A force of 2 kN is acting on a circular rod with diameter 25 mm. The stress in the rod can be
"\u03c3 = (20 103 N) \/ (\u03c0 ((20 10-3 m) \/ 2)2)\n = 127388535\\times 2 (N\/m2) \n = 127\\times 2 (MPa)\n= 381(MPa)"
The change of length can be calculated by:
"dl = \u03c3 l_o \/ E"
"= (381 106 Pa) (2 m) \/ (83 109 Pa) \n = 0.00381 m\n = 3.81 mm"
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