VAAL UNIVERSITY OF
TECHNOLOGY
Cover sheet for Test3
Faculty: Engineering
Department: Mechanical
Diploma: National Diploma: Engineering: Mechanical
Subject: Mechanical Engineering Design II
Date: 17 May 2021
Internal code: EMMED2A
Examiner: MD Mafoko
Moderator: AO Aniki
Hours: 120 min
Total Marks: 60 Full marks: 70
* Requirements: OPEN BOOK ASSESSMENT
* Instructions: Answer all the questions.
Signature of Examiner:…MD Mafoko……… Date: 8 June 2021
Signature of Moderator:…Abimbola Aniki….. Date:………………
, Q1.1 What is a spindle?
A spindle is a short non-rotating shaft. 1.
Q1.2 Circular shafts are for transmitting ……………….. and …………….. .
Power and torque 2.
Q1.3 How is the equivalent torque used in the general to the twisting moment and the
bending moment?
The square of the equivalent torque is directly proportional to the sum of the
squares of the twisting moment and bending moment. 2
Q1.4 In a belt drive mounted on a circular shaft, the effective force due to the
tensions in the belt causes bending on the shaft? True or false
True 1
Q1.5 If you designed a shaft with material BS970:709M40 and a factor of safety of 2
to be operated at room temperature, what would happen to the shaft, if it is
suddenly exposed to temperature of 565O C?
The allowable (working) tensile or compressive stress at temperature below
200O C would be half of the yield strength at 200O C (340 MPa). At a higher
temperature like 565O C, the yield is around 310 MPa. This means maximum
loading could cause an induced stress approaching 310 MPa and imminent
failure.
𝑆𝑦𝑡
𝛔a = 𝑓𝑜𝑠
680𝑀𝑃𝑎
𝛔a = = 340 𝑀𝑃
2
𝛔i = 340 𝑀𝑃 > Syt =310 MPa. 4.
Q1.6 Determine the diameters of a 500 mm long hollow shaft which is required
to transmit 120 kW at 1490 rpm. Make the shaft from a material with a
yield strength of 435 MPa. On the shaft is mounted a belt drive with
T1= 2,18T2. The outside diameter of the shaft is thrice the inner one
whereas the diameter of the pulley is 250 mm. Take the fos as 4.
Solution
The general torsion equation is
𝑇 𝜏 𝐺𝛳
= =
𝐽 𝑅 𝐿
Twisting moment
30𝑃 30 𝑥 120 𝑘𝑊
T= = 𝜋 𝑥 1490 𝑟𝑝𝑚
𝜋𝑁
T =769,07 Nm
Tensions in the belt
T = (T1 – T2) x R
T = (2,18 T2 -T2) x R = 1,18 T2 x R
T2 =769,07 Nm/(1,18 x 0,125)
T2 = 5214,034 N
TECHNOLOGY
Cover sheet for Test3
Faculty: Engineering
Department: Mechanical
Diploma: National Diploma: Engineering: Mechanical
Subject: Mechanical Engineering Design II
Date: 17 May 2021
Internal code: EMMED2A
Examiner: MD Mafoko
Moderator: AO Aniki
Hours: 120 min
Total Marks: 60 Full marks: 70
* Requirements: OPEN BOOK ASSESSMENT
* Instructions: Answer all the questions.
Signature of Examiner:…MD Mafoko……… Date: 8 June 2021
Signature of Moderator:…Abimbola Aniki….. Date:………………
, Q1.1 What is a spindle?
A spindle is a short non-rotating shaft. 1.
Q1.2 Circular shafts are for transmitting ……………….. and …………….. .
Power and torque 2.
Q1.3 How is the equivalent torque used in the general to the twisting moment and the
bending moment?
The square of the equivalent torque is directly proportional to the sum of the
squares of the twisting moment and bending moment. 2
Q1.4 In a belt drive mounted on a circular shaft, the effective force due to the
tensions in the belt causes bending on the shaft? True or false
True 1
Q1.5 If you designed a shaft with material BS970:709M40 and a factor of safety of 2
to be operated at room temperature, what would happen to the shaft, if it is
suddenly exposed to temperature of 565O C?
The allowable (working) tensile or compressive stress at temperature below
200O C would be half of the yield strength at 200O C (340 MPa). At a higher
temperature like 565O C, the yield is around 310 MPa. This means maximum
loading could cause an induced stress approaching 310 MPa and imminent
failure.
𝑆𝑦𝑡
𝛔a = 𝑓𝑜𝑠
680𝑀𝑃𝑎
𝛔a = = 340 𝑀𝑃
2
𝛔i = 340 𝑀𝑃 > Syt =310 MPa. 4.
Q1.6 Determine the diameters of a 500 mm long hollow shaft which is required
to transmit 120 kW at 1490 rpm. Make the shaft from a material with a
yield strength of 435 MPa. On the shaft is mounted a belt drive with
T1= 2,18T2. The outside diameter of the shaft is thrice the inner one
whereas the diameter of the pulley is 250 mm. Take the fos as 4.
Solution
The general torsion equation is
𝑇 𝜏 𝐺𝛳
= =
𝐽 𝑅 𝐿
Twisting moment
30𝑃 30 𝑥 120 𝑘𝑊
T= = 𝜋 𝑥 1490 𝑟𝑝𝑚
𝜋𝑁
T =769,07 Nm
Tensions in the belt
T = (T1 – T2) x R
T = (2,18 T2 -T2) x R = 1,18 T2 x R
T2 =769,07 Nm/(1,18 x 0,125)
T2 = 5214,034 N
