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 guarding against failure

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 [latex]F = 1000 [/latex] 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,

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
			Shear					Stress-
	Tensile	Compressive	Modulus	Modulus of		Endurance	Brinell	Concentration
ASTM	Strength	Strength	of Rupture	Elasticity, Mpsi		Limit*	Hardness	Factor
Number	[latex]S_{ut}[/latex], kpsi	[latex]S_{uc}[/latex], kpsi	[latex]S_{su}[/latex], kpsi	[latex]Tension^†[/latex]	Torsion	[latex]S_e[/latex], kpsi	[latex]H_B[/latex]	[latex]K_f[/latex]
20	22	83	26	9.6–14	3.9–5.6	10	156	1
25	26	97	32	11.5–14.8	4.6–6.0	11.5	174	1.05
30	31	109	40	13–16.4	5.2–6.6	14	201	1.1
35	36.5	124	48.5	14.5–17.2	5.8–6.9	16	212	1.15
40	42.5	140	57	16-20	6.4-7.8	18.5	235	1.25
50	52.5	164	73	18.8-22.8	7.2-8.0	21.5	262	1.35
60	62.5	187.5	88.5	20.4	23.5	24.5	30.2	1.5

[latex]S_{ut}= 31 \ kpsi \, \ S_{uc}= 109 \ kpsi / ,k_ak_bS'_e= 14 \ kpsi[/latex].

Since[latex] k_c [/latex] for axial loading is 0.9, then [latex]S_e= \left(k_ak_bS'_e ight) k_c=14\left(0.9 ight) =12.6[/latex] kpsi

From Table A–15–1,

Table A–15 Charts of Theoretical Stress-Concentration Factors K*

Figure A–15–1 Bar in tension or simple compression with a transverse hole.
[latex]\sigma_0= {F}/{A} \ , \ where \ A = \left(w − d ight) \ t [/latex] and t is the thickness.

[latex]A=t\left(\omega -d ight) =0.375\left(1-1.25 ight) =0.281 \ in^2 \ , \ {d}/{\omega }={0.25}/{1}=0.25 \ , \ and \ K_t=2.45[/latex].

The notch sensitivity for cast iron is 0.20 (see p. 296), so

[latex]K_f=1+q\left(K_t-1 ight) =1+0.20\left(2.45-1 ight) =1.29[/latex]

(a) [latex]\sigma _a=\frac{K_fF_a}{A} =\frac{1.29\left(0 ight) }{0.281} =0[/latex] [latex]\sigma _m=\frac{K_fF_m}{A} =\frac{1.29\left(1000 ight) }{0.281}\left(10^{-3} ight) =4.59[/latex]

and

[latex]n=\frac{S_{ut}}{\sigma _m} =\frac{31.0}{4.59} =6.75[/latex]

(b) [latex]F_a=F_m=\frac{F}{2} =\frac{1000}{2} =500[/latex] lbf

[latex]\sigma _a=\sigma _m=\frac{K_fF_a}{A} =\frac{1.29\left(500 ight) }{0.281} (10^{-3})=2.30[/latex] lbf

[latex]r=\frac{\sigma _a}{\sigma _m} =1[/latex]

From Eq. (6–52),

[latex]S_a= \frac{rS_{ut}+S_e}{2} \left[-1+\sqrt{1+\frac{4rS_{ut}S_e}{\left(rS_{ut}+S_e ight)^2 } } ight] [/latex]

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

[latex]n=\frac{S_a}{\sigma _a} =\frac{7.63}{2.30} =3.32[/latex]

(c) [latex]F_a=\frac{1}{2} \left|300-\left(-1000 ight) ight| =650[/latex] ibf

[latex]\sigma _a=\frac{1.29\left(650 ight) }{0.281} \left(10^{-3} ight) =2.98[/latex]kspi

[latex]F_m=\frac{1}{2} \left[300+\left(-1000 ight) ight] =-350[/latex] ibf

[latex]\sigma _m=\frac{1.29\left(-350 ight) }{0.281} \left(10^{-3} ight) =-1.61[/latex] kpsi

[latex]\frac{\sigma _a}{\sigma _m} =\frac{3.0}{-1.61} =1.86[/latex]

From Eq. (6–53)

[latex]S_a=S_e+\left(\frac{S_e}{S{ut}}-1 ight) S_m \ \ \ -S{ut} \leq S_m \leq0[/latex](for cast iron)

[latex] S_{a}=S_{e}+\left(S_{e} / S_{u t}-1 ight) S_{m} ext { and } S_{m}=S_{a} / r ext {. It follows that }[/latex]

[latex]S_a=\frac{S_e}{1-\frac{1}{r}\left(\frac{S_e}{S_{ut}-1} ight) } =\frac{12.6}{1-\frac{1}{-1.86}\left(\frac{12.6}{31}-1 ight) } =18.5[/latex] kpsi

[latex]n=\frac{S_a}{\sigma _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 2.6.13: A grade 30 gray cast iron is subjected to a load F applied t...

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