a guide to designing reinforced concrete water tanks · oct-15 dr. a.helba civ 416 e 1 a guide to...
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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
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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
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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)
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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
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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:
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)
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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
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Dr. A.Helba CIV 416 E 7
Control of crack width ECP Code 203
Recommendations
Control of crack width
ECP Code 203 Recommendations
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Dr. A.Helba CIV 416 E 8
Control of crack width ECP Code 203 Recommendations
Control of crack width ECP Code 203 Recommendations
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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
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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
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
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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
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
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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)
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
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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
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Dr. A.Helba CIV 416 E 14
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 R 2( )cu
c
f
uR M bd
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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
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
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Dr. A.Helba CIV 416 E 16
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
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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
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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
0.6T
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
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Dr. A.Helba CIV 416 E 19
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
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Dr. A.Helba CIV 416 E 20
• 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
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Dr. A.Helba CIV 416 E 21
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’)
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Dr. A.Helba CIV 416 E 22
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
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Dr. A.Helba CIV 416 E 23
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
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Dr. A.Helba CIV 416 E 24
2- Case of M = 30 kN.m
Solution:
Assume twall = 50 𝑀 = 50 30 = 274 mm
Try twall = 300 mm
Check tensile stress:
Calculate fct :
fct = fctN + fctM = T/Ac + M/Zc = T/(b x t) + M/(bt2/6)
= 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
Design of Uncracked Sections – Case # 2
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
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Dr. A.Helba CIV 416 E 25
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
Check tensile stress:
Calculate fct :
fct = fctN + fctM = T/Ac + M/Zc = T/(b x t) + M/(bt2/6)
= 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
Design of Uncracked Sections – Case # 3
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
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Dr. A.Helba CIV 416 E 26
4- Case of T = 150 kN and M = 15 kN.m
Solution:
Design of Uncracked Sections – Case # 4
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 :
fct = fctN + fctM = T/Ac + M/Zc = T/(b x t) + M/(bt2/6)
= 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 _