Question

5-36 This problem illustrates that the factor of safety for a machine element depends on the particular point selected for analysis. Here you are to compute factors of safety, based upon the distortion- energy theory, for stress elements at A and B of 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

At point A: The shear force creates a bending moment at the support of the bar. The bending moment (Fl) creates a normal tensile stress at point A. The tensile force creates a tensile stress at point A, so the total normal stress at point A will be the sum of bending stress due to load (F) and tensile stress due to axial force (P) in x-direction. Calculate the normal stress at point A. O ) Os = 0 + 0; 32F1 4P - ad ad 32(0.55 kN)(100 mm), 4(4 kN) 2 (15 mm) 2(15 mm)? 1 m 32 32(0.55 kN)[1000N)(100 mm) (100mm) 444 kN)(1000 N aſ(15 mm)(1000mm) a (15 mm)(1000mm) _32(550 N)(0.1 m) 4(4000 N) 7(0.015 m) 2(0.015 m)? =188.62x10 Pa 1000 N 1 kN 1000 mm =- -+

Calculate the shear stress at point A. 16T Tay=fayd 16(25 N-m) A (15 mm) 16(15 N·m) af 15 mm) 1000 mm) = (37.72x100 Pa) 1 MPa = 37.72 MPa The value of yield strength for AISI 1006 cold-drown steel is Sy = 280 MPa. The equation for factor of safety corresponding to Distortion-Energy theory (DE) is Here, d' is von Mises stress. The equation for von Mises stress is O=(( -0,04 +02 +373 )"

Here, Txy is shear stress in xy-plane and 0,0, are the stresses in x and y-planes. Substitute ( 62 -050, +02, +3133)" for o'. n = n= " (0 -00,+ 0 + 31, ) 280 MPa ((188.62 MPa)? – (188.62 MPa)(0 MPa)+(0 MPa) + 3(37.72 MPa)")" 280 MPa 199.61 MPa =1.40 Thus, factor of safety of point A is 1.40. At point B:

Calculate the total normal stress at the point A. O = 0 4P ad 4(4 kN) 7 (15 mm) 4(4 kN) 1000 N KN 1 KN 7 (15 mm)_1m) 1000 mm pa = (22.63x10 Pa) 10 Pa) = (22.63x10 Pa) = 22.63 MPa Calculate the shear stress at point B. qyr = 161 4 F +- sy id 3 A _ 161 4 4F adi * 3 ad? 16(25 N·m) 44(0.55 kN) 2 (15 mm)" * 37 (15 mm) - +-

16(25 N·m) 44(0.55 kN) ) = 41.87 MPa Calculate the factor of safety. n =- (o? -0,0,+0. +333 )" 280 MPa ((22.63 MPa) – (22.63 MPa)(0 MPa)+(0 MPa) +3(41.87 MPa)")" = 3.68 Thus, factor of safety of point B is 3.68