3 An aluminum rectangular column is subjected to an axial lo

3. An aluminum, rectangular column is subjected to an axial load. The column is fixed at a) Determine the critical buckling load of the member in Figure 2. Sketch the buckled b) Lateral braces are placed at the midspan of the member in the weak axis direction only both ends column making it clear in which direction it buckles. as shown in Figure 3. Determine the critical buckling load of the member and sketch the buckled column Cross-Section 12 in 20 ft 2 in E 10,000 ksi Figure 2: Structure for Problem 3a FIXED Cross-Section 10 ft 12 in 2 in 10 ft E = 10,000 ksi FIXED Figure 3: Structure for Problem 3b.

Solution

3a) The Euler Buckling loaf for a fixed fixed column is given by P_cr = 4*pi²*EI/L²

Moment of inertia about minor axis of section = 12*2³/12= 8 in⁴

Modulus of elasticity of Aluminum = 10000 ksi

Lenght of column=20 ft=20*12=240 in

Buckling load = 4*pi²*10000*8/240² = 54.83 kips

b) Let us deteremine the buckling load about both the axis

Major axis :

I=2*12³/12=288 in⁴

Length = 20 ft=240 in

critical load=4*pi²*10000*288/240² = 1973.9 kips

Minor axis:

Moment of inertia = 8 in⁴

length=20/2=10 ft=120in

critical load = 4*pi²*10000*8/120² = 219.3 kips

Therefore buckling load of member = min(1973.9,219.3)=219.3 kips