The 1018 steel strap of Fig. 9–22 has a repeatedly applied load of 2000 lbf (Fa = Fm = 1000 lbf). Determine the fatigue factor of safety fatigue strength of the weldment.

Question

The 1018 steel strap of Fig. 9–22 has a repeatedly applied load of 2000 lbf ([latex]F_{a} = F_{m}[/latex] = 1000 lbf). Determine the fatigue factor of safety fatigue strength of the weldment.

Accepted Answer

From Table 6–2, p. 280, [latex]k_{a} = 39.9(58)^{−0.995}[/latex] = 0.702.

Table 6–2 Parameters for Marin Surface Modification Factor, Eq. (6–19)

[latex]k_{a} = aS^{b}_{ut}[/latex] (6–19)

Exponent b	Factor a		Surface Finish
Exponent b	[latex]S_{ut}[/latex], MPa	[latex]S_{ut}[/latex], kpsi	Surface Finish
−0.085	1.58	1.34	Ground
−0.265	4.51	2.70	Machined or cold-drawn
−0.718	57.7	14.4	Hot-rolled
−0.995	272.	39.9	As-forged

From C.J. Noll and C. Lipson, “Allowable Working Stresses,” Society for Experimental Stress Analysis, vol. 3, no. 2, 1946 p. 29. Reproduced by O.J. Horger (ed.) Metals Engineering Design ASME Handbook, McGraw-Hill, New York. Copyright © 1953 by The McGraw-Hill Companies, Inc. Reprinted by permission.

[latex]A = 2(0.707)0.375(2) = 1.061 in^{2}[/latex]

For uniform shear stress on the throat [latex]k_{b}[/latex] = 1.

From Eq. (6–26), p. 282, [latex]k_{c}[/latex] = 0.59. From Eqs. (6–8), p. 274, and (6–18), p. 279,

[latex]k_{c} =\begin{cases} 1 & bending \ 0.85 & axial \ 0.59 & torsion \end{cases}[/latex] (6-26)

[latex]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}[/latex] (6-8)

[latex]S_{e} = k_{a}k_{b}k_{c}k_{d}k_{e}k_{f} S′_{e}[/latex] (6–18)

[latex]S_{se} = 0.702(1)0.59(1)(1)(1)0.5(58) = 12.0 kpsi[/latex]

From Table 9–5, [latex]K_{f s}[/latex] = 2. Only primary shear is

present:

Table 9–5 Fatigue Stress-Concentration Factors, [latex]K_{fs}[/latex]

[latex]K_{fs}[/latex]	Type of Weld
1.2	Reinforced butt weld
1.5	Toe of transverse fillet weld
2.7	End of parallel fillet weld
2.0	T-butt joint with sharp corners

[latex]τ^{′}_{a} = τ^{′}_{m} =\frac {K_{f s} F_{a}}{A} =\frac {2(1000)}{1.061} = 1885 psi[/latex]

From Eq. (6–54), p. 309, [latex]S_{su} \dot{=}0.67S_{ut}[/latex] . This, together with the Gerber fatigue failure criterion for shear stresses from Table 6–7, p. 299, gives

[latex]S_{su} = 0.67S_{ut}[/latex] (6–54)

[latex]n_{f} =\frac {1}{2} \left(\frac {0.67S_{ut}}{τ_{m}} ight)^{2} \frac {τ_{a}}{S_{se}} \left[ -1+ \sqrt{ 1+ \left(\frac {2τ_{m} S_{se}}{0.67S_{ut} τ_{a}} ight)^{2} } ight] [/latex]

[latex]n_{f} =\frac {1}{2} \left(\frac {0.67(58)}{1.885} ight)^{2} \frac {1.885}{12.0} \left[ -1+ \sqrt{ 1+ \left(\frac {2(1.885) 12.0}{0.67(58) 1.885} ight)^{2} } ight]= 5.85 [/latex]

Table 6–7 Amplitude and Steady Coordinates of Strength and Important Intersections in First Quadrant for Gerber and Langer Failure Criteria

Intersection Coordinates

Intersecting Equations

[latex]S_{a} =\frac {r^{2}S^{2}_{ut}}{2S_{e}} \left[-1+\sqrt{1+(\frac {2S_{e}}{r S_{ut}})^{2})} ight][/latex]

[latex]S_{m} =\frac {S_{a}}{r}[/latex]

[latex]\frac {S_{a}}{S_{e}} +(\frac {S_{m}}{S_{ut}})^{2}= 1[/latex]

Load line [latex]r =\frac {S_{a}}{S_{m}}[/latex]

[latex]S_{a} =\frac {r S_{y}}{1 + r}[/latex]

[latex]S_{m} =\frac {S_{y}}{1 + 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_{m} =\frac {S^{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]

[latex]\frac {S_{a}}{S_{e}} +(\frac {S_{m}}{S_{ut}})^{2}= 1[/latex]

[latex]\frac {S_{a}}{S_{y}} +\frac {S_{m}}{S_{y}}= 1[/latex]

Fatigue factor of safety

[latex]n_{f} =\frac {1}{2} \left[\frac {S_{ut}}{σ_{m}} ight)^{2} \frac {σ_{a}}{S_{e}} \left[ -1+ \sqrt{ 1+ \left(\frac {2σ{m} S_{e}}{S_{ut} σ_{a}} ight)^{2} } ight] σ_{m} > 0 [/latex]

Question 9.6: The 1018 steel strap of Fig. 9–22 has a repeatedly applied l...

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