MECHANICAL TECHNOLOGY
EXAMINATION GUIDELINES
GRADE 12
2021

TABLE OF CONTENTS  Page 
1. INTRODUCTION  3
 2. COGNITIVE LEVELS  4
 3. ELABORATION OF CONTENT FOR GRADE 12 (CAPS)  5
4. STRUCTURE OF THE QUESTION PAPER 
4.1 Fitting and Machining – Structure of question paper 
4.2 Automotive – Structure of question paper 
4.3 Welding and Metalwork – Structure of question paper
 6
7
13
16
5. CONCLUSION  19


1. INTRODUCTION
The Curriculum and Assessment Policy Statement (CAPS) for Mechanical Technology outlines the nature and purpose of the subject Mechanical Technology. This guides the philosophy underlying the teaching and assessment of the subject in Grade 12.
The purpose of these Examination Guidelines is to:

  • Provide clarity on the depth and scope of the content to be assessed in the Grade 12 National Senior Certificate Examination in Mechanical Technology.
  • Assist teachers to adequately prepare learners for the examinations.

This document deals with the final Grade 12 external examinations. It does not deal in any depth with the school-based assessment (SBA), performance assessment tasks (PATs) or final external practical examinations as these are clarified in a separate PAT document which is updated annually.
This guideline should be read in conjunction with:

  • National Curriculum Statement (NCS) Curriculum and Assessment Policy Statement (CAPS): Mechanical Technology
  • National Protocol of Assessment: An addendum to the policy document, the National Senior Certificate: A qualification at Level 4 on the National Qualifications Framework (NQF), regarding the National Protocol for Assessment (Grades R–12)
  • National policy pertaining to the programme and promotion requirements of the National Curriculum Statement, Grades R–12

These guidelines should be used in conjunction with the content outline per term and Section 4 of the CAPS document for Grade 12 Mechanical Technology specialisation. The duration of the final examination paper will be 3 hours with a maximum mark of 200. Summative assessment (examinations) will cater for a range of cognitive levels and abilities of learners.
THIS IS A GUIDELINE DOCUMENT AND NOT A WORK SCHEDULE.

2. COGNITIVE LEVELS
Blooms Taxonomy consists of six levels as shown below.
COGNITIVE

Bloom's Taxonomy  Bloom's Revised Taxonomy  Description 
Evaluation  Creating Generating, planning, producing 
Synthesis  Evaluating  Critiquing, judging, justifying, recommending 
Analysis  Analysing  Differentiating, organising, attributing, solving 
Application  Applying  Executing, implementing, preparing, using 
Understanding Understanding Interpreting, exemplifying, classifying, summarising, inferring, comparing, explaining 
Knowledge Remembering Recognising, recalling, labelling, naming 

The following cognitive levels and weighting thereof are applicable to the Mechanical Technology specialisations:

Cognitive Levels   Weighting 
Lower order   Knowledge: memorise and recall information: arrange, define, label, list, outline, repeat, order  30%  
Comprehension: (understanding) interpret information in one's own words: describe, indicate, restate, review, summarise, classify 
Medium order   Application: apply knowledge to new situations: apply, calculate, draw, explain, identify, illustrate, prepare, operate, practise, solve, sketch, use 50%  
Analysis: break down knowledge into parts and show relationship among parts: analyses, categorise, compare, distinguish, discuss, examine, investigate, test 
Higher order Synthesis: bring together parts of knowledge to form a whole; build relationships for new situation: arrange, compose, formulate, organise, plan, assemble, construct, problem-solving  20%  
Evaluation: make judgements on basis of criteria: appraise, assess, comment, critically analyse, evaluate, conclude, interrogate, judge, predict, compare, score 

3. ELABORATION OF CONTENT FOR GRADE 12 (CAPS)
BASIC SKILLS LINKED TO THE SUBJECT:
The following skills are measured in the question paper. The visibility of these skills gives an indication of the overall skills required in the subject:

  • Ability to follow instructions
  • Identifying labels/labelling/making drawings/diagrams/schematic representations
  • Plotting and interpretation of graphs/data
  • Working out and interpreting calculations
  • Organising/Recording and categorising data
  • Extraction and/or manipulation and/or evaluation of data

NOTE:
Calculations
Generally, the criteria used for calculations is as follows:

  • Correct formula (Manipulation)
  • Substitution of values
  • Simplifying of values
  • Answer and correct units

4. STRUCTURE OF THE QUESTION PAPER
This examination guidelines document is compiled with reference to the Mechanical Technology CAPS document that focuses on specialisation.
The INSTRUCTIONS AND INFORMATION part is the same for all three specialisations and this must be brought to the attention of the candidates.
The GENERIC questions for each of the specialisations are the same and have the same weighting.
The SPECIFIC questions focus only on content applicable for that specialisation.
Use the following FORMULA SHEET ANNEXURES for the specific specialisation:
4.1.1 ANNEXURE A (Fitting and Machining)
4.1.2 ANNEXURE B (Automotive)
4.1.3 ANNEXURE C (Welding and Metalwork)
4.1 FITTING AND MACHINING

QUESTION  CONTENT  MARKS  TIME IN MINUTES 
  Generic     
1 Multiple-choice questions   6  6
2 Safety   10  10
3 Materials   14  14
  Specific    
4 Multiple-choice Questions   14  14
5 Terminology (Lathe and Milling Machine)   18  15
6 Terminology (Indexing)   28  24
7 Tools and Equipment 13 12
8 Forces 33 31
9 Maintenance 18 15
10 Joining Methods 18 15
11 Systems and Control (Drive Systems) 28 24
TOTAL 200 180


ANNEXURE A (Fitting and Machining)

FORMULA SHEET FOR MECHANICAL TECHNOLOGY:
FITTING AND MACHINING

1 BELT DRIVES

1.1 Belt speed = πDN
                             60

1.2 Belt speed = π(D+t)xN
                                60
(t = belt thickness)

1.3 Belt mass = Area Length x Density

(A =thickness x width)

1.4 Speed ratio = Diameter of driven pulley
                            Diameter of driver pulley

1.5 Belt length (flat) – [(D+d)x1,57]+(2x centre distance)

1.6 Open belt length =π(D+d)(D + d)2 + 2c
                                        2             4c

1.7 Crossed belt length=π(D+d)(D + d)2 + 2c
                                           2             4c

1.8 Power (P) = (T1-T2 ) πDN
                                 60

Where:

T1 = force in the tight side

T2 = force in the slack side

T1 -T2 -effective tensile force (Te)

1.9 Ratio between tight side and slack side = T1
                                                                        T2

1.10 Power (P)= 2πNT
                             60

1.11 Width =                        T1                     
                        Permissable tensile force

1.12 NDR x DDR = NDN X DDN

2 STRESS AND STRAIN

2.1 Ashaft = πd2
                     4

2.2 Apipeπ(D2 - d2)
                        4

2.3 Safety factor = Maximum stress/Break stress
                                   Safe working stress

2.4 Stress = Force   OR σ - F
                     Area                A

2.5 Strain = Change in length       OR ε = ΔL
                     Original length                      L

2.6 Young's modulus = Stress         OR  E = σ
                                      Strain                       ε

3. HYDRAULICS

3.1 Pressure = Force             OR  P = F
                         Area                            A

3.2 Volume = Area x Stroke length  (l or s)

3.3 Work done = Force x Distance

3.4 PA =PB

3.5 FAFB
      AA     AB

4 GEAR DRIVES

4.1 Power (P) = 2πNT
                            60

4.2 Gear ratio = Product of teeth on driver gear        OR Speed ratio = NINPUT
                          Product of teeth on driven gear                                    NOUTPUT

4.3 NINPUTProduct of the number of teeth on driven gears
      NOUTPUT Product of the number of teeth on driver gears

4.4 Nx TA = NB x TB

4.5 Torque = Force x Radius

4.6 Torque transmitted = Gear ratio x Input torque

4.7 Module - Pitch circle diameter         OR m = PCD
                      Number of teeth                              T

4.8 Pitch circle diameter = Circlular pitch x Number of teeth  OR PCD = CP x T
                                                                π                                                    π

4.9 Outside diameter (OD)- PCD + 2(m)

4.10 Addendum = Module OR a = m

4.11 Dedendum (b)=1,157(m) OR  Dedendum (b) = 1,25(m)

4.12 Cutting depth (h)= 2,157 (m) OR Cutting depth (n) = 2,25 (m)

4.13 Clearance (c)-0,157 (m) OR Clearance (c)= 0,25 (m)

4.14 Circular pitch (CP)=mxπ

4.15 Working depth (WD) = 2 xm

5 PULLEYS

5.1 NDR x DDR = NDN x DDN

5.2 Power (P) = 2πNT
                            60

5.3 Velocity ratio=Diameter of driven pulley 
                             Diameter of driver pulley

6. KEYWAYS

6.1Width of key = Diameter of shaft 
                                       4

6.2 Thickness of key - Diameter of shaft
                                              6

6.3 Length of key = 1,5 x Diameter of shaft

6.4 Standard taper for taper key: 1 in 100 or 1: 100

7 CINCINNATI DRIVING HEAD TABLE FOR MILLING MACHINE

Hole Circles        
 Side 1  24 25  28  30  34  37  38  39 41 42 43
 Side 2  46  47  49 51 53 54 57 58 59 62 66
 Change gears       
 Gears  24 x 2 28 32 40 44 48 56 64 72 86 100


7.1 Indexing = 40 (n = number of divisions)
                         n
7.2 DrA - n 40   OR Dr = (A-n) x 40
      Dn      A        1          Dn                 A
Where

A= Chosen number of divisions
n = real number of divisions

8 DOVE TAILS

Where:

R = Radius of precision roller
y = Distance from top edge of dovetail in relation to bottom corner of dovetail
x = Distance from middle of precision roller to bottom corner of dovetail
θ= Dovetail included angle (normally 60°)
h = Height of dovetail
w = Minimum width distance of dovetail
W = Maximum width distance of dovetail
m = Distance between rollers
M = Distance over rollers

9 TAPERS

9.1 tan θD - d
            2     2 x l

(l = Taper length)

9.2  θ = L(D - d)
       2       2 x l

L=Distance between centres

10. SCREW THREADS

10.1 Mean diameter= Outside diameter -(½ x Pitch) OR Dm = OD - P
                                                                                                           2

10.2 Effective diameter (Deff) = Pitch diameter (Dp) = Mean diameter (Dm)

10.3 Lead = Pitch x Number of starts

10.4 Height of screw thread=0,866 Pitch (P)

10.5 Depth of screwthread = 0,613 x Pitch (P)

10.6 Helix angle: Tanθ =Lead
                                       πxDm

10.7 Leading angle = 90° - (Helix angle + Clearance angle)

10.8 Following angle = 90° + (Helix angle - Clearance angle)

4.2 AUTOMOTIVE

QUESTION  CONTENT  MARKS  TIME IN MINUTES 
  Generic     
1 Multiple-choice questions   6  6
2 Safety   10  10
3 Materials   14  14
  Specific    
4 Multiple-choice Questions   14  10
5 Tools and Equipment  23  20
6 Engines  28  25
7 Forces 32 25
8 Maintenance 23 20
9 Systems and Control (Automatic Gearbox) 18 20
10 Systems and Control (Axles, Steering Geometry and Electronics) 32 30
TOTAL 200 180


ANNEXURE B (Automotive)
FORMULA SHEET FOR MECHANICAL TECHNOLOGY: AUTOMOTIVE

  1. F=mxa
    Where:
    m = mass a acceleration
  2. Work done = Forcex Displacement OR W=F x s
  3. Power=  Force x Displacement    OR  P = F x s
                            Time                                       t
  4. Torque = Force x Radius OR T=F x r
  5. IP = P x L x A x N x n
    Where:
    IP - Indicated power
    P = Mean effective pressure
    L-Stroke length
    A- Area of piston crown
    N-Number of power strokes per second
    n-Number of cylinders
  6. BP = 2πNT
    Where:
    BP - Brake power
    N - Revolutions per second
    T = Torque
  7. Brake power with Prony brake = 2 x π x N x F x R
    Where:
    BP - Brake power
    N = Revolutions per second
    F -Force
    R = Brake arm length
  8. Mechanical efficiency = BP x 100%
                                          IP
  9. Compression ratio = SV + CV 
                                         CV
    Where:
    SV = Swept volume
    CV - Clearance volume
  10. SV = πD2 x L
              4
    Where:
    D = Bore diameter
    L = Stroke length
  11. CV =   SV  
            CR - 1
  12. Gear ratio = Product of teeth on driven gears
                        Product of teeth on driver gears |

4.3 WELDING AND METALWORK

QUESTION  CONTENT  MARKS  TIME IN MINUTES 
  Generic     
1 Multiple-choice questions   6  6
2 Safety   10  10
3 Materials   14  14
  Specific    
4 Multiple-choice Questions   14  10
5 Terminology (Lathe and Milling Machine)   23  20
6 Tools and Equipment 18 15
7 Forces 45 42
8 Joining Methods (Inspection of Weld) 23 20
9 Joining Methods (Stresses and Distortion) 18 15
10 Maintenance 8 8
11 Terminology (Development) 21 20
TOTAL 200 180


ANNEXURE C (Welding and Metalwork)
FORMULA SHEET FOR MECHANICAL TECHNOLOGY:
WELDING AND METALWORK

1. STRESS AND STRAIN

1.1 Ashaft = πd2
                     4

1.2 Apipeπ(D2 - d2)
                        4

1.3 Safety factor = Maximum stress/Break stress
                                   Safe working stress

1.4 Stress = Force   OR σ - F
                     Area                A

1.5 Strain = Change in length       OR ε = ΔL
                     Original length                      L

1.6 Young's modulus = Stress         OR  E = σ
                                      Strain                       ε

2. PYTHAGORAS THEOREM AND TRIGONOMETRY

pyth

2.1 Sin θ = 
                    r

2.2 Sin θ = 
                    r

2. 3 Sin θ = 
                    x

2.4 r2 = x2 + y2

3 TEMPLATES AND DEVELOPMENTS

3.1 Mean Ø = Outside Ø - Plate thickness

OR Mean Ø = Inside Ø + Plate thickness

3.2 Mean circumference = π x Mean Ø

(where Ø = diameter)

5.CONCLUSION
This Examination Guidelines document is meant to articulate the assessment aspirations espoused in the CAPS document. It is therefore not a substitute for the CAPS document which teachers should teach to.
Qualitative curriculum coverage as enunciated in the CAPS cannot be over-emphasised.

Last modified on Wednesday, 23 June 2021 11:06