311 C H18

• 1.
  • 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.

2.

• DEFINITIONS :

• AXIAL COMPRESSIVE LOAD - A force

• which points in the same direction as

• thelongitudinal axisof the member and

• ispushingon the ends of the member.

• STRUCTURAL MEMBER - Any object

• which transmits (or carries) a force.

3.

• EXAMPLES OF AXIAL COMPRESSION

• MEMBERS:

• HYDRAULIC PISTON

• BUILDING COLUMN

• SKI POLE

4.

• WHAT EFFECTS THE ABILITY OF A

• MEMBER TO CARRY COMPRESSION

• LOADS?

• 1.)

• 2.)

• 3.)

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

5.

• 1.)MEMBER LENGTH

• As you increase the length of a member,

• you dramatically decrease its ability to

• carry compression force.

6. 0 100 200 300 400 500 600 0 5 10 15 20 25 30 40 LENGTH (ft) LOAD (kips) 7.

• 2.)CROSS-SECTIONAL DIMENSIONS

• OF MEMBER

• As you increase the cross-sectional area,

• the axial compression capacity (P a )

• increase proportionally.

8.

• 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:

9.

• 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.

10.

• 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

11.

• 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.

12. 1.) Elastic Buckling 2.) Inelastic buckling 3.) Short Column Material Failure

• B.) Possible Modes of Column Failure

13.

• DESIGN FORMULAS(Includes Safety Factor)

• American Institute of Steel Construction

• American Concrete Institute

• Aluminum Construction Manual

• Alcoa Stuctural Handbook

• National Forest Products Association

14.

• SUMMARY

• Length

• Cross-sectional area

• Material

• End Fixity

• Design Formulas and Factor of Safety

15.

• 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 ) 16.

• 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 belowProportional Limit

17.

• D.) AISC COLUMN ANALYSIS

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

• 1.) Compute:KL/r) xandKL/r) y

• 2.) Compute C c= (2 2 E/ y )1/2

18.

• 3.) If KL/r > C cand 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

19.

• 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

20.

• 3.) Analyze this column

• a.) Compute:KL/r) xandKL/r) y

• b.) Compute C c= (2 2 E/ y )1/2

21.

• c.) If KL/r > C cand 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

22.

• 4.) Analyze this column

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

• If P A < P Select a larger column