MECHANICAL TECHNOLOGY: WELDING AND METALWORK
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. ✔
• Remove all flammable material from the area. ✔ 
• Secure the work piece. ✔ 
• (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 / bright light ✔ 
• To protect your eyes from smoke ✔
• (Any 2 x 1) (2) 

2.3 PPE for Hydraulic press: 

•  Overall ✔ 
• Safety shoes ✔ 
• Safety goggle ✔ 
• Leather gloves ✔ 
• Leather apron ✔ 
• 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 ✔ 
• 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 produce 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 B ✔ (1)
4.2 A ✔ (1)
4.3 B ✔ (1)
4.4 B ✔ (1)
4.5 A ✔ (1)
4.6 B ✔ (1)
4.7 D ✔ (1)
4.8 D ✔ (1)
4.9 C ✔ (1)
4.10 C ✔ (1)
4.11 A ✔ (1)
4.12 D ✔ (1)
4.13 B ✔ (1)
4.14 B ✔ (1)

TOTAL QUESTION 4: [14]

QUESTION 5: TERMINOLOGY (TEMPLATES) (SPECIFIC) 
5.1 Template loft: 
The template loft is separated from the workshop because… 

•  it is quieter. ✔ 
• the lighting is better. ✔ 
• all equipment is at hand. ✔ 
• it is a permanent base. ✔ 
• marking on the floor enhance accuracy. ✔
• (Any 2 x 1) (2) 

5.2 Purpose of purlins: 

•  The purlins support ✔ the roof covering ✔ 
• Stabilizes ✔ the trusses. ✔
• (Any 1 x 2) (2) 

5.3 A steel ring calculation: 
5.3.1 Dimensions of the required material: 

• Mean diameter = Outside diameter - plate thickness
  = 880 - 50
  = 830mm
• Mean circurmference = π × Meandiameter = ×
  = π × 830 
  = 2607,52mm ✔
• 2608 mm of 50 x 50 mm ✔ square steel bar is required to  fabricate the ring. (7) 

5.3.2 

5.4 Resistance weld symbols: 
5.4.1 Spot weld ✔ (1)
5.4.2 Seam weld ✔ (1) 
5.5 Welding symbols: 

• A. Tail ✔
• B. Weld symbol (Fillet weld) ✔
• C. Pitch of weld ✔
• D. Site weld ✔
• E. Arrow ✔
• F. Weld all round ✔ (6)

TOTAL QUESTION 5: [23]

QUESTION 6: TOOLS AND EQUIPMENT (SPECIFIC) 
6.1. Working Principles: 
6.1.1 Guillotine: 

•  A bottom cutting blade is fixed horizontally. ✔ 
• With a top cutting blade moving downwards. ✔ 
• It is driven by an electric motor, flywheel, gearbox and  axle ✔ by eccentric motion / action / hydraulic action. ✔
• OR 
• It is activated manually by foot ✔ with lever action. ✔ (4) 

6.1.2 Bending rolls: 

•  A bending roll has two fixed rollers next to each other  rotating in unison (Manually or Electrical motor). ✔ 
• A third roller is adjustable, moving in between the two  rollers. ✔ 
• The third roller applies downward pressure onto the  metal. ✔ 
• That causes the metal to deflect and ultimately form the  round shape desired. ✔ (4) 

6.2. Regulators on gas cylinders: 

• Regulators reduce ✔ the cylinder pressure ✔ to operating or working  pressure. ✔ (3) 

6.3 Press machine: 

• The press machine is used for installing ✔ or removing ✔ components on mechanical devices / machines. ✔ 
• To press ✔ profiles ✔ onto material ✔ 
• (Any 1 x 3) (3) 

6.4 MIGS/MAGS welding process: 

• A – Weld pool / weld bead / molten metal ✔
• B – Electrode wire / electrode ✔
• C – Gas shroud / electrical contact / nozzle / contact tip ✔
• D – Shielding gas ✔ (4)

TOTAL QUESTION 6: [18]

QUESTION 7: FORCES (SPECIFIC) 
7.1 Forces in members: 
SCALE: Vector diagram 1 mm = 5 N 
(20) 

NOTE: 

• Use a tolerance of 2 mm + and – on the vector diagram.
• = a tolerance of 10 N + and – on the answer.  

7.2 Bending moments: 

7.2.1 Moments about RR 

• RL × 10  = (8× 8)+ (4× 5) +(6× 2)
  RL  = 96 
           10
  =  RL 9,6kN ✔
  Moments about RL
• RR ×  10= (6× 8) +(4 × 5) +(8×2)
  RR  =  84
             10
  = RR 8,4kN   (8) 

7.2.2 Bending moments at point A, B, C, D and E: 

• Scale2 mm = 1 kN.m 
  Momentat A   =  0 kN.m 
  B = RL × 2  = 19,2 kN.m
  C = (RL × 5)  - (8×3) = 24 kN.m
  D =(RL×8) -(8×6) - (4×3) = 16,8 kN.m
  E = (RL×10) -(8×8) - (4×5) - (6×2) = 0 kN.m   (5) 

7.2.3 
(5) 

NOTE: 

• Use a tolerance of 2 mm + and – on the bending moment diagram.  

7.3 Stress and strain: 

• A = πd²
         4    
  A = π(0,02)²
             4
  A = 0,314 ×10^-3m²   
• Stress  = Load 
                  Area
  Load = Stress c  Area
  Load = (80 × 10⁶) ×  (0,314 × 10^-3)
  Load = 25,133kN ✔(7) 

TOTAL QUESTION 7: [45] 

QUESTION 8: JOINING METHODS (WELD INSPECTION) (SPECIFIC) 
8.1 Factors to be observed during oxy-acetylene welding: 

•  Correct flame for the work on hand. ✔ 
• Correct angle of welding torch and welding rod. ✔ 
• Depth penetration and amount of fusion. ✔ 
• The rate of progress along the joint. ✔
• The distance of the nozzle from the parent metal. ✔
• (Any 2 x 1) (2) 

8.2 Abbreviation 'HAZ': 

• Heat Affected Zone ✔ (1)

8.3 Causes of weld defects: 
8.3.1 Spatter: 

• Disturbance in the molten weld pool. ✔ 
• Too low welding voltages. ✔ 
• Too high welding current / amps. ✔
• Inadequate shielding gas flow. ✔
• Too fast travel speed ✔
• Arc length too long ✔ 
• Wet electrode ✔ 
• Wrong polarity ✔ 
• Arc length too short ✔ 
• Wrong included electrode angle ✔ 
• Wrong electrode used ✔ 
• Arc blow ✔
• (Any 2 x 1) (2) 

8.3.2 Undercutting: 

•  Too fast travel speed ✔ 
• Rapid solidification ✔ 
• Too low arc voltage ✔
• Arc length too long ✔ 
• Excessive welding current ✔ 
• Too slow movement over weld ✔
• Current / amps too high ✔ 
• Electrode too big ✔ 
• Wrong electrode ✔ 
• Wrong included electrode angle ✔ 
• Excessive weaving ✔ 
• Wrong joint design ✔
• (Any 2 x 1) (2)

8.3.3 Incomplete penetration: 

• Welding current too low ✔ 
• Too fast travel speed ✔ 
• Incorrect electrode angle ✔
• Poor edge preperation ✔ 
• Insufficient root gap ✔ 
• Electrode too big ✔ 
• Wrong electrode ✔ 
• No pre-heating done ✔ 
• Wrong shielding gas used ✔ 
• Too long arc ✔
• (Any 2 x 1) (2) 

8.4 Types of cracks: 
8.4.1 Transverse cracks: 

• Pre-heating the base metal ✔ 
• Using lower strength consumables / welding rod ✔ 
• Slow cooling after welding ✔ 
• Use clamping device. ✔ 
• Weld toward the unrestrained side of the weld. ✔
• (Any 2 x 1) (2) 

8.4.2 Centreline cracks: 

•  Ensure that width-to-depth ratio is 1:1. ✔ 
• Decrease the current to decrease excess penetration. ✔ 
• Decreasing welding voltage setting or slowing travel  speed to achieve a flat to convex weld surface. ✔ 
• Use clamping device. ✔
• (Any 2 x 1) (2) 

8.5 Differences between non-destructive and destructive tests: ∙ Non-destructive test does not destroy the welded joint. ✔ 

•  Destructive test destroys the welded joint. ✔ (2) 

8.6 Ultrasonic test: 

•  No defects will occurs during a ultrasonic test ✔✔ 
• Detect internal ✔ flaws as well as surface flaws. ✔ 
• Porosity ✔✔ 
• Slag inclusions ✔✔
• Cracks ✔✔
• Any 1 x 2) (2)

8.7 Nick break test for internal defects: 

• Slag inclusion ✔ 
• Porosity ✔ 
• Lack of fusing ✔ 
• Oxidised metal ✔ 
• Burned metal ✔
• (Any 2 x 1) (2) 

8.8 Machinability test: 

•  To determine the hardness ✔ and strength ✔of the welded joint. 
• To determine ✔ the machinability. ✔
• Any 1 x 2) (2) 

8.9 Visual requirements of welds: 

•  Shape of the profile ✔
• Uniformity of the surface ✔ 
• Overlap ✔
• Free from any external defects ✔ 
• Penetration bead ✔
• Root groove ✔
• (Any 2 x 1) (2) 

TOTAL QUESTION 8: [23]

QUESTION 9: JOINING METHODS (STRESSES AND DISTORTION) (SPECIFIC) 
9.1 Residual stress: 

• Residual stresses are stresses that exist ✔ in a metal after cooling /  welding. ✔ (2) 

9.2 Factors affecting grain size: 

• The amount of cold work. ✔ 
• The temperature and time of annealing process. ✔ 
• The composition and constitution. ✔ 
• The recrystallisation temperature of cold worked metal. ✔ 
• The melting point. ✔
• (Any 2 x 1) (2) 

9.3 Quenching medias: 

•  Oil ✔ 
• Water ✔ 
• Sand ✔ 
• Air ✔ 
• Brine / Salt water ✔ 
• Lime ✔ 
• Liquid salts ✔
• Molten lead ✔ 
• Ash ✔
• (Any 2 x 1) (2) 

9.4 Weld distortion: 

•  Distortion in a weld results from the uneven expansion and  contraction (warping) ✔ of the weld metal ✔ and adjacent base metal  ✔ during the heating and cooling cycle ✔ of the welding process. (4) 

9.5 Factors that affect distortion and residual stress: 

•  If the expansion that occurs when metal is heated is resisted ✔ then  deformation will occur. ✔ 
• When contraction that occurs on cooling is resisted ✔then a stress  will be applied. ✔ 
• If this applied stress causes movement ✔ then distortion occurs. ✔ 
• If the applied stress does not cause movement ✔ then there will be  residual stress in the welded joint. ✔
• (Any 2 x 2) (4) 

9.6 Result when metal is cooled rapidly: 

•  Rapid cooling of metal results in large temperature differences ✔ between the internal and external areas ✔ of the metal that set up  stresses, ✔ which cause cracks ✔ on the surface. 
• It will harden ✔✔ and the grain structure ✔ will change. ✔
• (Any 1 x 4) (4) 

TOTAL QUESTION 9: [18]

QUESTION 10: MAINTENANCE (SPECIFIC) 
10.1 Reasons maintenance: 

•  Promote cost saving ✔ 
• Improves safety ✔ 
• Increases equipment efficiency ✔
• Fewer equipment failure ✔ 
• Improves reliability of equipment ✔
• (Any 2 x 1) (2) 

10.2 Lockout on machines: 

• To ensure that nobody can turn on the machine ✔ while maintenance is  being carried out. ✔ (2) 

10.3 Reasons for service records: 

•  Assist in the monitoring of the condition of the machines. ✔ 
• Assist in upholding warrantees. ✔ 
• Assist in keeping a history of maintenance and repairs. ✔
• (Any 2 x 1) (2) 

10.4 Methods of reducing friction: 

•  By reducing both drill speed and feed speed. ✔ 
• By applying lubrication. (cutting fluid) ✔ 
• Use the correct drill bit ✔
•  Drill a pilot hole ✔
• (Any 2 x 1) (2) 

TOTAL QUESTION 10: [8]

QUESTION 11: TERMINOLOGY (DEVELOPMENT) (SPECIFIC) 
11.1 Use of transformers: 

• Transformers are used to connect ✔ ducting sections of dissimilar ✔ shapes to each other. ✔ (3) 

11.2
@1
11.3 Truncated cone: 

11.3.1 Base circumference: 

• Circumference =  π × Basediameter
  =  π × 1400
  = 4398,23 mm ✔(3) 

11.3.2 Main radius (AC): 

• Triangles ABC and CED has the same shape:
  AC : DC BC :EC
  Thus AC = BC
           DC    EC
  From where AC = BC × DC 
                                   EC
  and CE = Base Dia -  800
                               2 
  = 1400 - 800
            2
  CE = 300 mm
  For : DC
  DC² = DE² + CE²
  DC = √1200² + 300²
  DC = 1236,93mm
  rounded = 1237mm 
  AC = BC × DC
              EC
  = 700 × 1237
          300
  = 2886,17mm ✔ ✔
  rounded 2886mm (10)

11.3.3 Small radius (AD): 

• AD = AC - DC
  = 2886 - 1237
  AD = 1649 mm(1649,24mm) ✔ (3) 

TOTAL QUESTION 11: [21]
GRAND TOTAL: 200