The state of stress at a critical point in a steel member (E=200 GPa, G=79 GPa,

Question

The state of stress at a critical point in a steel member (E=200 GPa, G=79 GPa, ?y=250GPa ) is ?xx=?30MPa , ?yy=90Mpa , ?xy=?40MPa , and all other stress components are zero. Determine the safety factor of using a) the maximum shearing stress criterion (Tresca) and (b) the maximum energy of distortion criterion (von Mises).

Explanation / Answer

a) maximum shearing stress criterion= sigma 1,2= (?yy+ ?xx)/ 2 +- root[( sigma x -sigma yy)/2 +?xy^2]........... putting all values........simga 1=156.47 and sigma 2 = 96.47......................... now fos= max of ( sigma1 /2 , sigma 2/2 , 1-2/2 = 78.235now, fos= (?y/2)/ 78.235=1.59...........as...........?max >= ? limit / 2 ?max is the greatest of abs (?12, ?23, ?13) where: ....... ?12 = (?1 - ?2) / 2; ?23 = (?2 - ?3 ) / 2; ?13 = (?1 - ?3) / 2........... ?1, ?2, ?3 are the principal stresses in descending order. The Factor of safety (FOS) is given by: FOS = ? limit / (2 * ?max )............. b) maximum energy of distortion criterion = (?y/fos)^2 = sigma 1 ^2 sigma 2^2 -sigma 1 * sigma 2............................then fos= 250/136.728= 1.82.................

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