as type 3 or 4 in CODE Table (4-8).

• The maximum tensile stress of concrete is given by CODE Eq. (4-69) as follows :

( ) ( )[ ] /ct ct N ct M ctrf f f f

Oct-15


fct(N) is the tensile stress due to

unFactored axial tension

(+ve sign for tension and –ve sign for comp.)

where :

As = all steel area

and Assume

( )ct N

c s

Tf

A nA

10s

c

En

E

• fct(M) is the tensile stress due to unFactored moment

(+ve sign for tension and –ve sign for comp.)

( ) 2 2

6

/ 6ct M

M M Mf

Z bt b t

Oct-15


Tensile Strength of Concrete according to Egyptian Code

• is a reductuon coefficient given in code table (4-16) and depends on the ideal (virtual) thickness of the section (tv) , where

0.6 cuctrff

( )

( )

[1 ]ct N

v

ct M

ft t

f

Values of coefficient Table (4-16) Code

tv(mm) 100 200 400 ≥ 600

1 1.3 1.6 1.7

fcu Values of (N/mm2)

20 2.68 2.06 1.68 1.58

25 3 2.31 1.88 1.76

30 3.29 2.53 2.05 1.93

/ctrf

Oct-15


Calculation of steel rft. As required in Design

of R.C. Sections - I - Under T only - II - Under M only - III - Under T & M (e=M/T =big ecc.)

- IV - Under T & M (e = small ecc.)

Use Steel to resist all tension (neglect concrete resistance in tension)

us

y

cr

s

TA

f

I - Case of axial tension ( T only)

1.4

1.6

u f

f

f

T T

for water pressure

for other loads

where

T

1

2sA

1

2sA

1

2sA

1

2sA

TIE section

Oct-15


Consider a Cylindrical Wall of Water Tank

Wall SEC. PLAN Wall SEC. ELEV.

D H r Water Pressure

T T

r

R

T = r R

1 m

t

I - Case of pure tension ( T only)

R

Ring tension in Cylindrical

Wall t h

Oct-15


Ring Steel in

Cylindrical Wall

T T

r

D 1 m

t

1

2sA

1

2sA

I - Case of Ring tension ( T only)

T = r R = r D/2

r = wh

h

II - Case of pure flexure (M only)

,

12

us

y

cr

s

M aA

f dd

M sA

.cross Sec

1 1 3 R2( ) cu

c

f

uR M bd

Oct-15


Calculate Mus = Tu(e + t/2 - d) or = Mu – Tu(d – t/2)

Cases of eccentric tension (M &T)

u

u

M e =

T 2 III

tCase of

M

sA T

T(eccentric)

e

.cross Sec

III – Case of big eccentric tension

u

u

M e=

T 21

2

us us

y y

cr cr

s s

M T tA if

f fd

Where Mus = Mu – Tu(d – t/2)

sA T

T(eccentric)

e

.cross Sec

Oct-15


u

u

M e =

T 2

tC se oV fI a

M

T T(eccentric)

e

1sA

2sA

2se

1se

'd

'd d

.cross Sec

IV – Case of small eccentric tension

1 21 2,

u us s

y y

cr cr

s s

T TA A

f f

1sA

2sA

Teccentric T e

Calculate Tu1 = Tu / 2 + Mu / (d – d’)

Tu2 = Tu / 2 - Mu / (d – d’)

u

u

M e =

T 2

tC se oV fI a

.cross Sec

Oct-15


•Steps :

1- assume t and check fct

2- calculate As

Design of Uncracked Sections

Step (1) for Slabs and Walls • b = 1 m = 1000 mm Assume t as follows :

If T (in kN) only mm

• If M only M in (kN.m) mm

If M & T mm

0.6t T

50t M

50 50t M

For practical considerations tmin = 150 mm

Oct-15


Step (1) for Sections (b X t) (Beams) • Assume t as follows :

- for T only [ T in (N) ] mm

- for M only

M in (N.mm) mm

- For M & T mm

0.6T

tb

1.6M

tb

1.6 50M

tb

Step (1)b

• Check fct (tensile stress) For any case :

2

( ) ( )

( ) ,

( )/ 6

[ ] /

ct

c

ct

ct ct N ct M ctr

Tf N

A

Mf M

bt

f f f f

Oct-15


Step (2) • Calculate As as follows :

• case (I) T only mm2

• case (II) M only

us

y

cr

s

TA

f

12

us

y

cr

s

MA

fd

21 1 3 ,

/

u

cu c

Mawhere R R

d bd f

Oct-15


• Check m ≥ mmin (steps 1 to 4 in sequence)

Note:

If mcal ≥ mmin (steps 1 and 2) , No need to do step 3)

min

0.2251 #

max. #* 1.1

2 #

3 1.3 /

max. * 4* 0.25% . .240 / 350

4* 0.15% . .360 / 520

cu

y

y

s

f

fthe of

the smaller of

f

A bd

the of st Gr

or st Gr

m

Minimum Val RFT. for Walls

Oct-15


Step (2) • Calculate As as follows :

• case (III) T & M and e ≥ t/2

12

us us

y y

cr cr

s s

M TA

f fd

21 1 3 ,

/

us

cu c

MaR R

d bd f

Where Mus = Mu – Tu(d – t/2)

Step (2) • case (IV) T & M and e < t/2

1 21 2,

u us s

y y

cr cr

s s

T TA A

f f

Where Tu1 = Tu / 2 + Mu / (d – d’)

Tu2 = Tu / 2 - Mu / (d – d’)

Oct-15


Appendix

Solved Examples

on Design of different Uncracked Sections

With conc. of fcu=25 N/mm2 & steel grade 360/520 Design tank wall sections to resist the following cases:

1- T = 150 kN

2- M = 30 kN.m

3- T = 150 kN and M = 60 kN.m

4- T = 150 kN and M = 15 kN.m

Solution:

for concrete fctr = 0.6 fcu = 3 N/mm2

for wall b = 1 m = 1000 mm

Examples on Design of Uncracked Sections

Oct-15


1- Case of T = 150 kN

Solution:

Assume twall = 0.6 T = 90 mm < tmin = 150 mm

Chosen twall = 150 mm

Check tensile stress:

Calculate fct :

fct = fctN + fctM = T/Ac + M/Zc = T/(b x t) + M/(bt2/6)

= 150000/(1000 x 150) + 0 = 1 + 0

= 1 N/mm2

Calculate fctr/ :

tv = t (1 + fctN / fctM) = ∞ then = 1.7

fctr/ = 3 / 1.7 = 1.76 N/mm2

Check: as fct < fctr/ chosen t = 150 mm is O.k

Design of Uncracked Sections – Case # 1

Calculation of As in Case of T = 150 kN

As = Tu / (cr fy/s) - [ for bars 10 mm - cr = 0.93]

= 1.4 x 150 x 1000 / (0.93 x 360/1.15)

= 721.33 mm2

Use 5 10 /m’ on each side (As= 2 x 395 = 790 mm2)

Check As min

As min = 0.15 % Ac = 0.15 x b x t /100 = 1.5 t (for walls)

As > As min O.k

Oct-15


2- Case of M = 30 kN.m

Solution:

Assume twall = 50 𝑀 = 50 30 = 274 mm

Try twall = 300 mm


Calculate fct :


= 0 + 6 x 30 x 106 / (1000 x 3002) = 0 + 2

= 2 N/mm2

Calculate fctr/ :

tv = t (1 + fctN / fctM) = t = 300 mm from table = 1.45

fctr/ = 3 / 1.45 = 2.07 N/mm2

Check: as fct < fctr/ chosen t = 300 mm is O.k


Calculation of As in Case of M = 30 kN.m

Calculate As :

As = Mu / [(cr fy/s)(0.95d)] [ use a = 0.1 d]

= 1.4 x 30 x 106 / [(0.93 x 360/1.15)(0.95 x 260)] = 584 mm2

Use 8 10 /m’ on tension side (As= 632 mm2)

Check As min

As min = 0.15 % Ac = 1.5 t = 450 mm2

As > As min O.k

Oct-15


3- Case of T = 150 kN and M = 60 kN.m

Assume twall = 0.6 T = 90 mm

Assume twall = 50 𝑀 + 50= 50 60 + 50 = 437 mm

Try twall = 500 mm


Calculate fct :


= 150 x 103 /(1000 x 500) + 6 x 60 x 106 / (1000 x 5002)

= 0.3 + 1.44 = 1.74 N/mm2

Calculate fctr/ :

tv = t (1 + fctN / fctM) = t (1 + 0.3 / 1.44) = 1.21 x 500 = 605 mm > 600 mm then = 1.7

fctr/ = 3 / 1.7 = 1.76 N/mm2

Check: as fct > fctr/ chosen t = 500 mm is O.k


Calculation of As in Case of T = 150 kN and M = 60 kN.m

e = M / T = 400 mm > t / 2 = 250 mm Large ecc.

Calculate Ms :

Ms = T (e – t / 2 + d’)

= 150 (400 – 250 + 40) / 1000

= 28.5 kN. m

Calculate As :

As = Mus / [(cr fy/s)(0.95d)] + T1u / (cr fy/s) [ use a = 0.1 d]

= 1.4 x 28.5 x 106 / [(0.75 x 360/1.15)(0.95 x 460)]

+ 1.4 x 150 x 1000 / (0.75 x 360/1.15) = 388.9 + 894.4 = 1283 mm2

Use 7 16 /m’ on tension side

Oct-15


4- Case of T = 150 kN and M = 15 kN.m

Solution:


Assume twall = 0.6 T = 90 mm

Assume twall = 50 𝑀 + 50= 50 15 + 50 = 244 mm

Try twall = 250 mm

Check tensile stress: Calculate fct :


= 150 x 103 /(1000 x 250) + 6 x 15 x 106 / (1000 x 2502)

= 0.6 + 1.44

= 2.04 N/mm2

Calculate fctr/ :

tv = t (1 + fctN / fctM) = t (1 + 0.6 / 1.44) = 1.42 x 250 = 355 mm then = 1.53

fctr/ = 3 / 1.55 = 1.94 N/mm2

Check: as fct > fctr/ Not O.k increase t (try t = 300 mm)

Recheck: as fct > fctr/ chosen t = 300 mm is O.k

Calculation of As in Case of T = 150 kN and M = 15 kN.m

e = M / T = 100 mm < t / 2 = 150 mm Small ecc.

Calculate T1 and T2 :

T1 = 150/2 + 15/(0.26 – 0.04) = 143.2 kN

T2 = 150/2 - 15/(0.26 – 0.04) = 6.8 kN

Calculate As1 and As2 : As1 = T1u / (cr fy/s)

= 1.4 x 143.2 x 103 / (0.93 x 360/1.15) = 689 mm2

Use 9 10 /m’ or 7 12 /m’ on tension side ❶(As=711 mm2)

As2 = T2u / (cr fy/s)

= 33 mm2 < Asmin = 1.5 t

Use Asmin 5 10 /m’ on tension side ❷(As= 395 mm2)

❶

❷ T1 = T / 2 + M / (d – d’)

2 _

a guide to designing reinforced concrete water...

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