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.] |
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Fatigue |
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Shear |
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Stress- |
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Tensile |
Compressive |
Modulus |
Modulus of |
Endurance |
Brinell |
Concentration |
| ASTM |
Strength |
Strength |
of Rupture |
Elasticity, Mpsi |
Limit* |
Hardness |
Factor |
| Number |
S_{ut}, kpsi |
S_{uc}, kpsi |
S_{su}, kpsi |
Tension^† |
Torsion |
S_e, kpsi |
H_B |
K_f |
| 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 |
S_{ut}= 31 \ kpsi \, \ S_{uc}= 109 \ kpsi / ,k_ak_bS'_e= 14 \ kpsi.
Since k_c for axial loading is 0.9, then S_e= \left(k_ak_bS'_e\right) k_c=14\left(0.9\right) =12.6 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.
\sigma_0= {F}/{A} \ , \ where \ A = \left(w − d\right) \ t and t is the thickness. |
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A=t\left(\omega -d\right) =0.375\left(1-1.25\right) =0.281 \ in^2 \ , \ {d}/{\omega }={0.25}/{1}=0.25 \ , \ and \ K_t=2.45.
The notch sensitivity for cast iron is 0.20 (see p. 296), so
K_f=1+q\left(K_t-1\right) =1+0.20\left(2.45-1\right) =1.29
(a) \sigma _a=\frac{K_fF_a}{A} =\frac{1.29\left(0\right) }{0.281} =0 \sigma _m=\frac{K_fF_m}{A} =\frac{1.29\left(1000\right) }{0.281}\left(10^{-3}\right) =4.59
and
n=\frac{S_{ut}}{\sigma _m} =\frac{31.0}{4.59} =6.75
(b) F_a=F_m=\frac{F}{2} =\frac{1000}{2} =500 lbf
\sigma _a=\sigma _m=\frac{K_fF_a}{A} =\frac{1.29\left(500\right) }{0.281} (10^{-3})=2.30 lbf
r=\frac{\sigma _a}{\sigma _m} =1
From Eq. (6–52),
S_a= \frac{rS_{ut}+S_e}{2} \left[-1+\sqrt{1+\frac{4rS_{ut}S_e}{\left(rS_{ut}+S_e\right)^2 } } \right]
S_a=\frac{\left(1\right)31+12.6 }{2} \left[-1+\sqrt{1+\frac{4\left(1\right)31\left(12.6\right) }{\left[\left(1\right)31+12.6 \right]^2 } } \right] =7.63 kpsi
n=\frac{S_a}{\sigma _a} =\frac{7.63}{2.30} =3.32
(c) F_a=\frac{1}{2} \left|300-\left(-1000\right) \right| =650 ibf
\sigma _a=\frac{1.29\left(650\right) }{0.281} \left(10^{-3}\right) =2.98kspi
F_m=\frac{1}{2} \left[300+\left(-1000\right) \right] =-350 ibf
\sigma _m=\frac{1.29\left(-350\right) }{0.281} \left(10^{-3}\right) =-1.61 kpsi
\frac{\sigma _a}{\sigma _m} =\frac{3.0}{-1.61} =1.86
From Eq. (6–53)
S_a=S_e+\left(\frac{S_e}{S{ut}}-1 \right) S_m \ \ \ -S{ut} \leq S_m \leq0(for cast iron)
S_{a}=S_{e}+\left(S_{e} / S_{u t}-1\right) S_{m} \text { and } S_{m}=S_{a} / r \text {. It follows that }
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
n=\frac{S_a}{\sigma _a} =\frac{18.5}{2.98} =6.20
Figure 6–31b shows the portion of the designer’s fatigue diagram that was constructed.
