VAAL UNIVERSITY OF
TECHNOLOGY
Cover sheet for RETEST3
Faculty: Engineering
Department: Mechanical
Diploma: National Diploma: Engineering: Mechanical
Subject: Mechanical Engineering Design II
Date: 30 JULY 2021`
Internal code: EMMED2A
Examiner: MD Mafoko
Moderator: AO Aniki
Hours: 90 min
Total Marks: 60 Full marks: 60
* Requirements: OPEN BOOK ASSESSMENT
* Instructions: Answer all the questions. Show all your work.
Signature of Examiner:…MD Mafoko……… Date: 27 JULY 2021
Signature of Moderator:…Abimbola Aniki….. Date:………………
, Q1. A knuckle joint is loaded with a 90 kN tensile force. The joint is made from material
with allowable shear stress of 20 MPa. If this load causes shearing of the single eye whose
thickness is 1,5 times the solid rod diameter, determine the diameter of the pin using
standard sizes.
Given F= 90 K
Allowable shear stress τ = 20 MPa
Thickness of eye teye= 1,5 x diameter of pin
According to standard sizes
Fs=2 x dpin x teye x τ
Standard sizes diameter of pin = diameter of solid rod
dpin = ds
90 kN= dpin x ds x τ
90 kN = dpin x 1,5dpin x τ
90 kN =2 x 1,5d2pin x 20 MPa OR
90 𝑘𝑁
= 3d2pin OR
20 𝑀𝑃𝑎
90 𝑘𝑁
= d2pin OR
3 𝑥 20 𝑀𝑃𝑎
d2pin = 1,5 x 10 -3 m2
dpin = 0,03873 m
dpin = 38,73 mm
Standard size is 39 mm
5 marks.
TECHNOLOGY
Cover sheet for RETEST3
Faculty: Engineering
Department: Mechanical
Diploma: National Diploma: Engineering: Mechanical
Subject: Mechanical Engineering Design II
Date: 30 JULY 2021`
Internal code: EMMED2A
Examiner: MD Mafoko
Moderator: AO Aniki
Hours: 90 min
Total Marks: 60 Full marks: 60
* Requirements: OPEN BOOK ASSESSMENT
* Instructions: Answer all the questions. Show all your work.
Signature of Examiner:…MD Mafoko……… Date: 27 JULY 2021
Signature of Moderator:…Abimbola Aniki….. Date:………………
, Q1. A knuckle joint is loaded with a 90 kN tensile force. The joint is made from material
with allowable shear stress of 20 MPa. If this load causes shearing of the single eye whose
thickness is 1,5 times the solid rod diameter, determine the diameter of the pin using
standard sizes.
Given F= 90 K
Allowable shear stress τ = 20 MPa
Thickness of eye teye= 1,5 x diameter of pin
According to standard sizes
Fs=2 x dpin x teye x τ
Standard sizes diameter of pin = diameter of solid rod
dpin = ds
90 kN= dpin x ds x τ
90 kN = dpin x 1,5dpin x τ
90 kN =2 x 1,5d2pin x 20 MPa OR
90 𝑘𝑁
= 3d2pin OR
20 𝑀𝑃𝑎
90 𝑘𝑁
= d2pin OR
3 𝑥 20 𝑀𝑃𝑎
d2pin = 1,5 x 10 -3 m2
dpin = 0,03873 m
dpin = 38,73 mm
Standard size is 39 mm
5 marks.