425-Chp11-Slender Column and Two Way Slabs

Slender Columns and Slender Columns and Two-way SlabsTwo-way Slabs

Lecture GoalsLecture Goals

Slender Column DesignOne-way and two-way slabSlab thickness, h

Design of Long Columns- Design of Long Columns- ExampleExample

A rectangular braced column of a multistory frame building has floor height lu =25 ft. It is subjected to service dead-load moments M2= 3500 k-in. on top and M1=2500 k-in. at the bottom. The service live load moments are 80% of the dead-load moments. The column carries a service axial dead-load PD = 200 k and a service axial live-load PL = 350 k. Design the cross section size and reinforcement for this column. Given A = 1.3 and B = 0.9. Use a d’=2.5 in. cover with an sustain load = 50 % and fc = 7 ksi and fy = 60 ksi.


Compute the factored loads and moments are 80% of the dead loads

u D L

1u D L

2u D L

1.2 1.6 1.2 200 k 1.6 350 k

800 k

1.2 1.6 1.2 2500 k-in 1.6 0.8 2500 k-in

6200 k-in.

1.2 1.6 1.2 3500 k-in 1.6 0.8 3500 k-in

8680 k-in.

P P P

M M M

M M M


Compute the k value for the braced compression members

Therefore, use k = 0.81

A B

min

0.7 0.05 0.7 0.05 1.3 0.9

0.81 1.0

0.85 0.05 0.85 0.05 0.9

0.895 1.0

k

k

Design of Long Columns- Design of Long Columns- Example Example

Check to see if slenderness is going to matter. An initial estimate of the size of the column will be an inch for every foot of height. So h = 25 in.

n0.81 25 ft 12 in./ft

32.4r 0.3 25 in.

6200 k-in.32.4 34 12 25.43

8680 k-in.

kl

We need to be concerned with slender columns


So slenderness must be considered. Since frame has no side sway, M2 = M2ns, s = 0 Calculate the minimum M2 for the ratio computations.

2,min u

2

0.6 0.03 800 k 0.6 0.03 25 in.

1080 k-in. 8680 k-in.

M P h

M


Compute components of concrete

The moment of inertia of the column is

1.51.5c c

6 3

33 33 150 7000

5.07x10 psi 5.07x10 ksi

E w f

33

g

4

25 in. 25 in.

12 12

32552 in

bhI


Compute the stiffness, EI

3 4c g

d

7 2

0.4 5.07x10 ksi 32552 in0.4

1 1 0.5

4.4x10 k-in

E IEI


The critical load (buckling), Pcr, is

2 7 22

cr 2 2

u

4.4x10 k-in

12 in.0.81 25 ft

ft

7354.3 k

EIP

kl


Compute the coefficient, Cm, for the magnification coefficient

1m

2

0.6 0.4

6200 k-in.0.6 0.4 0.89 0.4

8680 k-in.

MC

M


The magnification factor

mns

u

cr

0.89

800 k1 10.75 0.75 7354.3 k

1.04 1.0

C

P

P


The design moment is

Therefore, the design conditions are

c ns 2 1.04 8680 k-in. 9027.2 k-in.M M

c c800 k & M 9027.2 k-in.

9027.2 k-in.e 11.28 in.

800 k

P


Assume that the = 2.0 % or 0.020

Use 14 # 9 bars or 14 in2

2 2s 0.02 25 in. 12.5 inA

2s

2cs

7.0 in

7.0 in

A

A


The column is compression controlled so c/d > 0.6. Check the values for c/d = 0.6

1

0.6 0.6 22.5 in. 13.5 in.

0.7 13.5 in. 9.45 in.

c d

a c


Check the strain in the tension steel and compression steel.

s1 cu

cs1 s s1

cs1

13.5 in. 2.5 in.0.003

13.5 in.

0.00244

29000 ksi 0.00244

70.76 ksi 60 ksi

c d

c

f E

f


The tension steel is

s cu

s s s

22.5 in. 13.5 in.0.003

13.5 in.

0.002

29000 ksi 0.002

58 ksi

d c

c

f E


Combined forces are

c c

2s1 cs cs c

2s s

0.85 0.85 7 ksi 25 in. 9.45 in.

1405.7 k

0.85 7 in 60 ksi 0.85 7 ksi

378.35 k

7 in 58 ksi

406.0 k

C f ba

C A f f

T A f


Combined force is

n c s1

1405.7 k 378.35 k 406.0 k

1378.05 k

P C C T


Moment is

n c s12 2 2 2

9.45 in.1405.7 k 12.5 in.

2

378.35 k 12.5 in. 2.5 in.

406.0 k 22.5 in. 12.5 in.

18773 k-in.

h a h hM C C d T d


The eccentricity is

Since the e = 11.28 in. < 13.62 in. The section is in the compression controlled region = 0.65. You will want to match up the eccentricity with the design.

n

n

18773 k-in

1378.05 k

13.62 in.

Me

P


We need to match up the eccentricity of the problem. This done varying the c/d ratio to get the eccentricity to match. Check the values for c/d = 0.66

1

0.66 0.66 22.5 in. 14.85 in.

0.7 14.85 in. 10.395 in.

c d

a c


Check the strain in the tension steel and compression steel.

s1 cu

cs1 s s1

cs1

14.85 in. 2.5 in.0.003

14.85 in.

0.00249

29000 ksi 0.00249

72.35 ksi 60 ksi

c d

c

f E

f


The tension steel is

s cu

s s s

22.5 in. 14.85 in.0.003

14.85 in.

0.00155

29000 ksi 0.00155

44.82 ksi

d c

c

f E


Combined forces are

c c

2s1 cs cs c

2s s

0.85 0.85 7 ksi 25 in. 10.395 in.

1545.26 k

0.85 7 in 60 ksi 0.85 7 ksi

378.35 k

7 in 44.82 ksi

313.74 k

C f ba

C A f f

T A f


Combined force is

n c s1

1546.26 k 378.35 k 313.74 k

1610.9 k

P C C T


Moment is

n c s12 2 2 2

10.395 in.1545.26 k 12.5 in.

2

378.35 k 12.5 in. 2.5 in.

313.74 k 22.5 in. 12.5 in.

18205.2 k-in

h a h hM C C d T d


The eccentricity is

Since the e 11.28 in. The reduction factor is equal to = 0.65. Compute the design load and moment.

n

n

18205.2 k-in

1610.9 k

11.30 in.

Me

P


The design conditions are

The problem matches the selection of the column.

u n

u n

0.65 1610.9 k

1047.1 k 800 k OK!

0.65 18205.2 k-in

11833.4 k-in. 9027.2 k-in. OK!

P P

M M


Design the ties for the column

Provide #3 ties, spacing will be the minimum of:

Therefore, provide #3 ties @ 18 in. spacing.

stirrup

bar

48 48 0.375 in. 18 in.

smallest 16 16 1.128 in. 18 in. controls

h 25 in.

d

s d

Using Interaction Using Interaction Diagrams Diagrams

Determine eccentricity.Estimate column size required base on axial load.Determine e/h and required Pn/Ag, Mn/(Agh)Determine which chart to use from fc, fy and . Determine from the chart.

Select steel sizes.Check values.Design ties by ACI codeDesign sketch

Two-way SlabsTwo-way Slabs

Comparison of One-way and Comparison of One-way and Two-way slab behaviorTwo-way slab behavior

One-way slabs carry load in one direction.

Two-way slabs carry load in two directions.


One-way and two-way slab action carry load in two directions.

One-way slabs: Generally, long side/short side > 1.5


Flat slab

Two-way slab with beams


For flat plates and slabs the column connections can vary between:


Flat Plate

Waffle slab

Comparison of One-way and Comparison of One-way and

Two-way slab behaviorTwo-way slab behavior

The two-way ribbed slab and waffled slab system: General thickness of the slab is 2 to 4 in.

Comparison of One-way and Comparison of One-way and Two-way slab behavior Two-way slab behavior Economic Choices Economic Choices

Flat Plate suitable span 20 to 25 ft with LL= 60 -100 psf

Advantages Low cost formwork Exposed flat ceilings Fast

Disadvantages Low shear capacity Low Stiffness (notable deflection)


Flat Slab suitable span 20 to 30 ft with LL= 80 -150 psf

Advantages Low cost formwork Exposed flat ceilings Fast

Disadvantages Need more formwork for capital and panels


Waffle Slab suitable span 30 to 48 ft with LL= 80 -150 psf

Advantages Carries heavy loads Attractive exposed ceilings Fast

Disadvantages Formwork with panels is expensive


One-way Slab on beams suitable span 10 to 20 ft with LL= 60-100 psf Can be used for larger spans with relatively higher

cost and higher deflections One-way joist floor system is suitable span 20 to 30 ft with LL= 80-120 psf Deep ribs, the concrete and steel quantities are

relative low Expensive formwork expected.


ws =load taken by short direction

wl = load taken by long direction

A = B

Rule of Thumb: For B/A > 2, design as one-way slab

EI

Bw

EI

Aw

384

5

384

5 4l

4s

ls4

4

l

s 162A BFor wwA

B

w

w

Two-Way Slab DesignTwo-Way Slab DesignStatic Equilibrium of Two-Way Slabs

Analogy of two-way slab to plank and beam floor

Section A-A:

Moment per ft width in planks

Total Moment

ft/ft-k 8

21wlM

ft-k 8

21

2f

lwlM

Two-Way Slab DesignTwo-Way Slab Design

Static Equilibrium of Two-Way Slabs

Analogy of two-way slab to plank and beam floor

Uniform load on each beam

Moment in one beam (Sec: B-B) ft-k 82

221

lb

lwlM

k/ft 2

1wl

Two-Way Slab DesignTwo-Way Slab Design

Static Equilibrium of Two-Way Slabs

Total Moment in both beams

Full load was transferred east-west by the planks and then was transferred north-south by the beams;

The same is true for a two-way slab or any other floor system.

ft-k 8

22

1

lwlM

General Design General Design ConceptsConcepts

(1) Direct Design Method (DDM)

Limited to slab systems to uniformly distributed loads and supported on equally spaced columns. Method uses a set of coefficients to determine the design moment at critical sections. Two-way slab system that do not meet the limitations of the ACI Code 13.6.1 must be analyzed more accurate procedures

General Design General Design ConceptsConcepts

(2) Equivalent Frame Method (EFM)

A three-dimensional building is divided into a series of two-dimensional equivalent frames by cutting the building along lines midway between columns. The resulting frames are considered separately in the longitudinal and transverse directions of the building and treated floor by floor.

Equivalent Frame Method Equivalent Frame Method

(EFM)(EFM)

Longitudinal equivalent frame

Transverse equivalent frame

Equivalent Frame Method Equivalent Frame Method

(EFM)(EFM)

Elevation of the frame Perspective view

Method of AnalysisMethod of Analysis

(1) Elastic Analysis

Concrete slab may be treated as an elastic plate. Use Timoshenko’s method of analyzing the structure. Finite element analysis

Method of AnalysisMethod of Analysis(2) Plastic Analysis

The yield method used to determine the limit state of slab by considering the yield lines that occur in the slab as a collapse mechanism.

The strip method, where slab is divided into strips and the load on the slab is distributed in two orthogonal directions and the strips are analyzed as beams.

The optimal analysis presents methods for minimizing the reinforcement based on plastic analysis

Method of AnalysisMethod of Analysis

(3) Nonlinear analysis

Simulates the true load-deformation characteristics of a reinforced concrete slab with finite-element method takes into consideration of nonlinearities of the stress-strain relationship of the individual members.

Column and Middle Column and Middle StripsStrips

The slab is broken up into column and middle strips for analysis

Minimum Slab Thickness for Minimum Slab Thickness for Two-way ConstructionTwo-way Construction

The ACI Code 9.5.3 specifies a minimum slab thickness to control deflection. There are three empirical limitations for calculating the slab thickness (h), which are based on experimental research. If these limitations are not met, it will be necessary to compute deflection.


22.0 m (a) For

2.0536

200,0008.0

m

yn

fl

h

fy in psi. But not less than 5 in.


m2 (b) For

936

200,0008.0 y

n

fl

h

fy in psi. But not less than 3.5 in.


2.0m (c) For

Use the following table 9.5(c)


Slabs without interior beams spanning between supports and ratio of long span to short span < 2

See section 9.5.3.3 For slabs with beams spanning between supports on all sides.

Minimum Slab Thickness for Minimum Slab Thickness for two-way constructiontwo-way construction

The definitions of the terms are:

h = Minimum slab thickness without interior beams

ln =

m=

Clear span in the long direction measured face to face of column

the ratio of the long to short clear span

The average value of for all beams on the sides of the panel.

Definition of Beam-to-Slab Definition of Beam-to-Slab Stiffness Ratio, Stiffness Ratio,

Accounts for stiffness effect of beams located along slab edge reduces deflections of panel

adjacent to beams.

slab of stiffness flexural

beam of stiffness flexural

Definition of Beam-to-Slab Definition of Beam-to-Slab Stiffness Ratio, Stiffness Ratio,

With width bounded laterally by centerline of adjacent panels on each side of the beam.

scs

bcb

scs

bcb

E

E

/4E

/4E

I

I

lI

lI

slab uncracked of inertia ofMoment I

beam uncracked of inertia ofMoment I

concrete slab of elasticity of Modulus E

concrete beam of elasticity of Modulus E

s

b

sb

cb

Beam and Slab Sections for Beam and Slab Sections for calculation of calculation of

Beam and Slab Sections for Beam and Slab Sections for calculation of calculation of

Definition of beam cross-section

Charts may be used to calculate


Slabs without drop panels meeting 13.3.7.1 and 13.3.7.2,

tmin = 5 in

Slabs with drop panels meeting 13.3.7.1 and 13.3.7.2,

tmin = 4 in

Example - SlabExample - Slab A flat plate floor system with panels 24 by 20 ft is supported on 20 in. square columns. Determine the minimum slab thickness required for the interior and corner panels. Use fc = 4 ksi and fy = 60 ksi

Example - SlabExample - Slab

Slab thickness, from table 9.5(c) for fy = 60 ksi and no edge beams

nmin

n

min

30

20 in. 1 ft.24 ft. 2 22.33 ft.

2 12 in.

12 in.22.33 ft.

1 ft.8.93 in. 9 in.

30

lh

l

h

Example - Slab Example - Slab

Slab thickness, from table 9.5(c) for fy = 60 ksi and no edge beams for = m = 0 (no beams)

nmin

min

3312 in.

22.33 ft.1 ft.

8.12 in. 8.5 in.33

lh

h

Example – Example – CalculationsCalculationsThe floor system consists of solid slabs and beams in two directions supported on 20-in. square columns. Determine the minimum slab thickness, h, required for the floor system. Use fc = 4 ksi and fy = 60 ksi

Example – Example – CalculationsCalculations

The cross-sections are:


To find h, we need to find m therefore Ib, Islab and for each beam and slab in long short direction. Assume slab thickness h = 7 in. so that x = y < 4 tf

f22 in. 7 in. 15 in. 4 4 7 in. 28 in.t

e 16 in. 2 15 in. 46 in.b


Compute the moment of inertia and centroid

b h Ai (in2) yi (in) yiAi (in

3) I (in4) d (in) d2A (in4)

Flange 7 46 322 3.5 1127 1314.833 -4.69751 7105.442Beam 15 16 240 14.5 3480 4500 6.302491 9533.135

562 4607 5814.833 16638.58

ybar = 8.197509 in

I = 22453.41 in4

4beam

33slab

4

22453 in

1 1 12 in.20 ft 7 in.

12 12 1 ft.

6860 in

I

I bh


Compute the coefficient for the long direction

Short side of the moment of inertia

4beam

long 4slab

22453 in

6860 in

3.27

EI

EI

33slab

4

1 1 12 in.24 ft 7 in.

12 12 1 ft.

8232 in

I bh


Compute the coefficient for short direction

The average m for an interior panel is

4beam

short 4slab

22453 in

8232 in

2.73

EI

EI

long shortavg

2 2 2 3.27 2 2.73

4 43.0


Compute the coefficient

Compute the thickness for m > 2

Use slab thickness, 6.5 in. or 7 in.

long

short

20 in. 1 ft.24 ft. 2

2 12 in.1.22

20 in. 1 ft.20 ft. 2

2 12 in.

l

l

yn

12 in. 600000.8 22.33 ft. 0.8200000 1 ft. 200000

36 9 36 9 1.22

6.28 in.

fl

h


Compute the moment of inertia and centroid for the L-beam

b h Ai (in2) yi (in) yiAi (in

3) I (in4) d (in) d2A (in4)

Flange 7 27 189 3.5 661.5 771.75 -5.36585 5441.761Beam 15 12 180 14.5 2610 3375 5.634146 5713.849

369 3271.5 4146.75 11155.61

ybar = 8.865854 in

I = 15302.36 in4

4L-beam

33slab

4

15302 in

1 1 12 in.10 ft 7 in.

12 12 1 ft.

3430 in

I

I bh


Compute the m coefficient for long direction

Short side of the moment of inertia

4L-beam

long 4slab

15302 in

3430 in

4.46

EI

EI

33slab

4

1 1 12 in.12 ft 7 in.

12 12 1 ft.

4116 in

I bh


Compute the m coefficient for the short direction

4L-beam

short 4slab

15302 in

4116 in

3.72

EI

EI


Compute the m coefficient for the edges and corner

m

4.46 2.73 3.27 2.73

43.30

m

3.72 3.27 2.73 3.27

43.25


Compute the m coefficient for the edges and corner

m

3.72 4.46 2.73 3.27

43.55


Compute the largest length ln of the slab/beam, edge to first interior column.

n

20 in. 1 ft. 12 in. 1 ft.24 ft.

2 12 in. 2 12 in.

22.67 ft.

l


Compute the thickness of the slab with m > 2

The overall depth of the slab is 7 in.

Use slab thickness, 6.5 in. or 7 in.

yn

12 in. 600000.8 22.67 ft. 0.8200000 1 ft. 200000

36 9 36 9 1.22

6.37 in.

fl

h

425-Chp11-Slender Column and Two Way Slabs

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