A steel shaft in bending has an ultimate strength of 690 MPa and a shoulder with a fillet radius of 3 mm connecting a 32-mm diameter with a 38-mm diameter. Estimate K_{f} using:
(a) Figure 6–20.
(b) Equations (6–33) and (6–35).
K_{f} = 1 + \frac {K_{t} − 1}{1 + \sqrt {a/r}} (6–33)
\sqrt {a} = 0.245 799 − 0.307 794(10^{−2})S_{ut} + 0.150 874(10^{−4})S^{2}_{ut} − 0.266 978(10^{−7})S^{3}_{ut} (6–35)
From Fig. A–15–9, using D/d = 38/32 = 1.1875, r/d = 3/32 = 0.093 75, we read the graph to find K_{t} \dot{=} 1.65.
(a) From Fig. 6–20, for Sut = 690 MPa and r = 3 mm, q \dot {=} 0.84. Thus, from Eq. (6–32)
K_{f} = 1 + q(K_{t} − 1) or K_{f s} = 1 + q_{shear}(K_{ts} − 1) (6–32)
K_{f} = 1 + q(K_{t} − 1) \dot{=} 1 + 0.84(1.65 − 1) = 1.55
(b) From Eq. (6–35) with S_{ut} = 690 MPa = 100 kpsi, \sqrt {a} = 0.0622\sqrt {in} = 0.313\sqrt {mm}. Substituting this into Eq. (6–33) with r = 3 mm gives
K_{f} = 1 +\frac {K_{t} − 1}{1 + \sqrt {a/r}} \dot {=} 1 +\frac {1.65 − 1}{1 +\frac {0.313}{\sqrt {3}}} = 1.55
