A 1035 steel has a tensile strength of 70 kpsi and is to be used for a part that sees 450°F in service.
Estimate the Marin temperature modification factor and $\left(S_e\right) _{450◦}$
(a) The room-temperature endurance limit by test is $\left(S_{e}^{\prime}\right)_{70^{\circ}}=39.0 \mathrm{kpsi}$ .
(b) Only the tensile strength at room temperature is known.

Question

A 1035 steel has a tensile strength of 70 kpsi and is to be used for a part that sees 450°F in service.
Estimate the Marin temperature modification factor and [latex]\left(S_e\right) _{450◦}[/latex]
(a) The room-temperature endurance limit by test is [latex]\left(S_e\right) _{70◦}=39.0[/latex] kpsi.
(b) Only the tensile strength at room temperature is known.

Accepted Answer

(a) First, from Eq. (6–27),
[latex]k_d=0.975+0.432\left(10^{-3} ight) T_F-0.115\left(10^{-5} ight) T^2_F+0.104\left(10^{-8} ight) T^3_F-0.595\left(10^{-12} ight)F^4_F [/latex]

[latex]k_d=0.975+0.432\left(10^{-3} ight) \left(450 ight) -0.115\left(10^{-5} ight) \left(450 ight) ^2 +0.104\left(10^{-8} ight) \left(450 ight)^3-0.595\left(10^{-12} ight)\left(450 ight)^4=1.007 [/latex]

Thus,
[latex]\left(S_e ight) _{450◦}k_d\left(\acute{S}_e ight) _{70◦}=1.007\left(39.0 ight) =39.3[/latex]

(b) Interpolating from Table 6–4 gives

Table 6–4 Effect of Operating Temperature on the Tensile Strength of Steel. * ([latex]S_T[/latex] = tensile strength at operating temperature; [latex] S_{RT}[/latex] = tensile strength at room temperature; 0.099 ≤ ˆ σ ≤ 0.110)
Temperature, °C	[latex]S_T/S_{RT}[/latex]	Temperature, °F	[latex]S_T/S_{RT}[/latex]
20	1.000	70	1
50	1.01	100	1.008
100	1.02	200	1.02
150	1.025	300	1.024
200	1.02	400	1.018
250	1	500	0.995
300	0.975	600	0.963
350	0.943	700	0.927
400	0.9	800	0.872
450	0.843	900	0.797
500	0.768	1000	0.698
550	0.672	1100	0.567
600	0.549

[latex]\left({S_T}/{S_{RT}} ight)_{450◦} =1.018+\left(0.995-1.018 ight) \frac{450-400}{500-400} =1.007[/latex]

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

[latex]\left(S_{ut} ight) _{450◦}=\left({S_T}/{S_{RT}} ight)_{450◦} \left(S_{ut} ight) _{70◦}=1.007\left(70 ight) =70.5[/latex]

From Eq. (6–8) then,

[latex]S_{e}^{\prime}=\left\{\begin{array}{ll} 0.5 S_{u t} & S_{u t} \leq 200 \mathrm{kpsi}(1400 \mathrm{MPa}) \100 \mathrm{kpsi} & S_{u t}>200 \mathrm{kpsi} \700 \mathrm{MPa} & S_{u t}>1400 \mathrm{MPa}\end{array} ight.[/latex] (6–8)

[latex]\left(S_{ut} ight) _{450◦}=0.5\left(S_{ut} ight) _{450◦} =0.5\left(70.5 ight) =35.2[/latex]

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

Question 2.6.5: A 1035 steel has a tensile strength of 70 kpsi and is to be ...

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