Question

A hollow cylindrical steel tube of 38 mm external diameter and 2.5 mm thick is used as a column of 2.3 m long with both ends fixed. The slenderness ratio is

Answer 1

Answer:

1. The external and internal diameters of the hollow cast iron column are 5cm and 4cm respectively. If the length of this column is 3m and both of its ends are fixed, determine the crippling load using Rankine’s formula. Take σc= 550N/mm2 and a=1/1600.

Step 1: Data:

D = external diameter = 50mm

d= internal diameter = 40mm

Rankine's constant=a=1/1600.

length of column= 3m = 3000mm

condition=both of its ends are fixed

crippling load =??

critical stress = σc= 550N/mm2

Step 2: Calculation of the area of the column

A= π/4(D2-d2)

A= π/4(502-402)

A = 706.85 mm2

Step 3 :Calculation of the moment of inertia of the column

I = π/64(D4-d4)

I = π/64(504-404)

I = 181.04X103 mm4

Step 4:Calculation of Effective length

condition=both of its ends are fixed

therefore,

Le=l / 2

Le = 3000/ 2

Le = 1500mm

Step 5:Calculation of radius of gyration

K= √(I/A)

K = √(181.04X103 /706.85)

K = 16.00 mm

Step 6:Calculation of crippling load

P= (σc A)/(1+a(Le/k)2)

P = ((550) (706.85))/(1+(1/1600)(1500/16)2)

P = 60.33KN