Question

A Hollow column of external diameter 300 mm is 20 mm thick. If crushing strength of material is is 550 MPa, What is safe load if F O S = 5 A. 1200.82kN B. 1935.29kN C. 9676.03kN D. 1495kN​

Answer 1

Answer 6 x 5.0

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 oc= 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 = oc= 550N/mm2

Step 2: Calculation of the area of the column

A= TT/4(D2-d2)

A= TT/4(502-402)

A = 706.85 mm2

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

I = TT/64(D4-d4)

I = TT/64(504-404)

I = 181.04X103 mm4

Step 4: Calculation of Effective length

condition=both of its ends are fixed

therefore,

Le=1 / 2

Le = 3000/ 2

Le = 1500mm

Step 5: Calculation of radius of gyration

K = V ( 1 / A )

K = V(181.04X103 /706.85)

K = 16.00 mm

Step 6: Calculation of crippling load

P= (oc A)/(1+a(Le/k)2)

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

P = 60.33KN

Explanation:

This is a similar answer .