*2.2.1 Calculation*

Material of shaft = Mild steel, it has tough material and it was absorbing vibration that was caused by the motor.

**Figure 2.** *Basic design for producing foam [10].*

Length of the shaft = 315 mm. Diameter of the shaft from top 225 mm = 20 mm. Diameter of the shaft at lower 80 mm length = 19.98 mm. Dimension of thread = 10 � 1.5 mm.

Considering the load at the top, i.e. load of the pulley mounted on the shaft.

$$\begin{aligned} \text{Slenderness ratio of the shaft} &= \frac{\text{L}}{\text{K}}\\ &= \frac{315}{10} \\ &= 31.5 \end{aligned} \tag{1}$$

Thus, the shaft is a medium type (**Figure 3**).

### *2.2.2 Rankine formula for calculating the crippling load*

$$\mathbf{W\_{cr}} = \frac{\sigma\_{\mathbf{c}} \times \mathbf{A}}{\mathbf{1} + \frac{\sigma\_{\mathbf{c}} \times \mathbf{A} \times \mathbf{L}^2}{\pi^2 \times \mathbf{E} \times \mathbf{l}}} \tag{2}$$

where Wcr = crippling load, σ<sup>c</sup> = crushing stress or yield stress, A = cross-sectional area, L = length of the shaft, E = modulus of elasticity, I = moment of inertia.

$$\mathbf{W\_{cr}} = \frac{\sigma\_{\mathbf{c}} \times \mathbf{A}}{\mathbf{1} + \mathbf{a} \left(\frac{\mathbf{L}}{\mathbf{K}}\right)^2} \tag{3}$$

*2.2.3 Considering the weight of pulley*

are required.

**91**

**Figure 4.**

**Figure 3.** *Shaft.*

**Figure 5.**

*Shaft in horizontal position.*

*Shaft in vertical position.*

**Figure 4** shows the vertical position of the shaft, and **Figure 5** shows the horizontal position of the shaft. In **Figure 6** power is transmitted from the driver pulley to driven pulley. For designing a shaft and pulley, the following calculations

*Design and Development of Manufacturing System Design for Producing Metallic Foam*

*DOI: http://dx.doi.org/10.5772/intechopen.92879*

$$\mathbf{a} = \text{Rankine's constant} = \frac{\sigma\_{\mathbf{c}}}{\pi^2 \mathbf{E}} \tag{4}$$

$$\mathbf{a} = \frac{1}{7500}$$

For one end fixed and another end free.

$$\begin{aligned} \text{L} &= 2 \text{L} \\\\ &= 2 \times 315 \\\\ &= 630 \text{ mm} \end{aligned}$$

$$\text{K} = 10 \text{ mm}$$

$$\frac{\text{L}}{\text{K}} = \frac{630}{10} = 63$$

$$\text{W}\_{\text{cr}} = \frac{320 \times \frac{\pi}{4} (20)^2}{1 + \frac{1}{7500} (63)^2}$$

$$\sigma\_{\text{c}} = 320.$$

$$\text{W}\_{\text{cr}} = 65.707 \text{ N}$$

*Design and Development of Manufacturing System Design for Producing Metallic Foam DOI: http://dx.doi.org/10.5772/intechopen.92879*

### *2.2.3 Considering the weight of pulley*

**Figure 4** shows the vertical position of the shaft, and **Figure 5** shows the horizontal position of the shaft. In **Figure 6** power is transmitted from the driver pulley to driven pulley. For designing a shaft and pulley, the following calculations are required.

Length of the shaft = 315 mm.

*Foams - Emerging Technologies*

Dimension of thread = 10 � 1.5 mm.

Thus, the shaft is a medium type (**Figure 3**).

*2.2.2 Rankine formula for calculating the crippling load*

For one end fixed and another end free.

**90**

Diameter of the shaft from top 225 mm = 20 mm.

Diameter of the shaft at lower 80 mm length = 19.98 mm.

Considering the load at the top, i.e. load of the pulley mounted on the shaft.

Slenderness ratio of the shaft <sup>¼</sup> <sup>L</sup>

Wcr <sup>¼</sup> <sup>σ</sup><sup>c</sup> � <sup>A</sup>

Wcr <sup>¼</sup> <sup>σ</sup><sup>c</sup> � <sup>A</sup> <sup>1</sup> <sup>þ</sup> <sup>a</sup> <sup>L</sup> K

<sup>a</sup> <sup>¼</sup> Rankine's constant <sup>¼</sup> <sup>σ</sup><sup>c</sup>

<sup>a</sup> <sup>¼</sup> <sup>1</sup> 7500

¼ 2 � 315

¼ 630 mm

<sup>10</sup> <sup>¼</sup> <sup>63</sup>

1 <sup>7500</sup> ð Þ <sup>63</sup> <sup>2</sup>

320 � π <sup>4</sup> ð Þ <sup>20</sup> <sup>2</sup>

1 þ

σ<sup>c</sup> ¼ 320*:*

Wcr ¼ 65*:*707 N

K ¼ 10 mm

L <sup>K</sup> <sup>¼</sup> <sup>630</sup>

Wcr ¼

L ¼ 2L

area, L = length of the shaft, E = modulus of elasticity, I = moment of inertia.

<sup>1</sup> <sup>þ</sup> <sup>σ</sup>c�A�L<sup>2</sup> π2�E�I

where Wcr = crippling load, σ<sup>c</sup> = crushing stress or yield stress, A = cross-sectional

K

<sup>¼</sup> <sup>315</sup> 10

(1)

(2)

¼ 31*:*5

<sup>2</sup> (3)

<sup>π</sup>2E (4)

$$\begin{aligned} \text{weight of the pulley} &= 400 \text{ g} \\ &= \frac{400 \times 9.81}{1000} \\ \text{W}\_{\text{p}} &= 3.9 \text{ N} \end{aligned}$$

The shaft bears more load; hence, the design is safe.

*2.2.3.1 Power of motor*

$$\mathbf{P} = \frac{2\pi \mathbf{N} \mathbf{T}\_0}{60} \tag{5}$$

T1 = 0.33 N mm. R = T1 + T2. R = 0.489 N mm.

Mt = T0 = 7.18 N mm.

Load is steady hence.

*2.2.4 Equivalent bending moment*

Me <sup>¼</sup> <sup>1</sup> 2

Determine the equivalent bending moment:

made from the catalog of a manufacturer (**Figure 7**).

*2.3.1 Striebeck formula for the strength of a single ball in compression*

Me <sup>¼</sup> <sup>1</sup> 2

Km = 1.5. Kt = 1.0.

design was safe.

K = 7200.

**93**

d = diameter of ball. Fe = 7200 � <sup>7</sup><sup>2</sup>

Fe = 352,800 N/mm<sup>2</sup>

.

.

**2.3 Selection of bearing**

Mb = T � L = 0.489 � 315 = 154.03 N mm.

*DOI: http://dx.doi.org/10.5772/intechopen.92879*

Km = Combined shock and fatigue factor for bending. Kt = Combined shock and fatigue factor for torsion.

*Design and Development of Manufacturing System Design for Producing Metallic Foam*

Km � Mb þ

1*:*5 � 154*:*035 þ

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð Þ <sup>1</sup>*:*<sup>5</sup> � <sup>154</sup>*:*<sup>035</sup> <sup>2</sup> <sup>þ</sup> ð Þ <sup>7</sup>*:*<sup>18</sup> <sup>2</sup>

<sup>2</sup> <sup>þ</sup> ð Þ Kt � <sup>T</sup> <sup>2</sup>

(8)

ð Þ Km � Mb

� � q

� � q

<sup>32</sup> <sup>σ</sup><sup>b</sup> � <sup>d</sup><sup>3</sup>

*:*

*:*

Fe <sup>¼</sup> Kd2 (9)

Me ¼ 231*:*108 N mm*:*

<sup>σ</sup><sup>b</sup> <sup>¼</sup> <sup>231</sup>*:*<sup>108</sup> � <sup>32</sup> <sup>π</sup> � ð Þ <sup>20</sup> <sup>2</sup> *:*

<sup>σ</sup><sup>b</sup> <sup>¼</sup> <sup>0</sup>*:*294 N*=*mm<sup>2</sup>

This was very less than the allowable tensile or compressive stress. Hence the

In order to select a most suitable ball bearing, first, the basic dynamic load was calculated. It was multiplied by the service factor to get the basic dynamic load capacity. After finding the basic dynamic load capacity, the selection of bearing was

<sup>231</sup>*:*<sup>108</sup> <sup>¼</sup> <sup>π</sup>

T0 = Torque.

<sup>746</sup> <sup>¼</sup> <sup>2</sup>�π�1380�T0�*:*<sup>72</sup> <sup>60</sup> . T0 = 7.18 N mm. T = (T1 � T2) � 44.

$$\frac{\text{T1}}{\text{T2}} = \mathbf{e}^{\mu 0} \tag{6}$$

T1 = Tension in the tight side.

T2 = Tension in the slack side.

μ = Coefficient of friction.

The distance between the driven and driven pulley is 330 mm:

θ = 180-2α.

θ = Angle of contact.

$$\text{Sin}\alpha = \frac{\mathbf{r}\_2 - \mathbf{r}\_1}{\mathbf{330}} = \frac{60 - 44}{\mathbf{330}} = 0.484\tag{7}$$

r1 & r2 = radii of the driving and driven pulley. α = 2.778° . θ = 180–5.55. θ = 174.44° = 3.03 rad. μ = 0.25. T1 T2 <sup>¼</sup> e0*:*25�3*:*<sup>02</sup> T1 = 2.12 T2. (2.12–1)T2 � 44 = 7.18. T2 = 0.159 N mm.

*Design and Development of Manufacturing System Design for Producing Metallic Foam DOI: http://dx.doi.org/10.5772/intechopen.92879*

T1 = 0.33 N mm. R = T1 + T2. R = 0.489 N mm. Mb = T � L = 0.489 � 315 = 154.03 N mm. Mt = T0 = 7.18 N mm.

Load is steady hence. Km = 1.5. Kt = 1.0. Km = Combined shock and fatigue factor for bending. Kt = Combined shock and fatigue factor for torsion.

*2.2.4 Equivalent bending moment*

weight of the pulley ¼ 400 g

<sup>P</sup> <sup>¼</sup> <sup>2</sup>πNT0

T1

<sup>330</sup> <sup>¼</sup> <sup>60</sup> � <sup>44</sup>

The distance between the driven and driven pulley is 330 mm:

Sin<sup>α</sup> <sup>¼</sup> r2 � r1

r1 & r2 = radii of the driving and driven pulley.

The shaft bears more load; hence, the design is safe.

*2.2.3.1 Power of motor*

*Power transmitted by rope.*

*Foams - Emerging Technologies*

**Figure 6.**

T0 = Torque.

θ = 180-2α.

α = 2.778°

μ = 0.25. T1

**92**

T2 <sup>¼</sup> e0*:*25�3*:*<sup>02</sup> T1 = 2.12 T2.

θ = Angle of contact.

. θ = 180–5.55.

θ = 174.44° = 3.03 rad.

(2.12–1)T2 � 44 = 7.18. T2 = 0.159 N mm.

<sup>746</sup> <sup>¼</sup> <sup>2</sup>�π�1380�T0�*:*<sup>72</sup>

T0 = 7.18 N mm. T = (T1 � T2) � 44.

<sup>60</sup> .

T1 = Tension in the tight side. T2 = Tension in the slack side. μ = Coefficient of friction.

<sup>¼</sup> <sup>400</sup> � <sup>9</sup>*:*<sup>81</sup> 1000

<sup>60</sup> (5)

T2 <sup>¼</sup> <sup>e</sup>μθ (6)

<sup>330</sup> <sup>¼</sup> <sup>0</sup>*:*<sup>484</sup> (7)

Wp ¼ 3*:*9 N

$$
\mathbf{M}\_{\mathbf{e}} = \frac{1}{2} \left[ \mathbf{K}\_{\mathbf{m}} \times \mathbf{M}\_{\mathbf{b}} + \sqrt{\left( \mathbf{K}\_{\mathbf{m}} \times \mathbf{M}\_{\mathbf{b}} \right)^{2} + \left( \mathbf{K}\_{\mathbf{t}} \times \mathbf{T} \right)^{2}} \right] \tag{8}
$$

$$
\mathbf{M}\_{\mathbf{e}} = \frac{1}{2} \left[ \mathbf{1.5} \times \mathbf{154.035} + \sqrt{\left( \mathbf{1.5} \times \mathbf{154.035} \right)^{2} + \left( \mathbf{7.18} \right)^{2}} \right]
$$

$$
\mathbf{M}\_{\mathbf{e}} = \mathbf{231.108 N mm} \text{.}
$$

Determine the equivalent bending moment:

$$\begin{aligned} \textbf{231.108} &= \frac{\pi}{32} \sigma\_{\textbf{b}} \times \textbf{d}^3. \\\\ \sigma\_{\textbf{b}} &= \frac{231.108 \times 32}{\pi \times (20)^2}. \\\\ \sigma\_{\textbf{b}} &= 0.294 \text{ N/mm}^2. \end{aligned}$$

This was very less than the allowable tensile or compressive stress. Hence the design was safe.
