Question 6.5: A 1035 steel has a tensile strength of 70 kpsi and is to be ...

(a) First, from Eq. (6–27),

k_{d }= 0.975 + 0.432(10^{−3})T_{F} −  0.115(10^{−5})T^{2}_{F}+ 0.104(10^{−8})T^{3}_{F} − 0.595(10^{−12})T^{4}_{F}                            ( 6–27)

k_{d} = 0.975 + 0.432(10^{−3})(450) − 0.115(10^{−5})(450^{2}) + 0.104(10^{−}8)(450^{3}) − 0.595(10^{−12})(450^{4}) = 1.007

Thus,

(Se)_{450◦} = k_{d} (S′_{e})_{70◦} = 1.007(39.0) = 39.3   kpsi

(b) Interpolating from Table 6–4 gives

(S_{T} /S_{RT} )_{450◦} = 1.018 + (0.995 − 1.018) \frac {450 − 400}{500 − 400} = 1.007

Thus, the tensile strength at 450°F is estimated as

(S_{ut} )_{450◦} = (S_{T} /S_{RT} )_{450}◦ (S_{ut} )_{70◦} = 1.007(70) = 70.5  kpsi

From Eq. (6–8) then,

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})_{450◦} = 0.5 (S_{ut} )_{450◦} = 0.5(70.5) = 35.2   kpsi

Part a gives the better estimate due to actual testing of the particular material.

Table 6–4   Effect of Operating Temperature on the Tensile Strength of Steel.* (S_{T} = tensile strength at operating temperature; S_{RT} = tensile strength at room temperature; 0.099 ≤ \hat {σ} ≤ 0.110)