Chapter 3 (rib)MOSTAFA
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Transcript of Chapter 3 (rib)MOSTAFA
Rib design
Rib design
Chapter 3
3.1 Introduction:
In reinforced concrete construction, slabs are used to provide flat, useful surface. A reinforced concrete slab is aboard, flat plate, usually horizontal, with top and bottom surfaces parallel or nearly so. It may be supported by reinforced concrete beams (and is usually cast monolithically with such beams), by masonry or reinforced concrete walls, by structural steel members, columns, or continuously by the ground.
The most common type in Jordan is the ribbed slabs (one way or two way). Ribs are one of the major and an important structural member in the slab, this importance due to its function in carrying loads, which is the first member in the structure, will carry loads and convert it to the beams which transfer it to columns, then columns dispose these loads by loading it on the foundation The solid slab technique is a good choice when there are very heavy loads or lots of cross piping etc in the floor or when extra fire resistance is required.
Figure (29): Side view for slab.
3.2 slab structural system
3.3 RIB 1:
Loading layout:
Shear Diagram
Moment Diagram
Steel Diagram
Design of R1:
Design of positive reinforcement at span1:
+Mu=14KN.m.
f c¿=25MPa
f y=420MPa
H = 300mm
bw = 150mm
tf =60mm
bf = effective width of the flange.
This width is determined using the ACI code
b f is the smallest of the following
bf = bw + 8 tf = 150+ 8× 60 =630 mm
bf = 550 mm. center line distance between ribs.
so bf =550 mm.
Effective depth (d)
Assume ds = 8mm and ∅ db=14mm
d = h−con .cover−ds−0.5db
d = 300−25−10−122
= 259 mm take d = 260
To determine wither the neutral axis falls in the flange or in the web, the moment due to compression force in the flange should be calculated.
C flange=0.85× f c' ×bf × tf
1000=0.85×25×550× 60
1000=701.25KN
T he moment caused frommt he flang Mf=∅ ×Cf (D−hf2 )×10−3
¿0.9×701.25(300−602 )×10−3=170.4KN .m
As Mu < Mf, the section well be designed as rectangular section.
Mu= 14 KN.m
β1=0.85−0.007× ( f c'−30 )
β1=0.85−0.007× (25−30 )=0.85
ρmax=0.85 β1 f c
'
f y× 3
8=0.85×0.85×25
420× 3
8=0.0161
M u=∅ f ybd2 ρ(1−0.95ρ
f yf c' )
Rh= Mu×106
0.9×f y×bf ×d2 =
14×106
0.9×420×550×2602=¿0.000996
Rh=ρ−0.59f yf c' ρ
2
0.000996=ρ−0.59 42025
ρ2
0.000996=ρ−9.912ρ2
ρ=0.001
As=0.001×550×260=143mm2
ρmin=14× √ f c'f y
=14× √25
420=0.00298,
ρmin=1.4f y
= 1.4420
=0.00333
Take ρ the minimum = 0.00333
Asmin=ρ×bw×d
¿0.00333×150×260=128.7mm2
So use 2Φ10 that provides 157mm 2 .
The positive steel in span 2
Mu=8 KN.mArea of steel= 128.7mm2
Use 2Φ10 that provides 157mm2
Design of the negative steel in span 1-2:
The ultimate positive moment=18.8KN.m.
f c¿=25MPa
f y=420MPa
H = 300mm
bw = 150mm
Effective depth (d)
Assume ds = 8mm and ∅ db=14mm
d = h−con .cover−ds−0.5db
d = 400−25−10−122
= 260 mm
β1=0.85−0.007× ( f c'−30 )
β1=0.85−0.007× (25−30 )=0.85
ρmax=0.85 β1 f c
'
f y× 3
8= 0.85×0.85×25
420× 3
8=0.0161
M u=∅ f ybd2 ρ(1−0.95ρ
f yf c' )
Rh= Mu×106
0.9×f y×bf ×d2 =
18.8×106
0.9×420×150×2602=¿0.0049
Rh=ρ−0.59f yf c' ρ
2
0.0049=ρ−0.59 42025
ρ2
0.0049=ρ−9.912ρ2
ρ=0.00516
ρmin=14× √ f c'f y
=14× √25
420=0.00298,
ρmin=1.4f y
= 1.4420
=0.00333
Take ρ = 0.00516
Asmin=ρ×bw×d
¿0.00516×150×260=201.24mm2
So use 2Φ12 that provides 226mm 2 .
3.4RIB 2:
Loading layout
Shear Diagram
Moment Diagram
Steel Diagram
Design of R2:
Design of positive reinforcement at span1:
+Mu=14KN.m.
f c¿=25MPa
f y=420MPa
H = 300mm
bw = 150mm
tf =60mm
bf = effective width of the flange.
This width is determined using the ACI code
b f is the smallest of the following
bf = bw + 8 tf = 150+ 8× 60 =630 mm
bf = 550 mm. center line distance between ribs.
so bf =550 mm.
Effective depth (d)
Assume ds = 8mm and ∅ db=14mm
d = h−con .cover−ds−0.5db
d = 300−25−10−122
= 259 mm take d = 260
To determine wither the neutral axis falls in the flange or in the web, the moment due to compression force in the flange should be calculated.
C flange=0.85× f c' ×bf × tf
1000=0.85×25×550× 60
1000=701.25KN
T he moment caused frommt he flang Mf=∅ ×Cf (D−hf2 )×10−3
¿0.9×701.25(300−602 )×10−3=170.4KN .m
As Mu < Mf, the section well be designed as rectangular section.
Mu= 14 KN.m
β1=0.85−0.007× ( f c'−30 )
β1=0.85−0.007× (25−30 )=0.85
ρmax=0.85 β1 f c
'
f y× 3
8=0.85×0.85×25
420× 3
8=0.0161
M u=∅ f ybd2 ρ(1−0.95ρ
f yf c' )
Rh= Mu×106
0.9×f y×bf ×d2 =
14×106
0.9×420×550×2602=¿0.000996
Rh=ρ−0.59f yf c' ρ
2
0.000996=ρ−0.59 42025
ρ2
0.000996=ρ−9.912ρ2
ρ=0.001
As=0.001×550×260=143mm2
ρmin=14× √ f c'f y
=14× √25
420=0.00298,
ρmin=1.4f y
= 1.4420
=0.00333
Take ρ the minimum = 0.00333
Asmin=ρ×bw×d
¿0.00333×150×260=128.7mm2
So use 2Φ10 that provides 157mm 2 .
The positive steel in span 2
Mu=6 KN.mArea of steel= 128.7mm2
Use 2Φ10 that provides 157mm2
Design of the negative steel in span 1-2:
The ultimate positive moment=18.45KN.m.
f c¿=25MPa
f y=420MPa
H = 300mm
bw = 150mm
Effective depth (d)
Assume ds = 8mm and ∅ db=14mm
d = h−con .cover−ds−0.5db
d = 400−25−10−122
= 260 mm
β1=0.85−0.007× ( f c'−30 )
β1=0.85−0.007× (25−30 )=0.85
ρmax=0.85 β1 f c
'
f y× 3
8= 0.85×0.85×25
420× 3
8=0.0161
M u=∅ f ybd2 ρ(1−0.95ρ
f yf c' )
Rh= Mu×106
0.9×f y×bf ×d2 =
18.45×106
0.9×420×150×2602=¿0.0048
Rh=ρ−0.59f yf c' ρ
2
0.0048=ρ−0.59 42025
ρ2
0.0048=ρ−9.912ρ2
ρ=0.00516
ρmin=14× √ f c'f y
=14× √25
420=0.00298,
ρmin=1.4f y
= 1.4420
=0.00333
Take ρ = 0.00516
Asmin=ρ×bw×d
¿0.00516×150×260=201.24mm2
So use 2Φ12 that provides 226mm 2 .
Design of the negative steel in span 2-3:
Mu=6 KN.mArea of steel= 128.7mm2
Use 2Φ10 that provides 157mm2
3.5RIB 3:
Loading layout
Shear Diagram
Moment Diagram
Steel Diagram
Design of R3:
Design of positive reinforcement at span3 and 4:
+Mu=15KN.m.
f c¿=25MPa
f y=420MPa
H = 300mm
bw = 150mm
tf =60mm
bf = effective width of the flange.
This width is determined using the ACI code
b f is the smallest of the following
bf = bw + 8 tf = 150+ 8× 60 =630 mm
bf = 550 mm. center line distance between ribs.
so bf =550 mm.
Effective depth (d)
Assume ds = 8mm and ∅ db=14mm
d = h−con .cover−ds−0.5db
d = 300−25−10−122
= 259 mm take d = 260
To determine wither the neutral axis falls in the flange or in the web, the moment due to compression force in the flange should be calculated.
C flange=0.85× f c' ×bf × tf
1000=0.85×25×550× 60
1000=701.25KN
T he moment caused frommt he flang Mf=∅ ×Cf (D−hf2 )×10−3
¿0.9×701.25(300−602 )×10−3=170.4KN .m
As Mu < Mf, the section well be designed as rectangular section.
Mu= 15 KN.m
β1=0.85−0.007× ( f c'−30 )
β1=0.85−0.007× (25−30 )=0.85
ρmax=0.85 β1 f c
'
f y× 3
8= 0.85×0.85×25
420× 3
8=0.0161
M u=∅ f ybd2 ρ(1−0.95ρ
f yf c' )
Rh= Mu×106
0.9×f y×bf ×d2 =
15×106
0.9×420×550×2602=¿0.001067
Rh=ρ−0.59f yf c' ρ
2
0.001067=ρ−0.59 42025
ρ2
0.001067=ρ−9.912ρ2
ρ=0.001
As=0.001×550×260=143mm2
ρmin=14× √ f c'f y
=14× √25
420=0.00298,
ρmin=1.4f y
= 1.4420
=0.00333
Take ρ the minimum = 0.00333
Asmin=ρ×bw×d
¿0.00333×150×260=128.7mm2
So use 2Φ10 that provides 157mm 2 .
The positive steel in span 2 and 5
Mu=6 KN.mArea of steel= 128.7mm2
Use 2Φ10 that provides 157mm2
Design of the negative steel in span 3-4:
The ultimate positive moment=28.69KN.m.
f c¿=25MPa
f y=420MPa
H = 300mm
bw = 150mm
Effective depth (d)
Assume ds = 8mm and ∅ db=14mm
d = h−con .cover−ds−0.5db
d = 400−25−10−122
= 260 mm
β1=0.85−0.007× ( f c'−30 )
β1=0.85−0.007× (25−30 )=0.85
ρmax=0.85 β1 f c
'
f y× 3
8= 0.85×0.85×25
420× 3
8=0.0161
M u=∅ f ybd2 ρ(1−0.95ρ
f yf c' )
Rh= Mu×106
0.9×f y×bf ×d2 =
28.69×106
0.9×420×150×2602=¿0.00746
Rh=ρ−0.59f yf c' ρ
2
0.00746=ρ−0.59 42025
ρ2
0.00746=ρ−9.912ρ2
ρ=0.00811
ρmin=14× √ f c'f y
=14× √25
420=0.00298,
ρmin=1.4f y
= 1.4420
=0.00333
Take ρ = 0.00811
Asmin=ρ×bw×d
¿0.00811×150×260=316mm2
So use 2Φ16 that provides 402mm 2 .
Design of the negative steel in span 2-3 and 4-5:
Mu=21 KN.mArea of steel= 220mm2
Use 2Φ12 that provides 226mm2
Design of the negative steel in span 1-2 and 5-6:
Mu=6 KN.mArea of steel= 128.7mm2
Use 2Φ10 that provides 157mm2
3.6RIB 4:
Loading layout
Shear Diagram
Moment Diagram
Steel Diagram
Design of R4:
Design of positive reinforcement at span2 and 3:
+Mu=15KN.m.
f c¿=25MPa
f y=420MPa
H = 300mm
bw = 150mm
tf =60mm
bf = effective width of the flange.
This width is determined using the ACI code
b f is the smallest of the following
bf = bw + 8 tf = 150+ 8× 60 =630 mm
bf = 550 mm. center line distance between ribs.
so bf =550 mm.
Effective depth (d)
Assume ds = 8mm and ∅ db=14mm
d = h−con .cover−ds−0.5db
d = 300−25−10−122
= 259 mm take d = 260
To determine wither the neutral axis falls in the flange or in the web, the moment due to compression force in the flange should be calculated.
C flange=0.85× f c' ×bf × tf
1000=0.85×25×550× 60
1000=701.25KN
T he moment caused frommt he flang Mf=∅ ×Cf (D−hf2 )×10−3
¿0.9×701.25(300−602 )×10−3=170.4KN .m
As Mu < Mf, the section well be designed as rectangular section.
Mu= 15 KN.m
β1=0.85−0.007× ( f c'−30 )
β1=0.85−0.007× (25−30 )=0.85
ρmax=0.85 β1 f c
'
f y× 3
8=0.85×0.85×25
420× 3
8=0.0161
M u=∅ f ybd2 ρ(1−0.95ρ
f yf c' )
Rh= Mu×106
0.9×f y×bf ×d2 =
15×106
0.9×420×550×2602=¿0.001067
Rh=ρ−0.59f yf c' ρ
2
0.001067=ρ−0.59 42025
ρ2
0.001067=ρ−9.912ρ2
ρ=0.001
As=0.001×550×260=143mm2
ρmin=14× √ f c'f y
=14× √25
420=0.00298,
ρmin=1.4f y
= 1.4420
=0.00333
Take ρ the minimum = 0.00333
Asmin=ρ×bw×d
¿0.00333×150×260=128.7mm2
So use 2Φ10 that provides 157mm 2 .
The positive steel in span 1 and 4
Mu=12 KN.mArea of steel= 128.7mm2
Use 2Φ10 that provides 157mm2
Design of the negative steel in span 2-3:
The ultimate positive moment=27.77KN.m.
f c¿=25MPa
f y=420MPa
H = 300mm
bw = 150mm
Effective depth (d)
Assume ds = 8mm and ∅ db=14mm
d = h−con .cover−ds−0.5db
d = 400−25−10−122
= 260 mm
β1=0.85−0.007× ( f c'−30 )
β1=0.85−0.007× (25−30 )=0.85
ρmax=0.85 β1 f c
'
f y× 3
8= 0.85×0.85×25
420× 3
8=0.0161
M u=∅ f ybd2 ρ(1−0.95ρ
f yf c' )
Rh= Mu×106
0.9×f y×bf ×d2 =
28.69×106
0.9×420×150×2602=¿0.00722
Rh=ρ−0.59f yf c' ρ
2
0.00722= ρ−0.59 42025
ρ2
0.00722= ρ−9.912 ρ2
ρ=0.00782
ρmin=14× √ f c'f y
=14× √25
420=0.00298,
ρmin=1.4f y
= 1.4420
=0.00333
Take ρ = 0.00782
Asmin=ρ×bw×d
¿0.00782×150×260=305mm2
So use 2Φ14 that provides 308mm 2 .
Design of the negative steel in span 1-2 and 3-4:
Mu=24 KN.mArea of steel= 250mm2
Use 2Φ14that provides 308mm2