STEM Problem Detail

Problem #69

Physics Hard

Problem Statement

Prompt

Provide me the solution to Example : 32.4

Solution Rendering...

## Example 32.4 — Plain Carbon Steel Centre Crankshaft Design

### Given:

\( D = 400\ \text{mm},\ L = 600\ \text{mm},\ N = 200\ \text{rpm},\ p_m = 0.5\ \text{N/mm}^2 \)
\( p = 2.5\ \text{N/mm}^2,\ W = 50\ \text{kN},\ T_1+T_2 = 6.5\ \text{kN} \)
\( \theta = 35^\circ,\ p' = 1\ \text{N/mm}^2,\ l/r = 5 \)

### 1. Crank at Dead Centre

#### (a) Piston gas load

\[
F_p = \frac{\pi}{4}D^2 p = \frac{\pi}{4}(400)^2(2.5) = 314200\ \text{N} = 314.2\ \text{kN}
\]
Assume \( b = 2D = 800\ \text{mm} \), \( b_1 = b_2 = b/2 = 400\ \text{mm} \).
Reactions at bearings 1 & 2:
\[
H_1 = H_2 = \frac{F_p \times b_1}{b} = \frac{314.2\times400}{800} = 157.1\ \text{kN}
\]
Assume \( c_1 = c_2 = c/2 \).
Vertical reactions due to flywheel weight \( W \):
\[
V_2 = V_3 = \frac{W}{2} = 25\ \text{kN}
\]
Horizontal reactions due to belt pull:
\[
H_2' = H_3' = \frac{T_1+T_2}{2} = 3.25\ \text{kN}
\]

#### (b) Design of crankpin

Bending moment at crankpin centre:
\[
M_c = H_1\cdot b_2 = 157.1\times400 = 62840\ \text{kN·mm}
\]
Using \( \sigma_b = 75\ \text{N/mm}^2 \):
\[
M_c = \frac{\pi}{32}d_c^3\sigma_b = 7.364\times10^{-3}d_c^3\ \text{kN·mm}
\]
\[
d_c^3 = \frac{62840}{7.364\times10^{-3}} = 8.53\times10^6,\quad d_c = 204.35\ \text{mm}\ (\text{take } 205\ \text{mm})
\]
Length of crankpin (\( p_b = 10\ \text{N/mm}^2 \)):
\[
l_c = \frac{F_p}{d_c\cdot p_b} = \frac{314.2\times10^3}{205\times10} = 153.3\ \text{mm}\ (\text{take } 155\ \text{mm})
\]

#### (c) Left-hand crank web

Thickness:
\[
t = 0.65d_c + 6.35 = 0.65\times205 + 6.35 = 139.6\ \text{mm}\ (\text{take } 140\ \text{mm})
\]
Width:
\[
w = 1.125d_c + 12.7 = 1.125\times205 + 12.7 = 243.3\ \text{mm}\ (\text{take } 245\ \text{mm})
\]
Bending moment on web:
\[
M = H_1\left(b_2 - \frac{l_c}{2} - t\right) = 157.1\left(400 - 77.5 - 140\right) = 39668\ \text{kN·mm}
\]
Section modulus:
\[
Z = \frac{1}{6}w t^2 = \frac{1}{6}\times245\times140^2 = 800\times10^3\ \text{mm}^3
\]
Bending stress:
\[
\sigma_b = \frac{M}{Z} = \frac{39668}{800\times10^3} = 49.6\ \text{N/mm}^2
\]
Direct compressive stress:
\[
\sigma_c = \frac{H_1}{w t} = \frac{157.1\times10^3}{245\times140} = 4.58\ \text{N/mm}^2
\]
Total stress:
\[
\sigma_{\text{total}} = \sigma_b + \sigma_c = 54.18\ \text{N/mm}^2 < 75\ \text{N/mm}^2\quad\text{(Safe)}
\]

#### (d) Shaft under flywheel

Assume main bearing lengths equal:
\[
l_1 = l_2 = l_3 = 2\left(400 - \frac{155}{2} - 140\right) = 365\ \text{mm}
\]
With flywheel width 300 mm, \( c = 365 + 300 = 665\ \text{mm} \), take \( c = 800\ \text{mm} \), so \( c_1 = c_2 = 400\ \text{mm} \).
Bending moments:
\[
M_W = V_3c_1 = 25\times400 = 10000\ \text{kN·mm} = 10\times10^6\ \text{N·mm}
\]
\[
M_T = H_3'c_1 = 3.25\times400 = 1300\ \text{kN·mm} = 1.3\times10^6\ \text{N·mm}
\]
Resultant:
\[
M_S = \sqrt{M_W^2 + M_T^2} = \sqrt{(10\times10^6)^2 + (1.3\times10^6)^2} = 10.08\times10^6\ \text{N·mm}
\]
Using \( \sigma_b = 42\ \text{N/mm}^2 \):
\[
M_S = \frac{\pi}{32}d_s^3\sigma_b = 4.12d_s^3
\]
\[
d_s^3 = \frac{10.08\times10^6}{4.12} = 2.45\times10^6,\quad d_s = 134.7\ \text{mm}\ (\text{take } 135\ \text{mm})
\]

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

#### (a) Piston load at \(\theta = 35^\circ\)

\[
F_p' = \frac{\pi}{4}D^2 p' = \frac{\pi}{4}(400)^2(1) = 125680\ \text{N} = 125.68\ \text{kN}
\]
\[
\sin\phi = \frac{\sin\theta}{l/r} = \frac{\sin35^\circ}{5} = 0.1147,\quad \phi = 6.58^\circ
\]
Thrust in connecting rod:
\[
F_Q = \frac{F_p'}{\cos\phi} = \frac{125.68}{\cos6.58^\circ} = 126.5\ \text{kN}
\]
Tangential force:
\[
F_T = F_Q\sin(\theta+\phi) = 126.5\sin41.58^\circ = 84\ \text{kN}
\]
Radial force:
\[
F_R = F_Q\cos(\theta+\phi) = 126.5\cos41.58^\circ = 94.6\ \text{kN}
\]
Reactions due to \(F_T\):
\[
H_{T1} = H_{T2} = \frac{F_T\times b_1}{b} = \frac{84\times400}{800} = 42\ \text{kN}
\]
Reactions due to \(F_R\):
\[
H_{R1} = H_{R2} = \frac{F_R\times b_1}{b} = \frac{94.6\times400}{800} = 47.3\ \text{kN}
\]

#### (b) Crankpin at max twist

Bending moment:
\[
M_C = H_{R1}\times b_2 = 47.3\times400 = 18920\ \text{kN·mm}
\]
Twisting moment:
\[
T_C = H_{T1}\times r = 42\times300 = 12600\ \text{kN·mm}
\]
Equivalent twisting moment:
\[
T_e = \sqrt{M_C^2 + T_C^2} = \sqrt{(18920)^2+(12600)^2} = 22740\ \text{kN·mm} = 22.74\times10^6\ \text{N·mm}
\]
Using \( \tau = 35\ \text{N/mm}^2 \):
\[
T_e = \frac{\pi}{16}d_c^3\tau = 6.873d_c^3
\]
\[
d_c^3 = \frac{22.74\times10^6}{6.873} = 3.3\times10^6,\quad d_c = 149\ \text{mm}
\]
But earlier \(d_c = 205\ \text{mm}\) governs.

#### (c) Shaft under flywheel at max twist

Twisting moment on shaft:
\[
T_S = F_T\times r = 84\times300 = 25200\ \text{kN·mm} = 25.2\times10^6\ \text{N·mm}
\]
Bending moment \(M_S = 10.08\times10^6\ \text{N·mm}\) (same as before).
Equivalent twisting:
\[
T_e = \sqrt{M_S^2 + T_S^2} = \sqrt{(10.08\times10^6)^2+(25.2\times10^6)^2} = 27.14\times10^6\ \text{N·mm}
\]
Check with \(d_s = 135\ \text{mm}\):
\[
T_e = \frac{\pi}{16}(135)^3\tau = 483156\tau
\]
\[
\tau = \frac{27.14\times10^6}{483156} = 56.17\ \text{N/mm}^2 > 35\ \text{N/mm}^2\quad\text{(Not safe)}
\]
Recalculate with \(\tau = 35\ \text{N/mm}^2\):
\[
27.14\times10^6 = 6.873d_s^3
\]
\[
d_s^3 = \frac{27.14\times10^6}{6.873} = 3.95\times10^6,\quad d_s = 158\ \text{mm}\ (\text{take } 160\ \text{mm})
\]

#### (d) Shaft at juncture of right-hand crank arm

Resultant force at bearing 1:
\[
R_1 = \sqrt{H_{T1}^2 + H_{R1}^2} = \sqrt{42^2 + 47.3^2} = 63.3\ \text{kN}
\]
Bending moment:
\[
M_{S1} = R_1\left(b_2 + \frac{l_c}{2} + \frac{t}{2}\right) - F_Q\left(\frac{l_c}{2}+\frac{t}{2}\right)
\]
\[
= 63.3\left(400+77.5+70\right) - 126.5\left(77.5+70\right)
\]
\[
= 34700 - 18700 = 16000\ \text{kN·mm} = 16\times10^6\ \text{N·mm}
\]
Twisting moment \(T_{S1} = T_S = 25.2\times10^6\ \text{N·mm}\).
Equivalent:
\[
T_e = \sqrt{(16\times10^6)^2 + (25.2\times10^6)^2} = 29.85\times10^6\ \text{N·mm}
\]
Using \(\tau = 42\ \text{N/mm}^2\):
\[
T_e = \frac{\pi}{16}d_{s1}^3\tau = 8.25d_{s1}^3
\]
\[
d_{s1}^3 = \frac{29.85\times10^6}{8.25} = 3.62\times10^6,\quad d_{s1} = 153.5\ \text{mm}\ (\text{take } 155\ \text{mm})
\]

#### (e) Right-hand crank web

Bending due to radial component:
\[
M_R = H_{R2}\left(b_1 - \frac{l_c}{2} - \frac{t}{2}\right) = 47.3\left(400 - 77.5 - 70\right) = 11.94\times10^6\ \text{N·mm}
\]
\[
M_R = \sigma_{bR}\times Z = \sigma_{bR}\times\frac{1}{6}w t^2 = 800\times10^3\sigma_{bR}
\]
\[
\sigma_{bR} = \frac{11.94\times10^6}{800\times10^3} = 14.9\ \text{N/mm}^2
\]
Bending due to tangential component:
\[
M_T = F_T\left(r - \frac{d_{s1}}{2}\right) = 84\left(300 - 77.5\right) = 18.69\times10^6\ \text{N·mm}
\]
\[
M_T = \sigma_{bT}\times\frac{1}{6}t w^2 = 1.4\times10^6\sigma_{bT}
\]
\[
\sigma_{bT} = \frac{18.69\times10^6}{1.4\times10^6} = 13.35\ \text{N/mm}^2
\]
Direct compressive stress:
\[
\sigma_d = \frac{F_R}{2w t} = \frac{94.6\times10^3}{2\times245\times140} = 1.38\ \text{N/mm}^2
\]
Total compressive stress:
\[
\sigma_c = \sigma_{bR} + \sigma_{bT} + \sigma_d = 14.9 + 13.35 + 1.38 = 29.63\ \text{N/mm}^2
\]
Twisting moment on arm:
\[
T = H_{T2}\left(b_1 - \frac{l_c}{2}\right) = 42\left(400 - 77.5\right) = 13.545\times10^6\ \text{N·mm}
\]
Shear stress:
\[
\tau = \frac{4.5T}{w t^2} = \frac{4.5\times13.545\times10^6}{245\times140^2} = 12.7\ \text{N/mm}^2
\]
Max combined stress (Rankine’s formula):
\[
(\sigma_c)_{\max} = \frac{\sigma_c}{2} + \frac{1}{2}\sqrt{\sigma_c^2 + 4\tau^2}
\]
\[
= \frac{29.63}{2} + \frac{1}{2}\sqrt{29.63^2 + 4\times12.7^2} = 14.815 + 19.5 = 34.315\ \text{N/mm}^2\quad\text{(Safe)}
\]

#### (f) Bearing pressure check (bearing 2)

Total reaction at bearing 2:
\[
R_2 = \frac{F_p}{2} + \frac{W}{2} + \frac{T_1+T_2}{2} = \frac{314.2}{2} + 25 + 3.25 = 185.35\ \text{kN} = 185350\ \text{N}
\]
Bearing pressure:
\[
\frac{R_2}{l_2\cdot d_{s1}} = \frac{185350}{365\times155} = 3.276\ \text{N/mm}^2\quad\text{(Safe, < 5–8 N/mm}^2\text{)}
\]

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