An open ductile cylinder of 60 mm bore is made from a material whose design stress is 400 MPa.

1. Use the maximum shear stress failure theory to determine an outside diameter suitable to withstand simultaneous internal and external pressures of 125 and 250 MPa.
2. Repeat (a) with internal and external pressures of 360 and 180 MPa respectively.
3. Repeat (b) but use the distortion energy failure theory.

   (a)

   Quickly assess whether thin or thick by means of thin cylinder theory ( ii) with S_y/n = σ_d = 400 MPa :
   t   =   Δ_p D_i / 2 σ_d   =   | 125 - 250 | x 60 / 2 x 400   =   9.4mm
   Since the corresponding t/D_i is 0.156, it may be seen from the graph above that the error due to the thin cylinder approximation approaches 30%. This is unacceptably rough, so repeat with thick cylinder theory.
   Using thick cylinder theory, from ( 2) with as yet unknown proportions, γ_o :
   σ_m   =   ( 125 - 250 γ_o ) / ( γ_o - 1 )   =   - 125 ( 2 γ_o - 1 ) / ( γ_o - 1 )   MPa
   σ_v     =   ( 125 - 250 ) γ_o / ( γ_o - 1 )   =   - 125 γ_o / ( γ_o - 1 )                 MPa

   At the critical bore ( γ = 1 ), from ( 3) :
   σ_t   =   σ_m + σ_v   =   - 125 ( 3 γ_o - 1 ) / ( γ_o - 1 )   =   σ_min
   σ_r   =   σ_m   - σ_v   =   - 125     (of course; it must equal the negative of the internal pressure )
   σ_a   =   0     (open)   =   σ_max

   Applying the maximum shear stress criterion :
   σ_e   =   σ_max - σ_min   =   0 - [ - 125 ( 3 γ_o - 1 ) / ( γ_o - 1 ) ]     and   =   400 MPa
   Solving,     γ_o   =   11   and so from the definition of γ_o,     D_o   =   √11 D_i   =   199 mm

   (b)

   The design equation ( viii) will be used, to demonstrate its directness compared to the method of part (a).
   Considering the application limits of ( viii),     p_o / p_i   =   180/360   =   0.5     and   <   ( 1 + 1/γ_o ) / 2 for all γ_o.
   So, for this range of applicability, ( γ_o - 1) σ*   =   2 γ_o ( p_i - p_o )     ie.     400 ( γ_o - 1 )   =   2 γ_o ( 360 -180 )     MPa
   Solving,     γ_o   =   10       and so from the definition of γ_o,     D_o   =   √10 D_i   =   190 mm
   The advantages of the design equation should be noted - although the correct application range cannot always be chosen a priori as here, and a couple of trials may be necessary.

   (c)

   The distortion energy criterion is not piecewise defined like the maximum shear stress theory, so in principle the design equation ( vii) is more immediately applicable than ( viii). Inserting values into ( vii) :
   ( γ_o - 1 )² σ*²   =   ( p_i - γ_o p_o )² + 3 γ_o² ( p_i - p_o )²     becomes
   ( γ_o - 1 )² 400²   =   ( 360 - 180 γ_o )² + 3 γ_o² ( 360 - 180 )²     which leads to the quadratic :
   19γ_o² - 119γ_o + 19   =   0,   from which   γ_o   =   6.10   and   D_o   =   √6.10 D_i   =   148 mm

   As can be seen the second order distortion energy equation poses its own problems.

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       Copyright 1999-2005 Douglas Wright,   doug@mech.uwa.edu.au
         last updated May 2005