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Chapter 12 · 4 hours

Material Joining Processes

Practice questions

Practice questions and answers

3 exam-style questions on this chapter, written for this site from the official syllabus. We haven’t found past IOE papers for this subject yet; if you have some, share them in the community.

  • Practice · 5 marks

Classify material joining processes. Compare welded joints with mechanical (riveted/bolted) connections and state when brazing and soldering are preferred over welding.

Answer

Joining is the bringing together of two or more parts to make a single assembly, either permanently or so that it can be taken apart.

Classification

ClassPrincipleProcesses
WeldingParent metals melted (with or without filler) or forced together to form an atomic bondFusion: arc, gas, laser; solid state: friction, resistance, forge
Brazing and solderingFiller metal melts, parent metal does not; filler flows by capillarityBrazing (above 450 degree C); soldering (below 450 degree C)
Adhesive bondingPolymer adhesive bonds surfacesEpoxy, cyanoacrylate
Mechanical fasteningParts held by fasteners or interferenceBolts, rivets, screws, press fits, seaming

Welded and mechanical connections

PointWelded jointRiveted / bolted joint
Joint efficiencyHigh, up to 100 % of parent metal60-80 % (holes weaken plate)
WeightLighter; no plates or headsHeavier; cover plates, fasteners
Leak tightnessGoodNeeds caulking, sealing
DisassemblyNot possible (permanent)Possible (bolts)
Cost and speedLower labour, fasterMore labour, drilling/punching
Residual stress, distortionPresent, HAZ changes propertiesNone
Skill, inspectionSkilled welder; NDT neededSimple inspection
FatigueNotch at weld toeStress concentration at holes

Brazing and soldering

They are preferred when parts are thin or heat-sensitive, when dissimilar metals (steel to copper, carbide tip to steel shank) or very small assemblies are joined, when low distortion and a neat appearance are needed, and for electrical joints (solder). Welding is preferred where high strength and heavy sections are required.

  • Practice · 6 marks

Describe the metallurgy of a fusion weld, showing the zones of the weld and the heat-affected zone (HAZ) in steel. List the common weld defects, and compare the characteristics of the energy sources used in welding.

Answer

Metallurgy of a fusion weld

During welding the parent metal and filler melt in the weld pool and then freeze. The freezing begins at the unmelted grains of the parent metal (epitaxial growth) and columnar dendrites grow toward the weld centre, in the direction of heat flow. The rapid cooling gives a fine, cast structure with segregation of solute.

 Weld bead
  _______________________
 | fusion zone (weld metal)   cast, columnar grains
 |------ fusion boundary ----
 | HAZ: coarse grain (near the boundary)
 |      fine grain (normalised)
 |      partly transformed
 |------------------------------
 | unaffected parent metal

Zones (from the weld centre outward):

  1. Fusion zone - melted and resolidified metal, with the composition of filler and parent.
  2. Fusion (partially melted) boundary.
  3. Heat-affected zone (HAZ) - not melted, but heated high enough to change the structure: coarse-grained region next to the fusion line (hard, brittle, with martensite in hardenable steels), then a fine-grain normalised region, then a partially transformed region.
  4. Parent metal - unaffected.

The HAZ properties depend on peak temperature, time at temperature and cooling rate. High carbon equivalent steels form martensite in the HAZ, giving hydrogen-induced cold cracks; this is avoided by preheating, low-hydrogen electrodes, and controlled heat input.

Common defects

  • Porosity (gas trapped), slag inclusion
  • Lack of fusion and incomplete penetration
  • Undercut and overlap
  • Cracks - hot cracks (solidification), cold cracks (hydrogen)
  • Distortion and residual stress

Energy sources compared

SourceTypical power densityFeatures
Oxy-fuel flameabout 10310^3 W/cm2^2Wide heat, slow, large HAZ, low cost
Electric arc10410^4-10510^5 W/cm2^2Most common; portable; moderate HAZ
Plasma arcabout 10510^5-10610^6 W/cm2^2Narrow, deep penetration
Laser beam, electron beam10610^6-10810^8 W/cm2^2Keyhole welding; narrow, deep weld; small HAZ; expensive

A higher heat intensity gives deeper and narrower welds, a smaller HAZ, less distortion and higher speed.

  • Practice · 3+3+3 marks

(a) Describe the three types of oxyacetylene flame and their uses. (b) Explain the principle of shielded metal arc welding (SMAW) and the functions of the electrode coating. (c) In arc welding at 24 V and 200 A the travel speed is 5 mm/s. The heat transfer efficiency is 0.8. Find the heat input per unit length of weld, and the cross-sectional area of weld metal that can be melted if the melting factor is 0.5 and the unit melting energy of the steel is 10 J/mm3^3.

Answer

(a) Oxyacetylene flames

Acetylene burns with oxygen in a torch. Flame has an inner cone (about 3100 degree C) and an outer envelope. The ratio of oxygen to acetylene decides the type:

FlameO2_2 : C2_2H2_2AppearanceUse
Neutralabout 1 : 1Clear inner coneWelding steel, cast iron, most work
Carburising (reducing)Excess acetyleneFeather around the coneHard facing, aluminium, monel
OxidisingExcess oxygenShort, pointed coneBrass, bronze (prevents zinc loss), cutting

(b) SMAW

A consumable flux-coated electrode is held in a holder and an arc (typically 20-30 V, 50-300 A) is struck between the electrode tip and the work. Arc heat melts the electrode core and the base metal, and the molten core wire is transferred to the pool as droplets. The coating burns and forms slag, which covers the weld. Current is AC or DC (electrode negative or positive).

Functions of the coating:

  1. Produces shielding gas (CO2_2, H2_2O) that protects the metal from oxygen and nitrogen.
  2. Forms slag that protects the cooling weld and slows the cooling.
  3. Stabilises the arc (potassium, sodium compounds).
  4. Adds alloying elements and iron powder (increasing deposition).
  5. Acts as a deoxidiser and refines the metal; shapes the bead.

(c) Heat input and melted area

Arc power:

P=VI=24×200=4800 WP = VI = 24\times200 = 4800\ \text{W}

Heat input per unit length of weld, with transfer efficiency f1=0.8f_1 = 0.8:

H=f1 VIv=0.8×48005=768 J/mmH = \frac{f_1\,VI}{v} = \frac{0.8\times4800}{5} = 768\ \text{J/mm}

Heat used in melting is f2=0.5f_2 = 0.5 of the transferred heat. Energy balance for melting:

f2 H=Um Awf_2\,H = U_m\,A_w Aw=0.5×76810=38.4 mm2A_w = \frac{0.5\times768}{10} = 38.4\ \text{mm}^2

Answer: heat input =768= 768 J/mm (0.768 kJ/mm); melted weld area =38.4= 38.4 mm2^2.

Written from the official syllabus. Questions and answers are written for this site; check them against your class notes.

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