Lecture25,Design of Two Way Slab

Lecture 25 - Design of Two-Way Floor Slab System

August 6, 2003

CVEN 444

Lecture Goals

Example of DDM

Panel Design

Example 1 Design an interior panel of the two-way

slab for the floor system.The floor

consists of six panels at each direction,

with a panel size 24 ft x 20 ft. All

panels are supported by 20 in square

columns. The slabs are supported by

beams along the column line with cross

sections. The service live load is to be

taken as 80 psf and the service dead

load consists of 24 psf of floor

finishing in addition to the self-weight.

Use fc = 4 ksi and fy = 60 ksi

Example 1 –Previous Example

The cross-sections are:

h = 7 in.

The resulting cross section:

The thickness was calculated in an earlier example.

Generally, thickness of the slab is calculated at the

for the external corner slab. So use h = 7 in.

Example 1- Loading

The weight of the slab is given as.

lb 1 ft lb lb24 7.0 in. 150 111.5

ft 12 in. ft ft

lb lb1.2 1.6 1.2 111.5 1.6 80

lb kips262 0.262

w DL LL

Example 1 – calculation d

Compute the average depth, d for the slab. Use an

average depth for the shear calculation with a #4 bar

(d = 0.5 in)

bcover / 2

7.0 in. 0.75 in. 0.5 in./ 2 6.0 in.

Example 1 – One-way shear

The shear stresses in the slab are not

critical. The critical section is at a

distance d from the face of the beam.

Use 1 ft section.

beam width12 ft. 1 ft.

1 ft16 in.

1 ft12 in.0.262 k/ft 12 ft. 6 in. 1 ft.

2 12 in.

2.84 k

Example 1 – One-way shear

The one way shear on the face of the beam.

1 kip0.75 2 3000 12 in. 6 in.

1000 lb

5.92 k 2.84 k OK.

V f bd

Example 1 – Strip size

Determine the strip sizes for the column and middle

strip. Use the smaller of l1 or l2 so l2 = 20 ft

in 168ft 14 ft 52 ft 24

in 120ft 10 ft 52 ft 20

Therefore the column strip b = 2( 5 ft) = 10 ft (120 in)

The middle strips are

Example 1 – Strip Size

Calculate the strip sizes

Example 1 – Static Moment Computation

Moment Mo for the two directions.

long direction

short direction

20 in. 1 ft.24 ft. 2 22.333 ft.

2 12 in.

0.262 k/ft 20 ft. 22.333 ft.M

326.7 k-ft

20 in. 1 ft.20 ft. 2 18.333 ft.

2 12 in.

0.262 k/ft 24 ft. 18.333 ft.M

264.2 k-ft

Example 1 – Internal Panel Moment distribution

Interior panel

0.35Mo

0.65Mo

Example 1 – Moments (long)

The factored components

of the moment for the

beam (long).

Negative - Moment

Positive + Moment

0.65 326.7 k-ft 212.4 k-ft

0.35 326.7 k-ft 114.4 k-ft

Example 1- - Moment (long)

Coefficients

The moments of inertia about beam, Ib = 22,453 in4 and

Is = 6860 in4 (long direction) are need to determine the

distribution of the moments between the column and

middle strip.

71.28333.0*27.3

in 6860

in 22453

8333.0

Example 1- Moment (long) Factors (negative)

5.08333.0

0.15.0

75.09.09.0

Factor

Need to interpolate to

determine how the

negative moment is

distributed.

Example 1 - Moment (long) Factors (positive)

5.08333.0

0.15.0

75.09.09.0

Factor

Need to interpolate

to determine how

the positive moment

is distributed.

Example 1 - Moment (long) column/middle strips

Components on the beam (long).

Negative – Moment

Positive + Moment

Column Strip

0.80 212.4 k-ft 169.9 k-ft

0.80 114.4 k-ft 91.5 k-ft

Negative – Moment

Positive + Moment

Middle Strip

0.20 212.4 k-ft 42.5 k-ft

0.20 114.4 k-ft 22.9 k-ft

Example 1 - Moment (long)-beam/slab distribution (negative)

When 1 (l2/l1) > 1.0, ACI Code Section 13.6.5 indicates

that 85 % of the moment in the column strip is assigned

to the beam and balance of 15 % is assigned to the slab

in the column strip.

Beam Moment

Slab Moment

Column Strip - Negative Moment (169.9 k-ft)

0.85 169.9 k-ft 144.4 k-ft

0.15 169.9 k-ft 25.5 k-ft

Example 1 - Moment (long)-beam/slab distribution (positive) When 1 (l2/l1) > 1.0, ACI Code Section 13.6.5

indicates that 85 % of the moment in the column strip

is assigned to the beam and balance of 15 % is

assigned to the slab in the column strip.

Beam Moment

Slab Moment

Column Strip - Positive Moment (91.5 k-ft)

0.85 91.5 k-ft 77.8 k-ft

0.15 91.5 k-ft 13.7 k-ft

Example 1- Moment (short)

The factored components

of the moment for the

beam (short).

Negative – Moment

Positive + Moment

0.65 264.2 k-ft 171.7 k-ft

0.35 264.2 k-ft 92.5 k-ft

Example 1 - Moment (short) coefficients

The moments of inertia about beam, Ib = 22,453 in4 and

Is = 8232 in4 (short direction) are need to determine the

distribution of the moments between the column and

middle strip.

b b1 4

24 ft1.22222

22453 in2.73

8232 in

2.73* 1.2222 3.333

Example 1 - Moment (short) Factors (negative)

6833.0

0.12222.1

0.20.1

45.075.075.0

Factor

Need to interpolate

to determine how

the negative

moment is

distributed.

Example 1 - Moment (short) Factors (positive)

Need to interpolate

to determine how

the positive

moment is

distributed.

6833.0

0.12222.1

0.20.1

45.075.075.0

Factor

Example 1- Moment (short) column/middle strip

Components on the beam (short).

Negative – Moment

Positive + Moment

Column Strip

0.683 171.7 k-ft 117.3 k-ft

0.683 92.5 k-ft 63.2 k-ft

Negative – Moment

Positive + Moment

Middle Strip

0.317 171.7 k-ft 54.4 k-ft

0.317 92.5 k-ft 29.3 k-ft

Example 1 - Moment (short) beam/slab distribution (negative)

When 1 (l2/l1) > 1.0, ACI Code Section 13.6.5 indicates

that 85 % of the moment in the column strip is assigned

to the beam and balance of 15 % is assigned to the slab

in the column strip.

Beam Moment

Slab Moment

Column Strip - Negative Moment (117.3 k-ft)

0.85 117.3 k-ft 99.7 k-ft

0.15 117.3 k-ft 17.6 k-ft

Example 1 - Moment (short) beam/slab distribution (positive)

When 1 (l2/l1) > 1.0, ACI Code Section 13.6.5

indicates that 85 % of the moment in the column strip

is assigned to the beam and balance of 15 % is

assigned to the slab in the column strip.

Beam Moment

Slab Moment

Column Strip - Positive Moment (63.2 k-ft)

0.85 63.2 k-ft 53.7 k-ft

0.15 63.2 k-ft 9.5 k-ft

Example 1 - Summary

Example 1- Reinforcement calculation

Use same procedure to do the reinforcement on the

concrete. Calculate the bars from the earlier

version of the problem.

Example 1 - Reinforcement calculation

Computing the reinforcement uses:

R*1.747.170.1

R*1.770.159.01 R

Example 1 - Reinforcement calculation for long –middle strip (negative)

Compute the reinforcement need for the negative moment

in long direction. Middle strip width b =120 in. (10 ft),

d =6 in. and Mu = 42.5 k-ft

12 in.42.5 k-ft

1 ft0.118 ksi

bd 120 in. 6 in.

1.7 0.118 ksi1.70 0

0.9 3 ksi

Compute the reinforcement need for the negative

moment in long direction. Middle strip width b =120 in.

(10 ft) d =6 in. and Mu = 42.5 k-ft

1.70 1.7 4 0.74330.0449

0.0449 3 ksi0.00225

60 ksi

The area of the steel reinforcement for a strip width

b =120 in. (10 ft), d = 6 in., and h = 7 in.

0.00225 120 in. 6 in. 1.62 in

0.0018 0.0018 120 in. 7 in. 1.52 in

b =120 in. (10 ft), d = 6 in., and As = 1.62 in2. Use a

#4 bar (Ab =0.20 in2 )

Maximum spacing is 2(h) or

18 in.

So 13.33 in < 14 in. OK! Use 10 #4

1.62 in# bars 8.08 Use 9 bars

0.2 in

120 in.13.33 in.

Example 1 – Long Results

The long

direction using

# 4 bars

Negative Positive Negative Positive

Moment (k-ft) 25.5 13.7 42.5 22.9

b (in) 120 120 120 120

d (in) 6 6 6 6

h (in) 7 7 7 7

fy (ksi) 60 60 60 60

fc (ksi) 3 3 3 3

Ru (ksi) 0.07083 0.03806 0.11806 0.06361

w 0.02665 0.01421 0.04491 0.02390

0.00133 0.00071 0.00225 0.00119

As (in2) 0.96 0.51 1.62 0.86

As(min) (in2) 1.51 1.51 1.51 1.51

# bars req 7.56 7.56 8.08 7.56

spacing (in) 15.00 15.00 13.33 15.00

# bars (#4) 10 10 10 10

spacing (in) 12 12 12 12

Column Strip Middle Strip

Example 1 – Long summary

The long direction

using # 4 bars

Example 1 – Short Results

The short

direction

using # 4

Moment (k-ft) 17.6 9.5 54.4 29.3

b (in) 120 120 168 168

d (in) 6 6 6 6

h (in) 7 7 7 7

fy (ksi) 60 60 60 60

fc (ksi) 3 3 3 3

Ru (ksi) 0.04889 0.02639 0.10794 0.05813

w 0.01830 0.00983 0.04096 0.02181

0.00092 0.00049 0.00205 0.00109

As (in2) 0.66 0.35 2.06 1.10

As(min) (in2) 1.51 1.51 2.12 2.12

# bars req 7.56 7.56 10.58 10.58

spacing (in) 15.00 15.00 15.27 15.27

# bars (#4) 10 10 14 14

spacing (in) 12 12 12 12

Example 1 – Short Summary

The short direction

using # 4 bars

Example –Two-way Slab (Panels)

Using the direct design method,

design the typical exterior flat-

slab panel with drop down

panels only. All panels are

supported on 20 in. square

columns, 12 ft long. The slab

carries a uniform service live

load of 80 psf and service dead

load that consists of 24 psf of

finished in addition to the slab

self-weight. Use fc = 4 ksi and

fy = 60 ksi

The thickness of the slab is

found using

in 8.0in 44.7

in 12*24

panels Noin 12.8

in 12*24

From the ACI Code limitation:

For panels with discontinuous edges, end beams

with a minimum equal to 0.8 must be used;

otherwise the minimum slab thickness calculated

by the equations must be increased by at least 10%.

When drop panels are used without beams, the

minimum slab thickness may be reduced by 10 %.

The drop panels should extend in each direction

from the centerline of support a distance not less

than one-sixth of the span length in that direction

between center to center of supports and also

project below the slab at least h/4.

Example –Two-way Slab (Panels) From the ACI Code limitation:

Regardless of the values obtained for the equations,

the thickness of two-way slabs shall not be less

than the following:

For slabs without beams or drop panels, 5 in.

for slabs without beams but with drop

panels, 4 in.

for slabs with beams on all four sides with

m > 2.0, 3.5 in. and for m < 2.0, 5 in. (ACI

Code 9.5.3)

Therefore, the panel thickness is

The panel half width are at least L/6 in length.

in. 10

in. 8 in. 8

ft 3.5 ft 33.3

Therefore, the drop down panel thickness is 10 in.

and has 7 ft x 8 ft.

The load on the slab is given as:

The load on the panel is

1 ftSlab load 24 lb/ft 8 in. 150 lb/ft 124 lb/ft

12 in.

1.2 124 lb/ft 1.6 80 lb/ft 276.8 lb/ftw

1 ftPanel load 24 lb/ft 10 in. 150 lb/ft 149 lb/ft

12 in.

1.2 149 lb/ft 1.6 80 lb/ft 306.8 lb/ftw

The drop panel length is L/3 in each direction, then the

average wu is

2 1276.8 lb/ft 306.8 lb/ft

286.8 lb/ft

The punch out shear at center column is

in. 8.75

in. 5.0in. 75.0 in. 10

in. 151

in. 8.75 in. 204 o

The punch out shear at center column is

1 ft0.287 k/ft 24 ft 20 ft 28.75 in.

12 in.

136.1 k

0.75 4 4000 115 in. 8.75 in.

190.9 k OK.

V f b d

The punch out shear at panel is

in. 6.75

in. 5.0in. 75.0 in. 8

in. 387

in. 6.75

in. 12ft. 72

in. 6.75

in. 12ft. 82 o

The punch out shear at panel is

24 ft 20 ft

0.287 k/ft 1 ft 1 ft102.75 in. 90.75 in.

12 in. 12 in.

119.2 k

0.75 4 4000 387 in. 6.75 in.

495.6 k OK.

V f b d

One way shear is not critical.

Moment Mo for the two directions are:

direction

0.287 k/ft 20 ft 22.33 ft357.9 k-ft

direction

0.287 k/ft 24 ft 18.33 ft289.4 k-ft

The column strip will be 10 ft. (20 ft /4 = 5ft),

therefore the middle strips for long section is 10 ft and

the middle strip for the short section will be 14 ft.

The average d for

the panel section in. 8.5

in. 5.1 in. 10

The average d for

the slab section in. 6.5

in. 5.1 in. 8

The factored

components of the

moment for the beam

(long) is similar to an

interior beam.

Negative – Moment

Positive + Moment

0.65 357.9 k-ft 232.6 k-ft

0.35 357.9 k-ft 125.3 k-ft

Components on the beam (long) interior.

Negative – Moment

Positive + Moment

Column Strip

0.75 232.6 k-ft 174.5 k-ft

0.60 125.3 k-ft 75.2 k-ft

Components on the beam (long) interior.

Negative – Moment

Positive + Moment

Middle Strip

0.25 232.6 k-ft 58.2 k-ft

0.40 125.3 k-ft 50.2 k-ft

Computing the reinforcement uses:

R*1.747.170.1

R*1.770.159.01 R

Compute the reinforcement need for the internal moment

in long direction. Strip width b =120 in. (10 ft) d =8.5 in.

and Mu = 174.5 k-ft

12 in.174.5 k-ft

1 ft0.242 ksi

bd 120 in. 8.5 in.

1.7*0.242 ksi1.70 0

0.9 4 ksi

Compute the reinforcement need for the internal moment

in long direction. Strip width b =120 in. (10 ft) d =8.5 in.

and Mu = 174.5 k-ft

1.70 1.7 4 0.11410.0700

0.0700 4 ksi0.00466

60 ksi

b =120 in. (10 ft), d = 8.5 in., and h = 10 in.

0.00466 120 in. 8.5 in. 4.76 in

0.0018 0.0018 120 in. 10 in. 2.16 in

b =120 in. (10 ft), d = 8.5 in., and As = 4.76 in2. Use

a #5 bar (Ab = 0.31 in2 )

Maximum spacing is 2(h)

or 18 in.

So 7.5 in. < 18 in. OK

4.76 in# bars 15.3 Use 16 bars

0.31 in

120 in.s 7.5 in.

The long

direction

Moment (k-ft) 174.5 75.2 58.2 50.2

b (in) 120 120 120 120

d (in) 8.5 6.5 6.5 6.5

h (in) 10 8 8 8

fy (ksi) 60 60 60 60

fc (ksi) 4 4 4 4

Ru (ksi) 0.24152 0.17799 0.13775 0.11882

w 0.06997 0.05097 0.03917 0.03367

0.00466 0.00340 0.00261 0.00224

As (in2) 4.76 2.65 2.04 1.75

As(min) (in2) 2.16 1.73 1.73 1.73

# bars req 15.35 13.25 10.18 8.75

spacing (in) 7.50 8.57 10.91 13.33

# bars 16 #5 15 #4 10 #4 10 #4

spacing (in) 7.5 8 12 12

The short

direction

Moment (k-ft) 141.1 60.8 47.1 40.5

b (in) 120 120 168 168

d (in) 8.5 6.5 6.5 6.5

h (in) 10 8 8 8

fy (ksi) 60 60 60 60

fc (ksi) 3 3 3 3

Ru (ksi) 0.19529 0.14391 0.07963 0.06847

w 0.07570 0.05508 0.03002 0.02575

0.00379 0.00275 0.00150 0.00129

As (in2) 3.86 2.15 1.64 1.41

As(min) (in2) 2.16 1.73 2.42 2.42

# bars req 12.45 10.74 12.10 12.10

spacing (in) 9.23 10.91 12.92 12.92

# bars 16 #5 12 #4 14 #4 14 #4

spacing (in) 7.5 10 12 12

Lecture25,Design of Two Way Slab

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