PSI - Issue 5

Hyung-Kyu Kim et al. / Procedia Structural Integrity 5 (2017) 63–68 Author name / Structural Integrity Procedia 00 (2017) 000 – 000

65

3

The potential energy, Π , for the deformed shape of Fig. 1 can be derived as

      r A B   2

  

C

  





2

2

2

,

(2)

p



   

2

9

3







3

r





  2 3   s

E E s

2 1

2 1

E E

 1 1 4 1   

  

s

t

s

   3 E E s t

  2 1  

       

A

 B E t

 C E t

1  

,

,

,

2

121



where E s , E t , E , ν are the secant modulus, t angent modulus, Young’s modulus and Poisson ’ s ratio in order. When the minimum potential energy theorem is applied to Eq. (2), we can obtain

3

r t



  

  

 p AE s

  2  

,



 

.

(3)

2

41

If the elastic region is assumed as perfectly linear, E s = E t = E , thus A = 1, μ = ν . As a result, p in Eq. (3) is written as

3

r t

  

   

   2

p E 

cr p

.

(4)

41 

Eq. (4) is a generally used formula of the critical buckling pressure, p cr . It implies that a tube is collapsed (buckled) when the external pressure exceeds p cr . For further conservatism, the minimum thickness ( t min ) and maximum radius ( r max ) are usually to be substituted instead of t and r in Eq. (4). On the other hand, it is necessary to incorporate an irregular shape of the cross section if the tube does not have a perfectly circular cross section at its initial state, which is practical owing to the tolerance in a tube manufacturing. It is readily anticipated that the critical buckling pressure of a tube of a perfectly circular cross section will be reduced if the tube has an irregular cross section. This is solved by Timoshenko and Gere (1961). In this problem, the initial irregular shape of the cross section is assumed as an ellipse (i.e., oval shape) which can be also represented in Fig. 1. Ovality is defined as δ 1 / r where δ 1 is a radial deviation from a perfect circle at its initial state. Now, the maximum stress, σ max , under an external pressure, p , is obtained as

t pr

pr



6

1

 



.

(5)

max

2

cr p p

1



t

It may be noted in Eq. (5) that the moment of inertia of the cross section perpendicular to and along the tube axis is t 3 /12 (the tube length is set as unity due to a plane strain condition assumed). When the limit of σ max is set as σ ys , the yield strength of the tube material, we can obtain p y , the critical buckling pressure of an oval tube, from the following equation.

       ys m 



  

ys



2

p

mn p p

p

1 6

0





(6)

y

cr y

cr

m

where m = r / t , n = δ 1 / r .