Question

4 of 9 to a 8000 in E:29000 ksi andoy=36 00 ksi Plion 1445.06 Previous Answer

Answer 1

Section details:

The actual column length is L = 8.75 ft

E=29000 ksi

Area (A)= πd²/4 = 3.14*8/4 =50.26548 in²

Moment of inertia of the section (I)= πd⁴/64=201.0619in⁴

EI =29000*201.0619 = 5.8308E+06 kip-in².

PART: B

Crushing load

Crushing load= yield stress * area= 36*50.26548=1809.5573 kip

Euler buckling load

For an ideal column, the load that causes a column to buckle (called the Euler buckling load) is given by: P_cr = π² EI/(KL)².

Since the column is pinned connection at both ends, the effective length factor K = 1.000.   

Substituting into the Euler buckling load equation gives P_cr = π ² × 5.8308E+06 kip-in² ÷ (1.000 × 8.75 ft × 12 in/ft)² = 5219.7451 kip

Maximum allowable force in the member due to compression

= minimum {Crushing load, Euler buckling load}

= minimum {1809.5573 kip, 5219.7451 kip }

= 1809.5573 kip (ANS)

Thus the load due to yielding in compression is the governing criteria in this case.

PART C

column is fixed at one end and free at other end, the effective length factor K = 2

The crushing load does not change with the end conditions. However, the buckling load is dependent on the end condition.

Substituting into the Euler buckling load equation gives P_cr = π ² × 5.8308E+06 kip-in² ÷ (2 × 8.75 ft × 12 in/ft)² = 1304.9363 kip

Maximum allowable force in the member due to compression

= minimum {Crushing load, Euler buckling load}

= minimum {1809.5573 kip, 1304.9363 kip }

= 1304.9363 kip (ANS)

in this case the Euler buckling load is the governing criteria