MECHANICAL TECHNOLOGY: FITTING AND MACHINING
GRADE 12
NATIONAL SENIOR CERTIFICATE EXAMINATION
MEMORANDUM
MAY/JUNE 2021

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

QUESTION 2: SAFETY (GENERIC)
2.1 First aid basic treatment:

• Examination ✓
• Diagnosis ✓
• Treatment ✓ (3)

2.2 Drill press (Already been switched on):

• Never leave the drill unattended while in motion.✓
• Switch off the drill when leaving. ✓
• Use a brush or wooden rod to remove chips. ✓
• When reaching around a revolving drill, be careful that your clothes do not get caught in the drill or drill chuck. ✓
• Don't stop a revolving chuck with your hand. ✓
• Don't adjust the drill while working. ✓
• Don't open any guard while in motion. ✓
• Keep hands away from action points. ✓
• Do not force the drill bit into the material.✓
• Apply cutting fluid if required. ✓
  (Any 2 x 1) (2)

2.3 Isolation of electrode holder:
To prevent electric shock. ✓ (1)
2.4 Disadvantages of the process layout:

• Production is not always continuous. ✓
• Transportation costs between process departments may be high. ✓
• Additional time is spent in testing and sorting as the product moves to the different departments. ✓
• Damage to fragile goods may result from extra handling. ✓
  (Any 2 x 1) (2)

2.5 Advantages of the product layout:

• Handling of material is limited 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]

QUESTION 3: MATERIALS (GENERIC)
3.1 Heat-treatment:

• Heat the metal slowly to a certain temperature. ✓
• Soak the metal for a certain period to ensure a uniform temperature. ✓
• Cool the metal at a certain rate to room temperature. ✓ (3)

3.2 Quenching mediums:

• Water ✓
• Brine✓
• Liquid salts ✓
• Oil ✓
• Soluble oil and water ✓
• Sand ✓
• Molten lead ✓
• Air ✓
• Lime ✓
  (Any 3 x 1) (3)

3.3 Annealing:

• To relieve internal stresses of the steel ✓
• Soften steel to make machining possible ✓
• Make steel ductile ✓
• Refine grain structure ✓
• Reduce brittleness ✓
  (Any 1 x 1) (1)

3.4 Carbon steels:

• Low carbon steel ✓
• Medium carbon steel ✓
• High carbon steel ✓ (3)

3.5 Iron-carbon equilibrium diagram:

1. Percentage carbon / carbon content ✓
2. Temperature in °C ✓
3. AC3 line / Higher critical temperature ✓
4. AC1 line / Lower critical temperature ✓ (4)

[14]

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

QUESTION 5: TERMINOLOGY (LATHE AND MILLING MACHINE) (SPECIFIC)
5.1 Disadvantages of compound slide method:

• The automatic feed of the machine cannot be used.✓
• Causes poor finish. ✓
• Only short tapers can be cut. ✓
• It causes fatigue in the operator. ✓
  (Any 3 x 1) (3)

5.2 Taper calculations:
5.2.1 Diameter of taper:
tan θ = D - d
      2     2 x l
tan 10  = 165 - d
       2       2 x 10
420 tan5º = 165 - d
d =165 - 36,75
d =128,25 mm (4)

5.2.2 Tailstock set-over:
x = L(D - d)
       2 x l
x = 325(165 - 128,25)
             2 x 210
x = 28,44 mm (3)

5.3 Calculation of parallel key:
5.3.1  Width = ^D/₄
= ⁵⁵/₄
= 13,75 mm (2)

5.3.2 Thickness = ^D/₆
= ⁵⁵/₆
= 9,17 mm (2)

5.3.3 Length = 1,5 diameterof shaft
= 1,5 x 55
= 82,5 mm (2)
5.4 Advantages of up-cut milling:

• Heavier cuts can be taken. ✓
• When hard steels are cut, the total cutting pressure is absorbed by the material at the back of the edge. ✓
• When milling material with a hard scale, the cut is started under the scale where material is softer, extending the life of the cutter✓
• A courser feed can be used. ✓
• The strain on the cutter and arbor is less. ✓
• Less vibration experienced on machine. ✓
  (Any 2 x 1) (2)

[18]

QUESTION 6: TERMINOLOGY (INDEXING) (SPECIFIC)
6.1 Gear calculations:
6.1.1 Number of teeth:
Module = PCD
                  T
T = PCD
        m
=136
    4
= 34 teeth (2)

6.1.2 Dedendum:
Dedendum = 1,157(m)   = 1,25(m)
= 1,157 x 4  OR             = 1,25 x 4
= 4,63 mm                      = 5mm (2)

6.1.3 Outside diameter:
OD = PCD + 2(m)    = m(T + 2)
= 136 + 2(4)    OR   = 4(34 + 2)
= 144 mm                 = 144 mm (2)

6.1.4 Circular pitch:
CP = m x π
= 4 x π
= 12.57mm (2)

6.2 Dove tail calculations:
w = 190 – 2(DE)
M = w + 2 (AC) + 2 (R) or M = w + 2 ( AC +R)

6.2.1 Minimum width of dove tail (w):
Calculate DE:
tan a = DE 
            AD
DE = ADtan a
= 38 tan30º
= 21,94mm

OR
tan θ = AD 
            ED
tan60º = 38 
              ED
ED =   38   
         tan60º
= 21,94 mm

w = 190 - 2(DE)
= 190 - 2(21,94)
= 190 - 43,88
= 146,12mm (6)

6.2.2 Distance over the rollers (M):
Calculate AC:
tan α = BC 
            AC
AC =  BC  
         tan a
=   15   
  tan30º
= 25,98mm
M = w + 2(AC) + 2(R)
= 146,12 + 2(25,98) + 2(15)
= 146,12 + 51,96 + 30
= 228,08mm

OR
tan θ = CA 
            BC
CA = BCtanθ 
= 15 tan60º
= 25,98 mm
M = w + 2(AC + R)
= 146,12 + 2(25,98) + 2(15)
= 146,12 + 81,96
= 228,08mm(6)

6.3 Milling of spur gear:
6.3.1 Indexing:
Indexing = ⁴⁰/_n
Indexing = ⁴⁰/_A
= ⁴⁰/₁₆₀
= ¼ x ⁶/₆
= ⁶/₂₄ 
Approximate indexing:
No full turns and 6 holes on a 24-hole circle ✓
OR
No full turns and 7 holes on a 28-hole circle ✓ (3)

6.3.2 Change gears:
D _DR = (A - n) x ⁴⁰/_A
D_DN
D _DR = (160 - 163) x ⁴⁰/₁₆₀
D_DN
= -3 x ⁴⁰/₁₆₀
=-120  
   160
= ³/₄ x ⁸/₈
D _DR = ²⁴/₃₂
D_DN (5)
[28]

QUESTION 7: TOOLS AND EQUIPMENT (SPECIFIC)
7.1 Reading:
Reading = 7,90 mm (2)
7.2 Brinell hardness test:

• Select the desired load to apply to the specimen. ✓
• The specimen is raised to be in contact with the Brinell ball by turning the hand wheel. ✓
• The load is then applied for about 15 - 30 seconds ✓
• Release the load from the specimen. ✓
• Measure the diameter of the impression.✓
• Determine the Brinell hardness number. ✓ (6)

7.3 The tensile tester:

• Yield stress ✓
• Ultimate / maximum tensile stress ✓
• Elongation percentage ✓
• Break stress ✓
• Limit of proportionality✓
• Elastic limit ✓
• Strain ✓
• Ductility ✓
  (Any 3 x 1) (3)

7.4 Screw thread micrometer:
Identify:
7.4.1 Screw thread micrometer ✓ (1)
Function:
7.4.2 Measure the pitch diameter ✓ of a screw thread. (1)
[13]

QUESTION 8: FORCES (SPECIFIC)
8.1 Magnitude and direction of the equilibrant:

8.1.1 Sum of the horizontal components (HC):
ΣHC = 280cos45º + 120cos0º - 150cos90º - 250cos30º
= 197,99 + 120 - 0 - 216,51
= 101,48 Ν

OR

Related Items

(4)
8.1.2 Sum of the vertical components (VC):
ΣVC = 280sin45º + 120sin0º - 150sin90º 250sin30º
= 197,99 + 0 - 150 - 125
= -77,01 N

OR

(4)
8.1.3 Magnitude of the equilibrium force:
E² = VC² + HC²
E = √(77,01)² + (101.48)²
= 127,39 N (3)
8.1.4 Direction of the equilibrium force:

tanθ = VC 
           HC
tanθ = 77.01 
          101,48
θ = 37,19º
E = 127,39 N at 37,19° N of W (3)
8.2 Magnitudes of the reactions in supports A and B:

Calculate A:
Take moments about B:
ΣCWM = ΣACM
A x 6 = (285 x 2,7) + (165 x 4,2) + (345 x 5,4)
A x 6 = 769,5 + 693 + 1863
A x 6 = 3325,5
A = 3325,5
           6
A = 554,25 N

Calculate B:
Take moments about A:
ΣCWM = ΣACM
(345 x 0,6) + (165 x 1,8) + (285 x 3,3) = 6 x B
207 + 297 + 940,5 = 6 x B
1444,5 = 6 x B
1444,5 = B
    6
= 240,75 N = B (8)

8.3 Stress and Strain:
8.3.1 The resistance area of the bush:
A = π(D² - d²) 
            4
A = π(0,058² - 0,042²) 
A  =1,26 x 10^-3 m² (2)

8.3.2 The stress in the material:
σ = ^F/_A
=  50 x 10 ³
   1,26 x 10^-3
= 39682539,68 Pa
= 39,68 MPa (3)

8.3.3 Strain:
ε = ^Δl/_l
= 0,975
     68
= 14,34 x 10^-3
(If any unit indicated, then NO mark for final answer) (3)

8.3.4 Young's modulus:
E = ^σ/_ε
= 39,68 x 10⁶
   14,34 x 10
= 2,77 x 10⁹ Pa
= 2,77 GPa (3)
[33]

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

• Risk of injury or death. ✓
• Financial loss.✓
• Damage to parts.✓
• Loss of production time. ✓
  (Any 2 x 1) (2)

9.2 Malfunctioning of chain drives:

• Uncovered chain drives not cleaned. ✓
• Tensioning device is not working efficiently.✓
• Chain is not inspected regularly for elongation. ✓
• Chain drive is not aligned. ✓
• Wear and tear of chain. ✓
• Wear of sprocket teeth.✓
• Lack of lubrication. ✓
• Chain drive has been overloaded. ✓
  (Any 2 x 1) (2)

9.3 Wear on a gear drive system:

• Checking and replacement of lubrication levels. ✓
• Ensuring that gears are properly secured to shaft. ✓
• Cleaning and replacement of oil filter. ✓
• Reporting excessive noise, wear, vibration and overheating for expert attention. ✓
• Cleaning of gears regularly. ✓
  (Any 2 x 1) (2)

9.4 Property of materials:
9.4.1 Polyvinyl chloride (PVC):

• Can be re-heated and re-shaped ✓
• Flexible ✓
• Rubber like substance and makes a dull sound when dropped.✓
• Can be modified to suit most applications. ✓
• Can be welded (plastic welding). ✓
• Can be bonded with an adhesive. ✓
• Weather resistant ✓
• Water proof ✓
• Easy to work with. ✓
• Light weight ✓
• Recyclable ✓
• Corrosion resistant ✓
  (Any 2 x 1) (2)

9.4.2 Carbon fibre:

• Cannot be re-heated and re-shaped ✓
• Tough and strong material. ✓
• Light weight ✓
• Weather resistant✓
• Heat resistant✓
• Enhance strength of plastic by entrenchment✓
• Highly electrically conductive✓
  (Any 2 x 1) (2)

9.4.3 Bakelite:

• Electrically non-conductor (electrical insulator) ✓
• Heat resistant ✓
• Well moulded into specific shapes ✓
• Weather resistant✓
• Cannot be re-heated and re-shaped ✓
  (Any 2 x 1) (2)

9.5 Thermoplastic composites or thermo-hardened (thermosetting) composites:
9.5.1 Vesconite:
Thermoplastic ✓ (1)
9.5.2 Glass fibre:
Thermo-hardened/Thermosetting ✓ (1)
9.5.3 Carbon fibre:
Thermo-hardened/Thermosetting ✓ (1)
9.6 Uses of materials.
9.6.1 Teflon:

• Orthopaedic and prosthetic appliances✓
• Hearing aids ✓
• Joints✓
• Upholstery✓
• Mechanical parts (e.g., taps and bearings)✓
• Electrical insulation✓
• Non-stick coatings✓
  (Any 1 x 1) (1)

9.6.2 Carbon fibre:

Sporting and leisure equipment like: Tennis rackets, squash rackets, badminton rackets, golf clubs, hockey sticks 

• Model airplanes ✓
• Bicycle frames✓
• Ski’s✓
• Surf boards ✓
• Boat masts ✓
• Compressor blades✓
• Self- lubricating gears ✓
• Artificial satellites ✓
• Helicopter blades ✓
• Car bodies✓
• Airplane parts (fuselage) ✓
  (Any 1 x 1) (1)

9.6.3 Nylon:

• Bushes ✓
• Gears ✓
• Pulleys ✓
• Fishing line✓
• Ropes ✓
  (Any 1 x 1) (1)

[18]

QUESTION 10: JOINING METHODS (SPECIFIC)
10.1 Square Thread:
10.1.1 Mean diameter:
Pitch =        Lead          
           Numberof starts
= ⁴⁰/₂
= 20 mm
D_m = OD - ^P/₂
= 85 - ²⁰/₂
= 75 mm (4)

10.1.2 Helix angle of the thread:
tanθ =  Lead  
           π x D_M
=    40    
   π x 75
θ = tan^-1(0,169765272)
= 9,63º or 9 38' (4)

10.1.3 Leading tool angle:
Leading tool angle = 90° -( helix + clearance angle)
= 90° -( 9,63° + 3)
= 77,37° or 77 22' (2)

10.1.4 Following tool angle:
Following tool angle = 90° +( helix angle - clearance angle)
= 90° +( 9,63° - 3°)
= 96,63° or 96 38' (2)

10.2 Screw thread label:

1. Pitch diameter/mean/effective ✓
2. Helix angle ✓
3. Pitch / Lead ✓
4. Root/Root length ✓ (4)

10.3 Uses of square thread:

• Vice screws ✓
• Brake screws ✓
• Lead screws of lathe machines ✓
• Scissor jacks✓
• Milling machine table feed screws ✓
• Hydraulic jacks (Adjustable top) ✓
  (Any 2 x 1) (2)

[18]

QUESTION 11: SYSTEMS AND CONTROL (DRIVE SYSTEMS) (SPECIFIC)
11.1 Hydraulic calculations:
11.1.1 The fluid pressure in MPa:
Area:
A_A = πD²_A
           4
= π(0.025)² 
        4
= 0,49 10 m OR 4,9 1 10 m
Pressure:
P = ^F/_A
= 1,32 x 10 ³
   0,49 x 10^-3
= 2,69 x 10⁶Pa
=  2,69 MPa (4)

11.1.2 The diameter of piston B:
P_B = P_A
F_B= F_A
A_B    A_A
6,45 x 10³ = 1,32 x 10³
     A_B           0,49 x 10^-3
6,45 x 10³ = 2,69 x 10⁶
      A_B 
A_B = 6,45 x 10³
        2,69 x 10⁶
A_B = 2,40 x 10^-3
A_B  = πD_B²  
            4
D_B = √ 4A_B 
            π
= √4(2,40 x 10^-3)
              π
= 0,05528m
= 55,28 mm (6)

11.2 Advantages of chain drive system over belt drive systems:

• No slipping or creep occurs. ✓
• Higher efficiency.✓
• Longer life span.✓
• Does not generate heat.✓
• Does not undergo the same degrading effects of what time has on belts.✓
• Much stronger✓
• Faster speeds can be obtained.✓
  (Any 2 x 1) (2)

11.3 Functions of hydraulic reservoir:

• A fluid storage tank. ✓
• Promotes air separation from the fluid. ✓
• Support for the pump and electric motor. ✓
• Promotes heat dispersion. ✓
• Acts as a base plate for mounting control equipment.✓
• It allows for expansion or contraction of the hydraulic system. ✓
  (Any 2 x 1) (2)

11.4 Application for hydraulic systems:

• Machine tools ✓
• Clutch systems ✓
• Brake systems ✓
• Aircraft ✓
• Jacks ✓
• Missiles ✓
• Ships✓
• Earth moving equipment ✓
• Punch machines ✓
• Turbines ✓
• Tractor lifts ✓
• Car lifts ✓
• Machine vices ✓
• Jaws of life ✓
• Trains ✓
  (Any 1 x 1) (1)

11.5 Belt drive:
11.5.1 Rotational frequency:
N_DR x D_DR = N_DN x D_DN
N_DR x 95 = 85 x 255
N_DR = 85 x 255
               95
N_DR =228,16 r/min
OR
N = 3,8 r/sec (3)

11.5.2 Speed ratio:
Speed ratio = Diameter of driven pulley
                       Diameter of driver pulley
Speed ratio = 255 
                       95
Speed ratio = 2,68 : 1

OR
Speed ratio = Frequency of driven pulley
                      Frequency of driver pulley
Speed ratio = 228
                       85
Speed ratio = 2,68 : 1 (3)

11.6 Gear drive:
11.6.1 Rotation frequency:
N_A = Product of the numberof teeth on driven gears
N_F    Product of the numberof teeth on driving gears
N_F = Product of the numberof teeth on driving gears
N_A    Product of the numberof teeth on driven gears
N_F = T_A x T_C x T_E x N_A 
            T_B x T_D x T_F
= 30 x 20 x 50 x 2500
         40 x 60 x 70
= 446,43 r/min
OR
7,44 r/sec  (4)

11.6.2 Gear ratio:
GearRatio = Product of the numberof teeth on driving gears
                     Product of the numberof teeth on driven gears
= 40 x 60 x 70
  30 x 20 x 50
= 168000
    30000
= 5,6 : 1

OR
Speed ratio = N_inoput 
                      N_output
=  2500  
  446,43
= 5,6:1 (3)
[28]
TOTAL: 200