(a) (i) Briefly explain the deterioration of concrete caused by leaching action, and (ii) chemical interaction. (10 marks) (b)…

(a)(i) Leaching action is the removal of soluble constituents of cement paste by percolating or flowing water. The chief soluble product is calcium hydroxide, Ca(OH)₂. Soft, acidic, or CO₂-bearing water dissolves Ca(OH)₂ and carries it away; CO₂ may convert Ca(OH)₂ first to CaCO₃ and then to soluble Ca(HCO₃)₂, accelerating removal. This increases porosity, reduces alkalinity, and exposes aggregates. The concrete becomes weak, permeable, and may show efflorescence, stalactites, and softening. Loss of Ca(OH)₂ lowers pH, destroys the passive film on reinforcement, and promotes corrosion. Leaching is severe in dams, canals, retaining walls, and structures subjected to continuous water flow. Prevention requires dense, low water-cement ratio concrete, proper compaction and curing, surface coatings, and use of pozzolanic materials to consume free lime.

(a)(ii) Chemical interaction is deterioration caused by aggressive chemicals in water, soil, or atmosphere reacting with cement hydration products or aggregates. Sulphate attack forms gypsum and ettringite, causing expansion, cracking, and spalling. Chlorides penetrate concrete and initiate corrosion of steel reinforcement. Acids dissolve the cement paste. Alkali-aggregate reaction occurs when reactive silica in aggregate reacts with alkalis from cement, forming a hygroscopic gel that absorbs water, swells, and cracks concrete. Carbonation, the reaction of CO₂ with Ca(OH)₂, reduces pH and depassivates steel. Seawater attack combines sulphate and chloride actions. Control measures include sulphate-resisting cement, low water-cement ratio, pozzolans, impermeable concrete, protective coatings, and avoiding reactive aggregates.

(b) General precautions during brick masonry construction:

• Use well-burnt, uniform, sharp-edged bricks, free from cracks, lime nodules, and excessive warping. Soak bricks in water for at least 12 hours before laying and remove surface water before use.
• Prepare mortar of specified mix and workable consistency. Use it within about 30 minutes; do not add water later to revive stiff mortar.
• Ensure the foundation is level, compacted, and capable of bearing the load. Clean the surface before starting masonry.
• Adopt proper bond. English bond is generally used for walls more than one brick thick; Flemish bond may be used for one-brick walls; stretcher bond for half-brick walls. Avoid continuous vertical joints.
• Construct quoins at corners first with selected good bricks. Provide bond stones or through stones at regular intervals, especially in thick walls.
• Lay bricks on a full bed of mortar. Fill vertical joints fully from the top. Keep mortar joints uniform, normally 10 mm thick (8–12 mm).
• Keep each course truly horizontal and the wall truly vertical. Check with spirit level, plumb bob, and line and pins. Do not allow the wall to lean or bulge.
• Do not raise masonry more than about 1.5 m in a day, so that mortar can set.
• Provide toothing for future extension; avoid straight or unbonded joints.
• Cure masonry for at least 7 days, preferably 10–14 days, by keeping it moist.
• Use independent scaffolding, not resting on green masonry. Protect masonry from rain, frost, and strong sun.
• Leave openings with proper jambs and provide lintels, arches, and damp-proof course where required. A neat sketch should show staggered vertical joints, uniform mortar joints, proper quoins, and toothing. These ensure strength, stability, and durability of the brick masonry.

(c) Design of tie bar for longitudinal joint of cement concrete pavement: Given:

• Pavement thickness, h = 20 cm = 0.20 m
• Total width = 7.2 m, two lanes with longitudinal joint at centre. Therefore distance from longitudinal joint to outer edge, W = 7.2/2 = 3.6 m
• Allowable working stress in tension for steel, σ_s = 1400 kg/cm²
• Bond strength with concrete, τ_bd = 18 kg/cm² Assume:
• Unit weight of concrete, γ = 2400 kg/m³
• Coefficient of friction between slab and subgrade, f = 1.5

Method: Tie bars are designed to resist the tensile force caused by friction between the slab and the subgrade when the two lanes tend to separate. The force per metre length of longitudinal joint is: T = W h γ f T = 3.6 × 0.20 × 2400 × 1.5 = 2592 kg/m

Area of steel required per metre length: A_s = T/σ_s = 2592/1400 = 1.851 cm²/m

Adopt 12 mm diameter tie bars. Area of one bar: a = π d²/4 = π × 1.2²/4 = 1.131 cm²

Spacing of tie bars: Spacing = 100 × a / A_s = 100 × 1.131 / 1.851 = 61.1 cm Adopt 60 cm centre-to-centre.

Length of tie bar: The bar is embedded L/2 in each adjacent slab. Bond force on one side = π d (L/2) τ_bd. Tensile force in bar = (π d²/4) σ_s. Equating: (π d²/4) σ_s = π d (L/2) τ_bd L = (d σ_s)/(2 τ_bd) L = (1.2 × 1400)/(2 × 18) = 46.67 cm Adopt 50 cm total length.

Final: Provide 12 mm diameter tie bars at 60 cm centre-to-centre and 50 cm long across the longitudinal joint, placed at mid-depth and perpendicular to the joint.

(d) Let the proposed gradient be 1 in G. For a ruling gradient of 1 in 200, the grade resistance is: R_ruling = 1000/200 = 5 kg/tonne

For M.G. track, curve resistance is taken as 0.03% per degree of curve. For a 6° curve: R_curve = 0.03 × 6 = 0.18% = 0.18 × 10 = 1.8 kg/tonne

Grade resistance for proposed gradient 1 in G: R_grade = 1000/G kg/tonne

Given: grade resistance + curve resistance = resistance due to ruling gradient. 1000/G + 1.8 = 5 1000/G = 3.2 G = 1000/3.2 = 312.5

Final: The gradient should be 1 in 312.5, say 1 in 312 or 1 in 313 on the 6° curve for the M.G. track. This is the limiting gradient so that total resistance equals that of the ruling gradient of 1 in 200.

(e) Orientation of a plane table means setting up the table at a station so that all lines already plotted on the drawing sheet are parallel to the corresponding lines on the ground. It is essential before plotting new details; otherwise rays will not intersect correctly and the plotted map will be distorted. The table must be oriented at every new station.

Various methods of orienting the plane table:

• Orientation by trough compass: The trough compass is placed on the sheet and the table is rotated until the magnetic needle points to the north line marked on the sheet. This method is simple and used when no previously plotted lines are available. It is not accurate where local magnetic attraction exists or where magnetic declination changes.
• Orientation by back sighting: This is the most accurate and commonly used method when the table is set at a station already plotted. Suppose the table is at B and A is the previous station. Place the alidade along the line ba on the sheet, sight towards A, and rotate the table until A is bisected. Clamp the table. Then line ba becomes parallel to ground line BA, and all other plotted lines are automatically oriented.
• Orientation by two-point problem: This is used when the table is to be set at a new station from which two known points are visible. The table is first roughly oriented by compass or estimation. Rays are drawn to the two known points. By trial and error, the table is rotated until the plotted positions of the two known points and the station lie on the corresponding rays. This orients as well as locates the station.
• Orientation by three-point problem: This is used when three well-defined known points are visible from the new station. Methods include trial and error, Lehmann’s rules, Bessel’s method, and the mechanical method. The table is oriented so that the rays drawn from the plotted known points intersect at a single point representing the new station. This method is powerful but requires careful work. In practice, orientation by back sighting is preferred because it is free from magnetic errors. Resection methods are used when the station is new and no line from a previously plotted station can be sighted.