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## Design of Aluminum Members in Compression

### Strength reduction factor and safety factor

LRFD design:

Strength reduction factors-building type structures

f = 0.9

ASD design for building-type structures

W = 1.65

ASD design for bridge-type structures

W = 1.85

### Design Strength of a compression member

The design strength of a compression member is the lesser of member buckling strength and local buckling strength.  The member buckling strength depends on the slenderness ratio of the member. The local buckling strength depends on weighted average of width to thickness ratio of components of the member.  In addition, interaction between member buckling and local buckling also need to be checked.

Design procedure:

1. Calculate buckling strength of member

2. Calculate local buckling strength of each member

3. Calculate total buckling strength based on weighted average of local buckling strength of each member

4. Check interaction between member and local buckling.

### Normal (Global) Buckling Strength of Member

Global buckling is bucking of the member as a whole, the normal buckling strength is calculated as

Pn = Fc Ag                                                                    

Ag: Gross section area

Fc is buckling strength determined as follows:

If kL/r < Cc   Fc = 0.85(Bc-Dc kL/r)  £   Fcy                 

If kL/r < Cc   Fc = 0.85 p2 E/(kL/r)2                             

E is elastic modulus,

k is slenderness ratio coefficient

L is unbraced length

r is radius of gyration

Fcy is compressive yield strength

Bc is buckling constant intercept for compression in columns and beam flanges

Dc is buckling constant slop for compression in columns and beam flanges.

Cc is buckling constant intersection for compression in columns and beam flanges

Bc, Dc, and Cc are calculated based on equations in Table B.4.1 and B.4.2 in Aluminum Design Manual as follows:

For aluminum alloy temper designations beginning with O,H, T1, T2, T3, or T4 and welded-affected zones of all tempers.

Bc = Fcy [1+(Fcy/1000 ksi)1/2             

Dc = Bc (6 Bc/E)1/2/20                          

Cc = (2/3)Bc/Dc                                     

For aluminum alloy temper designations beginning with T5, T6, T7, T8, or T9

Bc = Fcy [1+(Fcy/2250 ksi)1/2             

Dc = Bc (Bc/E)1/2/10                             

Cc = 0.41Bc/Dc                                      

#### Torsional and Flexural-Torsional Buckling

For member shape that torsional and flexural-torsional buckling is a concern, an equivalent slenderness ratio shall be calculated based on section E.3.2 of Aluminum Design Manual.

#### Design Aid:

Aluminum Design Manual Part VI provide simpler equations for calculating Cc (listed as S2) and Pn for different types of alloy and temper for allowable strength design.

### Local Buckling Strength of Member

#### Nominal buckling strength of member based on local Buckling strength

Local buckling is buckling of components of a member such as flange, web, etc.  The buckling strength of the member is a weight average of all the components.  The total buckling is the total of compressive strength of each component multiply by the area of each component. Fci is compressive strength of individual element

Ai is area of individual element

Ag is gross section area of member

#### Local buckling strength of components

The buckling strength of components depends on its width to thickness retio, b/t.  There are five cases:

1. Flat element supported on one edge such as flange of a wide-flange section.

2. Flat element supported on both edges such as web of a wide flange section.

3. Flat element supported on one edge with stiffener on the other edge

4. Flat element supported on both edges with an intermediate stiffener

5. Curved element supported on both edges In each case, the compressive strength are calculated in three stress range, yielding, inelastic buckling, and post buckling. The equations are listed in Section B.5.4.

For example, in Case 1: Flat element supported on one edge such as a flange of an I beam, the compressive strength is calculated as follows:

(a) Yielding, Fc = Fcy    when b/t £ S1 = Bp-Fcy/(5 Dp                                              

(b) Inelastic buckling, Fc = Bp – 5 Dp(b/t)   when S1 < b/t < S2 = k1Bp/(5 Dp) 

(c) Post-buckling, Fc = k2(BpE)2/(5 b/t)  when b/t ≥ S2                                               

For aluminum alloy temper designations beginning with O,H, T1, T2, T3, or T4 and welded-affected zones of all tempers.

Bp = Fcy [1+(Fcy/440 ksi)1/2              

Dp = Bc (6 Bp/E)1/2/20                         

For aluminum alloy temper designations beginning with T5, T6, T7, T8, or T9

Bp = Fcy [1+(Fcy/1500 ksi)1/2            

Dp = Bp (Bp/E)1/2/10                            

### Design aids:

These equations are very complicated.  To ease calculation effort, Aluminum Design Manual Part VI provides simplified equations for allowable strength design.  For example, for an U shape aluminum channel without stiffener at edges, the simplified equations for 6063-T5 are as follows:

Axial compression for member buckling:

Fc/W= 8.9-0.037 S for S = kL/r < S2 = 99                   

Fc/W = 51,352/S2 for S = kL/r ≥ S2                             

Elements – Uniform compression

Flat elements supported on one edge in columns whose buckling axis is not an axis of symmetry.

Fc/W = 9.7 for S= b/t £ S1 = 8.2                                              

Fc/W = 11.8-0.26 S if S1 < S < S2 = 19                                   

Fc/W = 2,417/S2 if S ≥ S2                                                        

Flat elements supported on both edges

Fc/W = 9.7 for S= b/t £ S1 = 25.6                                            

Fc/W = 11.8-0.083 S if S1 < S < S2 = 50                                 

Fc/W = 382/S if S ≥ S2                                                                     

### Example:

An aluminum column with pinned supports at top and bottom Design data:

Length of column: 5 ft

Alloy-temper: 6063-T5

Column section: C6x4.03

Depth: D = 6 in

Width: B = 3.25 in

Flange thickness: tf =0.35 in

Web thickness: tw = 0.21 in

Fillet radius: R = 0.3 in

Area: 3.43 in2.

Radius of gyration in y axis: ry = 1.05 in

Requirement: Determine allowable compressive strength of the column.

Solution:

Determine member-buckling strength:

Slenderness coefficient: k = 1 (pinned at top and bottom)

Slenderness ratio: S = kL/ry = (1)(5)(12)/1.05 = 57.1  < S2 = 99

Therefore, nominal compressive strength: Fc/W = 8.7-0.035(57.1) = 6.6 ksi

Allowable compressive strength of member, Pn/W = (6.6)(3.43) = 22.6 kip

Determine compressive strength based on elements:

Flange:  b = 3.25-0.3-0.21 = 2.74 in

Slenderness ratio, S = b/t = 2.74/0.35 = 7.8  < S1 = 8.2

Compressive strength of flange: Fc/W = 9.7 ksi

Area of flange: Af = (2.74)(0.35) = 0.959 in2.

Web: b = 6 – (2)(0.35)-(2)(0.3) = 4.7 in

Slenderness ratio, S = b/t = 3.7/0.21 = 17.6  < S1 = 25.6

Compressive strength of flange: Fc/W = 9.7 ksi

Area of web: Aw = (4.7)(0.21) = 0.987 in2.

Compressive strength of 6063-T5, Fcy = 16 ksi

Allowable compressive strength of member based on elements:

Pn/W = (2)(0.959)(9.7)+(0.987)(9.7)+[3.43-2(0.959)-0.987] = 36.6 kip

Therefore, allowable compressive strength of member,

Pn/W = 22.6kip

For LRFD design, the compressive strength of the member

fPn = (22.6)(1.65)/0.9 = 41.4 kip