Question 21.8: A full journal bearing has a diameter 75 mm with unilateral ......

Before solving the problem, we will discuss how to obtain the minimum and maximum clearances if such unilateral tolerances are given.

(i) Let the diameter of the bearing be specified as D_b{ }_{-b}^{+0}, where b is the lower tolerance. The upper tolerance is zero in this case. Similarly, let the bush or the journal be specified as D_{j_{-0.0}}^{+i} where j is the upper tolerance of the journal. The smallest radial clearance is obtained as

C_{r \min }=\frac{\left(D_j-0\right)-\left(D_b-0\right)}{2}=\frac{D_i-D_b}{2}

The maximum radial clearance is obtained as

C_{r_{\max }}=\frac{\left(D_j+j\right)-\left(D_b-b\right)}{2}=\frac{D_j-D_b}{2}+\frac{j+b}{2}

Now according to the problem, the dimensions are shown in Figure 21.25.

C_{r \min }=\frac{D_j-D_b}{2}=\frac{75.05-75}{2}=0.025 \mathrm{~mm}

C_{r \max }=\frac{D_j-D_b}{2}+\frac{j+b}{2}=\frac{75.05-75}{2}+\frac{0.05+0.1}{2}=0.1 \mathrm{~mm}

The same can also be obtained as follows

Minimum diameter of bearing, D_{b \min }=D_b-b=75.00-0.1=74.9 \mathrm{~mm}

Maximum diameter of the journal, D_{j \max }=D_j+j=75.05+0.05=75.10 \mathrm{~mm}

\text { Maximum radial clearance }=\frac{D_{j \max }-D_{b \min }}{2}=\frac{75.10-74.90}{2}=0.1 \mathrm{~mm}

C_r=0.1 \mathrm{~mm}

Similarly the minimum clearance is obtained as

maximum diameter of bearing, D_{b \max }=75.00+0=75.00 \mathrm{~mm}

minimum diameter of the journal, D_{j \min }=75.05-0.00=75.05 \mathrm{~mm}

\text { minimum radial clearance }=\frac{D_{j \min }-D_{b \max }}{2}=\frac{75.05-75.00}{2}=0.025 \mathrm{~mm}

Clearance range = 0.1 – 0.025 = 0.075 mm

(ii) Minimum oil film thickness

Bearing pressure based on average bearing area

p=\frac{W}{L D}=\frac{3500}{75 \times 75}=0.622 \mathrm{~N} / \mathrm{mm}^2

Sommerfeld number is (based on minimum clearance)

S=\left(\frac{\mu N^*}{p}\right)\left(\frac{r}{C_r}\right)^2=\left(\frac{0.011 \times 30}{0.622 \times 10^6}\right)\left(\frac{37.5}{0.025}\right)^2=1.194

From Table 21.7, \frac{h_0}{C_r} \text { corresponding to } S=1.194 is

\frac{h_0}{C_r}=0.9-\frac{0.9-0.8}{1.33-0.631}(1.33-1.194)=0.881

h_0=0.881 \times 0.025=0.022 \mathrm{~mm}

Sommerfeld number is (based on the maximum clearance)

S=\left(\frac{\mu N^*}{p}\right)\left(\frac{r}{C_r}\right)^2=\left(\frac{0.011 \times 30}{0.622 \times 10^6}\right)\left(\frac{37.5}{0.1}\right)^2=0.0746

From Table 21.7, corresponding to S = 0.0746

\frac{h_0}{C_r}=0.4-\frac{0.4-0.2}{0.121-0.0446}(0.121-0.0746)=0.279

h_0=0.279 \times 0.1=0.0278 \mathrm{~mm}

Hence minimum oil film thickness is 0.022 mm based on the minimum clearance.

(iii) Power lost

The coefficient of friction can be obtained from the friction variable. The friction variable corresponding to S = 1.194 is

\left(\frac{r}{C_r}\right) f=26.4-\frac{26.4-12.8}{1.33-0.631}(1.33-1.194)=23.754

f=23.754 \times \frac{C_r}{r}=23.754 \times \frac{0.025}{37.5}=0.0158

V=\frac{\pi D N}{60 \times 1000}=\frac{\pi \times 75 \times 1800}{60 \times 1000}=7.068 \mathrm{~m} / \mathrm{s}

Power lost =f W V=0.0158 \times 3500 \times 7.068=391 \mathrm{~W}

(iv) Side flow

\frac{Q_s}{Q} corresponding to S = 1.194 is obtained as follows.

\frac{Q_s}{Q}=0.15+\frac{0.28-0.15}{1.33-0.631}(1.33-1.194)=0.175

Side flow, Q_s=0.175 Q

Q, the quantity of oil, is obtained from the flow variable.

The flow variable corresponding to S = 1.194 is

\frac{Q}{r C_r N^* L}=3.37+\frac{3.59-3.37}{1.33-0.631}(1.33-1.194)=3.413

Quantity of oil

Q=3.413 \times r C_r N^* L=3.413 \times 37.5 \times 0.025 \times 30 \times 75=7198.9 \mathrm{~mm}^3 / \mathrm{s}

Side flow

Q_s=0.175 \times 7198.9=1259.9 \mathrm{~mm}^3 / \mathrm{s}

(v) Maximum film pressure

p / p_{\max }=0.54-\frac{0.54-0.539}{1.33-0.631}(1.33-1.194)=0.5399

p_{\max }=\frac{p}{0.5399}=\frac{0.622}{0.5399}=1.152 \mathrm{~N} / \mathrm{mm}^2