STEM Problem Detail

Problem #70

Physics Hard

Problem Statement

Prompt

Provide me the solution to Example : 32.5

Solution Rendering...

## Example 32.5 — Side Crankshaft Design

### Given:

\( D = 250\ \text{mm},\ L = 300\ \text{mm},\ r = 150\ \text{mm},\ W = 30\ \text{kN},\ p = 2.1\ \text{N/mm}^2,\ p' = 0.9\ \text{N/mm}^2,\ l/r = 4.5 \)

### 1. Crank at Dead Centre

#### (a) Crankpin

Piston gas load:
\[
F_p = \frac{\pi}{4}D^2 p = \frac{\pi}{4}(250)^2(2.1) = 103\times10^3\ \text{N}
\]
Assume \( l_c = 0.8d_c,\ p_b = 10\ \text{N/mm}^2 \):
\[
F_p = d_c l_c p_b = d_c(0.8d_c)(10) = 8d_c^2
\]
\[
d_c^2 = \frac{103\times10^3}{8} = 12875,\quad d_c = 113.4\ \text{mm}\ (\text{take } 115\ \text{mm})
\]
\[
l_c = 0.8\times115 = 92\ \text{mm}
\]
Check bending stress:
\[
M = \frac{3}{4}F_p l_c = \frac{3}{4}\times103\times10^3\times92 = 7.107\times10^6\ \text{N·mm}
\]
\[
Z = \frac{\pi}{32}d_c^3 = \frac{\pi}{32}(115)^3 = 149\times10^3\ \text{mm}^3
\]
\[
\sigma_b = \frac{M}{Z} = \frac{7.107\times10^6}{149\times10^3} = 47.7\ \text{N/mm}^2 < 60\ \text{N/mm}^2\quad\text{(Safe)}
\]

#### (b) Bearing 1

Thickness of crank web: \( t = 0.6d_c = 0.6\times115 = 69\ \text{mm}\ (\text{take } 70\ \text{mm}) \)
Length of bearing: \( l_1 = 1.7d_c = 195.5\ \text{mm}\ (\text{take } 200\ \text{mm}) \)
Bending moment at centre of bearing 1:
\[
M = F_p(0.75l_c + t + 0.5l_1) = 103\times10^3(0.75\times92 + 70 + 100) = 24.6\times10^6\ \text{N·mm}
\]
Using \( \sigma_b = 60\ \text{N/mm}^2 \):
\[
M = \frac{\pi}{32}d_1^3\sigma_b = 5.9d_1^3
\]
\[
d_1^3 = \frac{24.6\times10^6}{5.9} = 4.2\times10^6,\quad d_1 = 161.3\ \text{mm}\ (\text{take } 162\ \text{mm})
\]

#### (c) Crank web

Let \( w = \text{width} \).
Bending moment:
\[
M = F_p(0.75l_c + 0.5t) = 103\times10^3(0.75\times92 + 35) = 10.7\times10^6\ \text{N·mm}
\]
\[
Z = \frac{1}{6}w t^2 = \frac{1}{6}w(70)^2 = 817w\ \text{mm}^3
\]
\[
\sigma_b = \frac{M}{Z} = \frac{10.7\times10^6}{817w} = \frac{13\times10^3}{w}
\]
Direct compressive stress:
\[
\sigma_d = \frac{F_p}{w t} = \frac{103\times10^3}{70w} = \frac{1.47\times10^3}{w}
\]
Total stress:
\[
\sigma_T = \sigma_b + \sigma_d = \frac{14.47\times10^3}{w} = 60\ \text{N/mm}^2
\]
\[
w = \frac{14.47\times10^3}{60} = 241\ \text{mm}\ (\text{take } 245\ \text{mm})
\]

#### (d) Shaft under flywheel

Assume flywheel width \( 250\ \text{mm},\ l_1=l_2=200\ \text{mm} \).
Distance between bearings:
\[
b = 250 + \frac{200}{2} + \frac{200}{2} + 20 = 470\ \text{mm}
\]
\[
a = 0.75l_c + t + 0.5l_1 = 0.75\times92 + 70 + 100 = 239\ \text{mm}
\]
Horizontal reactions due to \( F_p \):
\[
H_1 = \frac{F_p(a+b)}{b} = \frac{103\times10^3(239+470)}{470} = 155.4\times10^3\ \text{N}
\]
\[
H_2 = \frac{F_p a}{b} = \frac{103\times10^3\times239}{470} = 52.4\times10^3\ \text{N}
\]
Vertical reactions due to flywheel weight (\( W=30\ \text{kN} \)):
\[
V_1 = V_2 = \frac{W}{2} = 15\times10^3\ \text{N}
\]
Horizontal bending moment at flywheel location:
\[
b_2 = \frac{b}{2} = 235\ \text{mm}
\]
\[
M_1 = F_p(a+b_2) - H_1 b_2 = 103\times10^3(239+235) - 155.4\times10^3\times235
\]
\[
= 48.8\times10^6 - 36.5\times10^6 = 12.3\times10^6\ \text{N·mm}
\]
Vertical bending moment:
\[
M_V = \frac{W b_1 b_2}{b} = \frac{W b}{4} = \frac{30\times10^3\times470}{4} = 3.525\times10^6\ \text{N·mm}
\]
Resultant bending moment:
\[
M_R = \sqrt{M_1^2 + M_V^2} = \sqrt{(12.3\times10^6)^2 + (3.525\times10^6)^2} = 12.8\times10^6\ \text{N·mm}
\]
Using \( \sigma_b = 60\ \text{N/mm}^2 \):
\[
M_R = \frac{\pi}{32}d_s^3\sigma_b = 5.9d_s^3
\]
\[
d_s^3 = \frac{12.8\times10^6}{5.9} = 2.17\times10^6,\quad d_s = 129\ \text{mm}
\]
But \( d_s \) should be > \( d_1 = 162\ \text{mm} \), so take \( d_s = 200\ \text{mm} \).

### 2. Crank at Max Twisting Moment (\(\theta = 35^\circ\))

Piston load:
\[
F_p' = \frac{\pi}{4}D^2 p' = \frac{\pi}{4}(250)^2(0.9) = 44200\ \text{N}
\]
\[
\sin\phi = \frac{\sin\theta}{l/r} = \frac{\sin35^\circ}{4.5} = 0.1275,\quad \phi = 7.32^\circ
\]
Thrust in connecting rod:
\[
F_Q = \frac{F_p'}{\cos\phi} = \frac{44200}{\cos7.32^\circ} = 44565\ \text{N}
\]
Tangential force:
\[
F_T = F_Q\sin(\theta+\phi) = 44565\sin42.32^\circ = 30\times10^3\ \text{N}
\]
Radial force:
\[
F_R = F_Q\cos(\theta+\phi) = 44565\cos42.32^\circ = 33\times10^3\ \text{N}
\]
Reactions due to \( F_T \):
\[
H_{T1} = \frac{F_T(a+b)}{b} = \frac{30\times10^3(239+470)}{470} = 45\times10^3\ \text{N}
\]
\[
H_{T2} = \frac{F_T a}{b} = \frac{30\times10^3\times239}{470} = 15.3\times10^3\ \text{N}
\]
Reactions due to \( F_R \):
\[
H_{R1} = \frac{F_R(a+b)}{b} = \frac{33\times10^3(239+470)}{470} = 49.8\times10^3\ \text{N}
\]
\[
H_{R2} = \frac{F_R a}{b} = \frac{33\times10^3\times239}{470} = 16.8\times10^3\ \text{N}
\]

#### (a) Crank web

Bending due to tangential force:
\[
M_{bT} = F_T\left(r - \frac{d_1}{2}\right) = 30\times10^3\left(150 - \frac{180}{2}\right) = 1.8\times10^6\ \text{N·mm}
\]
\[
\sigma_{bT} = \frac{6M_{bT}}{t w^2} = \frac{6\times1.8\times10^6}{70\times245^2} = 2.6\ \text{N/mm}^2
\]
Bending due to radial force:
\[
M_{bR} = F_R(0.75l_c + 0.5t) = 33\times10^3(0.75\times92 + 35) = 3.43\times10^6\ \text{N·mm}
\]
\[
\sigma_{bR} = \frac{6M_{bR}}{w t^2} = \frac{6\times3.43\times10^6}{245\times70^2} = 17.1\ \text{N/mm}^2
\]
Direct compressive stress:
\[
\sigma_d = \frac{F_R}{w t} = \frac{33\times10^3}{245\times70} = 1.9\ \text{N/mm}^2
\]
Total compressive stress:
\[
\sigma_c = \sigma_{bT} + \sigma_{bR} + \sigma_d = 2.6 + 17.1 + 1.9 = 21.6\ \text{N/mm}^2
\]
Twisting moment due to \( F_T \):
\[
T = F_T(0.75l_c + 0.5t) = 30\times10^3(0.75\times92 + 35) = 3.12\times10^6\ \text{N·mm}
\]
Shear stress:
\[
\tau = \frac{4.5T}{w t^2} = \frac{4.5\times3.12\times10^6}{245\times70^2} = 11.7\ \text{N/mm}^2
\]
Max combined stress:
\[
\sigma_{\max} = \frac{\sigma_c}{2} + \frac{1}{2}\sqrt{\sigma_c^2 + 4\tau^2}
\]
\[
= \frac{21.6}{2} + \frac{1}{2}\sqrt{21.6^2 + 4\times11.7^2} = 10.8 + 15.9 = 26.7\ \text{N/mm}^2 < 60\quad\text{(Safe)}
\]

#### (b) Shaft at junction of crank

Let \( d_{s1} = d_1 = 180\ \text{mm} \) (assume).
Bending moment:
\[
M = F_Q(0.75l_c + t) = 44565(0.75\times92 + 70) = 6.2\times10^6\ \text{N·mm}
\]
Twisting moment:
\[
T = F_T r = 30\times10^3\times150 = 4.5\times10^6\ \text{N·mm}
\]
Equivalent twisting moment:
\[
T_e = \sqrt{M^2 + T^2} = \sqrt{(6.2\times10^6)^2 + (4.5\times10^6)^2} = 7.66\times10^6\ \text{N·mm}
\]
Check shear stress:
\[
\tau = \frac{16T_e}{\pi d_{s1}^3} = \frac{16\times7.66\times10^6}{\pi(180)^3} = 6.72\ \text{N/mm}^2 < 30-40\quad\text{(Safe)}
\]

#### (c) Shaft under flywheel

Horizontal bending moment:
\[
M_H = F_p'(a+b_2) - \sqrt{H_{R1}^2 + H_{T1}^2}\times b_2
\]
\[
= 44200(239+235) - \sqrt{(49.8\times10^3)^2 + (45\times10^3)^2}\times235
\]
\[
= 20.95\times10^6 - 15.77\times10^6 = 5.18\times10^6\ \text{N·mm}
\]
Vertical bending moment:
\[
M_V = \frac{W b_1 b_2}{b} = \frac{30\times10^3\times235\times235}{470} = 3.53\times10^6\ \text{N·mm}
\]
Resultant bending moment:
\[
M_R = \sqrt{M_H^2 + M_V^2} = \sqrt{(5.18\times10^6)^2 + (3.53\times10^6)^2} = 6.27\times10^6\ \text{N·mm}
\]
Twisting moment \( T = 4.5\times10^6\ \text{N·mm} \).
Equivalent twisting moment:
\[
T_e = \sqrt{M_R^2 + T^2} = \sqrt{(6.27\times10^6)^2 + (4.5\times10^6)^2} = 7.72\times10^6\ \text{N·mm}
\]
Using \( \tau = 30\ \text{N/mm}^2 \):
\[
T_e = \frac{\pi}{16}d_s^3\tau = 5.9d_s^3
\]
\[
d_s^3 = \frac{7.72\times10^6}{5.9} = 1.31\times10^6,\quad d_s = 109\ \text{mm}
\]
But \( d_s > d_1 = 162\ \text{mm} \), so take \( d_s = 200\ \text{mm} \).

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