Question 9.5: The 1018 steel strap of Fig. 9–21 has a 1000-lbf, completely...

From C.J. Noll and C. Lipson, “Allowable Working Stresses,” Society for Experimental Stress Analysis, vol. 3, no. 2, 1946 p. 29. Reproduced by O.J. Horger (ed.) Metals Engineering Design ASME Handbook, McGraw-Hill, New York. Copyright © 1953 by The McGraw-Hill Companies, Inc. Reprinted by permission.
Table A–20 Deterministic ASTM Minimum Tensile and Yield Strengths for Some Hot-Rolled (HR) and Cold-Drawn (CD) Steels [The strengths listed are estimated ASTM minimum values in the size range 18 to 32 mm ( \frac {3}{4} to 1\frac {1}{4}in). These strengths are suitable for use with the design factor defined in Sec. 1–10, provided the materials conform to ASTM A6 or A568 requirements or are required in the purchase specifications. Remember that a numbering system is not a specification.] Source: 1986 SAE Handbook, p. 2.15

For a uniform shear stress on the throat, k_{b} = 1.

From Eq. (6–26), p. 282, for torsion (shear),

k_{c} =\begin{cases} 1 & bending \\ 0.85 & axial \\ 0.59 & torsion ^{17} \end{cases}    (6-26)

17: Use this only for pure torsional fatigue loading. When torsion is  combined with other stresses, such as bending, k_{c} = 1 and the combined loading is managed by using the effective von Mises stress as in Sec. 5–5. Note: For pure torsion, the distortion energy predicts that (k_{c})_{torsion} = 0.577.

k_{c} = 0.59            k_{d} = k_{e} = k_{f} = 1

From Eqs. (6–8), p. 274, and (6–18), p. 279,

S′_{e}=\begin{cases}0.5S_{ut}& S_{ut} ≤ 200  kpsi (1400 MPa) \\100  kpsi & S_{ut} > 200  kpsi \\700  MPa &   S_{ut} > 1400  MPa \end{cases}      (6-8)

S_{e} = k_{a}k_{b}k_{c}k_{d}k_{e}k_{f} S′_{e}              (6–18)

S_{se} = 0.702(1)0.59(1)(1)(1)0.5(58) = 12.0  kpsi
K_{f s} = 2.7      F_{a} = 1000  lbf        F_{m} = 0

Only primary shear is present:

τ^{′}_{a} =\frac {K_{f s} F_{a}}{A} = \frac {2.7(1000)}{1.061} =2545  psi          τ^{′}_{m} = 0  psi

In the absence of a midrange component, the fatigue factor of safety n_{f} is given by

n_{f} =\frac {S_{se}}{τ^{′}_{a}} =\frac {12 000}{2545} = 4.72