Question

A column has fixed boundary conditions at the bottom ends for weak-axis buckling and pinned boundary conditions at both ends for strong-axis buckling. Is a HSS10x8x1/2 (A500 Gr. B, Fy=46 ksi adequate for L=20’ and Pu=500 kips?

Answer 1

Given:

L=20 ft

L_e=0.707L=14.14 ft (As column is fixed at one end and hinged at the other end)

Ultimate load P_u=500 kips

Fy=46 ksi

Step 1:

Calculate the slenderness ratio of the HSS beam

KL/r_y=14.14x12/3.14

KL/r_y=54.03<200

4.71(E/Fy)^0.5=4.71(29000/50)^0.5=118

As,KL/r<4.71(E/Fy)^0.5, use AISC equation E3-2

Step 2:

Calculate the critical stress

F_e=²E/(KL/r)²

F_e=3.14x3.14x29000/(54.03)²

F_e=97.94 ksi

F_cr=0.658^Fy/Fe

F_cr=(0.658^46/97.94x46)=37.94 ksi

Step 3:

Calculate the nominal strength of HSS

P_n=F_crA_g

P_n=37.94x15.3

P_n=580.5 kips

As per LRFD design capacity=0.9x580.5=522 kips

The HSS10x8x0.5 is enough to resist a load of 500 kips