MECHANICAL TECHNOLOGY: FITTING AND MACHINING
GRADE 12
NSC EXAMS
PAST PAPERS AND MEMOS NOVEMBER 2018

MEMORANDUM 

QUESTION 1: MULTIPLE-CHOICE QUESTIONS (GENERIC) 
1.1 A ✔ (1)
1.2 C ✔ (1)
1.3 A ✔ (1)
1.4 B ✔ (1)
1.5 D ✔ (1)
1.6 A ✔ (1)

TOTAL QUESTION 1: [6]

QUESTION 2: SAFETY (GENERIC) 
2.1 Angle grinder: (Before using) 

• The safety guard must be in place before starting. ✔ 
• Protective shields must be placed around the object being grinded  to protect the people around. ✔ 
• Use the correct grinding disc for the job. ✔ 
• Make sure that there are no cracks in the disc before you start. ✔
• Protective clothing and eye protection are essential. ✔ 
• Check electrical outlets and cord/plugs for any damages. ✔ 
• Ensure that lockable switch is disengaged. ✔
• Ensure that the disc and the nut are well secured. ✔ 
• Ensure that the removable handle is secured. ✔ 
• (Any 2 x 1) (2) 

2.2 Welding goggles: 

• To protect your eyes against sparks ✔ 
• To protect your eyes against heat ✔ 
• To be able to see where to weld ✔ 
• To protect your eyes from UV rays ✔
• (Any 2 x 1) (2) 

2.3 PPE for Hydraulic Press: 

•  Overall ✔ 
• Safety shoes / boots✔ 
• Safety goggle ✔ 
• Leather gloves ✔
• Face shield ✔
• (Any 2 x 1) (2) 

2.4 Workshop layouts: 

•  Process layout ✔ 
• Product layout ✔ (2) 

2.5 Employer’s responsibility regarding first-aid: 

•  Provision of first-aid equipment ✔ 
• First aid training ✔ 
• First-aid services by qualified personnel ✔
• Any first aid procedures / treatment ✔ 
• Display first aid safety signs ✔ 
• First aid personnel must be identified by means of arm bands or  relevant personal signage ✔
• (Any 2 x 1) (2) 

TOTAL QUESTION 2: [10]

QUESTION 3: MATERIALS (GENERIC) 
3.1 Bending test:  

•  Ductility ✔✔ 
• Malleability ✔✔ 
• Brittleness ✔✔
• Flexibility ✔✔
• (Any 1 x 2) (2) 

3.2 Heat-treatment:  
3.2.1 Annealing: 

•  To relieve internal stresses ✔ 
• To soften the steel ✔ 
• To make the steel ductile ✔ 
• To refine the grain structure of the steel ✔ 
• To reduce the brittleness of the steel ✔
• (Any 2 x 1) (2) 

3.2.2 Case hardening: 

•  To require a wear resistant surface ✔ and it must be  tough enough internally ✔ at the core to withstand the  applied loads. 
• Hard case ✔ and tough core. ✔
• (Any 1 x 2) (2) 

3.3 Tempering process: 

•  To reduce ✔ the brittleness ✔ caused by the hardening process. 
• Relieve ✔ strain ✔ caused during hardening process. 
•  Increase ✔ the toughness of the steel. ✔
• (Any 1 x 2) (2) 

3.4 Factors for heat-treatment processes: 

•  Heating temperature / Carbon content ✔ 
• Soaking (Time period at temperature) / Size of the work piece ✔ 
• Cooling rate / Quenching rate ✔ (3) 

3.5 Hardening of steel: 

•  Steel is heated to 30 – 50°C above the higher critical temperature.  (AC3) ✔ 
• It is then kept at that temperature to ensure (soaking) that the whole  structure is Austenite. ✔ 
• The steel is then rapidly cooled by quenching it in clean water, brine  or oil. ✔ (3) 

TOTAL QUESTION 3: [14]

QUESTION 4: MULTIPLE-CHOICE QUESTIONS (SPECIFIC) 
4.1 C ✔ (1)
4.2 A ✔ (1)
4.3 D ✔ (1)
4.4 A ✔ (1)
4.5 B ✔ (1)
4.6 A ✔ (1)
4.7 B ✔ (1)
4.8 B ✔ (1)
4.9 D ✔ (1)
4.10 C ✔ (1)
4.11 B ✔ (1)
4.12 D ✔ (1)
4.13 D ✔ (1)
4.14 C ✔ (1)

TOTAL QUESTION 4: [14]

QUESTION 5: TERMINOLOGY (LATHE AND MILLING MACHINE) (SPECIFIC) 
5.1 Advantages of using the tailstock to cut an external taper:

• Long an accurate taper can be cut. ✔ 
• The automatic feed can be used which result in a good finish. ✔ (2)

5.2 Calculate the compound slide set-over:  

• Tan θ = D - d 
         2      2L
  Tan θ = 60 - 28 
         2    2 × 85
  = 0,188 ✔ ✔
  θ = 10,66 º
  2
  OR
• 
• X = D - d 
          2 
   = 60 - 28 
           2 
  = 16 mm 
• Tan θ = 16 
         2    85
  θ = 10,66 º  (5) 
  2 

5.3 Centre gauge: 

•  To measure the form and angle of the screw cutting tool angle while  grinding the tool ✔ 
• To set the screw cutting tool square/perpendicular to the axis of the  work piece ✔ (2) 

5.4 Parallel key: 
Length: 

• Length = 1,5  × diameter
  = 1,5 × 42 ✔ ✔
  = 63 mm ✔ (3) 

5.5 Advantages of up-cut milling: 

•  Deeper cuts can be made as the cutting pressure on the cutter is  lower than down cut milling. ✔
• The process enables hard steel to be cut, because the total cutting  pressure is absorbed by the material at the back of the edge. ✔ 
• Metal with hard scale, such as castings or forgings, the cut is started  under the scale where the material is softer which extends the life of  the cutter. ✔ 
• A quicker/course feed can be used. ✔ 
• The strain on the cutter and arbour will be less. ✔ 
• Vibration is limited ✔ 
• Good finish ✔ 
• Low noise level ✔
• (Any 2 x 1) (2) 

5.6 Disadvantage of down-cut milling: 

•  Vibration in the arbour is unavoidable. ✔ 
• A fine feed must be used. ✔ 
• When milling a material with hard scale the milling cutter will be  damaged. ✔ 
• Process takes time because of slower feed. ✔ 
• Noisy process. ✔ 
• Bad finish because of vibration. ✔ 
• (Any 2 x 1) (2) 

5.7 Methods of centring a milling cutter: 

•  Square and ruler method. ✔ 
• Set-over method by milling machine dial. ✔ 
• Dial indicator method ✔ 
• Using reference points on digital read out equipment ✔ 
• (Any 2 x 1) (2)

TOTAL QUESTION 5: [18]

QUESTION 6: TERMINOLOGY (INDEXING) (SPECIFIC) 
6.1 Spur gear: 
Chordal tooth thickness: 
(4) 
6.2 Calculate simple indexing: 

• Simple Indexing  = 40
                                 N
  = 40 
     13
   = 3 1 
        13 
   = 3 1  × 3
        13    3
• = 3  3  
        39
  3 full turns and 3 holes in a 39 hole circle  (4) 

6.3 Differential indexing:
6.3.1 Indexing required: 

•  Indexing  = 40  = 40
                      n      127
  = 40  = 40  ÷  5   
     A      125     5
  =  8  
     25
  Indexing 8 holes on the 25 hole circle  (3) 

6.3.2 Change gears required: 

• Dr = A-n  ×  40
  Dn    A          1
  =  125 - 127 × 40  
            125        1
  =  2  ×  40
    125     1
  = -80 ÷ 5
     125   5
  = -16  ×  4
      25      4
  = -64
     100   (5) 

6.3.3 Direction of rotation of index plate: 

• The index plate will turn the opposite ✔ direction as the index  crank handle. (1) 

6.4 Calculate distance “x” between rollers:

• "x"=150 +2(AB)− 2(CD)− 2r
  tanθ  = BC
              AB
  AB = BC 
          tanθ
  =   35   
  tan 60º
  = 20,207 mm
  = 20,21 mm
• tanθ = DE 
             CD
  CD =  15  
           tanθ
  CD =  15  
           tan30º
  = 25,98 mm 
• 
• " x" = 150 + 2(AB) - 2(CD) - 2r
  = 150 + 2 (20,21) -  2 (25,98) - 2 (15)
  = 150 + 40,42  - 51,96 - 30
  = 108,454m m
   = 108,45 m m  ✔  (9) 

6.5 Reasons for balancing work piece on a centre lathe: 

•  Prevent unnecessary bearing loads ✔ 
• Prevent excessive vibration ✔ 
• To obtain a good finish ✔ 
• To prevent clatter on the gear teeth ✔ 
• To prevent the spindle from bending ✔ 
• (Any 2 x 1) (2)

TOTAL QUESTION 6: [28] 

QUESTION 7: TOOLS AND EQUIPMENT (SPECIFIC) 
7.1 Hardness testers: 

• Brinell-hardness tester ✔ 
• Rockwell-hardness tester ✔ 
• Vickers ✔
• (Any 2 x 1) (2) 

7.2 Moment tester: 

• To determine the reactions ✔ on either side of a simply loaded beam. ✔ (2) 

7.3 Tensile test: 

• A piece of material is subjected to an increasing axial load ✔ while  measuring ✔ the corresponding elongation ✔ of the material. (3) 

7.4 Depth micro-meter: 

• Reading = 100 + 11,00 + 0,50 + 0,09  ✔ 
   = 111,59 mm (5) 

7.5 Measure depth: 

• Vernier calliper ✔ (1)

TOTAL QUESTION 7: [13]

QUESTION 8: FORCES (SPECIFIC) 
8.1 Forces: 
(15)

• E² = HC² + VC² 
  √E² = √(53,6² + 372,31²)
  E = 402,76N
  Tanθ = VC
              HC
  = 372.31
    153,62
  θ = 67,58º

Equilibrant = 402,76N en 67,58º North from East (15)
8.2 Moments: 

Calculate “x”: 
Take moments about O. 

• ∑RHM = ∑LHM
  700 × "x" 2800 × 1
  700 × "x" 2800
  "×" = 2800
           700
  "x"  = 4m ✔ ✔ (4) 

8.3 Stress and Strain: 
8.3.1 Type of stress: 

• Compressive stress ✔   (1)

8.3.2 Stress: 

• A  = π(D² -d²)
               4
  = π(0,04² - 0,03²)
                 4
  = A = 0,55 × 10^σ  m² 
  σ  = F
         A
  = 50 × 10^σ  
     0,55 ×10^σ 
  σ = 90,91 × 10^σ  Pa
  σ = 90,91 MPa ✔ 

(NO UNIT – NO MARK) (5) 
8.3.3 Change in length: 

• E = σ
         ε
  ε = σ
        E
  = 90,91 × 10⁶✔
        90 × 10⁹ 
  = 1,01 ×  10^-3 ✔ 

(IF ANY UNIT IS GIVEN – NO MARK) 

• ε  = ΔL
          L
  ΔL =  ε × L
  = (1,01 × 10^-3) × 80
  = 0,08 mm ✔ (5) 

8.3.4 Safety factor: 

•  Safety factor  =       Break stress         
                            Safe workingstress 
  Safe workingstress = Break stress
                                    Safety factor
  = 600 ×  10⁶
             4
  = 150 ×  10⁶ Pa
  = 150 MPa ✔(3) 

TOTAL QUESTION 8: [33] 

QUESTION 9: MAINTENANCE (SPECIFIC) 
9.1 Lack of preventative maintenance: 

•  Risk of injury or death. ✔
• Financial loss due to damage suffered as a result of part failure and  the waste of material. ✔ 
• Loss of valuable production time. ✔ (3) 

9.2 Causes for the malfunctioning of chain drive systems: 

•  Lack of or incorrect lubrication ✔ 
• Lack of maintenance ✔ 
• Overloading ✔ 
• Misalignment of sprockets ✔
• Incorrect chain tension ✔ 
• Contamination of chain drive system such as dust or sand ✔
• (Any 2 x 1) (2) 

9.3 Procedures to reduce the physical wear on a belt drive system:

• Check the belt alignment. ✔ 
• Checking the belt tension. ✔
• Prevent overloading of the system. ✔ 
• Keep the pulleys and belt clean. ✔ 
• Check that all covers are secure. ✔
• (Any 2 x 1) (2) 

9.4 Procedures to replace the belt on a belt drive system: 

•  Ensure that the machine is switched off ✔ 
• Release the tension on the belt ✔
• Remove the belt from the pulleys ✔
• Fit the correct size replacement belt onto the pulleys ✔ 
• Check the pulley alignment ✔ 
• Apply adequate tension according to specification and lock the  system ✔
• (Any 5 x 1) (5)

9.5 Properties of materials: 
9.5.1 Poly vinyl chloride (PVC): 

• Flexible ✔ 
• Rubber-like substance ✔
• Makes a dull sound when dropped ✔ 
• Tough ✔ 
• Act as an insulator ✔ 
• It is durable ✔ 
• Highly resistant to oxidative material ✔
• Oil, water and chemical resistant ✔
• (Any 1 x 1) (1) 

9.5.2 Carbon fibre: 

• Strong ✔ 
• Tough ✔ 
• Light weight ✔ 
• Good electrical conductor ✔
• (Any 1 x 1) (1) 

9.6 Difference between “Thermoplastic” and “Thermo hardened  (thermosetting)” composites: 

• Thermoplastics can be reheated and deformed. / Recyclable ✔
• Thermo hardened cannot be reheated. / Non-recyclable ✔ (2) 

9.7 Examples of thermo hardened composites: 

• Carbon fibre or (Any application) ✔ 
• Glass fibre or (Any application) ✔ 
• Bakelite or (Any application) ✔ 
• Teflon or (Any application) ✔ 
• (Any 2 x 1) (2)

TOTAL QUESTION 9: [18]

QUESTION 10: JOINING METHODS (SPECIFIC)
10.1 Square thread: 
10.1.1 The lead of the thread: 

• Lead = pitch × no of starts
  = 5 × 2 ✔
  = 10 mm ✔ (2) 

10.1.2 The helix angle of the thread: 
(5) 
10.1.3 The leading tool angle: 

• Leadingtoolangle 90º helix angle + clearanceangle
  = 90º - (2,29º + 3º)
  = 84,71º /84º 42'36"
  OR
• Leadingtoolangle 90º helix angle + clearanceangle
  = 90º -  (7,17º +  3º) 
  79,83º / 79º 49'48"   (2) 

10.1.4 The following tool angle: 

• Followingtoolangle = 90º + (helix angle - clearanceangle)
  = 90º + (2,29º -  3º)
  = 89,29º /89º 17'24"
  OR
• Following toolangle = 90º + (helix angle - clearanceangle)
  = 90º + ( 7,17º -  3º)
  = 94,17º / 94º 10'12" (2) 

10.2 Measurements of a screw thread : 
10.2.1 Metric screw thread ✔ (1)
10.2.2 Crest / Major / External / Basic / Nominal / Outside diameter ✔ (1)
10.2.3 Pitch ✔ (1) 
10.3 Angles of a square thread cutting tool: 

• A – Helix angle ✔
• B – Clearance angle ✔
• C – Leading tool angle ✔
• D – Following tool angle ✔ (4)

TOTAL QUESTION 10: [18] 

QUESTION 11: SYSTEMS AND CONTROL (DRIVE SYSTEMS) (SPECIFIC) 
11.1 Advantages of a belt drive system compared to a chain drive system:

• Silent operation ✔ 
• Less expensive ✔ 
• Drive can take place over a longer distance ✔ 
• No lubrication needed ✔
• (Any 2 x 1) (2) 

11.2 Hydraulics: 
11.2.1 Fluid pressure: 
(4) 

11.2.2 Diameter of the ram: 
(6) 
11.3 Hydraulic symbols: One-way valve 
(1) 

11.4 Belt drives: 
Rotation frequency of the drive pulley: 

• N_DRD_DR = N_DND_DN 
  N_DR = N_DN ×  D_DN
                   D_DR
   = 80 × 240
          75
  = 256 r/min ✔ ✔ ✔ (4) 

11.5 Gear drives: 
11.5.1 Rotation frequency of the output: 

• N_A = Product of Driven gears
  N_D    Product of Driver gears 
  N_D = T_A × T_C
  N_A    T_B × T_D
  N_D  = T_A × T_C × N_A 
                T_B × T_D
  = 20 × 25 × 3000
         35 × 30 
  N_D = 1428,57 r/min 
              60
  = 23,81 r/sec  ✔
  OR
• (6) 

11.5.2 Gear ratio: 

•  Gear ratio = Product of the number of teeth on driven gears✔
                       Product of the number of teeth on driver gears
  =35 × 30
    20    25
  = 2,1 : 1 (3) 

11.6 Work done: 

• Work done F ×  s
  = 250 ×  15
  = 3750 Jouleor N.m ✔(2) 

TOTAL QUESTION 11: [28] 
TOTAL: 200