A 1.5-in. rod grade aisi 1040 steel with a sanded finish and a strength properties of 110 ksi, subjected in rotational bending, is given the endurance strength kakbkckdkekfSe′ in accordance with the query.
Heat is the result of the movement of kinetic energy within a material or an item, or from an alternative fuel to a material or an object. Irradiation, conduction, and convection are the three mechanisms through which such energy can be transferred.
d=1.5 in,Sut
=110 kpsi
Eq. (6-8): S_{e}^{\prime}=0.5(110)=55 kpsiSe′
=0.5(110)=55 kpsi
Table 6-2: a=2.70, b=-0.265a=2.70,b=−0.265
Eq. (6-19): k_{a}=a S_{u t}^{b}=2.70(110)^{-0.265}=0.777ka =aSutb
=2.70(110)
−0.265
=0.777
Since the loading situation is not specified, we’ll assume rotating bending or torsion so Eq. (6-20) is applicable. This would be the worst case.
k_{b}=0.879 d^{-0.107}=0.879(1.5)^{-0.107}=0.842kb
=0.879d
−0.107
=0.879(1.5)
−0.107
=0.842
Eq. (6-18): S_{e}=k_{a} k_{b} S_{e}^{\prime}=0.777(0.842)(55)=36.0 kpsiSe
=kakbSe′
=0.777(0.842)(55)=36.0 kpsi
Eq. (6-8): S_{e}^{\prime}= \begin{cases}0.5 S_{u t} & S_{u t} \leq 200 kpsi (1400 MPa ) \\ 100 kpsi & S_{u t}>200 kpsi \\ 700 MPa & S_{u t}>1400 MPa \end{cases}Se′
= 0.5S ut100kpsi700MPa Sut ≤200kpsi(1400MPa)S ut >200kpsiSut >1400MPa
Eq. (6-19): k_{a}=a S_{u t}^{b}k
a =aS utb
Eq. (6-20): k_{b}= \begin{cases}(d / 0.3)^{-0.107}=0.879 d^{-0.107} & 0.11 \leq d \leq 2 \text { in } \\ 0.91 d^{-0.157} & 2<d \leq 10 \text { in } \\ (d / 7.62)^{-0.107}=1.24 d^{-0.107} & 2.79 \leq d \leq 51 mm \\ 1.51 d^{-0.157} & 51<d \leq 254 mm \end{cases}kb
=
(d/0.3) −0.107 =0.879d
−0.107 0.91d −0.157 (d/7.62) −0.107 =1.24d −0.107 1.51d −0.157 0.11≤d≤2 in 2<d≤10 in 2.79≤d≤51mm51<d≤254mm
Eq. (6-18): S_{e}=k_{a} k_{b} k_{c} k_{d} k_{e} k_{f} S_{e}^{\prime}Se
=kakbkckdkekfSe′
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