STEM Problem Detail

Problem #68

Physics Hard

Problem Statement

Prompt

Provide me the solution to Example : 32.8

Solution Rendering...

## Example 32.8 — Valve Gear Mechanism Design

### Given:

\( D = 140\ \text{mm},\ L = 270\ \text{mm},\ P = 8.25\ \text{kW},\ N = 475\ \text{rpm},\ p = 3.5\ \text{N/mm}^2 \)
Valve opens \( 33^\circ \) before ODC, closes \( 1^\circ \) after IDC.
Rocker arm length each side = \( 150\ \text{mm} \), included angle \( \theta = 160^\circ \), valve weight \( w = 3\ \text{N} \).

### 1. Size of valve port

Piston area:
\[
a = \frac{\pi}{4}D^2 = \frac{\pi}{4}(0.14)^2 = 0.0154\ \text{m}^2
\]
Mean piston velocity:
\[
v = \frac{2LN}{60} = \frac{2\times0.27\times475}{60} = 4.275\ \text{m/s}
\]
Take gas velocity through port \( v_p = 40\ \text{m/s} \).
Using continuity equation:
\[
a_p v_p = a v
\]
\[
\frac{\pi}{4}d_p^2 \times 40 = 0.0154\times4.275
\]
\[
31.42d_p^2 = 0.0658
\]
\[
d_p^2 = 2.09\times10^{-3},\quad d_p = 0.045\ \text{m} = 45\ \text{mm}
\]

### 2. Maximum lift of valve

Take seat angle \( \alpha = 45^\circ \):
\[
h = \frac{d_p}{4\cos\alpha} = \frac{45}{4\cos45^\circ} = 15.9\ \text{mm}\ (\text{take } 16\ \text{mm})
\]

### 3. Thickness of valve head

Take \( k = 0.42,\ \sigma_b = 56\ \text{N/mm}^2 \):
\[
t = k\cdot d_p\sqrt{\frac{p}{\sigma_b}} = 0.42\times45\sqrt{\frac{3.5}{56}} = 4.72\ \text{mm}
\]

### 4. Valve stem diameter

\[
d_s = \frac{d_p}{8} + 6.35 = \frac{45}{8} + 6.35 = 11.97\ \text{mm}\ (\text{take } 12\ \text{mm})
\]

### 5. Valve head diameter

Valve seat width = \( 0.06d_p = 2.7\ \text{mm} \).
\[
d_v = d_p + 2\times2.7 = 45 + 5.4 = 50.4\ \text{mm}\ (\text{take } 51\ \text{mm})
\]

### 6. Forces on rocker arm

Gas load (taking back pressure \( p_e = 0.4\ \text{N/mm}^2 \)):
\[
P_1 = \frac{\pi}{4}d_v^2 p_e = \frac{\pi}{4}(51)^2(0.4) = 817\ \text{N}
\]
Total load on valve:
\[
P = P_1 + w = 820\ \text{N}
\]
Initial spring force (taking suction pressure \( p_s = 0.025\ \text{N/mm}^2 \)):
\[
F_s = \frac{\pi}{4}d_v^2 p_s - w = \frac{\pi}{4}(51)^2(0.025) - 3 = 48\ \text{N}
\]
Crank angle for valve open = \( 33 + 180 + 1 = 214^\circ \).
Camshaft angle for valve open = \( 214/2 = 107^\circ \).
Half of this \( = 53.5^\circ \) for lift of 16 mm.
Half of that for constant acceleration \( = 26.75^\circ \) for lift of 8 mm.
Camshaft speed = \( N/2 = 237.5\ \text{rpm} \).
Angle turned per second = \( \frac{237.5}{60}\times360 = 1425\ \text{deg/s} \).
Time for constant acceleration:
\[
t = \frac{26.75}{1425} = 0.0188\ \text{s}
\]
Valve acceleration:
\[
s = \frac{1}{2}at^2 \quad\Rightarrow\quad 8 = \frac{1}{2}a(0.0188)^2
\]
\[
a = \frac{16}{3.534\times10^{-4}} = 45274\ \text{mm/s}^2 = 45.274\ \text{m/s}^2
\]
Force due to acceleration:
\[
F_a = m a + w = \frac{3}{9.81}\times45.274 + 3 = 16.84\ \text{N}
\]
Maximum load on rocker arm:
\[
F_e = P + F_s + F_a = 820 + 48 + 16.84 = 884.84\ \text{N}\ (\text{take } 885\ \text{N})
\]
Since arms equal, \( F_c = F_e = 885\ \text{N} \).
Reaction at fulcrum:
\[
R_F = \sqrt{F_e^2 + F_c^2 - 2F_e F_c\cos160^\circ} = 1743\ \text{N}
\]

### 7. Design of fulcrum pin

Let \( d_1 = \text{diameter},\ l_1 = 1.25d_1,\ p_b = 5\ \text{N/mm}^2 \).
Bearing check:
\[
R_F = d_1 l_1 p_b = 6.25d_1^2
\]
\[
d_1^2 = \frac{1743}{6.25} = 279,\quad d_1 = 16.7\ \text{mm}\ (\text{take } 17\ \text{mm})
\]
\[
l_1 = 1.25\times17 = 21.25\ \text{mm}\ (\text{take } 22\ \text{mm})
\]
Shear check (double shear):
\[
\tau = \frac{R_F}{2\times\frac{\pi}{4}d_1^2} = \frac{1743}{2\times\frac{\pi}{4}(17)^2} = 3.84\ \text{N/mm}^2\quad\text{(Safe)}
\]
External boss diameter:
\[
D_1 = 2d_1 = 34\ \text{mm}
\]
With 3 mm bronze bush, hole diameter:
\[
d_b = d_1 + 6 = 23\ \text{mm}
\]
Bending stress in boss:
\[
M = F_e l = 885\times150 = 132750\ \text{N·mm}
\]
\[
Z = \frac{1}{12}\times22\left(34^3 - 23^3\right) = 2927\ \text{mm}^3
\]
\[
\sigma_b = \frac{M}{Z} = 45.3\ \text{N/mm}^2\quad\text{(Safe)}
\]

### 8. Forked end (roller pin)

Let \( d_2 = \text{diameter},\ l_2 = 1.25d_2,\ p_b = 7\ \text{N/mm}^2 \).
Bearing check:
\[
F_c = d_2 l_2 p_b = 8.75d_2^2
\]
\[
d_2^2 = \frac{885}{8.75} = 101.14,\quad d_2 = 10.06\ \text{mm}\ (\text{take } 11\ \text{mm})
\]
\[
l_2 = 1.25\times11 = 13.75\ \text{mm}\ (\text{take } 14\ \text{mm})
\]
Shear check:
\[
\tau = \frac{F_c}{2\times\frac{\pi}{4}d_2^2} = \frac{885}{2\times\frac{\pi}{4}(11)^2} = 4.66\ \text{N/mm}^2\quad\text{(Safe)}
\]
Thickness of each eye:
\[
t_2 = \frac{l_2}{2} = 7\ \text{mm}
\]
Bending moment in pin:
\[
M = \frac{5}{24}F_c l_2 = \frac{5}{24}\times885\times14 = 2581\ \text{N·mm}
\]
\[
Z = \frac{\pi}{32}d_2^3 = \frac{\pi}{32}(11)^3 = 131\ \text{mm}^3
\]
\[
\sigma_b = \frac{M}{Z} = 19.7\ \text{N/mm}^2\quad\text{(Safe)}
\]
Overall eye diameter:
\[
D_2 = 2d_2 = 22\ \text{mm}
\]
Roller outer diameter = \( 28\ \text{mm} \).
Length of fork:
\[
l_3 = l_2 + 2\times\frac{t_2}{2} + 3 = 14 + 7 + 3 = 24\ \text{mm}
\]

### 9. Rocker arm cross-section (rectangular)

Bending moment near boss:
\[
M = 885\left(150 - \frac{34}{2}\right) = 117705\ \text{N·mm}
\]
Take \( B = D_1 = 34\ \text{mm} \) at section.
Section modulus:
\[
Z = \frac{1}{6}t_1 B^2 = \frac{1}{6}t_1(34)^2 = 193t_1
\]
Using \( \sigma_b = 70\ \text{N/mm}^2 \):
\[
70 = \frac{M}{Z} = \frac{117705}{193t_1}
\]
\[
t_1 = \frac{117705}{193\times70} = 8.7\ \text{mm}\ (\text{take } 10\ \text{mm})
\]

### 10. Tappet screw

Compressive load \( F_e = 885\ \text{N},\ \sigma_c = 50\ \text{N/mm}^2 \).
Core diameter:
\[
F_e = \frac{\pi}{4}d_c^2\sigma_c = 39.3d_c^2
\]
\[
d_c^2 = \frac{885}{39.3} = 22.5,\quad d_c = 4.74\ \text{mm}\ (\text{take } 5\ \text{mm})
\]
Nominal diameter:
\[
d = \frac{d_c}{0.84} = 6.25\ \text{mm}\ (\text{take } 6.5\ \text{mm})
\]
Circular end diameter \( D_3 = 13\ \text{mm} \), depth \( t_4 = 13\ \text{mm} \).

### 11. Valve spring

Initial load \( W_1 = F_s = 48\ \text{N} \).
Load at full lift (stiffness \( s = 8\ \text{N/mm} \)):
\[
W_2 = h s = 16\times8 = 128\ \text{N}
\]
Total load:
\[
W = W_1 + W_2 = 176\ \text{N}
\]

**(a) Mean coil diameter**

Assume \( C = D/d = 6 \). Wahl factor:
\[
K = \frac{4C-1}{4C-4} + \frac{0.615}{C} = 1.2525
\]
Using \( \tau = 420\ \text{N/mm}^2 \):
\[
\tau = K\frac{8WC}{\pi d^2}
\]
\[
420 = 1.2525\times\frac{8\times176\times6}{\pi d^2} = \frac{3368}{d^2}
\]
\[
d^2 = \frac{3368}{420} = 8.02,\quad d = 2.83\ \text{mm}\ (\text{use SWG 11: } d = 2.946\ \text{mm})
\]
\[
D = Cd = 6\times2.946 = 17.676\ \text{mm}
\]
Outer diameter:
\[
D_o = D + d = 20.622\ \text{mm}
\]

**(b) Number of turns**

Stiffness \( s = W/\delta = 8 \), so \( \delta/W = 1/8 \). With \( G = 84\times10^3\ \text{N/mm}^2 \):
\[
\frac{\delta}{W} = \frac{8C^3 n}{G d}
\]
\[
\frac{1}{8} = \frac{8\times6^3 n}{84\times10^3\times2.946} = \frac{6.98n}{10^3}
\]
\[
n = \frac{10^3}{8\times6.98} = 17.9\ (\text{take } 18)
\]
Total turns (squared and ground):
\[
n' = n + 2 = 20
\]

**(c) Free length**

Max compression under \( W \):
\[
\delta_{\max} = \frac{16}{128}\times176 = 22\ \text{mm}
\]
Free length:
\[
L_F = n'd + \delta_{\max} + 0.15\delta_{\max} = 20\times2.946 + 22 + 3.3 = 84.22\ \text{mm}\ (\text{take } 85\ \text{mm})
\]

**(d) Pitch**

\[
\text{Pitch} = \frac{L_F}{n'-1} = \frac{85}{19} = 4.47\ \text{mm}
\]

### 12. Cam design

Camshaft diameter (empirical):
\[
D' = 0.16\times\text{bore} + 12.7 = 0.16\times140 + 12.7 = 35.1\ \text{mm}\ (\text{take } 36\ \text{mm})
\]
Base circle diameter = \( 36 + 3 = 39\ \text{mm}\ (\text{take } 40\ \text{mm}) \).
Cam width = roller width = 14 mm.
Empirical width:
\[
w' = 0.09\times\text{bore} + 6 = 0.09\times140 + 6 = 18.6\ \text{mm}\ (\text{take } 18\ \text{mm})
\]
Prime circle radius:
\[
= \frac{\text{Base circle dia}}{2} + \frac{\text{Roller dia}}{2} = 20 + 14 = 34\ \text{mm}
\]
Angular opening = \( 53.5^\circ \), closing = \( 53.5^\circ \).
Cam profile constructed graphically using displacement diagram with uniform acceleration/deceleration.

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