A Guide to Designing Reinforced Concrete Water A Guide to Designing Reinforced Concrete Water Tanks

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Text of A Guide to Designing Reinforced Concrete Water A Guide to Designing Reinforced Concrete Water Tanks

  • 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

    Dr. A.Helba CIV 416 E 2

    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

    Dr. A.Helba CIV 416 E 3

    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

    Dr. A.Helba CIV 416 E 4

    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

    Dr. A.Helba CIV 416 E 5

    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:

    ctr ct

    f f

     

     ct ct(N) ct(M)

    T M f = 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

    Dr. A.Helba CIV 416 E 6

    Control of crack width

    ECP Code 203 Recommendations

    Table 4-11 Code

    Control of crack width ECP Code 203 Recommendations

  • Oct-15

    Dr. A.Helba CIV 416 E 7

    Control of crack width ECP Code 203

    Recommendations

    Control of crack width

    ECP Code 203 Recommendations

  • Oct-15

    Dr. A.Helba CIV 416 E 8

    Control of crack width ECP Code 203 Recommendations

    Control of crack width ECP Code 203 Recommendations

  • Oct-15

    Dr. A.Helba CIV 416 E 9

    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

    Dr. A.Helba CIV 416 E 10

    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

    T f

    A nA 

    10s

    c

    E n

    E  

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

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

    ( ) 2 2

    6

    / 6 ct M

    M M M f

    Z bt b t   

  • Oct-15

    Dr. A.Helba CIV 416 E 11

    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 cuctr ff

      

    ( )

    ( )

    [1 ] ct N

    v

    ct M

    f t 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/mm 2)

    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

    Dr. A.Helba CIV 416 E 12

    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)

    u s

    y

    cr

    s

    T A

    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

    2 sA

    1

    2 sA

    1

    2 sA

    1

    2 sA

    TIE section

  • Oct-15

    Dr. A.Helba CIV 416 E 13

    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

    Dr. A.Helba CIV 416 E 14

    Ring Steel in

    Cylindrical Wall

    T T

    r

    D 1 m

    t

    1

    2 sA

    1

    2 sA

    I - Case of Ring tension ( T only)

    T = r R = r D/2

    r = wh

    h

    II - Case of pure flexure (M only)

    ,

    1 2

    u s

    y

    cr

    s

    M a A

    f d d

     

     

        

         

    M sA

    .cross Sec

    1 1 3    R 2( ) cu

    c

    f

    uR M bd

  • Oct-15

    Dr. A.Helba CIV 416 E 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

    t Case of

    M

    sA T

    T(eccentric) e

    .cross Sec

    III – Case of big eccentric tension

    u

    u

    M e=

    T 2 1

    2

    us u s

    y y

    cr cr

    s s

    M T t A if

    f f d

      

     

           

             

    Where Mus = Mu – Tu(d – t/2)

    sA T

    T(eccentric) e

    .cross Sec

  • Oct-15

    Dr. A.Helba CIV 416 E 16

    u

    u

    M e =

    T 2  

    t C 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 2 1 2,

       

                 

    u u s s

    y y

    cr cr

    s s

    T T A 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  

    t C se oV fI a

    .cross Sec

  • Oct-15

    Dr. A.Helba CIV 416 E 17

    •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

    Dr. A.Helba CIV 416 E 18

    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

    0.6 T

    t b

        

     

    1.6 M

    t b

    1.6 50 M

    t b

     

    Step (1)b

    • Check fct (tensile stress) For any case :

    2

    ( ) ( )

    ( ) ,

    ( ) / 6

    [ ] /

    ct

    c

    ct

    ct ct N ct M ctr

    T f N

    A

    M f M

    bt

    f f f f 

      

  • Oct-15

    Dr. A.Helba CIV 416 E 19

    Step (2) • Calculate As as follows : • case (I) T only

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