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# Panelled Beams
2016 03
Panelled Beams
Reinforced Concrete Design 2
2
)80-150(
Hollow Block Slab 2
Solid Slab
)Panelled Frames(
00
)4-3 (
Two Way Solid Slab b
a b
–7
=/ = / =1.4∗+. ∗+. +1.6. Ws
= \
780
Grashoff
. > / α = 1 +1 = 1 +
78.
2
3
M
. α=0. ≤ 50./ 15 = 0.35
–4
r
%20
0 +21 + +2 = 6 1+2 3 Moments Equation
By: Karim Sayed
1 = 24
2= 7
2
Panelled Beams
Reinforced Concrete Design 2
= ∗ = = ∗ → & → =
5
6
Design OF Panelled Beams
=
= =
–7
2
Grashoff Grashoff
++ = > + + ∗ =+ . ∗∗ . . ∗∗ + + ∗ + ∗∗ =+ )Ls×L(
m
n
Hollow Block
By: Karim Sayed
7
3
Panelled Beams
Reinforced Concrete Design 2
] = [1.1.4∗+. +∗+. +1. 6 .. ∗ ∗ +1. 4 ℎ ℎ 2 2 ( ) 1.1.4 ∗=/∗ =/ = /∗ Wrib
= \ = → α= 1 + & = 1 +1
–3
Reduction of Moment
–4
90 = \2 = ∗ 90
Design of Panelled Beams
6
)Simply Supported(
d
::
=40
ℎ 16+ = 5 + ∗ T - Section
=30
ℎ 6+ = 10 + L - Section
= ∗ → & → = ∗ 10 =0.24 4 / − =0.70 0 / = ∗ < < < > 5 8/ = 2 Check of Shear
Use Minimum Stirrups
Increase Dimensions
By: Karim Sayed
7
4
7
Panelled Beams
Reinforced Concrete Design 2
8
)
(
Design of Edge Beams 7
= 10
Simply Supported
= 12 = = 300 . =. ∗ [ ] ∗
Continuous
2
3
Panelled
Wavg )L * Ls(
Panelled
By: Karim Sayed
7
5
Reinforced Concrete Design 2
Panelled Beams
Sheet)The Given Plan Shows general layout of a Panelled Beams slab floor
Givens
== // // ..=/ = . /
Floor Height = 3.50m
Design The Shown Slab
1st : Design The Slabs
200045 =44 = 200040 =50 | = : == +. .. =. ∗∗ ∗. +. = / =. ∗+. ∗+. ∗. +. / ∵ , << . α=0.50.15 0.35 = → = \ = 0.0.776∗26∗2 =1 →α=0.35 & =0.35 1 – Assume Assume Concrete Thickness
2- Calculate Ws
3 – Calculate Calculate values of
&
→=\ =0.80.0.7∗2887∗27∗2 = 1 → α = 0.35 & = 0.35 → = \ = 0.76∗2 = 1.145 → α = 0.4225 & = 0.267
By: Karim Sayed
7
6
Panelled Beams
Reinforced Concrete Design 2
4 – Take Take Strips on both Directions and drawing B.M.D
Strip 1
Strip 2
4. 2 ∗2 1= 24 = 245.1∗2=1. 4 | 2= 24 1= 24 = 3.=1.224∗24 =1.1 | 2= 24 = 4.224∗2 = =1. 7 24 ++ 2114 + 2 ∗ 2 = 661.4+1.7 ++ 21214 + 2 ∗ 2 = 661.1.4+1.1 2
1 ∗ 2 + 22.4 +|2 ∗=. 2 = 6 1.4.+ 1.4 =. 1 ∗ 2 + 22.4 +|2 ∗=. 2 = 6 1.7.+ 1.7 =. Strip 3
Strip 4
. ∗ . ∗ = = =. =. | = = 1= 24 = 4.224∗2 =1.4 | 2= 24 = 5. 124∗2 ++ 2114 + 2 ∗ 2 = 661.4+1.1 1 =1.7
1 ∗2 +22 24 + 2 ∗ 2 = 6 1.1 + 1.1 =. . | = . By: Karim Sayed
++ 22114 + 2 ∗ 2 = 661.1.4+1.7 =. 1 ∗ 2 + 22.4 +|2 ∗=. 2 = 6 1.7.+ 1.7 7
Panelled Beams
Reinforced Concrete Design 2
=== = ==∗ . . ∗ = ∗ → = ∗ =. →=. . ∗ = = = ∗.∗ = / 5 – Design Design of Critical Sections
/\
2nd – Design Design of Panelled Beams
= | = = =≈ 1. 1 . 4 ∗∗ ∗+∗ ∗+∗ ∗ + ∗∗ =+ 1∗+∗ ∗+∗ ∗ 25 = 15.15.9 / =12+ 1.4∗0.25∗ 0.0.650.12∗10 /
1 – Assume Assume Beams Dimensions
2 – Calculate Calculate Average Loads W avg avg
3- Calculating
&
By: Karim Sayed
For Beams
7
8
= ∗ =. ∗ 1 = α = 1 + =. 1 + =. W=Wavg∗a∗α=15. 9 ∗2∗0. 6 75 = 21.456456 KN/m 21. 4 56∗10 = 8 = 8 = 268.8.26 5 ..
Panelled Beams
Reinforced Concrete Design 2
4- Design Panelled Beams For B1,B2,B3 (
= 2 & 2= 6→ = = 26 ∗90=30 =. ∗ = . . =65050=600 25 2 5 =2000 =0. 2 4 = 0.0. 98 9 8 / / 1.52525 = 1850 0 165++=2250 =0. 7 = 2.2. 85 8 5 / / − 1. 5 ∗10 =1 ∗1000 134. 1 34. 3 ∗10 600=1 25∗1850134.3 →1=11. 1 3 &=0. 8 26 107. 3 ∗1000 ∗10 = = 0.0. 715 7 1 5 / / = = = 360∗0.826∗600 = 753 600∗250 ∵ < / 0.225∗= 0.√ 225225∗5 √ ∗ = 360 ∗250∗600=469 Calculation of Reduction Factor For Moment FOR B1
Design For Moment
Design For Shear
Take
No Need For Stirrups , USE (
Check Minimum As
Stirrups Hungers :
By: Karim Sayed
7
9
Panelled Beams
Reinforced Concrete Design 2
= 4 & 2= 6→ = = 46 ∗90=60 =. ∗ = . . =65050=600 25 2 5 =2000 =0. 2 4 = 0.0. 98 9 8 / / 1. 5 = 1850 0 165++=2250 107.600∗250 3∗1000 = 0.0.715715 / = / ∵ < 232. 2 32. 5 ∗10 / 600=1 25∗1850 1=8. 4 6 & =0. 8 26 232. 5 ∗10 = = = 360∗0.826∗600 = 1303 1303 0.225∗= 0.√ 225225∗5 √ ∗ = 360 ∗250∗600=469 Calculation of Reduction Factor For Moment FOR B2
Design For Moment
Design For Shear
No Need For Stirrups , USE (
]
Take Check Minimum As
Stirrups Hungers = 0.2 As =
By: Karim Sayed
7
71
Panelled Beams
Reinforced Concrete Design 2
Calculation of Reduction Factor For Moment FOR B3 There is NO reduction factor for Beam B3 ,as it Located in the middle of Horizontal Beams
=65050=600 =2000 = 1850 0 165++=2250 268. 2 68. 5 ∗10 600=1 25∗1850 1=7. 8 7 & =0. 8 26 268. 5 ∗10 = = ==360∗0. 8 26∗600 1505 1505 0.225∗= 0.√ 225225∗5 √ ∗ = 360 ∗250∗600=469 Design For Moment
3∗1000 = 0.0.715715 / = 107.600∗250 / ∵ < /
Design For Shear
No Need For Stirrups , USE (
Take Check Minimum As
Stirrups Hungers = 0.2 As =
= ∗ ∗ = 15.9 ∗ 2 ∗ 0.325 = 10.4 / = 8 = 10.48∗12 = 187.7.18 2 .. = 2 & 2 = 5 → = = 25 ∗90=36 For B4,B5 (
Calculation of Reduction Factor For Moment FOR B4
By: Karim Sayed
7
77
=. ∗ = . . = 650 7=2000 0 = 580 25 2 5 =0. 2 4 = 0.0. 98 9 8 / / 1. 5 1 6 + = 1 8 5 0 0 5 +=2650 . . ∗10 580=1 25∗1850 →1=11. 9 &=0. 8 26 . ∗10 4∗1000 = 0.0.430430 / = = = 360∗0.826∗580 = 638 = 62.580∗250 / ∵ < / 0.225∗= 0.√ 225225∗5 √ ∗ = 360 ∗ 250 ∗ 580 = 4 53
Panelled Beams
Reinforced Concrete Design 2
Design For Moment
Design For Shear
Take
No Need For Stirrups , USE (
Check Minimum As
Stirrups Hungers :
= 4 & 2= 5→ = = 45 ∗90=72 =. ∗ = . . = 650 70 = 580
Calculation of Reduction Factor For Moment FOR B 5
Design For Moment
By: Karim Sayed
Design For Shear
=0.24 1.25255 = 0.0.9898 / / 7
72
=2000 = 1850 0 165++=2650 . . ∗10 580=1 25∗1850 →1=9. 3 5 & =0. 8 26 . ∗10 = = ==360∗0. 8 26∗580 1033 1033 0.225∗= 0.√ 225225∗5 √ ∗ = 360 ∗ 250 ∗ 580 = 453 Reinforced Concrete Design 2
4∗1000 = 0.0.430430 / = 62.580∗250 / ∵ < /
Panelled Beams
No Need For Stirrups , USE (
Take Check Minimum As
Stirrups Hungers :
3rd – Design Design of Edge Beams
= 300 & = 650 ] .= 1.4=1.∗ 0.34∗0∗.5[5] ∗ ∗ 25 = 5.8 / = .. / 1- Assume Beams Dimensions Dimensions
2- Calculate Loads action on Beams
Design of Beam B6
By: Karim Sayed
7
73
Panelled Beams
Reinforced Concrete Design 2
1=. +6∑+ 1 ∗ 1=5.8 + 2 6 ∗ 5 ∗ 15.5.9 = 52.2 // Load for Moment= Load for Shear
S.F.D Using Imperial Calculations
B.M.D Using Imperial Calculations
Design of Critical Sections 1- Design of L-Sec
= = ℎ & = . . & = / 6+=6∗100+300=900 = 10 + = 0.8∗6000 + 3 0 0 = 7 8 0 10 { =6000 . . ∗10 =1 ∗ => 600= 1 25∗780 1=6. 4 1 →=0. 8 26 . ∗10 = = = 360∗0.826∗600 =958 25∗360√ 2525 ∗ 250 ∗ 600 = 563 > = 0.225∗√ ∗ = 0.225∗√ ℎ= ∗18958⁄4 ~4 => => Check Minimum As
By: Karim Sayed
7
74
Panelled Beams
Reinforced Concrete Design 2
= =/ = & = . . & . . ∗10 =1 ∗ => 600= 1 25∗300 1=3. 6 →=0. 7 86 ∗10 =1230 = = = 360∗0..826∗600 25∗360√ 2525 ∗ 250 ∗ 600 = 563 = 0.225∗>√ ∗ = 0.225∗√ ℎ= ∗181230⁄4 ~5 => => = 188∗1000 = 1. 1 /2 / 2 >=0. >= 0. 9 8 / 2 600∗300 240 2∗50. 3 ∗ 2 ==8.7~9 : →0.691= 8/300∗1.15 = 115 → = 114.7 .
2- Design of R-Sec
Check Minimum As
Check Shear::
Nmumber of Stirrups =
------* ------ *------* ------ *------* ------ *------* ------ *------* ------ *------* ------ *------* ------ *------* ------ *------* ------ *------* ------ *------* ------ *------
Design of Beam B7
Load for Moment= Load for Shear
By: Karim Sayed
7
75
1=. +1∑ ∗ ∗5∗5 2 1=5.8 + 5 ∗ 15.5.9 = 45.5.6 / /
Panelled Beams
Reinforced Concrete Design 2
S.F.D Using Imperial Calculations
B.M.D Using Imperial Calculations
Design of Critical Sections 1- Design of L-Sec
= = ℎ & = . . & = / 6+=6∗100+300=900 = 10 + = 0.8∗5000 + 3 0 0 = 7 0 0 10 { =5000 103.103.6∗10 =1 ∗ => 600= 1 25∗700 1=7. 8 →=0. 8 26 103. 6 ∗10 = = = 360∗0.826∗600 =581 25∗360√ 2525 ∗ 250 ∗ 600 = 563 > = 0.225∗√ ∗ = 0.225∗√ ℎ= ∗16581⁄4 ~4 => => = =/ = & = . & 126. 1 26. 7 ∗10 =1 ∗ => 600= 1 25∗300 1=4. 6 2 →=0. 8 21 ∗10 =714 = = = 360∗0.126.8726∗600 Check Minimum As
2- Design of R-SEC
By: Karim Sayed
7
76
Panelled Beams
Reinforced Concrete Design 2
25∗360√ 2525 ∗ 250 ∗ 600 = 563 = 0.225∗>√ ∗ = 0.225∗√ ℎ= ∗16714⁄.4 ~4∗ => => = ∗ = 0.76/ 6/22 < <0.0.98 / :5 8/8/ Check Minimum As
Check Shear::
)2016
()
(
.
-
–
2007
-
– )
(
.
-
)engineer-underconstruction.blogspot.com/p/obour.html(
By: Karim Sayed
-
7
7
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