34.5k views
1 vote
The W18 x 40 is used as a structural A992 steel column that can be assumed pinned at top and fixed at the base. Determine the largest axial force P that can be applied without causing it to buckle. For W18 x 40, A = 11.8 in². lx = 612 in. ly = 19.1 inª. Yield strength oy for A992 is 50 ksi. E = 29,000 ksi.

1 Answer

6 votes

The largest axial force (P) that can be applied without causing the W18 x 40 steel column to buckle is approximately 147,259.88 lb.

To determine the largest axial force (P) that can be applied without causing the W18 x 40 steel column to buckle, we need to calculate the critical buckling load.

The critical buckling load can be determined using the Euler's buckling formula:
P_critical = (pi^2 * E * I) / (l^2)

Where:
- P_critical is the critical buckling load
- pi is a mathematical constant approximately equal to 3.14
- E is the Young's modulus of elasticity, given as 29,000 ksi
- I is the moment of inertia of the column cross-section, which can be calculated as (b * h^3) / 12 for a rectangular cross-section
- l is the effective length of the column, which can be taken as the smaller of lx and ly in this case

First, let's calculate the moment of inertia (I) for the W18 x 40 column:
I = (b * h^3) / 12
= (4.16 * 18^3) / 12
= 199.584 in^4

Next, let's determine the effective length (l):
l = min(lx, ly)
= min(612 in, 19.1 in)
= 19.1 in

Now we can calculate the critical buckling load (P_critical):
P_critical = (pi^2 * E * I) / (l^2)
= (3.14^2 * 29,000 * 199.584) / (19.1^2)
= 3.14^2 * 29,000 * 199.584 / 365.21
= 3.14^2 * 29,000 * 199.584 / 365.21
≈ 147,259.88 lb

User Jens Pettersson
by
8.2k points