Column Analysis and Design

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University of Michigan, TCAUP Structures II Slide 2/19

Architecture 324Structures II

Column Analysis and Design

• Failure Modes• End Conditions and Lateral Bracing• Analysis of Wood Columns• Design of Wood Columns• Analysis of Steel Columns• Design of Steel Columns

University of Michigan, TCAUP Structures II Slide 3/19

Leonhard Euler (1707 – 1783)

Euler Buckling (elastic buckling)

– A = Cross sectional area (in2)– E = Modulus of elasticity of the material (lb/in2)– K = Stiffness (curvature mode) factor– L = Column length between pinned ends (in.)– r = radius of gyration (in.)

crcr F

rKL

Ef

2

2

2

2

rKL

AEPcr

A

Ir

Source: Emanuel Handmann (wikimedia commons)

University of Michigan, TCAUP Structures II Slide 4/19

Failure Modes

• Short Columns – fail by crushing(“compression blocks or piers” Engel)

– fc = Actual compressive stress

– A = Cross-sectional area of column (in2)– P = Load on the column – Fc = Allowable compressive stress per codes

• Intermediate Columns – crush and buckle(“columns” Engel)

• Long Columns – fail by buckling(“long columns” Engel)

– E = Modulus of elasticity of the column material– K = Stiffness (curvature mode) factor– L = Column length between pinned ends (in.)– r = radius of gyration = (I/A)1/2

cc FA

Pf

crcr F

rKL

Ef

2

2

University of Michigan, TCAUP Structures II Slide 5/19

Slenderness Ratio

• Radius of Gyration: a geometric property of a cross section

– r = Radius of Gyration– I = Moment of Inertia– A = Cross-sectional Area

• Slenderness Ratios:

The larger ratio will govern.

Try to balance for efficiency

A

Ir 2ArI

x

x

r

L

y

y

r

L

rx = 0.999

ry = 0.433

University of Michigan, TCAUP Structures II Slide 6/19

End Support Conditions

K is a constant based on the end conditionsl is the actual length

le is the effective length

le = Kl

K= 0.5

K= 2.0

K= 0.7

K= 1.0

Both ends fixed.

One end pinned, one end fixed.

Both ends pinned.

One end free, one end fixed.

University of Michigan, TCAUP Structures II Slide 7/19

Analysis of Wood Columns

Data:• Column – size, length• Support conditions• Material properties – Fc , E

Required:• Pcrit for buckling and crushing

• Calculate slenderness ratio; largest ratio governs.

• Check slenderness against upper limit.• Calculate Pcrit for buckling using Euler’s

equation:• Calculate Pmax for crushing:

Pmax = Fc A

• Smaller of Pcrit or Pmax will fail first.

University of Michigan, TCAUP Structures II Slide 8/19

Example Problem : Analysis

Data: section 3”x7” Full DimensionFc = 1000 psiE = 1,400,000 psi

Find: Pcritical for buckling and crushing.Determine the mode of failure for the wood column.

University of Michigan, TCAUP Structures II Slide 9/19

Example Problem : Analysis (cont.)

1. Calculate slenderness ratiosfor each axis.

The larger (more slender) controls.

2. Upper limits are usually given by codes.

University of Michigan, TCAUP Structures II Slide 10/19

Example Problem : Analysis (cont.)

3. Calculate critical Euler buckling load.

4. Calculate crushing load.

5. Smaller of the two will fail first and control.

University of Michigan, TCAUP Structures II Slide 11/19

Analysis of Steel Columnsby Engel

Data:• Column – size, length• Support conditions• Material properties – Fy

• Applied load - Pactual

Required:• Pactual < Pallowable

• Calculate slenderness ratios.The largest ratio governs.

• Check slenderness ratio against upper limit of 200

• Use the controlling slenderness ratio to find the critical Euler buckling stress, fcr.

• Apply some Factor of Safety (like 3) to fcr.

• Determine yield stress limit, Fy.

• Fallowable is the lesser stress: (fcr / F.S.) or Fy

• Compute allowable capacity: Pallowable = Fallow A.

• Check column adequacy:Pactual < Pallowable

2

2

rKL

Efcr

University of Michigan, TCAUP Structures II Slide 12/19

Design of Steel Columnsby Engel

Data:• Column – length• Support conditions• Material properties – Fy

• Applied load - Pactual

Required:• Column – section

• Use the Euler equation to solve for Ar2 which is equal to I for both x and y axis.

• Enter the section tables and find the least weight section that satisfies BOTH Ix and Iy.

• Check the slenderness ratios are both < 200.

• Calculate the actual Euler stress fcr for the final section.

• Fallowable is the lesser stress: fcr / F.S. or Fy

• Compute allowable capacity: Pallowable = Fallow A.

..)(

2

2

SFE

lKPI xxx

..)(

2

2

SFE

lKPI yyy

University of Michigan, TCAUP Structures II Slide 13/19

Example Problem : Design

Select a steel section that can carry the given load.

University of Michigan, TCAUP Structures II Slide 14/19

Example Problem : Design (cont.)

University of Michigan, TCAUP Structures II Slide 15/19

Example Problem : Design (cont.)

• Determine the controlling slenderness (larger controls)

• Find the actual buckling stress, fcr

• Compare to allowable stress, Fallowable is lesser of : fcr/F.S. or Fy

• Determine safe allowable load, Pallowable = Fallowable A

University of Michigan, TCAUP Structures II Slide 16/19

Determining K factorsby AISC

Sidesway Inhibited:Braced frame1.0 > K > 0.5

Sidesway Uninhibited:Un-braced frameunstable > K > 1.0

If Ic/Lc is largeand Ig/Lg is smallThe connection is more pinned

If Ic/Lc is smalland Ig/Lg is largeThe connection is more fixed

Source: American Institute of Steel Construction, Manual of Steel Construction, AISC 1980

University of Michigan, TCAUP Structures II Slide 17/19

Steel Frame Construction

University of Michigan, TCAUP Structures II Slide 18/19

Analysis of Steel Columnsby AISC-ASD

Data:• Column – size, length• Support conditions• Material properties – Fy

• Applied load - Pactual

Required:• Pactual < Pallowable

• Calculate slenderness ratios. largest ratio governs.

• In AISC Table look up Fa for given slenderness ratio.

• Compute: Pallowable = Fa A.• Check column adequacy:

Pactual < Pallowable

Source: American Institute of Steel Construction, Manual of Steel Construction, AISC 1980

University of Michigan, TCAUP Structures II Slide 19/19

Design of Steel Columnswith AISC-ASD Tables

Data:• Column – length• Support conditions• Material properties – Fy

• Applied load - Pactual

Required:• Column Size

1. Enter table with height. 2. Read allowable load for each section to

find the smallest adequate size.3. Tables assume weak axis buckling. If

the strong axis controls the length must be divide by the ratio rx/ry

4. Values stop in table (black line) at slenderness limit, KL/r = 200

Source: American Institute of Steel Construction, Manual of Steel Construction, AISC 1980