MECHANICAL TECHNOLOGY: WELDING AND METALWORK
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 (SPECIFIC)
4.1 D ✓ (1)
4.2 B ✓ (1)
4.3 A ✓ (1)
4.4 B ✓ (1)
4.5 D ✓ (1)
4.6 B ✓ (1)
4.7 D ✓ (1)
4.8 C ✓ (1)
4.9 A or B ✓ (1)
4.10 C ✓ (1)
4.11 A or B ✓(1)
4.12 B ✓ (1)
4.13 A ✓(1)
4.14 C ✓ (1)
[14]

QUESTION 5: TERMINOLOGY (TEMPLATES) (SPECIFIC)
5.1 Template loft:

• To save time in marking out. ✓
• Promotes accuracy.✓
  (Any 1 x 1) (1)

5.2 Purlins:

• To support roof covering. ✓✓
• To link the roof trusses. ✓✓
• Makes the roof structure stronger. ✓✓
  (Any 1 x 2) (2)

5.3 Roof truss:

1. Rafter ✓
2. Cleat ✓
3. Purlin ✓
4. Gusset plate ✓
5. Tie beam/Main tie✓(5)

5.4 Material calculation:
MeanØ = InsideØ + Thickness
= 230 + 16
= 246 mm 
Mean circumferance = π x MeanØ
= π x 246
= 772,83 mm
= Round off to 773 mm (6)
5.5 Welding symbols:

1. Tail ✓
2. Weld symbol / Fillet weld on the other side / Weld symbol on the other side / Fillet weld ✓
3. Pitch of weld✓
4. Site weld ✓
5. Arrow✓
6. Weld all round✓ (6)

5.6
(3)
[23]

QUESTION 6: TOOLS AND EQUIPMENT (SPECIFIC)
6.1 Plasma cutter:

• Creating an electrical channel of ionised gas (plasma), ✓ from the plasma cutter itself through the work piece that is being cut.
• It forms a completed electric circuit ✓via a grounding clamp.
• Compressed air is blown toward the work piece through a focused nozzle at high speed. ✓
• A high frequency, electrical arc is then formed within the gas between an electrode near or integrated into the gas nozzle and the work piece itself. ✓ (4)

6.2 Hydraulic press:

• For removing bearings or bushes✓
• Fitting of bearings or bushes. ✓
• To shape material.✓
• Testing of welded joints ✓
  (Any 2 x 1) (2)

6.3 Internal thread cutting process:

• Drill the required core diameter. ✓
• Use the three taps in order – taper / intermediate / plug. ✓
• Check thread with thread pitch gauge/bolt when complete.✓ (3)

6.4 Power saw:
To cut sections of metal / material. ✓ (1)
6.5 Gas welding:
6.5.1 Oxygen regulator / Acetylene regulator / regulator ✓(1)
6.5.2

1. Gauge✓
2. Outlet ✓
3. Inlet ✓
4. Pressure adjusting knob ✓(4)

6.6 Acetylene gas cylinder:
Red / maroon ✓ (1)
6.7 Flashback arrestor:
To prevent ✓ back feeding / flashback of flame ✓ (2)
[18]

QUESTION 7: FORCES (SPECIFIC)
7.1 Define:
7.1.1 Stress:
The internal resistance ✓ in a body to an external force or load. ✓(2)
7.1.2 Hooke's law:
Strain is directly proportional to the stress it causes; ✓ provided the limit of elasticity is not exceeded✓ (2)
7.2 Frameworks:
7.2.1. Space diagram: (4)

7.2.2 Vector diagram: (5)

NOTE: ±2mm tolerance on scale drawing. Marks awarded for scale accuracy.
7.2.3 Magnitude and nature of members:

(6)
7.3 Beam:
7.3.1 Calculate RL:
Taking moment about right reaction (RR)
RL x 10 = (25 x 2) + (30 x 6,5) + (15 x 8)
= 50 + 195 + 120
= 365
    10
RL = 36,5 N

Calculate RR:
Taking moment about left reaction (RL)
RL x 10 = (15 x 2) + (30 x 6,5) + (25 x 8)
= 30 + 105 + 200
= 335
    10
RL = 33,5 N (6)

7.3.2 Shear forces at point A, B and C:
SF_A = 36,5 15
= 21,5 N
SF_B = 36,5 - 15 - 30
= -8.5N
SF_C = 36,5 - 15 - 30 - 25
= -33.5N (6)

7.3.3 Shear force diagram: (6)

7.4 Stress and strain:
7.4.1 Stress:
Stress =  Load   But Area = πD²  
                Area                      4
Area = πD²  
            4
= π(0,03)²
       4
= 0,71 x 10^-3 m² or 7,07 x 10^-4 m²

Stress = Force 
               Area
=    80 x 10³ N 
    0,71 x 10^-3m²
=112,68 x 10⁶ Pa
= 112,68 MPa

OR
Stress = Force 
               Area
=    80 x 10³ N 
    0,71 x 10^-3m²
= 113154172,6 Pa
= 113,15 MPa (6)

7.4.2 Strain:
Strain = ΔL 
             OL
=  0,06  
   3000
= 0,02 x 10^-3
(If any unit indicated, then NO mark awarded for final answer) (2)
[45]

QUESTION 8: JOINING METHODS (INSPECTION OF WELDS) (SPECIFIC)
8.1 Welding defects (Causes):
8.1.1 Slag inclusion:

• Included angle too narrow. ✓
• Rapid chilling.✓
• Welding temperature to low / current too low.✓
• High viscosity of molten metal.✓
• Slag not removed from previous weld run. ✓
• Incorrect welding technique.✓
• Surface contamination. ✓
• Too big weaving action. ✓
• Too slow speed along the weld joint. ✓
• Too short arc length.✓
  (Any 2 x 1) (2)

8.1.2 Incomplete penetration:

• Speed too fast.✓
• Poor welding technique. ✓
• Electrode too large. ✓
• Current too low. ✓
• Joint preparation not prepared correctly.✓
• Weldability of parent metal not good.✓
  (Any 2 x 1) (2)

8.2 Welding defects (Prevention):
8.2.1 Porosity:

• Use correct current. ✓
• Hold a longer arc.✓
• Use correct electrodes.✓
• Check for impurities.✓
• Ensure adequate shielding gas. ✓
• Correct welding technique. ✓
• Check that electrode/ filler metal did not rust.✓
  (Any 2 x 1) (2)

8.2.2 Lack of fusion:

• Use correct included angle.✓
• Use the correct size of electrode. ✓
• Use the correct current setting. ✓
• Prepare the plate bevel/V-groove accordingly. ✓
  (Any 2 x 1) (2)

8.3 Destructive and non-destructive tests:
8.3.1 Free-bend:

• Used to determine the percentage of elongation of the welded metal.✓
• To determine the ductility of the weld metal and heat affected area. ✓
  (Any 1 x 1) (1)

8.3.2 X-ray test:

• To determine whether there has been full depth penetration.✓
• Determine if correct fusion between welded pieces took place. ✓
• To detect internal defects like pin holes, slag inclusions, cracks etc.✓
  (Any 1 x 1) (1)

8.4 Welding cracks:

• Heat affected zone (HAZ) cracks. ✓
• Centre line / longitude cracks. ✓
• Crater cracks.✓
• Transverse cracks✓
  (Any 3 x 1) (3)

8.5 Oxy-acetylene welding process:

• Correct flame for the work on hand.✓
• Correct angle of nozzle.✓
• Correct angle of rod. ✓
• Depth of fusion.✓
• The amount of penetration. ✓
• The rate of progress along the joint.✓
  (Any 2 x 1) (2)

8.6 Nick-break test:

• Each side of the weld is slotted by means of a saw.✓
• Place the specimen on two steel supports / In a bench vice.✓
• Break the specimen ✓ by striking it with a hammer. ✓
• Inspect the weld metal for exposed defects. ✓ (5)

8.7 Non-destructive tests:

• It does not involve the destruction/damage of the test piece ✓
• The test piece can still be used after test is done. ✓
  (Any 1 x 1) (1)

8.8 Machinability test:

• To determine the ease of machining ✓
• To determine the quality of the finish ✓ (2)

[23]

QUESTION 9: JOINING METHODS (STRESSES AND DISTORTION) (SPECIFIC)
9.1 Cold worked steel:

• Melting point ✓
• Its composition and constitution ✓
• The amount of cold work✓
• Annealing time✓ (4)

9.2 Shrinkage in a welded joint:
9.2.1 Electrode type:
Thermal properties have a greater potential to cause deformation.✓ (1)
9.2.2 Electrode size:
The larger the electrode diameter the higher the current the greater the deformation.✓(1)
9.2.3 Welding current:
The higher welding current the higher the welding temperature the higher the deformation. ✓ (1)
9.3 Factors that determine the cooling rate:

• Size of work piece ✓
• Weld thickness ✓
• Thermal conductive properties of parent metal ✓
  (Any 2 x 1) (2)

9.4 Definition:
9.4.1 Distortion:
Weld distortion is the warping of the base metal ✓ caused by heat from the welding arc/flame. ✓ (2)
9.4.2 Shrinkage:
Weld shrinkage is a form of plastic deformation ✓ where the metal has deformed as a result of contraction on cooling.✓ (2)
9.5 Factors affecting distortion and residual stress:

• When the metal is heated and expansion is resisted then deformation will occur. ✓
• When cooling occurs and contraction is resisted, then stress will occur. ✓
• If applied stress causes movement, the distortion occurs. ✓
• If applied stress does not cause movement then there will be residual stress in the welded joint. ✓
  (Any 3 x 1) (3)

9.6 Causes of residual stress:

• During welding, the welds and Heat Affected Zone (HAZ) are heated to temperatures well above those of the surrounding material✓
• The weld and HAZ deform plastically because their thermal expansion is restricted by the surrounding material✓
• As the weld cools and contracts, tensile stresses develop elastically.✓
• Welds develop tensile stresses that approach yield stress✓
  (Any 2 x 1) (2)

[18]

QUESTION 10: MAINTENANCE (SPECIFIC)
10.1 Overloading:
10.1.1 Shearing machines:

• Dulling or breaking blades. ✓
• Putting strain on the motor and drive mechanism. ✓
  (Any 1 x 1) (1)

10.1.2 Drill press:

• Damage / breakage to the drill bit.✓
• It puts strain on the drive components. ✓
  (Any 1 x 1) (1)

10.2 Friction:
10.2.1 Guillotine:
Excessive wear / damage to moving parts. ✓ (1)
10.2.2 Horizontal band saw:

• Overheating of the cutting blade✓
• Damage to the cutting blade. ✓
• Excessive wear to moving parts. ✓
  (Any 1 x 1) (1)

10.3 Maintenance of a power saw:

• Check the mains electrical switches. ✓
• Check the wiring and conduits for cracks. ✓
• Check for broken control mechanisms. ✓
• Check electrical connections.✓
• Check for loose electrical components. ✓
• Check that cutting fluid does not come in contact with electrical wiring and switches. ✓
  (Any 2 x 1) (2)

10.4 Methods to reduce friction:

• Applying cutting fluid. ✓
• Applying oil.✓
• Prevent excessive pressure / Apply adequate pressure.✓
• Ensure that the drill bit is sharp.✓
• Ensure to use correct speed for the size of drill bit✓
• Use the correct drill bit. ✓
  (Any 2 x 1) (2)

[8]

QUESTION 11: TERMINOLOGY (DEVELOPMENT) (SPECIFIC)
11.1 Square to square off centre hopper:

11.1.1 True length of A-2:
True length (A - 2) = √240² + 280² + 400² 
= √57600 + 78400 + 160000
= √296000
= 544,06 mm ≈ 544 mm (5)

11.1.2 True length of C-3:
True length (C - 3) = √220² + 60² + 400² 
= √48400 + 3600 + 160000
= √212000
= 460,43mm ≈ 460mm
11.2 Truncated cone:

11.2.1 True length of A-B:
True length(A - B) = πD 
                                 12
= π x 600 
      12
= 1884,96
       12
= 157,08 mm ≈ 157 mm (5)
11.2.2 Circumference of the top circle:
Circumference of top circle =  π x D
= π x 400
= 1256,64mm ≈ 1257mm (4)
11.2.3 600 mm. (2)
[21]
TOTAL: 200