Similar to One way slab design by Sabuj Chowdhury, Lecturer, Department of CIvil Engineering, Ahsanullah University Of Science and Technology AUST (20)
One way slab design by Sabuj Chowdhury, Lecturer, Department of CIvil Engineering, Ahsanullah University Of Science and Technology AUST
1. DESIGN OF ONE-WAY SLAB
Sabuj Chowdhury
Lecturer
Department of Civil Engineering
Ahsanullah University of Science and Technology
2. One Way Slab Design
Sabuj Chowdhury, Lecturer, Department of CE, AUST
Plan view of one-way slab may be---
1. Supported on two opposite edges (Shown in Fig 01)
2. Supported on all edges (L/B ≥ 2) (Shown in Fig 02)
Load Transfer
LoadTransfer
Beam
Beam
Beam
Beam
Beam
Fig 01 : One-way Slab
In Fig 01 slab is supported on two opposite sides only. In this case the structural
action of the slab is essentially one way.
3. One Way Slab Design
Sabuj Chowdhury, Lecturer, Department of CE, AUST
Fig 02 : One-way Slab
In Fig 02 there are beams on all four sides with a intermediate beam. Now if L & B ratio
is 2 or greater, slab is one way even though supports are provided on all sides.
L/B ≥ 2
4. One Way Slab Design
Sabuj Chowdhury, Lecturer, Department of CE, AUST
One way slab can be
1. Solid
2. Hollow or
3. Ribbed
One-way Slab
HollowSolid Ribbed
5. Design Drawing
Sabuj Chowdhury, Lecturer, Department of CE, AUST
12‘
12‘
6‘
27‘27‘
Load Transfer in Shorter Direction
L/B=2.25
L/B=2.25L/B=2.25
L/B=2.25
Support in one
direction only
1
1
6. Design Data
Sabuj Chowdhury, Lecturer, Department of CE, AUST
Note:
Here it is the Example Data. Students have to analyse and design the slab by their own
data, which are supplied in class where only A is unknown where, A = (Student No.+8) ft
Design Data
Dimension, A 12 ft (Student No. :13.02.03.004)
Strength of Concrete, 3 ksi
Strength of Steel, 𝑓𝑦 40 ksi
Live Load on Floor 60 psf
Floor Finish(FF) 25 psf
𝑓𝑐
‘
7. One Way Slab Design
Step 01 : Determination of Thickness (From consideration of Deflection)
To control deflection, ACI Code 9.5.2.1 specifies minimum thickness values
for one-way solid slabs.
Here, l is the clear span
Multiplying Factor = 0.4 +
𝑓𝑦
100
; 𝑤𝑒𝑟𝑒, 𝑓𝑦 in ksi
If ,
Thickness < 6 inch then upper rounding is 0.25
Thickness ≥ 6 inch then upper rounding is 0.50
t = 5.2 in = 5.25 in
t = 5.3 in = 5.50 in
t = 5.6 in = 5.75 in
t = 5.8 in = 6.00 in
t = 6.2 in = 6.50 in
t = 6.3 in = 6.50 in
t = 6.6 in = 7.00 in
t = 6.8 in = 7.00 in
Sabuj Chowdhury, Lecturer, Department of CE, AUST
8. Example
10” Wall 10” Wall
12” Beam
12 ft 12 ft 6 ft
One end Continuous Both end Continuous Cantilever
𝑙 = 12 × 12 −
10
2
−
12
2
𝑙 = 133 in
𝑙 = 12 × 12 −
12
2
−
10
2
𝑙 = 133 in
𝑙 = 6 × 12 −
10
2
𝑙 = 67 in
M F = 0.4 +
𝑓𝑦
100
= 0.8As, 𝑓𝑦 = 40 ksi; MF ≠ 1
𝑡 𝑚𝑎𝑥 = 5.36 in = 5.5 in (upper-rounding in 0.25 as 𝑡 𝑚𝑎𝑥< 6 in)
t =
𝑙
24
× 𝑀𝐹 = 4.433𝑖𝑛 t =
𝑙
28
× 𝑀𝐹 = 3.80𝑖𝑛
t =
𝑙
10
× 𝑀𝐹
= 5.36𝑖𝑛
Sabuj Chowdhury, Lecturer, Department of CE, AUST
Section 1-1
9. One Way Slab Design
Step 02 : Load Calculation
Dead Load
Self Weight
Super-imposed
Dead Load
FF (Floor)
LC (Roof)
PW (Floor)
No PW (Roof)
Live Load
Total Load, 𝑤 𝑢 = 1.2 × 𝐷𝐿 + 1.6 × 𝐿𝐿
Sabuj Chowdhury, Lecturer, Department of CE, AUST
10. Example
Dead Load = Self Wt.+ FF(Floor) or LC(Roof)+PW(if Given)
Self Wt. =
𝑡
12
× 150 psf
=
5.5
12
× 150 𝑝𝑠𝑓
= 𝟔𝟖. 𝟕𝟓 𝒑𝒔𝒇
t is found in Step 01
FF = 25 psf (Given)
PW = 0 psf (As not given)
LL = 60 psf (Given for floor)
For Floor,
DL = Self Wt.+ FF(Floor) +PW(if Given)
= (68.75 + 25 + 0) psf
= 93.75 psf
LL = 𝟔𝟎 𝒑𝒔𝒇
Total Load, 𝑤 𝑢 = 1.2 × 𝐷𝐿 + 1.6 × 𝐿𝐿
= (1.2 × 93.75 + 1.6 × 60) psf
= 208.5 𝑝𝑠𝑓
= 0.2085 𝑘𝑠𝑓
≈ 𝟎. 𝟐𝟏 𝒌𝒔𝒇
Sabuj Chowdhury, Lecturer, Department of CE, AUST
11. One Way Slab Design
Step 03 : Moment Calculation
Sabuj Chowdhury, Lecturer, Department of CE, AUST
12. 12 ft 12 ft 6 ft
Moment Coefficients are given below :
0
1
11
1
12
1
12
1
11
1
11
1
2
𝑙 = 133 in
= 11.083 ft
𝑙 = 67 in
= 5.583 ft
𝑙 = 133 in
= 11.083 ft
From Step 01
Example
Sabuj Chowdhury, Lecturer, Department of CE, AUST
21. Example
12 ft 12 ft 6 ft
ckd bar remain at top when negative moment
ckd bar remain at bottom when positive moment
Sabuj Chowdhury, Lecturer, Department of CE, AUST
28. Step 07 : Detailing
One Way Slab Design
Sabuj Chowdhury, Lecturer, Department of CE, AUST
29. Example
Detailing in Plan View/Top View
a) #3 @ 7” c/c alt. ckd.
b) 1 #3 extra top between ckd. bar
c) 1 #4 extra top between ckd. bar
d) #3 @ 10” c/c
a
b
c
d t = 5.5 inch
Sabuj Chowdhury, Lecturer, Department of CE, AUST
1
1
2 2
30. #3 @ 10” c/c
12 ft 6 ft
#3 @ 7” c/c alt. ckd.
1 #3 extra top 1 #4 extra top
12 ft
𝑙 = 133 in
= 11.083 ft
𝑙 = 67 in
= 5.583 ft
𝑙 = 133 in
= 11.083 ft
1’-7” 2’-10” 2’-10” 2’-10”
3’-9” 3’-9” 3’-9”
𝑆𝑒𝑐𝑡𝑖𝑜𝑛 1 − 1
Example
Sabuj Chowdhury, Lecturer, Department of CE, AUST