A grade 30 gray cast iron is subjected to a load F applied to a 1 by 3/8 -in cross-section link with a 1/4 -in-diameter hole drilled in the center as depicted in Fig. 6–31a. The surfaces are machined. In the neighborhood of the hole, what is the factor of safety

Question

A grade 30 gray cast iron is subjected to a load F applied to a 1 by [latex]\frac {3}{8}[/latex] -in cross-section link with a [latex]\frac {1}{4}[/latex] -in-diameter hole drilled in the center as depicted in Fig. 6–31a. The surfaces are machined. In the neighborhood of the hole, what is the factor of safety guarding against failure under the following conditions:
(a) The load F = 1000 lbf tensile, steady.
(b) The load is 1000 lbf repeatedly applied.
(c) The load fluctuates between −1000 lbf and 300 lbf without column action.
Use the Smith-Dolan fatigue locus.

Accepted Answer

Some preparatory work is needed. From Table A–24, [latex]S_{ut} = 31 kpsi, S_{uc} = 109 kpsi, k_{a}k_{b}S′_{e} = 14 kpsi[/latex]. Since [latex]k_{c}[/latex] for axial loading is 0.9, then [latex]S_{e}= (k_{a}k_{b}S′_{e})k_{c} = 14(0.9) = 12.6 kpsi[/latex] . From Table A–15–1, A = t(w − d) = 0.375(1 − 0.25) = 0.281 [latex]in^{2}[/latex] , d/w = 0.25/1 = 0.25, and [latex]K_{t}[/latex] = 2.45. The notch sensitivity for cast iron is 0.20 (see p. 288), so

Table A–24 Mechanical Properties of Three Non-Steel Metals (a) Typical Properties of Gray Cast Iron [The American Society for Testing and Materials (ASTM) numbering system for gray cast iron is such that the numbers correspond to the minimum tensile strength in kpsi. Thus an ASTM No. 20 cast iron has a minimum tensile strength of 20 kpsi. Note particularly that the tabulations are typical of several heats.]

Fatigue Stress- Concentration Factor $_{[latex]K_{f}[/latex]}$	Brinell Hardness $_{[latex]H_{B}[/latex]}$	Endurance Limit* $_{[latex]S_{e}[/latex]}$ , kpsi	Modulus of Elasticity, Mpsi		Shear Modulus of Rupture	Compressive Strength $_{[latex]S_{uc}[/latex]}$ , kpsi	Tensile Strength $_{[latex]S_{ut}[/latex]}$ kpsi	ASTM Number
	Brinell Hardness $_{[latex]H_{B}[/latex]}$	Endurance Limit* $_{[latex]S_{e}[/latex]}$ , kpsi	Torsion	Tension†	Shear Modulus of Rupture	Compressive Strength $_{[latex]S_{uc}[/latex]}$ , kpsi	Tensile Strength $_{[latex]S_{ut}[/latex]}$ kpsi	ASTM Number
1.00	156	10	3.9–5.6	9.6–14	26	83	22	20
1.05	174	11.5	4.6–6.0	11.5–14.8	32	97	26	25
1.10	201	14	5.2–6.6	13–16.4	40	109	31	30
1.15	212	16	5.8–6.9	14.5–17.2	48.5	124	36.5	35
1.25	235	18.5	6.4–7.8	16–20	57	140	42.5	40
1.35	262	21.5	7.2–8.0	18.8–22.8	73	164	52.5	50
1.50	302	24.5	7.8–8.5	20.4–23.5	88.5	187.5	62.5	60

*Polished or machined specimens.
†The modulus of elasticity of cast iron in compression corresponds closely to the upper value in the range given for tension and is a more constant value than that for tension.

Table A–15 Charts of Theoretical Stress-Concentration Factors [latex]K*_{t}[/latex]

*Factors from R. E. Peterson, “Design Factors for Stress Concentration,” Machine Design, vol. 23, no. 2, February 1951, p. 169; no. 3, March 1951, p. 161, no. 5, May 1951, p. 159; no. 6, June 1951, p. 173; no. 7, July 1951, p. 155. Reprinted with permission from Machine Design, a Penton Media Inc. publication.

[latex]K_{f} = 1 + q(K_{t} − 1) = 1 + 0.20(2.45 − 1) = 1.29[/latex]

(a) [latex]σ_{a} =\frac {K_{f} F_{a}}{A} =\frac {1.29(0)}{0.281} = 0 σ_{m} =\frac {K_{f} F_{m}}{A} =\frac {1.29(1000)}{0.281} =(10^{−3}) = 4.59 kpsi[/latex]

and

[latex]n =\frac {S_{ut}}{σ_{m}} =\frac {31.0}{4.59} = 6.75[/latex]

(b) [latex]F_{a} = F_{m} =\frac {F}{2} =\frac {1000}{2} = 500 lbf[/latex]

[latex]σ_{a}=σ_{m} =\frac {K_{f} F_{a}}{A} =\frac {1.29(500)}{0.281} =(10^{−3}) = 2.30 kpsi[/latex]

[latex]r =\frac {σ_{a}}{σ_{m}} = 1[/latex]

From Eq. (6–52),

[latex]S_{a} =\frac {r S_{ut} + S_{e}}{2} \left[ -1+\sqrt{1 +\frac {4rS_{ut} S_{e}}{(r S_{ut} + S_{e})^{2}}} \right][/latex] (6-52)

[latex]S_{a} =\frac {(1) 31 + 12.6}{2} \left[ -1+\sqrt{1 +\frac {4(1)31(12.6) }{[(1) 31 + 12.6]^{2}}} \right]=7.63 kpsi [/latex]

[latex]n =\frac {S_{a}}{σ_{a}} =\frac {7.63}{2.30} = 3.32[/latex]

(c) [latex]F_{a} =\frac {1}{2}|300 − (−1000)| = 650 lbf σ_{a} =\frac{1.29(650)}{0.281}(10^{−3}) = 2.98 kpsi[/latex]

[latex]F_{m} =\frac {1}{2}|300 + (−1000)| = -350 lbf σ_{m} =\frac{1.29(-350)}{0.281}(10^{−3}) =-1.61 kpsi[/latex]

[latex]r =\frac {σ_{a}}{σ_{m}} =\frac {3.0}{−1.61} = −1.86[/latex]

From Eq. (6–53), [latex]S_{a} = S_{e} + (S_{e}/S_{ut} − 1)S_{m} and S_{m} = S_{a}/r[/latex] . It follows that

[latex]S_{a} = S_{e} +\left(\frac {S_{e}}{S_{ut}} − 1\right)S_{m} −S_{ut} ≤ S_{m} ≤ 0 (for cast iron)[/latex] (6–53)

[latex]S_{a} =\frac {S_{e} }{1 −\frac {1}{r} \left(\frac{S_{e}}{S_{ut}} −1\right)} =\frac {12.6}{1 −\frac {1}{-1.86} \left(\frac{12.6}{31} −1\right)}=18.5 kpsi[/latex]

[latex]n =\frac {S_{a}}{σ_{a}} =\frac {18.5}{2.98} = 6.20[/latex]

Figure 6–31b shows the portion of the designer’s fatigue diagram that was constructed.

Question 6.13: A grade 30 gray cast iron is subjected to a load F applied t...

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