1. The document describes the design of a one-way slab using moment coefficients from a figure based on certain conditions being met, such as approximately equal spans and uniformly distributed loads. 2. Limitations for one-way slab design according to the ACI code are provided, such as minimum thickness, shear check, concrete cover, reinforcement spacing, and requirements for shrinkage and temperature reinforcement. 3. An example problem is worked through showing the design of a simply supported one-way slab for a given live and dead load, using the provided moment coefficients and checking reinforcement, shear, and capacity.

1. One‐way slab
One way slab

3. 1m
1m

4. Design Moment using Coefficients
Design Moment using Coefficients
The conditions under which the moment coefficients for
continuous beams and slabs given in Fig. 9.3 should be used
can be summarized as follows:
1. Spans are approximately equal: Longer span  1.2 (shorter span)
1. Spans are approximately equal: Longer span  1.2 (shorter span)
2. Loads are uniformly distributed
3. The ratio (live load/dead load) is less than or equal to 3
4 F l b ith l th l t 3 3 ti b di t
4. For slabs with spans less than or equal to 3.3m, negative bending moment
at face of all supports is (1/12) wu l2
n
5. For an unrestrained discontinuous end at A, the coefficient is 0 at A and
(+1/11) t B
(+1/11) at B.
6. Shearing force at C is (1.15 wu ln /2) and at the face of all the support is
(wu ln /2)
2
7. Mu = (coefficient)(wu l2
n) and ln = clear span

6. Design Limitations According to the ACI Code
1. A typical imaginary strip 1m wide is assumed.

9. Design Limitations According to the ACI Code
2. The minimum thickness of one-way slabs using grade
400MPa steel are:

10. Design Limitations According to the ACI Code
3. It is preferable to choose slab depth to the nearest 10mm.
4 Shear should be checked although it does not usually
4. Shear should be checked, although it does not usually
control.
5. Concrete cover in slabs shall not be less than 20mm at
surfaces not expose to weather or ground. In this case,
d = h – 20mm – (half-bar diameter)
6. The minimum amount of reinforcement shall not be less
than that required for shrinkage and temperature
reinforcement.
reinforcement.
7. The maximum spacing of principal reinforcement is:
Smax = min (3h, 450mm)
max ( , )

11. Temperature and Shrinkage Reinforcement
8. Reinforcement for shrinkage and temperature stresses
normal to the principal reinforcement should be provided.
p p p
9. For 280-350 MPa steel  steel ratio sh = 0.2% (sh = bh)
For 400 MPa steel  steel ratio sh = 0.18%
sh
10. Maximum spacing of shrinkage and temperature steel:
Smax = min(5h, 450mm)

12. Reinforcement Details
In continuous one-way slabs, the steel area of the main
reinforcement is calculated for all critical sections, at
,
midspans, and at supports. The choice of bar diameter and
detailing depends mainly on the steel areas, spacing
i t d d l t l th
requirements, and development length.

13. Example
Design a 3.65m simply supported slab to carry a uniform dead
load (excluding self-weight) of 5.7kN/m2 and a uniform live
l d f 4 8kN/ 2 U f’ 21MP d f 400MP
load of 4.8kN/m2. Use f’c = 21MPa and fy = 400MPa.
Solution
Solution
1. Assume a slab thickness. For fy = 400MPa, the minimum depth
to control deflection is L/20 = 3.65/20 = 0.183m. Assume a
total depth of h = 190 mm and assume d = 160 mm.
2. Calculate factored load: weight of slab = 0.1924 = 4.56kN/m2
W 1 2DL 1 6LL 1 2(5 7 4 56) 1 6(4 8) 20 kN/ 2
Wu = 1.2DL + 1.6LL = 1.2(5.7 + 4.56) + 1.6(4.8) = 20 kN/m2
For a 1-m width of slab, Mu = WuL2/8
M = 20  3 652/8 = 33 3 kN m
Mu = 20  3.65 /8 = 33.3 kN-m

14. Solution
3 C l l A F M 33 3kN b 1000 d 160
3. Calculate As : For Mu = 33.3kN-m, b =1000mm, d = 160mm
2
0.85 4
1 1 0.00377
1 7
c u
f M
f f bd


 

   
 

 
 
1.7
y c
f f bd

 
 
max 1
0.003
0.85 0.0142
0 003 0 005
c
f
f
 
  
 
 

 
0.003 0.005
y
f 
 
1.4
; 31
c
y
f MPa
f
  


min max
; 31
4
y
c
c
y
f
f MPa
f


 

  



min
1.4
0.0035
400
  
min max OK!
  
  

15. Solution
3
3.
Use 8DB10 => A = 628 mm2
2
603.2
s
A bd mm

  
Use 8DB10 => As = 628 mm2
4. Check the moment capacity of the final section
628 400
s y
A f
a mm

0.85 0.85 21 1000
y
c
a mm
f b
     
  
2
n s y
a
M A f d
 
 
 
 
 
5. Calculate the secondary (shrinkage) reinforcement normal to the
main steel
6 Ch k h i t
2
 
6. Check shear requirements:
Vu at a distance d from the support = Wu (L/2 – d) = 33.3kN

16. Example
Design continuous slab and draw a detailed section. The
p
dead load on the slabs (DL) = 4.2 kN/m2 + Self-weight of
slabs. The live load on the slabs (LL) = 7.5 kN/m2
Gi f’ 25 MP d f 400 MP
•The cross section of a continuous one-way solid
l b i b ildi i h i Fi
Given: f’c = 25 MPa and fy = 400 MPa.
slab in a building is shown in Figure.
20cm 360cm 20cm
20cm
380cm