Lecture 7 – Page 1 of 13

Lecture 7 – Two-Way Slabs Two-way slabs have tension reinforcing spanning in BOTH directions, and may take the general form of one of the following:

Types of Two-Way Slab Systems

Lecture 7 – Page 2 of 13

The following Table may be used to determine minimum thickness of various two-way slabs based on deflection:

Minimum Suggested Thickness “h” for Two-Way Slabs Two-Way Slab System: Minimum Thickness h:

Flat plate Ln/30 Flat plate with spandrel beams Ln/33 Flat slab Ln/33 Flat slab with spandrel beams Ln/36 Two-way beam-supported slab Ln/33

Ln = clear distance in long direction Flat Plates

Flat plates are the most common type of two-way slab system. It is commonly used in multi-story construction such as hotels, hospitals, offices and apartment buildings. It has several advantages:

• Easy formwork • Simple bar placement • Low floor-to-floor heights

Direct Design Method of Flat Plates per ACI 318-02

Two-way slabs are inherently difficult to analyze by conventional methods of statics because of the two-way bending occurring. Accurately determining the moments on a two-way slab is typically accomplished by finite element computer analysis.

Computer analysis of two-way slab

Lecture 7 – Page 3 of 13

The ACI 318 code allows a direct design method that can be used in most typical situations. However, the following limitations apply:

1. Must have 3 or more continuous spans in each direction. 2. Slab panels must be rectangular with a ratio of the longer span to

shorter span(measured as centerline-to-centerline of support) not greater than 2.0.

3. Successive span lengths in each direction must not differ by more than 1/3 of the longer span.

4. Columns must not be offset by more than 10% of the span (in direction of offset) from either axis between centerlines of successive columns.

5. Loads must be uniformly distributed, with the unfactored live load not more than 2 times the unfactored dead load (L/D < 2.0).

Design Strips

a) If L1 > L2:

Mid

dle

Stri

p

L1

L2/4

Mid

dle

Stri

p

L2 L2

L2/4

Inte

rior C

olum

n S

trip

Ext

erio

r Col

umn

Stri

p

Inte

rior C

olum

n S

trip

L2/4

Column (typ.)

Lecture 7 – Page 4 of 13

b) If L2 > L1: Design Moment Coefficients for Flat Plate Supported Directly by Columns

End Span Interior Span 1 2 3 4 5

Slab Moments

Exterior Negative

Positive First Interior

Negative

Positive Interior Negative

Total Moment

0.26Mo 0.52Mo 0.70Mo 0.35Mo 0.65Mo

Column Strip

0.26Mo 0.31Mo 0.53Mo 0.21Mo 0.49Mo

Middle Strip

0 0.21Mo 0.17Mo 0.14Mo 0.16Mo

Mo = Total factored moment per span

Mo = 8

22 nu LLw where Ln = clear span (face-to-face of cols.) in the direction of analysis

1

Mid

dle

Stri

p L1

L1/4

Mid

dle

Stri

p

L2 L2

L1/4

Inte

rior C

olum

n S

trip

Ext

erio

r Col

umn

Stri

p

Inte

rior C

olum

n S

trip

L1/4

2 3 4 5

End Span Interior Span

Lecture 7 – Page 5 of 13

Bar Placement per ACI 318-02

The actual quantity of bars required is determined by analysis (see Example below). However, usage of the Direct Design Method prescribes bar placement as shown below:

Lecture 7 – Page 6 of 13

Example 1 GIVEN: A two-way flat plate for an office building is shown below. Use the following:

• Column dimensions = 20” x 20” • Superimposed service floor Dead load = 32 PSF (not including slab weight) • Superimposed service floor Live load = 75 PSF • Concrete f’c = 4000 PSI • #4 Grade 60 main tension bars • Concrete cover = ¾”

REQUIRED: Use the “Direct Design Method” to design the two-way slab for the design strip in the direction shown.

L2 = 16’-0” L2 = 16’-0”

Col. strip

Ln

Design Strip = 16’

L2 = 16’-0”

20’-0”

20’-0”

20’-0”

L2/4 L2/4 ½ Middle strip = ½(16’ – Col. strip) ½ Middle strip

= ½(16’ – Col. strip)

Lecture 7 – Page 7 of 13

Step 1 – Determine slab thickness h:

Since it is a flat plate, from Table above, use h = 30

nL

where Ln = clear span in direction of analysis = (20’-0” x 12”/ft) – 20” = 220” = 18.33’

h = 30

"220

= 7.333” Use 8” thick slab

Step 2 – Determine factored uniform load, wu on the slab:

wu = 1.2D + 1.6L = 1.2[(32 PSF) + (8/12)(150 PCF)] + 1.6[(75 PSF)] = 278.4 PSF = 0.28 KSF

Step 3 – Check applicability of “Direct Design Method”:

1) Must have 3 or more continuous spans in each direction. YES 2) Slab panels must be rectangular with a ratio of the longer span to

shorter span(measured as centerline-to-centerline of support) not greater than 2.0. YES

3) Successive span lengths in each direction must not differ by more than

1/3 of the longer span. YES

4) Columns must not be offset by more than 10% of the span (in direction of offset) from either axis between centerlines of successive columns. YES

5) Loads must be uniformly distributed, with the unfactored live load not

more than 2 times the unfactored dead load (L/D < 2.0). YES

Column size

Slab weight

Lecture 7 – Page 8 of 13

Step 4 – Determine total factored moment per span, Mo:

Mo = 8

22 nu LLw

= 8

)'33.18)('16)(28.0( 2KSF

Mo = 188 KIP-FT

Step 5 – Determine distribution of total factored moment into col. & middle strips:

Design Moment Coefficients for Flat Plate Supported Directly by Columns

End Span Interior Span 1 2 3 4 5

Slab Moments

Exterior Negative

Positive First Interior

Negative

Positive Interior Negative

Total Moment

0.26Mo = 48.9 0.52Mo = 97.8 0.70Mo = 131.6 0.35Mo = 65.8 0.65Mo = 122.2

Column Strip

0.26Mo = 48.9 0.31Mo = 58.3 0.53Mo = 99.6 0.21Mo = 39.5 0.49Mo = 92.1

Middle Strip

0 0.21Mo = 39.5 0.17Mo = 32.0 0.14Mo = 26.3 0.16Mo = 30.1

Mo = Total factored moment per span = 188 KIP-FT

Step 6 – Determine tension steel bars for col. & middle strips:

a) Column strip for region 1 :

Factored NEGATIVE moment = 48.9 KIP-FT (see Table above) = 586.8 KIP-IN = 586,800 LB-IN

b = 96”

d 8”

d = 8” – conc. cover – ½(bar dia.) = 8” – ¾” – ½(4/8”) = 7”

Lecture 7 – Page 9 of 13

22 )"7)("96)(9.0(800,586 INLB

bdM u −

=φ

= 138.6 PSI From Lecture 4 → Table 2:

Use ρmin = 0.0033

ρ = bdAs

Solve for As:

As = ρbd = (0.0033)(96”)(7”) = 2.22 in2

Number of bars required = barperA

A

s

s

__

= barperin

in_4_#_20.0

22.22

2

= 11.1 → Use 12 - #4 TOP bars

Lecture 7 – Page 10 of 13

b) Column strip for region 2 :

Factored POSITIVE moment = 58.3 KIP-FT (see Table above) = 699,600 LB-IN

22 )"7)("96)(9.0(600,699 INLB

bdM u −

=φ

= 165.2 PSI From Lecture 4 → Table 2:

Use ρ = 0.0033

As = 2.22 in2 (see calcs. above) Use 12 - #4 BOTTOM bars

d = 8” – conc. cover – ½(bar dia.) = 8” – ¾” – ½(4/8”) = 7”

b = 96”

d 8”

Lecture 7 – Page 11 of 13

c) Middle strip for region 2 :

Factored POSITIVE moment = 39.5 KIP-FT (see Table above) = 474,000 LB-IN

22 )"7)("96)(9.0(000,474 INLB

bdM u −

=φ

= 112.0 PSI From Lecture 4 → Table 2:

Use ρ = 0.0033

As = 2.22 in2 (see calcs. above) Use 12 - #4 BOTTOM bars Use 6 - #4 Bottom bars at each ½ Middle Strip

b = 96”

d 8”

d = 8” – conc. cover – ½(bar dia.) = 8” – ¾” – ½(4/8”) = 7”

Lecture 7 – Page 12 of 13

Step 7 – Draw “Summary Sketch” plan view of bars:

4’-0”

Col. strip for region 2 12 - #4 BOTTOM bars

16’-0” 16’-0”

Col. strip

16’ – 0”

16’-0”

20’-0”

20’-0”

20’-0”

4’-0”

½ Middle strip = 4’-0” ½ Middle strip = 4’-0”

Col. strip for region 1 12 - #4 TOP bars

½ Middle strip for region 2 6 - #4 BOTTOM bars

8” Thick concrete slab

Lecture 7 – Page 13 of 13

Example 2 GIVEN: The two-way slab system from Example 1. REQUIRED: Design the steel tension bars for design strip shown (perpendicular to those in Example 1).

Solution → Similar to the procedure shown in Example 1, except:

• Re-check slab thickness to verify that 8” is still acceptable • Re-calculate “M0” • Using new value of M0, determine “Design Moment Coefficients” • Design tension steel based on these moment coefficients

20’-0”

½ Middle strip = 6’-0”

½ Middle strip = 6’-0”

Col. strip = 8’-0”

16’-0” 16’-0” 16’-0”

20’-0”

20’-0”

20’-0”