BE Civil Engineering (IOE, TU) Civil Engineering Materials (IOE, CE 501) Question Paper 2079 Nepal

Ordinary Portland Cement (OPC): OPC is a hydraulic cement produced by finely grinding Portland cement clinker (formed by burning a properly proportioned mixture of calcareous and argillaceous materials at about $1450^\circ C$) together with a small amount (about 3–5%) of gypsum ($CaSO_4\cdot2H_2O$) to control the setting time.

Role of the four major Bogue compounds:

Bogue equations (for $A/F \geq 0.64$, which holds here since $6/3 = 2.0$):

$$C_3S = 4.071\,CaO - 7.600\,SiO_2 - 6.718\,Al_2O_3 - 1.430\,Fe_2O_3 - 2.852\,SO_3$$ $$C_2S = 2.867\,SiO_2 - 0.7544\,C_3S$$ $$C_3A = 2.650\,Al_2O_3 - 1.692\,Fe_2O_3$$ $$C_4AF = 3.043\,Fe_2O_3$$

No $SO_3$ is given, so take $SO_3 = 0$.

Step 1 — $C_3S$:

$$C_3S = 4.071(64) - 7.600(21) - 6.718(6) - 1.430(3)$$ $$= 260.544 - 159.600 - 40.308 - 4.290 = 56.346\%$$

Step 2 — $C_2S$:

$$C_2S = 2.867(21) - 0.7544(56.346) = 60.207 - 42.508 = 17.699\%$$

Step 3 — $C_3A$:

$$C_3A = 2.650(6) - 1.692(3) = 15.900 - 5.076 = 10.824\%$$

Step 4 — $C_4AF$:

$$C_4AF = 3.043(3) = 9.129\%$$

Summary:

Comment: The high $C_3S$ content (≈56%) indicates the cement will develop good early strength and generate appreciable heat of hydration. The relatively high $C_3A$ (≈10.8%) means it is not suited to sulphate-rich environments and is more typical of an ordinary/rapid-strength cement rather than a low-heat or sulphate-resisting cement.

Factors affecting workability of fresh concrete:

1. Water content — more water increases fluidity (but reduces strength/durability).
2. Water–cement ratio — higher ratio improves workability.
3. Aggregate properties — size, shape (rounded > angular), texture (smooth > rough), grading.
4. Aggregate–cement ratio — leaner mixes are less workable.
5. Use of admixtures — plasticizers/superplasticizers greatly improve workability.
6. Fineness of cement — finer cement needs more water.
7. Ambient temperature and time — high temperature/long delay reduces workability.

Given: Mix $1:1.8:3.2$ by mass, $w/c = 0.48$, cement bag $= 50\ \text{kg}$. Specific gravities: cement $3.15$, FA $2.65$, CA $2.70$. Take density of water $= 1000\ \text{kg/m}^3$.

Step 1 — masses per 50 kg bag of cement:

• Cement $= 50\ \text{kg}$
• Fine aggregate $= 1.8 \times 50 = 90\ \text{kg}$
• Coarse aggregate $= 3.2 \times 50 = 160\ \text{kg}$
• Water $= 0.48 \times 50 = 24\ \text{kg}$

Step 2 — absolute volume of each ingredient $\left(V = \dfrac{\text{mass}}{\text{SG} \times 1000}\right)$:

$$V_{cement} = \frac{50}{3.15 \times 1000} = 0.015873\ \text{m}^3$$ $$V_{FA} = \frac{90}{2.65 \times 1000} = 0.033962\ \text{m}^3$$ $$V_{CA} = \frac{160}{2.70 \times 1000} = 0.059259\ \text{m}^3$$ $$V_{water} = \frac{24}{1 \times 1000} = 0.024000\ \text{m}^3$$

Step 3 — theoretical density (no air voids):

$$\rho = \frac{\text{Total mass}}{\text{Total absolute volume}} = \frac{324}{0.133094} = 2434.7\ \text{kg/m}^3$$

Answers:

• (i) Absolute volumes per 50 kg bag: cement $= 0.01587\ \text{m}^3$, FA $= 0.03396\ \text{m}^3$, CA $= 0.05926\ \text{m}^3$, water $= 0.02400\ \text{m}^3$.
• (ii) Theoretical fresh density $\approx 2435\ \text{kg/m}^3$.

Timber as a building material: Timber is wood obtained from trees that has been converted into a form suitable for construction (beams, planks, posts, formwork, doors, windows). It is renewable, has a high strength-to-weight ratio, good thermal insulation, ease of working and a pleasing appearance, but it is anisotropic, combustible and prone to moisture movement, decay and insect attack.

(a) Structure of a tree trunk (cross-section, from outside to inside):

  Bark (outer + inner) -> Cambium layer -> Sapwood -> Heartwood -> Pith (medulla)
  with Medullary rays radiating from centre and Annual rings (growth rings)

• Pith (medulla): innermost soft core; first-formed wood.
• Heartwood: inner, dark, dead but strong and durable wood; gives structural strength.
• Sapwood: outer, lighter, living wood that conducts sap; less durable.
• Cambium layer: thin growing layer between sapwood and bark; produces new cells.
• Bark: outer protective covering.
• Medullary rays: radial lines binding the annual rings and storing/transporting food.
• Annual rings: concentric rings; one ring per year, used to estimate age.

(b) Seasoning of timber: Seasoning is the controlled process of reducing the moisture content of freshly felled (green) timber to a level in equilibrium with the surrounding atmosphere (usually 10–15%).

Objectives of seasoning:

1. Reduce weight for easier handling and transport.
2. Increase strength, stiffness and durability.
3. Minimise shrinkage, warping and splitting after use.
4. Improve resistance to decay and insect/fungal attack.
5. Make timber suitable for painting, polishing and gluing.

Methods: Natural (air) seasoning — stacking timber on raised platforms with spacers under shade; cheap but slow. Artificial (kiln) seasoning — drying in a controlled kiln with regulated temperature, humidity and air circulation; fast and uniform. Other methods: water seasoning, chemical seasoning, electrical seasoning.

(c) Four natural defects in timber:

1. Knots: Bases of branches embedded in the trunk; appear as dark, hard, roughly circular marks that interrupt the grain and reduce strength.
2. Shakes: Cracks/separations in the wood. Heart shake radiates from pith outward; cup (ring) shake follows the annual rings; star shake radiates in a star pattern from the centre.
3. Wane: Absence of wood or presence of bark along the edge/corner of a sawn timber piece, leaving a rounded/deficient edge.
4. Twisted/cross grain: Fibres run spirally or at an angle to the axis instead of straight, weakening the timber and causing warping.

(Other acceptable defects: rind galls, upsets/rupture, dead wood, foxiness.)

Classification of steel by carbon content:

Significance of the stress–strain curve of mild steel: When mild steel is loaded in tension the curve shows:

• Proportional limit / elastic region (O–A): stress ∝ strain (Hooke's law), slope = Young's modulus $E$.
• Elastic limit (A): beyond this permanent set begins.
• Upper & lower yield points (B–C): sudden yielding at nearly constant stress; the steel flows plastically.
• Strain hardening (C–D): stress rises again to the ultimate strength at D.
• Necking & fracture (D–E): local reduction of area, load falls, then the specimen fractures. The curve demonstrates ductility (large plastic strain before failure) and gives a clear, well-defined yield point — a key advantage of mild steel for structural use.

Given: $d_0 = 16\ \text{mm}$, $L_0 = 80\ \text{mm}$, yield load $P_y = 58\ \text{kN}$, ultimate load $P_u = 92\ \text{kN}$, final length $L_f = 98\ \text{mm}$, neck diameter $d_f = 11\ \text{mm}$.

Step 1 — original cross-sectional area:

$$A_0 = \frac{\pi}{4}d_0^2 = \frac{\pi}{4}(16)^2 = 201.062\ \text{mm}^2$$

Step 2 — (i) Yield stress:

$$\sigma_y = \frac{P_y}{A_0} = \frac{58\,000\ \text{N}}{201.062\ \text{mm}^2} = 288.5\ \text{N/mm}^2$$

Step 3 — (ii) Ultimate tensile strength:

$$\sigma_u = \frac{P_u}{A_0} = \frac{92\,000\ \text{N}}{201.062\ \text{mm}^2} = 457.6\ \text{N/mm}^2$$

Step 4 — (iii) Percentage elongation:

$$\%\,elongation = \frac{L_f - L_0}{L_0}\times100 = \frac{98 - 80}{80}\times100 = 22.5\%$$

Step 5 — (iv) Percentage reduction in area:

$$A_f = \frac{\pi}{4}(11)^2 = 95.033\ \text{mm}^2$$ $$\%\,reduction = \frac{A_0 - A_f}{A_0}\times100 = \frac{201.062 - 95.033}{201.062}\times100 = 52.7\%$$

Answers:

• (i) Yield stress $= \mathbf{288.5\ \text{N/mm}^2}$
• (ii) Ultimate tensile strength $= \mathbf{457.6\ \text{N/mm}^2}$
• (iii) Percentage elongation $= \mathbf{22.5\%}$
• (iv) Percentage reduction in area $= \mathbf{52.7\%}$

Definitions:

• Bitumen: A black or dark-brown, sticky, highly viscous (or near-solid) hydrocarbon material obtained as a residue from the fractional distillation of crude petroleum. It is fully soluble in carbon disulphide and is used as a binder in road construction and waterproofing.
• Asphalt: A mixture of bitumen (the binder) with mineral aggregate (and filler). In civil engineering practice, asphalt (bituminous concrete) is the compacted paving material; the term sometimes also refers to natural asphalt found as deposits.

Distinction:

Three laboratory tests on bitumen:

1. Penetration test: A standard needle under a 100 g load is allowed to penetrate the bitumen for 5 s at $25^\circ C$; the depth in tenths of a mm is the penetration value. It measures the consistency/hardness (grade) of the bitumen.
2. Softening point test (Ring and Ball): The temperature at which the bitumen softens enough for a steel ball to sink through a sample ring is found. It measures temperature susceptibility / suitability for hot climates.
3. Ductility test: A briquette of bitumen is stretched at $27^\circ C$ at 50 mm/min; the elongation in cm at break is the ductility. It measures the ability to stretch/flexibility (resistance to cracking). (Other tests: flash & fire point, viscosity, specific gravity, solubility.)

Numerical — theoretical maximum specific gravity $G_{mm}$:

Given: For 1000 kg of mix: bitumen $= 5.5\%$, aggregate $= 94.5\%$; $G_b = 1.02$, $G_{agg} = 2.65$.

Step 1 — masses:

• Bitumen $= 0.055 \times 1000 = 55\ \text{kg}$
• Aggregate $= 0.945 \times 1000 = 945\ \text{kg}$

Step 2 — using mass fractions $\left(P_b = 5.5,\ P_{agg} = 94.5\right)$:

$$G_{mm} = \frac{100}{\dfrac{P_{agg}}{G_{agg}} + \dfrac{P_b}{G_b}} = \frac{100}{\dfrac{94.5}{2.65} + \dfrac{5.5}{1.02}}$$

Step 3 — evaluate:

$$\frac{94.5}{2.65} = 35.6604, \qquad \frac{5.5}{1.02} = 5.3922$$ $$G_{mm} = \frac{100}{35.6604 + 5.3922} = \frac{100}{41.0526} = 2.4359$$

Check by volume (1000 kg): $V_b = 55/1.02 = 53.922\ \text{L}$, $V_{agg} = 945/2.65 = 356.604\ \text{L}$, total $= 410.526\ \text{L}$. $G_{mm} = 1000/410.526 = 2.436$. ✓

Answer: Theoretical maximum specific gravity $G_{mm} \approx \mathbf{2.44}$.

Classification of bricks by field characteristics:

Numerical — compressive strength:

Given: brick size $240 \times 115 \times 70\ \text{mm}$, crushing load $P = 310\ \text{kN} = 310\,000\ \text{N}$. Load is applied on the largest (bed) face $240 \times 115\ \text{mm}$.

Step 1 — loaded area:

$$A = 240 \times 115 = 27\,600\ \text{mm}^2$$

Step 2 — compressive strength:

$$\sigma = \frac{P}{A} = \frac{310\,000}{27\,600} = 11.23\ \text{N/mm}^2$$

Conclusion: Compressive strength $= \mathbf{11.23\ \text{N/mm}^2} > 10.5\ \text{N/mm}^2$, so the brick satisfies the minimum requirement for a first-class brick.

Fineness modulus (FM): It is an empirical index of the average particle size of an aggregate, obtained by adding the cumulative percentages of mass retained on each of the standard IS sieves and dividing by 100. A higher FM means a coarser aggregate.

Significance: It indicates the coarseness/fineness and helps in classifying sand (fine, medium, coarse), proportioning mixes, and assessing grading consistency for concrete.

Step 1 — set up the table (total sample $= 500\ \text{g}$):

Step 2 — sum of cumulative % retained (standard sieves, pan excluded):

$$\Sigma = 2.0 + 11.0 + 29.0 + 55.0 + 84.0 + 98.0 = 279.0$$

Step 3 — fineness modulus:

$$FM = \frac{\Sigma\ \text{cumulative \% retained}}{100} = \frac{279.0}{100} = 2.79$$

Answer: Fineness modulus $= \mathbf{2.79}$, which lies in the range 2.6–2.9, indicating a medium (Zone II) sand suitable for general concrete work.

Desirable qualities of a good building stone:

1. Strength — high compressive strength to carry loads.
2. Durability — resistance to weathering, frost and chemical action.
3. Hardness & toughness — resistance to abrasion and impact.
4. Low water absorption / low porosity — less than ~5% to resist frost damage.
5. Appearance & colour — uniform, pleasing colour for facing work.
6. Workability — easy to cut, dress and carve economically.
7. Fire resistance and good seasoning without developing cracks.
8. Sound (free from cavities/cracks) giving a clear ring when struck.

Numerical:

Given: dry mass $W_d = 2480\ \text{g}$, saturated mass $W_w = 2620\ \text{g}$, bulk volume $V = 1000\ \text{cm}^3$.

Step 1 — (i) water absorption:

$$\text{Water absorption} = \frac{W_w - W_d}{W_d}\times100 = \frac{2620 - 2480}{2480}\times100 = \frac{140}{2480}\times100 = 5.65\%$$

Step 2 — (ii) bulk density:

$$\rho = \frac{W_d}{V} = \frac{2480\ \text{g}}{1000\ \text{cm}^3} = 2.48\ \text{g/cm}^3 = 2480\ \text{kg/m}^3$$

Comment: The water absorption of 5.65% is slightly above the desirable limit of about 5% for a good building stone, so the stone is marginally porous and would be only moderately suitable for severe external/exposed use (risk of frost damage). Its bulk density of 2480 kg/m³ is satisfactory and typical of a fairly dense stone.

Classification of lime (based on IS / impurity content):

1. Class A — Eminently hydraulic lime: high clay content (~20–30%); sets under water; used for heavy/wet masonry and foundations.
2. Class B — Semi-hydraulic lime: moderate clay; partial hydraulic property; general masonry and plaster.
3. Class C — Fat (high calcium / non-hydraulic) lime: very pure, made from pure limestone; sets only by carbonation in air; used for plastering, whitewashing, finishing.
4. Class D — Magnesian/dolomitic lime and Class E — Kankar lime (impure, used locally).

Fat lime vs hydraulic lime:

Slaking of lime: It is the chemical reaction of quicklime (calcium oxide, $CaO$) with water to form slaked lime (calcium hydroxide), accompanied by heat and expansion:

$$CaO + H_2O \rightarrow Ca(OH)_2 + \text{heat}$$

The resulting hydrated lime can then be used to make mortar/putty.

Setting of lime mortar: Setting is the hardening of the lime mortar after placing.

• For fat lime it occurs by carbonation — slaked lime reacts with atmospheric carbon dioxide and reverts to calcium carbonate:

$$Ca(OH)_2 + CO_2 \rightarrow CaCO_3 + H_2O$$

This is slow and needs air, so fat lime cannot set under water.

• For hydraulic lime setting occurs by hydraulic action — the silicates/aluminates formed from the clay impurities react with water (similar to cement), allowing it to set and harden even under water.

Objectives of painting a surface:

1. To protect the surface (wood, metal, plaster) from moisture, corrosion, weathering and decay.
2. To give a decorative, pleasing appearance and desired colour.
3. To provide a smooth, hygienic, washable surface that resists dirt.
4. To increase durability and reduce maintenance.
5. To make the surface fire / weather resistant and, for metals, to prevent rusting.

Basic constituents of an oil paint and their functions:

Difference between paint and varnish:

Smart materials: Smart (intelligent) materials are engineered materials whose one or more properties (shape, stiffness, colour, electrical/optical behaviour) can change significantly and reversibly in a controlled way in response to an external stimulus such as stress, temperature, moisture, electric or magnetic field, or pH.

Two examples used in civil engineering:

1. Shape memory alloys (SMA) — e.g. Nitinol; recover their original shape on heating, used in seismic dampers and self-centring/retrofitting devices.
2. Piezoelectric materials — generate an electric charge under mechanical stress (and vice versa); used in structural health monitoring sensors and energy harvesting. (Other valid examples: self-healing concrete, electrochromic/smart glass, magnetorheological dampers.)

One important property of each:

• Glass: Transparency (transmits light) — making it ideal for windows, facades and daylighting (also rigid, hard and chemically inert).
• Plastics: Light weight combined with corrosion resistance (and easy moulding) — useful for pipes, fittings, cladding and waterproofing.