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Reinforced Concrete Design

One-Way Slabs
Types of Slabs
Load Paths & Flaming Concepts
One-Way Slab
One-Way Joist
Slab on Grade
Mongkol JIRAVACHARADET

SURANAREE

INSTITUTE OF ENGINEERING

UNIVERSITY OF TECHNOLOGY

SCHOOL OF CIVIL ENGINEERING

Types of Slab

One-way slab

Flat plate slab

One-way slab

Flat slab

Two-way slab

Grid slab
Load Path / Framing Possibilities

Ln = 3.2 m

Ln = 4.4 m

Ln = 3.6 m
Ln = 8.2 m

Think we’ll need some additional framing members???

Framing Concepts
Let’s use a simple example
for our discussion…
Column spacing 8 m c-c

Think about relating it to your
design project.

Plan
Framing Concepts
We can first assume that
we’ll have major girders
running in one direction
in our one-way system

Framing Concepts
We can first assume that
we’ll have major girders
running in one direction
in our one-way system
If we span between girders
with our slab, then we have
a load path, but if the spans
are too long…
Framing Concepts
We will need to shorten up
the span with additional beams
But we need to support the
load from these new beams,
so we will need additional
supporting members

Framing Concepts
Now let’s go back through with
a slightly different load path.
We again assume that we’ll
have major girders running in
one direction in our one-way
system.
This time, let’s think about
shortening up the slab span by
running beams into our girders.
Our one-way slab will transfer our load to the beams.
Two Load Path Options

Framing Concepts - Considerations
For your structure:
Look for a “natural” load path
Identify which column lines are best suited to having
major framing members (i.e. girders)
Assume walls are not there for structural support, but
consider that the may help you in construction (forming)
Example
Condo Floor Plan

ก

One-Way Slab :

ก

ก
: F

S

(L) >

F

F

S
L

S

t

L

Simple supports
on two long
edges only

S

(S)

S

การแอนตัวเกิดขึ้นบนดานสั้น
ก

ก

ก

ก

ˈ

:

กF
S

L

1m

t

F
m
1.0

ก
ก

ก F : F

ก: F
S
Sn

F

F

F 10-15

.

F

ก 2-3

F

.

F
F
L / 20

ก
F
Ln

F

F

L / 24

Ln

L / 28

F

F
Ln

L / 10
Ln
ก
ก

˂ ก ก
ก
F

ก
As

F

(
ก

ก

ก F )

Ag : As/Ag

1m

RB24 (fy = 2,400 ksc) . . . . . . . . . . . . . . 0.0025
t

DB30 (fy = 3,000 ksc) . . . . . . . . . . . . . . 0.0020
Ag = b × t = 100 t

DB40 (fy = 4,000 ksc) . . . . . . . . . . . . . . 0.0018

DB (fy > 4,000 ksc) . . . . . . . . . . . . . . . . 0.0018 × 4,000 ≥ 0.0014
fy
Spacing ≤ 3 t ≤ 45 cm

Main Steel (short direction):
As ≥ ∅ 6 mm
Max. Spacing ≤ 3 t ≤ 45 cm
Min. Spacing ≥ f main steel ≥ 4/3 max agg. ≥ 2.5 cm

F

9.1 ก

ก

S1
ก
F ก 50 ก.ก./ .2 ก
fy = 2,400 ก.ก./ .2

ก F

300 ก.ก./ .2 WSD
F f c = 210 ก.ก./ .2

F
1)

F
F

S1

F

S2

F

fy = 2,400 ksc

0.4 +

2,400
= 0.74
7,000

t min = 0.74 ×
S1

2.7 m

ก

tmin = L/24

270
= 8.3 cm
24

t = 10 cm

ก
ก
ก

= 0.1 x 2,400 = 240 kg/m2
= 50 kg/m2
= 300 kg/m2
ก
F

= 240 + 50 + 300 = 590

F

ก

FF

.

:

.ก

F

+M = 590 × 2.7 2 /14 = 307.2 kg-m

.

ก:
F ก

−M = 590 × 2.72 / 24 = 179.2 kg-m

ก

fs = 0.5x2,400 = 1,200 ksc
134
≈9
210

n=

R=

k=

ก.10

−M = 590 × 2.72 / 9 = 477.9 kg-m

:

F

WSD
kg/m2

:
fc = 0.45x210 = 94.5 ksc

1
1
=
= 0.293
fs
1,200
1+
1+
9 × 94.5
n fc

j = 1 – 0.293/3 = 0.902

1
1
fc k j = × 94.5 × 0.293 × 0.902 = 12.49 ksc
2
2

ก

F

F

F

ก RB9 . .

F

2

.

WSD

d = 10 - 0.45 - 2 = 7.55 cm

FF

ก :
Mc = R b d2 = 12.49 × 100 × 7.5522 = 71,196 kg-cm

ก

= 712.0 kg-m > M

ก
ก

F ก :

As =

OK
RB9 : As = 0.636 cm2

M
fs jd

Spacing = 0.636×100/As

ก F : As,min = 0.0025 × 100 × 10 = 2.5 cm2
F

USE RB9 @ 0.25 m

ก

ก

5.85

RB9 @ 0.10

307.2

F

As

-477.9

ก

M

3.76

RB9 @ 0.16

-179.2

2.19

RB9 @ 0.25
0.29
WSD

: F

ก

ก

ก F

USE RB9 @ 0.25 m

RB9 @ 0.25 .

RB9 @ 0.25 .
0.7 .

RB9 @ 0.10 .

0.9 .

10

.

RB9 @ 0.16 .
2.7 .

F

9.1 ก

ก

S1
ก
F ก 50 ก.ก./ .2 ก
fy = 2,400 ก.ก./ .2

ก F

300 ก.ก./ .2 SDM
F f c = 210 ก.ก./ .2

F
1)

F
F

S1

F

S2

F

fy = 2,400 ksc

0.4 +

2,400
= 0.74
7,000

t min = 0.74 ×
S1

2.7 m

ก

tmin = L/24

270
= 8.3 cm
24

t = 10 cm

ก
ก
ก

= 0.1 x 2,400 = 240 kg/m2
= 50 kg/m2
= 300 kg/m2
SDM

ก
F

wu = 1.4×(240+50) + 1.7×300 = 916

F

ก

FF

.

:

.ก

F

ก.10

−Mu = 916 × 2.72 / 9 = 742.0 kg-m

:

.

kg/m2

+Mu = 916 × 2.72 /14 = 477.0 kg-m

ก:
ก

(

ก

ก

F

−Mu = 916 × 2.72 / 24 = 278.2 kg-m

ก.5):
F

F

ρmax = 0.0341

ก RB9 . .

F

2

.

d = 10 - 0.45 - 2 = 7.55 cm

ก
ρ =

ก

F ก :

Rn =

Mu
φ b d2

RB9 : As = 0.636 cm2

As
0.85 fc′ 
2Rn
=
1− 1−

bd
fy 
0.85 fc′






ก F : As,min = 0.0025 × 100 × 10 = 2.5 cm2
F

Spacing = 0.636×100/As

USE RB9 @ 0.25 m SDM

: F

ก

ก

4.75

RB9 @ 0.13

3.01

RB9 @ 0.21

-278.2

ก

ก

477.0

F

As

-742.0

ก

Mu

1.73

RB9 @ 0.25
0.36

ก F
RB9 @ 0.25 .

0.7 .

USE RB9 @ 0.25 m

RB9 @ 0.25 .
0.9 .

RB9 @ 0.10 .
10

RB9 @ 0.16 .
2.7 .

.
ก

ก
Top bars at
exterior beams

Top bars at
exterior beams
Temperature bars

Bottom bars
Exterior span

Interior span

(a) Straight top and bottom bars

Bent bar

Bent bars

Temperature bars

Bottom bars
Exterior span

Interior span

(b) Alternate straight and bent bars

F F F

F
RB9 @ 0.20

RB9 @ 0.10

ก
RB9 @ 0.10 m

F ก (ก

F ,

As = 6.36 cm 2

. . F F
F )

F
ก

L1
4

L1
3

Temp. steel
L1
8

L1

RB9@0.18

RB9@0.07
.13 .

RB9@0.10
1.0 .

1.3 .
4.0 .
RB9@0.14

RB9@0.18

.13 .
1.0 .

RB9@0.07
F F F
4.0 .

F 1.3 .

Floor Plan

1/14

1/16

1/16

1/24
1/12

1/12

1/12
F

ก

9.2

F

ก 500 ก.ก./ 2 ก

F f’c = 210 ksc
3 @ 12 m = 36 m

A

A

SDM

fy = 2,400 ksc
1) Minimum depth :
0.4 + 2,400/7,000 = 0.74
Min. h = 0.74×370/24 = 11.4 cm
USE h = 12 cm
Slab weight = 0.12×2,400 = 288 kg/m2
Assume beam + super DL :

Ln = 3.7 m

Ln = 3.7 m

Ln = 3.7 m

Service DL = 350 kg/m2
Service LL = 500 kg/m2
2) Factored Load :

Section A-A

wu = 1.4×350 + 1.7×500 = 1,340 kg/m2

3) Max. Moment :
Mu = 1,340 × 3.72 / 12 = 1,529 kg-m (Interior negative moment)
Max. reinforcement ratio (from Table ก.5) : ρmax = 0.0341
USE RB9 with 2 cm covering: d = 12-2-0.45 = 9.55 cm
Rn =

ρ =

Mu
1,529 × 100
=
= 18.63 kg/cm2
2
2
φb d
0.9 × 100 × 9.55
0.85 fc'
fy


2Rn
1− 1 −

0.85 fc'



 = 0.0082 < ρmax = 0.0341



OK

Required As = ρbd = 0.0082 × 100 × 9.55 = 7.83 cm2/m
RB9 : As = 0.636 cm2 → s = 0.636×100/7.83 = 8.12 cm
Select RB9@0.08 : As = 0.636×100/8 = 7.95 cm2/m > Required As

OK
Temp. steel = 0.0025 × 100 × 12 = 3.00 < 7.95 cm2/m

OK

Select RB9@0.20 : As = 0.636×100/20 = 3.18 cm2/m

RB9 @ 0.16 .
0.95 .

RB9 @ 0.20 . ก ก F
RB9 @ 0.08 .
1.25 .
F F F F +
12

0.55 .

RB9 @ 0.08 .
F F F F

0.95 .

3.7 .

ก

ก

.

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طه

  • 1. 16 Reinforced Concrete Design One-Way Slabs Types of Slabs Load Paths & Flaming Concepts One-Way Slab One-Way Joist Slab on Grade Mongkol JIRAVACHARADET SURANAREE INSTITUTE OF ENGINEERING UNIVERSITY OF TECHNOLOGY SCHOOL OF CIVIL ENGINEERING Types of Slab One-way slab Flat plate slab One-way slab Flat slab Two-way slab Grid slab
  • 2. Load Path / Framing Possibilities Ln = 3.2 m Ln = 4.4 m Ln = 3.6 m Ln = 8.2 m Think we’ll need some additional framing members??? Framing Concepts Let’s use a simple example for our discussion… Column spacing 8 m c-c Think about relating it to your design project. Plan
  • 3. Framing Concepts We can first assume that we’ll have major girders running in one direction in our one-way system Framing Concepts We can first assume that we’ll have major girders running in one direction in our one-way system If we span between girders with our slab, then we have a load path, but if the spans are too long…
  • 4. Framing Concepts We will need to shorten up the span with additional beams But we need to support the load from these new beams, so we will need additional supporting members Framing Concepts Now let’s go back through with a slightly different load path. We again assume that we’ll have major girders running in one direction in our one-way system. This time, let’s think about shortening up the slab span by running beams into our girders. Our one-way slab will transfer our load to the beams.
  • 5. Two Load Path Options Framing Concepts - Considerations For your structure: Look for a “natural” load path Identify which column lines are best suited to having major framing members (i.e. girders) Assume walls are not there for structural support, but consider that the may help you in construction (forming)
  • 6. Example Condo Floor Plan ก One-Way Slab : ก ก : F S (L) > F F S L S t L Simple supports on two long edges only S (S) S การแอนตัวเกิดขึ้นบนดานสั้น
  • 7. ก ก ก ก ˈ : กF S L 1m t F m 1.0 ก ก ก F : F ก: F S Sn F F F 10-15 . F ก 2-3 F . F F L / 20 ก F Ln F F L / 24 Ln L / 28 F F Ln L / 10 Ln
  • 8. ก ก ˂ ก ก ก F ก As F ( ก ก ก F ) Ag : As/Ag 1m RB24 (fy = 2,400 ksc) . . . . . . . . . . . . . . 0.0025 t DB30 (fy = 3,000 ksc) . . . . . . . . . . . . . . 0.0020 Ag = b × t = 100 t DB40 (fy = 4,000 ksc) . . . . . . . . . . . . . . 0.0018 DB (fy > 4,000 ksc) . . . . . . . . . . . . . . . . 0.0018 × 4,000 ≥ 0.0014 fy Spacing ≤ 3 t ≤ 45 cm Main Steel (short direction): As ≥ ∅ 6 mm Max. Spacing ≤ 3 t ≤ 45 cm Min. Spacing ≥ f main steel ≥ 4/3 max agg. ≥ 2.5 cm F 9.1 ก ก S1 ก F ก 50 ก.ก./ .2 ก fy = 2,400 ก.ก./ .2 ก F 300 ก.ก./ .2 WSD F f c = 210 ก.ก./ .2 F 1) F F S1 F S2 F fy = 2,400 ksc 0.4 + 2,400 = 0.74 7,000 t min = 0.74 × S1 2.7 m ก tmin = L/24 270 = 8.3 cm 24 t = 10 cm ก ก ก = 0.1 x 2,400 = 240 kg/m2 = 50 kg/m2 = 300 kg/m2
  • 9. ก F = 240 + 50 + 300 = 590 F ก FF . : .ก F +M = 590 × 2.7 2 /14 = 307.2 kg-m . ก: F ก −M = 590 × 2.72 / 24 = 179.2 kg-m ก fs = 0.5x2,400 = 1,200 ksc 134 ≈9 210 n= R= k= ก.10 −M = 590 × 2.72 / 9 = 477.9 kg-m : F WSD kg/m2 : fc = 0.45x210 = 94.5 ksc 1 1 = = 0.293 fs 1,200 1+ 1+ 9 × 94.5 n fc j = 1 – 0.293/3 = 0.902 1 1 fc k j = × 94.5 × 0.293 × 0.902 = 12.49 ksc 2 2 ก F F F ก RB9 . . F 2 . WSD d = 10 - 0.45 - 2 = 7.55 cm FF ก : Mc = R b d2 = 12.49 × 100 × 7.5522 = 71,196 kg-cm ก = 712.0 kg-m > M ก ก F ก : As = OK RB9 : As = 0.636 cm2 M fs jd Spacing = 0.636×100/As ก F : As,min = 0.0025 × 100 × 10 = 2.5 cm2 F USE RB9 @ 0.25 m ก ก 5.85 RB9 @ 0.10 307.2 F As -477.9 ก M 3.76 RB9 @ 0.16 -179.2 2.19 RB9 @ 0.25 0.29
  • 10. WSD : F ก ก ก F USE RB9 @ 0.25 m RB9 @ 0.25 . RB9 @ 0.25 . 0.7 . RB9 @ 0.10 . 0.9 . 10 . RB9 @ 0.16 . 2.7 . F 9.1 ก ก S1 ก F ก 50 ก.ก./ .2 ก fy = 2,400 ก.ก./ .2 ก F 300 ก.ก./ .2 SDM F f c = 210 ก.ก./ .2 F 1) F F S1 F S2 F fy = 2,400 ksc 0.4 + 2,400 = 0.74 7,000 t min = 0.74 × S1 2.7 m ก tmin = L/24 270 = 8.3 cm 24 t = 10 cm ก ก ก = 0.1 x 2,400 = 240 kg/m2 = 50 kg/m2 = 300 kg/m2
  • 11. SDM ก F wu = 1.4×(240+50) + 1.7×300 = 916 F ก FF . : .ก F ก.10 −Mu = 916 × 2.72 / 9 = 742.0 kg-m : . kg/m2 +Mu = 916 × 2.72 /14 = 477.0 kg-m ก: ก ( ก ก F −Mu = 916 × 2.72 / 24 = 278.2 kg-m ก.5): F F ρmax = 0.0341 ก RB9 . . F 2 . d = 10 - 0.45 - 2 = 7.55 cm ก ρ = ก F ก : Rn = Mu φ b d2 RB9 : As = 0.636 cm2 As 0.85 fc′  2Rn = 1− 1−  bd fy  0.85 fc′     ก F : As,min = 0.0025 × 100 × 10 = 2.5 cm2 F Spacing = 0.636×100/As USE RB9 @ 0.25 m SDM : F ก ก 4.75 RB9 @ 0.13 3.01 RB9 @ 0.21 -278.2 ก ก 477.0 F As -742.0 ก Mu 1.73 RB9 @ 0.25 0.36 ก F RB9 @ 0.25 . 0.7 . USE RB9 @ 0.25 m RB9 @ 0.25 . 0.9 . RB9 @ 0.10 . 10 RB9 @ 0.16 . 2.7 . .
  • 12. ก ก Top bars at exterior beams Top bars at exterior beams Temperature bars Bottom bars Exterior span Interior span (a) Straight top and bottom bars Bent bar Bent bars Temperature bars Bottom bars Exterior span Interior span (b) Alternate straight and bent bars F F F F RB9 @ 0.20 RB9 @ 0.10 ก RB9 @ 0.10 m F ก (ก F , As = 6.36 cm 2 . . F F F ) F
  • 13. ก L1 4 L1 3 Temp. steel L1 8 L1 RB9@0.18 RB9@0.07 .13 . RB9@0.10 1.0 . 1.3 . 4.0 . RB9@0.14 RB9@0.18 .13 . 1.0 . RB9@0.07 F F F 4.0 . F 1.3 . Floor Plan 1/14 1/16 1/16 1/24 1/12 1/12 1/12
  • 14. F ก 9.2 F ก 500 ก.ก./ 2 ก F f’c = 210 ksc 3 @ 12 m = 36 m A A SDM fy = 2,400 ksc 1) Minimum depth : 0.4 + 2,400/7,000 = 0.74 Min. h = 0.74×370/24 = 11.4 cm USE h = 12 cm Slab weight = 0.12×2,400 = 288 kg/m2 Assume beam + super DL : Ln = 3.7 m Ln = 3.7 m Ln = 3.7 m Service DL = 350 kg/m2 Service LL = 500 kg/m2 2) Factored Load : Section A-A wu = 1.4×350 + 1.7×500 = 1,340 kg/m2 3) Max. Moment : Mu = 1,340 × 3.72 / 12 = 1,529 kg-m (Interior negative moment) Max. reinforcement ratio (from Table ก.5) : ρmax = 0.0341 USE RB9 with 2 cm covering: d = 12-2-0.45 = 9.55 cm Rn = ρ = Mu 1,529 × 100 = = 18.63 kg/cm2 2 2 φb d 0.9 × 100 × 9.55 0.85 fc' fy  2Rn 1− 1 −  0.85 fc'    = 0.0082 < ρmax = 0.0341   OK Required As = ρbd = 0.0082 × 100 × 9.55 = 7.83 cm2/m RB9 : As = 0.636 cm2 → s = 0.636×100/7.83 = 8.12 cm Select RB9@0.08 : As = 0.636×100/8 = 7.95 cm2/m > Required As OK
  • 15. Temp. steel = 0.0025 × 100 × 12 = 3.00 < 7.95 cm2/m OK Select RB9@0.20 : As = 0.636×100/20 = 3.18 cm2/m RB9 @ 0.16 . 0.95 . RB9 @ 0.20 . ก ก F RB9 @ 0.08 . 1.25 . F F F F + 12 0.55 . RB9 @ 0.08 . F F F F 0.95 . 3.7 . ก ก .