Question

This problem illustrates that the factor of safety for machine element depends on the particular point selected for analysis. Here you are to compute factor of safety, based upon the distortion energy theory, for stress elements at A and B pf the member shown in the figure. This bar is made of AISI 1006 cold drawn steel and is loaded by the forces F=0.55 kN, P=4.0 kN, and T= 25 N.m.

Answer 1

Answer: Il & obtain the yield strength from Table, "Deterministic ASTM minimum Tenbile Test and field strengths for Some Hot - Rolled and cold-Drawn steels, for AISI loob CD steel. . 5y = 280 mpa Now, Consider the stress element A, - The stress element A experiences arial stress only. - Calculate the stress at a wsing relation, 4P One = 4 (4x163) P. 4 KN 24x13? Tid? -3 165212 D-15 mm=1510m On = 22.63 X10 N/m² - Given (n=22.63 MPa Plent, Calculate the shear stress by using relation, 1 СТ. Tay= + Tman Ta3 Tnya 1GT Ta3 >/ Tmy 4 (0.55x153) 16x25 TIX(15X16373 2 3 Ix (1s=103) 4 → Try = 41.87 x 10 N/m2 → Try= 41.87 MPa

Hent, Calculate the masunum general Phase stress by using the distortion energy theory. ra Jon+itis >rs 522.63+3 (41.87) r = 72.6768 MPa Now, Calculating the factor of safety, we have, 280 72.6768 =>n=3.8526 Hences → The factor of Safety for stress element A is 3.8526 Hent, Consider the element B, - The element B experiences buth bending and asual stresses So, Calculate the stress at B using, ..ona 32 Fl .49 + Td3 Tol2

on = 32x(0.55*102)x(100x103) Tx(15x103) 3 + 4X10²x (4) Tx(15*10-32 u= 188.6280 x 10 N/m² De = 188.6208 mpa Nents Calculate the shear strebt using, IGT Tay= Ta3 16 x 25 Try= T*(*15%) 3 > Tony = 37.2556X10 N/m2 > Ery=57.2556 MPa Now, Calculate the masunum general plane stress using the distortion energy theory. r = Jon²+5lny r's (188.6208)3+ 6208)2+3 (37.2560) ? ²6 =199.3538 mpa Tent, Calculate factor of Safety, 2.80 o' 199.3538 1.4045 Hence, > The facter of safety for stress element Bis 1.4045