16 Reinforced Concrete Design Design of Slabs - .xyzlibvolume3.xyz/.../designofslabs/designofslabstutorial2.pdf · 16 Types of Slabs Load Paths and Framing Concepts One-way Slabs

1616

� Types of Slabs

� Load Paths and Framing Concepts

� One-way Slabs

� Two-way Slabs

Mongkol JIRAVACHARADET

Reinforced Concrete DesignReinforced Concrete Design

S U R A N A R E E INSTITUTE OF ENGINEERING

UNIVERSITY OF TECHNOLOGY SCHOOL OF CIVIL ENGINEERING

Design of Slabs

Types of Slab

One-way slab Two-way slabOne-way slab

Flat plate slab Flat slab Grid slab

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

Load Path / Framing Possibilities

Ln = 4.4 m

Ln = 8.2 m

Ln = 3.2 m

Ln = 3.6 m

Plan

Framing Concepts

Let’s use a simple example

for our discussion…

Think about relating it to your

design project.

Column spacing 8 m c-c

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…

But we need to support the

load from these new beams,

so we will need additional

supporting members

Framing Concepts

We will need to shorten up

the span with additional beams

We again assume that we’ll

have major girders running in

one direction in our one-way

system.

Framing Concepts

Now let’s go back through with

a slightly different load path.

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

Assume walls are not there for structural support, but

consider that the may help you in construction (forming)

Identify which column lines are best suited to having

major framing members (i.e. girders)

ExampleExample

Condo Floor PlanCondo Floor Plan

1.0 m

LS

Main reinforcement

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One-way Slab

Design of one-way slabs is like design of parallel 1m beams.

Design of One-way Slab (L > 2S)

S

L 1 m

������������� ��������� 1 ���

S

w

Minimum Thickness (ACI)

Simply

supported

One end

continuous

Both ends

continuousCantilever

L/20 L/24 L/28 L/10

*multiplied by 0.4 + fy/7,000 for steel other than SD40

ACI Design ProvisionACI Design Provision

Shrinkage and temperature reinforcement

Spacing ≤≤≤≤ 5 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

For structural slabs only; not intended for soil-supported slabs on grade

Ratio of reinforcement As to gross concrete area Ag : As/Ag

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

DB30 (fy = 3,000 ksc) . . . . . . . . . . . . . . . . . 0.0020

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

DB (fy > 4,000 ksc) . . . . . . . . . . . . . . . . . . . 0.0018 4,0000.0014

yf

×≥

Effect of column width

b b

L

b/212

2wL

2

wL

A′A

b/2

B

w

Moment at A’:

( )

+−−=

−

+−=

8412

2

2/

2212

22

22

wbwLbwL

bwbwLwL

( )

+−−=

−−=

1261212

222wbwLbwLbLw

M

If A’ and B’ are fiexed against rotation,

B′

12

2wL

2

wL

Typical reinforcement in a oneTypical reinforcement in a one--way slabway slab

Exterior span

Bottom bars

Top bars at

exterior beams

Top bars at

exterior beams

Interior span

Temperature bars

(a) Straight top and bottom bars

Exterior span

Bottom bars

Bent bar Bent bars

Interior span

Temperature bars

(b) Alternate straight and bent bars

3 @

8 m

= 24 m

4 @ 12 m = 48 m

G1

A AS1 S2 S3

Example: Design one-way slab as shown below to carry the live

load 500-kg/m2 fc’ = 210 kg/cm2, fy = 2,400 kg/cm

2

0.4 + 2400/7000 = 0.74

min h = 400(0.74)/24 = 12.3 cm

USE h = 13 cm

DL = 0.13×2400 = 312kg/m2

wu = 1.4(312) + 1.7(500) = 1,286.8 kg/m2

clear span = 4 - 0.3 = 3.7 m

Mu = (1,286.8)(3.7)2/10 = 1,762 kg-m

ρmax = 0.75ρb = 0.75(0.0454) = 0.0341

USE RB9 with 2 cm covering: d = 13-2-0.45 = 10.55 cm

ksc 6.1755.101009.0

100176222=

××

×==

bd

MR u

n φ

0077.085.0

211

85.0'

'

=

−−=

c

n

y

c

f

R

f

fρ < ρmax OK

As = ρbd = 0.0077(100)(10.55) = 8.16 cm2/m

Select [email protected] (As = 9.28 cm2/m)

Temp. steel = 0.0025(100)(13) = 3.25 < 9.28 cm2/m OK

Select [email protected] (As = 3.53 cm2/m)

Detailing of one-way slab

3

1L

4

1L

8

1L

L1

Temp. steel

4.0 �.

.13 �.

1.0 �. 1.3 �.

[email protected]

[email protected] [email protected]

4.0 �.

.13 �.

1.0 �. 1.3 �[email protected]

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Design of Two-way Slab (L < 2S)

L

S

Min. Thickness:

t ≥ 9 cm ≥ Perimeter/180 = 2(L+S)/180

Reinforcement Steel:

As ≥ φ 6 mm ≥ Temp. steel

Max. Spacing ≤ 3 t ≤ 45 cm

Min. Spacing ≥ φ main steel ≥ 4/3 max agg. ≥ 2.5 cm

Load transfer from two-way Slab

45o 45o

45o 45o

A

D C

B

S

L

Short span (BC):

Floor load = w kg/sq.m

Tributary area = S2/4 sq.m

Load on beam = wS/4 wS/3 kg/m

Long span (AB): Span ratio m = S/L

Tributary area = SL/2 - S2/4 = sq.m

Load on beam kg/m

−m

mS 2

4

2

−2

3

3

2mwS

Moment Coefficient Method

S/4

S/4

S/2

L/4 L/2 L/4

%&'�(��

%&'��

%&'��

%&'�(�� %&'��%&'��

-Ms

-Ms

+Ms

+ML

-ML-ML

Middle strip moment: MM = CwS2

Column strip moment: MC = 2MM/3

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-/��0�)�) �� �

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1.0 0.9 0.8 0.7 0.6 0.5

0.033-

0.025

0.040-

0.030

0.048-

0.036

0.055-

0.041

0.063-

0.047

0.083-

0.062

0.033-

0.025

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-/��0�)�) �� �

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0.0410.0210.031

0.0480.0240.036

0.0550.0270.041

0.0620.0310.047

0.0690.0350.052

0.0850.0420.064

0.0410.0210.031

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0.0490.0250.037

0.0570.0280.043

0.0640.0320.048

0.0710.0360.054

0.0780.0390.059

0.0900.0450.068

0.0490.0250.037

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1.0 0.9 0.8 0.7 0.6 0.5

0.0580.0290.044

0.0660.0330.050

0.0740.0370.056

0.0820.0410.062

0.0900.0450.068

0.0980.0490.074

0.0580.0290.044

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-0.0330.050

-0.0380.057

-0.0430.064

-0.0470.072

-0.0530.080

-0.0550.083

-0.0330.050

Bar detailing in slab

L1 L2

L1/7 L1/4

L1/3

L2/4

L2/3����������������

Bar detailing in beam

L1 L2

L1/8

L1/3 L2/3

L1/8

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L/5

L/5

L = ���������������

Example: Design two-way slab as shown below to carry the live

load 300-kg/m2 fc’ = 240 kg/cm2, fy = 2,400 kg/cm

2

3.80

4.00

4.805.00

Floor plan

0.10

0.50

0.20 0.20

Cross section

Min h = 2(400+500)/180 = 10 cm

DL = 0.10(2,400) = 240 kg/m2

wu = 1.4(240)+1.7(300) = 846 kg/m2

m = 4.00/5.00 = 0.8

As,min = 0.0018(100)(10) = 1.8 cm2/m

Short span

Moment coeff. C

-M(������������) +M -M(���������)

0.032 0.048 0.064

Max.M = C w S 2 = 0.064 × 846 × 4.02 = 866 kg-m/1 m width

d = 10 - 2(covering) - 0.5(half of DB10) = 7.5 cm

2

2 2

86,60017.11 kg/cm

0.9 100 7.5

un

MR

bdφ= = =

× ×

0.85 21 1 0.0045

0.85

c n

y c

f R

f fρ

′= − − = ′

As = 0.0045(100)(7.5) = 3.36 4�.2 > As,min

(� �63��(7�1���8� [email protected] (As=3.90 4�.2)

Long span

Moment coeff. C

-M(������������) +M -M(���������)

0.025 0.037 0.049

Max.M = C w S 2 = 0.049 × 846 × 4.02 = 663 kg-m/1 m width

d = 10 - 2(covering) - 1.5(half of DB10) = 6.5 cm

2

2 2

66,30017.44 kg/cm

0.9 100 7.5

un

MR

bdφ= = =

× ×

0.85 21 1 0.0046

0.85

c n

y c

f R

f fρ

′= − − = ′

As = 0.0046(100)(6.5) = 2.97 4�.2 > As,min

(� �63��(7�1���8� [email protected] (As=3.90 4�.2)

����� 8�9�"����8:��;� � �� ��7�

%��9� 9(2�: Vu = wuS/4 = (846)(4.0)/4

= 846 ��./�.

�<�(8��8'%��9� � ��2� φVc = 0.85(0.53) (100)(7.5)

= 5234 ��./�. OK

240

0.50

0.10

0.20 4.80 0.20

[email protected] � ���������[email protected] ������#$

[email protected] ������#$

���6��&��

0.70

1.20

1.20

1.60

0.50

0.10

0.20 3.80 0.20

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[email protected] ������#$

���6�����

0.95

1.300.55

0.95