diseño de marcos trabe columna

7/28/2019 diseño de marcos trabe columna

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Lecture 15 – of 12

Lecture 15 – Columns (cont.)

In the previous lecture, we talked about columns having small eccentricity (i.e.,small applied moment). While this may be the case for interior columns withoffsetting moments, the majority of concrete columns do experience appliedmoments.

Concrete framed buildings typically have columns that are poured monolithicallywith beams and slabs, thus creating a statically-indeterminate frame such as the

one shown below:

The analysis of such a frame is usually quite complex and requires computer software such as STAAD or approximate analysis methods such as the PortalMethod and others discussed in Structural Theory.




The compression capacity of a reinforced column is reduced by the bendingstresses on the column and vice-versa. A graph of the axial load capacity of a

column vs. the moment capacity of a typical column is shown below (fromLecture 14):

Determining points along the curve is quite laborious and typically not done usinghand calculations. Instead, computer programs or design guides are used toperform column analysis and design.

Below are some “Column Interaction Diagrams” that are used for column

analysis and design.

A x

i a l c o m p r e s s i o n c a p a c i t y

Bending moment capacity

Pure compression (no

applied moment)

Pure bending (no

applied axial load)




Rectangular TIED Column Interaction Diagrams

R4-60.75

“R” = Rectangular cross-section“4” = f’c = 4 KSI

“60” = Grade 60 vertical bars

“75” = γ




Circular SPIRAL Column Interaction Diagrams




“Short” Column Design:

Short columns are not considered to be susceptible to theeffects of buckling as are long columns. The ACI dictates that

short columns satisfy the slenderness ratio as shown below:

Short column IF 22min

≤r KL

Where: K = end fixity factor

L = unbraced length in inches

r min = least radius of gyration, inches= 0.3h for rectangular or square columns

= 0.25dout for circular columns where dout = outer dia.




ExampleGIVEN: A 20” x 20” square interior tied column is shown below. Use concrete f’c

= 4000 PSI and 8 - #8 grade 60 vertical bars and #3 ties. Concrete cover = 1½”. All loads are factored and includes beam weight. Assume the beams are “fixed”to the columns.

REQUIRED:

1) Determine the applied factored axial load, Pu on the middle column. Besure to add column weight. Assume the loads from the roof above are75% of the floor load.

2) Determine the slenderness ratiominr

KLusing K = 0.7 and determine if

column qualifies as “short.”3) Determine total factored moments Mtotal applied to the column.

4) Determine if the column is acceptable based on “Column InteractionDiagram.”

wu = 1.7 KLF wu = 2.5 KLF

46 K46 K

14’-0”

30’-0”22’-0”

Beam 1

20” x 20” Middle

column

Beam 2




Step 1 – Determine the total factored axial load, Pu on the middle column:

a) Beam 1:

End reaction = ½(46 K + 46 K + 1.7 KLF(22’-0”))= 64.7 KIPS

b) Beam 2:

End reaction = ½(2.5 KLF(30’-0”))= 37.5 KIPS

c) Column weight:

Weight = (1.2) )150)("0'14(/144

"20"2022

PCF ft in

x−

= (1.2)5833 Lbs.= 7.0 KIPS

Total Pu = Pfloor + Proof

= [(64.7K + 37.5K + 7.0K)] + [0.75(64.7K + 37.5K + 7.0K)]

Total Pu = 191.1 KIPS

Step 2 - Determine the slenderness ratiominr

KLusing K = 0.7:

minr

KL=

)"20(3.0

)/"12"0'14)(7.0( ft x−

minr

KL= 19.6

Since KL/r min < 22→ it is a “short” column

75%

ACI Dead

Load factor




Step 3 – Determine total factored moment Mtotal applied to the column:

a) Beam 1:

To determine the moment applied to the column, we mustdetermine the “Fixed End Moment.” A useful table obtained

from the AISC Manual can be used to determine themoment.

wu = 1.7 KLF

46 K46 K

22’-0”

Ln = 20.33’

Fixed end Fixed end




Mcolumn = Maximum negative moment acting on column

= Muniform + Mpoint

= (bPL)unif + (bPL)point

= [(0.083)(1.7 Kips/ft)(20.33 ft)(20.33 ft)]+[(0.222)(46 Kips)(20.33 ft )]

= 58.6 Kip-ft + 207.6 Kip-ft

Beam 1 → Mcolumn = 266.2 Kip-ft

Point Loads

Unif. Load




b) Beam 2:

Mcolumn = Maximum negative moment acting on column

= Muniform

= (bPL)unif

= [(0.083)(2.5 Kips/ft)(28.33 ft)(28.33 ft)]

= 166.5 Kip-ft

Beam 2 → Mcolumn = 166.5 Kip-ft

c) Determine Mtotal:

Since these moments are offsetting each other,

Mtotal = MBeam 1 – MBeam 2

= (266.2 Kip-ft) – (166.5 Kip-ft)

Mtotal = 99.7 KIP-FT

wu = 2.5 KLF

30’-0”

Ln = 28.33’

Fixed end Fixed end




Step 4 - Determine if the column is acceptable based on “ColumnInteraction Diagram.”

a) Determineh

eratio:

e = eccentricity

=u

total

P

ft M )/"12(

= KIPS

ft FT KIP

1.191

)/"12(7.99 −

e = 6.26”

"20"26.6=h

e

h

e= 0.31 > 0.10 → CANNOT use small eccentricity formula

b) Determine γ:

X = concrete cover + stirrup dia. + ½(vert. bar dia.)

= 1½” + "8

3+ ½

"8

8

= 2.375”

γ h = 20” – (X + X)

= 20” – (2.375” + 2.375”)= 15.25”

γ (20”) = 15.25”

γ = 0.76 → USE γ = 0.75

XX λh

20”

h = 20”

8 - #8 vertical bars




c) Use Interaction Diagram R4-60.75:

g

u

g

n

A

P

A

P =

φ

="20"20

1.191 KIPS

g

n

A

P φ = 0.48 KSI

h

e x

A

P

h

e x

A

P

g

u

g

n =φ

= (0.48 KSI)(0.31)

h

e x A

P

g

nφ = 0.15 KSI

ρg = g

s

A

A

="20"20

) _ 8 _# _ 79.0( _ 8 2

x

bar per inbars = 0.016

Since ρg = 0.016 > 0.01→ col. is acceptable

Use ρg = 0.01

diseño de marcos trabe columna

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