425-chp11-slender column and two way slabs
DESCRIPTION
Slender Column and Two Way SlabsTRANSCRIPT
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Slender Columns and Slender Columns and Two-way SlabsTwo-way Slabs
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Lecture GoalsLecture Goals
Slender Column DesignOne-way and two-way slabSlab thickness, h
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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.
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Design of Long Columns- Design of Long Columns- ExampleExample
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
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Design of Long Columns- Design of Long Columns- ExampleExample
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
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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
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Design of Long Columns- Design of Long Columns- Example Example
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
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Design of Long Columns- Design of Long Columns- Example Example
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
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Design of Long Columns- Design of Long Columns- ExampleExample
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
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Design of Long Columns- Design of Long Columns- ExampleExample
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
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Design of Long Columns- Design of Long Columns- ExampleExample
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
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Design of Long Columns- Design of Long Columns- ExampleExample
The magnification factor
mns
u
cr
0.89
800 k1 10.75 0.75 7354.3 k
1.04 1.0
C
P
P
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Design of Long Columns- Design of Long Columns- ExampleExample
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
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Design of Long Columns- Design of Long Columns- ExampleExample
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
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Design of Long Columns- Design of Long Columns- ExampleExample
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
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Design of Long Columns- Design of Long Columns- ExampleExample
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
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Design of Long Columns- Design of Long Columns- ExampleExample
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
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Design of Long Columns- Design of Long Columns- ExampleExample
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
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Design of Long Columns- Design of Long Columns- ExampleExample
Combined force is
n c s1
1405.7 k 378.35 k 406.0 k
1378.05 k
P C C T
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Design of Long Columns- Design of Long Columns- ExampleExample
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
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Design of Long Columns- Design of Long Columns- ExampleExample
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
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Design of Long Columns- Design of Long Columns- ExampleExample
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
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Design of Long Columns- Design of Long Columns- ExampleExample
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
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Design of Long Columns- Design of Long Columns- ExampleExample
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
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Design of Long Columns- Design of Long Columns- ExampleExample
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
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Design of Long Columns- Design of Long Columns- ExampleExample
Combined force is
n c s1
1546.26 k 378.35 k 313.74 k
1610.9 k
P C C T
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Design of Long Columns- Design of Long Columns- ExampleExample
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
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Design of Long Columns- Design of Long Columns- ExampleExample
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
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Design of Long Columns- Design of Long Columns- ExampleExample
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
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Design of Long Columns- Design of Long Columns- ExampleExample
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
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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
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Two-way SlabsTwo-way Slabs
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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.
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Comparison of One-way and Comparison of One-way and Two-way slab behaviorTwo-way slab behavior
One-way and two-way slab action carry load in two directions.
One-way slabs: Generally, long side/short side > 1.5
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Comparison of One-way and Comparison of One-way and Two-way slab behaviorTwo-way slab behavior
Flat slab
Two-way slab with beams
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Comparison of One-way and Comparison of One-way and Two-way slab behaviorTwo-way slab behavior
For flat plates and slabs the column connections can vary between:
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Comparison of One-way and Comparison of One-way and Two-way slab behaviorTwo-way slab behavior
Flat Plate
Waffle slab
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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.
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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)
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Comparison of One-way and Comparison of One-way and Two-way slab behavior Two-way slab behavior Economic Choices Economic Choices
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
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Comparison of One-way and Comparison of One-way and Two-way slab behavior Two-way slab behavior Economic Choices Economic Choices
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
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Comparison of One-way and Comparison of One-way and Two-way slab behavior Two-way slab behavior Economic Choices Economic Choices
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.
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Comparison of One-way and Comparison of One-way and Two-way slab behaviorTwo-way slab behavior
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
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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
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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
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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
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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
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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.
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Equivalent Frame Method Equivalent Frame Method
(EFM)(EFM)
Longitudinal equivalent frame
Transverse equivalent frame
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Equivalent Frame Method Equivalent Frame Method
(EFM)(EFM)
Elevation of the frame Perspective view
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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
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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
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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.
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Column and Middle Column and Middle StripsStrips
The slab is broken up into column and middle strips for analysis
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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.
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Minimum Slab Thickness for Minimum Slab Thickness for Two-way ConstructionTwo-way Construction
22.0 m (a) For
2.0536
200,0008.0
m
yn
fl
h
fy in psi. But not less than 5 in.
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Minimum Slab Thickness for Minimum Slab Thickness for Two-way ConstructionTwo-way Construction
m2 (b) For
936
200,0008.0 y
n
fl
h
fy in psi. But not less than 3.5 in.
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Minimum Slab Thickness for Minimum Slab Thickness for Two-way ConstructionTwo-way Construction
2.0m (c) For
Use the following table 9.5(c)
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Minimum Slab Thickness for Minimum Slab Thickness for Two-way ConstructionTwo-way Construction
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.
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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.
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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
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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
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Beam and Slab Sections for Beam and Slab Sections for calculation of calculation of
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Beam and Slab Sections for Beam and Slab Sections for calculation of calculation of
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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
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Minimum Slab Thickness for Minimum Slab Thickness for Two-way ConstructionTwo-way Construction
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
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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
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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
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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
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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
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Example – Example – CalculationsCalculations
The cross-sections are:
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Example – Example – CalculationsCalculations
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
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Example – Example – CalculationsCalculations
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
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Example – Example – CalculationsCalculations
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
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Example – Example – CalculationsCalculations
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
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Example – Example – CalculationsCalculations
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
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Example – Example – CalculationsCalculations
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
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Example – Example – CalculationsCalculations
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
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Example – Example – CalculationsCalculations
Compute the m coefficient for the short direction
4L-beam
short 4slab
15302 in
4116 in
3.72
EI
EI
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Example – Example – CalculationsCalculations
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
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Example – Example – CalculationsCalculations
Compute the m coefficient for the edges and corner
m
3.72 4.46 2.73 3.27
43.55
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Example – Example – CalculationsCalculations
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
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Example – Example – CalculationsCalculations
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