Unit 9: Mechanical Design
LO2: Be able to select appropriate engineering
materials to achieve design solutions
Material Selection
Instructions and answers for teachers
These instructions should accompany the OCR resource ‘Material Selection’ activity which supports
Cambridge Technicals in Engineering Level 3.
The Activity:
This activity offers an
This activity offers an
opportunity for English
opportunity for maths
skills development.
skills development.
!
Suggested
timings:
1 hour
Version 1
!
Activity 1: Material selection justification based on material’s properties
Teachers might begin this activity with a basic introduction to the concepts involved, and with an
explanation of the formulae. This could be related to Unit 2: Science for engineering and Unit 3:
Principles of mechanical engineering.
The density value shown in the table is not required in the solution. The reason for placing it in the table
is to show that in real life situations the learner might be given a set of experimental data to solve a
problem with no indication of which values will be useful in finding the solution.
The Excel spread sheet LE3_Teacher_Template_v11 enables teachers to quickly obtain solutions as
well as show learners how changes of given properties and dimension influence the results obtained.
In terms of material selection it is worth stressing that even though the yield stress of the titanium alloy is
not surpassed it cannot be used due to its low modulus of elasticity value. Steel alloy is the only material
that will not fail due to the bending stresses exceeding the materials yield strength as well as maximum
deflection being below the maximum allowable value. Results are shown in the table below (red
indicates where allowable values are exceeded).
Material
Maximum Stresses (MPa)
Maximum Deflection (mm)
Steel alloy
427
11.4
Aluminium alloy
427
30.4
Titanium alloy
427
50.6
Magnesium alloy
427
28.4
The correct solution is therefore steel alloy as this is the only material that meets the design
requirements.
This activity provides the opportunity for learners to develop skills at producing their own spreadsheet in
order to solve the problem.
Version 1
Maximum stress and deflection calculations (worked example for Steel Alloy):
Material: Steel
h = 5mm
w = 45mm
L = 200mm
F = 400N
E = 200 Gpa
𝐼=
𝑤ℎ3
12
𝐹𝐿3
δ=
3𝐸𝐼
6𝐹𝐿
σ=
𝑤ℎ2
hence:
𝐼=
0.045 × 0.0053
= 4.687 × 10−10 (𝑚4 )
12
δ=
σ=
400 × 0.23
9
3 × 200 × 10 × 4.687 × 10
−10
= 11.38 × 10−3 (𝑚)
6 × 400 × 0.2
= 426.67 × 106 (𝑃𝑎)
0.045 × 0.0052
Version 1
We’d like to know your view on the resources we produce. By clicking on ‘Like’ or ‘Dislike’ you can help us to ensure that
our resources work for you. When the email template pops up please add additional comments if you wish and then just
click ‘Send’. Thank you.
If you do not currently offer this OCR qualification but would like to do so, please complete the Expression of Interest Form
which can be found here: www.ocr.org.uk/expression-of-interest
OCR Resources: the small print
OCR’s resources are provided to support the teaching of OCR specifications, but in no way constitute an endorsed teaching method that is required by the Board,
and the decision to use them lies with the individual teacher. Whilst every effort is made to ensure the accuracy of the content, OCR cannot be held responsible
for any errors or omissions within these resources.
© OCR 2015 - This resource may be freely copied and distributed, as long as the OCR logo and this message remain intact and OCR is acknowledged as the
originator of this work.
OCR acknowledges the use of the following content: English and Maths icon. Steel I beam image © Nattakit Jeerapatmaitree.
Please get in touch if you want to discuss the accessibility of resources we offer to support delivery of our qualifications: resources.feedback@ocr.org.uk
Version 1