# Mechanical Technology Grade 12 NSC June 2022 Memo (English)

> Grade 12 Mechanical Technology memorandum from the NSC June Exam 2022 (English), 26 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/mechanical-technology-may-june-2022-fitting-and-machining-mg-eng/40ab52e8-9112-4eac-b99a-8946ba552c0c
- Updated: 2026-10-29
- Subject: Mechanical Technology
- Grade: Grade 12
- Document type: Memorandum
- Year: 2022
- Exam period: NSC June Exam
- Curriculum: CAPS
- Pages: 26

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## About this document

Mechanical Technology · Grade 12 · NSC June Exam · 2022. Memorandum. Read online or download the PDF.

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             SENIOR CERTIFICATE EXAMINATIONS/
         NATIONAL SENIOR CERTIFICATE EXAMINATIONS

           MECHANICAL TECHNOLOGY: FITTING AND MACHINING

                                          2022

                                MARKING GUIDELINES

                                                                      MARKS: 200

                      These marking guidelines consist of 26 pages.

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QUESTION 1: MULTIPLE-CHOICE QUESTIONS (GENERIC)

1.1       A                                                                   (1)

1.2       B                                                                   (1)

1.3       C                                                                   (1)

1.4       D                                                                   (1)

1.5       A                                                                   (1)

1.6       C                                                                   (1)
                                                                               [6]

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QUESTION 2: SAFETY (GENERIC)

2.1       Rated speed of a grinding wheel:
           Because the wheel could burst/break if it turns faster than its revolution
             range. / Avoid an accident. 
           Effectiveness of the grinding process will be compromised.  (Any 1 x 1)           (1)

2.2       Safety precautions of a band saw in operation:
           Never leave the band saw unattended. 
           Use a push stick when cutting. 
           Hold the work piece firmly and flat on the table. 
           Don’t adjust the machine while working. 
           Don’t open any guard while the machine is on. 
           Make relief cuts before cutting tight curves. 
           Don’t force the material into the blade. 
           Keep hands clear from the action point. 
           Keep hands braced against the table. 
           Keep your hands on either sides of the blade and not in line with the
              cutting line and the blade. 
           Keep loose clothing clear from action point.             (Any 2 x 1)              (2)

2.3       Stages in which first aid is applied:
           Examination 
           Diagnosis 
           Treatment                                                                         (3)

2.4       Causes of accidents:
           Unsafe acts 
           Unsafe conditions                                                                 (2)

2.5       TWO advantages of the product layout:
           Handling of material is kept to a minimum. 
           Time period of manufacturing cycle is less. 
           Production control is almost automatic. 
           Control over operations is easier. 
           Greater use of unskilled labour is possible. 
           Less total inspection is required. 
           Less total floor space is needed per unit of production.     (Any 2 x 1)          (2)
                                                                                               [10]

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QUESTION 3: MATERIALS (GENERIC)

3.1       Tempering:
          Tempering is a process generally applied to steel to relieve the
          strains/brittleness/improve ductility  induced during the hardening
          process.                                                                             (2)

3.2       Annealing:
           To relieve internal stresses  that may have been set up during working
             of metal.
           To soften steel  in order to facilitate the machining process.
           To refine their grain structure. 
           Reduce brittleness. 
           Make the steel ductile.                                        (Any 3 x 1)         (3)

3.3       Normalising temperature:
           Above  higher/upper critical temperature 
           Above  AC3 line.                                             (Any 1 x 2)          (2)

3.4       Spark pattern for carbon steels:

          3.4.1       High-carbon steel                                                        (1)

          3.4.2       Low-carbon steel / Mild steel                                            (1)

          3.4.3       Cast-iron                                                                (1)

3.5       Carbon diagram:
          A. Temperature range / °C 
          B. AC3 line / Higher/upper critical temperature line 
          C. AC1 line / Lower critical temperature line 
          D. Carbon content / % carbon                                                         (4)
                                                                                                [14]

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QUESTION 4: MULTIPLE-CHOICE QUESTIONS (SPECIFIC)

4.1       B                                                                   (1)

4.2       A                                                                   (1)

4.3       A                                                                   (1)

4.4       C                                                                   (1)

4.5       C                                                                   (1)

4.6       D                                                                   (1)

4.7       B                                                                   (1)

4.8       A                                                                   (1)

4.9       C                                                                   (1)

4.10      D                                                                   (1)

4.11      B                                                                   (1)

4.12      C                                                                   (1)

4.13      A                                                                   (1)

4.14      A                                                                   (1)
                                                                               [14]

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QUESTION 5: TERMINOLOGY (LATHE AND MILLING MACHINE) (SPECIFIC)

5.1       Disadvantages of using compound slide method:
           The automatic feed of the machine cannot be used. 
           Only short tapers can be cut. 
           It causes fatigue in the operator. 
           Poor finish. 
           Take longer time to cut. 
                                                                  (Any 2 x 1)     (2)

5.2       Taper:

          5.2.1       Length of taper:

                             θ Dd
                       tan    
                             2 2

                            12 72  40
                      tan             
                             2   2

                                    32 
                       2 
                                  tan 6 

                                  304,46
                                       
                                    2

                              = 152,23 mm                                       (5)

         5.2.2       Tailstock set-over:
                                  LD - d
                     Set - over 
                                    2
                                                   
                              x
                                 70  152,2372 - 40
                                       2152,23

                              x  23,36 mm                                       (3)

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5.3       Parallel key:

          5.3.1       Width:
                                D
                      Width 
                                4

                                102
                      Width        
                                 4

                             25,50 mm                                                 (2)

          5.3.2       Thickness:
                                     D
                      Thickness 
                                     6

                                     102
                      Thickness         
                                      6

                                     17 mm                                            (2)

          5.3.3       Length:
                      Length = 1,5 x diameter of shaft

                               = 1,5 x 102 

                               = 153 mm 
                                                                                        (2)

5.4       Disadvantages of gang milling:
           The gang of cutters place more stress on the machine spindle
              bearings. 
           The cutters make the milling machine work harder.                           (2)
                                                                                         [18]

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QUESTION 6: TERMINOLOGY (INDEXING) (SPECIFIC)

6.1       Spur gear:

          6.1.1       Number of teeth:
                                         PCD
                            Module 
                                          T

                                         168
                            Module          
                                          42

                            Module  4                                                           (2)

          6.1.2       Circular pitch:

                      CP = m x π                                         PCD  π
                                                                  CP 
                         =4xπ                                             T
                         = 12,57 mm 
                                                    OR                   168  π
                                                                                
                                                                           42

                                                                      12,57 mm 
                                                                                                  (2)

          6.1.3       Outside diameter:

                      OD = PCD + 2(m)                               OD = m(T + 2)
                         = 168 + 2(4)                   OR               = 4(42 + 2) 
                         = 176 mm                                        = 176 mm 
                                                                                                  (2)

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6.2       W  135  2 y 
          m  W  2 x   2R 
          m  W  2 x  R 

          6.2.1       Maximum width distance of dove tail (W):

                      Calculate y:

                                   h
                      Tan θ 
                                   y                                        y
                                                                  tanθ 
                                                                           35
                             h
                      y               
                           tan θ                 OR                  y  tanθ  h 

                             35                                      y  tan30  35 
                      y           
                           tan 60
                                                                       20,21 mm 
                      y  20,21 mm 

                      W = 135 + 2(y) 
                         = 135 + 2(20,21) 
                         = 135 + 40,42
                         = 175,42 mm 
                                                                                                 (6)

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          6.2.2       Distance between the rollers (m):

                      Calculate x :

                                   R
                         Tan α 
                                   x

                                                                            x
                               R                                  tanθ 
                         x                                               12
                              Tan α

                                                                     x  tan  12 
                              12                    OR
                         x         
                            Tan 30
                                                                     x  tan60  12 

                                12
                         x                                            20,78 mm 
                              0,5774

                         x  20,78 mm 

                        m = W – (2 x ) – (2R) 

                          = 175,42 – (2 x 20,78) – (2 x 12) 
                          = 175,42 – 41,56 – 24
                          = 109,86 mm 

                                 OR

                        m = W – 2( x +R) 

                        m = 175,42 – 2(20,78 + 12) 
                        m = 175,42 – 65,56
                        m = 109,86 mm 
                                                                                             (6)

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6.3       Milling of spur gear:

          6.3.1       Indexing:
                                   40
                      Indexing 
                                   n

                                    40
                      Indexing 
                                   113

                                    40
                      Indexing        
                                   110

                                    40 5
                      Indexing       
                                   110 5

                                   8   3
                      Indexing            
                                   22 3

                                   24
                      Indexing       
                                   66
                      Approximate indexing: 24 holes on a 66-hole circle 
                                                                                          (4)

          6.3.2       Change gears:
                      Dr             40
                          A  n 
                      Dn             A

                      Dr                  40
                          110  113      
                      Dn                 110

                      Dr     - 3  40
                          
                      Dn       110

                      Dr     - 120 5
                                
                      Dn      110   5

                      Dr   - 24 2
                                 
                      Dn    22   2

                      Dr   48
                         
                      Dn   44
                                                                                          (4)

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6.4       Unbalanced work piece:
             Unnecessary bearing loads 
             Excessive vibration 
             A bad finish 
             Work that is not perfectly round 
             Danger to the operator 
             Clatter on the gear teeth 
             A tendency to bend the spindle 
                                                                  (Any 2 x 1)      (2)
                                                                                   [28]

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QUESTION 7: TOOLS AND EQUIPMENT (SPECIFIC)

7.1       Tensile tester:

              Yield stress 
              Ultimate tensile stress / Maximum stress 
              Elongation percentage / Ductility 
              Limit of proportionality 
              Limit of elasticity 
              Break stress 
                                                                        (Any 3 x 1)      (3)

7.2       Brinell hardness tester:

          A - Work piece / Test piece 
          B - Steel Carbide ball / Indenter 
          C - Indentation diameter                                                      (3)

7.3       Calculate cutting depth:

          Cutting depth = 0,613 x pitch
                         = 0,613 x 1,75 
                         = 1,07 mm                                                      (2)

7.4       Depth micrometer reading:

           
          17,36 mm                                                                       (2)

7.5       Function of moment tester:

          To determine the reactions  on either side  of a simply loaded  beam.       (3)
                                                                                         [13]

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QUESTION 8: FORCES (SPECIFIC)

8.1       System of forces:

               HORIZONTAL COMPONENT:

                HC  50cos30  85cos0  75cos70
                                  
                HC  43,30  85  25,65

                HC   16,05 N 

               VERTICAL COMPONENT:

                VC  50sin30  25sin90  75sin70

                                  
                VC  25  25  70,48

                VC   70,48 N 
                                        OR
       Force          θ     VC/y = Fsinθ                   HC/x = Fcosθ
        50 N         30º    VC = 50sin30º          25 N  HC = 50cos30º                     43,30 N 
        85 N         180º   VC = 85sin180º         0N      HC = 85cos180º                    -85 N 
        25 N         270º   VC = 25sin270º        -25 N  HC = 25cos270º                      0N
        75 N         290º   VC = 75sin290º      -70,48 N HC = 75cos290º                    25,65 N 
                                        Total   -70,48 N                                  -16,05 N 

               R 2  VC 2  HC2                                       VC
                                                         tan θ 
                                                                      HC

                R 2  ( 70,48)2 ( 16,05)2                                  70,48
                                                          θ  tan 1(                 )
                                                                               16,05
                R 2  5225,033
                                                          θ  tan 1(4,39)
               R  72,28 N 
                                                                  
                                                          θ  77,17 OR 77 10' 12" 

                                      
                                                            W
                                                                   θ
                                                                  R       S
                                                                                                         (15)

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8.2   Moments:

          8.2.1       Point load for UDL:

                      6 x 11 = 66 kN                                                 (1)

          8.2.2       Calculate A
                      Take moments about B:
                      ∑ RHM = ∑LHM

                      A  11 (12  2,5)  66  5,5   75  11 

                         11A  30  363  825

                                 1218
                           A         
                                  11

                           A  110,73 kN 
                                                                                      (3)

          8.2.3       Calculate B
                      Take moments about A:
                      ∑ LHM = ∑RHM

                     (B  11) (75  0) (66  5,5) (12  8,5) 

                         11B  0  363  102

                                 465
                            B
                                  11 

                            B  42,27 kN 

                                    OR

                            A  B  75  66  12 

                           B  153  110,73 

                           B  42,27 kN                                              (3)

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8.3.1     The resistance area:

              πD 2
          A
                4
              π(0,05) 2
          A            
                   4
          A  0,00196 m 2 or 1,96  10 3 m 2                                 (2)

8.3.2     The stress in the material in MPa:

               F
          σ
               A

              50  10 3 
          σ
             1,96  10 3 

          σ  25510204,082 Pa

          σ  25,51 MPa                                                       (3)

8.3.3     Strain in material:

             ΔL
          ε
              L
             0,005
          ε       
                3
          ε  0,0017 OR 1,67  10  3 
          (No unit! If unit indicated, answer is wrong)                        (2)

8.4       Label stress and strain diagram:

          A – Limit of proportionality 
          B – Elastic limit 
          C – Yield point 
          D – Maximum stress 
                                                                                (4)
                                                                                [33]

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QUESTION 9: MAINTENANCE (SPECIFIC)

9.1       Main purpose of maintenance:
          To ensure that the machinery always operates at an optimal level.               (1)

9.2       Preventative maintenance:

          9.2.1       Chain drives:
                       Cleaning uncovered drives. 
                       Check sprocket teeth and link plate wear. 
                       Check lubricants. 
                       Checking the functioning of tensioning devices. 
                       Inspecting chains regularly for elongation. 
                                                                         (Any 2 x 1)       (2)

          9.2.2       Belt drives:
                       Checking the wear and tear on belt. 
                       Checking belt alignment. 
                       Checking the tension setting. 
                       Checking the tensioning devices. 
                                                                         (Any 2 x 1)       (2)

9.3       Properties of EACH of the following nylon products:

          9.3.1       Polyethylene (PET):
                       Stiff 
                       Strong 
                       Hard 
                       Chemical resistance 
                                                                         (Any 2 x 1)       (2)

          9.3.2       Polypropylene (PP):
                       Good fatigue resistance 
                       Heat resistance 
                       Tough 
                       Semi rigid 
                       Good chemical resistant 
                                                                         (Any 2 x 1)       (2)

          9.3.3       Polystyrene (PS):
                       Good insulation properties 
                       Water resistant 
                       Odourless properties 
                       Light weight 
                                                                         (Any 2 x 1)       (2)

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9.4       Factors influencing coefficient of friction:
           Surface roughness/texture 
           Temperature 
           Type of lubricant 
           Type of friction 
                                                                        (Any 3 x 1)      (3)

9.5       Thermo-hardened composites:
          This type of plastic goes through a chemical change during moulding that
          prevents reheating.                                                           (1)

9.6       Types of belt drive systems:
           V-belt 
           Wedge belt / Cog belt 
           Flat belt 
           Multi-groove belt 
           Toothed / Timing belt 
           Circular belt / Round belt 
           Poly V-belt 
                                                                        (Any 3 x 1)      (3)
                                                                                         [18]

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QUESTION 10: JOINING METHODS (SPECIFIC)

10.1      Square Thread:

          10.1.1      Pitch:

                      Lead  Pitch  Number of starts

                                    Lead
                      Pitch                    
                                Numberof starts

                                36
                      Pitch       
                                3

                      Pitch  12 mm                                           (3)

          10.1.2      Pitch diameter:

                                   P
                      Dp  OD 
                                   2

                                  12
                      Dp  70       
                                   2

                      Dp  64 mm 
                                                                               (2)

          10.1.3      Helix angle of the thread:

                                 Lead
                      Tan θ 
                                π  PD

                                  36 
                      Tan θ 
                                π  64 

                      Tan θ  0,179049311

                          θ  10,15  OR 10 9 ' 
                                                                               (3)

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          10.1.4      Leading angle:

                      Leading angle = 90° - (helix angle + clearance angle)
                                     = 90° - (10°9’ + 3°) 
                                     = 76°51’ or 76,85º 
                                                                                             (2)

          10.1.5      Following angle:

                      Following angle = 90° + (helix angle - clearance angle)
                                       = 90° + (10°9’ - 3°) 
                                       = 97°9’ or 97,15º                                    (2)

10.2      Square thread cutting tool angles:

          10.2.1      Square thread                                                         (1)

          10.2.2      A - Clearance angle 
                      B - Leading angle 
                      C - Following angle 
                      D - Helix angle                                                       (4)

10.3      Included angle of Metric V-thread:

          60°                                                                               (1)
                                                                                             [18]

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                                    SC/NSC Marking – Guidelines

QUESTION 11: SYSTEMS AND CONTROL (DRIVE SYSTEMS) (SPECIFIC)

11.1      Hydraulic calculations:

          11.1.1      The fluid pressure in the hydraulic system in MPa:

                                πd 2
                      A RAM 
                                 4

                             π(0,120) 2          
                          A
                                  4

                          A  11,31 10  3 m2       

                      NOTE: Force of Ram = 4 500 kg x 10 m/s2
                                         = 45 x 103 N

                                       F
                                Ρ
                                       A

                                        
                                    45  103
                                P
                                   11,3  10-3

                                P  3,98 N                                             (4)

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                                    SC/NSC Marking – Guidelines

          11.1.2      Force to be applied on the plunger:

                                     πd 2
                         A Plunger 
                                      4

                                     π(0,032) 2        
                                A
                                          4

                                A  0,80  10  3 m2       

                                     F
                                P
                                     A

                                F  P A 

                                F  3,98  106  0,80  10 3 

                                F  3184 N 

                                                       OR

                                FA  F
                                    B
                                A A AB

                                              
                            FA       45  103
                                  
                        0,8  103 11,31  103 
                            

                                                 
                                    45  103 0,8  103
                                FA 
                                                               
                                         11,31  103

                                      3184 N                                  (5)

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                                    SC/NSC Marking – Guidelines

11.2      Function hydraulic non-return valve:
          Used to provide one-directional oil flow and the fluid is not allowed to flow
          back.                                                                            (1)

11.3      Spring-loaded double-action control valve is used in a water line:
           Protect water supplies from back flow. 
           Prevent water contamination. 
           Safety device if one valve is jammed. 
                                                                     (Any 2 x 1)            (2)

11.4      Pressure gauges:
           Bourdon tube gauge. 
           Schrader gauge.                                                                (2)

11.5      Belt drive:

          11.5.1        The rotational frequency:

                      NDN  DDN  NDR  DDR

                              NDR  DDR
                      NDN                
                                DDN

                              1320  0,085
                      NDN                    
                                 0,375

                              299,2
                      NDN          
                               60

                      NDN  4,99 r/sec                                                     (4)

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                                    SC/NSC Marking – Guidelines

          11.5.2      Power transmitted in Watt:

                        (T  T2 )  D N
                      Ρ 1
                             60
                                                  
                      Ρ (275  120) π  0,375  4,99

                      P  911,199 Watt 

                             OR

                      P  T1  T2 π DN

                                                  
                        
                          275  120  π  0,085  1320
                                           60

                         910,59 Watt                                         (3)

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                                    SC/NSC Marking – Guidelines

11.6      Gear drive:

          11.6.1      Number of teeth on TA:

                         Ninput        Product of the numberof teeth on driven gears
                                  
                        Noutput       Product of the number of teeth on driving gears

                            NF TA  TC  TE
                               
                            NA   TB  TD  TF

                           720 TA  45  50 
                              
                           320 20  25  20 

                                      2250  TA
                           2,25                
                                       10000

                      2250 TA  2,25  10000

                                      22500
                             TA 
                                      2250

                             TA  10                                                        (4)

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                                    SC/NSC Marking – Guidelines

          11.6.2      Gear ratio:

                                        Product of the number of teeth on driven gears
                        Gear ratio 
                                        Product of the number of teeth on driving gears

                                        20  25  20 
                        Gear ratio 
                                        10  45  50 

                        Gear ratio  0,44: 1 

                                             OR

                                        NA
                      Speed Ratio 
                                        NF

                                        320 
                                    
                                        720 

                                     0,44 : 1                                                (3)
                                                                                               [28]

                                                                                 TOTAL:        200

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