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Chapter 1 · 2 hours

Introduction to Civil Engineering Material

IOE past exam questions

Past questions and answers

40 questions set from this chapter, 17 of them more than once; 10 are most repeated (set, or a close variant set, in 3 or more exams). Most repeated first.

  • Most repeated · 9 of 26 exams
  • Asked 9 times
  • 2076 Chaitra · 3 marks
  • 2080 Bhadra · 2 marks
  • 2080 Baisakh · 2 marks
  • 2078 Kartik · 2 marks
  • 2079 Bhadra · 1 mark
  • 2075 Asoj · 2 marks
  • 2073 Shrawan · 2 marks
  • 2081 Baisakh · 2 marks
  • 2081 Baisakh (new course) · 1 mark

What are the factors to be considered while selecting construction materials? Explain.

Answer

The selection of a construction material is a decision that balances performance, cost and availability so that the structure is safe, durable and economical.

  1. Strength and stability: the material must carry the expected loads (compression, tension, bending, shear) with a proper factor of safety.
  2. Durability: it should resist weathering, moisture, chemical attack, insects and fire for the design life.
  3. Cost and economy: both initial cost (purchase, transport, labour) and life-cycle cost (maintenance, repair) should be low.
  4. Availability: locally available materials reduce transport cost and construction time. In Nepal, river sand, bricks and stone are preferred near the site.
  5. Workability: the material should be easy to cut, shape, place and finish with the available tools and skill.
  6. Appearance and finish: important for exposed surfaces such as facing stone, tiles and paint.
  7. Thermal and acoustic properties: insulation, thermal conductivity, sound absorption and fire resistance affect comfort and safety.
  8. Weight: lighter materials reduce dead load and foundation cost.
  9. Maintenance: materials needing little upkeep are preferred.
  10. Environment and standards: the material should comply with relevant standards (Nepal Standards, IS, ASTM), be eco-friendly, and suit the local seismic and climatic conditions as per NBC.
  • Most repeated · 8 of 26 exams
  • Asked 8 times
  • 2075 Asoj · 2 marks
  • 2073 Shrawan · 1 mark
  • 2069 Chaitra · 2 marks
  • 2071 Chaitra · 1 mark
  • 2081 Baisakh (new course) · 1 mark
  • 2065 Shrawan (old course) · 8 marks
  • 2059 Poush (old course) · 9 marks
  • 2057 Chaitra (old course) · 9 marks

Explain the important (mechanical and physical) properties of civil engineering materials. Why is it necessary to study these properties?

Answer

Properties of a material are its measurable characteristics that decide how it behaves under load, heat, water and weather. They are grouped as physical and mechanical properties.

Physical properties

  • Density and specific gravity: mass per unit volume; relates to self weight of the structure.
  • Porosity: ratio of void volume to total volume. High porosity lowers strength and durability.
  • Permeability: ability to let water or fluid pass through the pores.
  • Hygroscopicity: tendency to absorb moisture from air.
  • Thermal conductivity: rate at which heat flows through the material. Low conductivity gives good insulation.
  • Fire resistance: ability to keep its strength and shape when exposed to fire.
  • Durability / weathering resistance: resistance to the action of rain, frost, sun and chemicals.

Mechanical properties

  • Strength: capacity to resist load without failure (compressive, tensile, shear, flexural).
  • Elasticity: property of regaining original shape after the load is removed. Plasticity is the opposite, permanent deformation.
  • Ductility: ability to be drawn into wires or to elongate considerably before fracture (mild steel).
  • Malleability: ability to be hammered into thin sheets without cracking (lead, gold).
  • Brittleness: fracture without appreciable deformation (cast iron, glass, brick).
  • Toughness: energy absorbed up to fracture (area under the stress-strain curve).
  • Resilience: energy stored up to the elastic limit.
  • Hardness: resistance to scratching, abrasion or indentation.
  • Fatigue: failure under repeated or cyclic loading at stress below the ultimate strength.
  • Creep: slow, continuous deformation under constant load over a long time.

Why these properties must be studied

  1. To select the right material for a given use (steel for tension, concrete and stone for compression).
  2. To design safe and economical members using correct permissible stresses.
  3. To predict behaviour under load, temperature, moisture and fire so that failures are avoided.
  4. To decide testing and quality control methods on site.
  5. To compare alternative materials on cost, performance and durability.
  6. To plan proper storage, handling and maintenance.
  • Most repeated · 8 of 26 exams
  • Asked 8 times
  • 2071 Chaitra · 2 marks
  • 2076 Asoj · 1 mark
  • 2069 Chaitra · 1 mark
  • 2072 Chaitra · 1 mark
  • 2068 Baisakh · 3 marks
  • 2059 Poush (old course) · 3 marks
  • 2070 Chaitra (old course) · 5 marks
  • 2057 Chaitra (old course) · 4 marks

Describe the importance and scope of the study of civil engineering materials for engineers.

Answer

Civil engineering materials is the study of the origin, manufacture, properties, testing and uses of the materials that make up buildings, roads, bridges, dams and other structures.

Importance

  • A structure is only as good as its materials. Knowing their properties lets the engineer choose safe materials.
  • It helps achieve economy by choosing the cheapest material that satisfies the requirements and by reducing wastage.
  • It supports quality control: engineers can supervise tests (brick, cement, aggregate, steel) and reject poor material on site.
  • It improves durability because the causes of decay (efflorescence, corrosion, weathering) are understood.
  • It prepares the engineer to use new and local materials and to follow codes and standards.
  • It is the base for later subjects such as structural design, concrete technology and construction management.

Scope

  • Study of natural materials: stone, timber, sand, clay.
  • Manufactured materials: brick, tile, lime, cement, mortar, glass, paint, varnish.
  • Metals and alloys: steel, cast iron, aluminium, copper.
  • Bituminous and miscellaneous materials: bitumen, tar, asphalt, plastics, geosynthetics, insulating materials.
  • Testing methods, standards and specifications.
  • Selection, storage, handling and economics of materials.
  • Modern developments: high performance concrete, composites, green and sustainable materials.
  • Most repeated · 8 of 26 exams
  • Asked 8 times
  • 2078 Bhadra · 1 mark
  • 2068 Chaitra · 1 mark
  • 2080 Bhadra · 1 mark
  • 2074 Asoj · 1 mark
  • 2072 Chaitra · 0.5 marks
  • 2076 Asoj · 0.5 marks
  • 2078 Kartik · 0.5 marks
  • 2066 Shrawan (old course) · 2 marks

Define creep as a property of materials.

Answer

Creep is the slow, time-dependent and permanent deformation of a material under a constant load (stress) maintained for a long period, even when the stress is below the yield strength.

It is more important at high temperature and in materials such as concrete, steel in boilers, timber and lead. Creep strain adds to the elastic strain, so long-term deflections in beams and loss of prestress in prestressed concrete members are caused by it.

  • Most repeated · 4 of 26 exams
  • Asked 4 times
  • 2080 Bhadra · 1 mark
  • 2079 Bhadra · 1 mark
  • 2076 Asoj · 0.5 marks
  • 2072 Chaitra · 0.5 marks

Define hygroscopicity as a property of materials.

Answer

Hygroscopicity is the property of a material to absorb moisture from the surrounding air and hold it. Timber, lime and cement are hygroscopic, so they swell, shrink or lose strength with changes in humidity.

  • Most repeated · 3 of 26 exams
  • Asked 3 times
  • 2081 Bhadra · 1 mark
  • 2078 Bhadra · 1 mark
  • 2068 Chaitra · 1 mark

Explain briefly the property porosity.

Answer

Porosity is the ratio of the volume of voids (pores) in a material to its total volume, usually expressed as a percentage.

n=VvV×100n = \frac{V_v}{V}\times 100

A porous material has low density, low strength, high water absorption and low resistance to frost. Brick and stone with high porosity are less durable, while dense materials such as granite have very low porosity.

  • Most repeated · 3 of 26 exams
  • Asked 3 times
  • 2079 Bhadra · 1 mark
  • 2074 Chaitra · 2 marks
  • 2068 Baisakh

Define fatigue as a property of materials.

Answer

Fatigue is the failure of a material under repeated or fluctuating (cyclic) stresses, even when the maximum stress is much lower than the ultimate tensile strength.

Failure starts as a tiny crack at a point of stress concentration, grows with every load cycle and ends in sudden fracture. The stress below which a material can bear an infinite number of cycles is called the endurance (fatigue) limit. Fatigue is important for bridges, crane girders, railway tracks and machine parts.

  • Most repeated · 3 of 26 exams
  • Asked 3 times
  • 2074 Chaitra · 2 marks
  • 2070 Chaitra (old course) · 1 mark
  • 2066 Shrawan (old course) · 2 marks

Define resilience as a property of materials.

Answer

Resilience is the capacity of a material to store strain energy when deformed within the elastic limit and to give it back on unloading. It is measured by the area under the elastic part of the stress-strain curve.

Modulus of resilience=σe22E\text{Modulus of resilience} = \frac{\sigma_e^2}{2E}

where σe\sigma_e is the stress at the elastic limit (or proof stress) and EE is Young's modulus. Its unit is N/mm2\text{N/mm}^2 (N·mm/mm3^3) or J/m3^3. Steel springs have high resilience.

  • Most repeated · 3 of 26 exams
  • Asked 3 times
  • 2074 Asoj · 1 mark
  • 2070 Chaitra (old course) · 1 mark
  • 2072 Chaitra · 0.5 marks

Define hardness as a property of materials.

Answer

Hardness is the resistance of a material to scratching, abrasion, cutting or indentation (penetration) by another body. It is measured by the Mohs scale (minerals and stone) and by Brinell, Rockwell and Vickers tests (metals). Granite and diamond are hard; lead and gypsum are soft.

  • Most repeated · 3 of 26 exams
  • Asked 3 times
  • 2074 Asoj · 1 mark
  • 2070 Chaitra (old course) · 1 mark
  • 2068 Baisakh

Define malleability as a property of materials.

Answer

Malleability is the property of a material by which it can be hammered or rolled into thin sheets without cracking or breaking. It depends on plasticity under compression. Gold, silver, lead, copper and aluminium are highly malleable, and heating usually increases malleability.

  • Asked 2 times
  • 2074 Chaitra · 2 marks
  • 2068 Baisakh · 1 mark

How are civil engineering materials classified?

Answer

Civil engineering materials can be classified in several ways.

1. On the basis of origin

  • Natural materials: stone, timber, sand, clay, bamboo, bitumen (natural asphalt).
  • Manufactured (artificial) materials: brick, tile, cement, lime, glass, steel, paint, plastics.

2. On the basis of composition

  • Metallic: steel, cast iron, aluminium, copper, brass (ferrous and non-ferrous).
  • Non-metallic: stone, brick, cement, timber, glass, bitumen, plastics.

3. On the basis of use

  • Structural: concrete, steel, brick, stone, timber.
  • Finishing: paint, varnish, tiles, plaster.
  • Protective and insulating: bitumen, damp-proofing and thermal insulation materials.

4. On the basis of structure

  • Crystalline: metals and many rocks.
  • Amorphous: glass, bitumen, plastics.

5. On the basis of behaviour

  • Ductile / brittle, elastic / plastic, organic / inorganic materials.
  • Asked 2 times
  • 2080 Baisakh · 1 mark
  • 2078 Kartik · 0.5 marks

Define permeability as a property of materials.

Answer

Permeability is the property of a material that permits water, air or other fluid to pass through it under pressure. It depends on the size and interconnection of the pores. Sand and gravel are highly permeable, while dense concrete and clay are nearly impermeable. Low permeability is needed in dams, water tanks and basements.

  • Asked 2 times
  • 2066 Shrawan (old course) · 2 marks
  • 2068 Baisakh

Define toughness as a property of materials.

Answer

Toughness is the ability of a material to absorb energy and deform plastically before it fractures. It is the total area under the stress-strain curve up to fracture (modulus of toughness), measured in N/mm2\text{N/mm}^2 or J/m3^3.

A tough material (mild steel) resists shock and impact loads, whereas a brittle material (glass, cast iron) has low toughness even if it is strong. Toughness is tested by Charpy and Izod impact tests.

  • Asked 2 times
  • 2078 Bhadra · 1 mark
  • 2068 Chaitra · 1 mark

Explain briefly the property brittleness.

Answer

Brittleness is the property of a material to fracture suddenly with little or no plastic deformation and no warning. Brittle materials have a short plastic range and are weak in tension and under impact.

Examples: cast iron, glass, brick, stone and concrete. In the stress-strain curve, fracture occurs almost at the elastic limit.

  • Asked 2 times
  • 2081 Baisakh · 2 marks
  • 2076 Asoj · 0.5 marks

Write short notes on tenacity and elasticity as properties of materials.

Answer

Tenacity

Tenacity is the property of a material to resist fracture under a tensile (pulling) load, i.e. its tensile strength. It is the maximum tensile stress a material can bear before it breaks. Wrought iron, mild steel and timber along the grain are tenacious.

Elasticity

Elasticity is the property of a material to regain its original shape and size completely after the external load is removed. Within the elastic limit, stress is proportional to strain (Hooke's law):

E=stressstrainE = \frac{\text{stress}}{\text{strain}}

EE is the modulus of elasticity. Steel is nearly perfectly elastic up to its elastic limit, whereas clay and lead are almost plastic.

  • Asked 2 times
  • 2072 Chaitra · 0.5 marks
  • 2076 Asoj · 0.5 marks

Define soundness and specific heat capacity as properties of materials.

Answer

Soundness

Soundness is the ability of a material (mainly cement, lime or aggregate) to resist volume change, cracking or expansion after setting. Unsound cement expands because of free lime or magnesia. It is tested by Le Chatelier and autoclave tests.

Specific heat capacity

Specific heat capacity is the quantity of heat required to raise the temperature of unit mass of a material by one degree. Its unit is J/(kg·K).

c=Qm ΔTc = \frac{Q}{m\,\Delta T}

Materials with high specific heat (water, concrete) heat up slowly and are good for thermal storage.

  • Asked 2 times
  • 2070 Chaitra (old course) · 3+2+3 marks
  • 2066 Shrawan (old course) · 3+2+3 marks

Define 'yield' and 'strain hardening' related to the tensile strength test for steel. How is the yield point of a brittle material found out from the stress/strain diagram? Draw the stress/strain diagram for ductile and brittle materials.

Answer

Yield

Yield is the stage in a tensile test of steel at which the specimen begins to deform plastically without any noticeable increase in load. The stress at this point is the yield stress. Mild steel shows an upper and a lower yield point; after the lower yield point, the strain rises rapidly at nearly constant stress.

Strain hardening

After yielding, the dislocations in the metal interlock, so more stress is needed to produce further strain. The curve rises again until the ultimate stress is reached. This region is called strain hardening (work hardening). Cold working of steel bars (TOR steel) uses this effect to raise yield strength.

Yield point of a brittle material

A brittle material has no clear yield point on its stress-strain diagram. The yield point is found by the offset (proof stress) method: draw a line parallel to the initial straight portion from a strain offset of 0.2% on the strain axis. The stress at which this line cuts the curve is taken as the 0.2% proof stress (yield strength).

Stress-strain diagrams

 Ductile (mild steel)         Brittle (cast iron)
 stress                       stress
   |      U__                   |        . X (fracture)
   |  Y2 /   `-.                |      .
   | Y1 /---'    \ F            |    .
   |   /          X             |  .
   |  /                         | .
   | /                          |.
   +---------------- strain     +---------------- strain
 E: elastic limit, Y: yield, U: ultimate, F: fracture

The ductile curve shows a long plastic region and necking before fracture, whereas the brittle curve is almost linear and ends suddenly with a small strain.

  • 2070 Chaitra (old course) · 8 marks

A mild steel specimen of 10 mm diameter and 300 mm long, which resists the maximum tensile load of 250 kN at 2 mm diameter. If the material resists 120 kN yield load by elongating 8 mm, then what is the modulus of toughness (printed 'roughness') and resilience?

Similar questions: Modulus of toughness and resilience (300 kN) (2066 Shrawan (old course))

Answer

Given: d=10d = 10 mm, L=300L = 300 mm, Pu=250P_u = 250 kN, Py=120P_y = 120 kN, elongation δ=8\delta = 8 mm.

Assumptions: the stresses are based on the original area A0A_0 (the 2 mm diameter at maximum load is the neck diameter and does not change A0A_0). The 8 mm elongation is taken as the strain to which the curve is evaluated. Modulus of resilience is the elastic energy per unit volume to yield, and the modulus of toughness is taken as the approximate area under the curve (average of yield and ultimate stress times strain).

A0=π4(10)2=78.54 mm2σy=120×10378.54=1527.89 N/mm2σu=250×10378.54=3183.10 N/mm2ε=8300=0.02667\begin{aligned} A_0 &= \frac{\pi}{4}(10)^2 = 78.54\ \text{mm}^2\\ \sigma_y &= \frac{120\times10^3}{78.54} = 1527.89\ \text{N/mm}^2\\ \sigma_u &= \frac{250\times10^3}{78.54} = 3183.10\ \text{N/mm}^2\\ \varepsilon &= \frac{8}{300} = 0.02667 \end{aligned}

Modulus of resilience (elastic limit taken at yield):

Ur=12σyε=12(1527.89)(0.02667)=20.37 N/mm2U_r = \tfrac12\sigma_y\varepsilon = \tfrac12(1527.89)(0.02667) = 20.37\ \text{N/mm}^2

Modulus of toughness:

Ut≈σy+σu2 ε=1527.89+3183.102(0.02667)=62.81 N/mm2U_t \approx \frac{\sigma_y+\sigma_u}{2}\,\varepsilon = \frac{1527.89+3183.10}{2}(0.02667) = 62.81\ \text{N/mm}^2

Answer: Modulus of resilience = 20.37 N/mm² (MJ/m³); modulus of toughness ≈ 62.81 N/mm² (MJ/m³).

  • 2066 Shrawan (old course) · 8 marks

A mild steel specimen of 10 mm diameter and 400 mm long, which resists the maximum tensile load of 300 kN at 2 mm diameter. If the material resists 150 kN yield load by elongating 10 mm, then what is the modulus of toughness (printed 'roughness') and resilience?

Similar questions: Modulus of toughness and resilience (250 kN) (2070 Chaitra (old course))

Answer

Given: d=10d = 10 mm, L=400L = 400 mm, Pu=300P_u = 300 kN, Py=150P_y = 150 kN, elongation δ=10\delta = 10 mm.

Assumptions: stresses are based on the original area. The elastic energy is taken up to yield, and the modulus of toughness is taken as the approximate area under the curve (average of yield and ultimate stress times strain).

A0=π4(10)2=78.54 mm2σy=150×10378.54=1909.86 N/mm2σu=300×10378.54=3819.72 N/mm2ε=10400=0.025\begin{aligned} A_0 &= \frac{\pi}{4}(10)^2 = 78.54\ \text{mm}^2\\ \sigma_y &= \frac{150\times10^3}{78.54} = 1909.86\ \text{N/mm}^2\\ \sigma_u &= \frac{300\times10^3}{78.54} = 3819.72\ \text{N/mm}^2\\ \varepsilon &= \frac{10}{400} = 0.025 \end{aligned}

Modulus of resilience:

Ur=12σyε=12(1909.86)(0.025)=23.87 N/mm2U_r = \tfrac12\sigma_y\varepsilon = \tfrac12(1909.86)(0.025) = 23.87\ \text{N/mm}^2

Modulus of toughness:

Ut≈σy+σu2 ε=1909.86+3819.722(0.025)=71.62 N/mm2U_t \approx \frac{\sigma_y+\sigma_u}{2}\,\varepsilon = \frac{1909.86+3819.72}{2}(0.025) = 71.62\ \text{N/mm}^2

Answer: Modulus of resilience = 23.87 N/mm² (MJ/m³); modulus of toughness ≈ 71.62 N/mm² (MJ/m³).

  • 2078 Bhadra · 1 mark

State the scope of construction materials in transportation.

Answer

Materials are the basis of transport infrastructure.

  • Roads: aggregates, bitumen, asphalt, cement concrete, soil and geotextiles for pavement layers.
  • Railways: steel rails, concrete or timber sleepers, ballast stone.
  • Bridges and tunnels: structural steel, reinforced and prestressed concrete, bearings and cables.
  • Airports and ports: rigid concrete pavements, piles, steel and protective coatings.
  • Drainage and markings: pipes, culverts, paints and signboards.

Knowledge of these materials ensures strength, riding quality, durability and low maintenance cost.

  • 2081 Kartik (new course) · 1 mark

How does weathering affect civil engineering materials?

Answer

Weathering is the combined action of rain, sun, wind, frost, temperature change and chemicals on materials. Its effects are:

  • Loss of strength and surface decay: stone and concrete crack, scale and disintegrate.
  • Corrosion: steel rusts in moist air and loses section.
  • Frost damage: water freezes in pores and expands, causing spalling.
  • Colour fading and chalking: paint, plastics and bitumen degrade under ultraviolet rays.
  • Swelling, shrinkage and rotting: timber warps and decays.
  • Efflorescence: soluble salts appear on brick and masonry surfaces.

Materials for exposed use must therefore be weather resistant or protected.

  • 2081 Kartik (new course) · 1 mark

What is the major difference between ductile and brittle failure of materials?

Answer

A ductile failure occurs after large plastic deformation and necking, so it gives visible warning and absorbs a lot of energy. A brittle failure occurs suddenly with very little plastic deformation and no warning.

BasisDuctile failureBrittle failure
DeformationLarge plastic strainVery small strain
WarningVisible neckingSudden, none
Energy absorbedHighLow
Fracture surfaceRough, cup-and-coneFlat, shiny
ExampleMild steelCast iron, glass
  • 2081 Chaitra (new course) · 3 marks

What is the difference between hardness, toughness and strength of a material? Explain with example.

Answer

  • Hardness is resistance to scratching or indentation of the surface.
  • Toughness is the energy absorbed up to fracture (area under the stress-strain curve).
  • Strength is the maximum stress a material can carry before failure.
BasisHardnessToughnessStrength
MeaningSurface resistance to scratch or indentationEnergy absorption before fractureLoad-carrying capacity per unit area
MeasureMohs, Brinell, RockwellCharpy or Izod impact test, area under curveStress in N/mm2\text{N/mm}^2
Depends onSurface/atomic bondingStrength together with ductilityMaterial and section
ExampleGlass is hard but not toughMild steel is toughHigh-tensile steel is strong

For example, a diamond is very hard but brittle, rubber is tough but not hard, and concrete has high compressive strength but low toughness.

  • 2081 Bhadra · 2 marks

Explain briefly the properties: Strength and Ductility.

Answer

Strength

Strength is the ability of a material to resist the applied load without failure or excessive deformation. It is expressed as stress (force per unit area) and may be compressive, tensile, shear or flexural strength. Steel has high tensile strength, concrete and stone high compressive strength.

Ductility

Ductility is the ability of a material to undergo large plastic deformation under tension, so that it can be drawn into wires before fracture. It is measured by percentage elongation and percentage reduction in area.

Elongation %=Lf−L0L0×100\text{Elongation \%} = \frac{L_f - L_0}{L_0}\times 100

Mild steel, copper and aluminium are ductile. Ductility gives warning before failure and is very important for earthquake-resistant structures.

  • 2080 Baisakh · 1 mark

Define fire resistance [printed 'first resistance'] as a property of materials.

Answer

Fire resistance is the ability of a material or building element to withstand fire without losing its strength, shape or stability for a stated time, and to prevent the spread of fire. It is expressed as the number of hours it can stand a standard fire test (for example a 2-hour rated wall). Brick, stone (to a limit), concrete and fire clay are fire resistant; timber burns and steel loses strength at about 500 to 600 °C.

  • 2066 Shrawan (old course) · 2 marks

Define thermal conductivity as a property of materials, with sketch where necessary.

Answer

Thermal conductivity (kk) is the rate of heat flow through a unit thickness of a material, per unit area, per unit temperature difference between its faces. By Fourier's law:

Q=k A (T1−T2) tLQ = \frac{k\,A\,(T_1-T_2)\,t}{L}

Its unit is W/(m·K). A low value means a good insulator (timber, cork, air-filled bricks); a high value means a good conductor (steel, copper).

  T1 (hot)  +---------------+  T2 (cold)
   ======>  |  area A       |  ======>
   heat Q   +---------------+
            <------ L ------>
  • 2066 Shrawan (old course) · 4 marks

Illustrate the Charpy Impact Test with figures.

Answer

The Charpy impact test measures the energy a material absorbs when fractured by a sudden blow, and so gives its toughness and notch brittleness.

Specimen

A standard bar 10 × 10 × 55 mm with a 2 mm deep V-notch (or U-notch) at the middle.

Apparatus

A pendulum hammer of known mass raised to a fixed height, an anvil with two supports 40 mm apart, and a scale to read the energy.

        pivot
          o
          |\
          | \ hammer
   h1 ....|..O
          |
   =======+=====  <- specimen
   |__ ___|__ __|
   supports (anvil), notch faces away from the hammer

Procedure

  1. The specimen is placed simply supported on the anvil with the notch on the side opposite to the striking edge.
  2. The pendulum is released from the initial height h1h_1.
  3. It strikes the specimen behind the notch, breaks it and swings up to a lower height h2h_2.
  4. The energy absorbed is
E=m g (h1−h2)E = m\,g\,(h_1 - h_2)

It is read directly from the dial in joules.

Use

The test is repeated at different temperatures to find the ductile-brittle transition temperature. A tough material absorbs high energy; a brittle one absorbs little.

  • 2059 Poush (old course) · 3+4 marks

Draw a neat stress-strain curve for a ductile material and explain the significant points in the curve.

Answer

A ductile material such as mild steel is tested in a universal testing machine and the stress σ=P/A0\sigma = P/A_0 is plotted against the strain ε=ΔL/L0\varepsilon = \Delta L/L_0.

 stress
   |            C
   |        B ,-'''-. D
   |   A  ,.'/        \
   | ,-'  / yield      \ E (fracture)
   |/ P  /              x
   +--------------------- strain
   O
 P proportional limit, A elastic limit,
 B upper yield, C ultimate, D necking, E fracture

Significant points

  1. Proportional limit (P): up to this point stress is proportional to strain (Hooke's law). The slope is Young's modulus EE.
  2. Elastic limit (A): the maximum stress up to which the material returns to its original length on unloading. It is very close to the proportional limit.
  3. Upper and lower yield points (B): the stress drops slightly and the material elongates at nearly constant stress. This is the start of plastic flow. The yield stress is used as the design strength.
  4. Strain hardening region: after yielding, the stress rises again with strain because of work hardening.
  5. Ultimate stress (C): the maximum stress on the curve, equal to the maximum load divided by original area.
  6. Necking (D): the cross-section reduces locally, so the engineering stress falls although the true stress keeps rising.
  7. Fracture point (E): the specimen breaks. The area under the curve to this point is the toughness.

Percentage elongation and reduction in area measure the ductility.

  • 2057 Chaitra (old course) · 2+3+5 marks

Define true stress-strain. Draw a typical stress strain curve for structural steel and explain the significant points in the curve.

Answer

Definition of true stress and true strain

True stress is the load divided by the instantaneous (actual) cross-sectional area, σt=P/Ai\sigma_t = P/A_i. True strain is the sum of all small strains based on the instantaneous length:

εt=∫L0LdLL=ln⁡LL0\varepsilon_t = \int_{L_0}^{L}\frac{dL}{L} = \ln\frac{L}{L_0}

Up to the start of necking: σt=σ(1+ε)\sigma_t = \sigma(1+\varepsilon) and εt=ln⁡(1+ε)\varepsilon_t = \ln(1+\varepsilon), where σ\sigma and ε\varepsilon are engineering stress and strain.

Stress-strain curve of structural steel

 stress
   |            C
   |        B ,-'''-. D        true curve rises
   |   A  ,.'/        \ - - - - -> T
   | ,-'  /             \  E
   |/    /               x  engineering curve
   +------------------------ strain
 A elastic limit, B yield, C ultimate, D necking, E fracture

Significant points

  1. A, proportional/elastic limit: linear region where σ=Eε\sigma = E\varepsilon and deformation is recoverable.
  2. B, yield point: large plastic strain at almost constant stress. This stress is the yield strength (for example 250 MPa for Fe 250, 415 MPa for Fe 415).
  3. Strain hardening: stress rises again with strain.
  4. C, ultimate tensile strength: the highest engineering stress.
  5. D, necking: local reduction of area; the engineering curve falls because the load decreases.
  6. E, fracture: specimen breaks. The true stress curve continues to rise up to fracture because the actual area keeps decreasing.
  • 2070 Chaitra (old course) · 4 marks

Define true stress and engineering stress.

Answer

Engineering stress

Engineering (nominal) stress is the applied load divided by the original cross-sectional area:

σe=PA0\sigma_e = \frac{P}{A_0}

Engineering strain is εe=(L−L0)/L0\varepsilon_e = (L-L_0)/L_0. It is used in design because A0A_0 is known.

True stress

True stress is the load divided by the instantaneous cross-sectional area at that moment:

σt=PAi=σe (1+εe)\sigma_t = \frac{P}{A_i} = \sigma_e\,(1+\varepsilon_e)

True strain is εt=ln⁡(1+εe)\varepsilon_t = \ln(1+\varepsilon_e).

BasisEngineeringTrue
Area usedOriginal A0A_0Instantaneous AiA_i
After neckingFallsKeeps rising
ValueLowerHigher (in tension)
UseDesignMaterial research
  • 2059 Poush (old course) · 4 marks

Write a short note on elastic and plastic behaviours of materials.

Answer

Elastic behaviour

A material is elastic if it fully regains its original shape and size on removal of the load. Stress is proportional to strain (Hooke's law), σ=Eε\sigma = E\varepsilon, and the deformation is caused by stretching of atomic bonds without any slip of atoms. It occurs below the elastic limit. Steel and rubber show it.

Plastic behaviour

Beyond the elastic limit (yield point), the deformation remains permanently after unloading. It results from slipping of atomic planes by the movement of dislocations. Stress is no longer proportional to strain. Clay, lead and mild steel above yield are plastic.

 stress
   |        ,-----  plastic
   |      ,'
   |    /  <- yield
   |  /  elastic
   +----------- strain
 on unloading, the line returns parallel to the elastic slope,
 leaving a permanent (plastic) strain

Design of structures is done in the elastic range, while plasticity gives ductility and warning before failure.

  • 2057 Chaitra (old course) · 4 marks

Write a short note on fracture mode of materials.

Answer

Fracture is the separation of a body into two or more parts under stress. There are two main modes.

Ductile fracture

  • Occurs after extensive plastic deformation and necking.
  • Absorbs large energy and gives warning.
  • Surface is dull and fibrous, in a cup-and-cone shape in tension.
  • Mechanism: voids form in the neck, join and the rest shears at 45 degrees. Example: mild steel, aluminium.

Brittle fracture

  • Occurs suddenly with very little or no plastic deformation.
  • Crack grows rapidly, almost at the speed of sound in the material.
  • Surface is flat, bright and granular, perpendicular to the tensile stress.
  • It occurs by cleavage (breaking of atomic bonds along crystal planes) or along grain boundaries. Example: cast iron, glass, concrete, steel at low temperature.

Other modes

Fatigue fracture (cyclic loading) and creep fracture (long-term high temperature).

Factors that promote brittle fracture: low temperature, high strain rate, notches and sharp corners (stress concentration).

  • 2057 Chaitra (old course) · 4 marks

Write a short note on Griffith theory.

Answer

Griffith theory (A. A. Griffith, 1920) explains why the actual fracture strength of brittle materials is far below the theoretical strength. It states that every real material contains microscopic cracks or flaws, which magnify the local stress at their tips.

Principle

A crack extends only if the elastic strain energy released by its growth is at least equal to the surface energy needed to create the new crack surfaces. For a thin plate with an internal crack of length 2a2a, the critical (fracture) stress is

σc=2Eγπa\sigma_c = \sqrt{\frac{2E\gamma}{\pi a}}

where EE is the modulus of elasticity, γ\gamma is the surface energy per unit area and aa is half crack length.

Conclusions

  • Larger flaws give lower fracture strength.
  • Brittle materials (glass, ceramics, concrete) fail by crack propagation under tension, and are much stronger in compression, where cracks close.
  • For ductile materials, plastic work at the crack tip also adds to the energy needed (Irwin-Orowan modification).

The theory is the base of fracture mechanics.

  • 2070 Chaitra (old course) · 4 marks

Write a short note on chemical bond.

Answer

A chemical bond is the force of attraction that holds atoms or ions together in a molecule or solid. Atoms bond to attain a stable (octet) electron configuration with lower energy. The type of bond decides the strength, melting point, ductility and conductivity of a material.

Primary (strong) bonds

  1. Ionic bond: transfer of electrons from a metal to a non-metal, giving oppositely charged ions that attract each other (NaCl, MgO). Hard, brittle, high melting point, poor conductors in solid state.
  2. Covalent bond: atoms share pairs of electrons (diamond, silica, polymers). Very hard and strong, directional, poor conductors.
  3. Metallic bond: positive ions in a sea of free electrons (iron, copper). Ductile, malleable, good conductors.

Secondary (weak) bonds

  • Van der Waals and hydrogen bonds between molecules. They are weak, found in polymers, bitumen, water and ice.
BondStrengthExample material
IonicStrongCeramics, cement minerals
CovalentVery strongSilicates, diamond
MetallicModerate to strongSteel, aluminium
Van der WaalsWeakPlastics, bitumen
  • 2065 Shrawan (old course) · 4 marks

Differentiate between metallic and covalent bond.

Answer

BasisMetallic bondCovalent bond
FormationAttraction between positive metal ions and a sea of free electronsSharing of electron pairs between two atoms
ElectronsDelocalised, free to moveLocalised between the bonded atoms
DirectionNon-directionalDirectional (fixed bond angles)
ConductivityGood electrical and thermal conductorGenerally poor conductor
Mechanical natureDuctile and malleableHard and brittle
Melting pointModerate to highVery high (diamond) or low (molecular solids)
LustreMetallic lustreUsually none
ExamplesIron, copper, aluminiumDiamond, silica, polymers
  • 2066 Shrawan (old course) · 4 marks

Write a short note on metallic bond.

Answer

A metallic bond is the bond formed in metals in which the outer (valence) electrons of the atoms leave their atoms and move freely throughout the whole lattice. The metal consists of positive ions arranged in a regular lattice and held together by the electrostatic attraction of this "sea" of free electrons.

  + + + + +      + = metal ion
  + + + + +      . = free electron cloud
  + + + + +  (electrons move through the lattice)

Properties that result

  • High electrical and thermal conductivity because free electrons carry charge and heat.
  • Ductility and malleability: layers of ions can slide over each other without breaking the bond, because the electron cloud is non-directional.
  • Metallic lustre and opacity.
  • Moderate to high strength and melting point, which increase with number of free electrons (iron vs sodium).

Examples: iron, copper, aluminium, zinc and their alloys.

  • 2059 Poush (old course) · 2+4+3 marks

Define bonding. Explain metallic bonding with example. Write the important characteristics of metallic materials.

Answer

Bonding

Bonding is the attraction between atoms, ions or molecules that holds them together to form a stable substance. It arises from the tendency of atoms to reach a stable electron configuration. The main types are ionic, covalent, metallic and secondary (van der Waals, hydrogen) bonds.

Metallic bonding with example

In metallic bonding, each metal atom gives up its valence electrons to a common pool. The resulting positive ions form a regular lattice, held together by the attraction of the freely moving electron cloud. The bond is non-directional and not tied to particular atoms.

Example: in iron or copper, the valence electrons are shared by the whole crystal. When a load is applied, layers of ions slip over each other and the electron cloud keeps holding them, so the metal deforms without breaking.

  (+) (+) (+) (+)
   .  .  .  .  .   electron sea
  (+) (+) (+) (+)

Important characteristics of metallic materials

  1. High strength and stiffness.
  2. Ductile and malleable, so they can be drawn and rolled.
  3. Good conductors of heat and electricity.
  4. Metallic lustre and opacity.
  5. High density and fairly high melting point.
  6. Tough, and can withstand impact and fatigue (steel).
  7. Can be alloyed, welded, cast and heat treated to modify properties.
  8. Prone to corrosion in moist air, so they need protection.
  9. Recyclable.
  • 2065 Shrawan (old course) · 12 marks

The following observations were made during a tensile test on a mild steel specimen 40 mm diameter and 200 mm long. Given: 2 mm elongation with 40 kN load (within limit of proportionality); yield load = 160 kN; maximum load = 240 kN; length of specimen at fracture = 250 mm. Determine the modulus of toughness and resilience.

Answer

Given: d=40d = 40 mm, L0=200L_0 = 200 mm; in the elastic range P=40P = 40 kN gives δ=2\delta = 2 mm; Py=160P_y = 160 kN; Pu=240P_u = 240 kN; Lf=250L_f = 250 mm.

Step 1: area and stresses

A0=π4(40)2=1256.64 mm2σy=160×1031256.64=127.32 N/mm2σu=240×1031256.64=190.99 N/mm2\begin{aligned} A_0 &= \frac{\pi}{4}(40)^2 = 1256.64\ \text{mm}^2\\ \sigma_y &= \frac{160\times10^3}{1256.64} = 127.32\ \text{N/mm}^2\\ \sigma_u &= \frac{240\times10^3}{1256.64} = 190.99\ \text{N/mm}^2 \end{aligned}

Step 2: modulus of elasticity (from the elastic reading)

σ=40×1031256.64=31.83 N/mm2,ε=2200=0.01,E=31.830.01=3183.1 N/mm2\sigma = \frac{40\times10^3}{1256.64} = 31.83\ \text{N/mm}^2,\quad \varepsilon = \frac{2}{200} = 0.01,\quad E = \frac{31.83}{0.01} = 3183.1\ \text{N/mm}^2

Step 3: modulus of resilience (up to yield)

Ur=σy22E=(127.32)22(3183.1)=2.546 N/mm2U_r = \frac{\sigma_y^2}{2E} = \frac{(127.32)^2}{2(3183.1)} = 2.546\ \text{N/mm}^2

Step 4: modulus of toughness

Strain at fracture: εf=250−200200=0.25\varepsilon_f = \dfrac{250-200}{200} = 0.25. Taking the average of yield and ultimate stress over the whole strain:

Ut≈σy+σu2 εf=127.32+190.992(0.25)=39.79 N/mm2U_t \approx \frac{\sigma_y+\sigma_u}{2}\,\varepsilon_f = \frac{127.32+190.99}{2}(0.25) = 39.79\ \text{N/mm}^2

Answer: Modulus of resilience = 2.55 N/mm² (MJ/m³); modulus of toughness ≈ 39.79 N/mm² (MJ/m³).

  • 2059 Poush (old course) · 4 marks

A mild steel rod 3 m long having a cross-sectional area of 4 cm2^2 is subjected to an axial pull of 1500 kg. If EE for steel is 2.1×1062.1\times10^{6} kg/cm2^2, find stress, strain and elongation of the rod.

Answer

Given: L=3L = 3 m =300= 300 cm, A=4 cm2A = 4\ \text{cm}^2, P=1500P = 1500 kg, E=2.1×106 kg/cm2E = 2.1\times10^6\ \text{kg/cm}^2.

Stress σ=PA=15004=375 kg/cm2Strain ε=σE=3752.1×106=1.786×10−4Elongation δ=εL=1.786×10−4×300=0.0536 cm\begin{aligned} \text{Stress } \sigma &= \frac{P}{A} = \frac{1500}{4} = 375\ \text{kg/cm}^2\\ \text{Strain } \varepsilon &= \frac{\sigma}{E} = \frac{375}{2.1\times10^{6}} = 1.786\times10^{-4}\\ \text{Elongation } \delta &= \varepsilon L = 1.786\times10^{-4}\times300 = 0.0536\ \text{cm} \end{aligned}

Answer: Stress = 375 kg/cm² (about 36.8 N/mm²); strain = 1.786 × 10⁻⁴; elongation = 0.0536 cm = 0.536 mm.

  • 2057 Chaitra (old course) · 4 marks

The yield stress of a medium carbon steel is found to be 415 N/mm2^2. Its ultimate stress is 20% more than the yield stress. If the steel fractures at 2% of strain, find the modulus of toughness of the steel.

Answer

Given: σy=415 N/mm2\sigma_y = 415\ \text{N/mm}^2, σu=1.20×415=498 N/mm2\sigma_u = 1.20\times415 = 498\ \text{N/mm}^2, fracture strain εf=2%=0.02\varepsilon_f = 2\% = 0.02.

The modulus of toughness is the area under the stress-strain curve up to fracture. Taking it as the average of the yield and ultimate stress times the fracture strain:

Ut≈σy+σu2 εf=415+4982×0.02=456.5×0.02=9.13 N/mm2U_t \approx \frac{\sigma_y+\sigma_u}{2}\,\varepsilon_f = \frac{415+498}{2}\times0.02 = 456.5\times0.02 = 9.13\ \text{N/mm}^2

Answer: Modulus of toughness ≈ 9.13 N·mm/mm³ = 9.13 MJ/m³.

Questions from Old Question Collection (CE 506) (IOE BCE exam papers CE 506 / EG463CE, 2057 to 2081 (23 papers)) and Old Question Collection (CE 506) (New course (2080 batch) CE 103 / ENCE 103 papers, 2081 Baisakh, Kartik, Chaitra). Answers are written for this site; check them against your class notes.

Chapter titles and hours from the IOE syllabus ↗