MECHANICAL TECHNOLOGY: FITTING AND MACHINING
GRADE 12
NATIONAL SENIOR CERTIFICATE
NOVEMBER 2019
MEMORANDUM 

QUESTION 1: MULTIPLE-CHOICE (Generic)
1.1 B ✓(1)
1.2 C ✓ (1)
1.3 A ✓ (1)
1.4 C ✓ (1)
1.5 A ✓ (1)
1.6 C ✓ (1)
[6]

QUESTION 2: SAFETY (Generic)
2.1 Machine safety rule:

• Know how to switch the machine off / emergency stop.✓
• Wear personal protective equipment (PPE). ✓
• Know how to use the machine.✓
• Ensure that all guards are in place. ✓
• No tools lying on the machine. ✓
• Work piece is properly secured. ✓
• Check the condition of the machine. ✓
• Follow manufacture's specifications before operating a machine. ✓
• Operator must have authorization to working on a machine. ✓
• Make sure the machine is not locked out. ✓
• Ensure that the machine setup is correct and safe. ✓
• Ensure that the machine area is clean and safe. ✓
  (Any 1 x 1) (1)

2.2 Drill press safety precautions:

• To prevent injuries. ✓
• To improve accuracy. ✓
• To prevent work piece rotating/moving. ✓
• To prevent the drill bit from breaking. ✓
  (Any 1 x 1) (1)

2.3 Hydraulic press safety rules:

• Make sure the press is in a good working condition. ✓
• Take notice of the pre-determined maximum pressure of the hydraulic press. ✓
• Make sure the area around the press is clean and free of oil, grease and water. ✓
• Ensure that the platform is rigid and square to the cylinder. ✓
• Ensure that suitable jigs and prescribed equipment is available. ✓
• Check hydraulic pipes for leaks or cracks. ✓
• Check supporting pins are not worn out and fitted properly. ✓
• Check fluid levels. ✓
• Compressive force must be applied at 90° to the object.✓
• Check cable and pulleys on the platform if equipped. ✓
• Correct PPE. ✓
• Pressure gauge must be checked and calibrated. ✓
• Ensure that all guards are in place. ✓
  (Any 2 x 1) (2)

2.4 Reasons for wearing surgical gloves:

• To prevent HIV/AIDS or any blood related infections being transmitted✓
• To prevent contamination of the open wounds✓ (2)

2.5 Safe handling of portable electrical equipment:

• Ensure the electrical cord and plug, are in a good condition. ✓
• Ensure all safety guards are in place. ✓
• Ensure that the correct attachments (drill bits, blades etc.) are fixed in the correct way. ✓
• Do not force the machine/equipment.✓
• Operate according to manufacturer instructions. ✓
• Avoid contact with water.✓
• Keep the cable away from heat, oil, sharp edges and moving parts. ✓
• Make sure that the wires don't wrap around each other. ✓
• Avoid dropping the machine. ✓
• Check the condition of the equipment. ✓
  (Any 2 x 1) (2)

2.6 Responsibility of employer:

• Provide and maintain working systems, work area, equipment and tools in a safe condition. ✓
• Eliminate or reduce any potential hazard. ✓
• Produce, handle, store and transport goods safely.✓
• Ensure that every person employed complies with the requirements of this OHS Act. ✓
• Enforce measures if necessary in the interest of health and safety. ✓
• Appoint a person who is trained and who have the authority to ensure that the employee takes precautionary measures. ✓
• Inform employees of the hazards to his health and safety attached to any duty or work situation. ✓
• Provide first aid equipment. ✓
  (Any 1 x 1) (1)

2.7 Responsibility of employee:

• Pay attention to their own and other people's health and safety. ✓
• Co-operate with the employer regarding the OHS Act. ✓
• Carry out a lawful order given to them. ✓
• Report any situation that is unsafe or unhealthy.✓
• Report all incidents and accidents. ✓
• Not to interfere with any safety equipment or misuse such equipment.✓
• Obey all safety rules. ✓
  (Any 1 x 1) (1)

[10]

QUESTION 3: MATERIAL (Generic)
3.1 Filing test:

• Use the right ✓ filing skills. ✓
• File on the tip or edge ✓✓ of the metal.
• By applying chalk ✓ to the file surface. ✓
  (Any 1 x 2) (2)

3.2 Purpose of heat treatment of steel:
Heat treatment of steel is done to change ✓ the properties/grain structure ✓ of steel. (2)
3.3 Reasons for tempering hardened steel:

• To reduce ✓ the brittleness ✓ caused by the hardening process.
• To relieve ✓ strain ✓ caused during hardening process.
• To increase ✓ the toughness ✓ of the steel.
• To give hardened work piece a more ✓ fine-grained structure. ✓
  (Any 2 x 2) (4)

3.4 Heat treatment processes on steel:
3.4.1 Annealing:

• The steel is heated to the prescribed temperature. ✓
• The steel is soaked at that temperature for the required time. ✓
• The steel is then cooled very slowly to produce maximum softness.✓ (3)

3.4.2 Hardening:

• The steel is heated slightly higher than the upper critical temperature. (AC₃) ✓
• The steel is soaked at that temperature for the required time. ✓
• The steel is then rapidly cooled by quenching in rapid cooling medium. ✓(3)

[14]

QUESTION 4: MULTIPLE-CHOICE QUESTIONS (Specific)
4.1 A ✓ (1)
4.2 D ✓ (1)
4.3 B/A ✓ (1)
4.4 A ✓ (1)
4.5 B/A ✓ (1)
4.6 B ✓ (1)
4.7 B✓ (1)
4.8 A ✓ (1)
4.9 C ✓ (1)
4.10 B ✓ (1)
4.11 D ✓ (1)
4.12 B ✓ (1)
4.13 A ✓ (1)
4.14 B ✓ (1)
[14]

QUESTION 5: TERMINOLOGY (Lathe and Milling Machine) (Specific)
5.1 Taper turning:
5.1.1 Small diameter:

Tan θ = AB 
       2   300
AB = Tan 4º x 300
= 20.98mm
d = D - 2(AB)
= 200 - 2(20,98)
= 158,04 mm

OR
Tan θ = D - d 
       2      2L
= 200 d 
2(300)
tan4º x 600 = 200 - d
d = 200-(tan4º x 600)
d = 158,04 mm

OR
d = D - 2AB
= 200 - 2(300 tan4º)
= 158,04 mm (5)
5.1.2 Setting over of tailstock:
Setting over : 20,98 mm over 300 mm
Thus "X" mm over 400 mm
300"X" = 20,98 x 400 
"X" = 20,98 x 400
               300
"X" 27,97mm
Set over = L(D- d)
                     2l
= 400(200 - 158,04)
           2(300)
= 27,97 mm (3)
5.2 Parallelkey:
5.2.1 Width: D 
                    4
=42
   4
= 10,5 mm (2)
5.2.2 Thickness:
Thickness  D 
                   6
=42
   6
= 7 mm (2)
5.3 Advantages of down cut milling:

Related Items

• A better finish is obtained. ✓
• Friction is reduced. ✓
• High speed cutting is possible. ✓
• Less power required. ✓
• Coolant is carried down to the teeth to where it is required. ✓
• Tends to force the work piece onto the machine table there for deeper cuts can be made. ✓
• Less vibration. ✓
  (Any 2 x 1) (2)

5.4 Straddle milling: (4)

[18]

QUESTION 6: TERMINOLOGY (Indexing) (Specific)
6.1 Spur gear terminology:
6.1.1 Module:
Module = PCD
                  T
= 126 
    42
= 3 (2)
6.1.2 Working depth:
WD=  2 x m
= 2 x 3
= 6mm (2)
6.1.3 Cutting depth:
Cuttingdepth 2,157 x m    = 2,25 x m
= 2,157 x 3        or            = 2,25 x 3
= 6,47mm                         = 6,75mm (2)
6.2 Angular indexing:
Indexing = n  = 34º
                 9º     9º
= 3⁷/₉ x ⁶/₆
= 3⁴²/₅₄
3 full turns and 42 holes on the 54 hole circle. (4)
6.3 Indexing:
6.3.1 Differential indexing:
Indexing = 40 
                  N
= 40 
  121
Chosen divisions = = 40 
                                   A
= 40 
  120
= ¹/₃ x ⁸/₈
= 8 
  24
No full turns, 8 holes on the 24 hole circle. 
OR
No full turns, 10 holes on the 30 hole circle.
OR
No full turns, 13 holes on the 39 hole circle. 
OR
No full turns, 14 holes on the 42 hole circle. 
OR
No full turns, 17 holes on the 51 hole circle. 
OR
No full turns, 18 holes on the 54 hole circle. 
OR
No full turns, 19 holes on the 57 hole circle. 
OR
No full turns, 22 holes on the 66 hole circle. (5)
6.3.2 Change gears:
 Dr  =A - N x 40 
 Dn      A        1
=120 - 121 x 40
         120       1
=  -1   x 40 
  120      1
= -40 
   120
= ^-1/₃ x ²⁴/₂₄
 Dr  = 24 
 Dn    72    (5)
6.4 Calculate distance Z between rollers:

z =150 + 2x - 2(y + r)
Tan40º= 25 
               x
x =   25    
    Tan 40
= 29,79 mm
r = ^d/₂
= ¹⁰/₂
= 5mm

Tan20º = ^r/_y
y =     5     
      Tan20º
= 13,74 mm

z = 150 + 2x - 2(y + r)
= 150 + 2(29,79)-2(13,74 + 5)
= 150 + 59,58 - 37,48
= 172,1 mm (8)
[28]

QUESTION 7: TOOLS AND EQUIPMENT (Specific)
7.1
(4)
7.2 Tensile test:

• Tensile strength✓
• Elasticity ✓
• Ductility ✓
• Plasticity ✓
  (Any 2 x 1) (2)

7.3 Depth micro-meter reading:
50,00✓
16,00 ✓
0,00 ✓
0,33 ✓
66,33 mm ✓ (5)
7.4 Screw thread ✓ micro meter ✓ (2)
[13]

QUESTION 8: FORCES (Specific)
8.1Equilibrant:

ΣHC = 30 + 80cos30º - 20cos60º - 60cos60º
30° + 69,28 - 10 - 30
= 59,28N
= 5,36 N

ΣVC = 20sin60º + 80sin30º - 60sin60º
= 17,32 + 40 - 51,96
= 5,36N
OR

E² = HC² + VC² 
E = √59,28² + 5,36² 
E = 59,52 N
Tanθ = VC 
            HC
= 5,36 
  59,28
θ = 5,17º
E = 59,52 N at 5,17° (South from West) or
E = 59,52 N at 84,83° (West from South) or
E = 59,52 N at a bearing of 264,83° (14)
8.2 Moments:

Distributed load:
= 100 N/m X 4,4 m
= 440 N
Calculate A:
Moments about B:
ΣRHM = ΣLHM
(A x 4,4) = (350 x 1,2) + (440 x 2,2) + (600 x 3,6)
4,4A = 3548
 4.4       4.4
A = 806,36N
Calculate B:
Moments about A.
ΣLHM = ΣRHM
(B x 4,4) = (600 x 0,8) + (440 x 2,2) + (350 x 3,2)
4,4B = 2568
 4,4       4,4
B = 583,64N (8)
8.3 Stress-strain:
8.3.1 Compressive stress  (1)
8.3.2 Stress:
A =π(D² - d²)
           4
= π(0,04² - 0,025²)
               4
A = 0,77 x 10^-3 m²
σ = F 
      A
=      600      
   0,77 x 10^-3
σ = 779220,78 Pa or
= 0,78 x 10⁶ Pa or
= 0,78 MPa (4)
8.3.3 Change in length:
E = σ 
      ε
ε = σ 
     E
= 0,78 x 10⁶
    90 x 10⁹
ε = Δl 
      ol
Δl = ε x ol
= (8,67 x 10^-6) x 100
= 0,87 x 10^-3 mm (6)
[33]

QUESTION 9: MAINTENANCE (Specific)
9.1 Types of maintenance:

• Preventative✓
• Predictive ✓
• Reliable centred ✓ (3)

9.2 Malfunctioning of belt drives:

• Lubrication between belt and pulley causing belt slip. ✓
• Pulleys not properly secured to shafts. ✓
• Incorrect pulley alignment. ✓
• Overloading the system. ✓
• Incorrect belt tension. ✓
• Worn belts. ✓
• Faulty/damaged tensioner pulley. ✓
• Lack of maintenance. ✓
  (Any 2 x 1) (2)

9.3 Replace the chain on a chain drive system:

• Release the tension on the chain and remove from sprocket. ✓
• Check the condition and alignment of the sprockets. ✓
• Fit the new specified chain and lubricate. ✓
• Apply adequate tension to the chain. ✓
• Check for proper operation. ✓ (5)

9.4 Wear on a gear drive system:

• Check and replenish of lubrication levels. ✓
• Ensuring the gears are properly secured to shafts. ✓
• Cleaning and replacement of oil filters. ✓
• Reporting excessive noise, wear, vibration and overheating for expert attention. ✓
  (Any 2 x 1) (2)

9.5 Material:
9.5.1 Nylon:

• Bushes ✓
• Gears ✓
• Pulleys ✓
• Fishing line ✓
• Clothing ✓
• Sails ✓
• Ropes ✓
• Sport equipment ✓
• Powder coating ✓
  (Any 1 x 1) (1)

9.5.2 Glass fibre:

• Used in boats ✓
• Motor vehicle bodies ✓
• Transparent roof sheets ✓
• Petrol tanks ✓
• Swimming pools ✓
• Furniture ✓
• Fruit and salad bowls ✓
• Ornaments ✓
• Fishing rods ✓
• Sporting equipment ✓
  (Any 1 x 1) (1)

9.6 Thermoplastic or Thermo hardened composites:
9.6.1 Teflon:
Thermoplastic ✓ (1)
9.6.2 Bakelite:
Thermo hardened / Thermo setting ✓ (1)
9.7 Coefficient of friction:

• Contact pressure ✓
• Surface roughness ✓
• Temperature ✓
• Sliding velocity ✓
• Type (amount) of lubricant ✓
• Type of material ✓
  (Any 2 x 1) (2)

[18]

QUESTION 10: JOINING METHODS (Specific)
10.1 Screw thread diameters:

A = Pitch diameter/Effective diameter ✓
B = Minor diameter/Root diameter ✓
C = Major diameter/Crest diameter/Outside diameter/Nominal diameter/Basic diameter ✓(4)
10.2 Lead of a screw thread:
The lead is the distance a thread ✓ will move axially✓ in one full revolution✓ (3)
10.3 Square screw thread:
10.3.1 Screw thread lead:
Lead = pitch x noof starts
= 4 x 3
= 12 mm (2)
10.3.2 Mean/pitch circumference:
Mean/pitchcircumference = π(OD - ^P/₂)
= π(68 - ⁴/₂)
=207,35 mm
10.3.3 Helix angle:
Helix angle tanθ =                 lead                 
                             mean/pitch circumference
=   12   
  207,35
θ = 3,31º (2)
10.3.4 Leading angle:
Leadingtoolangle = 90º - (helix angle + clearance angle)
= 90º - (3,31º - 3)
= 83,69 (2)
10.3.5 Following angle:
Followingtoolangle = 90º + (helix angle x clearance angle)
= 90º + (3,31º - 3º)
= 90,31º (2)
[18]

QUESTION 11: SYSTEMS AND CONTROL (Drive Systems) (Specific)
11.2.2 Advantages of a gear drive:

• Compact assembly ✓
• More power can be transmitted/Stronger ✓
• No slip occurs ✓
• Less maintenance ✓
  (Any 2 x 1) (2)

11.2 Hydraulics:
11.2.1 Fluid pressure:
A_B = πD_B²
           4
= π(0,2)²
      4
= 31,42 x 10^-3 m²
P = F_B 
      A_B
=   15 x 10  ³
  31,42 x 10^-3
=477,40 x 10³Pa
= 477,40 kPa
11.2.2 Distance 'X':
A_A = πD_A²
          4
= π(0,075)²
         4
=4,42 x 10^-3 m²
V_B = V_A 
A_B x L_B = A_A x L_A
L_B = A_A x L_A 
            A_B
= 16,88 x 10^-3 m
= 16,88mm (6)
11.3 Advantages of pneumatics:

• Compressed air is easy and cheap to generate. ✓
• Leakages are not messy – no oil spills. ✓
• Positive and negative pressure can be generated. ✓
• More compact. ✓
• Easily maintain due to fewer working parts. ✓
  (Any 2 x 1) (2)

11.4 Belt-drive system:
11.4.1
(2)
11.4.2 Belt speed:
v = πDn
= π x 0,25 24
v = 18,85m.s^-1OR 18,85m/s (3)
11.4.3 Power transmitted:
P =(T₁ - T₂)v
= (300-120)18,85
= 180 x 18,85
= 3393Watt
= 3,39 kW
OR
P =(T₁ - T₂)πDn
= (300-120 π 0,25 24
= 180 x 18,85
= 3393Watt
= 3,39 kW (4)
11.5 Gear drive system:
Number of teeth on gear C:
N_A =T_B x T_D 
N_D T_A x T_CA
N_A = T_B x T_D x N_D
             T_A x T_C
= 80 x 60 x 120
        30 x 40
= 480 r/min
OR
N_C T_C = N_D T_D
N_C = N_D x T_D
             T_C 
=120 x 60
      40
= 180 r/min

N_A T_A = N_B T_B
N_A = N_B x T_B
             T_A
=180 x 80
      30
= 480 r/min (4)
[28]
TOTAL: 200