ๅฐŠๆ•ฌ็š„ ๅพฎไฟกๆฑ‡็Ž‡๏ผš1ๅ†† โ‰ˆ 0.046089 ๅ…ƒ ๆ”ฏไป˜ๅฎๆฑ‡็Ž‡๏ผš1ๅ†† โ‰ˆ 0.04618ๅ…ƒ [้€€ๅ‡บ็™ปๅฝ•]
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REVISION
sections
Water section
UN CRACKED SECTION
CHECK STRESS DESIGN
Air section
CRACKED SECTION
DESIGN
WHERE IS THE SECTION ?
๏‚š CHECK STRESS WORKING
๏‚š DESIGN ULTIMATE
(Moment + Tension) (Moment + comp)
EXAMPLE ONE
Solution
B.MD โ€“ N.F.D
Section 1-1 ( Air section)
Section 2-2 ( water section )
B= 100 cm M= 45 kN.m T=52.5 kN tw=30cm water section
1- check tensile stress (ft)
Ft =
๐‘‡
๐‘โˆ—๐‘ก๐œ”
+
6๐‘š
๐‘โˆ—๐‘ก๐œ”2
Ft =
52.5โˆ—1000
1000โˆ—300
+
6โˆ—45โˆ—10^6
1000โˆ—300^2
=0.175 + 3= 3.175 N/mmยฒ
tv=t(1+
๐‘“๐‘ก(๐‘)
๐‘“๐‘ก(๐‘€)
) =300(1+
0.175
3
)=317.5 ๐œ‚=1.5
FT allowable =
0.6 ๐‘“๐‘๐‘ข
๐œ‚
=
0.6 25
1.5
= 2 N/mmยฒ ๐œ‚ = 1.7 ( T ONLY )
FT > Fct ALLOWABLE ( UNSAFE )
INCREASE DIMINSIONS ( TW = 40 Cm )
Ft =
๐‘‡
๐‘โˆ—๐‘ก๐œ”
+
6๐‘š
๐‘โˆ—๐‘ก๐œ”2
Ft =
52.5โˆ—1000
1000โˆ—400
+
6โˆ—45โˆ—10^6
1000โˆ—400^2
= 1.82 N/mmยฒ
FT < FT ALLOWABLE ( SAFE )
DESIGN
โ…‡ =
๐‘€
๐‘
= = 0.85 > t/2 โ€ฆ(t/2) = 0.4 /2 = 0.2
BIG ECC
es = e โ€“ (t/2) + Cover
es = 0.85 โ€“ 0.2 + 0.05=0.7
Mus = Tu * ES = 1.5 * 52.5*0.7
Mus = 55.125 KN.m
D =c1
๐‘€๐‘ข๐‘ 
๐น๐‘๐‘ข.๐‘
C = 7.45 J = 0.826
As =
1
ฮฒ๐‘๐‘Ÿ
(
๐‘€๐‘ข๐‘ 
๐ฝ โˆ—๐‘‘โˆ—๐น๐‘ฆ
+
๐‘‡๐‘ข๐‘™
๐น๐‘ฆ / ฦณ ๐‘ 
) =
1
0.85
(
55.125 ๐‘ฅ 106
0.826 โˆ—350โˆ—360
+
1.5 ๐‘ฅ 52.5 ๐‘ฅ 103
360 / 1.15
)=920 mmยฒ use 10 12
45
52.5
EXAMPLE Two :-
Mworking= 55 KN.m , Tworking = 800 KN , fcu= 30 N/mmยฒ
SOLUTION
1- Assume t t =
๐‘€
0.3 ๐‘
+ 40 =
55 ๐‘ฅ 106
0.3 ๐‘ฅ 1000
+ 40 =468.2 mm
Take t = 500 mm
2- check tensile stress (ft)
ft =
๐‘‡
๐‘โˆ—๐‘ก๐œ”
+
6๐‘€
๐‘โˆ—๐‘ก๐œ”2 =
800 ๐‘ฅ103
1000 ๐‘ฅ 500
+
6 ๐‘ฅ 55 ๐‘ฅ 106
1000 ๐‘ฅ 5002 = 1.6 +1.32 =2.92
N/mmยฒ
tv=t(1+
๐‘“๐‘ก(๐‘)
๐‘“๐‘ก(๐‘€)
) = 500 1 +
๐Ÿ.๐Ÿ”
๐Ÿ.๐Ÿ‘๐Ÿ
= 1106 ๐œ‚=1.7
Fct allowable =
0.6 ๐‘“๐‘๐‘ข
๐œ‚
=
0.6 30
1.7
= 1.9 N/๐‘š๐‘š2
Ft > Fct allowable UNSAFE increase t
Take t=700 mm
ft =
๐‘‡
๐‘โˆ—๐‘ก๐œ”
+
6๐‘€
๐‘โˆ—๐‘ก๐œ”2 =
800 ๐‘ฅ103
1000 ๐‘ฅ 700
+
6 ๐‘ฅ 55 ๐‘ฅ 106
1000 ๐‘ฅ 7002 = 1.14+0.67=1.8 N/mmยฒ
Ft < Fct allowable SAFE
MU.L.= 55 x 1.5=82.5 KN.m
TU.L. = 800 x 1.5=1200 KN
โ…‡ =
๐‘€
๐‘
=
82.5
1200
= 0.068 <
๐‘ก
2
โ€ฆ.
๐‘ก
2
= 0.7 /2 = 0.35
Small Ecc
es1 =
๐‘ก
2
- e โ€“ cover = 0.35 โ€“ 0.068 - 0.05=0.232
es2 =
๐‘ก
2
- e โ€“ cover = 0.35 + 0.068 - 0.05=0.368
T1 =
๐‘‡๐‘ข๐‘™ ๐‘ฅ ๐‘’๐‘ 2
๐‘’๐‘ 1+๐‘’๐‘ 2
=
1200๐‘ฅ 0.368
0.6
=736 KN
T2=
๐‘‡๐‘ข๐‘™ ๐‘ฅ ๐‘’๐‘ 1
๐‘’๐‘ 1+๐‘’๐‘ 2
=
1200๐‘ฅ 0.232
0.6
=464 KN
As1 =
๐‘‡1
ฮฒ๐‘๐‘Ÿ ๐‘ฅ๐‘“๐‘ฆ
ฮณ๐‘ 
=
736 ๐‘ฅ 103
0.85 ๐‘ฅ360
1.15
= 2766 mmยฒ 26 12
As2 =
๐‘‡2
ฮฒ๐‘๐‘Ÿ ๐‘ฅ๐‘“๐‘ฆ
ฮณ๐‘ 
=
464 ๐‘ฅ 103
0.85 ๐‘ฅ360
1.15
= 1743 mmยฒ 16 12
Example Three
Force in Tie = T = W * L = 15 * 4.5 = 67.5 KN
Design Tie = Design Water Section with tension only
1- Assume square cross section (30*30)
2- Check tensile stress
Ft = T / (b*t) = (67.5 * 1000) / (300*300) = 0.75 N/mm2
Fct all. =
0.6 ๐‘“๐‘๐‘ข
๐œ‚
=
0.6 25
1.7
= 1.76 N/mm2
AS = Tu / (Fyโ€œ/ ฦณs) = 1.5 * 67.5*1000 /( 0.85 * (360 /1.15))=380.5 mmยฒ
EXAMPLE Four
B= 1000 mm M = 45 KN.m water section
1 โ€“ assume dimension
t =
๐‘€๐‘คโˆ—10^6
0.3โˆ—๐‘
=
45โˆ—10^6
0.3โˆ—1000
= 387.3 =400 mm
2- check tensile stress (ft)
Fct( =
6๐‘€โˆ—10^6
๐‘โˆ—๐‘ก^2
=
6โˆ—45โˆ—10^6
1000โˆ—(400)^2
= 1.68N/mm2
FcT allowable =
0.6 ๐‘“๐‘๐‘ข
๐œ‚
=
0.6 25
1.6
= 1.87 N/๐‘š๐‘š2
๐œ‚ = from chart
tv = t[1+
๐น๐‘๐‘ก(๐‘)
๐น๐‘๐‘ก(๐‘€)
] = t = 400 mm ; Fct(N) = 0
FcT allowable > Fct (M) (safe)
3-DESIGN
Mu =1.5 Mw = 45*1.5= 67.5 KN.m
d = c1
๐‘€๐‘ข โˆ—10^6
๐‘“๐‘๐‘ขโˆ—๐‘
350 = c1
67.5 โˆ—10^6
25โˆ—1000
C1 = 6.7 J= 0.826
As =
๐‘€๐‘ขโˆ—10^6
รŸ๐‘๐‘Ÿโˆ—๐น๐‘ฆโˆ—๐ฝโˆ—๐‘‘
=
67.5โˆ—10^6
0.85โˆ—360โˆ—.826โˆ—350
= 763 mm2
Use 7 ร˜ 12 /m
รŸ๐‘๐‘Ÿ = 0.85 from table by using ร˜ 12
Example Five
M= 80 kN.m N= -1100 kN b=100 Cm water section
1- assume tw
t =
๐‘€
0.3๐‘
- t =
80 ๐‘ฅ 106
0.3โˆ—1000
โˆ’ 30 โˆ’ 50 = 500 mm
2- check stress
Ft=fct (N) + ft (M)
Ft =
๐ถ
๐‘โˆ—๐‘ก๐œ”
+
6๐‘€
๐‘โˆ—๐‘ก๐œ”2
Ft =
โˆ’1100โˆ—1000
1000โˆ—500
+
6โˆ—80โˆ—106
1000โˆ—500^2
= - 0.28 N/mmยฒ
Ft< ft allowable safe
3- design
๏‚š Mu = 80 * 1.5 = 120 kN.m
๏‚š Pu = 1100* 1.5 = 1650 kN
๏‚š e= M/p = 0.07 > 0.05 t
๏‚š e < t/2
Use I.D
Pu / bt = ( 1650*1000)/(30*1000*500) = 0.11
Mu / bt2 = 120*10^6 / 1000*500^2 = 0.48
๐œŒ =
๐œ‚ = ๐œŒ * fcu 8 10 ^-4 = ๐œŒ *30*10-4 =
AS = ๐œ‚ * B * T
ASโ€™ = AS
CHECK ASMIN

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design of water section concrete structures

  • 1. REVISION sections Water section UN CRACKED SECTION CHECK STRESS DESIGN Air section CRACKED SECTION DESIGN
  • 2. WHERE IS THE SECTION ?
  • 3. ๏‚š CHECK STRESS WORKING ๏‚š DESIGN ULTIMATE
  • 4. (Moment + Tension) (Moment + comp)
  • 7. Section 1-1 ( Air section) Section 2-2 ( water section ) B= 100 cm M= 45 kN.m T=52.5 kN tw=30cm water section 1- check tensile stress (ft) Ft = ๐‘‡ ๐‘โˆ—๐‘ก๐œ” + 6๐‘š ๐‘โˆ—๐‘ก๐œ”2 Ft = 52.5โˆ—1000 1000โˆ—300 + 6โˆ—45โˆ—10^6 1000โˆ—300^2 =0.175 + 3= 3.175 N/mmยฒ tv=t(1+ ๐‘“๐‘ก(๐‘) ๐‘“๐‘ก(๐‘€) ) =300(1+ 0.175 3 )=317.5 ๐œ‚=1.5 FT allowable = 0.6 ๐‘“๐‘๐‘ข ๐œ‚ = 0.6 25 1.5 = 2 N/mmยฒ ๐œ‚ = 1.7 ( T ONLY )
  • 8. FT > Fct ALLOWABLE ( UNSAFE ) INCREASE DIMINSIONS ( TW = 40 Cm ) Ft = ๐‘‡ ๐‘โˆ—๐‘ก๐œ” + 6๐‘š ๐‘โˆ—๐‘ก๐œ”2 Ft = 52.5โˆ—1000 1000โˆ—400 + 6โˆ—45โˆ—10^6 1000โˆ—400^2 = 1.82 N/mmยฒ FT < FT ALLOWABLE ( SAFE )
  • 9. DESIGN โ…‡ = ๐‘€ ๐‘ = = 0.85 > t/2 โ€ฆ(t/2) = 0.4 /2 = 0.2 BIG ECC es = e โ€“ (t/2) + Cover es = 0.85 โ€“ 0.2 + 0.05=0.7 Mus = Tu * ES = 1.5 * 52.5*0.7 Mus = 55.125 KN.m D =c1 ๐‘€๐‘ข๐‘  ๐น๐‘๐‘ข.๐‘ C = 7.45 J = 0.826 As = 1 ฮฒ๐‘๐‘Ÿ ( ๐‘€๐‘ข๐‘  ๐ฝ โˆ—๐‘‘โˆ—๐น๐‘ฆ + ๐‘‡๐‘ข๐‘™ ๐น๐‘ฆ / ฦณ ๐‘  ) = 1 0.85 ( 55.125 ๐‘ฅ 106 0.826 โˆ—350โˆ—360 + 1.5 ๐‘ฅ 52.5 ๐‘ฅ 103 360 / 1.15 )=920 mmยฒ use 10 12 45 52.5
  • 10.
  • 11. EXAMPLE Two :- Mworking= 55 KN.m , Tworking = 800 KN , fcu= 30 N/mmยฒ SOLUTION 1- Assume t t = ๐‘€ 0.3 ๐‘ + 40 = 55 ๐‘ฅ 106 0.3 ๐‘ฅ 1000 + 40 =468.2 mm Take t = 500 mm
  • 12. 2- check tensile stress (ft) ft = ๐‘‡ ๐‘โˆ—๐‘ก๐œ” + 6๐‘€ ๐‘โˆ—๐‘ก๐œ”2 = 800 ๐‘ฅ103 1000 ๐‘ฅ 500 + 6 ๐‘ฅ 55 ๐‘ฅ 106 1000 ๐‘ฅ 5002 = 1.6 +1.32 =2.92 N/mmยฒ tv=t(1+ ๐‘“๐‘ก(๐‘) ๐‘“๐‘ก(๐‘€) ) = 500 1 + ๐Ÿ.๐Ÿ” ๐Ÿ.๐Ÿ‘๐Ÿ = 1106 ๐œ‚=1.7 Fct allowable = 0.6 ๐‘“๐‘๐‘ข ๐œ‚ = 0.6 30 1.7 = 1.9 N/๐‘š๐‘š2 Ft > Fct allowable UNSAFE increase t
  • 13. Take t=700 mm ft = ๐‘‡ ๐‘โˆ—๐‘ก๐œ” + 6๐‘€ ๐‘โˆ—๐‘ก๐œ”2 = 800 ๐‘ฅ103 1000 ๐‘ฅ 700 + 6 ๐‘ฅ 55 ๐‘ฅ 106 1000 ๐‘ฅ 7002 = 1.14+0.67=1.8 N/mmยฒ Ft < Fct allowable SAFE
  • 14. MU.L.= 55 x 1.5=82.5 KN.m TU.L. = 800 x 1.5=1200 KN โ…‡ = ๐‘€ ๐‘ = 82.5 1200 = 0.068 < ๐‘ก 2 โ€ฆ. ๐‘ก 2 = 0.7 /2 = 0.35 Small Ecc es1 = ๐‘ก 2 - e โ€“ cover = 0.35 โ€“ 0.068 - 0.05=0.232 es2 = ๐‘ก 2 - e โ€“ cover = 0.35 + 0.068 - 0.05=0.368 T1 = ๐‘‡๐‘ข๐‘™ ๐‘ฅ ๐‘’๐‘ 2 ๐‘’๐‘ 1+๐‘’๐‘ 2 = 1200๐‘ฅ 0.368 0.6 =736 KN T2= ๐‘‡๐‘ข๐‘™ ๐‘ฅ ๐‘’๐‘ 1 ๐‘’๐‘ 1+๐‘’๐‘ 2 = 1200๐‘ฅ 0.232 0.6 =464 KN As1 = ๐‘‡1 ฮฒ๐‘๐‘Ÿ ๐‘ฅ๐‘“๐‘ฆ ฮณ๐‘  = 736 ๐‘ฅ 103 0.85 ๐‘ฅ360 1.15 = 2766 mmยฒ 26 12 As2 = ๐‘‡2 ฮฒ๐‘๐‘Ÿ ๐‘ฅ๐‘“๐‘ฆ ฮณ๐‘  = 464 ๐‘ฅ 103 0.85 ๐‘ฅ360 1.15 = 1743 mmยฒ 16 12
  • 16. Force in Tie = T = W * L = 15 * 4.5 = 67.5 KN Design Tie = Design Water Section with tension only 1- Assume square cross section (30*30) 2- Check tensile stress Ft = T / (b*t) = (67.5 * 1000) / (300*300) = 0.75 N/mm2 Fct all. = 0.6 ๐‘“๐‘๐‘ข ๐œ‚ = 0.6 25 1.7 = 1.76 N/mm2 AS = Tu / (Fyโ€œ/ ฦณs) = 1.5 * 67.5*1000 /( 0.85 * (360 /1.15))=380.5 mmยฒ
  • 17. EXAMPLE Four B= 1000 mm M = 45 KN.m water section 1 โ€“ assume dimension t = ๐‘€๐‘คโˆ—10^6 0.3โˆ—๐‘ = 45โˆ—10^6 0.3โˆ—1000 = 387.3 =400 mm
  • 18. 2- check tensile stress (ft) Fct( = 6๐‘€โˆ—10^6 ๐‘โˆ—๐‘ก^2 = 6โˆ—45โˆ—10^6 1000โˆ—(400)^2 = 1.68N/mm2 FcT allowable = 0.6 ๐‘“๐‘๐‘ข ๐œ‚ = 0.6 25 1.6 = 1.87 N/๐‘š๐‘š2 ๐œ‚ = from chart tv = t[1+ ๐น๐‘๐‘ก(๐‘) ๐น๐‘๐‘ก(๐‘€) ] = t = 400 mm ; Fct(N) = 0 FcT allowable > Fct (M) (safe)
  • 19. 3-DESIGN Mu =1.5 Mw = 45*1.5= 67.5 KN.m d = c1 ๐‘€๐‘ข โˆ—10^6 ๐‘“๐‘๐‘ขโˆ—๐‘ 350 = c1 67.5 โˆ—10^6 25โˆ—1000 C1 = 6.7 J= 0.826 As = ๐‘€๐‘ขโˆ—10^6 รŸ๐‘๐‘Ÿโˆ—๐น๐‘ฆโˆ—๐ฝโˆ—๐‘‘ = 67.5โˆ—10^6 0.85โˆ—360โˆ—.826โˆ—350 = 763 mm2 Use 7 ร˜ 12 /m รŸ๐‘๐‘Ÿ = 0.85 from table by using ร˜ 12
  • 20. Example Five M= 80 kN.m N= -1100 kN b=100 Cm water section 1- assume tw t = ๐‘€ 0.3๐‘ - t = 80 ๐‘ฅ 106 0.3โˆ—1000 โˆ’ 30 โˆ’ 50 = 500 mm 2- check stress Ft=fct (N) + ft (M) Ft = ๐ถ ๐‘โˆ—๐‘ก๐œ” + 6๐‘€ ๐‘โˆ—๐‘ก๐œ”2 Ft = โˆ’1100โˆ—1000 1000โˆ—500 + 6โˆ—80โˆ—106 1000โˆ—500^2 = - 0.28 N/mmยฒ Ft< ft allowable safe
  • 21. 3- design ๏‚š Mu = 80 * 1.5 = 120 kN.m ๏‚š Pu = 1100* 1.5 = 1650 kN ๏‚š e= M/p = 0.07 > 0.05 t ๏‚š e < t/2 Use I.D Pu / bt = ( 1650*1000)/(30*1000*500) = 0.11 Mu / bt2 = 120*10^6 / 1000*500^2 = 0.48 ๐œŒ = ๐œ‚ = ๐œŒ * fcu 8 10 ^-4 = ๐œŒ *30*10-4 = AS = ๐œ‚ * B * T ASโ€™ = AS CHECK ASMIN
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