ALLOWABLE AXIAL COMPRESSIVE LOADS ON STRUCTURAL MEMBERS A.) INTRODUCTION OBJECTIVE : Determine exactly how much axial compressive load (lb) a given structural component can safely hold without bowing.

ALLOWABLE AXIAL COMPRESSIVE

LOADS ON STRUCTURAL MEMBERS

A.) INTRODUCTION

OBJECTIVE :

Determine exactly how much axial

compressive load (lb) a given structural

component can safely hold without bowing.

DEFINITIONS : AXIAL COMPRESSIVE LOAD - A force which points in the same direction as the longitudinal axis of the member and is pushing on the ends of the member. STRUCTURAL MEMBER - Any object which transmits (or carries) a force.

DEFINITIONS :

AXIAL COMPRESSIVE LOAD - A force

which points in the same direction as

the longitudinal axis of the member and

is pushing on the ends of the member.

STRUCTURAL MEMBER - Any object

which transmits (or carries) a force.

EXAMPLES OF AXIAL COMPRESSION MEMBERS: HYDRAULIC PISTON BUILDING COLUMN SKI POLE

EXAMPLES OF AXIAL COMPRESSION

MEMBERS:

HYDRAULIC PISTON

BUILDING COLUMN

SKI POLE

WHAT EFFECTS THE ABILITY OF A MEMBER TO CARRY COMPRESSION LOADS? 1.) 2.) 3.) 4.) How the ends are attached (end fixity)

WHAT EFFECTS THE ABILITY OF A

MEMBER TO CARRY COMPRESSION

LOADS?

1.)

2.)

3.)

4.) How the ends are attached (end fixity)

1.) MEMBER LENGTH As you increase the length of a member, you dramatically decrease its ability to carry compression force.

1.) MEMBER LENGTH

As you increase the length of a member,

you dramatically decrease its ability to

carry compression force.

0 100 200 300 400 500 600 0 5 10 15 20 25 30 40 LENGTH (ft) LOAD (kips)

2.) CROSS-SECTIONAL DIMENSIONS OF MEMBER As you increase the cross-sectional area, the axial compression capacity (P a ) increase proportionally.

2.) CROSS-SECTIONAL DIMENSIONS

OF MEMBER

As you increase the cross-sectional area,

the axial compression capacity (P a )

increase proportionally.

More importantly, as you increase the distance of the area of a cross-section from its centroid (center for symmetric sections), you increase the axial compression dramatically. P a = f (Ad 2 ) = f ( I ) Example:

More importantly, as you increase the

distance of the area of a cross-section

from its centroid (center for symmetric

sections), you increase the axial

compression dramatically.

P a = f (Ad 2 ) = f ( I )

Example:

3.) MATERIAL WHICH MEMBER IS MADE OF The compressive load capacity of a member is directly proportional to the Modulus of Elasticity of the material it is made of.

3.) MATERIAL WHICH MEMBER IS

MADE OF

The compressive load capacity of a

member is directly proportional to the

Modulus of Elasticity of the material

it is made of.

The Modulus of Elasticity is a material property which indicates how much a members will deform under a given load. Steel: E = 30,000,000 psi Aluminum: E = 10,000,000 psi Wood: E = 1-2,000,000 psi Plastic: E = 10,000-2,000,000 psi

The Modulus of Elasticity is a material

property which indicates how much a

members will deform under a given load.

Steel: E = 30,000,000 psi

Aluminum: E = 10,000,000 psi

Wood: E = 1-2,000,000 psi

Plastic: E = 10,000-2,000,000 psi

4.) END FIXITY PINNED END - free to rotate FIXED END - cannot rotate A column with both ends fixed can hold twice as much compressive load as one with two pinned ends.

4.) END FIXITY

PINNED END - free to rotate

FIXED END - cannot rotate

A column with both ends fixed can hold

twice as much compressive load as

one with two pinned ends.

1.) Elastic Buckling 2.) Inelastic buckling 3.) Short Column – Material Failure B.) Possible Modes of Column Failure

B.) Possible Modes of Column Failure

DESIGN FORMULAS (Includes Safety Factor) American Institute of Steel Construction American Concrete Institute Aluminum Construction Manual Alcoa Stuctural Handbook National Forest Products Association

DESIGN FORMULAS (Includes Safety Factor)

American Institute of Steel Construction

American Concrete Institute

Aluminum Construction Manual

Alcoa Stuctural Handbook

National Forest Products Association

SUMMARY Length Cross-sectional area Material End Fixity Design Formulas and Factor of Safety

SUMMARY

Length

Cross-sectional area

Material

End Fixity

Design Formulas and Factor of Safety

C. THE COLUMN BUCKLING FORMULA Leonard Euler, 1757 P e =  2 E I L 2 P e = load at which a slender column will buckle (lb)  = 3.14 E = Modulus of Elasticity (psi) I = Moment of Inertia (in 4 )

C. THE COLUMN BUCKLING FORMULA

Leonard Euler, 1757

P e =  2 E I

P e =  2 El__ (KL) 2 - K = effective length factor (Table 18-1)  e = P e =  2 E l_ =  2 E_ A (KL) 2 A (KL/r) 2 - where r = ( I /A) 1/2 - does not include a Factor of safety - Only valid when stress is below Proportional Limit

P e =  2 El__

(KL) 2

- K = effective length factor (Table 18-1)

 e = P e =  2 E l_ =  2 E_

A (KL) 2 A (KL/r) 2

- where r = ( I /A) 1/2

- does not include a Factor of safety

- Only valid when stress is below Proportional Limit

D.) AISC COLUMN ANALYSIS - the allowable stress depends on the slenderness ratio (KL/r) 1.) Compute: KL/r) x and KL/r) y 2.) Compute C c = (2  2 E/  y ) 1/2

D.) AISC COLUMN ANALYSIS

- the allowable stress depends on the slenderness ratio (KL/r)

1.) Compute: KL/r) x and KL/r) y

2.) Compute C c = (2  2 E/  y ) 1/2

3.) If KL/r > C c and KL/r < 200 , elastic buckling occurs  A = 12  2 E _ (FS = 23/12) 23(KL/r) 2 If KL/r < C c ,inelastic buckling occurs  A = Formulas 18-8 and 18-9 4.) P A =  A A

3.) If KL/r > C c and KL/r < 200 , elastic buckling occurs

 A = 12  2 E _ (FS = 23/12)

23(KL/r) 2

If KL/r < C c ,inelastic buckling occurs

 A = Formulas 18-8 and 18-9

4.) P A =  A A

E.) AISC COLUMN DESIGN 1.) Estimate the required Area, by using an approximate allowable stress of 16 ksi A req = P 16ksi 2.) Select a column with A > A req

E.) AISC COLUMN DESIGN

1.) Estimate the required Area, by using an approximate allowable stress of 16 ksi

A req = P

16ksi

2.) Select a column with A > A req

3.) Analyze this column a.) Compute: KL/r) x and KL/r) y b.) Compute C c = (2  2 E/  y ) 1/2

3.) Analyze this column

a.) Compute: KL/r) x and KL/r) y

b.) Compute C c = (2  2 E/  y ) 1/2

c.) If KL/r > C c and KL/r < 200 , elastic buckling occurs  A = 12  2 E _ (FS = 23/12) 23(KL/r) 2 If KL/r < C c ,inelastic buckling occurs  A = Formulas 18-8 and 18-9 c.) find allowable stress from App. J d.) P A =  A A

c.) If KL/r > C c and KL/r < 200 , elastic buckling occurs

 A = 12  2 E _ (FS = 23/12)

23(KL/r) 2

If KL/r < C c ,inelastic buckling occurs

 A = Formulas 18-8 and 18-9

c.) find allowable stress from App. J

d.) P A =  A A

4.) Analyze this column If P A > P (the applied load) , column is OK If P A < P Select a larger column

4.) Analyze this column

If P A > P (the applied load) , column is OK

If P A < P Select a larger column