A hallow alloy tube 5m long with external and internal diameters 40mm and 25mm was found to extend 6.4mm under the tensile load of 60KN. Find the buckling load for the tube of column with both ends pinned. Also find the safe load for the tube, taking FOS=4.

  A hallow alloy tube 5m long with external and internal diameters 40mm and 25mm was found to extend 6.4mm under the tensile load of 60KN. Find the buckling load for the tube of column with both ends pinned. Also find the safe load for the tube, taking FOS=4.

Solution
Step 1: Data

Length of the column = 5000mm

External diameter = 40mm

Internal diameter =25mm

Extension = 6.4mm

Tensile load = 60KN

FOS = 4

Buckling load =??

Safe load =??

Condition = Both ends fixed

Step 2: Calculation of strain

e = dl / l

e = 6.4/ 5000

e = 1.28X10^-3 

Step 3: Calculation of Area of cross section 

A = π (D²- d²)/4

    = π ((40)²- (25)²) / 4

 A= 765.76mm²

Step 4: Calculation of stress

σ = P/A

σ = 60X10³/765.76

σ = 78.35 N/mm²

Step 5: Calculation of Young's modulus

E = σ /e

E = 78.35 / 1.28X10^-3 

E = 0.198X10⁵ N/mm²

Step 6: Calculation of  Moment of inertia

I= π (D⁴- d⁴⁾/64

  = π ((40)⁴- (25)⁴) /64

 I = 1.065X10⁵mm⁴

Step 7: Calculation of Buckling load

Condition =  both  ends  fixed

P = 4Π ²E I/ L²

P =4Π ²(0.198X10⁵) (1.065X10⁵)/ (5000)²

P = 4.182 KN