STEM Problem Detail

Problem #73

Physics Hard

Problem Statement

Prompt

Provide me the solution to Example : 32.7

Solution Rendering...

## Example 32.7 — Rocker Arm Design

### Given:

\( d_v = 80\ \text{mm},\ h = 25\ \text{mm},\ r = 12.5\ \text{mm},\ m = 0.4\ \text{kg},\ \alpha = 110^\circ,\ N = 1500\ \text{rpm} \)
\( p_c = 0.4\ \text{N/mm}^2,\ p_s = 0.02\ \text{N/mm}^2,\ l = 180\ \text{mm},\ \theta = 135^\circ \)

#### Forces on rocker arm

Gas load:
\[
P_1 = \frac{\pi}{4}d_v^2 p_c = \frac{\pi}{4}(80)^2(0.4) = 2011\ \text{N}
\]
Weight:
\[
w = mg = 0.4\times9.8 = 3.92\ \text{N}
\]
Total load on valve:
\[
P = P_1 + w = 2014.92\ \text{N}
\]
Initial spring force:
\[
F_s = \frac{\pi}{4}d_v^2 p_s - w = \frac{\pi}{4}(80)^2(0.02) - 3.92 = 96.6\ \text{N}
\]
Camshaft speed \( = N/2 = 750\ \text{rpm} \).
Angle turned by camshaft per second \( = \frac{750}{60}\times360 = 4500\ \text{deg/s} \).
Time to open/close valve:
\[
t = \frac{\alpha}{4500} = \frac{110}{4500} = 0.024\ \text{s}
\]
Max acceleration of valve:
\[
a = \omega^2 r = \left(\frac{2\pi}{t}\right)^2 r = \left(\frac{2\pi}{0.024}\right)^2\times0.0125 = 857\ \text{m/s}^2
\]
Force due to acceleration:
\[
F_a = ma + w = 0.4\times857 + 3.92 = 346.72\ \text{N}
\]
Max load on rocker arm:
\[
F_e = P + F_s + F_a = 2014.92 + 96.6 + 346.72 = 2458.24\ \text{N}\ (\text{take } 2460\ \text{N})
\]
Since arms are equal, \( F_c = F_e = 2460\ \text{N} \).
Reaction at fulcrum:
\[
R_F = \sqrt{F_e^2 + F_c^2 - 2F_e F_c\cos\theta}
\]
\[
= \sqrt{(2460)^2 + (2460)^2 - 2\times2460\times2460\times\cos135^\circ} = 4545\ \text{N}
\]

#### 1. Fulcrum pin

Let \( d_1 = \text{diameter},\ l_1 = 1.25d_1,\ p_b = 5\ \text{N/mm}^2 \).
Bearing check:
\[
R_F = d_1 l_1 p_b = d_1(1.25d_1)(5) = 6.25d_1^2
\]
\[
d_1^2 = \frac{4545}{6.25} = 727,\quad d_1 = 26.97\ \text{mm}\ (\text{take } 30\ \text{mm})
\]
\[
l_1 = 1.25\times30 = 37.5\ \text{mm}
\]
Shear check (double shear):
\[
R_F = 2\times\frac{\pi}{4}d_1^2\tau
\]
\[
\tau = \frac{4545}{2\times\frac{\pi}{4}(30)^2} = 3.2\ \text{N/mm}^2\quad\text{(Safe)}
\]
External diameter of boss:
\[
D_1 = 2d_1 = 60\ \text{mm}
\]
With 3 mm bronze bush, hole diameter:
\[
d_b = d_1 + 6 = 36\ \text{mm}
\]
Bending stress in boss:
\[
M = F_e l = 2460\times180 = 443\times10^3\ \text{N·mm}
\]
\[
Z = \frac{1}{12}\times37.5\left(60^3 - 36^3\right) = 17640\ \text{mm}^3
\]
\[
\sigma_b = \frac{M}{Z} = 25.1\ \text{N/mm}^2\quad\text{(Safe)}
\]

#### 2. Roller pin

Let \( d_2 = \text{diameter},\ l_2 = 1.25d_2,\ p_b = 7\ \text{N/mm}^2 \).
Bearing check:
\[
F_c = d_2 l_2 p_b = d_2(1.25d_2)(7) = 8.75d_2^2
\]
\[
d_2^2 = \frac{2460}{8.75} = 281,\quad d_2 = 16.76\ \text{mm}\ (\text{take } 18\ \text{mm})
\]
\[
l_2 = 1.25\times18 = 22.5\ \text{mm}\ (\text{take } 24\ \text{mm})
\]
Shear check:
\[
\tau = \frac{F_c}{2\times\frac{\pi}{4}d_2^2} = \frac{2460}{2\times\frac{\pi}{4}(18)^2} = 4.83\ \text{N/mm}^2\quad\text{(Safe)}
\]
Thickness of each eye:
\[
t_2 = \frac{l_2}{2} = 12\ \text{mm}
\]
Max bending moment in pin:
\[
M = \frac{5}{24}F_c l_2 = \frac{5}{24}\times2460\times24 = 12300\ \text{N·mm}
\]
\[
Z = \frac{\pi}{32}d_2^3 = \frac{\pi}{32}(18)^3 = 573\ \text{mm}^3
\]
\[
\sigma_b = \frac{M}{Z} = 21.5\ \text{N/mm}^2\quad\text{(Safe)}
\]
Outer diameter of eye:
\[
D_2 = 2d_2 = 36\ \text{mm}
\]
Outer diameter of roller = \( 42\ \text{mm} \).
Length of fork:
\[
l_3 = l_2 + 2\times\frac{t_2}{2} + 3 = 24 + 12 + 3 = 39\ \text{mm}
\]

#### 3. Rocker arm cross-section (I-section)

Bending moment at section near boss:
\[
M = 2460\left(180 - \frac{60}{2}\right) = 369\times10^3\ \text{N·mm}
\]
Assume proportions: width = \( 2.5t \), depth = \( 6t \), web thickness = \( t \), flange thickness = \( t \).
Section modulus:
\[
Z = \frac{1}{12}\left[2.5t(6t)^3 - 1.5t(4t)^3\right]\frac{1}{3t} = 12.33t^3
\]
Using \( \sigma_b = 70\ \text{N/mm}^2 \):
\[
\sigma_b = \frac{M}{Z} = \frac{369\times10^3}{12.33t^3} = \frac{29.93\times10^3}{t^3} = 70
\]
\[
t^3 = \frac{29.93\times10^3}{70} = 427.6,\quad t = 7.5\ \text{mm}\ (\text{take } 8\ \text{mm})
\]
Width of flange \( = 2.5t = 20\ \text{mm} \)
Depth of web \( = 4t = 32\ \text{mm} \)
Depth of section \( = 6t = 48\ \text{mm} \)

#### 4. Tappet screw

Compressive load \( F_e = 2460\ \text{N},\ \sigma_c = 50\ \text{N/mm}^2 \).
Core diameter \( d_c \):
\[
F_e = \frac{\pi}{4}d_c^2\sigma_c = 39.3d_c^2
\]
\[
d_c^2 = \frac{2460}{39.3} = 62.6,\quad d_c = 7.9\ \text{mm}\ (\text{take } 8\ \text{mm})
\]
Nominal diameter:
\[
d = \frac{d_c}{0.84} = 9.52\ \text{mm}\ (\text{take } 10\ \text{mm})
\]
Diameter of circular end \( D_3 = 20\ \text{mm} \), depth \( t_4 = 20\ \text{mm} \).

#### 5. Valve spring

Initial load \( W_1 = F_s = 96.6\ \text{N} \).
Load at full lift (assuming stiffness \( s = 10\ \text{N/mm} \)):
\[
W_2 = h\times s = 25\times10 = 250\ \text{N}
\]
Total load:
\[
W = W_1 + W_2 = 346.6\ \text{N}
\]

**(a) Mean coil diameter**

Assume \( C = D/d = 8 \). Wahl factor:
\[
K = \frac{4C-1}{4C-4} + \frac{0.615}{C} = \frac{31}{28} + 0.0769 = 1.184
\]
Using \( \tau = 420\ \text{N/mm}^2 \):
\[
\tau = K\frac{8WC}{\pi d^2}
\]
\[
420 = 1.184\times\frac{8\times346.6\times8}{\pi d^2} = \frac{8360}{d^2}
\]
\[
d^2 = \frac{8360}{420} = 19.9,\quad d = 4.46\ \text{mm}\ (\text{use SWG 7: } d = 4.47\ \text{mm})
\]
\[
D = Cd = 8\times4.47 = 35.76\ \text{mm}
\]
Outer diameter \( D_o = D + d = 40.23\ \text{mm} \).

**(b) Number of turns**

Stiffness \( s = W/\delta = 10\ \text{N/mm} \), so \( \delta/W = 1/10 \). Using \( G = 84\times10^3\ \text{N/mm}^2 \):
\[
\frac{\delta}{W} = \frac{8C^3 n}{G d}
\]
\[
\frac{1}{10} = \frac{8\times8^3 n}{84\times10^3\times4.47} = \frac{10.9n}{10^3}
\]
\[
n = \frac{10^3}{10.9\times10} = 9.17\ (\text{take } 10)
\]
Total turns (squared and ground ends):
\[
n' = n + 2 = 12
\]

**(c) Free length**

Max compression under \( W = 346.6\ \text{N} \):
\[
\delta_{\max} = \frac{25}{250}\times346.6 = 34.66\ \text{mm}
\]
Free length:
\[
L_F = n'd + \delta_{\max} + 0.15\delta_{\max} = 12\times4.47 + 34.66 + 5.2 = 93.5\ \text{mm}
\]

**(d) Pitch of coil**

\[
\text{Pitch} = \frac{L_F}{n'-1} = \frac{93.5}{11} = 8.5\ \text{mm}
\]

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