Safety Factor against Fatigue Failure of a Tensile Link A tensile link of thickness t with two fillets is subjected to a load fluctuating between Pmin and Pmax (Figure 7.15). Calculate the factor of safety n for unlimited life based on the Goodman criterion. Given: D = 120 mm, d = 80 mm, r = 16 mm,

Question

Safety Factor against Fatigue Failure of a Tensile Link

A tensile link of thickness [latex]t[/latex] with two fillets is subjected to a load fluctuating between [latex]P_{\min}[/latex] and [latex]P_{\max}[/latex] (Figure 7.15). Calculate the factor of safety [latex]n[/latex] for unlimited life based on the Goodman criterion.

Given: [latex]D[/latex] = 120 mm, [latex]d[/latex] = 80 mm, [latex]r[/latex] = 16 mm,

[latex]t[/latex] = 15 mm, [latex]P_{\min}[/latex] = 90 kN, [latex]P_{\max}[/latex] = 210 kN.

Design Decisions: The link is made of steel with [latex]S_{u}[/latex] = 700 MPa. The fillets and adjacent surfaces are ground. A reliability of 99.9% is desired.

Accepted Answer

The minimum cross-sectional area equals [latex]A[/latex] = 80 × 15 = 1200 mm². The maximum mean and range stresses are

[latex] \begin{array}{c} \sigma_m=\frac{(210+90)\left(10^3 ight)}{2\left(1200 imes 10^{-6} ight)}=125 MPa \ \sigma_a=\frac{(210-90)\left(10^3 ight)}{2\left(1200 imes 10^{-6} ight)}=50 MPa \end{array} [/latex]

Referring to Figure C.1, for [latex]D/d[/latex] = 1.5 and [latex]r/d[/latex] = 0.2, we obtain [latex]K_{t}[/latex] = 1.72. Inasmuch as the given 16 mm fillet radius is large, we use the value of the notch sensitivity for the steel having [latex]S_{u}[/latex] = 700 MPa, corresponding to [latex]r[/latex] = 4 mm in Figure 7.9a; that is, [latex]q[/latex] = 0.85. Hence, [latex] K_f=1+0.85(1.72-1)=1.61 .[/latex]

For axial loading, there is no size factor, [latex]C_{s}[/latex] = 1.0. Corresponding to a round surface finish, from Equation 7.7 and Table 7.2,

[latex] C_f=A S_u^b [/latex] (7.7)

[latex] \begin{aligned} C_f &=A\left(S_u ight)^b \ &=1.58(700)^{-0.085}=0.91 \end{aligned} [/latex]

By Table 7.3, for 99.9% material reliability, [latex]C_{r}[/latex] = 0.75. The temperature is not elevated, [latex]C_{t}[/latex] = 1.0. The modified endurance limit, using Equations 7.3 and 7.6, is

[latex] \begin{aligned} S_e &=C_f C_r C_s C_t\left(1 / K_f ight)\left(0.45 S_u ight) \ &=(0.91)(0.75)(1.0)(1.0)(1 / 1.61)(0.45 imes 700) \ &=133.5 MPa \end{aligned} [/latex]

The factor of safety, applying Equation 7.22, is therefore

[latex] n=\frac{S_u}{\sigma_m+\frac{S_u}{S_e} \sigma_a} [/latex] (7.22)

[latex] n=\frac{700}{125+\frac{700}{133.5}(50)}=1.81 [/latex]

Comments: If the load is well controlled and there is no impact, this factor guards the link against the fatigue failure.

Table 7.2 Surface Finish Factors [latex]C_{f}[/latex]
	[latex]A[/latex]
Surface Finish	MPa	ksi	[latex]b[/latex]
Ground	1.58	1.34	−0.085
Machined or cold drawn	4.51	2.7	−0.265
Hot rolled	57.7	14.4	−0.718
Forged	272.0	39.9	−0.995

Table 7.3 Reliability Factors
Survival Rate (%)	[latex]C_{r}[/latex]
50	1.00
90	0.89
95	0.87
98	0.84
99	0.81
99.9	0.75
99.99	0.70

Question 7.5: Safety Factor against Fatigue Failure of a Tensile Link A te...

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