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)



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


2

0.0049=ρ−0.59 42025

ρ2

0.0049=ρ−9.912ρ2

ρ=0.00516


=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


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 = bw + 8 tf = 150+ 8× 60 =630 mm


so bf =550 mm.

Effective depth (d)



d = 300−25−10−122

= 259 mm take d = 260



1000=0.85×25×550× 60

1000=701.25KN


¿0.9×701.25(300−602 )×10−3=170.4KN .m


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


2

0.000996=ρ−0.59 42025

ρ2

0.000996=ρ−9.912ρ2

ρ=0.001

As=0.001×550×260=143mm2


=14× √25

420=0.00298,

ρmin=1.4f y

= 1.4420

=0.00333


Asmin=ρ×bw×d

¿0.00333×150×260=128.7mm2


The positive steel in span 2





f c¿=25MPa

f y=420MPa

H = 300mm

bw = 150mm

Effective depth (d)



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


2

0.0048=ρ−0.59 42025

ρ2

0.0048=ρ−9.912ρ2

ρ=0.00516


=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





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 = bw + 8 tf = 150+ 8× 60 =630 mm


so bf =550 mm.

Effective depth (d)



d = 300−25−10−122

= 259 mm take d = 260



1000=0.85×25×550× 60

1000=701.25KN


¿0.9×701.25(300−602 )×10−3=170.4KN .m


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


2

0.001067=ρ−0.59 42025

ρ2

0.001067=ρ−9.912ρ2

ρ=0.001

As=0.001×550×260=143mm2


=14× √25

420=0.00298,

ρmin=1.4f y

= 1.4420

=0.00333


Asmin=ρ×bw×d

¿0.00333×150×260=128.7mm2


The positive steel in span 2 and 5





f c¿=25MPa

f y=420MPa

H = 300mm

bw = 150mm

Effective depth (d)



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


2

0.00746=ρ−0.59 42025

ρ2

0.00746=ρ−9.912ρ2

ρ=0.00811


=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


Design of the negative steel in span 2-3 and 4-5:

Mu=21 KN.mArea of steel= 220mm2





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 = bw + 8 tf = 150+ 8× 60 =630 mm


so bf =550 mm.

Effective depth (d)



d = 300−25−10−122

= 259 mm take d = 260



1000=0.85×25×550× 60

1000=701.25KN


¿0.9×701.25(300−602 )×10−3=170.4KN .m


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


2

0.001067=ρ−0.59 42025

ρ2

0.001067=ρ−9.912ρ2

ρ=0.001

As=0.001×550×260=143mm2


=14× √25

420=0.00298,

ρmin=1.4f y

= 1.4420

=0.00333


Asmin=ρ×bw×d

¿0.00333×150×260=128.7mm2


The positive steel in span 1 and 4





f c¿=25MPa

f y=420MPa

H = 300mm

bw = 150mm

Effective depth (d)



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


2

0.00722= ρ−0.59 42025

ρ2

0.00722= ρ−9.912 ρ2

ρ=0.00782


=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



Mu=24 KN.mArea of steel= 250mm2

Use 2Φ14that provides 308mm2

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Chapter 3 (rib)MOSTAFA