Answer:
(a) 126.66 kN (b) 31.665 kN (c) 258.49 kN (d) 506.64 kN
Explanation:
Solution
Given
A HSS152.4 × 101.6 × 6.4 structural steel is used as a column
Actual length of the column , L= 6 m
The elasticity modules, E = 200 GPa
The factor of safety with respect to failing buckling . F.S =2
Geometric properties of structural steel shapes for, A HSS152.4 × 101.6 × 6.4
the moment of inertial about y axis Iy =4 .62 * 10^ 6 mm ^4
For
(a) If the end condition is pinned - pinned
The effective length factor, K =1
The critical buckling load , Pcr = π²EI/(KL)²
Pcr = π² * 200 * 10 ^3 * 4.62 * 10 ^6/( 1* 6* 10 ^3)
= 253319.85N
= 253.32N
The maximum safe load , Pallow = 253.32 /2 = 126.66kN
hence, the maximum safe for the column is 126.66kN
(b)If the end condition is fixed free-free
the effective length factor, K= 2
The critical buckling load , Pcr = π²EI/(KL)²
Pcr = π² * 200 * 10 ^3 * 4.62 * 10 ^6/( 2 * 6 * 10 ^3)²
= 63329.96N
=63.33kN
The maximum safe load, P allow = Pcr/F.S
P allow = 63.33/2
31.665 kN
Therefore the maximum safe for the column is 31.665 kN
(c) If the end condition is fixed- pinned
The effective length factor K =0.7
The critical buckling load , Pcr = π²EI/(KL)²
Pcr = π² * 200 * 10 ^3 * 4.62 * 10 ^6/( 0.7 * 6 * 10 ^3)²
=516979.2 8N
=516.98 kN
The maximum safe load, P allow = Pcr/F.S
P allow = 516.98 kN/2
=258.49 kN
Therefore the maximum safe for the column is 258.49 kN
(d) If the end condition is fixed -fixed
The effective factor, K =0.5
The critical buckling load , Pcr = π²EI/(KL)²
Pcr = π² * 200 * 10 ^3 * 4.62 * 10 ^6/( 0.5 * 6 * 10 ^3)²
=1013279.4 N
=1013.28 kN
The maximum safe load, P allow = Pcr/F.S
P allow = 1013.28 / 2
= 506.64 kN
P allow = 506.64 kN
Therefore the maximum safe for the column is 506.64 kN
