Mark: %
MAIN ASSIGNMENT
DATE: 20 JULY 2020
SUBJECT: MECHANICAL ENGINEERING DESIGN I CODE: MED21A
MAXIMUM MARKS: 50
INSTRUCTIONAL PROGRAMME: DIP.ENG.TECH: MECHANICAL ENGINEERING
N.DIP ENGINEERING: MECHANICAL
ASSESSOR: DR. L. MUGWAGWA
Question: 1 2 3 Total
Allocated marks:
/25 /15 /10 /50
STATIONARY:
Programmable alphanumerical calculators are NOT allowed
Textbook: Mechanical Engineering Design – Peter RN Childs
INSTRUCTIONS:
Answer all questions in this question paper
Show all calculations with the necessary steps
Work written in pencil will not be marked.
No half marks will be allocated
Round off to 3 decimal places
Student to complete this block:
STUDENT NUMBER:
SURNAME AND INITIALS:
(Write in Block Letters)
SIGNATURE OF STUDENT:
(You have read the instructions on this page)
THIS PAPER CONSISTS OF 10 PAGES INCLUDING THIS COVER PAGE.
, Page |2
QUESTION 1: BEARING DESIGN [25]
A full journal bearing has a nominal diameter of 50 mm and bearing length 25 mm. The bearing
supports a load of 1450 N and the journal design speed is 2500 rpm. The radial clearance is 0.025mm.
An SAE 10 oil has been chosen and the lubricant supply temperature is 50°C. Using the standard
Raimondi and Boyd design charts, find the temperature rise of the lubricant, taking ΔT = 20°C as the
initial assumption. (NB: Carry out the necessary number of iterations) [25]
Given data:
D=50.0 mm, L=25 mm, W=1450 N, N=2500 rpm, c=0.025 mm, SAE 10, and T1=50 °C
Assuming lubricant temperature rise ∆𝑇 across the bearing to be ∆𝑇 = 20 °C,
∆𝑇 20
𝑇𝑎𝑣 = 𝑇1 + = 50 + = 60 °𝐶 (1 𝑚𝑎𝑟𝑘)
2 2
From Figure 4.7 for SAE 10 at 60 oC, µ=0.014 Pa s. (1 mark)
2500
𝑁𝑠 = = 41.67 𝑟𝑝𝑠 (1 𝑚𝑎𝑟𝑘)
60
𝐿 25
= = 0.5 (1 𝑚𝑎𝑟𝑘)
𝐷 50
𝑊 1450
𝑃= = = 1.16 × 106 𝑁/𝑚2 (1 𝑚𝑎𝑟𝑘)
𝐿𝐷 0.025 × 0.05
The next step is to determine the Somerfield number (bearing characteristic number)
𝑟 2 µ𝑁𝑠
𝑆=( )
𝑐 𝑃
2
25 × 10−3 0.014 × 41.67
=( ) = 0.5029 (1 𝑚𝑎𝑟𝑘)
0.025 × 10−3 1.16 × 106
From Figure 4.12 with S = 0.5029 and L/D = 0.5; (𝑟/𝑐)𝑓 = 12 (1 mark)
From Figure 4.13, with S = 0.5029 and L/D = 0.5; 𝑄/𝑟𝑐𝑁𝑠 𝐿 = 4.6 (1 mark)
From Figure 4.14, with S = 0.5029 and L/D = 0.5; 𝑄𝑠 /𝑄 = 0.65 (1 mark)
MAIN ASSIGNMENT
DATE: 20 JULY 2020
SUBJECT: MECHANICAL ENGINEERING DESIGN I CODE: MED21A
MAXIMUM MARKS: 50
INSTRUCTIONAL PROGRAMME: DIP.ENG.TECH: MECHANICAL ENGINEERING
N.DIP ENGINEERING: MECHANICAL
ASSESSOR: DR. L. MUGWAGWA
Question: 1 2 3 Total
Allocated marks:
/25 /15 /10 /50
STATIONARY:
Programmable alphanumerical calculators are NOT allowed
Textbook: Mechanical Engineering Design – Peter RN Childs
INSTRUCTIONS:
Answer all questions in this question paper
Show all calculations with the necessary steps
Work written in pencil will not be marked.
No half marks will be allocated
Round off to 3 decimal places
Student to complete this block:
STUDENT NUMBER:
SURNAME AND INITIALS:
(Write in Block Letters)
SIGNATURE OF STUDENT:
(You have read the instructions on this page)
THIS PAPER CONSISTS OF 10 PAGES INCLUDING THIS COVER PAGE.
, Page |2
QUESTION 1: BEARING DESIGN [25]
A full journal bearing has a nominal diameter of 50 mm and bearing length 25 mm. The bearing
supports a load of 1450 N and the journal design speed is 2500 rpm. The radial clearance is 0.025mm.
An SAE 10 oil has been chosen and the lubricant supply temperature is 50°C. Using the standard
Raimondi and Boyd design charts, find the temperature rise of the lubricant, taking ΔT = 20°C as the
initial assumption. (NB: Carry out the necessary number of iterations) [25]
Given data:
D=50.0 mm, L=25 mm, W=1450 N, N=2500 rpm, c=0.025 mm, SAE 10, and T1=50 °C
Assuming lubricant temperature rise ∆𝑇 across the bearing to be ∆𝑇 = 20 °C,
∆𝑇 20
𝑇𝑎𝑣 = 𝑇1 + = 50 + = 60 °𝐶 (1 𝑚𝑎𝑟𝑘)
2 2
From Figure 4.7 for SAE 10 at 60 oC, µ=0.014 Pa s. (1 mark)
2500
𝑁𝑠 = = 41.67 𝑟𝑝𝑠 (1 𝑚𝑎𝑟𝑘)
60
𝐿 25
= = 0.5 (1 𝑚𝑎𝑟𝑘)
𝐷 50
𝑊 1450
𝑃= = = 1.16 × 106 𝑁/𝑚2 (1 𝑚𝑎𝑟𝑘)
𝐿𝐷 0.025 × 0.05
The next step is to determine the Somerfield number (bearing characteristic number)
𝑟 2 µ𝑁𝑠
𝑆=( )
𝑐 𝑃
2
25 × 10−3 0.014 × 41.67
=( ) = 0.5029 (1 𝑚𝑎𝑟𝑘)
0.025 × 10−3 1.16 × 106
From Figure 4.12 with S = 0.5029 and L/D = 0.5; (𝑟/𝑐)𝑓 = 12 (1 mark)
From Figure 4.13, with S = 0.5029 and L/D = 0.5; 𝑄/𝑟𝑐𝑁𝑠 𝐿 = 4.6 (1 mark)
From Figure 4.14, with S = 0.5029 and L/D = 0.5; 𝑄𝑠 /𝑄 = 0.65 (1 mark)
