STEM Problem Detail

Problem Statement

Prompt

Provide me the solution to Example : 32.3

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## Example 32.3

### Given:

\( N=1800\ \text{rpm},\ p=3.15\ \text{N/mm}^2,\ D=100\ \text{mm},\ m_R=2.25\ \text{kg} \)

### 1. I-section of connecting rod

Assume proportions: \( B=4t,\ H=5t \).
\[
A = 2(4t\cdot t) + 3t\cdot t = 11t^2
\]
\[
I_{xx} = \frac{1}{12}[4t(5t)^3 - 3t(3t)^3] = \frac{419}{12}t^4
\]
\[
I_{yy} = 2\cdot\frac{1}{12}t(4t)^3 + \frac{1}{12}(3t)t^3 = \frac{131}{12}t^4
\]
\[
\frac{I_{xx}}{I_{yy}} = \frac{419}{131} = 3.2\ \text{(OK)}
\]
Max gas load:
\[
F_C = \frac{\pi D^2}{4}p = \frac{\pi\times100^2}{4}\times3.15 = 24740\ \text{N}
\]
Buckling load \( W_B = F_C\times F.S. = 24740\times6 = 148440\ \text{N} \)
\[
k_x = \sqrt{\frac{I_{xx}}{A}} = 1.78t,\quad L = l = 380\ \text{mm}
\]
Using Rankine’s formula:
\[
148440 = \frac{\sigma_c A}{1+a\left(\frac{L}{k_x}\right)^2},\quad \sigma_c=320,\ a=\frac{1}{7500}
\]
\[
148440 = \frac{320\times11t^2}{1+\frac{1}{7500}\left(\frac{380}{1.78t}\right)^2}
\]
\[
464(t^2+6.1) = 11t^4
\]
\[
t^4 - 42.2t^2 - 257.3 = 0
\]
\[
t^2 = \frac{42.2\pm\sqrt{42.2^2+4\times257.3}}{2} = 47.6
\]
\[
t = 6.9\ \text{mm} \ (\text{take } 7\ \text{mm})
\]
Dimensions:
\[
B = 4t = 28\ \text{mm},\quad H = 5t = 35\ \text{mm}
\]
Near big end: \( H_1 = 1.2H = 42\ \text{mm} \)
Near small end: \( H_2 = 0.85H = 30\ \text{mm} \)

### 2. Big end bearing

\( l_c = 1.3d_c,\ p_{bc} = 10\ \text{N/mm}^2 \)
\[
F_C = d_c l_c p_{bc} = 1.3d_c^2\times10 = 13d_c^2
\]
\[
13d_c^2 = 24740,\quad d_c^2 = 1903,\quad d_c = 43.6\ \text{mm} \ (\text{take } 44\ \text{mm})
\]
\[
l_c = 1.3\times44 = 57.2\ \text{mm} \ (\text{take } 58\ \text{mm})
\]

### 3. Small end bearing

\( l_p = 2d_p,\ p_{bp}=15\ \text{N/mm}^2 \)
\[
F_C = d_p l_p p_{bp} = 2d_p^2\times15 = 30d_p^2
\]
\[
30d_p^2 = 24740,\quad d_p^2 = 825,\quad d_p = 28.7\ \text{mm} \ (\text{take } 29\ \text{mm})
\]
\[
l_p = 2\times29 = 58\ \text{mm}
\]

### 4. Big end bolts

Inertia force at TDC:
\[
F_I = m_R \omega^2 r\left(1+\frac{r}{l}\right)
\]
\[
\omega = \frac{2\pi\times1800}{60} = 188.5\ \text{rad/s},\quad r = 0.095\ \text{m}
\]
\[
F_I = 2.25\times188.5^2\times0.095\left(1+\frac{0.095}{0.38}\right) = 9490\ \text{N}
\]
Two bolts, \( \sigma_t = 60\ \text{N/mm}^2 \):
\[
F_I = 2\times\frac{\pi}{4}d_{cb}^2\sigma_t = 94.26d_{cb}^2
\]
\[
d_{cb}^2 = \frac{9490}{94.26} = 100.7,\quad d_{cb} = 10.03\ \text{mm}
\]
\[
d_b = \frac{d_{cb}}{0.84} = 11.94\ \text{mm} \ (\text{take } 12\ \text{mm})
\]

### 5. Big end cap thickness

Bolt centers distance:
\[
x = d_c + 2\times3 + d_b + 3 = 44+6+12+3 = 65\ \text{mm}
\]
Bending moment:
\[
M_c = \frac{F_I x}{6} = \frac{9490\times65}{6} = 102810\ \text{N·mm}
\]
\[
b_c = l_c = 58\ \text{mm},\quad \sigma_b = 80\ \text{N/mm}^2
\]
\[
Z_c = \frac{b_c t_c^2}{6} = 9.7t_c^2
\]
\[
\sigma_b = \frac{M_c}{Z_c},\quad 80 = \frac{102810}{9.7t_c^2}
\]
\[
t_c^2 = 132.5,\quad t_c = 11.5\ \text{mm}
\]

### 6. Whipping stress check

Mass per unit length:
\[
m_1 = A\rho = 11t^2\rho = 11\times0.007^2\times8000 = 4.312\ \text{kg/m}
\]
Max bending moment due to whipping:
\[
M_{\max} = m_1\omega^2 r \frac{l^2}{9\sqrt{3}}
\]
\[
M_{\max} = 4.312\times188.5^2\times0.095\times\frac{0.38^2}{9\sqrt{3}} = 134.7\ \text{N·m} = 134700\ \text{N·mm}
\]
Section modulus:
\[
Z_{xx} = \frac{I_{xx}}{H/2} = \frac{419t^4/12}{5t/2} = 13.97t^3 = 13.97\times343 = 4792\ \text{mm}^3
\]
Induced stress:
\[
\sigma_{b(\max)} = \frac{M_{\max}}{Z_{xx}} = \frac{134700}{4792} = 28.1\ \text{N/mm}^2 < 80\ \text{OK}
\]

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