1. 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 1 of 13

2. 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 2 of 13

3. 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:
L2 L2
Column
(typ.)
Exterior Column Strip
Interior Column Strip
Interior Column Strip
Middle Strip
Middle Strip
L1
L2/4 L2/4 L2/4
Lecture 7 – Page 3 of 13

4. b) If L2 > L1:
L2 L2
Exterior Column Strip
Interior Column Strip
Interior Column Strip
Middle Strip
Middle Strip
L1
L1/4 L1/4 L1/4
Design Moment Coefficients for Flat Plate Supported Directly by Columns
Slab End Span Interior Span
Moments 1 2 3 4 5
Exterior Positive First Positive Interior
Negative Interior Negative
Negative
Total 0.26Mo 0.52Mo 0.70Mo 0.35Mo 0.65Mo
Moment
Column 0.26Mo 0.31Mo 0.53Mo 0.21Mo 0.49Mo
Strip
Middle 0 0.21Mo 0.17Mo 0.14Mo 0.16Mo
Strip
Mo = Total factored moment per span
End Span Interior Span
1 2 3 4 5
2
wu L2 Ln
Mo = where Ln = clear span (face-to-face of cols.) in the direction of analysis
8
Lecture 7 – Page 4 of 13

5. 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 5 of 13

6. 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” L2 = 16’-0”
20’-0”
Ln 20’-0”
20’-0”
L2/4 L2/4
½ Middle strip
= ½(16’ – Col. strip) ½ Middle strip
Col. strip = ½(16’ – Col. strip)
Design Strip = 16’
Lecture 7 – Page 6 of 13

7. Step 1 – Determine slab thickness h:
Ln
Since it is a flat plate, from Table above, use h =
30
where Ln = clear span in direction of analysis
= (20’-0” x 12”/ft) – 20” Column size
= 220” = 18.33’
220"
h=
30
= 7.333”
Use 8” thick slab
Step 2 – Determine factored uniform load, wu on the slab:
wu = 1.2D + 1.6L Slab weight
= 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
Lecture 7 – Page 7 of 13

8. Step 4 – Determine total factored moment per span, Mo:
2
wu L2 Ln
Mo =
8
(0.28 KSF )(16' )(18.33' ) 2
=
8
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
Slab End Span Interior Span
Moments 1 2 3 4 5
Exterior Positive First Positive Interior
Negative Interior Negative
Negative
Total 0.26Mo = 48.9 0.52Mo = 97.8 0.70Mo = 131.6 0.35Mo = 65.8 0.65Mo = 122.2
Moment
Column 0.26Mo = 48.9 0.31Mo = 58.3 0.53Mo = 99.6 0.21Mo = 39.5 0.49Mo = 92.1
Strip
Middle 0 0.21Mo = 39.5 0.17Mo = 32.0 0.14Mo = 26.3 0.16Mo = 30.1
Strip
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”
8”
d
d = 8” – conc. cover – ½(bar dia.)
= 8” – ¾” – ½(4/8”)
= 7”
Lecture 7 – Page 8 of 13

9. Mu 586,800 LB − IN
=
φbd 2
(0.9)(96" )(7" ) 2
= 138.6 PSI
From Lecture 4 → Table 2:
Use ρmin = 0.0033
As
ρ=
bd
Solve for As:
As = ρbd
= (0.0033)(96”)(7”)
= 2.22 in2
As
Number of bars required =
As _ per _ bar
2.22in 2
=
0.20in 2 _ per _#4 _ bar
= 11.1 → Use 12 - #4 TOP bars
Lecture 7 – Page 9 of 13

10. b) Column strip for region 2 :
Factored POSITIVE moment = 58.3 KIP-FT (see Table above)
= 699,600 LB-IN
b = 96”
8” d
d = 8” – conc. cover – ½(bar dia.)
= 8” – ¾” – ½(4/8”)
= 7”
Mu 699,600 LB − IN
=
φbd 2
(0.9)(96" )(7" ) 2
= 165.2 PSI
From Lecture 4 → Table 2:
Use ρ = 0.0033
As = 2.22 in2 (see calcs. above)
Use 12 - #4 BOTTOM bars
Lecture 7 – Page 10 of 13

11. c) Middle strip for region 2 :
Factored POSITIVE moment = 39.5 KIP-FT (see Table above)
= 474,000 LB-IN
b = 96”
8” d
d = 8” – conc. cover – ½(bar dia.)
= 8” – ¾” – ½(4/8”)
= 7”
Mu 474,000 LB − IN
=
φbd 2
(0.9)(96" )(7" ) 2
= 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
Lecture 7 – Page 11 of 13

12. Step 7 – Draw “Summary Sketch” plan view of bars:
16’-0” 16’-0” 16’-0”
Col. strip for region 1
12 - #4 TOP bars
½ Middle strip for
region 2 20’-0”
6 - #4 BOTTOM bars
Col. strip for region 2
12 - #4 BOTTOM bars 20’-0”
8” Thick
concrete slab 20’-0”
4’-0” 4’-0”
½ Middle strip = 4’-0” ½ Middle strip = 4’-0”
Col. strip
16’ – 0”
Lecture 7 – Page 12 of 13

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).
16’-0” 16’-0” 16’-0”
20’-0”
½ Middle strip = 6’-0”
20’-0” Col. strip = 8’-0”
½ Middle strip = 6’-0”
20’-0”
20’-0”
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
Lecture 7 – Page 13 of 13