Lecture 15 – Page 1 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 with offsetting moments, the majority of concrete columns do experience applied moments. Concrete framed buildings typically have columns that are poured monolithically with 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 Portal Method and others discussed in Structural Theory.

Lecture 15 – Page 2 of 12

The compression capacity of a reinforced column is reduced by the bending stresses 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 (from Lecture 14):

Determining points along the curve is quite laborious and typically not done using hand calculations. Instead, computer programs or design guides are used to perform column analysis and design. Below are some “Column Interaction Diagrams” that are used for column analysis and design.

Axi

al c

ompr

essi

on c

apac

ity

Bending moment capacity

Pure compression (no applied moment)

Pure bending (no applied axial load)

Lecture 15 – Page 3 of 12

Rectangular TIED Column Interaction Diagrams

R4-60.75 “R” = Rectangular cross-section “4” = f’c = 4 KSI “60” = Grade 60 vertical bars “75” = γ

Lecture 15 – Page 4 of 12

Circular SPIRAL Column Interaction Diagrams

Lecture 15 – Page 5 of 12

“Short” Column Design:

Short columns are not considered to be susceptible to the effects of buckling as are long columns. The ACI dictates that short columns satisfy the slenderness ratio as shown below:

Short column IF 22min

≤rKL

Where: K = end fixity factor

L = unbraced length in inches rmin = least radius of gyration, inches = 0.3h for rectangular or square columns = 0.25dout for circular columns where dout = outer dia.

Lecture 15 – Page 6 of 12

Example GIVEN: 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. Be sure to add column weight. Assume the loads from the roof above are 75% of the floor load.

2) Determine the slenderness ratio minr

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 Interaction

Diagram.”

wu = 1.7 KLF wu = 2.5 KLF 46 K 46 K

14’-0”

30’-0” 22’-0”

Beam 1

20” x 20” Middle column

Beam 2

Lecture 15 – Page 7 of 12

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

PCFftin

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 ratio minr

KLusing K = 0.7:

minrKL

= )"20(3.0

)/"12"0'14)(7.0( ftx−

minrKL

= 19.6

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

75%

ACI Dead Load factor

Lecture 15 – Page 8 of 12

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

a) Beam 1:

To determine the moment applied to the column, we must determine the “Fixed End Moment.” A useful table obtained from the AISC Manual can be used to determine the moment.

wu = 1.7 KLF

46 K 46 K

22’-0”

Ln = 20.33’

Fixed end Fixed end

Lecture 15 – Page 9 of 12

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

Lecture 15 – Page 10 of 12

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

Lecture 15 – Page 11 of 12

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

a) Determine he

ratio:

e = eccentricity

= u

total

PftM )/"12(

= KIPS

ftFTKIP1.191

)/"12(7.99 −

e = 6.26”

"20"26.6

=he

he

= 0.31 > 0.10 → CANNOT use small eccentricity formula

b) Determine γ:

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

= 1½” + "83

+ ½

"

88

= 2.375” γh = 20” – (X + X) = 20” – (2.375” + 2.375”) = 15.25” γ(20”) = 15.25” γ = 0.76 → USE γ = 0.75

X X λh

20”

h = 20”

8 - #8 vertical bars

Lecture 15 – Page 12 of 12

c) Use Interaction Diagram R4-60.75:

g

u

g

n

AP

AP

=φ

= "20"20

1.191xKIPS

g

n

APφ

= 0.48 KSI

he

xAP

he

xAP

g

u

g

n =φ

= (0.48 KSI)(0.31)

he

xAP

g

nφ = 0.15 KSI

ρg = g

s

AA

= "20"20

)_8_#_79.0(_8 2

xbarperinbars = 0.016

Since ρg = 0.016 > 0.01 → col. is acceptable

Use ρg = 0.01