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Architecture 324 Structures 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 2/19

Leonhard Euler

(1707 – 1783) Euler Buckling (elastic buckling) – – – – –

P cr

  2

AE

  

KL r

   2

r



I A

A = Cross sectional area (in 2 ) E = Modulus of elasticity of the material (lb/in 2 ) K = Stiffness (curvature mode) factor L = Column length between pinned ends (in.) r = radius of gyration (in.)

f cr

  2

E

  

KL r

   2 

F cr

University of Michigan, TCAUP Structures II Source: Emanuel Handmann (wikimedia commons) Slide 3/19

• • •

Failure Modes

– – – –

Short Columns

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

f c



P A



F c

f c = Actual compressive stress A = Cross-sectional area of column (in 2 ) P = Load on the column F c = Allowable compressive stress per codes

Intermediate Columns

(“columns” Engel) – crush and buckle – – – –

Long Columns

– fail by buckling (“long columns” Engel)

f cr

  2

E

2 

F cr KL r

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 University of Michigan, TCAUP Structures II Slide 4/19

Slenderness Ratio

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

r



I I



A

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

Ar

2 r x = 0.999

• Slenderness Ratios:

L x r x L y r y

The larger ratio will govern.

Try to balance for efficiency r y = 0.433

University of Michigan, TCAUP Structures II Slide 5/19

End Support Conditions

l

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

l

e is the effective length

l

e = K

l

K= 1.0

Both ends pinned.

K= 0.7

One end free, one end fixed.

K= 2.0

K= 0.5

Both ends fixed.

One end pinned, one end fixed.

University of Michigan, TCAUP Structures II Slide 6/19

Analysis of Wood Columns

• • •

Data:

Column – size, length Support conditions Material properties – F c , E •

Required:

P crit for buckling and crushing 1.

2.

3.

4.

5.

Calculate slenderness ratio; largest ratio governs.

Check slenderness against upper limit.

Calculate P crit for buckling using Euler’s equation: Calculate P max P max = F c A for crushing: Smaller of P crit or P max will fail first.

University of Michigan, TCAUP Structures II Slide 7/19

Example Problem :

Analysis Data: section 3”x7” Full Dimension F c = 1000 psi E = 1,400,000 psi Find: P critical for buckling and crushing.

Determine the mode of failure for the wood column.

University of Michigan, TCAUP Structures II Slide 8/19

Example Problem :

Analysis (cont.) 1.

Calculate slenderness ratios for each axis.

The larger (more slender) controls.

2.

Upper limits are usually given by codes.

University of Michigan, TCAUP Structures II Slide 9/19

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 10/19