Given a part with Sut= 151 kpsi and at the critical location of the part, Se= 67.5 kpsi.For the loading of Fig. 6–33, estimate the number of repetitions of the stress-time blockin Fig. 6–33 that can be made before failure.

Question

Given a part with [latex]S_{ut}= 151[/latex] kpsi and at the critical location of the part, [latex]Se= 67.5[/latex] kpsi.
For the loading of Fig. 6–33, estimate the number of repetitions of the stress-time block in Fig. 6–33 that can be made before failure.

Accepted Answer

From Fig. 6–18, p. 285, for [latex]S_{ut}= 151 \ kpsi \ , \ f = 0.795[/latex].

From Eq. (6–14),

[latex]a={\left(fS_{ut} ight)^2 } / {S_e} [/latex] (6–14)

[latex]a=\frac{\left(fS_{ut} ight)^2 }{S_e}=\frac{\left[0.795\left(151 ight) ight]^2 }{67.5} =213.5 [/latex] kpsi

From Eq.(6–15),

[latex]b=- [\log \left({fS_{ut}}/{S_e} ight) /3 [/latex] (6–15)

[latex]b=-\frac{1}{3} \log \left(\frac{fS_{ut}}{S_e} ight) =-\frac{1}{3} \log \left[\frac{0.795\left(151 ight) }{67.5} ight] =-0.0833[/latex]

so.

[latex]S_f=213.5N^{-0.0833}[/latex] [latex]N=\left(\frac{S_f}{213.5} ight) ^{-{1}/{0.0833}}[/latex] (1),(2)

We prepare to add two columns to the previous table. Using the Gerber fatigue criterion,
Eq. (6–47), p. 306, with [latex]S_e= S_f \ , \ and \ n = 1[/latex], we can write

Gerber [latex] \frac{n\sigma _a}{S_e} +\left(\frac{n\sigma _m}{S_{ut}} ight)^2=1 [/latex] (6–47)

[latex]S_f=\begin{cases} \frac{\sigma _a}{1-\left(\frac{\sigma _m}{S _{ut}} ight) } & \sigma _m\gt 0 \S_e&\sigma _m\leq 0 \end{cases} [/latex] (3)

where [latex]S_f[/latex] is the fatigue strength associated with a completely reversed stress, [latex]\sigma _{rev}[/latex] ,
equivalent to the fluctuating stresses [see Ex. 6–12, part (b)].

Cycle 1: [latex]r={\sigma _a}/{\sigma _m}={70}/{10}=7[/latex] , and the strength amplitude from Table 6–7, p. 307

Table 6–7 Amplitude and Steady Coordinates of Strength and Important Intersections in First Quadrant for Gerber and Langer Failure Criteria
Intersecting Equations	Intersection Coordinates
[latex]\frac{S_a}{S_e} +\left(\frac{S_m}{S_{ut}} ight)^2 =1[/latex] Load line[latex]r=\frac{S_a}{S_m}[/latex]	[latex]S_a=\frac{r^2S^2_{ut}}{2S_e} \left[-1+\sqrt{1+\left(\frac{2S_e}{rS_{ut}} ight)^2 } ight] [/latex] [latex]S_m=\frac{S_a}{r} [/latex]
[latex]\frac{S_a}{S_y} +\frac{S_m}{S_y} =1[/latex] Load line[latex]r=\frac{S_a}{S_m}[/latex]	[latex]S_a=\frac{rS_y}{1+r} [/latex] [latex]S_m=\frac{S_y}{1+r} [/latex]
[latex]\frac{S_a}{S_e} +\left(\frac{S_m}{S_{ut}} ight)^2 =1[/latex] [latex]\frac{S_a}{S_y} +\frac{S_m}{S_y} =1[/latex]	[latex]S_m=\frac{2S^2_{ut}}{2S_e} \left[1-\sqrt{1+\left(\frac{2S_e}{S_{ut}} ight)^2 \left(1-\frac{S_y}{S_e} ight) } ight][/latex] [latex]S_{a} = S_{y}- S_{m} , r_{crit}={S_a}/{S_m}[/latex]
Fatigue factor of safety [latex]n_f=\frac{1}{2} \left(\frac{S_{ut}}{\sigma _m} ight)^2\frac{\sigma _a}{S_e}\left[-1+\sqrt{1+\left(\frac{2\sigma _mS_e}{S_{ut}\sigma _a} ight)^2 } ight] \ \sigma _m\gt 0[/latex]

, is

[latex]S_a=\frac{7^2151^2}{2\left(67.5 ight) } \left\{-1+\sqrt{1+\left[\frac{2\left(67.5 ight) }{7\left(151 ight) } ight]^2 } ight\} =67.5[/latex] kpsi

Since [latex]\sigma _a\gt S_a[/latex] that is, [latex]70\gt 67.2[/latex], life is reduced. From Eq. (3),

[latex]S_f=\frac{70}{1-\left({10}/{151} ight)^2 } =70.3[/latex] kpsi

and from Eq. (2)

[latex]N=\left(\frac{70.3}{213.5} ight) ^{-{1}/{0.0833}}=619\left(10^3 ight) [/latex] cycles

Cycle 2: [latex]r={10}/{50}=0.2[/latex] and the strength amplitude is

[latex]S_a=\frac{0.2^2151^2}{2\left(67.5 ight) } \left\{-1+\sqrt{1+\left[\frac{2\left(67.5 ight) }{0.2\left(151 ight) } ight]^2 } ight\} =24.2[/latex] kpsi

Since [latex]\sigma _a\lt S_a [/latex] , that is [latex]10 \lt 24.2[/latex], then [latex]S_f = S_e[/latex] and indefinite life follows. Thus,[latex]N\longrightarrow \infty [/latex].

Cycle 3: [latex] r ={ 10}/{−30} = −0.333[/latex], and since [latex]\sigma _m \lt 0, S_f = S_e[/latex], indefinite life follows and [latex]N\longrightarrow \infty [/latex]

Cycle Number	[latex]S_{f}, kpsi[/latex]	N, cycles
1	70.3	[latex]619\left(10^3 ight) [/latex]
2	67.5	[latex]\infty [/latex]
3	67.5	[latex]\infty [/latex]

From Eq. (6–58)

[latex]D=\sum{\frac{n_i}{N_i} } [/latex]

the damage per block is

[latex]D=\sum{\frac{n_i}{N_i} } =N\left[\frac{1}{619\left(10^3 ight) }+\frac{1}{\infty }+ \frac{1}{\infty } ight] =\frac{N}{619\left(10^3 ight) } [/latex]

Setting [latex]D = 1[/latex] yields [latex]N= 619\left(10^3 ight) [/latex] cycles.

Table 6–7 Amplitude and Steady Coordinates of Strength and Important Intersections in First Quadrant for Gerber and Langer Failure Criteria
Intersecting Equations	Intersection Coordinates
$\frac{S_a}{S_e} +\left(\frac{S_m}{S_{ut}}\right)^2 =1$ Load line $r=\frac{S_a}{S_m}$	$S_a=\frac{r^2S^2_{ut}}{2S_e} \left[-1+\sqrt{1+\left(\frac{2S_e}{rS_{ut}} \right)^2 } \right]$ $S_m=\frac{S_a}{r}$
$\frac{S_a}{S_y} +\frac{S_m}{S_y} =1$ Load line $r=\frac{S_a}{S_m}$	$S_a=\frac{rS_y}{1+r}$ $S_m=\frac{S_y}{1+r}$
$\frac{S_a}{S_e} +\left(\frac{S_m}{S_{ut}}\right)^2 =1$ $\frac{S_a}{S_y} +\frac{S_m}{S_y} =1$	$S_m=\frac{2S^2_{ut}}{2S_e} \left[1-\sqrt{1+\left(\frac{2S_e}{S_{ut}} \right)^2 \left(1-\frac{S_y}{S_e} \right) } \right]$ $S_{a} = S_{y}- S_{m} , r_{crit}={S_a}/{S_m}$
Fatigue factor of safety $n_f=\frac{1}{2} \left(\frac{S_{ut}}{\sigma _m} \right)^2\frac{\sigma _a}{S_e}\left[-1+\sqrt{1+\left(\frac{2\sigma _mS_e}{S_{ut}\sigma _a} \right)^2 } \right] \ \sigma _m\gt 0$

Question 2.6.15: Given a part with Sut= 151 kpsi and at the critical location...

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