# Mechanical Technology Grade 12 NSC November 2016 Memo (Eng)

> Grade 12 Mechanical Technology memorandum from the NSC November Exam 2016 (English), 20 pages. Read online or download the free PDF.

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- Updated: 2026-09-27
- Subject: Mechanical Technology
- Grade: Grade 12
- Document type: Memorandum
- Year: 2016
- Exam period: NSC November Exam
- Curriculum: CAPS
- Pages: 20

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Mechanical Technology · Grade 12 · NSC November Exam · 2016. Memorandum, 20 pages. Read online or download the PDF.

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                                 NATIONAL
                            SENIOR CERTIFICATE

                                   GRADE 12

                          MECHANICAL TECHNOLOGY

                                NOVEMBER 2016

                                 MEMORANDUM

MARKS: 200

                       This memorandum consists of 21 pages.

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Mechanical Technology                2          DBE/November 2016
                             NSC – Memorandum

QUESTION 1: MULTIPLE-CHOICE QUESTIONS

1.1       A                                                      (1)

1.2       A                                                      (1)

1.3       C                                                      (1)

1.4       C                                                      (1)

1.5       B                                                      (1)

1.6       C                                                      (1)

1.7       D                                                      (1)

1.8       C                                                      (1)

1.9       A                                                      (1)

1.10      C                                                      (1)

1.11      C                                                      (1)

1.12      B                                                      (1)

1.13      C                                                      (1)

1.14      D                                                      (1)

1.15      C                                                      (1)

1.16      C                                                      (1)

1.17      C                                                      (1)

1.18      D                                                      (1)

1.19      A                                                      (1)

1.20      C                                                      (1)

                                                                  [20]

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Mechanical Technology                     3                         DBE/November 2016
                                  NSC – Memorandum

QUESTION 2: SAFETY

2.1      Surface Grinder
         • Do not force the work piece into the wheel. 
         • Do not clean the machine while it is in motion. 
         • Avoid large cuts. 
         • Use coolant. 
         • Know how to do an emergency stop. 
         • Keep an eye on the position of the work piece. 
         • Any coolant spilt on the floor must be clean immediately to prevent
            slipping. 
         • Always stop the machine before taking measurements or adjusting the
            machine or work piece.
         • Keep all tools clear of the worktable. 
         • Do not allow the wheel to become badly glazed before dressing it. 
         • Never operate the grinding machine at a speed higher than the
            recommended speed for a particular wheel. 
         • Do not leave the machine while it is in operation. 
         • Do not lean on the machine. 
                                                                        Any 2 x 1       (2)

2.2      Arc welding
         • Welding helmet 
         • Gloves 
         • Overall 
         • Apron
         • Boots 
         • Leg protection 
         • Ear protection 
         • Vaporising mask. 
                                                                        Any 2 x 1       (2)

2.3      Bearing puller
         • Make sure that the legs of the puller are not worn. 
         • Make certain that the legs are well secured. 
         • Use split cover. 
         • Do not work directly behind the puller. 
         • Always keep an eye on the puller when tightening so that other
            components are not damaged in the process. 
         • Use the right size of puller. 
         • Puller must be 90° to the bearing. 
                                                                 Any 2 x 1              (2)

2.4       3 mm                                                                         (1)

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Mechanical Technology                    4                         DBE/November 2016
                                 NSC – Memorandum

2.5       Rockwell tester
          • The area around the tester should be cordoned off.
          • Only one person inside the confined space.
          • Wear protective clothing.
          • Wear safety goggles.
          • The hardness tester should be mounted on a rigid support on a work
             table.
          • Specimen must be safely positioned. 
          • Do not exceed the prescribed load. 
          • Use the correct indenter.
          • Specimen must be shielded. 
          • Remove excess oil.
                                                                      Any 3 x 1       (3)
                                                                                      [10]

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Mechanical Technology                      5                          DBE/November 2016
                                   NSC – Memorandum

QUESTION 3: TOOLS AND EQUIPMENT

3.1      Beam Bending Test
         • To investigate the deflection of beams.
         • To see if the beam is capable to support the load.
         • To see if the beam is strong enough.                                         (2)

3.2      CO reading (Gas Analyser)
         • Rich mixture setting
         • Incorrect idle speed
         • Clogged air filter
         • Faulty choke creating a rich mixture (jammed in the closed position.)
         • Incorrect ignition timing. 
         • Faulty injectors 
         • Incorrect compression (rings and valves) 
         • Too high float level 
                                                                          Any 2 x 1      (2)

3.3       Advantages of MIG/MAGS welding
          • Can weld in any position.
          • Less skilled operator required.
          • Long (continuous) welds can be made without stops and starts.
          • Minimum post-weld cleaning/no slag removal is required.
          • Cause less deformation.
          • Gives better finish.
          • Faster than arc welding.
          • Less heat generated. 
          • High deposition rate. 
                                                                        Any 2 x 1        (2)

3.4       Multimeter
          To test:
          • DC voltage 
          • DC current 
          • AC current 
          • AC voltage 
          • Resistance
          • Transistor
          • Capacitor 
          • Diode
          • Battery(voltage) 
          • Continuity 
          • Temperature 
                                                                         Any 2 x 1       (2)

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Mechanical Technology                     6                           DBE/November 2016
                                  NSC – Memorandum

3.5       Compression tester:

         3.5.1 High tension lead
               • The ignition system will be disabled
               • To be able to remove spark plugs 
               • Prevent electrical shock
               • To protect the ECU
               • Reduce fire hazard
                                                                         Any 1 x 1      (1)

         3.5.2 Fuel injectors disconnected
               • To prevent unburned fuel entering the exhaust system.
               • To prevent fuel entering the tester.
                                                                       Any 1 x 1        (1)

         3.5.3 Throttle valve fully open
               • To obtain the correct amount of air entering the cylinder. 
               • To obtain a correct reading.
                                                                          Any 1 x 1     (1)

         3.5.4 Recording the readings
               • To compare the readings to the specification.
               • To note the differences in readings between the cylinders.
                                                                        Any 1 x 1       (1)

                                                                                         [12]

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Mechanical Technology                                       7                          DBE/November 2016
                                                    NSC – Memorandum

QUESTION 4: MATERIALS

4.1                       Characteristics of cementite
                          • Hard 
                          • Brittle                                                                        (2)

4.2                       Case hardening on camshaft
                          • To ensure hardness to the surface. 
                          • To ensure resistance against wear. 
                          • To ensure a soft and tough inside/core.
                                                                                             Any 1 x 2      (2)

4.3                       Tempering hardened steel
                          • To reduce brittleness. 
                          • To increase toughness. 
                                                                                             Any 1 x 2      (2)

4.4                       Iron carbon equilibrium diagram

                          1000
                                                                                  AC3

                           900

      TEMPERATURE - °C
                                        AC3
                                                       Austenite 
                           800
                                 AC1                                                  AC1
                          720
                          700
                                  Ferrite 
                          600                                 Pearlite 

                             0       0,2      0,4      0,6     0,8         1,0   1,2         1,4
                                                     % - CARBON CONTENT                                         (7)

                                                                                                             [13]

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Mechanical Technology                               8                      DBE/November 2016
                                            NSC – Memorandum

QUESTION 5: TERMINOLOGY

5.1       Gear calculation

                                     PCD
          5.1.1           Module =
                                      T
                                     90         
                                   =
                                     30
                                   =3                                                        (2)

          5.1.2         OD = m(T + 2 )
                            = 3( 90 + 2 )       
                            = 276 mm                                                          (2)

          5.1.3         PCD = m × T
                                                 
                             = 3× 90
                             = 270 mm                                                         (2)

          5.1.4         Dedendum = 1,25× m or 1,157 × m
                                 = 1,25 × 3 or 1,157 × 3       
                                 = 3,75 mm or 3,471 mm                                        (2)

                                     PCDA PCDB
          5.1.5                 Y=         +               
                                       2      2
                                     270 90                
                                   =     +
                                      2    2
                                   = 135 + 45
                                   = 180 mm                
                                                                                               (3)

                                    40
          5.1.6         Indexing =
                                    n
                                    40
                                  =                    
                                    33
                                     7 2
                                  =1 ×
                                     33 2
                                     14
                                  =1                   
                                     66

                      Indexing:
                      = one full turn of the crank and 14 holes in a 66 hole plate            (3)

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Mechanical Technology                       9                      DBE/November 2016
                                    NSC – Memorandum

5.2       Key calculations

                                             Diameter
          5.2.1              Width of key =
                                                4
                                           92                          
                                         =
                                            4
                                         = 23 mm                       
                                                                                       (2)

          5.2.2              Length of key = 1,5 × Diameter
                                                                       
                                           = 1,5 × 92
                                           = 138 mm                                   (2)

                                                 Diameter
          5.2.3              Thickness of key =
                                                      6
                                                 92                    
                                               =
                                                  6
                                                                       
                                               = 15,33 mm
                                                                                       (2)

                                                      138     
          5.2.4                          T2 = T1 −
                                                      100
                                              =15,33 −1,38    
                                              = 13,95 mm      
                                                                                       (4)

5.3       Milling Machine components

         5.3.1 Functions of the index plate
               • A plate with holes to be able to subdivide 1 turn of the crank
                  into fractions of a turn.                                          (2)

         5.3.2 Function of the sector arms
               • Used to ease the setting of the required number of holes
               • Do not have to recount the number of holes needed.                  (2)

5.4       External V-screw thread
          • Compound slide method 
          • Cross slide method                                                       (2)

                                                                                      [30]

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Mechanical Technology                      10                         DBE/November 2016
                                    NSC – Memorandum

QUESTION 6: JOINING METHODS

6.1       Causes of welding defects

          6.1.1       Undercutting
                      • Improper settings of equipment
                      • Welding speed too fast
                      • Current too high 
                      • Current too low 
                      • Arc length to long 
                      • Incorrect electrode size 
                                                                         Any 2 x 1      (2)

          6.1.2       Lack of fusion
                      • Weld speed too fast
                      • Welding current too low
                      • Weld joint very wide
                      • Incorrect electrode size
                                                                         Any 2 x 1      (2)

6.2       Ultrasonic test
          • A high frequency sound wave  is send into the metal
          • The same unit which used to send the sound wave is then used to
              receivethe wave.
          • The deflectionsin the wave received indicate defects in the weld
              metal.                                                                    (6)

6.3       Ultrasonic vs X-ray testing
          • It can measure the depth of the defect.
          • No additional material and equipment needed. 
          • No isolation needed. 
          • Safer than x-ray. 
          • It is cheaper. 
                                                                         Any 1 x 1      (1)

6.4       X-ray test on welded joint
          • Photographic film/paper/plate 
          • Monitor/TV/screen/computer                                Any 1 x 2       (2)

6.5       Dye penetrant test
          • To detect surface / external flaws on a welded joint.                     (2)

6.6       Destructive tests
          • Nick break test
          • Bend test or Free bend test or Nick bend test 
          • Machinability test
          • Tensile test 
                                                                         Any 2 x 1      (2)

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Mechanical Technology                      11                            DBE/November 2016
                                    NSC – Memorandum

6.7       MIG/MAGS shielding gases
          • Argon 
          • CO 2 
          • Helium 
          • Teral 
                                                                             Any 2 x 1      (2)

6.8       MIG/MAGS welding
          A. Molten weld pool / welding joint / weld metal 
          B. Arc/Weld wire
          C. Contact nozzle / nozzle / contact tube 
          D. Gas shroud / welding pistol / nozzle 
          E. Shielding gas
          F. Earth cable/ earth / earth wire / earth clamp / ground wire / 'skelm'         (6)

                                                                                            [25]

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Mechanical Technology                                      12                                DBE/November 2016
                                                    NSC – Memorandum

QUESTION 7: FORCES

7.1       Resultant
                                                                 300cos45°   300 N

                                                                             300sin45°
                             200 N             200cos25°

                                  200sin25°
                                                     25°       45°                  185 N

                                            100 N
                                             
           ∑ HC = 300 cos 45° − 200 cos 25° + 185
                = 212 ,13 − 181,26 + 185
                = 215,87 N 

                                           
            ∑ VC = 200 sin 25° − 100 + 300 sin 45°
                 = 84 ,52 − 100 + 212 ,13
                     = 196 ,65 N         
                                                            OR
              Horizontal                      Magnitude           Vertical        Magnitude
             component                                          component
           -200Cos25°                    - 181,26 N          200Sin25°     84,52 N
           300Cos45°                     212,13 N            300Sin45°     212,13 N
           185                           185 N               0              0N
           0                              0N                  -100          - 100 N
           TOTAL                          215,87 N           TOTAL          196,65 N

             R 2 = HC 2 + VC 2                                
                              2                 2
             R = 215,87 + 196 ,65
             R = 292 ,01 N                                    

                 VC
          Tan θ =                                   
                 HC
                 196 ,65
               =
                  215,87
             θ = 42 ,33°                            
            R = 292 ,01 N at 42 ,33° ( 42°20' ) north of east
                                                                                         
                OR Bearing 47,67° ( 47°40' ) East                                                               (13)

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Mechanical Technology                               13          DBE/November 2016
                                             NSC – Memorandum

7.2       Stress and Strain

          7.2.1       Diameter of shaft

                           F
                        σ=
                           A
                            F                          
                        A=
                            σ
                           40 × 103
                         =
                           20 × 106
                         = 2 × 10− 3 m 2               

                           πD 2
                        A=
                            4
                            4A                         
                        D=
                             π

                          =
                                (
                               4 2 × 10 −3   )         
                                    π
                          = 0,05046 m
                          = 50,46 mm                   
                                                                                   (5)

          7.2.2       Strain
                          σ                                
                       ε=
                          E
                                                           
                          20 × 106
                       ε=
                          90 × 109
                        = 0,22 × 10− 3                     
                                                                                   (3)

          7.2.3       Change in length
                             Δl
                         ε=
                             ol                            
                        Δl = ε × ol
                                (                )
                              = 0 ,22 × 10− 3 × 0 ,3       

                              = 0,07 × 10− 3 m
                              = 0,07 mm                                           (3)

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Mechanical Technology                            14                DBE/November 2016
                                          NSC – Memorandum

7.3       Moments

          Calculate A: Moments about B

                     ∑ RHM = ∑ LHM
          ( A × 8) + (300 × 2) = (550 × 6) + (800 × 3)        
                         8 A = 3300 + 2400 − 600
                         8 A 5100
                             =
                          8      8                            
                           A = 637 ,5 N

          Calculate B: Moments about A

           ∑ LHM = ∑ RHM
              (B × 8) = (550 × 2) + (800 × 5) + (300 × 10)    
                 8 B = 1100 + 4000 + 3000
                 8 B 8100
                     =
                  8      8
                                                              
                   B = 1012,5 N
                                                         OR
           ∑Up ward = ∑ Down ward
             637 ,5 + B = 300 + 550 + 800                     
                      B = 300 + 550 + 800 − 637 ,5
                      B =1012,5 N                             
                                                                                      (6)

                                                                                      [30]

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Mechanical Technology                       15                        DBE/November 2016
                                     NSC – Memorandum

QUESTION 8: MAINTENANCE

8.1       Effects – routine maintenance
          • Risk of injury or even death.
          • Financial loss due to damages.
          • Decrease the lifespan of equipment. 
          • Loss of valuable production time.                                                (3)

8.2       Preventive maintenance
          • Planned/scheduled maintenance 
          • Condition based maintenance                                                      (2)

8.3       Properties of oil

          8.3.1       Pour point
                      • Lowest temperature at which a liquid remains pourable.              (2)

          8.3.2       Flashpoint
                      • Lowest temperature at which the oil gives off vapour which
                         can ignite.                                                         (2)

8.4       Belt drive maintenance
          • Wear and tear of the belts.
          • The belts stretch.
          • To ensure the belt stays in a good condition 
                                                                          Any 1 x 2           (2)

8.5       Cutting fluid
          • Cutting tool is kept cool. 
          • Prolong life of a cutting tool. 
          • Swarf / shavings/ cuttings are washed away. 
          • Ensure a better finish. 
          • Higher cutting or feed speed. 
          • Increase productivity 
                                                                          Any 1 x 2           (2)

8.6       Chain-drive maintenance
          By doing the following regularly:
          • Set the tension 
          • Correct alignment
          • Check for wear and tear 
          • Clean system 
          • Lubrication of parts 
                                                                          Any 1 x 2           (2)

                                                                                          [15]

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Mechanical Technology                             16               DBE/November 2016
                                           NSC – Memorandum

QUESTION 9: SYSTEMS AND CONTROL

9.1       Belt Drives

          9.1.1       Rotation frequency of the driver pulley

                        V = πDn
                            V                          
                        n=
                            πD
                               36                      
                          =
                            π (0 ,23)
                                                       
                           = 49 ,82 r.s −1 × 60
                           = 2989,35 r/ min                                          (4)

          9.1.2       Power transmitted

                        T1                                 
                           = 2 ,5
                        T2
                         T1 = 2 ,5 × T2
                            = 2 ,5 × 140
                            = 350 N                        

                        P = (T1 − T2 )V                    
                        P = (350 − 140 ) × 36
                          = 7560 Watt or 7 ,56 kW          
                                                                                      (4)

9.2       Gear drives

          9.2.1       Rotation frequency of the output shaft

                        N INPUT TB × TD × TF
                                =
                        N OUTPUT TA × TC × TE
                                  T ×T ×T                      
                         N INPUT = B D F × N OUTPUT
                                  TA × TC × TE
                                  36 × 46 × 60                 
                         N INPUT =             × 160
                                  18 × 16 × 40
                                 =1380 r/ min                                            (3)

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Mechanical Technology                              17                DBE/November 2016
                                            NSC – Memorandum

          9.2.2       Velocity Ratio

                                 N INPUT
                         VR =
                                 N OUTPUT
                               1380                  
                             =
                                160
                             = 8,625 :1
                             = 8,63:1                                                  (2)
9.3       Hydraulics
          9.3.1       Fluid pressure

                                      2
                               πDA
                          AA =                               
                                 4
                               π 0.042
                             =
                                  4
                             = 1,26 × 10− 3 m 2              

                                 FA
                         PA =
                                 AA
                                   275                       
                             =                Pa
                               1,2566 × 10− 3
                             = 218844 Pa
                                                             
                             = 218,84 kPa                                               (4)
          9.3.2       Load on piston B in kg

                               πD 2
                          AB =
                                 4
                               π 0.0752
                             =
                                   4
                             = 4,42 × 10− 3 m2                   

                                 FB
                          PB =
                                 AB
                                                                 
                        FB    = PB × AB
                                 (            ) (
                             = 218,85 × 103 × 4,42 × 10− 3   )   
                             = 967 ,32 N
                               967 ,32 N
                        Mass =
                                  10
                             = 96,73 kg                          
                                                                                        (4)

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Mechanical Technology                     18                           DBE/November 2016
                                   NSC – Memorandum

9.4       Traction control

          •   Traction control system prevents  the wheels from spinning.               (2)

9.5       Air bags

          •   Driver and passengers do not  need to activate the air bags.
          •   It is automatically  activated 
                                                                           Any 1 x 2      (2)

                                                                                           [25]

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Mechanical Technology                       19                          DBE/November 2016
                                     NSC – Memorandum

QUESTION 10: TURBINES

10.1      Supercharger
          • Roots 
          • Twin screw 
          • Centrifugal 
          • Vane type 
                                                                           Any 2 x 1        (2)

10.2      Supercharger function
          • To improve the volumetric efficiency of an engine. 
          • To make the car faster. 
          • To increase the engine output / power. 
          • Increase pressure in cylinder above atmospheric pressure. 
                                                                        Any 1 x 2           (2)

10.3      Advantages of supercharger
          • Supercharger has no turbo lag. 
          • Supercharger is less expensive. 
          • Supercharger is less complicated to mount. 
          • Supercharger has less moving parts. 
          • More efficient at lower revs. 
          • No special shutdown procedure. 
                                                                           Any 2 x 1        (2)

10.4      Turbocharger
          • By the exhaust gases.                                                          (1)

10.5      Runaway speed
          • The runaway speed of a water turbine is its speed at full flow  and
            with no shaft load.                                                            (2)

10.6      Water turbines

          10.6.1      Reaction principle 
                      OR
                      Francis 
                      OR
                      Description of operation:
                      Reaction turbines are acted on by water which changes
                      pressure as it moves through the turbine and gives up its
                      energy. The turbine must be encased to contain the water
                      pressure or suction or the must be fully submerged in the water
                      flow. 
                                                                            Any 1 x 2       (2)

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### Page 20

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Mechanical Technology                       20                            DBE/November 2016
                                     NSC – Memorandum

          10.6.2      Impulse principle 
                      OR
                      Pelton 
                      OR
                      Description of operation:
                      The impulse turbine changes the velocity of the water jet. The jet
                      pushes on the turbine’s curved blades which changes the
                      direction of the flow. The resulting change in momentum
                      (impulse) causes a force on the turbine blades. 
                                                                              Any 1 x 2    (2)

          10.6.3      Reaction principle 
                      OR
                      Kaplan 
                      OR
                      Description of operation:
                      Reaction turbines are acted on by water which changes
                      pressure as it moves through the turbine and gives up its
                      energy. The turbine must be encased to contain the water
                      pressure or suction or the must be fully submerged in the water
                      flow. 
                                                                            Any 1 x 2      (2)

10.7      Gas turbine
          A. Intake / Inlet 
          B. Compression / Compressor 
          C. Combustion / Ignition 
          D. Exhaust / thrust / Outlet 
          E. Turbine / rotor / spinner                                                    (5)

                                                                                           [20]

                                                                               TOTAL:      200

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