1) Ribs are an important structural member in slabs that carry loads and transfer them to beams and columns. 2) The document provides details on the design of positive and negative reinforcement for two ribs (R1 and R2) in a slab. 3) The design includes calculating steel ratios and areas based on the ultimate moments, concrete properties, and code requirements. Reinforcement is selected to meet the calculated minimum area.

1. Rib
design
Ribdesign
Chapter 3

2. 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. 3.2 slab structural system

4. 3.3 RIB 1:
Loading layout:
Shear Diagram
Moment Diagram
Steel Diagram

5. 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 − 0.5𝑑 𝑏
d = 300−25 − 10 −
12
2
= 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 × 𝑓𝑐
′ × 𝑏𝑓 ×
𝑡𝑓
1000
= 0.85 × 25 × 550 ×
60
1000
= 701.25𝐾𝑁
𝑇ℎ𝑒 𝑚𝑜𝑚𝑒𝑛𝑡 𝑐𝑎𝑢𝑠𝑒𝑑 𝑓𝑟𝑜𝑚𝑚 𝑡ℎ𝑒 𝑓𝑙𝑎𝑛𝑔 𝑀𝑓 = ∅ × 𝐶𝑓 (𝐷 −
ℎ𝑓
2
) × 10−3

6. = 0.9 × 701.25 (300 −
60
2
) × 10−3
= 170.4 𝐾𝑁. 𝑚
As Mu < Mf, the section well be designed as rectangular section.
Mu= 14 KN.m
𝛽1 = 0.85 − 0.007 × ( 𝑓𝑐
′
− 30)
𝛽1 = 0.85 − 0.007 × (25 − 30) = 0.85
𝜌 𝑚𝑎𝑥 =
0.85𝛽1 𝑓𝑐
′
𝑓𝑦
×
3
8
=
0.85 × 0.85 × 25
420
×
3
8
= 0.0161
𝑀 𝑢 = ∅𝑓𝑦 𝑏𝑑2
𝜌(1 − 0.95𝜌
𝑓𝑦
𝑓𝑐
′
)
𝑅ℎ =
𝑀𝑢×106
0.9×𝑓𝑦×𝑏𝑓×𝑑2 =
14×106
0.9×420 ×550×2602 =0.000996
𝑅ℎ = 𝜌 − 0.59
𝑓𝑦
𝑓𝑐
′ 𝜌2
0.000996 = 𝜌 − 0.59
420
25
𝜌2
0.000996 = 𝜌 − 9.912𝜌2
𝜌 = 0.001
As=0.001 × 550 × 260 = 143𝑚𝑚2
𝜌 𝑚𝑖𝑛 =
1
4
×
√ 𝑓𝑐
′
𝑓𝑦
=
1
4
×
√25
420
= 0.00298,
𝜌 𝑚𝑖𝑛 =
1.4
𝑓𝑦
=
1.4
420
= 0.00333
Take 𝜌 the minimum = 0.00333
As min = 𝜌 × 𝑏 𝑤 × 𝑑
= 0.00333× 150 × 260 = 128.7𝑚𝑚2
So use 2Φ10 that provides 157mm2.

7.  The positive steel in span 2
Mu=8 KN.m
Area 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 − 0.5𝑑 𝑏
d = 400−25 − 10 −
12
2
= 260 mm
𝛽1 = 0.85 − 0.007 × ( 𝑓𝑐
′
− 30)
𝛽1 = 0.85 − 0.007 × (25 − 30) = 0.85
𝜌 𝑚𝑎𝑥 =
0.85𝛽1 𝑓𝑐
′
𝑓𝑦
×
3
8
=
0.85 × 0.85 × 25
420
×
3
8
= 0.0161
𝑀 𝑢 = ∅𝑓𝑦 𝑏𝑑2
𝜌(1 − 0.95𝜌
𝑓𝑦
𝑓𝑐
′
)
𝑅ℎ =
𝑀𝑢×106
0.9×𝑓𝑦×𝑏𝑓×𝑑2 =
18.8×106
0.9×420 ×150×2602 =0.0049

8. 𝑅ℎ = 𝜌 − 0.59
𝑓𝑦
𝑓𝑐
′ 𝜌2
0.0049 = 𝜌 − 0.59
420
25
𝜌2
0.0049 = 𝜌 − 9.912𝜌2
𝜌 = 0.00516
𝜌 𝑚𝑖𝑛 =
1
4
×
√ 𝑓𝑐
′
𝑓𝑦
=
1
4
×
√25
420
= 0.00298,
𝜌 𝑚𝑖𝑛 =
1.4
𝑓𝑦
=
1.4
420
= 0.00333
Take 𝜌 = 0.00516
As min = 𝜌 × 𝑏 𝑤 × 𝑑
= 0.00516× 150 × 260 = 201.24𝑚𝑚2
So use 2Φ12 that provides 226mm2.

9. 3.4RIB 2:
Loading layout
Shear Diagram
Moment Diagram
Steel Diagram

10. 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 − 0.5𝑑 𝑏
d = 300−25 − 10 −
12
2
= 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 × 𝑓𝑐
′ × 𝑏𝑓 ×
𝑡𝑓
1000
= 0.85 × 25 × 550 ×
60
1000
= 701.25𝐾𝑁
𝑇ℎ𝑒 𝑚𝑜𝑚𝑒𝑛𝑡 𝑐𝑎𝑢𝑠𝑒𝑑 𝑓𝑟𝑜𝑚𝑚 𝑡ℎ𝑒 𝑓𝑙𝑎𝑛𝑔 𝑀𝑓 = ∅ × 𝐶𝑓 (𝐷 −
ℎ𝑓
2
) × 10−3

11. = 0.9 × 701.25 (300 −
60
2
) × 10−3
= 170.4 𝐾𝑁. 𝑚
As Mu < Mf, the section well be designed as rectangular section.
Mu= 14 KN.m
𝛽1 = 0.85 − 0.007 × ( 𝑓𝑐
′
− 30)
𝛽1 = 0.85 − 0.007 × (25 − 30) = 0.85
𝜌 𝑚𝑎𝑥 =
0.85𝛽1 𝑓𝑐
′
𝑓𝑦
×
3
8
=
0.85 × 0.85 × 25
420
×
3
8
= 0.0161
𝑀 𝑢 = ∅𝑓𝑦 𝑏𝑑2
𝜌(1 − 0.95𝜌
𝑓𝑦
𝑓𝑐
′
)
𝑅ℎ =
𝑀𝑢×106
0.9×𝑓𝑦×𝑏𝑓×𝑑2 =
14×106
0.9×420 ×550×2602 =0.000996
𝑅ℎ = 𝜌 − 0.59
𝑓𝑦
𝑓𝑐
′ 𝜌2
0.000996 = 𝜌 − 0.59
420
25
𝜌2
0.000996 = 𝜌 − 9.912𝜌2
𝜌 = 0.001
As=0.001 × 550 × 260 = 143𝑚𝑚2
𝜌 𝑚𝑖𝑛 =
1
4
×
√ 𝑓𝑐
′
𝑓𝑦
=
1
4
×
√25
420
= 0.00298,
𝜌 𝑚𝑖𝑛 =
1.4
𝑓𝑦
=
1.4
420
= 0.00333
Take 𝜌 the minimum = 0.00333
As min = 𝜌 × 𝑏 𝑤 × 𝑑
= 0.00333× 150 × 260 = 128.7𝑚𝑚2
So use 2Φ10 that provides 157mm2.

12.  The positive steel in span2
Mu=6 KN.m
Area 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 − 0.5𝑑 𝑏
d = 400−25 − 10 −
12
2
= 260 mm
𝛽1 = 0.85 − 0.007 × ( 𝑓𝑐
′
− 30)
𝛽1 = 0.85 − 0.007 × (25 − 30) = 0.85
𝜌 𝑚𝑎𝑥 =
0.85𝛽1 𝑓𝑐
′
𝑓𝑦
×
3
8
=
0.85 × 0.85 × 25
420
×
3
8
= 0.0161
𝑀 𝑢 = ∅𝑓𝑦 𝑏𝑑2
𝜌(1 − 0.95𝜌
𝑓𝑦
𝑓𝑐
′
)
𝑅ℎ =
𝑀𝑢×106
0.9×𝑓𝑦×𝑏𝑓×𝑑2 =
18.45 ×106
0.9×420 ×150×2602 =0.0048

13. 𝑅ℎ = 𝜌 − 0.59
𝑓𝑦
𝑓𝑐
′ 𝜌2
0.0048 = 𝜌 − 0.59
420
25
𝜌2
0.0048 = 𝜌 − 9.912𝜌2
𝜌 = 0.00516
𝜌 𝑚𝑖𝑛 =
1
4
×
√ 𝑓𝑐
′
𝑓𝑦
=
1
4
×
√25
420
= 0.00298,
𝜌 𝑚𝑖𝑛 =
1.4
𝑓𝑦
=
1.4
420
= 0.00333
Take 𝜌 = 0.00516
As min = 𝜌 × 𝑏 𝑤 × 𝑑
= 0.00516× 150 × 260 = 201.24𝑚𝑚2
So use 2Φ12 that provides 226mm2.
 Design of the negative steel in span 2-3:
Mu=6 KN.m
Area of steel= 128.7mm2
Use 2Φ10 that provides 157mm2

14. 3.5RIB 3:
Loading layout
Shear Diagram
Moment Diagram
Steel Diagram

15. 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 − 0.5𝑑 𝑏
d = 300−25 − 10 −
12
2
= 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 × 𝑓𝑐
′ × 𝑏𝑓 ×
𝑡𝑓
1000
= 0.85 × 25 × 550 ×
60
1000
= 701.25𝐾𝑁
𝑇ℎ𝑒 𝑚𝑜𝑚𝑒𝑛𝑡 𝑐𝑎𝑢𝑠𝑒𝑑 𝑓𝑟𝑜𝑚𝑚 𝑡ℎ𝑒 𝑓𝑙𝑎𝑛𝑔 𝑀𝑓 = ∅ × 𝐶𝑓 (𝐷 −
ℎ𝑓
2
) × 10−3

16. = 0.9 × 701.25 (300 −
60
2
) × 10−3
= 170.4 𝐾𝑁. 𝑚
As Mu < Mf, the section well be designed as rectangular section.
Mu= 15 KN.m
𝛽1 = 0.85 − 0.007 × ( 𝑓𝑐
′
− 30)
𝛽1 = 0.85 − 0.007 × (25 − 30) = 0.85
𝜌 𝑚𝑎𝑥 =
0.85𝛽1 𝑓𝑐
′
𝑓𝑦
×
3
8
=
0.85 × 0.85 × 25
420
×
3
8
= 0.0161
𝑀 𝑢 = ∅𝑓𝑦 𝑏𝑑2
𝜌(1 − 0.95𝜌
𝑓𝑦
𝑓𝑐
′
)
𝑅ℎ =
𝑀𝑢×106
0.9×𝑓𝑦×𝑏𝑓×𝑑2 =
15×106
0.9×420 ×550×2602 =0.001067
𝑅ℎ = 𝜌 − 0.59
𝑓𝑦
𝑓𝑐
′ 𝜌2
0.001067 = 𝜌 − 0.59
420
25
𝜌2
0.001067 = 𝜌 − 9.912𝜌2
𝜌 = 0.001
As=0.001 × 550 × 260 = 143𝑚𝑚2
𝜌 𝑚𝑖𝑛 =
1
4
×
√ 𝑓𝑐
′
𝑓𝑦
=
1
4
×
√25
420
= 0.00298,
𝜌 𝑚𝑖𝑛 =
1.4
𝑓𝑦
=
1.4
420
= 0.00333
Take 𝜌 the minimum = 0.00333
As min = 𝜌 × 𝑏 𝑤 × 𝑑
= 0.00333× 150 × 260 = 128.7𝑚𝑚2
So use 2Φ10 that provides 157mm2.

17.  The positive steel in span2 and 5
Mu=6 KN.m
Area 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 − 0.5𝑑 𝑏
d = 400−25 − 10 −
12
2
= 260 mm
𝛽1 = 0.85 − 0.007 × ( 𝑓𝑐
′
− 30)
𝛽1 = 0.85 − 0.007 × (25 − 30) = 0.85
𝜌 𝑚𝑎𝑥 =
0.85𝛽1 𝑓𝑐
′
𝑓𝑦
×
3
8
=
0.85 × 0.85 × 25
420
×
3
8
= 0.0161
𝑀 𝑢 = ∅𝑓𝑦 𝑏𝑑2
𝜌(1 − 0.95𝜌
𝑓𝑦
𝑓𝑐
′
)
𝑅ℎ =
𝑀𝑢×106
0.9×𝑓𝑦×𝑏𝑓×𝑑2 =
28.69×106
0.9×420 ×150×2602 =0.00746

18. 𝑅ℎ = 𝜌 − 0.59
𝑓𝑦
𝑓𝑐
′ 𝜌2
0.00746 = 𝜌 − 0.59
420
25
𝜌2
0.00746 = 𝜌 − 9.912𝜌2
𝜌 = 0.00811
𝜌 𝑚𝑖𝑛 =
1
4
×
√ 𝑓𝑐
′
𝑓𝑦
=
1
4
×
√25
420
= 0.00298,
𝜌 𝑚𝑖𝑛 =
1.4
𝑓𝑦
=
1.4
420
= 0.00333
Take 𝜌 = 0.00811
As min = 𝜌 × 𝑏 𝑤 × 𝑑
= 0.00811× 150 × 260 = 316𝑚𝑚2
So use 2Φ16 that provides 402mm2.
 Design of the negative steel in span 2-3 and 4-5:
Mu=21 KN.m
Area of steel= 220mm2
Use 2Φ12 that provides 226mm2
 Design of the negative steel in span 1-2 and 5-6:
Mu=6 KN.m
Area of steel= 128.7mm2
Use 2Φ10 that provides 157mm2

19. 3.6RIB 4:
Loading layout
Shear Diagram
Moment Diagram
Steel Diagram

20. 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 − 0.5𝑑 𝑏
d = 300−25 − 10 −
12
2
= 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 × 𝑓𝑐
′ × 𝑏𝑓 ×
𝑡𝑓
1000
= 0.85 × 25 × 550 ×
60
1000
= 701.25𝐾𝑁
𝑇ℎ𝑒 𝑚𝑜𝑚𝑒𝑛𝑡 𝑐𝑎𝑢𝑠𝑒𝑑 𝑓𝑟𝑜𝑚𝑚 𝑡ℎ𝑒 𝑓𝑙𝑎𝑛𝑔 𝑀𝑓 = ∅ × 𝐶𝑓 (𝐷 −
ℎ𝑓
2
) × 10−3

21. = 0.9 × 701.25 (300 −
60
2
) × 10−3
= 170.4 𝐾𝑁. 𝑚
As Mu < Mf, the section well be designed as rectangular section.
Mu= 15 KN.m
𝛽1 = 0.85 − 0.007 × ( 𝑓𝑐
′
− 30)
𝛽1 = 0.85 − 0.007 × (25 − 30) = 0.85
𝜌 𝑚𝑎𝑥 =
0.85𝛽1 𝑓𝑐
′
𝑓𝑦
×
3
8
=
0.85 × 0.85 × 25
420
×
3
8
= 0.0161
𝑀 𝑢 = ∅𝑓𝑦 𝑏𝑑2
𝜌(1 − 0.95𝜌
𝑓𝑦
𝑓𝑐
′
)
𝑅ℎ =
𝑀𝑢×106
0.9×𝑓𝑦×𝑏𝑓×𝑑2 =
15×106
0.9×420 ×550×2602 =0.001067
𝑅ℎ = 𝜌 − 0.59
𝑓𝑦
𝑓𝑐
′ 𝜌2
0.001067 = 𝜌 − 0.59
420
25
𝜌2
0.001067 = 𝜌 − 9.912𝜌2
𝜌 = 0.001
As=0.001 × 550 × 260 = 143𝑚𝑚2
𝜌 𝑚𝑖𝑛 =
1
4
×
√ 𝑓𝑐
′
𝑓𝑦
=
1
4
×
√25
420
= 0.00298,
𝜌 𝑚𝑖𝑛 =
1.4
𝑓𝑦
=
1.4
420
= 0.00333
Take 𝜌 the minimum = 0.00333
As min = 𝜌 × 𝑏 𝑤 × 𝑑
= 0.00333× 150 × 260 = 128.7𝑚𝑚2
So use 2Φ10 that provides 157mm2.

22.  The positive steel in span 1 and 4
Mu=12 KN.m
Area 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 − 0.5𝑑 𝑏
d = 400−25 − 10 −
12
2
= 260 mm
𝛽1 = 0.85 − 0.007 × ( 𝑓𝑐
′
− 30)
𝛽1 = 0.85 − 0.007 × (25 − 30) = 0.85
𝜌 𝑚𝑎𝑥 =
0.85𝛽1 𝑓𝑐
′
𝑓𝑦
×
3
8
=
0.85 × 0.85 × 25
420
×
3
8
= 0.0161
𝑀 𝑢 = ∅𝑓𝑦 𝑏𝑑2
𝜌(1 − 0.95𝜌
𝑓𝑦
𝑓𝑐
′
)
𝑅ℎ =
𝑀𝑢×106
0.9×𝑓𝑦×𝑏𝑓×𝑑2 =
28.69×106
0.9×420 ×150×2602 =0.00722

23. 𝑅ℎ = 𝜌 − 0.59
𝑓𝑦
𝑓𝑐
′ 𝜌2
0.00722 = 𝜌 − 0.59
420
25
𝜌2
0.00722 = 𝜌 − 9.912𝜌2
𝜌 = 0.00782
𝜌 𝑚𝑖𝑛 =
1
4
×
√ 𝑓𝑐
′
𝑓𝑦
=
1
4
×
√25
420
= 0.00298,
𝜌 𝑚𝑖𝑛 =
1.4
𝑓𝑦
=
1.4
420
= 0.00333
Take 𝜌 = 0.00782
As min = 𝜌 × 𝑏 𝑤 × 𝑑
= 0.00782× 150 × 260 = 305𝑚𝑚2
So use 2Φ14 that provides 308mm2.
 Design of the negative steel in span 1-2 and 3-4:
Mu=24 KN.m
Area of steel= 250mm2
Use 2Φ14that provides 308mm2