Question

4. The following structure is made of AISI 1006 cold-drawn steel (Sy=280MPa) and it is loaded by the forces F=0.55 kN, P=8.0 kN and T=30 Nm. The factor of safety for a machine element depends on the particular point selected for the analysis. Using Tresca failure theory, determine the factor of safety for points A and B. 15 points -100 mm 20-mm D.

Answer 1

Given data: AISI 1006 CD steel, Force (F)=0.55kN=0.55x10' N. Force (P) = 8.0 kN = 8.0x10’N. Torque (T) = 30.0 N.m = 30x10²N.mm. Diameter of bar is(d)= 20 mm. Length of the bar is(1)= 100 mm . (S,)= 280 MPa Stress element A: The stress element A experiences both bending stress and axial stress. Therefore, stress on element A is the sum of bending stress and axial stress. Stress()= Bending Stress + AxialStress 32F1 Bending stress(0)= 32(0.55x10')(100) 0g =- (20) Os = 70.03 N/mm Ad P Axial stress(oa)=7d2 - 0 .- 4(8.0*10') A(20) On = 25.46 N/mm?

Thus, stress 0 = 0; +oa = 70.03+25.46 = 95.49 N/mm? Calculate the shear stress, twº 16T 16(30x103) (20) Tv = 19.09 N/mm² Calculate the principal stress at point A. 2.95.49+0 +, 95.49–0 ) +(19.09) 012 = 47.745–51.42 0 = 47.745 +51.42 0 = 99.165 N/mm 02 = 47.745 – 51.42 02=-3.675N/mm

Apply the Tresca criterion, to calculate the FOS at point A. S, 0-02 SM - 2x(FOS). *2*2(FOS). 99.165-(-3.675)= (FOS). (FOS), = 2.723 280 Stress element B: The stress element B experiences axial stress. Axial stress(Ca)= 4(8.010') 4P ad2 Out (20) 0x=0= 25.46N/mm² +- . 3 A - +- or stresse 161 4 F Shear stress, Tw= 16(30x10'), 4(0.55x103)*4 (20) 3 (20)? = 19.098+ 2.3342N/mm² Ty = 21.432N/mm²

Calculate the principal stress at point B. Oxuyt 2 + 01.2= - T . 2 25.46+0 4. 25.46–0)+(21.432)? 012 = 12.73 +24.93 0 =12.73 + 24.93 0 = 37.66 N/mm 02 = 12.73 – 24.93 02=-12.2 N/mm Apply the Tresca criterion, to calculate the FOS at point B. S. Tomas – 2x (FOS); 07.02 - 2x(FOS), 37.66-(–12.2) – (FOS), 280 (FOS), = 5.616