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Deformation of a Bar due to stress developed

Centrifugal force developed due to rotation

P = w2 r M where M = mass of length ( ½ – x)

Now   M  = p x A x (1/2 – x)

and  r = x + (1/2 – x) ½

p = pA w2 (1/2 – x) [ x + ( 1/2 - x) 1/2 ]

 

Applying formula δ =PL / AE , we have

dδ = p dx / AE

= P w2 / E  (1/2 – X) {x + 1/4 – x /2} dx

= 2 pw2 / E  (1/4 – x/2) ( x /2 + 1/4 ) dx

= 2 pw2 / E  ( t2 / 16 – x2 / 4 ) dx

= 2 pw2 / E [ t2 / 16 x ½ - 2 pw2 / E  x ¼ [ x3 / 3] 0  ½

= 2 pw2 [ t3 / 32 – 2 pw2 / 4E [ t3 / 24]

= pw2 / E [ t3 / 16 – t3 / 48]

= pw2 / E [ 3t3 – t3 / 48] = p w2 t3 / 24 E

 

Total extension of the bar = 2δ = p w2 t3 / 12 E