224 Budynas-Nisbett:Shigley's ll.Failure Prevention 5.Failures Resulting from T©The McGraw-Hil Mechanical Engineering Static Loading Companies,2008 Design,Eighth Edition 220 Mechanical Engineering Design Figure 5-12 Three Mohr circles,one for the uniaxial compression test,one for the test in pure shear,and one for the uniaxial tension test, are used to define failure by the Mohr hypothesis.The strengths S and S,are the compressive and tensile strengths, respectively;they can be used for yield or ultimate strength. Figure 5-13 Coulomb-Mohr failure line Mohr's largest circle for a general state of stress. B B O3C3 C2 Sc S 2 2 2 2 S,1+03 Se St 2 2 +2 2 Cross-multiplying and simplifying reduces this equation to 0103 -=1 (5-22) where either yield strength or ultimate strength can be used. For plane stress,when the two nonzero principal stresses are o >og,we have a situation similar to the three cases given for the MSS theory,Eqs.(5-4)to(5-6). That is, Case 1:A >og >0.For this case,o1 =oA and o3 =0.Equation (5-22) reduces to a failure condition of gA≥S (5-23) Case 2:A 0>B.Here,a1 A and o3 oB,and Eq.(5-22)becomes 受-贤1 (5-24④) Case 3:02A >Og.For this case,1 =0 and o3 =og,and Eq.(5-22)gives OB≤-Se (5-25)
Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition II. Failure Prevention 5. Failures Resulting from Static Loading 224 © The McGraw−Hill Companies, 2008 220 Mechanical Engineering Design or σ1 − σ3 2 − St 2 St 2 − σ1 + σ3 2 = Sc 2 − St 2 Sc 2 + St 2 Cross-multiplying and simplifying reduces this equation to σ1 St − σ3 Sc = 1 (5–22) where either yield strength or ultimate strength can be used. For plane stress, when the two nonzero principal stresses are σA ≥ σB , we have a situation similar to the three cases given for the MSS theory, Eqs. (5–4) to (5–6). That is, Case 1: σA ≥ σB ≥ 0. For this case, σ1 = σA and σ3 = 0. Equation (5–22) reduces to a failure condition of σA ≥ St (5–23) Case 2: σA ≥ 0 ≥ σB . Here, σ1 = σA and σ3 = σB , and Eq. (5–22) becomes σA St − σB Sc ≥ 1 (5–24) Case 3: 0 ≥ σA ≥ σB . For this case, σ1 = 0 and σ3 = σB , and Eq. (5–22) gives σB ≤ −Sc (5–25) Figure 5–12 Three Mohr circles, one for the uniaxial compression test, one for the test in pure shear, and one for the uniaxial tension test, are used to define failure by the Mohr hypothesis. The strengths Sc and St are the compressive and tensile strengths, respectively; they can be used for yield or ultimate strength. A B C D E St –Sc Figure 5–13 Mohr’s largest circle for a general state of stress. 1 St –Sc 3 O Coulomb-Mohr failure line B3 B2 B1 C C1 2 C3
Budynas-Nisbett:Shigley's ll.Failure Prevention 5.Failures Resulting from T©The McGraw-Hill 225 Mechanical Engineering Static Loading Companies,2008 Design,Eighth Edition Failures Resulting from Static Loading 221 Figure 5-14 Plot of the Coulomb-Mohr theory of failure for plane stress states. A plot of these cases,together with the normally unused cases corresponding to OB oA,is shown in Fig.5-14. For design equations,incorporating the factor of safety n,divide all strengths by n. For example,Eq.(5-22)as a design equation can be written as 01031 -=n (5-26) Since for the Coulomb-Mohr theory we do not need the torsional shear strength circle we can deduce it from Eq.(5-22).For pure shear t,o=-03 =t.The torsional yield strength occurs when tmax=Sy.Substituting o1=-03=Sy into Eq.(5-22) and simplifying gives Ssy= SyiSye (5-27刀 Syt Sye EXAMPLE 5-2 A 25-mm-diameter shaft is statically torqued to 230 N.m.It is made of cast 195-T6 aluminum,with a yield strength in tension of 160 MPa and a yield strength in com- pression of 170 MPa.It is machined to final diameter.Estimate the factor of safety of the shaft. Solution The maximum shear stress is given by 16T 16(230) t= 25(10-T=7510）NWm2=75MP n=- The two nonzero principal stresses are 75 and-75 MPa,making the ordered principal stresses o1 =75,02=0,and o3 =-75 MPa.From Eq.(5-26),for yield, 1 Answer n 01/Sr-3/Se=75/160-(-75/170=1.10 Alternatively,from Eq.(5-27), SyiSye 160(170) S+3=160+170 =82.4MPa
Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition II. Failure Prevention 5. Failures Resulting from Static Loading © The McGraw−Hill 225 Companies, 2008 Failures Resulting from Static Loading 221 A plot of these cases, together with the normally unused cases corresponding to σB ≥ σA, is shown in Fig. 5–14. For design equations, incorporating the factor of safety n, divide all strengths by n. For example, Eq. (5–22) as a design equation can be written as σ1 St − σ3 Sc = 1 n (5–26) Since for the Coulomb-Mohr theory we do not need the torsional shear strength circle we can deduce it from Eq. (5–22). For pure shear τ,σ1 = −σ3 = τ . The torsional yield strength occurs when τmax = Ssy . Substituting σ1 = −σ3 = Ssy into Eq. (5–22) and simplifying gives Ssy = Syt Syc Syt + Syc (5–27) Figure 5–14 Plot of the Coulomb-Mohr theory of failure for plane stress states. B A St –Sc –Sc St EXAMPLE 5–2 A 25-mm-diameter shaft is statically torqued to 230 N · m. It is made of cast 195-T6 aluminum, with a yield strength in tension of 160 MPa and a yield strength in compression of 170 MPa. It is machined to final diameter. Estimate the factor of safety of the shaft. Solution The maximum shear stress is given by τ = 16T πd3 = 16(230) π 25 10−3 3 = 75 106 N/m2 = 75 MPa The two nonzero principal stresses are 75 and −75 MPa, making the ordered principal stresses σ1 = 75, σ2 = 0, and σ3 = −75 MPa. From Eq. (5–26), for yield, Answer n = 1 σ1/Syt − σ3/Syc = 1 75/160 − (−75)/170 = 1.10 Alternatively, from Eq. (5–27), Ssy = Syt Syc Syt + Syc = 160(170) 160 + 170 = 82.4 MPa