Transcript
- 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
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