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# 311 C H18

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### 311 C H18

1. 1. <ul><li>ALLOWABLE AXIAL COMPRESSIVE </li></ul><ul><li>LOADS ON STRUCTURAL MEMBERS </li></ul><ul><li>A.) INTRODUCTION </li></ul><ul><li>OBJECTIVE : </li></ul><ul><li>Determine exactly how much axial </li></ul><ul><li>compressive load (lb) a given structural </li></ul><ul><li>component can safely hold without bowing. </li></ul>
2. 2. <ul><li>DEFINITIONS : </li></ul><ul><li>AXIAL COMPRESSIVE LOAD - A force </li></ul><ul><li>which points in the same direction as </li></ul><ul><li>the longitudinal axis of the member and </li></ul><ul><li>is pushing on the ends of the member. </li></ul><ul><li>STRUCTURAL MEMBER - Any object </li></ul><ul><li>which transmits (or carries) a force. </li></ul>
3. 3. <ul><li>EXAMPLES OF AXIAL COMPRESSION </li></ul><ul><li>MEMBERS: </li></ul><ul><li>HYDRAULIC PISTON </li></ul><ul><li>BUILDING COLUMN </li></ul><ul><li>SKI POLE </li></ul>
4. 4. <ul><li>WHAT EFFECTS THE ABILITY OF A </li></ul><ul><li>MEMBER TO CARRY COMPRESSION </li></ul><ul><li>LOADS? </li></ul><ul><li>1.) </li></ul><ul><li>2.) </li></ul><ul><li>3.) </li></ul><ul><li>4.) How the ends are attached (end fixity) </li></ul>
5. 5. <ul><li>1.) MEMBER LENGTH </li></ul><ul><li>As you increase the length of a member, </li></ul><ul><li>you dramatically decrease its ability to </li></ul><ul><li>carry compression force. </li></ul>
6. 6. 0 100 200 300 400 500 600 0 5 10 15 20 25 30 40 LENGTH (ft) LOAD (kips)
7. 7. <ul><li>2.) CROSS-SECTIONAL DIMENSIONS </li></ul><ul><li> OF MEMBER </li></ul><ul><li>As you increase the cross-sectional area, </li></ul><ul><li>the axial compression capacity (P a ) </li></ul><ul><li>increase proportionally. </li></ul>
8. 8. <ul><li>More importantly, as you increase the </li></ul><ul><li>distance of the area of a cross-section </li></ul><ul><li>from its centroid (center for symmetric </li></ul><ul><li>sections), you increase the axial </li></ul><ul><li>compression dramatically. </li></ul><ul><li>P a = f (Ad 2 ) = f ( I ) </li></ul><ul><li>Example: </li></ul>
9. 9. <ul><li>3.) MATERIAL WHICH MEMBER IS </li></ul><ul><li>MADE OF </li></ul><ul><li>The compressive load capacity of a </li></ul><ul><li>member is directly proportional to the </li></ul><ul><li>Modulus of Elasticity of the material </li></ul><ul><li>it is made of. </li></ul>
10. 10. <ul><li>The Modulus of Elasticity is a material </li></ul><ul><li>property which indicates how much a </li></ul><ul><li>members will deform under a given load. </li></ul><ul><li>Steel: E = 30,000,000 psi </li></ul><ul><li>Aluminum: E = 10,000,000 psi </li></ul><ul><li>Wood: E = 1-2,000,000 psi </li></ul><ul><li>Plastic: E = 10,000-2,000,000 psi </li></ul>
11. 11. <ul><li>4.) END FIXITY </li></ul><ul><li>PINNED END - free to rotate </li></ul><ul><li>FIXED END - cannot rotate </li></ul><ul><li>A column with both ends fixed can hold </li></ul><ul><li>twice as much compressive load as </li></ul><ul><li>one with two pinned ends. </li></ul>
12. 12. 1.) Elastic Buckling 2.) Inelastic buckling 3.) Short Column – Material Failure <ul><li>B.) Possible Modes of Column Failure </li></ul>
13. 13. <ul><li>DESIGN FORMULAS (Includes Safety Factor) </li></ul><ul><li>American Institute of Steel Construction </li></ul><ul><li>American Concrete Institute </li></ul><ul><li>Aluminum Construction Manual </li></ul><ul><li>Alcoa Stuctural Handbook </li></ul><ul><li>National Forest Products Association </li></ul>
14. 14. <ul><li>SUMMARY </li></ul><ul><li>Length </li></ul><ul><li>Cross-sectional area </li></ul><ul><li>Material </li></ul><ul><li>End Fixity </li></ul><ul><li>Design Formulas and Factor of Safety </li></ul>
15. 15. <ul><li>C. THE COLUMN BUCKLING FORMULA </li></ul><ul><li>Leonard Euler, 1757 </li></ul><ul><li>P e =  2 E I </li></ul>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. 16. <ul><li>P e =  2 El__ </li></ul><ul><li>(KL) 2 </li></ul><ul><li> - K = effective length factor (Table 18-1) </li></ul><ul><li> e = P e =  2 E l_ =  2 E_ </li></ul><ul><li> A (KL) 2 A (KL/r) 2 </li></ul><ul><li>- where r = ( I /A) 1/2 </li></ul><ul><li>- does not include a Factor of safety </li></ul><ul><li>- Only valid when stress is below Proportional Limit </li></ul>
17. 17. <ul><li>D.) AISC COLUMN ANALYSIS </li></ul><ul><li>- the allowable stress depends on the slenderness ratio (KL/r) </li></ul><ul><li>1.) Compute: KL/r) x and KL/r) y </li></ul><ul><li> </li></ul><ul><li>2.) Compute C c = (2  2 E/  y ) 1/2 </li></ul>
18. 18. <ul><li>3.) If KL/r > C c and KL/r < 200 , elastic buckling occurs </li></ul><ul><li>  A = 12  2 E _ (FS = 23/12) </li></ul><ul><li> 23(KL/r) 2 </li></ul><ul><li>If KL/r < C c ,inelastic buckling occurs </li></ul><ul><li> A = Formulas 18-8 and 18-9 </li></ul><ul><li>4.) P A =  A A </li></ul>
19. 19. <ul><li>E.) AISC COLUMN DESIGN </li></ul><ul><li>1.) Estimate the required Area, by using an approximate allowable stress of 16 ksi </li></ul><ul><li>A req = P </li></ul><ul><li> 16ksi </li></ul><ul><li>2.) Select a column with A > A req </li></ul>
20. 20. <ul><li>3.) Analyze this column </li></ul><ul><li>a.) Compute: KL/r) x and KL/r) y </li></ul><ul><li> </li></ul><ul><li>b.) Compute C c = (2  2 E/  y ) 1/2 </li></ul>
21. 21. <ul><li>c.) If KL/r > C c and KL/r < 200 , elastic buckling occurs </li></ul><ul><li>  A = 12  2 E _ (FS = 23/12) </li></ul><ul><li> 23(KL/r) 2 </li></ul><ul><li>If KL/r < C c ,inelastic buckling occurs </li></ul><ul><li> A = Formulas 18-8 and 18-9 </li></ul><ul><li>c.) find allowable stress from App. J </li></ul><ul><li>d.) P A =  A A </li></ul>
22. 22. <ul><li>4.) Analyze this column </li></ul><ul><li>If P A > P (the applied load) , column is OK </li></ul><ul><li>If P A < P Select a larger column </li></ul>