Question 6.13: A grade 30 gray cast iron is subjected to a load F applied t...

Some preparatory work is needed. From Table A–24, S_{ut} = 31    kpsi, S_{uc} = 109   kpsi, k_{a}k_{b}S′_{e} = 14  kpsi. Since k_{c} for axial loading is 0.9, then S_{e}= (k_{a}k_{b}S′_{e})k_{c} = 14(0.9) = 12.6  kpsi . From Table A–15–1, A = t(w − d) = 0.375(1 − 0.25) = 0.281 in^{2} , d/w = 0.25/1 = 0.25, and K_{t} = 2.45. The notch sensitivity for cast iron is 0.20 (see p. 288), so

Table A–24  Mechanical Properties of Three Non-Steel Metals (a) Typical Properties of Gray Cast Iron [The American Society for Testing and Materials (ASTM) numbering system for gray cast iron is such that the numbers correspond to the minimum tensile strength in kpsi. Thus an ASTM No. 20 cast iron has a minimum tensile strength of 20 kpsi. Note particularly that the tabulations are typical of several heats.]

Table A–15    Charts of Theoretical Stress-Concentration Factors K*_{t}

*Factors from R. E. Peterson, “Design Factors for Stress Concentration,” Machine Design, vol. 23, no. 2, February 1951, p. 169; no. 3, March 1951, p. 161, no. 5, May 1951, p. 159; no. 6, June 1951, p. 173; no. 7, July 1951, p. 155. Reprinted with permission from Machine Design, a Penton Media Inc. publication.

K_{f} = 1 + q(K_{t} − 1) = 1 + 0.20(2.45 − 1) = 1.29

(a) σ_{a} =\frac {K_{f} F_{a}}{A} =\frac {1.29(0)}{0.281} = 0              σ_{m} =\frac {K_{f} F_{m}}{A} =\frac {1.29(1000)}{0.281} =(10^{−3}) = 4.59  kpsi

and

n =\frac {S_{ut}}{σ_{m}} =\frac {31.0}{4.59} = 6.75

(b) F_{a} = F_{m} =\frac {F}{2} =\frac {1000}{2} = 500  lbf

σ_{a}=σ_{m} =\frac {K_{f} F_{a}}{A} =\frac {1.29(500)}{0.281} =(10^{−3}) = 2.30  kpsi

r =\frac {σ_{a}}{σ_{m}} = 1

From Eq. (6–52),

S_{a} =\frac {r S_{ut} + S_{e}}{2} \left[ -1+\sqrt{1 +\frac {4rS_{ut} S_{e}}{(r S_{ut} + S_{e})^{2}}} \right]              (6-52)

S_{a} =\frac {(1) 31 + 12.6}{2} \left[ -1+\sqrt{1 +\frac {4(1)31(12.6) }{[(1) 31 + 12.6]^{2}}} \right]=7.63  kpsi

n =\frac {S_{a}}{σ_{a}} =\frac {7.63}{2.30} = 3.32

(c) F_{a} =\frac {1}{2}|300 − (−1000)| = 650 lbf          σ_{a} =\frac{1.29(650)}{0.281}(10^{−3}) = 2.98  kpsi

r =\frac {σ_{a}}{σ_{m}} =\frac {3.0}{−1.61} = −1.86

From Eq. (6–53), S_{a} = S_{e} + (S_{e}/S_{ut} − 1)S_{m}  and  S_{m} = S_{a}/r . It follows that

S_{a} = S_{e} +\left(\frac {S_{e}}{S_{ut}} − 1\right)S_{m}      −S_{ut} ≤ S_{m} ≤ 0  (for  cast  iron)                (6–53)

S_{a} =\frac {S_{e} }{1 −\frac {1}{r} \left(\frac{S_{e}}{S_{ut}} −1\right)}  =\frac {12.6}{1 −\frac {1}{-1.86} \left(\frac{12.6}{31} −1\right)}=18.5  kpsi

n =\frac {S_{a}}{σ_{a}} =\frac {18.5}{2.98} = 6.20

Figure 6–31b shows the portion of the designer’s fatigue diagram that was constructed.