Budynas-Nisbett:Shigley's ll.Failure Prevention 5.Failures Resulting from T©The McGraw--Hill 21 Mechanical Engineering Static Loading Companies,2008 Design,Eighth Edition Failures Resulting from Static Loading 217 Thus,the shear yield strength predicted by the distortion-energy theory is Sy=0.577S (5-211 which as stated earlier,is about 15 percent greater than the 0.5 S,predicted by the MSS theory.For pure shear,txy the principal stresses from Eq.(3-13)are A =-0g =txy. The load line for this case is in the third quadrant at an angle of 45 from the oA.o axes shown in Fig.5-9. EXAMPLE 5-1 A hot-rolled steel has a yield strength of Sy=Syc=100 kpsi and a true strain at fracture of sf=0.55.Estimate the factor of safety for the following principal stress states: (a)70,70,0kpsi (b)30.70.0kpsi. (c)0,70.-30kpsi. (d)0,-30,-70kpsi. (e)30,30,30kpsi. Solution Sinces>0.05 and Sye and Sy are equal,the material is ductile and the distortion- energy (DE)theory applies.The maximum-shear-stress (MSS)theory will also be applied and compared to the DE results.Note that cases a to d are plane stress states. (a)The ordered principal stresses are A=a1=70,og =02=70.03=0 kpsi. DE From Eq.(5-13), a'=[702-70(70)+702]/2=70kpsi Answer n=S、100 =70=1.43 MSS Case 1,using Eq.(5-4)with a factor of safety, Answer n=S=100 =1.43 70 (b)The ordered principal stresses are oA=o1=70,og =02=30,03=0 kpsi. DE g'=[702-70(30)+302/2=60.8kpsi Answer n= S100 。=60.8 =1.64 MSS Case 1.using Eq.(5-4), Answer n=S100 A 70 =1.43Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition II. Failure Prevention 5. Failures Resulting from Static Loading © The McGraw−Hill 221 Companies, 2008 Failures Resulting from Static Loading 217 Thus, the shear yield strength predicted by the distortion-energy theory is Ssy = 0.577Sy (5–21) which as stated earlier, is about 15 percent greater than the 0.5 Sy predicted by the MSS theory. For pure shear, τxy the principal stresses from Eq. (3–13) are σA = −σB = τxy . The load line for this case is in the third quadrant at an angle of 45o from the σA, σB axes shown in Fig. 5–9. EXAMPLE 5–1 A hot-rolled steel has a yield strength of Syt = Syc = 100 kpsi and a true strain at fracture of εf = 0.55. Estimate the factor of safety for the following principal stress states: (a) 70, 70, 0 kpsi. (b) 30, 70, 0 kpsi. (c) 0, 70, −30 kpsi. (d) 0, −30, −70 kpsi. (e) 30, 30, 30 kpsi. Solution Since εf > 0.05 and Syc and Syt are equal, the material is ductile and the distortion￾energy (DE) theory applies. The maximum-shear-stress (MSS) theory will also be applied and compared to the DE results. Note that cases a to d are plane stress states. (a) The ordered principal stresses are σA = σ1 = 70, σB = σ2 = 70, σ3 = 0 kpsi. DE From Eq. (5–13), σ = [702 − 70(70) + 702 ] 1/2 = 70 kpsi Answer n = Sy σ = 100 70 = 1.43 MSS Case 1, using Eq. (5–4) with a factor of safety, Answer n = Sy σA = 100 70 = 1.43 (b) The ordered principal stresses are σA = σ1 = 70, σB = σ2 = 30, σ3 = 0 kpsi. DE σ = [702 − 70(30) + 302 ] 1/2 = 60.8 kpsi Answer n = Sy σ = 100 60.8 = 1.64 MSS Case 1, using Eq. (5–4), Answer n = Sy σA = 100 70 = 1.43